Heat transfer by dr. d.s kumar

e

4. CONDUCTION WITH HEAT GENERATION

5

wa CONTENTS

1. DEFINITIONS AND BASIC CONCEPTS Le
12 Maas and basis laws ol ea taser 1
111 Conduction 1
112 Convection 3
113 Radiaton 5
12 | Steady and unsteady heat transfer 7
13. Skgniicance of heat transfer 7
Kevin questions u
2. FOURIER EQUATION AND THERMAL CONDUCTIVITY iss
21 Fourier Equation a4
22 Thermal resistance "4
23 Thermal conductivity of materials =
24 General heat conduction equation e
241 Cartesian coordinates 4
242 Gylindnial co-ordinates 3
243 Sphercalco-ordinates A
25 Initial and boundary conditions Es
Revie questions E
3. STEADY STATE CONDUCTION 31-103
31. Conduction through a plane wal a
32 Conduction through composite wall 5
33. Heat flow between surface and surroundings:
cooling snd heating of fluids 48
34. Conduction through a cylindrical wall 57
35 Conduction through a multi ayer cylindrical wall a
36 Conduction through a sphere a
37 Effetof variable conductivity 7
38 Critical thickness of insulation 9
review questions 101

41. Plane wall with uniform heat generation
42. Cylinder with uniform heat generation
43 Sphere with uniform heat generation
Review questions

HEAT TRANSFER FROM EXTENDED SURFACES

51 Steady flow of hea along a rod

52. Heat dissipation from an infinitely long fin (> ©)
53 Heat dissipation from a fin insulated at the ip

54. Heat dissipation from a fn losing heat at the tip
55 Fin performance

56 Thermometric well

Review questions

‘TRANSIENT (UNSTEADY STATE) HEAT CONDUCTION

61 Transient conduction in solids ith into ee
(Lampe parameter ana eA coder +
62. Timeconstnt and response of thermocouple =
63 Tramient hat condition ins wh inte 1
conduction and convective resistance (9 .<100)
64 Transient conduction with given temperate dota e
Review questions FREE Seen =
RADIATION: PROCESSES AND PROPERTIES 190-219
74 Salient features and characteristics of radiation 190
72 Absorptivity, reflectivity and transmissivity am
73 Spectral and spatial energy distribution 1
74 Wavelength distribution of back body radiation; Planck's law 196
75 Total emissivo power Stefan Boltzman law m
76. Wien's displacement law an.
77 Kirchottstaw e
78 Gray body and selective emitters En
79 Intensity af radiation and Lamberts cosine law 2
710. Solar radiations ne
Review questions a
RADIATION EXCHANGE BETWEEN SURFACES 220-267
81 Heat exchange between black bodies: configuration factor 20

82 Shape farb and salle ue a De tape ao 2
83 tech tree nonin odie =
851 Ine paral planes =
832 _Inintlong concent older E
533 Salir bode >
834 Small etc m
84 Radation mike ES
85 ciento hat rante and radio combi wiheanvcion 27
86 Gaseous radiation E]
evi quesos =
CONVECTION: PROCESSES AND PROPERTIES 266-280
91 Free and fred convection za
92 amcaná tren ow 2s
93 — Newton-Rikhman law : convection rate equation 26
94. Dimensiones grouper comen m
95. Darcos groups for forcedcoveton 7
36 Since dmersonless groups =
Recon
EMPIRICAL CORRELATIONS FOR FREE AND FORCED CONVECTION 201-308
101. Pak temperature and mea fn temperate m
102. Local and average convective coset m
103, Canlaon or fee conecton
031 Horizontal ies eyes and wes =

1032 Vertical plates and large cylinders:

d

£ 1035. traine

E 1020, Morro is
dm

reo on pu at pasan wale
RAD Css de PPS

1043 Turbulent ow over flat plate

JURA Turn Pow in tates
1045 Turbulent Now over evingers

1046 Tarbulent lon over spheres
Reine ques
11, HYDRODYNAMIC AND THERMAL BOUNDARY LAYERS
MA. Hydrodynamic boundary layer fat plate
112 Thermal boundary Laver

113 Turtle boundary ayer flows: at plate
114 Reynoldsanalogy
115. Turbulent rate flow

12. CONDENSATION AND BOILING

Review questions
121. Condensation

122 Lamunar film condensation on a vertical plate
123 Turbulent film condensation
124 Convectivecoeffcient or film condensation on tubes
125 Boling
125 Boiling regimes
127 Comelanon of boiling heat transfer data
Revie questions
| 13, HEAT EXCHANGERS
131. Classification for heat exchangers
1311 Nature of heat exchange process
1312 Relative direction of motion of fluids
1313 Mechanical design of heat exchange surface

1314 Physical state of heat exchanging fluids (condensation and evaporation) 379

132 Performance analysis
153. Overall heat transfer coefficient
154. Logarithmic mean temperature difference
135 Effectiveness and number of transfer units (NTU)
1351 Effectiveness for the parallel low heat exchanger
1352 Biene forthe counterlow hot exchanger
1353 _ Limiting values of capacity ratio, C
Review questions ?
14. MASS TRANSFER
141. Mass transfer processes: clasificaiton
12 Fick's law
143 General equation of mass diffusion in stationary media
144. Equimolal diffusion
145 Diffusion of water vapours through air
Review questions pes à
BIBLIOGRAPHY
INDEX

short lingers amd blocks)

309-349.
309
35
37
329
346
ES
350-375
350
ES]
37
358
365
366
368
a

376-418
376
376
37
378

37
380

419-437
419

420

a

42

si

336

438
439-442

Definitions and Basic Concepts

Me me wore ho mt nn con,
Hiatt Sen ate ese er
to accomplish that duty. bi b =

1.1. MODES AND BASIC LAWS OF HEAT TRANSFER
The literature on heat transfer generally recognises thee distinct modes of heat transmission;
conduction, convection and radiation. These three modes are similar in that a temperature
differential’ must exist and the heat exchange is in the direcion of decreasing temperature
Each method has, however, different physical picture and different controling laws.

1.14. Conduction
Thermal conduction isa mechanism of heat propagation fom rein of higher temperature
toa segon of ow temperature within a mam el, Equ a ass) or vee ete
mediums in direct physical contact. Condi ye any movement of macroscopic
potions of mater rea tone anahet. The era energy ma be tans
Sean eh ar reto move through Le
Tie structure of the RAT In aon, or
altérmatively, it may be transferred as vibrational
energy in the lie structure respective of
the exact mechanism, the observable effet of
Conduction is an equalization of temperature.
‘Consider the flow of heat alo a metal ro
one end of which i placed adjacent toa fame
The elementary particles (molecules, toms,
electrons) composing the rod, and which are in
immediate vicinity ofthe flame, get heated.
1

Fig. 1.1. Conduction heat fou long a rod

A SEE

2 Basics of Hest and Mass Tensor
ure goth thi int
“ar vain aba hir m
he active molecules 88

e energy increases and this puts them
cause of the est tempera san postions, Consequenty,
Inventar af acton and ent to them, During collision,
these more active puros coli Wt Te thermal energy is imparted to them. The process

clo gt ented tT eT en ofthe rod is reached. Each,

some rch ween
ipa oper afer ne of cle eng oe empese

se € of heat flow between the two ends depends
See ea

aa ee aa
Les

steady flow of heat by conduction is prescribed by

ut
ort an

ere Qi he hen ne rt, A the ata of ea tante suce, e ise temperature
DS peels sane d andthe termal conductivity is cadera
ee mare bce temperature gradients negative nthe posite section
Sh mies ign equation ge ose Rat fa.

eu engi in he don of he low ar (= is the temperature ditference,
sen

gt) a2

‘The heat flux q s the heat conducted per unit time por unit area and is given by

ae Pe
Heat are in metal rods, in heat treatment of steel forgings and through the walls of

heat exchange equipment are some practical examples of heat conduction

EXAMPLE 1.1.2

AA 75 cm thick side wall ofan oven is primarily made of insulation with a thermal conductivity

£004 Wmk. Conditions on the inside of wall fix the temperature on that side at 420 K

‘The electric ois within the oven dissipate 365 watts of electrical energy to make up forthe

het loss through the wall. Calculate the wall surface ares, perpendicular to hat How, 50

that temperature on the other side ofthe wall does not exceed 310 K.

Solution: Under the stipulations of one-dimensional steady state heat conduction, the electrical

‘energy dissipation rate within the oven must equal the conduction heat transfer rate across the
wall That i

BA =)
Vo:
08.4 420-310)
x5 = A 54
se 587

Hence, the required wal! surface area,
A

365
Bap © 5218 mt
cometa Ed

A plane wall of 10cm thickness and 3 mé

area is made of a material whose conductivity 15.
faces are steady at 100°C and 30°C respectively.
low across the wall

83 Wink. The temperature fhe ath
Find the empesar aden and ha te

fee Bef

Ostos and Bose Concept 13
Suton: Temperature gradient in he dicton of eo os
a hou mom
a Bt Bae oc
() Het flow cas the wal gen by Fe has condom easton

a
aM 84003 (700) = 17850 W or 1785 KW
EXAMPLE 1.3
To effect a bond between two metal plats, 25 em and 15 cm thick, heat eniformly
applied through the thinner plate by a radiant heat source. The bonding epoxy mus be
hiela at 520K fora shor time. When the heat source in adjute to haves steady value of
435 KW, a thermocouple installed on the side ofthe thinner plate nex to source indicates
2 temperature of 345 K. Calculate the temperature gradient for heat conduction through
ininner plate and thermal conductivity of x material

AS has has
cm à 4

ze FF

Temperatur gradient, À ka

Invoking Fourier law, the heat conduction equation for this one-dimensional case can be
writtenas 39-320 2 25

Be y incre vy bt Top
re 32e

35x 10°
er 005

‘Thermal conductivity for the material of thin plate is,

1.1.2, Convection

Thermal convection is a process of energy transport affected by the circulation or mixing
of a fluid medium (gas, liquid or a powdery substance). Convection is possible only im a fluid
medium and is directly inked with the transport of medium self The effectiveness of heat

With respect to origin, two types of convection
free convection.

{IR satura or free convection) the circulation ofthe lid medium is caused by buoyancy,
cifecis, 12, by The Aiepence IR the ENS of the cold and heated particles. Consider beat \
flow from a ROCPIAT atmosphere: The sagan lye of arm the ummediate vicinity ofthe
Plate Gets thermal energy by conduction, The energy thus transferred serves to increase the
temperature and internal energy ofthe air articles. Because of temperature rise these particles
become less dense (and therefore lighter) than the surrounding air. The lighter air particles
move upwards lo a region ol low temperature where they mix with and transfer a part of their

y cotection depend largely upon the mixing motion ofthe Quid
si are distinguished forced and natbral or
x

4 / Basico of Haac and Mess Trenster

rg the cl pre Small: te cold ir pates end owed? 1
Sesto soe ap ae er
ar are tee rae ee E
Some amples of free convection are: al
tt coi aati
ores
E Tacs ane
See Ce iin id
el de |
ERE lala alone pe roer
E
saree es
o he ce inane
Sara rere
Lt et he cr

Regardless of the particular nature, the appropriate rate cquation for the convective heat

‘wansier between a surface and an adjacent fluid is presented by Newton's Iso of cooling.

Be A

‘consluctivity of the Maid), e geometry ofthe surace, the nature 01 Hui flow, and the prevailing
thermal conditions.

‘Convection mechanism involving phase changes leads to the important fields of boiling
(evaporation) and condensation
EXAMPLE 14, Z
‘An oil cooler in a high performance engine has an outside surface area 012 m? and a surface
temperature of 65°C. The air rushes over the surface of the cooler at a temperature of 30°C
and gives rise to a surface coefficient of heat transfer equal to 354 Wak, Caleulate the
heat transfer rate from the cooler,
Solution ? The conditions described imply a convective process, that is, heat transfer from a
solid surface (the ol cooler) to an adjacent moving fluid (the ae passing over the cooler) The
rate of heat transfer by convection from oll cooler to the ar is then,

Q = HA (I, = 1) = 454 © 012 (65 - 30) = 19068 W

ExampLe 15,

A wire 10 cm long and Lm in diameter is held taud between two conducting supports in
2 water tank and is submerged. A controlled amount ofcorent is made to pass through the
wire until the temperature of water becomes 100°C and it tarts builing, Make calculations
for the steady temperature of wire if 235 watt of electric power is consumed. Take convective
heat transfer coefficient to be 5000 W/nrsdep-

Solution When steady state is reached, the

eat loss which is given by

Power supplied to the wire equals the convective

re Definitions and Basic Concept 1 8
LA At na.
CRETE O
+ Surface temperature fe, PO BA (= 10) 3,74, 10)
Bs

$ 57 * 100 à asec
EXAMPLE 15.
then the uri STO mai pas of 035 aa a temperance of PE
Srl conection? acl a ent What tatoo ne DURE
pipa at convection ent confor t he aaa

Solution : Convective heat transfer coefcen,

A = 25 (ADP = 25 (6 - 2095 = 6257 W/m?
Heat lost by convection = h À ar Oe a

= 6287 « 025 x (60 - 20)» 6247 W
Heat lost by convection as faction of heat supple,

Tap "05239 or 209%,

‘The remaining 47.61% would be lost o he surroundings by radiation.
Radiation

‘Thermal radiation isthe tr

mision ofen inthe orm faa energy or ware tion
from one body to another across an intervening space. Unlike heat ranıla Ur ondaa on
convection ramon a thermal radon diese cy en condo and
betwen the hat Sure ad he sever An Inrena mn a ei
th ration can e flected though vaca ar ane See an ae Bad
change, in (a, curs mos caia rota mat as ene Ain
Source and the receiver would either reduce or elminae ec Ue propagan ol cies,
ad
Te mechs ofthe he Ay ition cons of Be dst pases:
(0) Conversion Oy Mél energy ofthe hot source nto electromagnetic waves “All bodies above
bolt ero tempera ar cub ol mung rada Cn En need
“Tiscrete) packets or quan of energy called phofons The photons are propgated
through the space a ae ment toe poles ee a
os tere

(i) Passage of cave motion through intervening space. The photons, as carriers of energy,
travel with unchanged frequency in straight paths with speed equal 1 that of light.
(Gi) Transformation of waves into eat: When the photons approach the cold receiving surface,
there occurs reconversion of wave motion ino thermal energy which spar absarbed,
reflected or transinitted through the receiving surface
The most vió evidence rain hat was is at represented y lar energy
Which passes through interstellar space (conditions cose 10 that for perfect vacuum) on its
Way to he earth surface Solar radiation plays an impartant part in the design of heating and
‘ventilating systems. Heat traste by aiaton encountered oler Faces bile ebeting
furnaces and other HR oT heat exchange apparatus. The design arf construcion of gngines

À
<

6 // Basics of Heat and Mass Transfer

ps tings nur estr and sla clitoris alo significo eluent ib lion
er equations fr rason heat anse ar sed on Stefan-Doltzman La
eat a)

ehe. Eps the enor rate pr uit ie, Tis the absolute temperature ofthe surface,
and 9 Scan ols constant
9, = 5.67 x 10% W/m? K* trs
sen 1.1 esta val fran idea! adator or Mack body-sufix b designates a
acetate TS St emy emite taro rae se than that or an ideal eme
ds vent
Escort 0.6)
ehe «Ha radiative sung of the surface and scaled emissivity its value depends
See crane i indicas Tuer electiva ihm muros cone
Se ne a an Heal or Hck body radiator. Normally a body radiating heat À
See seg hast em er podes adan Conse hat sure at temperature
Fra compl ende by ananer Back surface 2a temperature 7, The net radian het
Es

Arad TT) an
Lixeaize, the net rate of heat transfer between the real surface (called gray surface) at
temperature T, to a surrounding black surface al temperature T, is
Qa a A 8 (MT) as
‘The net exchange of heat between: the Iwo radiating surfaces is due to the fat hat one a
the higher temperature radiates more and receives les energy fr it absorption. An isolated
bode which remains at constant temperature cmite just as much energy by radiation as it
EXAMPLE 47.
radiator in a domestic heating system operates at a surface temperature of 60°C. Calculate
‘the heat lux at the surface of the radiator if it behaves as a black body.
Solution: The heat fl atthe surface isthe ate at which
per uni ars

at energy leaves the surface

for pure muele bot

est
a
a
a A
Ga T4 = 567 x 10° (273 + 60) = 6972 Wim?
EXAMPLE 1.8.

A cylindrical rod, 13 m long and 2 cm in diameter, is heated electrically and positioned in
3 vacuum furnace which has interior walls at 800 K temperature. A controlled amount of
current is passed through the rod and its surface is maintained at 1000 K. Calculate the
power supplied to the heating rod if its surface has an emissivity of 03.
Solution : For steady state conditions, the electric power supplied to the rod equals the radiant
eat loss from it. Further since the walls of the furnace completely enclose the heating rod, al
the radiant energy emitted by the surface ofthe rod is intercepted by the furnace walls. Thus
ara Ac (M - 73)
9, (ull) ET

Detintions and Basic Concepts 117
002 x 15) (09) (1000 - 500%

TA

ETE
0) E) |
= 5.67 (x * 002 x 1.5) (09) (10000 - 4096) = 2838 w
ec np tthe od mus egal Be 1
1.2 STEADY AND UNSTEADY HEAT TRANSFER
Heat exchange ben two systems may ae plas under dy tae
or under unsendy (unstable thermal sonda sey ae ml mo
int of the system remains constant in the course of time, and a fan pp
rome cite econ ny Spa

‘Thus the rate of el

fate
runs: no as

Steady stat results in a constant rte of heat exchange (heat inf equals heat eff, and

tere is no change in the internal energy ofthe system during such a procens. Typical ar
fof steady state heat transfer are o. pa

+ cooling of an electric bulb by the surrounding atmosphere,
+ heat flow from the products of combustion to water in the tubes ofa boiler, fom the

hot to cold fluid in a heat exchanger, and from a refrigerated space to cooling surface
of the evaporator.

Under unsteady thermal conditions, ten
time. Temperature is obviously a function ol

ature ofthe system changes continuously with

space and time co-ordinates,
a

ef 9: 20 am

sense results in hat trae ie which changes wih ine Further. a change in
temperature Indian a change of eral eng of the sytem, Energy Morges has apa
A parcel of unsteady heat tow. Pia examples of urtendy heat waar ae
+ warn period of furnaces
+ boilers and turbines
< cooling of castings in a fundey
+ heat restent and ies vleingof metal casings
A api Kind of unen proces ste tase sate wherein he pte td 10
cycle variaton in he temperature ls entranmen: The temperatur «prc pent
oF the system peums pedia tothe same value the ato est lw and energy Sage
ths underg penodl variation, Tamper ae; esting cong of the wae fan LE
mg esting or cooling ofthe wale of ing during the 2 hours ceo the dy
Further, the heat tante in a system may bein one no or mare diene Ina one
dimensional heat low ne singe predomina ection heh temperature feral
ono and viu ih fo taker pls a fw in he er two ins cn be
y reese, When te temperature 5.8 Kanon of te orinate heat fe 640
ment À ive dimensional hat How sults tat temperatura fneion of Bose
Coordina, nd corsequenty hut How ocur in all tice decors

1.3. SIGNIFICANCE OF HEAT TRANSFER

The discipline of the heat transfer encompasses a great many fascinating areas ke

|

8 1) Basics of Heat and Mass Transfer,

«Design af ten ert omen a Ng and or ere Fer

plant Inn, soar energy conversion.

lees ‚its, superhcater and condensers an
+1. engines ret re chal cines, The operation

many other cooing and eating SPP aes ea on the effective transfer of

e and airconditioning units depends greatly on ster of

Beat in condones and evaporator. Seal ‘een

. ystems for electric motors, generators and transformers in electrica
Drifting est ner nt
nes can te efecively dissipated. TNS sto avoid the conditions which
al cause overheating and damage the equipment.

« Evaporation condensation, heating nd cooling of fluids in chemical operations, Hardly
ona operation can be identified that does not involve heating or cooling of à
materia at some stage or the othe.

+ Construction of dams and other heavy structures, calculation of thermal expansion of
‘Ssyersion ridges and railway tacks, minimisation of building het losses by means
(of improved insulation techniques,

1 Proper functioning of valves and other controls operated by temperature changes,
thermal contre of space vehicle.

+ Heat treatment of metals where diffusion rate of carbon in see is required to be made
te estimate the period for which the steel component must be exposed to carburizing
atmosphere.

+ Dispersion of atmospheric pollutants: problem of thermal pollution associated with the
discharge of large amounts of waste heat fom a powerplant to environment Indus
haus! gaces laden with noxious pollotants are discharged high enough. This isto
sure chat bythe time polltantsdifuse downwards, their concentration falls below
Safe limits. An understanding of mass transfer is needed for accurate predicaion of
concentration at ground ofthe pollutants discharged from the chimney

‘An engineer utliss his knowledge of heat transfer either to transmit heat in the most
‘fective/economic way, or o protect his equipment against excessive heat gains or losses, The
Various engıneenug problems involving heat transfer can be categorised into Iwo groups

() Heat ow situations where maximum heat transfer is desirble with minimum possible
heat exchange aea. Gas turbine blades, wall of LC. engines and combustion chambers,
‘outer surface ofa space vehicle all depend for their durability on rapid removal of heat
rom their surfaces. The design of heat exchangers is considered to be optimum under
specified temperature conditions when maximum heat transfer occurs with minimum

Surface area,

(@ Heat low situations where heat transfer is undesirable and its flow isto be prevented.
‘The walls of centrally heated buildings, and the steam pipes in a steam power plant are
propery insulated to restrict heat losses.

With few exceptions, engineering problems involve more than one of the three modes of
heat transfer and this aspect results into a complicated heat exchange pattern. The significance
of heat anale andthe simltaneous ocorence ol diferent modes of heat tanser cn De
welljudged by citing the following examples

() Closed container filled with hot coffee and kept in a room whose air and walls are at
2 fixed temperature. Fig (13)

Definuons and Basie Concepts 1,8
Evrongs Cover
[ a
a
À coman Mescaco

Fig 13. Cooing ono coe

Al three models of heat transfer contribute towards cool
for energy transfer from coffee are

= free convection from the coffee to the flask, q,

= heat conduction through the flask, 25

= free convection from the Mask to the air, 4y

— radiation exchange between the outer surface of the flask and the inner surface of the

cover 94

= free convection from air to the cover, 94

— heat conduction through the cover, 4,

= free convection from th

ing of coffee, and diferent pas

€ cover to the room air, gy
— radiation exchange between the outer surface of the cover and the surroundings. qu

(6) Automobile engine with thermo-syphon cooling system. Here the relevant heat transfer
processes are

— free convection and radiation from hot combustion gases to cylinder walls

— conduction through cylinder walls

= free convection from cylinder walls to water and from water to radiator tubes

= conduction through walls of radiator tubes Ba

= conection from radiator tubes to surrounding Congo conectan

(Vi) Heat flow through a wall consisting of two plates _
separated by vacuum (Fig, 1), Er

Heat is convected from the fluid at temperature fy non
to plate A, conducted through plate A, radiated {rom E
Plate to plate B, conducted through plate B and finally —
Sonvected from plate B to the fluid at temperature 1,7 or

CA
Likewise the process of steam generation, the bol "eng aa rnser ug a

tubes receive heat from the products of combustion by
all the three modes of heat transfer.

EXAMPLE 1.9
(0) Cite an analogy that would be useful in fixing the concepts of heat conduction, convection
and radiation. (0) Does convection strictly comply with the definition of heat transfer?
Solution : (a) Consider that a house has caught fire which is being extinguished by people who
carry water from well to the site of fire. Let the water be analogous to heat and the people be

alogous to heat transfer medium. Then

composite wall

AO 4 Basics of Heat and Mass Tronster
water buche
AI ina 0
site through th

ecson nearest to him who deliv

apeoe ets
‘eather reaches the fire site. Here the
fe medium, This is analogous 16 heat

A pre atte wal er he
perry
a tl
Saison me

ne
e vet
Ep the he
oe pron wi
Pe ee a he media Te an 6
Soon oe gue
(Te mec tno

the medium) and caries

«comes the single runner (repres
ame convection in liquids and

he site: This corresponds to

Lip of 3 hose, directs the water from the well 10 the hause
‘ellected by thermal radiation in a

ction and radiation depend for their operation on the
ee temperature ference and as such constte the fundamental physical
heat mopar, Convection does not strictly comply with this definition of heat
Sa sex depends lr ls operation on mechanical mass transport also, Heat exchange
sold wall and a fluid depends not only on temperature difference but also on mass
ere Ad Convection may be kened o conduction through the film of ui (gas or
LR aise with tansterof energy due to mixing of uid particles, Further, convection
ne ne by geometry ofthe surface, leve of turbulence inthe fluid and various
Ania properties.
EXAMPLE 1.10.
[A'S cm diameter steel pipe maintained ata temperature of 60°C is kept in a large room
‘these the ar and wall temperatures are 25°C. Ifthe surface emissivity of the steel is 07,
{Meclate the total neat loss per unit length of pipe if convective het transfer coefficient
is 65 W/m deg, Comment on the resul
Solation : The pipe loss heat both by convection and radiation
Heat loss by convection, Qu =/ (nd) At
5 (6005 x 3) (60 = 25) = 3572 W
Heat loss by radiation, Qu = (Bl) 9 (Ti = Te)
Ty = 60+ 273 = 353K; 5 + 273 = 98K
Qu = 07 » (m x 005 x 1) » 567 x 10% (2334 2989

sn]
core (5005 x1) 567|(335) (xs) |

“rw

Total heat loss * Quy + Quy = 3572 + 3372 = 69:44 W

Comments: Both the convection and radiation heat loss are of almost equal amount. It
would be a serious mistake to neglect either of the two.

EXAMPLE 1.11.
A surface at 475 K convects and radiates heat to the surroundings at 335 K. If the surface
‘conducts this heat through a solid plate of thermal conductivity 125 W/m-deg, determine
the temperature gradient at the surface inthe solid. Take convective coefficient and radiation

{actor as 80 Wa-deg and 09 respectively,
Solution : Under steady sate conditions the hat conducted tough the plate equals the sum
‘of convection and radiation heat losses. That is úl úl

Define and Basic Concepts 111
eat conducted through the plate = convection heat louse + radon host

a
leo AW AA Th ee outre th
— toros real

i le
II

Heu cone rou mec
Fig. 18
“aking unit area and substituting the relevant data, we have

a
125 D 280 (475 = 335) + 09 à (547 » 10% (75 33
80 (475 - 335) + 09 x 5, o (2 BET

as) “io |

11200 + 1955 = 13155
‘Temperature gradient through the plate,

aT ms
oe we TL
REVIEW QUESTIONS

14 What i the driving force for heat transer ?

12 Distinguish between the conduction, convection and radiation modes of hear wane.

1.3 Discues the diferent modes by which heat can be transfered. Give stable cumples 1 strate

14. Cie few examples where conduction plays a major role

1S What Is the difference between natural and forced convection ? Does any convection process
involve condueton to some extent ? Explain

16 A person who sit in front of fireplace [els warm. Though what process or processes of heat
transfer does he receive heat ?

17 What wil be your response toa person who states that host can not be transferred in a vacuum ?

18 State by giving illustrations that in practice the transfer of heat is the combined effect of conduction,
convection and radiation.

19 Identify the different modes of heat transfer in the following systems/epeations

(steam raising in a steam boiler

(i) air/water cooling of an LC. engine cylinder

{i} condensation of steam in à condenser

(6) heat loss from a thermos flask

(0) heating of water in a bucket with an immersion beater

(vi) heat transfer from a zoom hester
(vi) thermo regulator system of human body

42 11 Basics of Heat and Mo

yuo
in
um

us

16

Wire the ate quais or e
the ns for ea

sss Transfer

hor modes ofa ae eine he me ed ave

et anse in various ks of engineering

te importance

a en, wall, 5 m long. x 4 m high * 0.25 m thick,
Da tr aren SFE Te wal ad ie
eh ey qual 1 7 W/E es
“ satis. Calculate surface temperature of
en ton 190 va ete
rcp tng ae eS cn be
‘en ver the "surface of pane is 125 W/0v-des.
= The incident radiant energy from the sun is

Sed and the alu
a ee is
tina he tempera

nd ee gm sy Se
stmt net re s

ce Shainin He emisiviy is 008.

0 en
= 2 x 950% 2 ars wim!
iat: Energy resadied * 395 Bsus

Tre surface o see
afore temperature 01

on inside surface ofthe plate. Take thera

5

478 ee 7 = 008 TEE DE}

5
I) sm
cta] >
measuring 09 m Yong * 06 m wide x 0025 m thick is maintained at a
on he pate Toes 250 walt by radiation Fair at 15°C temperature
ei Blows over the plate calculate the temperature
N conductivity of pate as 45 W/edeg

ra
Or" Gon Ca
en rauhen

EN
us
ES

9.08 20 (09 + 06) « (300 15) + 250

‘A ck metal plate 1 m in diameter and 1 cm ik, exposed to su’ rays. The plate heats up
6 such a temperature tha the rte a whch it receive solar energy on one face equals, the rate

‘which fone eat by radiation and by convection {rom both su

ces. Presuming the following

culte the he are by convection.

Net

Sola energy x = 200 W/m?
ao heat flax leaving the plate = 80 9/0
Piste temperature = 310 K, and surounding ar temperature = 300 K

990

Fourier Equation and Thermal
Conductivity

2.1. FOURIER EQUATION
‘Conduction is primarily a molecular phenomenon requiring temperature gradient as the driving
Gene Experimental evidence does indicate that steady-state one-dimensional flow of heat by
de sucton through a homogeneous material is given by the Fourier Law

a
Cn a

em

“The heat flux g (heat conducted per unit time per unit area) flows along normal to area À
in the direction of decreasing temperature. The units on each term are
Q : rate of heat flow, K)/Ar
A + area perpendicular to the direction of heat flow, m?
lds thickness of material along the path of heat flow, m
dt: temperature difference between the two surfaces across which heat is passing,
degree kelvin K or degree centigrade €.

The ratio dx represents the change in temperature per nit thickness, i, the temperature
gradient. The negative sign indicates that the heat Flow i in the direction of negative temperature
Gradient, and that serves to. make the heat flow positive. The proportionality factor ks called
the heat conductivity or thermal conductivity ofthe material through which the heat propagates.
Thermal conductivity provides an indication of the rate at which heat energy is transferred
through medium by the diffusion (conduction) process. Fora prescribed temperature gradient
and geometric parameters, the heat flow rate increase with increasing thermal conductivity.

‘The Fourier law is essentially based on the following assumptions

+ steady state conduction which implies thatthe time rate of heat flow between any two
selected points is constant with time. This also means that the temperature ofthe fixed
points within a heat conducting body does not change with time : £ # AD.

+ one-directional heat low only one space co-ordinate is required to describe the temperature
distribution within the heat conducting body; £ = f(x). The surfaces in the yand
directions are perfectly insulated.

+ bounding surfaces are isothermal in character, ie, constant and uniform temperatures
are maintained at the two faces

+ isotropic and homogeneous material, ie, thermal conductivity has a constant value in
all the directions.

+ constant temperature gradient and a linear temperature profile.

+ no internal heat generation.

B

14 // Basics of Heat and Mass Transfer
er elation are enumerated below

‚me essential features of the Fou!
= ne ‘hough a medium from à region of high

owner an pe tone

Read region low sempeäure
Pea ra mater regates fs sae sold quid or as.
a ann ding ht he os ate enema an soler

and ie in the direction of decreasing temperature
ea ayy itis a generalization bases un

Fourie ation a

ie Eu nd Therma Conduct 418
scale erat conductance sr resents he noun sted
rough sold wall of are A a D cece es feat cond

is maintained across the bounding surfaces ® Tune diferemce of unit degree

iid flow ne

wade et camaro dred rom As inch
pero
RR Ee the warsport propery ki the thermal conductivity ofthe
ns median
as tr Ing A Tan d= 1, we obtain
Ont

tence ermal conductivity may be defined a he amount of heat contd per unit timo
ea al tucks whet à temperatura dleence of Un degre la
a a ea ataca The magie thermal conduc tlt how
Tare ner comport energy ty conduction

en eee cont are worked ou rom the Fourie aw weiten in the form

The unit k/m-hr-deg could also be specified as J/m-s-deg or W/m-deg and this is actually
done while quoting the numerical values of thermal conductivity

22. THERMAL RESISTANCE e
Observations indicate that in systems involving flow of fluid, heat and electricity, the,
juantity is directly proportional t the driving potentiaLand inversely proportional to

tane® In a hydraulic circuit, the pressure along the path is the driving potential, and
roughness of the pipe 3 18€ flow resistance. The ques Naik ductor is governed by
the voltage potential and ceci] resistance of the material, Likewise, temperature difference
constitutes the driving force for heat conduction through a medium,

From Ohms law ve IR

current () =

voltage potential (dV)_

ketal resistance (Re
and from Fourier's law. =

eat flow rate (Q) =

thermal resistance (dx / KA)

‘Obviously there isa one-one correspondence between the flow of electric curr

and heat,

— electric current (amperes) is analogous to thermal heat flow rate (kJ/h).
— Desire Voltage (Vols) Corresponds to Ihermal temperature difference (degree Kelvin)

~ SSE NSTC (er) analogous 1 quant 14. This quant i calle thermal

Thermal resistance, , = (AA), is expressed in the units sdg/] or deg/W. The reciprocal

a od |e

Le

eau ese

159 No
aa
Fig, 24. Concept of ermal resistance
Sometimes the heat conducting capacity of a given physical system is expressed in terms

of unt thermal resistance and unt thermal conductance €
# agenda
pus à
eet
=

ee

u, Ti concept af heal resistance is advantageously applied we main computations
or heat flow.

2.3, THERMAL CONDUCTIVITY OF MATERIALS
Thermal conductivty isa property of the material and it depends essentially upon the m
fracture (chemical compositon,phy ica state and texture), mostur coment and density
‘Matera, and operating conditions of pressure and temperature The value o hermalcandu
may range from 00083 W/medes for gases sucias Frer-12 0 as grit as 410 W ence for
metals such as silver. Apparently, the pure silver has conductivity almost 5,000 times as rest
28 that of Freon2
Following remarks apply t the thermal conductivity and its variation fr different materials
and under different conditions
Y () Metals are the best conductors while liquids are generally poor conductors.
1 conductivity is always higher in SITE! TORY ea Alloying of meta

ME TRE cause an appreciable decrease I thermal SORAUsWY—

| Mechanical forming (Le, forging, drawing and bending) or heat treatment of metal
‘couse considerable variation in thermal conductivity. For instance, thermal conductivity of
hardened see lower than that of annealed sel

AB Basics of Hest ona Mass Transfer

tcreases with ten
(4) Termal sondustnit of most metals si

oo q robe
empece
tan increase e vated temperatures also results in greater frequency

O ada inciso in molecular exchange rats

es ih sing temperature due to deter
me 2220
A uma conduct on sr Sal de

satyre growth; aluminium

pendent on pressure for solids and for

tae Ps Independent of pressure for gases at pressure near standard
gende AT RTE
amp A

Est marea, the dependence el rms conduct on empernture i almost
et mu
= a+
acre Ls he thermal condctity PC temperate and P E constant whose value
ET Tus conca be native or gaie depending on whether,
thermal conductivity increases or decreases with temperature. The co-efficient B is usually
ae nc an sat natal ocepuon magnesite OEE] and gave lor
en ae 3laminiom and coin onerous alloys)

Sh ae amd ey increases wih temperature fr ses wie tends to
arcanes enpertte or most fuit water big stale rio.

(7 Non-metlie solids donot conduct heat a ficienty as ital. For many of
nd RTE SAT Conca sone trick glas wool ark et) the her cad}
Say Tay ran pe 6 sample due to versions In Sucre. comportó densly and
ER A
Peral conductivity of porous mates depend upon the type o as or Liquid existing
in the voids. Presence of air fled pores and cavities reduce thermal conductivity because then
Se ber has coer card oss many i pas
pened
© Fra conductivity of damp mater consider higher han she ermal onde
of ney natal ang ware ker RT

Dery is another parameter tht aft the thermal conductivity of materia, thermal
conduct increases with density growth. At densities 40 and 800 kg/m thermal conductivity
‘Sins of hot a 0108 nd OS W/m des repo, Therma Educ, of son Sale
‘Proportional to its density. With a density of 9.8 kg/m? thermal conductivity is 0.081 W/m-deg,
td wi density 538 kg/m hermal conductivity increase to D627 W/m, Since dens
fir ones with dc tempera te vario other end th temperature
Would as low te same pater Thi information helps to state he rac of ie formation
ana lake or cua

Matra having a rian score have high aus of era conduct Ian the
‘substances In amorphous form. # u

réel arrangement ofthe atoms in cise of amorphous sois inhibits the effectiveness
‘of heat transfer by molecular impact. E

(8) Majority of engineering materials ae isotropic, that is, tir properties and constitution
rection from that point. However, some

inthe neighbourhood of any point are invariant with

Four Eaton ard Thermal Conducto 47

tetera eid me nor cos

pyar Yapa

e dec pues cee

fran Other maten ch crates eto ca ae reci actos De

Se laminated metals. Conducting mates a a ls
60) Base on experimental real, Wiedemann and Cy nade ne o er

ue to dretonalpefereners cause by
<<) Thermal cond l mon
Compara o hat in nec sos us

concerning thermal and electrical onu of» tera
“Te ratio o the thermal and electrical conduces seme or el metal

ature and hal rato is directly proportional to the absolute tempera

Let and be the thermal and lil ét mea temperature degrees
absolute, then =

Lor
k
or GF = constant forall metas

The constant is referred to as Lorenz number with the value Ly = 245 10% W ohma/K?
The Wiedemann and Franz law does suggest that material that = RFO Sera conductors
(pure metals viz copper and silver) are good conductors of heat too

(10) Materials with large thermal conductivity are called thermal conductors. and those
‘with small thermal conductivity are called thermal insulators. Insulating materials are used for
obstructing the flow of heat between an enclosure and its surroundings.

(@) Low temperature insulation (cork, rock wool, glass wool, catle hair, slag woot and
thermocole ete) are used when the enclosure is ata temperature lower than the ambient
temperature and itis desired to prevent the enclosure from gaining heat.

(0) High temperature insulations (asbestos, diatomaceous earth, magnesia etc) are used
¡when it is desired to prevent an enclosure at a temperature higher than the ambient
from losing heat to its surroundings,

(©) Super insulators include powders, lores or multi-Jayer materials that have been evacuated
of all air.

‘The low conductivity of insulating material is due primarily to ar (a poorly conducing
1828) that is contained in the pores rather then the low conductivity of the solid substance.

Substances under low temperature conditions that have exceedingly high thermal conductivity
are known as super conductors. For example thermal conductvity of aluminium reaches a
value of 20000 W/m-deg at 10°K and this is over 100 times as large as the value that occurs
at room temperature.

(11) Pure metals possess the highest thermal conductivity (€ = 10 to 400 W/m-deg). Heat
insulating and building materials have a comparatively low thermal conductivity, (k= 0.023 to
29 W/medeg) and it ranges from 02 to 0.3 W/m-deg for liquids, Sull gases and vapours
possess the lowest thermal conductivity within the range 0.006 to 0.05 W/ m-deg, Evidently the
ratio of conductivity values between metals and good insulators is about 10* and it causes heat
loss to be an important factor in thermal systems. In contrast, the rato of electrical conductivity
Between good and poor conductors is about 10% and therefore the current loss through insulation
in electrical net works is negligible.

(12) Thermal conductivity of different materials decreases in the following order

A8 / Basics of Heat ond Moss Transfer

0 Poema
an
oom
(of space coordinate as well as time.
more pue fi il volume element oriented in à th
in dap 2 2 Te ses dd and e have ben ken para
a

nd amorphoos substances

ar

5 loe Ce
4

Fi. 22. Conduction ana in
Canale co rénales

Fig. 23. Change in temperature as à
functon of distance

Let t represent the temperature at the left face of the differential control volume, Since area

af is face can be made arbitrary small, the temperature may be assumed uniform over the

entire surface. The temperature changes along the x-direction and the rate of change is given.

by d1/0x. Then change of temperature through distance dx will be (1/0x)dx, This temperature

change has been graphically illustrated in Fig, 23. Therefore the & f
erefore the temperature on the right face,
phic lies ata distance dx from he left face will be [t+ (1/x)dr]. For non-sotrophic materials
ere ill also be a change in thermal conductivity as heat flows through the control volume.
„The general conduction equation can be set up by applying Fourier equation in each cartesian

‘direction, and then applying the energy conservation juirement ef thermal
Sonden at he et ace then Quant oi o he or

heat flowing into the control volume throu
face during time interval de is given by as ce te eee valine ees

Fourier Equetion and Thermal Conductivity) 19

Bo Qi 2 a

a»
During the same time interval the hat flow out ofthe ight face of the control volume wi
bo
Heat flex Q, + À em
ea Qt 20 a

Equation 27 simply slates thatthe component of heat transfer ete at + de fs quatro
value ofthis component as plus he amount by cs whe na 30 Wal

‘Accumulation of hat inthe elemental volume due fo hat ow nthe nahen gen
ty the france between heat nf and heat un Thos the sac DEE
Pw incio le

a
«0 (urge

a

%

(Q) dx
af a
= Aft san a)

ape

Likewise the heat accumulation in the control volume due to heat flow along the yrand
rections will be

a, ar)

40, > FB Seay ae at es)

es

Sum of heat accumulations as prescribed by equations 24, 25 and 26 gives the total heat
stored in the elemental volume due to het flow along all the co-ordinate ae.
Total or net accumulation of heat is equal to
af, HY, a0, a), a, &
2 (6 Ve E IS en
oe en
There may be heat sources inside the control volume due to nuclear fission, flow of electric
current in the coils of electric motors and generators, and ohmic heating of the material. U ,
is the heat generated per unit volume and per unit time, then the the total heat generated ih
the control volume equals 10
qy dx dy de dt es
The total het accumulated in th ate due 1e heat flow along al the eoordante aes
(Eq. 27) and the heat generated within the lettice (Eq. 28) together serve to increase the
thermal energy of the lattice. Ths increase in thermal energy is reflected by the time rate of
‘change in the heat capacity ofthe control volume and is given by

px dy

a
a 29
x” 4

a rrr un

20 Basics of Heat and Moss Transfer

here p is the density and is the specific heat ofthe material. Thus from energy balance

cities >

22) 20 au dy dz ar #9, a dy de de o dv dy dec Kar
ES ies |

Dividing both sis by a d dd

af, 2), 22,4) og = pe À 210)
2).263)- 262 gree 210)
cx, using the vector operator Y,

a
vam 2100
tin 210 represents a volumeti eat Balance which must be sisted at each point
forse generating unctody sn three dimensiona heat low tough a non sotopic moter
‘hs expresion. Known asthe general heat conduction equation, establishes diferent
form te relaonhip between the Une and space variation of temperature at any point ofthe
Sel tough which conduction takes poe I shouldbe noted tot the hat generation term
fy may bes function of postion or tine, or both
Homogeneous and isotropic materia A homogeneous material plis that the proper,
de. den, epeafie het and termal conductivity ofthe material are same everywhere In he
miel tem, Kotropc means ha hee properties ae not directional characteristics of te
| material de, ey ae independent o the orenain of he surface. Therefore for an iotepie
and Homogeneous maleta, thermal conductivity is same at every point and in all direction
Ita a E +, =f, =k and the diferentil equation 210 takes the form

at, wt dt D peat 18

E He HD om

a Tin ae a

The quan = £/ocis called te thermal duo, and represents a physical property
of the marea of which e solid element I composed. Thermal diffusivity isan important

characteristic quantity for unsteady conduction situations By using the Laplacian operator V2,
the equation 211 may be written as

met ia eis

Equation 2.11 governs the temperature distribution under unsteady heat flow through a
homogeneous and isotropic material

Different cases of particular interest are

() In many situations there is no dependence of temperature on time. Conduction then
per
‘occurs in the steady stat, and the heat flow equation reduces 10

de dt dt, 8
2,2%, ,%,%
EE

4,
or vrs «0 (Poisson's equation)
In the absence of internal heat

ex)
216 further reduces to:

generation or release of energy within the body, equation

Founer
Fu Equation and Thermal Conductivity I 27
q,
O
or DEN

(i) Unsteady state

(Laplace equation
eat flow with no mera! heat acne

ag
ped hed gaat ga
Be ae ae ds
E
Co extn ES
Mo

general conduction equation takes the form

£(#)
alfa)" rn (2.16)
Soluoncí thee eats o ny pectic ry cans wt
distribution in the conducting material. = Haute ones

th no internal heat generation, the

Thermal diffusivity: The following reflections can be made with regard to this physi
property ofthe conducting material sar to this physical

ia ATES ao» ‘ters the rato of 8 tena conduct the

ab storage Cafe BE The heat Worse each a en

capacitance or thermal inertia of the material, Y
NENAS

en ve. upgshnes o sonduct he A
resul either Ho à igh Value ot dermal
conauetivny oF rom Tow Vale Dr the

m capacity. igus hase a u thermal

ty high thermal iets and enge am thermal doses Mana

thermal conductivity: Tow thermal inertia and Rene Trge termal aes,

cl fus indicates the cats which heu ditbuted ina material, and

this rate depends not only on the conductivity bat also on the rate at Which peaterergy

an be storee-Anv sig into equation 211 would reveal thal large the heal diffusivity,
‘of

iger would be the rate of chargeafteriperatne at any po of outa eon
af merle al hon posed st a ihe le mar Fang ne eral
diffusivity: u

npetalure distibution in the unsteady state is being governed both by therm
ci 39 well By thertal storage Capac. In contas, during AeSOY sate
Coe (Eg. 212) thermal Coney ls the Sly property of ETO
iiénces the temperature distribution.

tude of thermal diffusivity isa mesure of he rapidity with which
the material responds 1 Ridden temperature Changes I He Ang Meal and
aso have flatively Lage Val of cand thei response fo Keira TOTS Ts
quite rapid. The poanhetallic solids and liquids respond slowly to temperature changes
Because of ther relire small Vale of berms user

24:2. Cylindrical Co-ordinates

When heat conduction occurs through systems having cylindrical geometries (eg, conduction
through rods and pipes) itis considered more convenient to work in the cylindrical co-ordiantes.

‘The general heat equation can be set up by considering an infinitesimal cylindrical volume
clement

ranstac
22 j Basics of Meat and Mass T

RE eg nd avt tos

and ern nergy Balance th
Assumptions
i ternal condo
enon,
(69 uniform beat generation atthe
(e) Radiat direction (r= € Plane)

pant pci hs fo the material do no vary with
density pa

ate ol y per unit volume per unit time,

eat into Q, == Kr de

à ayer
Hest an 64 * @* (DE
teat steed inthe element due to few of heat in the cial direction

40, ©Q, = Qu

E al (nla
sóc dy (Je

ru E

2 (ar rp de) ( jé (2.160)

a rar)

ue

Elemental

Urs

Fla. 24. Conducbon analysis in indica! co-ordinates

Founer Equation and Thermal Conauctivity 4 29
(6) Tangential direction (rt plane)

Heat influx Q,

Wray Ea

a

Heat eu Qu + 2 (09 0
O sn Qu + za (QD ro

Heat stored in the element due to flow im the tangential direction.
IÓN

a
Fe ad do

2 tendo

ok (ded
rl re

afıa),
Elia
a

= (tra ds)

ak (dr rdo E de en
(6) Axial direction (r - 9 plane)

ae

Heat efflux Q., 4. = Q, + 5 (Q)
Heat stored in the element due to heat flow in axial direcion,
40, =.= Q à

(rdo dr) 2 drs = à (ar ndo de) (218)
a}

(4) Heat generated within the control volume
= q, dv de (2.19)
() Rate of change of energy within the control volume

pave

(2.20)
From energy balance considera hin the control volume
equals 4

the rate of change o ener
total heat storage pls the hea generated. Terre,
MERA
EE
Diving Both sides by dV de
¡ETA

a

k ramos

de oe 1a

Ca

e

es Transfer
RES à direction, and with no internal heat

ie rad

For steady-state wn-tinection heat flow in 1
generation, equation 221 reduces ©

noord «omit ex

“The general heat conduction equation in spherical co-ordinates can be st up by considering

an names pei solar clement
” AV = (dr x rd0 * r sin 0 dé)

“and writing the heat Balance equation for the r, 8 and 6 directions,
Oe

Elemental vehine

io,
9.25, Conduction nas in spherical co-ordinates
Assumptions: () Thermal conductivity k, density p and specific heat «for Ihe material do
not vary with position, (i) uniform heat generation at the rate of 4, per unit volume per unit
time.
(6) Heat flow through +8 plane; direction

Heat inter Q == Kar od) So de

Heat efflux Q, « à = 0) + =.

a
are A sin Op

Heat stored in the elemental volume due to heat flow in the ¢-direction
dQ, =~ Quy

Fourier Equation end Thermal Conductivity 128
1a

Fundos (0) rn 0 de

“Taine à [klar

=H (dr 8 x rin 0 dp =

em
(8) Heat flow through 4 plane 8 direction
Heat nu Q == (dr ra 0 49 ae
Qu kr sin omy) die
et MEN
Heat eflox Ob = 5+ 7 (0) 9
Heat stored in the elemental volume dueto ha! flow in the rection
40) =~ Qo sm
a
TH (Co ne
2
an [raros
Lts 2 (ogg
IE
(de «180 sin 9 49) 22)
(6) Heat Move in the 0-9 plane; recon
Heat lox Q == (do sino do) a
Heat efx Q, , 4 = Q + D (Q) dr
Heat storage in the elemental volume dueto heat low in the direction
4Q, = Q,= Qu
a
dar
a a
À La dr sino agp ae] ae.
5 ren]
cross à
ek rx nox nods) 2202 es

‘ana Mass Transfer.
1 volume
atin te conto
a dv te (22

26 Basics of Moot
(0 ot gonad |
ena ot ange eer i e rl

ae ml
Iv rate of change of energy within the control Volume

e comer heat generation. Therefore,

a er
trage os de mem
120, 1 me Jara? a
SE ) 2
ving sides ade ;
fa %
am

part

ical co-ordinates.

eat generan uation 231 can be rewritten as

Fal om

The one-dimensional time dependent heat conduction equation can be written more compactly
sa simple equation

Baltes] aces om

where = ,1and 2 or rectangular, cylindrical and spherical co-ordinates respectively. Further,
hile using rectangular co-ordinates its customary to replace the r-variable by the x-variable,

2.5. INITIAL AND BOUNDARY CONDITIONS
‘The general heat conduction equation is of fist order in time co-ordinates, and of second order

in spatial co-ordinates, For the solution of single integration one constant is required, and for
the solution of each double integration two constants are required. Obviously for the complete
solution of general conduction equation, seven integration constants are needed. These constants

are evaluated through a set of intial and boundary conditions.
The initial conditions describe th ‘bation ina

of me, and these are needed only forthe transient (aime dependent) problems. The

initial conditions can be expressed as
At TsO t=t(x y 2) (2314)
For à um EEE dation, à simple but typical form of the above
ur RER ihn, a pl ut tpl fr of
ea

Att 20/6 y = constant

The boundary RAR TT FAST conditions existing atthe boundaries of he

medium, and spec the Hr
fan and sec the temperatur Se hs wa ar of he ad The SEE!

Fourier Equation ana
"Termal Conductivity 127:
(0 Presta sure temperature: The temperate driv
TE TOMER SU, prescribed at bounding

a DATES) =

Foc Bode un Ne ee pin 26 we a
ama 00)
yeO: tay De 0
Y ob: ya flay

Myst

0207]

1.000
Fig. 26. Boundsry condion-tt hind
(ip Preseribed heat flux: The heat Aux is prescribed atthe boundary surface, and is expressed

Fig. 27. Boundary condition 2nd un

233)

at a solid boundary when there is
id tat leaving the surface convection

== in Come condon “Thi ong a ecos
«ual etw bat tants th saco By ORÁN

ig. 28)

Aro
2348

„ea

28 1 Basics of Heat and Mass Transfer

BL

.

pese]

Fig. 28, Boundary condition nd kind

‘econvetve beundar conditions described above can be expressed inthe com pac form

238)

he specific convertion sont is he prescribed function and 3/0 he
ts AN

o RENE sation of general heat coructon in closed form due to
a ne peter Jd ae element y pes 6 numerical techniques

e proßem: Fre frees and fl
O lor the appronuimae soluto of heat conduction problem
EXAMPLE 21.
de) Explin the mechanism of hat conduction in metas and insulators.
10 tage vanne factors which influence the thermal conductivity of a substance. In
conducir feed bythe solid qui and gaseous phase ofthe substance?
1) catan some ofthe situations where poor conductivity of air helps to restrict the
bea transmision by conduclon.
Eoletions Im inalaters (glass wood, asbestos) conduction of heat takes place due to vibration
Sharon about thelr mean posos, When heat given to one part ofan insulating substance,
Some belonging to dat par are Put ina violent state ol agitation and start vibrating with
pester amputoves,Concequenly these more active particles cole wi less ative atoms
Fring nex fo them During colision, theless active alos also get exited ic, thermal energy
imported to them. The proces is repeated layer after layer of molecules/atoms until the
ther par ofthe Insultos reci
Im metals besides atomic vibrations, there are large numberof fre electrons which also
participate in the process of heat conduction. When a temperature difference exists between
he diferent parts ofthe metal a general di ofthese free electrons occurs in the direction of
decreasing temperature, Iti his drt of tee electrons which makes Ihe metals so much better
as conductors than other solids These fee electrons account forthe observed proportionality
tween the thermal an electrical conductivities of pure meals
“The icons do nt contribute t eat conductivity in insulators. Electrons in an insulators
are not re Du fixed in the valence band. According lo band structure of solis, the energy
gap between valence Band and conduction band is quite large and electron cannot move 10
Conducton band and contribute towards heal as well as elcrcal conductivity
Insulators have low value ofthe

al conductivity due o their porosity, which may contain

|
\
|
|

| fc ota ct
+ gaseous, liquid and solid phase in which the substance mesi. i

(e) Listed below are the applications where poor conductivity of air restricts the heat
«mission by conduction:

() Eskimos make double walled glass houses; air is enclosed betwen the walls and that
reduces the outflow of heat from the inside of houses.

(6) Woollen fibres are rough and hence have fine pores filled with air. Both wool and air
are bad conductors of heat and do not allow the body heat to flow to the atmosphere.

(di) Two thin blankets are warmer than a single blanket of double the thickness because the
two blankets enclose between them a layer of air. A simple blanket of double the
thickness does not have air entrapped in it, and so it does not provide as good an
insulation as the two thin blankets

(Gv) Birds often swell their feathers to enclose air and thus prevent the outilow of body
het

REVIEW QUESTIONS

21 White the Fourier ate equation or hea transfer by conduction. Give the units and physical sgifcance
Of each term appearing inthis equation

22 Define thermal conductivity thermal es stance and thermal conductance. What ls the approximate
range of thermal conduct of solids, quid and gases?

23. How are Fourier law and Ohm's lave similar?

24 Lit some good conductors of heat; st poor conductors,

25 State the effect of impurities on the thermal conductivity oa metal.

26 How thermal conductivity is affected by the nature of solid sate (crystalline or amorphous)?

27 Discuss the eect of temperature on thermal conductivity.

28

Name the various types of insulating materials used in engineering, and mention the specific
purpose for which they are used.

19
zu
zu

22
zu

zu

f
|
d
i
zu
a
2
zw
2
a
zu

fer
30 Basics of Heat and Mass Te“

ung mes oer than hat of meta
cms mt

wach my a ea a a
wet or rater
pin why feat than DS pe

Senger grue maes hautes of lodo ES

oem Ca
oe ‚plain its physical sigs
ne oy non a
oo idity of following stater

Cement pn ar swe on coats e

2 Son conduct is ine aby of
a ery a Js always higher tran tha of ts alloys.

Thera conduct 6 à PU y higher than that of gases and vapours

ost cam be expresos BY

ES conduct het and thermal fit 6 à

Pra
aso, ie and thermal dios, and represent
sehen and respectively denote emer
da dd ee stan co-ordinates fr a homogeneous
“ne mensional conducin equation in cesan «fr» home
row at th Sern and whet et generation gue bythe Laplace es
Ñ veo
2 posent the Lapin operator
re een he Lense steady se bes conducto though dl
De and al conducen equation In end cordiats for a homogeneous
Del sin Jr unreconal unsteady ate stem when est is
DR wan ita the ate of perm? of the mater, ; BR
a rites for tres dimensional unten
ss he geal feat equation In caren co |
Far aii anna! volume ment, Deduce therefrom the conduction
sons forte flowing coe “ M
andy sate ao dimensional low with het generation at uniform rte within the mates
(9 seady one dieron! flow without hat generation
o sen tre general conduction equation of nen state of hat low and with uniform eat
dinero for sang tem coordina,
Bore ne genera condocton equation fr ()elindsial co-ordinates (i) spherical coordinates
the syste bg with arora heat generation and unsteady ste
Wie agent conduction equation for ore dimensional est low ith unform heat generation
td rey sut for sonia ne and spherical eoordinate systems.
‘What re boundary and intl conions? How many boundary conditions are needed to solve à
cond order ional equation or het conduction

Sat and explain the diferen types of boundary conditions applied to heat conduction problems.

ago

Steady State Conduction

Free oer ir ers pl tati fe
ep ne
aer ed serlo sal ar
pati es ate peana cas a SS ner aie pe
Be sto es loca halo a oor LR ES lay ee
3,1. CONDUCTION THROUGH A PLANE WALL
ae onecimestonal bat conducon hough a homogeneous, bee wal of nse
ee ee ed
D nen tad sa Un EE vaca | ml tee mines à ei
A o e e ep

Let atention is focussed on an elementary strip of thickness dr located ata distance x from
the reference plane. The temperature fi

A] one to recon
ay empermun Se eal
dx. Since Q is constant in a steady state, us
Q=- KA (dt/da) may be separated and ”
3 > a
Qu «ws ja
gs ant
Ga
a
ore on

temperature distribution can be set u
by integrating the Fourier m
ven be mt à à 1
() x = 0 where the temperature is
stated to be 4
d to be purge Erk
(i) x = x where the temperature is to be worked out

Fig. 3.1 Steady sate conduction tough a plane wall

ronson
32 1 Basics of Het end Mass T
Ts. . .
oja =
f y 62
que tan
ex
and 32.
Comping te expressions 31
‘ate „MA
«th lux .
ne ner I ,
A CCE
fret fiw ao, ation 3 cam
o er on
A ace to heat low. Equivalent thermal circuit for heat low

Cero, EPA he rol resin fo
wee By fe val has bn nied in F

Fern |

À or af an ven ls ml
Taco ue oe wa ar 00 mm
e nc 03 Wade, For an 0

Re 1 es flaw andthe het lo pet SQUREE e

ga.

wed at a temperature of 850°C by means of à suitable
tained damn thick and are fabricated from a material of
de wall temperature of 250°C, workout the

Are of wall surface. Also calculate

en at point 20 a en
nes conc and ace A he thermal ren
pr
2.05 ,
E 05 1607 de
DA a

or one-dimensional steady state, the conduction heat flow through the wall is
= „ 850-25
eto

) x 02 = 610°C

EXAMPLE 22
À plane slab of thickness & = 60 cm is made of a material of thermal conductivity
= 173 Wim-deg, The left side of the slab absorbs à net amount of radiant energy from à
radiant source at the rte q = 530 wat. Ifthe right hand face of the slab is at a constan
temperatur f, 38°C, set up an expression for temperature distribution within the slab 3
2 function of relevant space co-ordinates. Therefrom work out the temperature at the mi
plane of he sab and the maximum temperature within the slab. Tt may be presumed it
‘he temperature distribution is steady and there ls no heat generation.

Steady State Conduction 133
Solution : Under stipulations of steady state and no he

o ie cat generation, the energy absorbed from

ot which à 1 conducted through the slate

Heat flux

7514-39)

590 -
0 06

Temperature at the left side of slab,

rs ‘
30.0.5 ma We
= ES à 38e ssare a e
E =
This also represents the maximum temperature within the lab. LL ©
From the expression for steady state temperature distribution, 2

EXAMPLE 3.3
A metal piece of length 60 cm has a cross-section corresponding to a sector of a circle of
radius 10 cm and included angle 60°C. Its ends are maintained at (Emperarare OFIES-C and

25°C, and the thermal conductivity of the material has a linear variation with temperature
in degree Celsius.

k = (100 - 0.01 1) Wm-deg

Find the heat flow rate through the metallic piece. Presume uni-directional heat conduction,
Le, neglect any variation of temperature in the 6 and directions.
Solution : Average thermal conductivity,

er

= 001 t, where 4 =

99.25 W/mdegz Ft

Aa
TT
ET

4) _ 99.25 5.23 10° x (25-25) |

where @ isthe INTRA Aigle in radians 7 = 7-19 1

CA
fo IES ogy caps;

Heat low Q
ae ‘

EXAMPLE 34

Derive an expression for temperature distribution and conduction heat flow in a circular
amical rod with diameter at any section given by D = ax where x is the distance measured
{fam the apex of the cone and a is a certain numerical constant. It may be presumed that

Is no internal heat generation, steady state conditions prevail and that the lateral

and Mass ander 7 al value for Meat Slow rate
a to obtain the namen ow va
tc PERES en
i an as meter a equals 020 and the Material op
e peratare of ey a 345 W/m-deg.
ge hem te UNION CPR fon,

ne direction for wi

341 Basics of Heat

om
way Den 070 Pe and pt IN

40 a

€ variables
Well insulated lateral Surface

Subetiuting for A
oka

since and Q are consta, negation From 319
Shun he conical rod velds

wii
t ao fé
aja- BJ
4 OR om
= 8m
Fig 34,
9
Erlen ofthe above expression atx = % where 171 te,
1)
4 “BE >) i)
‘The heat flow rate then works out as
m7)

nek
4/1 -1/0)
en substitution for Qs mude in expresion () we get the tem
Yx-1/% +
His pal ‘
Sobstuing the given data expression i) fr he flow rte

Q = ODER 5 142 y
4(1/0.06-1/0.3)

perature distribution as

EXAMPLE 2.5,
Heat flow occurs along the axis of a solid which has the shape
«ircumferential surface insulated. The base is at 300°C and the area of the 5
x measured from the base of the cone is given by

‘A= 12 (1-15 2) m? where x is in metre

of a truncated cone with
rection at distance

‘Steady State Conduction 1138

It the plan tx #02 mis matins a 107, determi

at 2 9 GI a and (€) the tepecaaro radios ioe en ee nes ion ea emma
"Take thermal conductivity ofthe solid material as 25 Me a.”

Sotelo: The isa Grenier ura Ing Bate Léman come

dimensional conduction in he redirection for which tation corresponds 10

a
CET WE) ates rece
pas
2 a
Sepaatng he variables and upon integration —>
de
e -120 ja
fist oan
Y cr Qu
ET
2015 4)
Heat ow, Q =
Log (1-15(2 x)
1.222 $x23(00 100)
Heat flow, Q = og [1502-00] ~ 285 W
(0) Let be the temperature at x = 02 m. Then
ya, _MEXISX25(200—1) | 450000
252385 = og [1150-00 "01625
1200 RS ac
(0 rom identity. Q = = À dé, e temperature gradient at section maybe writen as
2.2
au
(Atre0; A= 120-150" 1008
a, 2503.85
PRET ac
(i) Avr 02 m; A= 120-1500) "1m
MN na
ar” 25x10
(9 Atx = 02 m;A = 12 0-15 % 02) = 08m?
2503.84 +
BEAL = nos ga

3.2. CONDUCTION THROUGH À COMPOSITE WALL
‘A composite wall refers to a wall of a several heterogencous layers, e. wall of dwelin
houses where bricks are given a layer of plaster on either side, Likewise furnaces,
boilers and other het exchange devices const of several layers a ayer fr mecanica rene

|
|
|
|

Tronster
ang Moss.

ot eat N

ot

36 1 Basis

SE et hot ri
we appara tondinacy Beck!
Sr a ee ch
sn Re ame are a
corer a muta
STE engine

a lover of Tow thermal conductivity m

er layer for structural require al

rie ent fo

nar ca te AN

layers ol fern materias tig
ES Band thee termal conde
rh ice temperature OF Ee mal

ste wall ong As

Intertacos

a

Temperature prole

Fig.36. Seay state conducbon trough a composite wall

Under stedy state conditions, heat flow does not vary across the wall, fi is same for
every layer: Tos

kA

A La a
ann

Rewriting the above expresion In terms of temperature drop across each layer,

26 Q& 2%
ha TRON" BA iB Ba

‘Summation gives the overall temperature difference across the wall

(2, b,5
GA "La ra)
Then
ee A 03)
Gars Are BA” R,+R, +R, ‘
Distt

ig. 36).

‘buting of temperature ina plane Muller wall

represented by a polygons ie

‘Sceody State Conduction 127

In the above analysis is extended to y "
Whe laver composite wal, one obtains

Gen 7
Lana mr

where ZO/RA 1 sum oft thermal resistance 4 a
Anis othe con
oo RP drop Dee

EXAMPLE 36 1

An exterior yall of house may be approximated by 10 0m lye o common Wo
(5 075 Wade) followed by 4 em layer of gypsum plaster = DS Winde. ek
thishnes o loosely packed rock wool insulation fe = 0.088 Wim den PRÉ Mat
reduce the heat 1096 or gain trough the wall by 100

Solution; Thermal resistance for plane wal of cho 6, te À
Eis prescribed by the relation R= BER Consort

How necio,

nt layers composing the compost wal.

there is a perfect contac et
BENGE materia

thermal condcutivity
1 unit area of wall perpendicular 1 heat

ot
4 wa

0x
Resa otek work = 22 ons ag O
tan

Resistance of gypsum plas La
Sn 5x7

= 008 deg/w

Resistance of rock woo! insulation = X

x10
Saas” 01538 x deg/W

where x is thickness of rock woul insulation in em
Case 1: Rockwool

Case HL Ro

woo insulation is used
Heat loss or gain = (1 - 075) x 4695 al

se |
O)

1

SN a di

oF CI + 008 + 01588 N = = 09918
0.8518 -0.133-0.08
Thickness of rock wool, x = 289 ai em
ús 01538 m

EXAMPLE 3.7 1/7
A toroge chamber of interior dimensions 10 m x8 m x 2.5 m high has ls Inside maintained

Ah temperature of - 20°C whilst the outside is at 25°C. The walls and ceiling of the
chamber have three layers made of

60 um thick board (k = 0.2 W/m-deg) on the inside
30 mm thick insulation (K = 0.08 Wm-deg) at mid =

u wenster
30 Basica of Heat and Mast TF

me
Deane

‘i side
on the elt rate at which heat can flow
er ermine the 13
Neve ‘of heat throu
Sean den “
= e ad ges be nt
ee am An
dono en
cling thermal e

PES
4107, 20x10"
a "18
01578 deg/W
= 8.7 w
Heat ow
EXAMPLE 3.8 ES

a of hose i col region compis three ayer
Te ea mane rick work (k= 075 Warde)

asm Inner mooden paneling = 02 Wn-deg)

e anaes e De ut
woe ade Ind ouside temperate of he componite mall are 20°C and SC
Se écrire Le rate of heat lose per unit sea ofthe wall and the thermal
Conduct of the insulating material IO
Station Thermal resistance fora plane val of thickness 8 area À and thermal con
Tis pected bythe raión, = AA

Resistance of brick work, R, = = 02 deg/

125%10

Resistance ol wooden paneling &, = Eo

0.0625 deg/W
Resistance of insulating material, R, =2 x 02 = 04 deg/W

Total resistance Ry = SR, = 02 + 00625 + 04 = 0.6625 deg/W

a. 9-615)
feat loss Q = yr = = w
Han Cr qu S28
(Thermal conductivity of insulating material,
O os Winde
ea Gh RE oars wnes

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Steody State Conduction f 38
EXAMPLE 39
A furnace wall isa made up of sect plate 0 mam nick
ved sica Brick 150 mm thick (65 732 kmh gy ans pen Se ne om ini
bricks 200 mm thick (k » 18.84 Ky/mchr-dept The inside and outside surfen ey me an
at temperautre 650°C and 125°C respectively. Make calelations forthe bent ane fa at
Ares of the wall

Ati required thatthe heat loos be reduced to 10 Mhou by means of ss gap between
steel and magnesia bricks. Estimate the necessary with of si gap Y oral Pci
for air is 0126 Kifmhrdeg.

Solution : Thermal resistance fora plane wal of thickness

8, area A and thermal conductivity
Fis prescribed by the relation Ry = 8/LA

o
Resistance of steel plate R, = OT = 0009159 deg hr/kj

Resistance of iia bricks R, = 002049 dep ak} mil =
noes
he

0.20
Resistance of magnesia bricks R, = = 001061 deg hr/1

Total resistance of the composite wall,

ER = R +8, +8, = om deg hey vf >
0-105

Se Goma “Su EF

lue the eat los to 10 MJ/h, the thermal resistance shouldbe increased 10
we neh
TOR” = 00828 deg hr/

‘Thermal resistance for the air gap,

Heat loss from the wall =

FA * 0025 - 0.08126 = 0.02124 deg he/kf
Thickness of air gap, 8 = 0.02124 = 0.126 x 1

= 2676 x 10% m = 2.676 mm
EXAMPLE 3.10

A furnace wall com
iddle layer of ins

bythe dre
5m = 2m and the data on thermal conductivities en
Fire brick 212 Wine #
Insulating brick fy 2 O24 Winden
Red brick fy 2 088 Wines

Solution : The wall area (5 m x 2 m) = 10m?
The th

8

A

prises three layers: 13.5 cm thick inside layer of fire brick, 75 cm thick
ulating brick and 11.5 em thick outside layer of red brick The furnace
Sgen at 870°C and itis anticipated that the outside ofthis composite wall can be maintained.

ulation of air. Assuming close bonding of layers at their interfaces, find
Sie tate of heat loss from the furnace and the wall interface temperature. The wall measures

or
is constant for all layers of the composite wall,
ermal resistance R, ofa slab (thickness 8, conductivity k and area À) is given by

A ds el

renster
40 1 Basics of Heat ona Mass Irons

Resistance of fire Wick A"

rating beck Re *

sistance of io
AUS. 091358 dex/W
A ak EI
1
compost al
y Rae Ra + Rs
omace (temperature ly

ota restar of he Los da
yw toma idee = 870°C) to outside of the
engin all gemperatune fy * 1°C)

veto = YF
gh each ner i same, then fr in
y Eh ee
105935 = 9 61725 0.01
lve of re brick and insulating bre
ep me e E
© = 75082°C

SE. 105935 W

nu

side layer of fire brick

(Since heat flowing I

Similarity forthe mid layer of inslating brick,
. 750.82
= "9.05857
“insulating brick and the red brick
= 105935 x 0.05357 = 183.33°C

where fs the temperature atthe interface of

fy = 73082
Check: The temperature t could le be worked ou by considering the heat love trough
the oud lpr of md bck
hu | baad

105995 = 01357 ” 0.01357
D + 105935 x 001357

b
This is same as calculated above
EXAMPLE 341
Explain the concept of thermal contact resistance.
A fornce mal consists of an inside layer of silica brick 10 cm thick. (k= 6.28 KR
1e outside. The

followed by a 20 cm layer of magnesite brick (X = 20.95 kj/m-hr‘C) on th :
inside surface of the silica brick wall is maintained at 750°C whilst the outside surface e

magnesite sat 125C. The conact thermal resistance between the two walls at the interface
8 000076 bef per unit wall area. What i the rate of hat loss per unit area of the wal!

Also calelate the temperature drop atthe interface. di
Solation : Calculation of heat flow through a multi-layer composite wall are made on the
presumption that E per pe

(O thee s perfect contact between adjacent layers

pi ee

‘Steady State Conduction // 44

(i the temperature is continuous at the interface although there is discontinuty in

temperature gradient
(ii) there is no fall of temperature atthe interface, Le, the temperature of the layers are
same atthe plane of contact AR APE of De Wn
However in real systems, the contact surfaces touch only at discrete Locations due to surface
roughness, interspersed with void spaces which are usually filed with air Obviously. there is
Ro a single place of contact. This implies that the area for heat flow atthe interface will be
ol compared to the geometric area of the face. Due to his apparent decrease m the heat
flow area and ako due tthe presence of air voids, there occurs a large resistance Lo heat flow
ithe imerace This resistance 5 referred 1 4 thermal contact resistance ad it causes temperature

drop between two materials atthe interlace.

lo
(No temperature crop at imite (0) Temperature dro a meta

Fig. 37.

Let ty and fy represent the temperature at the theoretical plane interface obtained by
consideration of heat flow in materials on either side, Then for a given wall area A, the thermal

A resistance R, is defined as

a y
The value of metlliccontact resistance depends on the metals involved, the suface roughness,

DZ]

the contact pressure and temperature and the matter occupying void spaces. The values of
contact Ihermal resistance are obtained through experiments. or solid to oli interference, the
temperature drop is usually small and the effect of contact resistance is generally ignored.

Thermal resistance for a plane wall of thickness 3, area A and thermal conductivity k is
prescribed by the relation R, = 8/KA

nce of silica brick R,

Resist
Resistance of magnesite brick R,

R

Contact thermal rsistance

Total resistance of the composite wall,
ER = IR +R, +R)

(001592 + 0.009546 + 0.000716) = 0.02618

des

3 70-

as 2873 Kine

Heat loss form the xs

142 1) Basics of Heat and Mess Transfer

‘Te same heat ows thorugh sh a
trick

4 X

q 070-287

fy 2105 + 23873

Hence, temperature
eh)

“eo compost sistem. According, or Ue lic

E
23873 * 9.01592

001502 = 369.95°C

ns
207 = Ge
soompi = 35289°C

drop at the interface
Pi) # 36995 - 35289 = 17 06°C

EXAMPLE 3.12

Determine the heat transfer rate across à composi
in Fig, 38 The ent

with top and bottom as shown
Y. while the entire right hand face i at th
different materials ae stated ask, and ky,
thickness 8 are A, and A, respectively. Ste
the slab is so thin that any edge effects can

“and their areas as viewe

ite slab which is made of different material
ie left-hand face is held atthe temperature
ature Ty. The conductivities of the two
the direction of slab,

vagy state exists, there is no heat generation and
te neglected. Interpret the result in terms of an

rote
Pes.
i asthe,
& Ps

Solution: Applying Fourier law of heat condi
AT)
3

7 | a
370,

1,1
(Rae

Fig. 38.

luction separately to materials a and b , we obtain

RAT =P
y

1
a

1
“eg TD

‘Seandy Sete Conduction
“Apparently, the two thermal resistances de
3
Re a wal
ECS

Y ough the given composite slab will be as indicated un fit sae
EXAMPLE 3.13 te © ue
rte 2 20 gm kn of a im eal conte ai
Sres. If the extreme surfaces of the arrangement are at temperstarte of 2506 sea are
determine the heat flow through the composite system, the contact jest
Lane ns eee
(of the confact area) is due to either metal gp serene OM SE eat

lem and its equivalent thermal resistance
Matai A

1,2500: toro
A e A,
12500 yo
Fig.
The various thermal resistances to flow of heat are
3, 100x107 À
OR, ep 0 deg/
8 _ 0.02107
CR, = Goa = enon” 7 10000685 dew

44 Basics of Heat and Mess Transfer.

8, 0028107. oomos2 deg/W

O

Su, DMX „00000005 des /W

wo) Re

oR = Pot pd
yd thie equivalent resistance (8), à

“The meisten Ra. Ry and E, are in parallel a

2
wR,

1

1 sit
= Goo * O0IOEZ " 0.000000
2160000 + 9597 + 2000000 = 21609597

0462 x 10% deg/W

“This equivalent resistance is now in series with resistance A, and R,. The total thermal

resistance forthe entire circuit then becomes
EA, = 0.00625 + 0.462 x 10% + 0.0005

000675 deg/W
Hence, heat transfer rate through the system Is
at 250-30
7 = 30592 W
a 0005 ~ 92592

(0) Contact resistance = 0462 x 10° deg/W
Temperature drop in contact = Q x contact resistance
= 52592 « (0462 x 10) = 0.1505*C

EXAMPLE 3.14 47
‘Solution : A thermal circuit for heat flow in a com) site system has been illustrated in Fig. 3.10.
emos is the sum of the component resistances. For heat flow through layers arranged in

parallel the overall thermal conductance (reciprocal ofthe ls the sum of
the component thermal conductances, "OSs! of thermal resistance) equals

‘Steody State Conduction / 48
en im aoe
}

DE mom

Fig. 310. Thermal creut fora heat ow in à compose system
(0) Total resistance for the composite wall is
ER, Rire + Ru

RA ZA ha

‘The wall area (2 m x 0.5 m) = 1 m? is constant for all layers of the composite wall

ge 030 | 0.50
Mera taa ta

= 125 + 8571 + 0001 = 9822 deg/W

at | 150-25

Heat low (without bolt) = ¿La 190=25 „1372
(oithout boty + Se A 1272 w
When aluminium rods are inserted
Area of the bolt = 3009? = 1.256 x 10° me?
he bolt = E09? = 1.256 x 10
°
TN AAA me AAA Aaa AAAS eee
i Ed soe
Tenber ok sa

‘Thermal resistance due to bolt
length ofthe boit
Ühermal conduct

A x10?
ity waren " 205% (L256x10")

= 1942 deg/W

Area of wall less bolt area

= 12 1.256 x 10° = 099874 m?

mento. Steedy Stata Conduction ff 47

creases ee |
of insulati i 83. 008
tified thermal resistance | AER
Mo iO O° RAT aber "02 8/4
a EI nin pall with BE TENCE OF the iy Teal thermal resistance Ry 02 + 345 +02 = 585 deg/W
a na Lea rere ap
Ia eae on
compte all ee = Po a2 wet
E Re Re Hz (6) The arrangement and the thermal circuit for the joined
: - arrangement and the ermal ecu oc he system shen tb layers are
A Le nes Role is shown below in Fig. 312 77. Y
Re Ra + Se FT TA
Ate az deg/W $ +
janie
En
teat dx ow + a 701
EXAMPLE 3.15 = ise three layers. A 12
cy ar Condori purposes compris pers A 12 cm thik ss
Te ain Hrs che Been 3 em ick Layer of ply L + |
TB tang nad ih a wc des nt ny O
resistance to eat low E e sie ould the heat ux e affected if instead of gla, re] prenne]
ape tem sc ot À 220 WX) of 12 cm diam a a » 5
jr u
on em Tem, and dnde “ »
o ring uit a, thermal rsstanos for he respective layers are
cos
y= ft non deg/W
AE]
ia pers
The rstance Ry Rand ya caked above are

Re, 02 545 deg/W
Bed Con * 555 de8/

Fig. att.

R= 02 deg/W ; Ry = 545 deg/W and R,= 02 deg/W
‘These resistances are in series with total resistance
E, =02 + 545 + 02 + 585 deg/W
Ss _ 003401264
= 208401200 à so dew
Br

+8,
Resistance of a bolt = I
aot « E

‘There will be four such resistances in parallel with equivalent resistance
a ee y pl ad
"pa BET Da BH

nd R, are in parallel and therefore equivalent thermal resistance for the circuit

= 0.1709 + 02005 = 02714

8 Transfer

8 Basics of Hest and Mass Trans
Sans)

cat fw yor me ;

‘Tre bast oy (est fon D 543 We

X

+ AND SURROUNDING:

many surface; a thin bound ly
pe case velocity with rey
comes ino coma vi rn relativ veloc Wit rept à
spr ee ae a
sera Ina em and cometió POT nly due 10 convection The rte
oe fh congo fe a edn gen y Neto,
pra mena sl bou
See estan

I en
iman law ee mat“) en

of the hot moving

BETWEEN SURFAC
AND HEATING OF FLUIDS

33 MEATFLO
act with a sa

COOLING AND!

When a moving

cme ha Does
eat transfer. The ree (ee density, viscosiy, specific heat and

er eae
rig ea
Sma ss en A ag ute ner (fo

Heidi. Solid wall

ols uid

1 5
MR RU: R

A
Fi: 343. Heztconducton trough a wal separating two Muids

Steady State Conduction / 49

Under steady sate conditions te at fes can be expresed by the equations
Core
2

A 08)

where h, and I, represent the convective film coefficients, X 1 the thermal conductivity of the
Solid wall having thickness à. These expressions can be presented in the form.

rare
cea

O ARS ]LA VHA 6

‘The denominator (I/hA + B/KA + 1/14) isthe sum of thermal resistance of diferent
sections through which heat has to flow

Quite often, the heat flow through a composite section is writen in the form

QUA (= 510)
where Us the overall heat transfer coefficient. represents the intensity of heat transfer from
one Aid to another through a wall separating thom. Numerical it equals the quantity of heat
passing through unit area of wall surface in unit time at a temperature dierence of unit
degree. The coefficient U has dimensions of W/mädeg

‘Comparing equations 39 and 310.

1 1,5,1

UC, 6

Apparently the overall heat transfer coefficient is reciprocal of unit thermal resistance to
eat How.

‘The overall heat transfer coefficient depends upon the geometry of the separating wall, ts
thermal properties and the convective coefficient at the two surfaces. The overall het transfer
<onifcien is particularly useful in the ease of composite walls, such as in he design of structural
walls for boilers retigerators, airconditioned buildings, and inthe design of heat exchanges

EXAMPLE 3.16
‘An electric heater of exposed surface area 0.09 m? and output 600 watt is designed to operate
fully submerged in water. Calculate the surface temperature of the heater when the water
is at 37°C and the surface coefficient of heat transfer is 285.3 W/ml.deg, How this value will
be affected if the heater is mistakenly operated at 37°C in air with a surface covefficient of
85 Wim?-deg?

Solution : When the heater operates in water as per design,

Q=hAat
600 = 283,5 x 0097, - 37)
+ Surface temperature of the heater,

600
rar 605
a 77" OSC

0 Bess of Hon ane Mass Trans
ven pe ET
te pee
Surface temperature of 37 sac
alii

Tis eee
ce posts el
ce (k= 8 KJ/mhr-deg) when the ambieny

thick layer of ic er surface of
loss per square ki

ae required at that surface Is 0°C

|
|
|
i
|
|

ote

pp surface

Solo = Since water at
ing) the minimum temper

ein
iy

3
À

Fig. ath
Invoking Fourie’ Iw, the conduction beat transfer rte across the ce layer becomes
sas _ SACO 2000)» (0 (-5)}
25 où
Since steady ste conditions real, the conduction heat flou across the ice slab equa
the convective heat transfer beten he top surface of ice slab and surrounding atmosphere
Accordingly

5 x 108 Ki/he

Qahaat
5 + 10 =h x (1000 x 2000) » [-5-(-12.5)]

Heat wansfer coficient atthe upper surface,
so

Tage 7 9587 metres

EXAMPLE 318

The oven ofan elect sor, of tll outside surface area 2.9 m? dissipates electric ene)

at the rate of 600 wat. The samounding room air i at 20°C and the surface coeffce!

of heat taster between the room air and Ihe surface of the oven is estimated 10 de

BSS Mide Determine the nea steady sate temperature of the ouside Sue

SERRE Lat would be the inside surface temperature if wall thickness of stove is 38
Condeciviy of the stove material o 0059 Winden?

Solation: The electri energy is diss
Stove tote am a Ped as convective heat flow from the outside suse

Sxoody State Conduction 1151
anal
600 = 1135 4 29 0,2
Outside surface temperature of the stove
=
has m marc
Ambient
tee
Os saca
Sr mat
ano vise autice
Fans

“The electric energy is frst conducted across the wall of the oven. Invoking Fourier law
of heat conduction

= 526 (t,- 3822)

Inside surface temperature of the stove
600
SE +3822 = 152230

EXAMPLE 3.19
Hot gases at 980°C flow past the upper surface of the blade of a gas turbine and the lower
surface is cooled by air bled off the compressor. The convective heat transfer coefficient at
the upper and lower surfaces are estimated to be 2830 and 1415 W/adeg respectively. The
blade material has a thermal conductivity of 11.6. W/m-deg. If metallurgical considerations
limit the blade temperature at 870°C, workout the temperature of the cooling air. Consider
the blade apa flat plate 0.115 cm thick and presume that steady state conditions have been

a
E - Upper surace

sense LT pasta
& ne

52 7 Bence of Heat ond Mass Transfer

noe get
per umi an om a
st ow at per un a tom he 11300 w/m?
A o 870 E ed aces the blade
o ou be conducted 15 Then

ai i
estate conditions prev
gue ae Sucios

propuse
conn ds Pre}

ane upper surface of he blade ig

0 - 1)

blade

of the
Temperature a ee lower sure of

coolant by convection
‘The heat conducted aro th
de no = 145 1 9-1

ature ofthe coolant ait
+ Temper a
eus

= 619°C

perature set at 290°C, whereas the temperatura
Ttchen oven has its maximum operating temperature set at 290°C; :
A kitchen oven has its max

ivr oe i re
Sir Therefore for und area of the wall
A” Ek +17h
as
;

SUSO Inst ;
CI En
a ty ee tert, 2
be ide ck ae
En [Rte ee:
i rer MA Mi
The ties ol bre gls insulation wil be RE

Large for 4, = 30°C

0 "00975 m or 8.75 cm

Seancy Sata Conduction 83
exaMPLE 321
Consider» plane composite wall a in made
A, = 735 Kmhr-deg and ky = 165 kl/m-hr-deg and thickness 8, = 3 qm and Dun 23 oor
Material À dois sho aid a 150€ Yor witch, ak A decane es
A CCE ths cold lid 30€ and 05 Mn a ae tnd the mater
transfer through a wall which 2m high tnd 25 m mide te heit
Satatiom : The wall arca (2 m # 25 m) = 5 m? is constant for al the layer, Te
Fes of the composite syste sven by
(1 S42]
CON

Of two materials of thermal conductivities

thermal

111.005, 005

dass

a, 190-30
Rat of hea tamer = He + JOIE
(0) Heat transfer from the Hot fo cold uid can also be expressed as

Q=UAAt
where Us the overall reflet of heat transfer
In = Ux 3» (150 2

1 hr-deg,
A]- oo 5

win he

zn x
or Ur gang © 28518 ky/mt hres
EXAMPLE 3.22

‘The interior of a refrigerator has inside dimensions 60 cm = 45 cm base area and 120 cm
Wallis made of two 3 mm mild steel sheets (4 = 145 kWfarhr-deg) with

(a) Calculate the individual re
surfaces, and the over

tance of this composite wall and the resi
conductance.

(0) Draw the thermal circuit

(0 For the air temperature inside the refrigerator at 65°C and outside of 25°C, determine
sean wich heat must be removed from the refrigerator. Also, calculate he lemprctaso
‘on the outer surface of the metal sheet,
Solution : Neglecting corner effect, the area for heat flow is

A= 212 x 06 + 06 x 0.45 + 045 x 1.2) = 306 m

cay constant for al layers ofthe composite wall and the various thermal resistances to
flow of heat are offered by

1

(0) outside ai firm

1
E = 9248 x 109 deg-he
MA” 330308 ~ 6248 * 10° deg-hr/k]

0.008
CO) mild sheet sheet ee gg 7 6761 109 deghe/ i

184 7 Basics of Heat and Mass Tronster

wa

de 006 20100 dept

(a) gas woolinslation DEN

0.008,
= RO = 6761 x 10° dog
(io) mild sel sheet 1530 €
1 1 Er
Lg = 80097 x 107 deg-hr/k
(0) inside airfirm ATRAS "© ,

# resistances and the summation gives
“The toral resistance equals the sum of individual
ER, = 011857 deg-hr/k]

Overall conductance = 8484 Ki/hedeg,

25-65
IR “om
‘Since het flowing through each layer is same; ton for the outside ale fm
16 2h A = 1) = 923% 30605 - 0)
ere fs e temperature tthe oe surface of metal sheet. Solution gives
156
235%

= 156 J/ne

es

ac

EXAMPLE 323

‘The door of a domestic refrigerator has an area of 0.7 m? and is essentially made of a thin
metal sheet with à 25 em thick layer of insulation on the inside. The thermal conductivity
‘of the insolation i 025 W/mK and the heat transfer coefficients to the surrounding aif on

‘ach side ofthe door are both 10 W/m? K. Estimate the heat flow rate through the door and
the surface temperature ofthe metal sheet which is presumed to have negligible thermal

Stendy State Conduction 38

resistance. The inside of refrigerator (cold chamber) and the kitchen (where the refrigerator
placed) are to be maintained at O°C and 20°C respectively

Sotetion : Thermal resistance ofthe door of rfigener lo
228,21], 173,00
sali ERY oc"
a, 2-0
Beat «she BR aa
(9 The heat flow through each layer is same, Accordingly forthe cuides ar film
167 2hA (0 210% 07 (0-9
here 6 ih surface temperature of metal she. Seton gives
167
mxo7

ES

= pase
EXAMPLE 3.24
A 3 mm thick metal plate, having therm

conductivity k = 98.6 Wim-den, is exposed to
Vapour at 100°C on one side and cooling water at 30°C on the opposite side

“The heat transfer coefficients are +
a, = 14200 Wax?-deg. on the vapour side
h, = 2325 Wa?-deg on the water side
Determine the rate of heat transfer, the overall heat transfer coe
temperature at each side of heat transfer.

cient and the drop in

Solution: Thermal resistance of composite system is given by
578,1
MO!

The wall area A is constant for all the layers. Considering unit area and inserting appropriate
values

sr 2 (1.006, 11
nes * 38)
2708 x 105 + 304 + 10° + 4301 x 105
= 53.09 205 deg/W
Rate of heat transfer from the vapour to wate side,
a 130510 war
Het flow through the composite system can be writen inthe form

Q=UAat
Where U isthe overall heat transfer coefficient

1318 x 10% = U x 1 x (100 - 30)
2318x108

or u

= 1882 W/mPdeg,
The same heat flows through each layer of

‘Temperature drop in any layer = heat flow x thermal resistance of that layer

56 // Basics of Hoot ano Mass Transfer

a _ engin te te
en 5 if mild steel plate. TI

A name
plat ool tec of wall area, Assume that the temperature of the

Fire lay bic + 128 Wik
Steel. 40 Wak
Solution: Te tive outside (ir ie) heat transfer count.
ig 24+ 8 = 68 W/m'K
Toa resistance forthe Compost wall
TR, Roa * Rant * Row
CACA
"HA ka HA
he wall surface are is constant for all ars ofthe wall, Considering unit surface are,
veg, = OS BOS 5 1
ER" Ty5x1* 40x | 68x1
0.0652 + 00001625 + 041470 = 008 deg/ W
Heat fw (without bot) per
a er
DOS
‘The eat fw theough each layer is same
Accordingly for the outside ale im
THT =A (Ty = T
“arm.
Outside surface temperature of sel plate,
een ease
(0) Minimum length of tel bois
= thickness of composite wall
075 + 00065 = 0.0815 m

unit area of the wall,

7787 W

Steady State Conduction / 87

Area of 18 bolts = 18 x £(0.02)* = 000565 m?
“Thermal resistance ofthe bolts,
2.005
a * 020 dew
‘The thermal resistance ofthe bolis in parallel with resistance dueto brick and t
‘The equivalent resistance ofthese resistors is given as ine
MENE)
LCL)

0.3606 (0.06524 00001625) _ us
103606 (0.0652 +0.00n1625) ~ 105599 des/W

This equivalent resistance is now in series with R.,, and the total resistance for the circuit
becomes

IR, = 005533 + 00147 = 0.07 deg/W
Heat flux (with bots) per unit area of the wall,

at | 60-27
“ER og UN
Percentage increase in the heat flow,
3900-7787,
E ua

Further, corresponding to outside air film,
8900 =h, À (Ty ~ T,) = 681 x (T,- 27)
Outside surface temperature of steel plate,
E

IS

ently the steel surface temperature is decreased by
(15788 - 141.5) = 1638°C after putting the bolts.

34. CONDUCTION THROUGH A CYLINDRICAL WALL
Cylindrical metal tubes constitute an essential element of power stations, oll refineries and
most process industries. The boilers have tubes in them, the condensers contain bank of tubes,
the heat exchangers are tubular and all these units are connected by tubes. Evidently then the
radial heat transfer rate through a tube or any insulation which may surround itis quite
important

Consider heat conduction through a ylindrical te
Len The inside and outside surfaces of he tube ae at constant temperatures hand hand
al Rond of de tube material is constant within the given temperature range.
Bo ende ase pevtcayinculated, the eat flow is limited to radial tection ony. Further 1
temperature atthe lane surface greater than temperature atthe ute urine, heat
flowe radially outwards

of inner radius ry outer radius 7, and

EE a CES

Temperature pola

A,
1 andy ser conduction tou a tna! va

ze

Et
PER ET
unin the prescribed boundary co

separating the variables and integrating wi tons me |

chat “nd
BTS fa

ge (=
e

du, (ee) any
an
he thermal resistance takes tie form

evidently for conduction in a hollow eylinds.
nin
MANE
"The expression for steady state temperature distribution in a cylindrical walk can besetuf
by imegrating the Fourier rate equation between the limits
Or = y where the temperature is stated to be
(6) r= r where the temperature isto be worked out

8

Thus,

and

Story Stace Conduction 7 88

a Qi fe eae tay
: on
Compara. wh equation 312, we btn 4

san a F

ten. meen)

made Et = EAD om
N ee
ee Lampe
a te om

(9) Hollow cyinder (0) Pano vaa
Fig, 320. Concept of toga men ara
‘Comparing equations 311 and 312,

250

Re
le n7n PEAR 615)

where A and A, are the inner and outer surface areas ofthe cylindrical tube. The equivalent
area A, is Called the logarithmic mean area of the tube. Further,

Car

Obviously, logarithmie mean radius of the cylindrical tube is

Ania ie

010)

Transfer
60 1 Basics of Hast and Mass Y

exanrie 9.25 —
had ent ake of
AT aie nny tempe
nn a en
Seca a ies
Cel par ce pet
ne f wine Rear pe mL
so ne ovine nr 2 er ln
“ THERES 10 = 250 wim length

a generated equals the Heat transfer through the

Eu ong
een the inner and eet

re deren ot Ae ire, Make calculations for yt”
to fo i ermal conductivity of cement,

Heat generated *
Under sendy state conditions,
clinical element ;

ge mie

een)
Aaxkxıx]20
250 = jog. (10/005)
‘Thermal conductivity of cement, .
230108, (1.0/0.5)
omg (20/089) ,

0994 Wmdeg
nx

rte ¡meter and covered wi
fet ids xing comet uh ang ip of cm tr diameter and covered i
em hie neta I rene soning with some tneaation. Calclte te
te onethrd ofthe presen ae by Father covering

Sont tikes of vlan

Solation à 7, = 2 em and r= 4 em

Heat loss with existing insulation. Q, + Tag,

ESTE]
Totes 0/5
wheres he additional thickness ol nation

According tothe given condi: Q; =

Heat loss with addtional insulation, Q, =

Enr
ÉTÉ
e ala

vols) BIO)
RÉF een

EXAMPLE 228

A steam pipe, 75 em external diameter and 25 m re

Ata 7 nd 25m long, conveys 1000 Kg of steam pet
Decor of 20 bat. The seas eater the pipe it dss fea 2 ed a

‘Steedy State Condvetion / 61

required that the steam a exit fom the pipe most havea minimum dryness fraction of 036,
‘The tsk is accomplished y suitably lagging the pipe;

being 07 Ki/mhe-deg, Neglecting thermal resistance of
necessary conditions. The temperature at the outside surface of Tapsann be taken as were

the thermal conductivity of lagging
the steam pipe (na temperature drop
ness of lagging required to meet the

the steam pipe), determine the minimum thick

Also determine the temperature at the point half way between the inner and enter

surfaces
Solution : A reference to steam tables indicates that at 20 bar pressure

‘Saturation temperature of steam = 212420.
Latent heat of steam = 1890 KJ/kg,
+ Heat loss por kg of steam passing through the pipe,
= (098 - 0.96) 1890 = 378 4)
‘Total heat loss Q = 37.8 x 1000 = 37800 K)/he
In terms of thermal resistance and temperature drop, the heat loss is given by

Qu,
2242
are = FES 5 000496 degen /K}
For a cylindrical wall, the thermal resistance is
or) og
nz
log, (nf) = 0545 5 (fm) = 1.725

Y, #375 cm (given), and hence
2 "375 x 1725 = 647
‘Thickness of lagging = 6.47 - 375 = 272 em
Radius at mid-plane of the pipe,
pea

32754647

stm

2 2
saute temperature at a point of the cylindsical pipe wall can be obtained by using the
25224 © Tog, (6.7 73.75) “0567
#52124 ~ (za - 25) « 05673 = 1061'C
EXAMPLE 3.28

‘The hot combu:

stion gases at 150°C flow through a hollow cylindrical pipe of 10 em inner
diameter and 1

2 cm outer diameter. The pipe is located in a space at 30°C and the thermal

conductivity of ¡eglecting surface heat transfer coefficients,
ait length and the temperature at a point
hat should be the surface area normal to the

igh the pipe can be determined by
€ same thickness 2

the pipe material is 200 Wmk.
cat loss through the pipe per u
en the inner and outer surface. Wi
cat flow so that the heat transfer throws
material of the pipe as a plane wall of th

‘calculate the hi
halfway betwes
direction of hi
considering,

e
|
|
|

js Transfer
62 1 Sasso Heaton Mass rma resistance of à eyindrical pipe ig

108. 12/10)
78

4516 10% deg/W

3 = 826674 W
ES

ET

di) Radios a hay through the pipe wal
we

+3 „DR « em

ie = WH ay

E ane
Thermal resistace of cylindrical pipe up sts mid-p
1
Logan) on
= Biles)" aos

= 75884 x 10% deg/W

Since heat flow through each section is same

73894510"

e at the mid plane,
1 = 150 = 826674 x 75884 x 10°
Alternatively from the expression for temperature distribution

m

)
150 - 150 20) jog (12/10) * 8718

log. (37
(ii) The equivalent logarithmic mean area is
2n(0.12-0.10) «1

Bare = 0.689 0
log, (12/10)

heck

20 826800 W
a 6

This is approximately same as calculated above,

0.

3.5. CONDUCTION THROUGH A MULTELAYER CYLINDRICAL WALL
Multilayer cylinérica walls are frequent
meant for handling a hot uid. The
insolation, eg. a steam pipe used
may have cylindrical metal wall,
plaster. The arrangement is Cale

ly employed to reduce heat losses from metallic pipes
Pipe is generally wrapped in one or more layers of es
for conveying high pressure steam in a steam power plant
layer of insulating material and then a layer of protecting
led lagging ofthe pipe system

‘Steady State Conduction / 63

Temperature,

No heat ow
inths erection

Pa Re Po
Fig. 2.2. Conducton trough a compost cynical wal
Fig 321 shows conduction of heat through a composite cylindrical wall having three
layers of different materials. There is a pectect contact between the layers and so an equal
interface temperature for any two neighbouring layers. For steady sate conduction, the heat
flow through each layer is same and it can be described by the following set of equations
tnt)
ACTA

px ese equations hp to determine the temperature difference foreach layer ofthe composite

t= ty = Lo
= hs zone

From summation of these equalities, one obtains

&-w-0|

Le

Transfer
84 Basics of Most and Mass T

‘Steady State Conduction 188

poste indica! wal is

Ths the fot aero a CMP u abet =

za aie TEE] Bi "UN Thy log Ren Th legen nenne
AS

‘he sum ofthe thermal resistance ofthe different layer, À

“The quantity in the denom DELETE

ine me onde. ¿1 ao composi cylinder of concer ht,
When the ve nn set

| XAMPLE 3.30
eat How rate is given bY Ls
enveloping cach ber, the Beat fi E

Similar,

Mohn A seat pipe 10 em outside diameter is covered with two

- eo 3 Gas) | thickness of 25 cm. The average thermal condactivity of one maenal oy a Ban
Simpson] DS ‘ther and the surface temperatures of the Insulated seam pipe ar fied. Pee ae
rd à | rs ernst men

pe i heat dissipation from steam
results from that arrangement?
‚m pipe are
re Som: "75cm; r= 10m
Let k and 3 be the thermal conductivities of the two ins
material with low thermal conductivity is a better Insularor
@ Better insulator (material with low thermal conductivity) is placed inside,
the steam pipe.

heat transfer eoefcient forthe composite cinder dépit iq | fe to be minimum. What % age saving in heat dissipate
fig 8 we and resp, then theft thera resistance 10 eat Flow would be Solution: The geometrical dimensions ofthe lagged sea
ig. 32 are hd, respectively

CT

lating materials, Obviously the
he next to
Thermal resistance =

Tee |
|
|

Log, 20] . 02506
3er nl

AT _ atxnkt
Heat dissipation Q, = SE TE o sur

sto

(i) Better insulator is outside

centre 1 à
Thermal resistance = 310,
19.322. Conduction tough composta cynics wal ot traslr coins considero Tue
+ 2 og N: 0
Re DEC DIET TD
0
= m 5) Qi
O Aa lr 709
Introducing the concept of overall het transfer coefficient U, he heat flow rte can be
ten as: Q = UA (4 2°). Since the Mow area varies for a cylindrical tbe, become A re 75. op 20), 024
[necessary to specify the area on which Us based. Thus depending upon whether the inner OF * goal |3 18 te" ar
outer area is specified, two diferent values are defined Ne Ur
QU Au 6) = Usp A= 6) 62 Het dissipation Q = SE = AT x kt = 47905 anar
‘Comparing equations 3.19 and 320, we note that ae * Gara * FH!
re u + __ Q 39
Mara +172 Dog (177) 1 BnkzTloge(n m) 172 a “OS

ster
66 / Basics of Heat and Mess Tr

al ‘Steady State Conduction // 67
isipation is small m would be affected if the pipe is lagged with 5 em thick ins
022 Wim K 7
Solution : When the pipe is not insulated
r= Sem and
and the effective resistance to flow of heat is
ER, =R, +R, +R,

eval with low thermal cond,

Ihe material with sont
‘Obviousiy the heat di

ds placa! nest 2e the sam PS
‘Saving in haar disparo

ion of thermal conductivity

39 à 100 = 18.55%

De a oped au
ET
re rk
Sa nee te ichnss and conductivity ofthe pipe material are not given, ii by a eee a,

Heat loss = Se * Ga

meer covered with 4 cm thick layer of insulation
i diamamn, Determine the percentage change in the gat
‚ging (= 125 W/m-deg) is provided. Gast
fal the heat transfer coefficient for both wa

Case I: Resistance to heat low,
ERR, +R

= 578 W per metre length of pipe

(0) When the pipe is insulated
na Sem; rye 6em and ne6+5% tem
and the effective resistance to heat flow becomes
ZR, = R, +R, +8,

=
Tenth,

14 12, 1]. 1.409 deg/w
malos escama 7 a ett ig, Bt gg yl

\ SN
CEA 1 ag 008, peg Od

y 19 DIAZ * 2xx450x1 À 0.05 2x KO EXT 006 20 TRINO

(Case IL: Resistance to heat flow,
ER, +R, +R,

| = 265 010% + 0645 x 10% + 0.080 + 0438 = OSIS deg/W
à „1023
Heat toss = Sr = "gie = 164 W per metre length

Accordingly, addition of insulation reduces the heat loss from the steam by

a 1 578-1
2 [im Be chim eg! EEE 5 09 0716 or 71%
alos 13 oa ii
7x05 gt tar] A
a sh {oasis + 0.1235 0.5952] = 21169 dog/W The shell of an experimental boiling water reactor is cylindrical having inside radius 1m,
Fi zu Length 1.25 m and 10 cm wall thickness. The hell is made from alloy steel and a concrete

50 cm thick surrounds it. The thermal conductivities of alloy steel and concrete are
225 Wim-deg and 1.12 W/m-deg respectively.

The reactor operates at a power level of 6 KW, of which 4% is lost in heat transfer
1747-17 through the shell. If the inside water is at 150°C, workout the interface temperatures.
ar" 00206 or 2.06 % (temperatures on the inside and outside of concrete covering). Neglect any resistance to heat

flow between water and steel.
Solution : Heat loss through the reactor = 4/100 x 6000 = 240 W

au Ar
Ex "ue
2

Heat toss,

1m alar

Decrease in heat loss =

EXAMPLE 3.32
Saturated steam at 11

IPC flows inside a copper pipe (thermal conductivity 450 Wo K) nelom: gema y= 16m
NE tera cana om nda enrol dote of em the ua ei i

of the seam side le 1000 Wins Kae th aa Thermal resistance de o a india walt = Elo 0/1
Bra ke 200 Wi ad that onthe ande surface af pie 38 Wh resistance due to clinical wall = 548.62)

ipe if it is located in space at 25°C. How this heat 10

joss Transfer
188 7 Becks of Host and Mess

396 x 104 dey

ost ides |
net 5 we |
Resistance due to concrete covering” Fy |

Resstane due

x 104 = 431.3964 « 10“deg/
2 5396 104 6 426 1 transfer
Sera ce and tenir drop he hat SE rt ven
items of hermal tante
Fr}
150-15
DS RTE TE

the temperature an ie use surface of concrete
.. fy =150 = 240 x 491396 x 10% = 139.65°C
inc he same het pass though sc ayer ofthe reat
Seen mee eet hob 1504
rear TS
where is th temperature on he inside sure of concrete
un O 240 5396 » 10% = 14987
I ay be noted at temperature drop aces th tel shel i nel
Gc Tre empertr can abo be worked ut by considering the het Pasing tng
the concrete covering

then for the alloy see sh

20.

hito h-139065
W617 ar

fy 2 13965 + 426 « 104 x 240 = 199870

This is same as calculated above

EXAMPLE 3.34
A steam main 75 mm inside diameter and 90 mm outside diameter is lagged wit
successive layers of insulation, The layer in contact y
asbestos layer is covered with 25 mm thick magnesia i
inside and outside surfaces are 227 Win
is 375°C and the ambient temperata

240 =

me
the pipe is 38 mm asbestos andthe
ulation. The surface coefficients for
PK and 68 W/mK respectively If the steam temperature
e i 35, calculate the steady state loss of eat fom
eam for 60 m length of pipe Also workout the overall coefficient of heat transfer bused
Genen sie and outside surfaces ofthe lagged steam main. Comment upon the important
‘Beneralisation of your result. = a ‘ce E

Thermal conductas ae
Tipe mata 4 Wk
obeso | OM Wk
Magna ion} 007 Won
Selon: mm = 00578 m 72 0085 mn
F3 2008 +008 = ong
ae 000 o
smal stan Ao of hea ar fered by
(0) Inner steam film = i

FA

=3n7 «104

‘Steady State Conduction // 69

1 1 rosés
(i9 Pipe material = 3 lo. /n) =

Ki Bere gags) 01075 » 104

losa El go
E A

1
ESOS

(o) More elton = zul In) = (UE) eau 104

1
(9) Outside aie film = pe =

A = 36137 x 104

resistance of ser

es path is equal tothe sum of individual resistances and the summation
gives

ER, = 255.2315 » 10-4deg/W
Heat fost through conduction,
ge 5,
Sara IW
(9) The heat loss can also be expressed as
Q>=U,A,AT=1U,A,aT

and U, are the overall coeficie

ea A, respectively

and inside

5 of heat transfer based on the outside area A,

m

Mare]

0962 W/meK

ups at

(rx 0.0375 x 60)(375- 35) ” 9219 W/mk

‘Important Generalisations
(4) Resistance to heat flow rises mai

nly from the laggings and not from metal of the

() Total resistance is most strongly controlled by the film with lowest coefficient. Thus
ena Bain would be obtained ifthe stam side coefficient were increased buts large
‘gin would result from an increase in the air side coefficient
(© Resistance ofthe pipe material and inner steam ln could Rave been neglected without
Introducing any appersciable error
EXAMPLE 3.35
(E em diameter pipe at 100°C is losing heat at then
to the surroundiny
Insulation.The followin
Insulation A
Quantity = 3.15 x 10-%m per metre length of pipe
Thermal conductivity = 5 Wades,
Insulation &
‘Quantity

© of 100 W per metre length of pipe
a 20°C. This isto be reduced to a minimam value by proviging
8 insulation materials are available

4x 10m? per metre length of pipe

Transfer
70 1 Basics of Host and Mass Y

er ‚| works out as
Seite stance due

o pipe materia

4
= |
|
|
1

WO 06 deg/W

KE

qa
Bora pre wi

uns

DE he ay”

Arrangement: The insulation Is paced inside, ic next 10 the pipe
ston material A 1 placed
“ ‘ O5 m

« je 2315 109 à Fy #0065 m

ass ia 4x 10°

or BR =00482) xt 4109 à mr 00555 m
i

1 joy 2.0565
EEE

lo
a
8 + 00467 + 0.1629 = 09096 deg/W
y „10-20

uw
ER, 0.909 =

2nd Arrangement ; Ti insulation material Bis placed inside, Le, next to the pipe
1, =O015 m

Heat loss, Q =

Het sent; n= 007m
Rdg 2345 0103 ry = 00965 m
o ous
ER =08 + TITI "8 TITRETT ET Be 0.073

= 08 + 0252 + 0.0089 = 1.0609 deg/W
10-20
Heat loss, Q gg, "754 W

‘Obviously the heat los is small when the insu

Sos lation material B is placed next 10 PIP?

10-754
Ta *100 2.69%

Steasy State Conducuon 1 71
3.6. CONDUCTION THROUGH A SPHERE
{Consider heat conduction through a how sphere

of inner radius, outer radios; and made

of a material of constant thermal conductivity. The inne and outer surfaces are maintained at
ad 1, respectively. Geometrical symmetry indicates that
ther if temperature fy atthe sane surface

Constant but different temperatures,
{the heat flow i limited te radial direction only Fur
is greater than temperature fat the oo
surface, the heat flows radially outwards
Consider an infinitesimal thin
spherical element at rads ret thickness
Of this elementary ring be dr and the
Change of temperature acro it bed. Then
Invoking Fourier law of heat conduction

a

rta «tana

Separating the variables and

the prescribed

| “oe dei
i The expression for steady state temperature distribution in a spherical wall can be set
Q + m + |
and snk [dt |
i nn gee cht
That gives: E 62

TR I Basics of Heat and Mass Transfer

ntvantagcous
trough
ow ate area A TUS

ne heat flow equation through a sphere

to wets me Then thickness 3 will be equal ty

Quite often itis considera ane wall

inthe same form as that tor et

War m and the arcas A wil Bea SÍ

| = 2
e

Comparing equations 32 ant 978 om
penn

Further, Au = and = dann
‘Obviously geometric mean radius of the Sp
rene 02

tin the same way as composite walls and cylinders. The
erica shall will be:

ere shell is,

spherical compos ae del i
eat Bow ough relayer compost ph
(cer
COEUR
Farther he convective film coefiients athe it
spheneal sel are also considered, then:

STAT em

Inner and outer surfaces of the composite

(ht)
nl = a) den 90 A 1/4,
Tae Varna) nam

messe
See parana aan oder ad 3 hall spre made of he sme
D ee Temper ep eri wall kes
end © m and eae uae ae = 1m
the oui sorbate pomeranian te
yin atthe oe

Solution : Let sux 3 ad 2 refe to inside and outside surface respectively

For sphere PRET
[ei

or - 2 =028%m
27 Va
Foreylinder: A," 2k 1

‘The outside radius ofthe cylinder is same as that of sphere, Le, 1, 0282 m
1

Pa ro 56m
rent ht)
aan tot
io, N ln)
CETTE
nn
A 0565

= 1756

Gee” Pinto) © FORT og, (022/01)

Steady Stata Conduction / 73
EXAMPLE 3.37

A hollow sphere of inner radius 30 mm and outer radius 50 mm ix electrically heated at the
inner surface ata rate of 10° W/m?. At the outer surface, i dissipates heat by convection into

a fluid at 100°C and a heat transfer coefficient of 400 W/m? K. Determine the temperature at
the inside and outside surfaces of the sphere,

lt may be presumed that there is no energy gen:
material is constant at 15 W/mK.
Solution: 1, = 008 m and 1, = 005 m
Rate of heat dissipation Q = q x dr 108 (dr = 009) = 113044 W
o

he IIA + 1006
Lerman

jon and the thermal conductivity of the

Fig. 3.24

DEIN

Heat dissipation Q = Zt

where Ryresitanee to conduction) = EE * 0078 kJ

"rn

and K(sestance to convection) =

A = 007% K/W

h 1,100
‘Taras + 0.0796 ~ “01504
Temperature at inner surface ofthe sphere,
1, = 11204 01504 + 100 = 270°C
Under steady state conditions, heat flow through each section is same. Therefore
hot, 100

mo BM
‘Temperature at outer surface of the sphere,

1, = 11304 x 00796 + 100 = 18998°C

Then 11304

EXAMPLE 3.38,

‘The thermal coductivity of a material is to be determined by fabricating Ihe material into
the shape of a hollow sphere, placing an electric heater at the centre and measuring the
surface temperatures with thermocouples when steady state conditions have been attained.
The sphere has internal radius 3 cm, external radius 8 em and the corresponding temperatures
are 95°C and 85°C when an electrical input to heater is 10 watts, Determine the experimental
value of thermal conductivity and the temperature at a point halfway through the wall.

Solution: In terms of geometical parameters, thermal resistance of a sphere is
men | 008-003 1.6597
R= = ROS 008 A we
Anka A) a
Heat lowing through the sphere is equal to
at 95-85 ik

==. w
R © Loser /k DIN

74 1 Basics of Heat end Mass Transfer nz
er say ste coins tis est on SI © p
wi à = 16587 Whores
ES
D Radius a halfway t

0:

‘eva de wal
308

= 55cm

era resistance ofthe epi wal wp 85 mid FAME equal to

00 727 deg

ough each section is same,
5-1

Since heat lowing,

ws

om

kempertae a a pont Ray though be wal i
q en nor = 7730

Alternative from the expression for temperature dist
rum)
STO)

tion for a spherical wall

8 (55-3)
(55-3) « o7a72
a)

295 - (95 - 85) x 07272 = 87.728°C

1-88
35-95

eae th 0. thick: if il
3003 monde diante is insaned with 02 m thickness of insulation

O Wurde, The surface temperature al the ver is 195°C and
Sea a A Determine $

to he ow,

(O eat ln erm? based on inside and ouside ars, and

(tempers pants at hefner and outside sua,
Sota Te bat fw tough aspera ube wih rar and given by

artnm(t 1) _ 40% 0.04%0,25 x0.45(-195-10)
per 045-035
‘The negative sign indicates that heat flow Is in negative radial direction
2.7%
() Heat flow based on inside area = ¿Ly = TUE = - 73,79 wim?
o ar mW

Heat low based on outside area
La A
Ada] 2228 Wie
(e) The temperature gradients are worked out from the relation
swig: |
Mai aa

‘Steady State Conduction /! 78

(57.95)

Inside Ge pe
ar” 00h xan 0.257 © 49m

)
Outside À 5 11m
ar Donnan (ogy SAPO

Evidently the temperature gradient decreases along the radius, Further, the gradient is
postive and heat flow is in the opposite direction.
EXAMPLE 340
‘Two insulation materials A and B, in powder form, with thermal conductivities of
0.005 W/m-deg and 0.03 W/m-deg were purchased for use over a sphere of 40 cm diameter.
Material A was to form the first layer 4 cm thick and material B was to be the next layer 5
Mn (nick Due to oversight during installation, whole of material B was applied first and
Stbsequently there was a layer formed by material A. Investigate how the conduction heat
fransfer would be affected.

Solution : Caset ry 02m Om; 12029
Thermal resistance 10 heal flow,
Ra
___ 02-02 02-02
3Rx0.005x0.2x0.28 * Anx0.0x0.24x08

27 + 1906 = 15176 *

When the materials get interchanged, there would be change in radi also.

Case 1

Volume of ma

re
eri À = 020 -0.2) = 0028396 n°

) = 001255 m?

Volume of material 8

0-0
The new radi are then worked out as

ies Auf dan
oouass = 42(9-02); 7,

= 0.2688); r= 029m
a,
TA

0.2618 -0.2 029-0648

= Gx 0.03 x 0.2 «0.2608 * Tx 0.005 x 0.2648 x0.29
= 3.247 + 5.225 = 8472 °C/W
Heat transfer is inversely proportional 1 thermal resistance. As such the heat flow will

= 447%

anstor
‘TE y Bass of Heat ana Mass Tree!

cme
a aa
emi ee
eee caer Era
the pete Pre tik DS WHOSE
of th nena ance due o the oven meteria,

seh mann" 00+ 07

Ra eme

nina EE ceric heater is placed at
ets ne tens

3 temperatenrough the oven and the wattage

eat to rahe same heat transfer. Also dí

Se oe nal

mem
Beat
The |
eat
el
cern
Sages 0353 Wand te

m 07m 13 2 70 + 5 2 75 em 2 075 q

optan fora apte body en DY
Rn Geier

misphere, it equals half of this value
Te 1, 070-060

700 00009 de
esisance of fc bck = Ir a9 315x770) sw
1 075-070

EI = 00722 de
Resistance of magnesia = 5% 7250 0525x(075%070) sw

1
Rene of cutie sr fm = Fr
LL - 000038 dogs

Pa
otal esa, E, + 01365 deg/W
ar w-2

ds aw
Heat os from the oven = Sq ” EE = SN
Therefore the required filament watage is 594 KW
(9) Radius at mid plane ofthe fir brick lining

Sth DEN m

TZ À

Thermal resistance of fie Brick upto its mid plane |

1 265-060

= 0.0162 dog/W

2” rx 0-315 x (0.65% 0.60)

Since heat flowing through each section is same,

|

825-1 |

F3 © Gore ]

Temperature at the mid plane of the fre brick lining is = |

1 = 825 - 5941.3x 0.0162 = 728.75 °C
EXAMPLE 342

A storage tank fabricated from 20 mm thick 2 jets of a cyrind |
229m thik pyrex gaa (= A Wn) consists o ai
‘ston enh 2m and inner dance I m and to hemisphe end sections Fe |

‘Steody State Conduction 1177.
‘exposed to ambient air a 25°C and having convective coefficient 0 Wav K. The tank is used.
do store heated oil which keeps the inner surface of tank at 125°C. Determine the electric

ower that needs 10 be supplied fo the heater submerged inthe els tha te presrbed
Penditons can be maintained. Neplct the radiation ete soy

Solution: Refer Fig, 325 for the geometry of storage tank
=05m and 7, 05 +002 = 052 m

——— 0 —
ig. 328.

Under steady state conditions,
Oy + Opn = Qu
That is, the heat dissipation from the heater comprises two parts
(a). yy heat flow from the cylindrial section whose length 1 = 2 m,
inner radius r, = 05 m and outer radius r, = 052 m
Ra = resistance to conduction through cylinder

tog, (052/05) = 2229 «10°

and R, = resistance to convection from cylinder
a ee En >
BA,” ra" aaa RA

125-25

Then: =e Bs
(22294300107

= 18900 W

(0) Que Best lw from the two hemispherical ends or a phere with
05m and = 052
Ra = tance to conduction tough a sphere
nah, 052-08,
ann "asas
and Ry= resistance to convection Uwough a sphere

4375109

3889 x 10%

Then: Que 5 SW

Ratha” ITESO
Electric power that needs to be supplied to heater
= 18900 + 9742.8 = 2862.8 W = 28.64 KW

78 17 Basics of Meat and Mass Transfer
EXAMPLE 3.43,
À spherical tank of Sam internal Ha
GS Wymadeg) fs used to store ce wate A
by natural conection and ration with 2
Wane avec cocficient at the inner surface Of
0) the rate of heat transfer to the feed wate
(6) the amount of ce that melts during PO
prism and net

tec The tank loses heat o surrounding, y

internal día

he tank, and

+002 = 152m
Solution :

Ry

Fig, 326.

res of the tank are
an» (15) 2826 mi

San + (152)? = 2902 m?

‘The inner and outer surface a
Asien
Arien
‘The individual thermal resistances are

1

= 442 x 104 K/W

Ra (convection at the inner surface) -

BOX

through surface) » PA 4666 x 10% KW
(conduction trough sure) » RE
1
er surface) = I = DE 2223 © 104 K/W
Convection a he outer surfe) = 7.4, © 755392

[All the resistances are in series and the total resistance becomes.
ER, =Ry + Ry + Ra
= (442 + 04666 + 22.23) x 107
(a) Rate of heat transfer to iced water,
TT

ER,

27116 x 104 K/W

3-0
2 NEW

(8) Amount of heat transfer during 24-hours

Q = 9219.65 » 24 x 3600
796 «10°
7.96 x 10°]

It takes 334 kJ of energy to melt 1 kg of ice at 0°C, and therefore, the amount of ce thal

ina period of 24 hours

7.96108
E]

= 2383 kg,

made of 2em thick stainte

CU heat transfer coefficient of 155 Aja €
ect Se

N

ms.

od of 24 hours (latent Heat of ce = pay,
9 |

Steody Stete Conduction 1178
37. EFFECT OF VARIABLE CONDUCTIVITY

For most mati e dependence of tema conduct mm temperature snes
ree inet
ii to the teresa concert cae
Shoe teed
pela

To dt
ve ermal a CE

Fourier law of heat conduction through a plane wall

CT om

ing the variables and integrating within the limits t= à at x » O and 1 = fy at

oft sa vena

mean thermal conductivity evaluated atthe arithmetic
1
mean temperatures 4, +1)

‘Temperature variation

Fourier law of heat induction through a plane wall is
«a El
ea e)

For stes state heat conduction in the infinite wall, SU

y.
Flg.321.Temporatue prof wih
"anable thea CRI

Q/A is constant and so has to be the parameter [x
i Va)
Obviously then, the temperature variation in he wall

governed by the condition that parameter | 1

constant

Case | B= 0: We have
= ky (1 + Bl) = hy constant

80 1 Bosics of Heat ond Moss Transfer

sore and equals the constant value
jah temperature and sa lues,

em onda des ot eames
TEN conan ne SE mist be cos Thr

coring for he parameter Chop
Br, constant and Ihe temperature profe 3
the slope of temperature cor

Case 11 B> 0: We have

bite
et + hie do ir

bs À directly proportional to temperature; incre
avy he wall ate Ed ae the tempers

creases wih decren
occas y would also decrease. Accordingly 10 main

‘The eer con $
sah ns peat or
cion in er

a z case, Consequently the value of
al tte term E must ner y of slop
the parameter [1] constant =

U and Band hat means thatthe curve should go steeper from A toy,

increases fom point À ue af the temperature variation curve is OÍ convex nature

Evidently with post
Case IB <0: We bave, ;
ROUE fe = 7

smal candy ofthe wall atea e inver proprtonal 1 temperature y
doce ternal conduc reeves wth decreasing temperature, Si he
decease hea ae ermal condo oul ire inorder io mama

‘of temperature profile decreases. Evidently the temperature profile will be concave for negative
Value of coefficient.

EXAMPLE 3.44

‘The inner and outer surfaces ofa furnace wal, 25 em thick, are at 300°C and 30°C respectively
‘The thermal conductivity of the wall material varies with temperature and is prescribed by
the relation

k= (145 + 05 + 105 0) X/m-hr-deg
whore is the temperature in degree centigrade, Proceed from the basic princ
the heat loss per square metre of Ihe wall surface area.

Solution : Invoking Fourier law of heat conduction,

les to calculate

a
et

Q dx =- kAd = (145 + 05 x 10%
Integrating over the wall thickness 8,

‘Steady State Conduction // 81

Quo. b ne LA eg) a a « 3645

loss from the furnace,

26.45
as = 17458 mehr

EXAMPLE 3.45

À plane wall of thickness 6 has its surfaces maintained at temperatures T, and Ty The wall
i made of a material whose thermal conductivity varies with temperature according to the
relation k= kyl Set up an expression to workout the steady state heat conduction through
the wall, Further proceed to caleulate the temperature at which mean thermal conductivity
be evaluated so as to get the same heat flow by its substitution in the simplified Fourier
relation,

Solution: Invoking Fouricr's law of heat conduction.

a

a on
ara ura

Separating the variables and integrating within the prescribed boundary conditions, we
obtain

Alora
e 06 RA)

Mors 7
= SER) whichis the required relation

Mths hat flow sto be obtained by subtitling an average value 0 thermal code
in the simplified Fourier relation, we have ns 7 om

EXAMPLE 2.46

An infinite slab of $0 mm thickness and 0:1 m? cross-sectional area has its sides maintained
at temperatures of 300°C and 30°C respectively. Measurements indicate that when 1 KW of
‘nergy as heat flows through tthe temperature at its centre plane is 125°C. Set up an expression
for thermal conductivity ofthe slab material i it varies linear with temperature.
Solution; Consider a slab of thickness 8 and temperatures atthe boundary surfaces 1, and ,
From Fourier law of heat conduction, ae

a
CEE

Transfer

82 7 Basics of Moat and Mass

ng tar eatin on band a 10 29

Asam a
a--mrard
. wrescribed limits, we have

separating the variables and iterating between Me PT

2 jo = [imrroa

“eh

Beth]

= 125°C and 6 = 25 mm

22 [Ee 2?) e129]
| mes 10° |

or 2508371875 m+ 15e o

For right hal pet of lb : = 125"C, f = 30°C and 8 = 25 mm

oa
Then un [
or 250=15175 m + 8 a

Simutancous solution of deities ( ad (i) gives

im 0028
aná 2627
Tue expresion for thermal conductivity may then be writen a

fe .0m2 t+ 6273

EXAMPLE 347

Derive an expression for the heat loss per m? of the surface area for a furnace wall of

thickness 8 when the thermal conductivity varies with temperature according to the relation:
k= (a + 51%) Winwdeg where tis in °C.

Proceed to calculate the rate of heat transfer through the wall if 8 = 0.25 m, 1, = 250€,

t= 25°C, and a = 03 and b= 5 x 10%

Solution : Invoking Fourier law of heat conduction
COR ie
get ass À

begin tin he presi boundary conditions, wes
via fon

| Separating te arabe a
1

E
95=- [a

Steady Steve Conduction 1183

woe Seen)

|

which isthe required expression
Sulsttuting the given dato, the rate of heat transfer through the wall works out a.

250225, 4, 510 oops 1
ERE 1
0 RS |
Los, 200%, 1
= 900 [0.39 222 (60500 62 +6250)| = 37406 w

EXAMPLE 3.48
A plane brick wall 5 m long x 3 m high 250 mm thick has temperatures of 800°C and 20°C
“maintained on its bounding surfaces, Presuming that thermal conductivity of the brick
material is related to its temperature in degree celsius by the relation
“k= (1 + 00011) Windeg,

make calculations for the average thermal conductivity, thermal resistance and heat loss.
From the wall. What would be the temperature at 100 mm distance from the wall surface at
800°C?
Solution : As the dependence of thermal conductivity on temperature is linear the rate of heat
flow can be calculated from the formula for the constant thermal conductivity taken at mean
wall temperature.

Average thermal conductivity,

[a -0.00n( 320420)
140,001
L \

k

ran

5. 03
hermal resistance R = D = Te

3 = 001182 deg/W

FR)

Heat ons Q = O 66 x 108

001182
à [roo] ox

01

Ri Taso 0005) x (5x5)

0 pel names
CS

For steady state heat conduction, heat passing through each section of the wall is same.

184 1 Basics of Most and Mass Transfer

spe(1.4+0.008)) 218
aa st = MPD TOME
Doms À + 680 =0
Sota ei usr equation ges
PLE 3.43 say
‘are sons wth yo y nd e of el Be, Te ed ate
an Cal a e temperatur the layer interface, Thermal conduc
Torte wal mae ae
Fre day À 028 0000893) Wines,
Redoute | e078 Winden
Soto + Lt, denote the emperatare the layer netce
er onda ie cy

r (u)
028 [100.000 224)
U (ur)

1= $36.23°C

028 [ir noes (r20 1]
Fora plane wall thermal resistance is 8/44,

"Thermal resistance of the clay
1
ES ATAN PES
1

"28 0.0012501200 +)

Thermal resistance of red brick
05
= peg = 0067
ESTO
ar
Heat los from the wall = pe

Total resistance IR, 40.667

(1200-50)
3772.8 =000125(1200+1))+056
Under steady state conditions, the same amount of heat flows through each layer. Then
‘considering heat flow through the red brick
(4-30)
der
Equating the two expressions for hat loss,

e

_ 1150 50
1/(2.8+0.00125(1200+1,))+ 0.667 ~ 0.667.

ee B3[28+ 000252006 4] | 4-50

FRS TRAD) OT

AS0(O 000125: 44.3) _ 1,50
DORE +3867 0.007
or DOME + 286, - 491.6 20
Solution of this quadratic equation gives,
Temperature a the layer interface
955.3850

ES

Heat loss from wall =

= 1355.89 w

0667

EXAMPLE 3.50 —
À furnace wall comprises two layers

15 cm thick fire brick with & = (0.25 + 000025 1) W/n-deg where £is in °C.
50 em thick red brick with k = 08 Wiardeg
‘The inside surface temperature of fire brick is 1200°C and the outside of red brick work
is at SOC. At the contact surface, the temperature drops by 20°C due to uneven contact
Work out the heat loss per unit area of the furnace wall and temperature of the red brick
surface which is in contact with the fire brick.
Solution : Under steady state conditions and considering unit area of the composite wall

sie Fire brick
Tk
o “
itn epee
ce
Tis mond
5 05m b nn
4 u
025 + ous
iy 208 Wymdeg rite

m substitution of given data in expression (we

120044),
2

as

_ ju 22)

(1200

8
or + 0000252258) a0-1) = 015 $ (5-70)

Simpliicain gives
0000125 + + 049 1 4968 =0

86 // Basics of Heat ond Mess Trensfer

men ty #886 = 20* BC

Heat loss per unit area of wal
DESTA

2 1225.6 Whe?

Soyer mata A and 8 The If side ayer
is made of aer ol a or, and the righ se
een 04 O0, hee eens,
ee empor

nena 0 and DC
Nt eat x OUR

{An infinite composite a
10 cm thick of material wi
Sem thick of materia B hav
{denotes the temperature in
‘on the left and right side of
respectively, workout the stes
the wall

N nte Fg 33 ih compost sa. The et
Peres: ‘made of a material of thermal

Row through an infinie wall
Conductivity prescribed by the relation

RTE
is given by
a= [rimes en
For lb on ft Fig, 32.
82 01m; 90007
Ke 035 ec

Y, gemnperature at the interface)

| and
| 2, 93 rn] ww
mene: 4 Le 20000) 800

= sft + 00085 (500 + A 600 1)
= 31600 - 9 + 0.0085 (5002 - 1]
= 4125 - 3t- 00005 #

For sab on right:

2.04 [1,00 4,75] + 0.042 (a)
ee
E rubio

A125 ~ 3¢~ 001050 = 8 + 0.00% - 2025
or OU RH - 43275 =0

Steady State Conduction / 87
or Bo 7586 = ua = 0
Solution ofthis quadratic equation gives: = 57°C
‘Substituting this value of Fm expression (i) we get

pica tax, 03-30 253-0016 sn}
as 7 2 06 2210 Wt

EXAMPLE 3:52
Derive an expression forthe heat flow rate through a hollow sphere of inside radias r, and
outside radius ry, whose internal and external surfaces are maintained at temperatures ly
Bad f respectively. The thermal conductivity of the sphere material Ras 3 quadratic variation

with temperature :

= ka + at + pe
aw of heat condi
Att
Hz
where isthe area normal to radial direction r.
Subsituting A = dar? and k= KO + at + Be), we obtain

Q

Solution : Invoking Four

8 + a po der a

mur mia

CEE

«af

fir

or Peg
Which is the required expression
EXAMPLE 3.53,
A hollow sphere of inside and outside radi r, and r, respectively is heated such that its
inner and outer surfaces are maintained at uniform temperatures f, and ty If the material
of which the sphere is composed has a thermal conductivity which varies with temperature
according to the expression

find the heat flow rate through the sphere.
Solution : invoking Fouriers law for wni-direction steady state heat conduction;

‚ss Transfer

88 1 Basics of Heat and Mos

Q

os normal o rad

jal direction

shore Ais the à

sand ke ky +
Suteituting A = der and kh + Ka

des Iren

Later

x nd hence integration through the spherical wall gives
ve. Q is constant a

= [s tarsi

e. @

In stead

teeny en

EXAMPLE 3.54
A thin spherical container of 100 cm inside radius is covered with 20 em layer of asbesto
Insulation, The temperature at the inside and outside surfaces are maintained 4
185°C and 5°C respectively. The container stores liquid oxygen which boils at = 165° art
Ras latent heat of vaporization equal 10 2125 Kg, Determine the rate of evaporation Ÿ
liquid oxygen.

The thermal conductiviy of asbestos ins
values are sated as follows

0156 WimK at SC and 0125 Wink at - 185°C

ion varies linearly with temperature and

Steady Stace Conduction // 89

olution : Let the linear correlation between X and Y he prescribed as k mt +, Them from the
en data,
given data,

0156 =5m +6 and 0125 =~ 185 mee
‚ncous solution of then expressions gives

mo 163% 10% and c= 0155
Fora hollow sphere, the Fourier conduction equation s

a

armen

‘Separating the variables and upon integration within the given boundary conditions we have

Simul

ent sepa

o£ [E]-Fer-weeaew

Substituting the appropriate values, we obtain

2 [EL]. 2000 op 91 ous (1854
ge [ina] “SE te wor 1-015 cas 0
oe EE aay «ans
A
OT on
‘The negate sgh oe lc att ws om mer rc 1e nain

Rate of evaporation jy = À = PAGO

176 x 10" kg/s = 33.08 Kr

EXAMPLE 3.55
A hollow cylinder has inside radius r, outside radius r, and length L The cylinder is
subjected to steady heat transfer which results in constant surface temperature I and fy at
ri and r, respectively. For a linear dependence of thermal conductivity upon temy
‘expressed as k * ky (1 + Bi), obtain an expression for the heat transfer from the cylinder.

Workout this eat transfer per unit length ofa hollow cylinder of inside radius 7, =125 cm
Outside radius r, = 2 cm, The corresponding temperatures on the surfaces are S10°C and
290°C. The thermal conductivity of the eylinder material varies depending on temperature,
obeying the following equation

À =(372 0034 1) W/mdeg

Solution : Invoking Fourier law of heat conduction,

Q

a
“Ma

À ni ee ee TO aes

ranster
Basics of Host and Moss T
mE yr Subettuting À

vdi ka (1 + BA) ai
er Ate aon normal cdi
seer emg ve at }
annee ze
OH eae pont
E .
bn steady at 0/7 constant nd once

[o
los er

tegration through the cylinder al yyy

pete]

1
defines re average thermal conductivity al the mean temperature (hi #1) of cylinder surfaces

2 Host transfer from the cylinder

Rex A

ET

xo 0

tere Rs he era reino of he inde wall
De due of herald upon empero
à 2 oo 20:20

= 361.68 W/m-deg

1 20
to BE = 2069 x 104 deg
A “F135 des/W

EXAMPLE 3.56

Setup an equation representing temperature variation in terms of surface temperature for
‘one-dimensional steady slate heat conduction through a plane wall. There is no internal
heat generation and the thermal conductivity has a linear variation with temperature
Solution À linear dependence of thermal conductiv
Kay (+ Bs)

Invoking Fourier equation for heat conduction,

Stesoy Seace Conduction // 81

Upon integration

0x=[rbrlase

“the constant of integration € is evaluated by applying the condition £= £, at x = 0. That

ives
cen(nebe)s
arr- nf dr )aeufnete)a
ANA o
2-3] id
Applying the 2nd boundaty valve, de, #2 tat =

an Al(1,+8a)-(448 ) ©]

Leta 2")
Equating the expressions () and (i) and upon rearrangement, we obtain the following
«quadratic equation in

B

ot
Berria) TC

Solution for £ works out to be

Hope step 2)

|
alme Ham <a 003]

whichis the required expression for temperature distribu in terms o surface temperatures
ib sindy sm Heat conduction through plane wall of neral conduct having har

3.8. CRITICAL THICKNESS OF INSULATION
Contrary tothe common belief that addition of insulating material on a surface always brings
about a decrease in the heat transfer rate, there are instances when the addition of insulation
to the outside surfaces of cylindrical or spherical walls (geometries which have non-constant
cross-sectional areas) does not reduce the heat loss. In fact, under certain circumstances, it
actually increases. the heat Flow wpto a certain thickness of insulation. To establish this fact,

a A
era rahe min this cylinder is un
ERBE

ou me

Ancien of vanale tees

trio tem

+. |
os > Dumm
Lita
An Re Pa

EST

Fig, 230. Cc! thcaness ol ipo insu

With stipulations of
(i) steady state conditions
(i) one-dimensional het flow only in the radial direction
(Gi) negligible thermal resistance due to cylinder wall
(i) negligible radiation exchange between outer Surface of insulation and surroundings
the heat transmission can be expressed as
(con sn
Ta dj 17 2 =
‘here hand re the fm coefcient atte inner and outer surface respectively. The denominar
represents the sum of thermal resistance to eat low. The values off, rh and are constant
\herefore the total thermal resistance will depend upon thickness of insulation which species
the outer radius r of the arrangement. An examination of equation 3.51 would reveal that
increase in (ie, thichnkess of insulation), the thermal resistance of insulating increases bte
TEsistance due to convection coefficient at Ihe outer surface drops. The thermal resistance der
Late fl cocicient remains unasected with change in radius + Obviously, adden d
‘tsulation material can ether increase or decrease the rate of heat transmission depend
‘pon a change in the total resistance with outer radius y, ‘ 7
The effect of insulation thickness can be studie
With respect to rand setting the derivative equal y

by differentiating the total resistance À

which gives re
‘To determine whether the foregoing result maximises or minimises the total resistance, the
second derivative needs to be calculated
PR
CE

whichis indeed positive

Then r= K/h represents the condition for minimum resistance and consequently maximum

eat flow rate. The insulation radius at which resistance to heat flow is minimum 1 called U

critical radius. The critical radius, designated by r, is dependent only on the thermal quan

Kana h, Tus
x

en 83)
The fact that heat flow rate attains a maximum at 7 , is the result of the above mentioned
opposing effects; increasing increases the thermal resistance ofthe insulation layer but decreases
the thermal resistance of the surface area, At 7 = 7, the toll resistance reaches à minimum.
Apparently, a pipe castying a high temperature fluid wil lose more heat (compared to bare
Pipe) if the conductivity and thickness of insulation are improperly chosen, Dependence of
heat loss on the thickness of insulation has been shown in Fig 331
a

ES

2.
a Se
3
a4
coo
ln Seat al

Fig. 2.31. Dependence ol heat ss on Wickness of insulation

je Monster 7
84 1 Basics of Heat and Me
To cases of practical interest a
ire te mien olaa
eut ae until he outer CAG

San te range FSC
reaped we a Gites ve
= ‚ih addition:

int in Fig, 249) lato Incrasin
adecco. The sat
sees eres ome
Ton den Increase in cond
“ incio rene onsequenty the ent loss increase
reap, Te et eval s drop in ss causes the heat os 10 decreas om tis pea
"on fren nero in insulto TS on e represented by po 0) ls added the
coro eee ee PO
raat ass is still greater that ace the heat mere
emo apple 1 pipes and wire of sgh
mates of oot nation of electrical wirs and cates
ge te vet the prime objective 10 proue
tation te rate of heat dissipation can e
ved within sae temperature limits by a proper choice
in a ip coment ating apa ef ie

si wires

SA rc esca hazards. H
Free and the conductors maintained with
irae on NS. That permits 5

cable
> re effet of wall tic
sheathing of insulation then acts as

inates and the overall thermal resistance increas,
cts the flow of heat

Tagging and that obstructs th

aE o he man ocre in steam and refrigeration pies. For insulation foe

seh NH pow be th nn ma

for insulation
Following a similar app

ness domi

can be workes

roach the critical radius of insulation for a spher

POC E
nl 1 rh
2 _ 6 for maximum heat flow rate

ah,

63

EXAMPLE 3.57 ;
transfer

Addition of insulating material does not always bri
rate for geometries with nonconstant crosesecion area.” Comment upon the Y
this statement

"A pipe of outside diameter 20 mm isto be insulated with asbestos which has a mest
thermal eanductivityof1 W/m eg. The local coefficient of convective heat to the surrounding?
155 Wimtdeg. Comment upon the utility of asbestos as the insulating material.

What should be the minimum value of thermal conductivity of insulating materia! %
reduce heat tranefer?
Solution: The critical radius of insulation for optimum heat transfer from pipe is given bY

18 about a decrease in the heat

ity of

|
|

Steady State Conduction // 85.

Oh ae
; mor 20mm

1,=10mm

For insulation to be properly effective in restricting heat transmission, the pipe radius fr,
must be greater than or equal to the entical radius ,. Here 7, <1, and as such there 1 no point
saat ing absestos as the insulating material, Addition of asbestos insolation will increase the
at transfer rte and that isnot desirable. An insulating material wit smaller thermal conductivity

need to be employed.
Tor insulation to be effective, the pipe radius should be greater than the critical radius Le.

nen à omal
e428 1001 = 005 Wing
gpg ia Say ri ds Macs

EXAMPLE 3.58
cable of 10 mm outside is to be laid in an atmosphere of 25°C (4, = 125 W/m'-deg) and

As Surface temperature is likely to be 75°C due to heat generated within it How would the
hast flow from the cable be affected if it is insulated with rubber having thermal conductivity

Ke 0.5 Wimedeg?
Solution : The critical radius of insulation for the given

k 038
ret ig pg "02 m 22 mm

ble is,

Radius of the cable, 7, = 5 mm
ce, > fy the heat flow from the cable would be increased by insulating the cable

‘The heat flow from the cable when provided with critical insulation is given by

e-

for unit length of cable

= 2850 53.80 W per metre length

x = I
u)

EXAMPLE 3.50
A 2 mm diameter wire with 0.8 mm thick layer of insulation (k = 0.15 W/m-deg) is used in
à certain electric heating application. The insulated surface is exposed to atmosphere
‘convective heat transfer coefficient 40 W/m-deg. What percentage change in heat transfer
rate would occur if critical thickness of insulation is used ? It may be assumed that temperature
difference between surface of the wire and surrounding air remains unchanged.
Solution: Let At be the temperature difference between the surface of wire and air surrounding,

the nsulation,
f= mm r= 1 + 08 = 18 mm

Case

Heat loss from the wire,

96 / Basics of Heot ond Mass Transfer
A aniat
tte aca
Tog Bee” giebe() amines

atar 7 ñ
PAL 2 00562 x (21 a1
RS :

Case 11: The critical radius of insulation for the pipes,

1000875 m = 375 mm

Heat los from the wire when provided with critical layer of insulation,
Dalat) __ mar

a DE
Bog} (ara
RE » 00666 » x ay
TEs+68 Ra;
Y ag inne nt ons
LOS OR 99» 14.95%
002

EXAMPLE 3,60

A wire of radius 3 mm and 1.25 m length isto be maintained at 60°C by insulating iby
‘material of thermal conductivity 0.175 W/mK. The temperature of surrounding air is 20°C weg

‘heat transfer coefficient 85 Was°K. For maximum heat dissipation, determine :
(6) minimum thickness of insulation and the heat loss
(6) percentage increase in heat loss due to insulation

|
|

Solution : For maximum heat dissipation, the thickness of insulation corresponds to cie!

radius of insulation.

oi a,
F7 FP = 0020 m = 206 mm

Heat los with insulation Q, =

whee Fan da) = a bg
i ns
2mxO175x1.25 Be “3

1.403 K/W

Ry (resistance to convection) = 1
Ger,
EBEN:
20x 0.0206x1.5x85

= 0727 K/W

mn
Taasvorm "RR W
(9) Heat loss without insulation Q, = WA (= 19

= 85 (x x 0003 « 125) 69-20) = um W
‘Then percentage increase in heat dissipation due to insulation

zn

EXAMPLE 3.61
‘An electric cable of 5 mm radius is applied a uniform sheathing of plastic insulation
(k= 0.175 W/m-deg). The convective film coefficient on the surface of bare cable as well an
Insulated cable was estimated as 1145 Wimy-deg and a surface temperature of 35% mas
noted when the cable was directly exposed to ambient air at 15°C- For keeping the wire as
cool as possible, find the thickness of insulation. Also determine the surface temperature
of insulated cable ifthe intensity of current carried by the conductor remains unchanged,
Solution : For keeping the wire as cool as possible, the required condition corresponds to that
for critical radius of insulation, that is

k

0015 m= 15 mm

Thickness of insulation
=(,=1)=18-5=10 mm
(0) For a bare (uncinsulated) wire, the heat low is
Q = A at
1.65 x (2x x 0.005 = 1) * (55 - 15)
= 14.68 W per metre length of cable
For the sheathed (insulated) cable

ES

VETE

(4-18)
. er meter length of cable
ET meh
If the intensity of current carried by the conductor remains unchanged, then
CA ER
(9-15)
RETO

Hence surface temperature of the insulated cable
Le = 14.63 x 19108 + 15 = 45.95°C

EXAMPLE 3,62 ae
An electrical conductor of 10 mm diameter and having 2 mm thick insulation
(# = 0.18 W/mK) covering is located in aie at 25°C having convection heat transfer coefficient

98 ener of Het on a

js Transfer

For outside convection,

La
Ro" TA

‘Total thermal resistance,

ral? RRA

02976

A that Te

From he ration Q* PR,

1

Pod

1

2843
T

&-&-—
TR” Jos “Vp

Ace 0005 and p= 724 Qem=72% 108 Am
«Curent eapci of eonducr,

k

se

9
dog (7/5) = 27° egy

2843
Y degıw

3.1406

SERS aes/w

5
DE „1751 W/m length of conductor
Sage = 1751 W/m lent

=

‘Steady Stato Conduction 8
= shite. (ci 1 133

where Pa BIBI) lt = der

A fend

iin” a ee

24.835 W/m length of conductor
‘Then the maximum current that can flow through the conductor,

e (A e LO
tat E > fa E

AS
Percentage increase in current carrying capacity,
Low =U, 5209-4869

Te RE = 01908 or 19.08%

EXAMPLE 3.63
A tube with 20 mm outside diameter is covered with an insulation (k = 0.18 W/mK) to reduce
heat loss so as to maintain the tube at a prescribed uniform temperature. The dissipation of
heat from the outside surface of insulation occurs by convection into the ambient air with
convection coefficient 15 W/mK. Determine à

(a) the critical thickness of insulation
(6) the ratio of heat loss from the tube with insulation to that without solution for the

thickness of insulation equal to the critical thickness, and forthe thickness of insulation
20 mm thicker than the critical thickness. E

Solution : Critical radius of insulation,
k _ 038
vun
Critical thickness of insulation = 1, - 1, = 12 - 10 = 2 mm
(@) Heat loss from the tube with insulation

O12 m = 12 mm

MEDICA]
het),
Em à vs : *
‘where isthe uniform temperature ofthe tube and T, isthe temperature of ambien i.

Heat ios from the tube withoot insulation,
Qui © 2 FT = TD

an“

and Mags Transfer

400 1 Basics of Hest
Ñ

menu

Ge et fing, |
Qe 5 woe

Bay
Elis Bat) 1015
ale)

ae hat ios is ines y 13 pent sie fat hat 2 cm ck li cop

pangs @+20)=32 mm

2 [BH (2)

|
=32 x (4.102) = 0.78.

the het ns decreases by 2 percent when 22 mm thick insulation i provided

Lu
Raines

Daunız 364
Ben sy an net al umeler wire has a high cure carrying capacity thang
coveted wie?

ee ae of radis 05 mm is inulted uniformly with plastic = 05 Wa

e ih The wire is exposed to imosphere at SC and the Ouse su
sees ON. ind the mese safe current cared by the wire wo that eat
Sine lad phase abone °C. For copper |

“Thermal conductivity = 00 Wak

Spec era resistance a 2 # 10° ohm

‘oud the capaci ofthe wire o carey mor caren (diesipate more het) nene se
decree with farther ion ol asc ?

Solution: Te electra resistance ofthe wie

NE

AT RS

The heat o be disipated per uit length ofthe conductor,
Qué R= 002548 2

{Under steady state conditions, this must equal the heat flow through the plastic sheathing
per mare length ofthe conductor. Thun? _o

2 = 0.2548 Q per metre length

(>t)

000548 R = Sen
DR ¥1/ Pek log 7)

En

1/0:5x0%%3)+1/05xl08,10.5:10)/(0.5x10")]
MEN
ER 204
Masias ste cren por,

Steody Stave Conduction 1101

Le = 139 amperes
(o) Critical radius of plastic sheathing,
k 08
ur 0005 m= 625 mm
“Thickness of insulation = 625 - 05 = 62 mm.

Since the insulation is only upto 1.5 men, the current carrying capacity of the wire can be
substantially increased by adding more thickness of plastic insulation. A limit would, however,
ie imposed by the unacceptable temperature which may develop atthe centre ofthe wire and
for that an analysis of temperature distribution within the wire needs to be made

REVIEW QUESTIONS.

it. Wile the Fourier rte equation for het transfer by conduction. Give the physical significance of
Determine the steady het transfer per unit area th
vo faces maintained sl uniform lemperatues of
mater 19 10" RW) de

ugh 2 38 cm thick homogenous sab witht

"nd 25°C The thermal conductivity of wall

(Ana. 073 La)

32. Doping with a general conduction equain, make suitable assumptions te show thal tempe

dicas through 2 plane wall lene

Compute the het fax through a D o thik rick wal (k= 049 W/m-deg) which has temperatures

{of 15°C and IUT maintained on ke wo faces. How this eat Max would vary when dem Hick

layers af magnesia insulation are added both on the inside and outside faces ofthe wall? Alo
determine he interlace tenperstae forthe composite wall

Ans, 972 Wind, 1305 Wa, 931°C and 725°C)

rm thick see plate and is designed for a heat transfer ate of
800 KW. I heat wansfer i uniformly distributed over the beating surface and Degiecting any
effect due o corvatre, workout the temperature difference te be maintained between the tne
dnd outer surfaces ofthe plate Presume that thermal conductivity ofthe steel is 15 Wide and
eating surface hs a tl ares of 30 me

22. À steam boiler made from

34. tis necessary to insulate a plane wall 25 m long and 2 m high so that amount of eat Rowing
per hour through this wall will not exceed 8500 kj. The temperature of wall undemeath
{he insulation is 400°C and the outer surface of the insulation I at à temperature of 30°C
Determine: (0) thickness of insulation ler made of asbestos material of thermal conduc
2 0397 W/ aed: () temperature inthe ination layer ata distance of C5 m from the inner
Surface; and (6) temperature gradient of the insulation. (Ans. 01139 m, 246°C, 307237 des/a)
plain the analogy between heat transfer by conduction and flow of eectrcty though eh
resistance. state the concept by considering composite wall of Building
“The layers of mater of thermal conductivities Ay ky Ay and aches, BB are placed in
od contact Deduce from first principle an expresion for he heat low through Whe composite
Slab per unit surface ara in tems ofthe overall temperature difference actos the slab
38. À furnace wall comprises tuo layers: 8 cm of fie clay = 12 W/mK) next 1 the fire box and
(0.5 em of mil ste! (= 35 W/mk) on the outside. The inside surace of ik at 900 Kand the
steel surrounded by air at 300 K with an outside surface coefficient of 5 W/mK. Estimate the
heat flux through square metre of furaace wall and the outside Surface temperature of ste.
What would be the percentage increase in heat ax i in addition to the conditions specified
eighteen steel bolts of 1 em diameter pass through the composite wall per square metre of wall
ps ans, 6276 W, 39685, 1395 %)

39. The door of a domestic refrigerate has an area of 07 m and it basically consists of a thin metal
sheet with a 25 mm thick layer of insulation on the inside, The thermal conductivity of this

as.

108 7 Basics of Most ond Mess Transfer

er on enh side of the do
am ie 2 mag and et taser o
hon rn Och the door and the semperature of the
SERRE ISS de room are at O and 20 repartis, Net

mat

The alle of retorgerato truck consist of 1.5

lace 8 mn hak under Bing = LI W/ar
RSR Rem Te temperate at te me and aude srtaces of the rusk are

SEES O Croce Cokulne fencer ringertion capacity of the plant, (9) timber cork

ie temperature) error numduce y neglecting the resistance ofthe outer skin of the see
shel (Ans. 348 KW, 675°C, 0.003%)

Sat. Obtain dt/dr fr single layered cin
Statue this nto the Fourier law fo workout an expression fo

oe s 10 W/n?.deg. Determine
metal sheet The refrigerated
thermal resistance due tothe

Ans. 4662 W, 1320)

thick tel sheet (k= 18 W/-deg) at the outer

= deg) on the inner surface and 25 mm thick cork

ler fom the appropriate scady soe conduction equation
Pre for heat low through a cylindrical

wall
Re em internal diameter pipe is provided with 4 cm thickness of laggin for which coefficient
er Sonden i 0167 Ky/arhradeg. The lagging outside surface temperature is 35°C and the
Beine temperature at a section ine the pape t= 20°C. Calculate the het transfer 10 brine per

metre engin at this secon and the ese In temperature if the brine mass flow is 03 Kg/S. Take
Fi heat of Brine as 356 K)/kg K and neglect thermal resistance ofthe pie.
(ns. 126 he 08370)

A refigerat at = 30 flows inside a copper tube (k = 400 W/m-deg) of inside diameter 1 cm
And wall cles 2mm. To reduce Tosser, a 3 mm thick shell of thermocole type of material
{8 = 008 W/mdeg) i lagged onto the pipe. Calculat the heat leakage fo the refrigerant per mete
"and external fn coecient are 450 and 5 W/medeg respectively and.

san

length of pipe. The nern.
the outside ai temperature i 30€.

336. Determine the heat transfer berwcen two fluids separated by a copper condenser tube of 20 mm
‘ouside diameter 18 m in length and ol Duchess 25 mm ifthe outer (steam) temperature i
100€ and he inner (water) temperature is 15°C. Assume that the water side fl colin
360 kfm deg and on he sen side i 41025 M/mhe deg Comment upon the important
generlizain of your result.

347. À metal (k= 45 W/medeg) steam pipe 5 cm internal diameter and 65 cm external diameter
lange with 273 cn dal hana of hgh temperature rslaton having thermal cont
of 13 Wo des, The surface hut tanlercocticents fr inside and outside ae 4650 and
HS Waste. I he steam temperature 15 200 and the ambient temperature 25€, make
calculations for
(0 est loss per mete les of pipe 7
(2) temperature at the ninos and
A ra coin of het transfer refered to inside and ouside surfaces,

(Ane. $44.19 M 192250, 19875°C 150.46; 1930 and 825 Waste)

318, Derive an expresion fr heat flow through à composite cylinder taking int acount the film heat,
transfer coeficiente on the Inside and outside surface ofthe cylinder
Find the Log mean ara of the cylinder neglecting film heat manser cocfients

339. What isthe “log mean area” when problems of heat tansfer by conduction are considered for
Follow elinders 7 Do yu need 1 consider fm coefficients tthe surfaces to determine 7 Based
upon your reasoning ind the expressions or rate of het flow per unit length through a hollow
Sinder of inside radio, and ose radi If arithmetic mean area place the logarithme
‘ean are, find the expression for percentage error. Compute the ratio 7, /r, that will give 4%

324. Obiain an analytical expression fr temperature distribution in a plane wall uniform reference
temperate 1 ad y and nd recy and a hema cont ta ac ey
wi temperature, £ 4, (1 +), Dicuss the nature of temperature curve for #ve,=ve and 2650
Vases a. è N een fer see yea

325

32.

32.

a.

332

‘Steady State Conduction # 103

220 cm thick plane ll bu rom re Brick motera whene thermal conductivity ares
lineal with temperature = 4 (= Bi) where A = OMS Wire and D O/C Te
temperate of as OMG an FE omnes ro the wal an
temperate at 0 mm away trom the higher temperature suce, What woud be du er ot
temperature ala un Ihe bass ofan average Bert coma PO
What is meant by rial chnel insulanon >
À thin eylinder of radi ri aged to an

ue a7, with a aia materia of eral
Sony Show ha he Fumar casal Nat ane ta wh fy
Sere he heat canse ae apa ol value l been ad re I
Slider end tects A E
an the energy on increase with growth in the ic of nut being added ar
pipes? posible, under what condition wil this aa atc?

À thin walled copper tube of radius 5 mm has been given a sheathing of celular glas
{= 0085 W/m de), Estimate the thickness ina upto which Rea want mil mares,
{Take outside Ilm coccion = SN /nodeg m 0206 m

a ent
See lerne
un ee
ee ee dept Het,

aos

Conduction with Heat Generation

In many situations of practical importance, heat is generated internally at uniform rate within
the conducting medium ws, Notable examples are
(O resistance heating in electrical appliances: essentially it is the conversion of electrical
‘energy into thermal energy in the current carrying, medium
(in energy generated in the fuel element of a nuclear reactor
(i) libecation of energy due to some exothermic chemical reactions occuring within the
medium
(ic) drving and setting of concrete
‘The rate of heat generation has to be controlled one; otherwise the resulting, temperature
growth might result inthe failure of the medium. Undoubtedly temperature distribution within
the medium and the rate of heat dissipation 10 the surroundings assume great importance in
the design of thermal units,

4.1. PLANE WALL WITH UNIFORM HEAT GENERATION

conduction through a plane wall in which heat sources are uniformly distributed
Volume. The wall surfaces are maintained at temperature f, and fp and the wall
thickness 8 is small in comparison with other dimensions

, 4

FR] |
SD St

1. Steady state conditions

2. One-dimensional heat low

3. Constant thermal conductivity k

4. Uniform volumetric heat generation (9, per unit volume) within the wall

104

Q, (heat conducted in at distance x) = 44%
a az
Q (heat generated in th element) =A dr,

Ounae eat conducted out at a distance x + 44) = Q, +

From steady slate condition of heat flo,
QF nO |

a
= Q + Fur

a)

or Ada =
a a)

|
=
That gives appropriate heat flow equation as

(nd a ag
( alt a

dt oy

ak an
rating equation 4.1, we obtain the general solution as

42)
%

TIRA ee)
A. Both surfaces are maintained at a common temperature

From the prescribed toundary conditions,
at red

sn + ES
the constants of integration take the values

we Gets

Substituting these values of C, and C; in equation 43, the expression for temperature
Profile becomes oi E a

24M gre
E

%
ren

09)

Maximum temperature and its location within the wall can be worked out by differentiating,
the equation for temperature profile with respect to x and equating the derivative to zero

da
ri

er gees | |
106 y Basics of Heat and Mass Tronster |

since he factor LE #0
E
Sano of 80

ec anreisen
a

f 4
sens) te = BR the (45)
fan" [10 La E

Aus parabolic and symmetrical |
My 8/2 and it equals |

Heat transmission then occurs towards both surfaces, and foreach surface itis given by

oC |

“|

For both surface
Q= Ady,
elf of conducting med) hat generating cap)
eat conducted tothe wal surface is all spa tothe surrounding atmosphere at
temperature Then fr each surface
Bann ten ters on
Sang ths vale of wal emperturein oat $4, oe ge the temperate distrib
in term enperature 1, ih surrounding atmosphere

DER ER ET (43)
tate se (5x) (48)

Equation 48. applies equally well to plane walls which are perfectly insulated at one face
and maintained at à fixed temperature 4, on the other face.
The full hypothetical slab with thickness 25 can be considered
with the same bounding conditions +
fet, at r= 25

ated

The locaton x = 3 corresponds to mid-plane of the
| Aypotheia al wall or sut face ofthe given wall
The equation 44, for temperature distribu and the
equation 45. for maximum temperature at the mid-plane
(valga end of given wall) may be writen as

%
nei

Fig. 42. Conducion ina
Insulated wall

ney.
Ben,

Conduction with Heat Generation / 107
5. Temperatares of both the surfaces are different:
‘Wien the boundary conditions 1
fet, at x 0 08
teh axes
are applied 10 equation 43, the constants of integration take the values
Pre
ch,
EE
Subsituing these values of integration constant in equation 43, he expression for temperature
profile becomes:

Geh:

pu tease
x

Boxe

45,
tg 2 E pe
rad tit

ota Semen es «9

Further, rate of heat flow from any face of the wall is;
a

ra fit | 09)
Qui] 430)
Heat flow from the right hand face, Le, at x = à
not 4
ra [24% (4.208
% [5 | En

Let us now examine two cases
() Maximum temperature occurs within the wall; the heat flow will then be from both the
‘surfaces and the total heat flow will become
Q = Q+Q x
(i) Maximum temperature occurs at he left hand face, ie, l is maximum; the heat flow
will then be only towards the right, Le, in the direction of falling temperature
Q,=Q (only)

ul

408 1 Basics of Host and Mass Transfer

conslction equation @ = KA (ty = 19/8 for a plane wall without any internal gener

even

©. Cument carying electrics
‘When beat generated w

esperes in electrical terms

Ora; rn

&
= LE à neat generated por unit volume of conductor
4g 7 RE > heat generated pa

conductor :
Gin a material is due to passage of electric curtent, q, can be

here Tis the current, R is the electrical resistance, p is resstivi
and cross-sectional area of the conductors. Combi

, Land A are the length
ng these relations, we get

ero E ga
A AL (AP TE uy

Here iis the current density and electrical conductivity k, is the reciprocal of the resistivity
1 of the medium.

EXAMPLE 4.1

‘The rear window of an automobile is made of 5 cm thick glass of thermal conductivity
08 W/m-deg, To defrost this window, a thin transparent film type heating element has been
fixed to its inner surface, For the conditions given below, determine the electric power that
‘must be provided per unit area of window if a temperature S'C is maintained at its outer
surface

Interior air temperature and the corresponding surface coefficient, = 20°C and 12 W/m deg.

Surrounding air temperature and the corresponding surface coefficient, = - 15°C and
70 Win -deg.

Electric heater provides uniform heat flux.
Solution : Given: # = 20°C ; h = 12 W/mädeg,

For unit area, the heat balance provides

lt)

OS

A]
Substituting the given dat
(20-5) ouside
+ 0005 *% Taste
n°08
15

or ass + cons 4 “1400

Electric power to be provided,

Fig. 43.

15 2
Ig "1400 - Tonga = 1400 - 167.50 = 12325 W/m
EXAMPLE 42

A composite stab consists of 5 cm thick layer of steel (k = 14617 id
- ick layer of steel (k = 146 kj/m - hr-deg) on the Left side
and 2 6 em thick layer of bras (= 276 kfmchirdeg) on the right hand side, The outs

and hea gente rt 020 Wie
og exch tb. me
alas LE eue
co Une pitted sé
Print ms

Accordingly we may write
+O"
o
Considering unit surface area
6x» 100), 276.1 aan
00 a Tan eer
for 2920 (f, 100) + 4600 (, ~ 50) = 42 x 105

2 10° + 292 » 109 + 23 » 105 = 942 x 108
Temperature at the interface,

ARA oe ne
BA 125260

Heat transfer Uhrough the steel layer,
„ M61 (125,26~100)

x = 3759 M/s he
Heat transfer through the brass layer,
2761 (125.26- 50)
ar

os = 546196 mich
EXAMPLE 43

A square plate heater measuring 16 cm x 16 cm and
of rating 1 KW is inserted between two slabs. Slab
À is 2m thick (k= 60 W/m-deg) and slab Bis 1 cm
thick (k= 0.25 W/m-deg). The outside heat transfer
‘efficients onside A and side B are 200 Winden &
and 50 W/mtdeg respectively. If the surrounding,
air is at 20°C, make caleulations for the maximum Ÿ se
temperature in the system and outer surface

ES
temperature of two slabs. Also calculate the heat + —
transfer through each slab. há

Solutio

elf, be the mai temperature at
the heater section Then for steady state heat ov,
Q = heat flow through slab A (Q)) = =

‘+ heat flow through slab B(Q) MT? A

Fig. 45,

10 7 Basics of Host and Mass Transter

in ut a
et a o pr)
ARA RATA

‘Substituting the given data,

2000 = (216 x 016) (a, - 20) |

256 €, Dt, |
(2005 "ome

= 0.0256 (Ina, — 20) [187.617 + 16.667] = 5.2296 (ty, = 20)
2000
EAS
Considering the heat flow through slab A,
tm = #1)
5, CE)
200 (1, ~ 20)
= 200 +, - 4000
633600 + 4000
‘That gives , = 199250
Qu A 19

= 200 x (0.16 * 0.16) x (19925 - 20) = 917.76 W
Considering the heat flow through slab 8,
2 a ame 4)
+ A Male = 10
ssen2-+)
‘oor
52805 - 25 t,

= 50, = 20)
14-1000
5280.5 + 1000
m
Qe whl (la = 1)
= 50 x (016 x 0.16) x (8374 - 20)

¿ = 8159 W

Check + =O, +0,

= 91736 + 8159 = 999.35 W

; 1 1 HW which equals the rating of heater.

That gives h "rc

St nn One 14
Sk Woche ba a ts ear es

th sides ofthe slab at the same temperature,

relation
%
te Oost,
dh
a 72 6-2
a
Heat flow rate Q = - ka À
At the quarter plane AB:, x = 3 em = 003 m
10
Lo 012 - 003) « 008 + 150
= 675 + 150 = 15675
a. ae “a
E01 2000 = 150C/m nn
Fe zung 012-2 x 009 = 1000 / “i
Q=- 200 x 1 x 150 ne
~ 30,000 W/m? (unit area) =

‘The negative sign signifies that the heat low rate atthe quarter plane AB is in a direction
opposite to that of measurement of distance (ve adiccion)
The eat conducted ata section equals the heat generated from mid section to that section
eat conduction at section AB = Heat generate rom midsection Lo section AB
q,» (A= distance fom mid section to section AB)
Aion)
= 30,000 w/m?
At the quarter plane CD: x = 9 cm = 0.09 m and the diferent parameters would work out
tobe

EE 20000 WHat
Braut yey, e amen peu ae nin ea
PRA
saa foca = een
a FOS) EA a
«lt
Fa

2

(BAD +150 = 9 + 150 = 159°C

EXAMPLE 4.5

with oth of side
Consider a 1.2 m thick slab of poured concrete (k = 1.148 W/m-deg) with be

surfaces maintained at a temperature o 20°C. Daring lt curing, chemical engray a released
At the rate of 80 W/m, Presuming that temperature does not vary with time, workout the
‘maximum temperature of concrete.

412 7 Bosics of Hone ond Mass Transfer

(6) What maximum thickness of concrete can be poured without causing the temperature
gradient to exceed BEC per metre any where in the slab?
Solution With uniform heat generation and both sides of the slab maintained atthe same
temperature, the temperature distribution S prescrited by the relation,

Bares o

x(1.2}8 +20 = DC

Sa
(9 Ditferemiating expression ( with respect 0
ats
Las
aa
Apparently the temperature gradient largest at x 0, and
(#) 2%
¡MS

Substituting the given values,
so

a eee)
EZ © 7088
Maximum permissible thickness of slab,
5
5285 366 m

EXAMPLE 46
AA 25 mm thick meat slab (k = 1 W/m-deg) is roasted with the help of microwave heating,
For good quality roasting, itis desired that centre temperature of the slab be maintained a
200€ when the surrounding temperature is 25°C. What should be the heating capacity
in Wm? of the microwave if the heat transfer coefficient on the surface of meat slab is
20 W*deg? Also caleulate the surface temperature of the slab.

Solution: Maximum temperature occurs at Ihe centre of sab and itis prescribed by the relation

Inserting the appropriate values,

0.025 , 0.025? ]
100 = 25+4,| 0025 , 002 |
LE |

= 25 + 4, (0.00625 + 0.0000781)

= 25 + 00007081 4,

1
+ Heating capacity of microwave, q = 0 102599 W/m? = 1026 kW/m?

Conduction with Heat Generation // 113
Heat conducted to à section equals
w 3H the heat generate fom mid section te that section

eat conducted ta the surface = 4, (4 À

The heat conducted to the wal surface i finale

1, Then for each surface mip

ae
2.2 0025 = oar

EXAMPLE 47
A long stainless steel bar 20 mm x 20 mm in square cross-section a perfeily insulated om
three sides and is maintained at a temperature of 40°C on the remaining side, Dor

the maximum temperature in the bar when itis conducting a crono 100 saree ice
thermal and electrical conductivities of steel as 16 Maig and 13 2 ¡etico aed
neglect the edge effects. Also work out the heat flow Dom the ba

Solution : The he

rom the relation

generated per unit volume due to flow of electric current is worked out

LY
merlo) >
where ps resistivity in ohm-cm, he, reciprocal of electrical conductivity

„m, 2 7 Wea? 10
se (BBR) E 0070/0437 +10 00

"he tempat ong he a pre by e lon
[Los
[o
Mam tempera cs cono ct 8/2 ar regen nny
4 CE
nl)
Er DA ge. an = are

E Bx16

„Under steady state conditions, the het low through the bar equals the heat generated

= 4, volume of the bar
= (4.267 105 x (002 x 0.021)

167 x 10% W/m length of the bar

EXAMPLE 43,

A metallic rod of 65 mm diameter and 1.25 m length runs between two large bus bars which
Are at 25°C temperature, The lateral surface of the rod is insulated against the flow of heat and
Slectric current. Determine the maximum current the rod can carry if its temperature is not to
exceed 180°C at any point. For the material of rod,

«testi resistivity ©1.78 x 10% char em trade oe nace

ahermal conductivity = 250 Wink.

Satin onedimensona het low throug the ase we,

icon ende mamada e same temperate,

cnn deu precio bythe elton ==

A Fig

ear,
Farber, due to symmetry the maximum temperature would occur a the mid plane, 6

EE

tes,
aja

Substituting the appropriate values, we get

ER
„re

%
= gig ists
rn

Then heat generated per unit volume due to flow of current,
“(180 -25)x8%250

=

Rate of heat generation Q,

= 198400 W/m?

= que volume

198400 x 3 057 « 125 = 822 W
In terms of current flowing and resistance of rod,
„a
Qurrunı À
Given: Resistivity p = 1.75 x 10% oh cm » 175 x 10% ohm m
IO À
Er
Fr
Maximum current the fod can erry,

- BET «161 amp
Vaso >

Ten Qu 6595 «1048

EXAMPLE 49
A plane wall 0 m thick (k = 20 Wa-deg) is insulated on one side while the other side is
“maintained at a constant temperature of 350°C. The rate of heat generation within the wall
is 500 W/m. Determine the maximum temperature to which the wall will be subjected and
location of the plane where it occurs

Solution : With uniform heat generation and both sides of plane wall maintained at the same
temperature, the temperature distribution is prescribed by the relation

%
PRET

The above relation apps equally well at toa plane wall which is perfect
‘one face and maintained at fixed temperature onthe other face. À full hypoth
thickness 26 can be considered with the same bounding conditions

J 414 Basics of Heat and Mass Transfer

CConcvetion wich Meat Generation / 115
= he atx and atx = 25

he location x = $ corresponds to mid plane of the

Ry pothetical wall or insulated face of the given mal,

Wa fer an insulated wall of thicknes $, the emperanene

Gatribution conforms to the relation

route cn

%
tater Ses 0x

‘The maximum temperature occurs at mid plane
(slated face of the given wall where x =

EN
Laos

%

= AES? = 350 200 057 = ana

EXAMPLE 4.10
‘An 8 cm thick plane wall generates heat at the rate of 1.2 x 10° W/m One side of the wall
is exposed to environment at 90°C whilst the other side is insulated, The convective heat
transfer coefficient between the wall and environment is 560 W/m'-deg. Proceed from the
basic principles to determine the maximum temperature to which the wall will be subjected.
‘The thermal conductivity of the wall material may be taken as 0.15 Wiardeg.

Solution : The approy

ate form of-one-dimensional steady state conduction heat equation is

“

Another integration gives the general solution
for temperature distribution

A tostado
terre Fig. 49.

The constants of integration are determined from the relevant boundary conditions which

(9 Atx = 0, the conduction region is perfectly insulated and hence heat flow is zero, From
Fourier law Q = - KA (dt /dx), and accordingly the temperature derivative must de
Zero at x = 0, Hence using expression (), we get : Cy = 0

(i) Atx 8, the heat conduction equals the convective hea flow tothe surroundings. That

a él, «toa

h
e [El Brea «i
Asin fom expression ( and noting that Cy

|

416 Bases of Host and Mass Transfer

ls (ee)
rom expression tn) son 4)
we 88
+4 9

ss

ab ese val cu constants, the general solution tor temperature distribution becomes

liso
x DE Crus
‘he sims temperature occurs a he Inulate wal bordar, de, at = D

2210 «0.08
El

A zer
15

EXAMPLE 4.11
‘The plane wall A of thickness 5 cm and thermal conductivity 80 W/m-deg and
volumetric heat generation of 1.25 MW/m is insulated on one side while the other
in contact with surface of another wall B. The wall B has no heat generation, is made of a
material of thermal conductivity 160 W/m-deg and has a thickness of 25 cm. The non-
contact surface of wall Bis exposed to a cooling fluid at 25°C. Ifthe convective heat transfer
coefficient between wall B and the fluid is 1000 W/m -deg, calculate temperature at the
insulated surface and that at the cooled surface of this composite wall.

Solution : Considering unit surface area, the heat generated in layer A of the composite wall

Q, = (1.25 x 104) x (005 * 1)
= 62500 W per unit area of wall

8 i
gyn 125 at fi
KE 80 Mi, ¡

® B
Insuatdn ae dl

Fig. 610,

As the face at =D late, here sn low of any heat at this surface, Al the heat
generated would pass though wall 8 and finally dissipated o the surrounding cooling best
fluid. That is y de id id

Conduction wen Heat Generation 1117

Q

ES Dee

emos

BETEN)
TE
0.035

Temperature at contact surface between the two wal

LA
À 6400 (4, = 875)

62500

500 aos
1 = SS +875 = a6

‘The wall À represents an insulated slab of thickness 5 em with one end maintained at à
fixed temperature f, The temperature f atthe insulated surface 1 then given by
E
A (0057 + 9726 = 11679
42, CYLINDER WITH UNIFORM HEAT GENERATION
Consider heat conduction through a long and cylindrical cod of radius R and length L
Assumptione
1. Steady state conditions
2. One-dimensional radial conduction,
3. Constant thermal conductivity &
3. Uniform volumetric heat generation (4, per unit volume) within the solid,
‘The situation corresponds to a current carrying wire or a fet element ina nuclear reactor
‘The appropriate differential equation describing the temperature distribution can be obtained,
by making an energy balance on a cylindrieal shell of thickness dr and located at radivs +

Q, (heat conducted in at radius 1)

O, (heat generated in the element)

Q, à (heat conducted out at radius r + dr) = @ + À Q,) de

For steady state conduction of heat flow,

4
er dela, E [Ari ar
Que, à at)

This yields the appropriate form of heat equation as

ky “an

$

4107 Snes of Hose and Mass Transfer

Fueron 412 may also be obtained from

cun 225 by assuming steady state un

Sat heat tow in the radial direction
"Upon miegraton

els sie

+. Another integration gives the general
action for temperature distribution

eR scylog.r4 Ce
Per Giles r+Cs

410) Fig 411. Conduction si einer win unterm
“The constants of integration are to be heat generation

determine from the relevant boundary conditions which ae:
Ota teatraR
(a) Heat generated equals the het lost by conduction atthe surface

A
eso a co (£),,

Ace isn ss rm the ft o aso der he cn ns e Int
ea ta atone ee ger ase mt te
A ay won bundy cos
Histor

The tempera rate a ec

e ar A
2) ua
Des

E.

Mr de

(Eqn. $13)
Also

(second boundary condition)

E
or Gro
Applying the boundary condition # = 1, at = to equation 4.14
up
ER (because G = 0
tee ER HC, (ee = 0)

With these values of integration constants the general solution forthe temperature distribution

becomes,

on
PRET «9
Undoubtedly the temperature distribution i parabolic and the maximum temperature

‘occuring at the centre (r= 0) of the rod is given by =

Conduction wich Mase Generación / 118

Ine er HER? wo
contingent hip an nen
PR

(sy

(4) #9

ion Su RES men red cera que
Aedo

%

or tht ER un
‘substituting this value of surface temperature 1, in equation 4:18, one gets the temperature
distribution in terms of temperature 1, of the surrounding atmosphere,

tote Reel 419
and temperate fg, at 0
«Les
SS 020)

EXAMPLE 4:12

{A 25 mm diameter meat roll (k = 1 W/m-deg) is roasted with the help of microwave heating,
For good quality roasting, it is desired that temperature at the centre of rol is maintained.
at 100°C when the surrounding temperature is 25°C. What should be the heating capacity
in W/m of the microwave if the heat transfer coefficient on the surface of meat roll is
20 W/mt-deg,? Also claculate surface temperature of the roll.

Solution : Maximum temperature occurs atthe centre of the roll and it i presented by the
relation, a

à 228 à 4, 00125 + GONE «25 + ONIS y,
fig gai ram:
yo E « mano what «ata ae
(Te net which neg in genre white oid ls blanc by the at
at wet energy leaves Sol a tefl Duda ha
ay (RD =H GERD (o = 1

a
or tnt res + SES

wk ZS 00125 = 66.66°C

Ay” 420 // Basics of Heat and Mess Transfer

EXAMPLE 4.13

A casintens steel wir conductivity = 20 Wm-deg and resistivity = 70 micro ohm-<m) of

Arai nen ES hm is submerged in a oid a SOC and an electric current of
rea an spe snes trough iT conductance at the wire surface is 4 kW,

cos State temperature a he ent and atthe surface ofthe wire,

al. 7x10" 220

D ans
Sotation ; Electrical resistance of wire, 8, = À = LICOR 255 a
x025)

Heat generated, Q, = FR, = 300° 0.285 Watt

Volume of wire, Va Er e

generated per anit Volume, y

Radios of wire, R

The wire surface temperature is given by.
LEA
2615x10*
2x (8x4000)

Maximum temperature inthe wire occurs atts geometric centre line, and can be computed
from the relation,

150+ x 0.0125 = 50 + 408.59 = 458.59°C

4
a

ER
toe thy HER

+ Er
E

= 45859 +
DE]

(0.00125)?

58,59 + 51.07 = 510.66

EXAMPLE 4.14
A concrete column used in bridge construction is cylindrical in shape with a diameter of
1 metre. The column is completely poured in a short interval of time and the hydration of
concrete results in the equivalent of a uniform source strength of 0.7 W/kg. Determine the
temperature at the centre ofthe cylinder at a time when the outside surface temperature is,
75°C. The column is sufficiently long so that temperature variation along its length may be
neglected. For concrete? re re rene
‘Average thermal conductivity = 0.95 W/mK
Average density = 2300 kg/m,

Solution : Since the hydration of concrete results in uniform internal heat generation, the
maximum temperature occurs at the centre ofthe cylindrical column and is described by the

Conduction with Heat Generation // 124
«he
a“
4,787 Wig = 07 «2300 wap
‘Substituting the given values ye

ars + az
OS

EXAMPLE 4.15
Show that in a wire carrying elec
wire is given by

current the maximum temperature at the centre of

where dis the current density, ty

isthe surface temperature coesponding to outer radis
of the wire, A, and k ae the clctrical and thermal conducting of We cee mala

Timited to 10°C. The wire has a surface temperature
«conductivities of the copper are stated to be 5 x 107

mm and 20 Wak respectively.
Solution () Fora cyindrical wire conducir, the maximums temperature un te ee
tod is preserbed bythe relation,

+ tage
ar hy + GER? ©

Total volumetric heat generation is equal to

D

TAC
Ala dk

The elec onda is th epoca ety
Heat generation per unit volume, .

PR

ñ

Hocirica

(22) [e
lan ala) tí 0
The term = 1/4 defines the current density,
From () and (i)
i ay

“a

which is the required expression :
(0) Inserting the appropriate values in the expression (ii) derived above
tea tuläkk, | 10%8
Er

= 76 «108

Current density i = 872 x 108 ampere/m?
Voltage drop = 1,
a

ao Be at vas

gape Bancs of Heat and Mess Treaster sit La pe

Kimi _

REN Santa lo the surroundings is 132 W/mK, make calculations for:
{i surtace temperature of the transmission line

|
Perature distribution can be set up by making À
nes dr trade ji

nd
Q (heat generated in the element) = 4, x serie

(D rate of heat ration per unit volume of the wire Q, + az (heat conducted out at radin cs
(Em temperate the Une ana Onde
Te thermal cond of copper ls 380 Wink For steady state conduction of hes Now,
Solution: (Heat generated inthe transmision ine due to flow of eurent
FR
0% « 0075 Wk 5
$50? x0.075
= BRATS 5.57 W per metre length
LR ps oe tod EIN
Heat dissipated 10 surroundings by combined convection and radiation ee je ae
haus Dr
1126 002% 1 iy 29 W per mer length
{Under steady sate conditions
54187 = 42 (mr 0 à 2) (= 39)
Solving (wire surface temperature) = 876°C
D Ler, be he volumatri heat generated at uniform rate over the wire cross-section
(907.1
a
rim ww
(id) Maximum temperature in he wire vil car atthe geomet entre ine of the wie 0
and may be computed fom the relation, ren ie
En
% FOREN:
12608 oo 5 i
29676 + 172610! (yon? = 0876 + 001135 = 96.771
6 + LEO oy?» 98.7 + 00 . on
‘The small difference between surface and centre temperature results from the relatively a :
PER pe Ene ah et nie ae u JE nea ges he ga sli fo he sate at erate
43, SPHERE WITH UNIFORM HEAT GENERATION der
Consider heat conduction trough a solid sphere of radius e | EE =e “)
Assumptions Corresponding o centre ofthe sphere r = and that gives C= 0. Applying ths boundary
1 Steady state condi SEN expresión i) and ting that Ca = we Bt

2. One-dimensional radial conduction
3. Constant thermal conductivity &
4, Uniform volumetric heat generation

per unit volume) within the solid

424 p Goes of Most and Mass Transfer

ao ars ty

With ebene value integration constants, the general solution for temperature distribution

tot wae eet
rue
ur in
oe tote ir Gi)

{Undoubted the temperature distribution is parabolic and the maximum temperature occuring
at the centre (= 0) of the sphere is given by

de 6
toa tet SER a
Combining expresions (i) and (2), we obtain the temperature distribution in dimensionless
form
bh ry E
CRE 1-3)
‘The heat flow can be evaluated by using the Fourier relation,
N
e-ula),

untl,

ä am

%
aterm x ER
bra a 7:

ARI x gy

‘lue of sphere * heat generation capacity 7)
‘Undoubtedly the heat conducted is equal to heat generated. But for steady state condition,
the heat conducted (or generated) must be equal to heat convected from the outer surface of
the sphere,

vn
Sitting this value of ura temperature, in expresion ion es he per
dicton interme of temperature the sand a À
re
Bee oi
ETS [is )

Conduction wen Heat Generation # 128 |
and temperature Lu at r= 0

Lis Le Rte
à Ca

EXAMPLE 4.17

eee re a

SR Ta lr, sen trans rae
edn ng me ef crt a
= AR

= 18000 j/mm?-her = 2000010?

Solutia me

5000 JJ = 5000 W/m

x

98x10
08 1/mhrdeg = ER 0200 i mandeg = 022 Wimdeg

safe ete of he orang, temperature maximo and vl spree by the

»

ashy + HR? 265+ oon? nase

‘The heat conducted equals the heat generated

Som

o.

(004) 5000 = 134 3/5 = 482 Kye
EXAMPLE 4.18

A solid sphere of 8 cm radius has a uniform heat generation of 4 x 10° W/a. The outside
Surface is exposed to a fluid at 150°C with convective heat transfer coefficient of 750 Wav eg,
thermal conductivity of the solid material is 30 W/m -deg, determing; a) maximum temperature
and its location, (b) temperature at 5 cm radius

Solution : Heat generated in the sphere is convected over the surface. That is

4
ee ee
4
or 4 x 106 ZAS) = 750 x 4x (008)? x (= 150)

4x10 0.08
35750

‘Temperature at the surface of sphere, fy = 150 + 14222 = 292220

Maximum temperature occurs atthe centre and its value is prescribed by the relation

fra

CE

Temperature at any radius r can be worked out from the relation,

or 1.7180 = on

wieder? ne
Im tut GR? > 2

(0.08) = sac

J 1426.1 Bosics of Host and Mass Transfer

4s.

ss
1 2 000 (43445 - 29222) + 292.22 = SIC

REVIEW QUESTIONS

Mention some ofthe ua where het is generated internal’ uniorm rate inthe conducting

Show tha fora lane wll of thickness 2 a
Fame he temperature 4 at the mid plan

À pren

abut het generation 4, per un
À the ston

here tthe temperature on eier side of the wall and Ki the thermal condctety of wall
Considera plane wall of thiskness 21 with temperature 1, and 1 on the Bounding surface. The
Salts unir distrae best gencraion er un volume, Betermine an analy! ©
Tor he dimensiones temperature (= 5)/( 9 where fs the temperature a the ca
in wall Fortes show tha the temperature atthe cone i the mani temporäre when
SE postie and the minimum temperate when 4, negative

X plane wall of Uschness 10 cm and thermal conductivity 25 W/m deg has a solera heat
on of 03» 10° W/m: The wall insulated on one side and Ihe Ohr dei eos lo.
Fa at rc temperatre,Desermine the maximum temperature in he wa the convective hea
transercoeificien Between the wall and ud 1 500 M/m K

“An ceci current of 150 ampere pares through the 20 mam diameter wire of an ectricl resistance
{ester Determine the centre Une temperatore rise above the surface temperature. The wire hos
el rest P = 8 + 107 Gham and thermal conductivity k = 19 W/mdeg

(is. 23750)

‘Show tht in along cylinder of radivsR with uniformly distributed heat sources, the temperature

run is prescrted by the relation

were £ st temperature atthe outer surface o the liner and yi he temperature along
Be ie ani

‘An electric heater made from nichrome wire, measuring 2 mm diameter and 10 m long caries
{current of 25 ampere: Make calculations fr the rate of hea low from Im length of the heater,
and also the temperature a the surface and conto ne af he wire. Given thatthe specific resistance
of nichrome 9 = 101 olmo thermal conductivity of nichrame k= 17.5 W/m-dep, and the
old aris blown atthe heer a temperature , = 20°C giving local coefficient of et transfer
ftom the surface of ater the aie = 463 Wjmdeg (Ans. 2185 Way, 769 and 77°C)
A nuclear element inthe form of a hollow cylinder has Inner and outer radi 5 cm and 10 cm
respectively. The cylinder 15 insulated atthe Inner surface and the heat generated ts convert
‘over the outside surface of a did at SOC and surface conductance 100 W/m des I thera
conductivity of the cylinder materia is 50 W ade, make calculations forthe

50 that moxdmum temperature inthe system will ot end 200C

te of heat generated
(ns, 350 Ke)

St up expresos fo tempertre divi during steady state hat conduction in à sid
sphere with internal heat generation. un A Ñ

000

Heat Transfer from Extended
Surfaces

Im many engineering situations, means are often
ta race 10 its surroundings. The Newton Riktn

eat flow can be enha 1) reveals that the
‘SEhvectve heat flow can be enhanced by increasing the fm codicia he

Sought to improve heat dissipation from a
an relation a RA den
cor o temperature difference (1-4). The convective

‘he surface ares A
fluid properties and the flow rate

<cocficient sa function of the geometry,

Control of h through these parameters helps to Ob
x w tate, Con ites haps 10 obtain As
“plu value, With regard tothe effect of temperature excess (tf) dal be empire

(hen the ambient temperature , is 100 high particulary in hot weather conditions, The surface
rea exposed to the surroundings is frequently increased bythe etachment of protestors 0
the surfaces, and the arrangement provides a means by which heat transfer ae can Le substantially

improved. The protrusions are called fins or spines, and these extensions can tke a veiely of
forms, the most common types ate illustrated in Fig 3.1

x Lx
a y @

Fig. 5.1. Common type on configuration

a

A straight fin is an extended surface attached to a plane wall the cross-sectional area of
the fin may be uniform or it may vary with distance from the Wall (Fig 5.14, ). Annular fins
ae attached circumferentially to cylindrical surface and their cross-sectional area varies with

dis from the centre line ofthe cylinder (ig, 510) However, both the straight and annular
fins are of rectangular crossaection, whose area can be expressed as a product of the fin

vekness 6 and width b for straight fins or the circumferenee 2er for for annular fins. In
contrast a pin fin or a spine is an extended surface of circular cross-section (Fig, 5.14) which,

may be uniform or non-uniform. Thus a spine represents a thin cylindrical oF conical rod
Protruding from a wall

Common applications of finned surfaces are with
+ Beonomisers for steam power plants ;
*+ Convectors for steam and hot water heating of systems ;
+ Air cooled cylinders of aireraft engines, LC. engines and air compressors ;
+ Electrical transformers and motors à

+ Cooling coils and condenser coils in refrigerators and air conditioners;
+ Electronic equipment ct

m

A a

J 428 7 Basics of Host and Mass Trensfor

5.1. STEADY FLOW OF HEAT ALONG A ROD
(Governing diferent! equation)

Permet P= 20 + 8)

Cross sector
we rer)

= Surrounding thi at,

Fig. 52. Extondod surfaces wih uniform cross-sacton(a) Fin wit rectangular
(0) Fin win Greuas proto spine)

Consider a straight rectangular fin or a pin fin (spine) protruding from a wall surface (Fig
52). The characteristic dimensions of the fin are its length J, constant cross-sectional area À
and the circumferential parameter P. Thus for a rectangular fin

Abd; P=20+ 5
and for the spine

aha; pond
‘The temperature at the base ofthe fin is ty and the temperature of the ambient fluid into
‘which the rod extends is considered to be constant at temperature 1, The base temperature fy
is highest and the temperature along the fin length goes on diminishing (Fig. 53)
Analysis of heat flow form the finned surface is made with the following assumptions
+ thickness ofthe fi is small compared with the length and width; temperature gradients
over the cross-section are neglected and heat conduction treated one dimensional

+ homogeneous and isotropic fin material; the thermal conductivity kof the fin material
is constant

+ uniform heat transfer coefficient h over the entire fin surface

+ no heat generation within the fin itself

+ joint between the fin and the heated wall offers no bond resistance; temperature at root
A or base of the fin is uniform and equal to temperature 15 of the wall

On = NPB (=D)
Fig. 53. Heat transe trough af
+ negligible radiation exchange with the surroundings; radiation effects if any, are considered
35 included in the convection coefficient # d
+ steady state heat dissipation

um 2) 6.

eL a £2
(i) Heat convectd out ofthe element bien the plans x and (+50
O = CT 3)

Here temperature ofthe fin has been presumed to be uniform and non-varan fr the
infinitesimal element

A heat balance on the element gives:
Qu * Qu « a? On
a A fradtas)-npée (0
had ea ejerce
Upon rearrangement and simplification
ét ve

ae

64

190.4 Boston of Hoot ond Mess Transfer

Equation $4 i further simplified by transforming the de

pendent variable by defining the
pera ncoss 8 as

96) = 19 =f
Since the ambient temperature , is Constant, we get by differentiation
E

az

dede
Thos 63
where

Equations 54 and 53 provide a general form of the energy equation for one-dimensional
heat dissipation from an extended surface, Fora given fin, the parameter m is constant provided
the convective film coefficient I is constant over the entire surface and the thermal conductivity

ie constant within the considered temperature range. Then the general solution of this linear,
homogeneous second order differential equation is ofthe form

D = Coreen

69
The constant C, and C; ae to be determined with the aid of relevant Boundary conditions,
5.2. HEAT DISSIPATION FROM AN INFINITELY LONG FIN (=> =)

‘The relevant boundary conditions are

{) Temperature at the base of fin equals the temperature ofthe surface to which the fin
is attached,

In terms of excess

(i) Temperature atthe end of an infinitely long fin equals that ofthe surroundings.

ono
Substitution ofthe

‘boundary conditions in equation 5.6 gives
CR ©

Gems Genel ao ©
Since the term C, ei is ero, the equality is valid only iC, = 0, Then it follows from
relation (4) that Cy = 0, Substituting these values of constants C and Ci in equation 5.6, one
‘obtains the following expression for temperature distribution along the length of the fin

CRE 67
Fig. 5.4 shows the dependence of dimensionless temperature (+ = £)/(o = 1) along the
Length of fin for diferent values of parameter m (m, < ma © M). The pit indicates tha the

‚dimensionless temperature falls more with increase in factor m. With the fin length extending
10 infinity, x > =, all the curves approach (

Pen

1/(o = 1) = 0 asymptotically
The most important design variable for afin is the amount of heat that i can remove ftom
the heated wall and dissipate it to the surroundings

Howe Tate ;
of the heat fl fer from Excenced Surfaces 1] 184
von of ae fin ending ro

Que 7 HA Yard,

rom the expression for temperature distribution

tte

QUEUE

ur

Fig 54 Temperate ren
ss woe

Recalling that

ie

Qu [PER (ut)
Alternatively the heat flow ate
heat transport from the Infinite

688)
in be worked out by integrating the expression for convective
al element of the fin surface tthe surroundings

= hat

Fire peser
rn
Plot à note

ER
ST APRA tty~ 1)

‘which is the same as evaluated above (Equation

Equations 57 and 3: are reasonable approximations of temperature distribution and heat
flow rate in a finite fin sits length E

I very large compared to ite crocesectional ar,
The temperature distribution (Equation 57) would suggest that

the tip ofthe fin. Hence area near the fin tip i ot uilized to the extent as the lateral
the base, Obviously an increase inthe in length beyond a certain point does not pay
much regarding an increase in the heat dissipation. A tapered fin is then considered to Le à
Detter design as it has more lateral arca near the base whet

the difference in temperature is
high ve

+ temperature drops

Heat flow rates through solids can be compared by having an arrangement consisting
essentially of a box to which sods of different material are attached (Ingen-Hausz experiment)
The tods are of same length and ates of cross-section (same size and shape); their outer
surfaces are electroplated with the same material and are equally polished. This is to ensure

{hat for each rod, the surface heat transer wil be same. The procedure would involve coating,
the rods with wax and filling the box with be same. Heat flow (rom the box along the rod
Would melt the wax for a distance which would depend upon the rod material. Let

38 Guns of Haat and Mons Manor
exc of teatre ofthe hot bth above the ambient emperture
SEIS rata ot meting point of war above the ambien

pa

Au ty = tengo upto which war mets.
‘Then fr different rods (renting each a in of Infinite Length)
nen
“on
“nem
‘Obviously then

‘Thus, the thermal conductivity of the material of the rod is directly proportional to the
re of the length upto which the wax melts on the rod.

Pr
EXAMPLE 5:1
‘Two long rods of the same diameter, one made of brass (k = 85 W/m-deg) and the other of
‘copper (k= 375 Wm-deg), have one of their ends inserted into a furnace. At a section 105 cm
away from the furnace, the temperature of the brass rod is 120°C, At what distance from the
furnace end, the same temperature would be reached in the copper rod. Both rods are exposed
to the same environment,

Solution: Treating the rods as
by the relation,

inte long fins, the temperature distribution is pres

For brass rod
For copper rod

where I, and I, are the lengths upto which same temperature is reached in both the rods
Obviously from expressions () and (i), we get

mala = mh,

or [an

à Ay

‘Since the rods are exposed to the same envir
(y= Ps and A, = A) we get

=H + (to = 1) em o

120 =, + (=) em @

ronment (hy = h) and are of the same diameter

AMPLE 52
E ‘rods, one made of silver (k = 420 Wy

tah rod.

Satin = Let fy fy and fy be the eng upto which wax wil mel on each rod, Them |
ku |
CA

© @ Gi)
From () and (i) we get,

From (i) and (ii) we get,

Estimate the energy input required to solder together two very long pieces of bare copper wire
(01625 cm in diameter with a solder that melts at 195°C. The wires are positioned vertically in
air at 24°C and the heat transfer coefficient on the wire surface is 17 Wmtdeg, For the wire
alloy, take the thermal conductivity 335 W/mdeg.

Solution : The physical situation approximates as two infinie fins with a base temperature of
195°C in an environment at 24°C with the given value of surface coefficient

==

Coppoñre
Fiss

Crono ana, A, À LOGS) 207 «104 mi

à Da nd = x x 000625

Perime 0051 m

Ar:

1170

dy Tong fis

Most duo fon an im
EAL mig)

RS à 2073 € 10 117 (195

“Therefore the energy input required for two wires is 2652 W

EXAMPLE 54
A rod of 10 mm square section and 160 mm length
“protrudes from a furnace wal a 200°C, and is exposed to as
ED W/m deg, Make calculations for the heat convected upto 80 mm and 158 mm lengths ang

‘Comment on the result. Adopt Tong fin model for the arrangement
seipation for an infinitely long fin is
Qeram art,

thermal conductivity of 50 Wincdeg
30°C with convection coccion

Solution : He

[E72

1200

she me (ka, ~ ESCUETO

39 «0.03 © AI) © 12649 < (20M = 30) = 1075 W

Jel the temp

For the long +

me = 12.619 x 0.08

Louse

703035 » (200 - 30) +

OS

Heat conducted upto any length is worked out by taking the difference of total heat and
heat conducted at that section.
Heat convected upto 0.08 m length
#1075 = RAM (loos = 4)
= 10.75 - 50 À (001 » 001) x 12.649 (91.8 - 20) = 6.84 W

3% of total heat dissipation.

m length

FAC Mm (lease = 2)

= 1075 - 50 x (OA x 01) x 12.649 (63.04 - 30) = 9.293 W

Meet Tranater (rom Extended Surtaces / 138

9.293
which ie 2293100 = 864% of ota
0.78 À heat dissipation,

ar

‘Applying these boundary conditions to equation 5.6
SEEN, Q

Further
tte Ge + em

mc 2 = m

mye cen

= Geno Pr

Substituting these values of constant C, and C; in equation 56, one obtains the following
expression for temperature distribution along the length of fin

,
Expressing in terms of hyperbolic functions,

o ct

ES coo sil
The rate of heat flow from the fin is given by

Qu aL)

From the expression for temperature distribution
cost =)

Etoo och
at gg SP
em

a

F381 2008 of Hoa ond Mo

|

—
a
Drum
Qu “BA, mg = 1 tan mt
2 PER =) tn mt 610)
camps ss

A carton steel rod (k = 55 Win-dep) has been attached to a plane wall which Is maintained
at a temperature of 350°C. The rod is 8 cm long and has the cross-section of an equilaleza

pation from the rod if itis exposed to
se conductance 100 W/mdeg, Consider end

tríamgle with each side 5 mm. Determine the heat diss

a convection environment at 25°C with unit surface

surface loss to be negligible.

Solution : For a fn of triangular cross-section
Pde

à
Fi # econ at x
Fig 86.

HE | 100x(3x0.05)

RE din

A "| 55% 2 x(0.005)?

For a fin with end loss negligible (tip insulated)
Q =k Am (fy = 1) tanh mi

me 5029 met

5 x 22 0005} 50,19 x (850 - 40) tanh (50.19 x 0.08) = 9.26 W

EXAMPLE 5.6

‘Which of the following arrangement of pin fins will give higher heat transfer rate from a hot
Surface?

6 fins of 10 em length
12 fins of 5 cm length
‘The base temperature ofthe fin is maintained at 200°C and the fin is exposed toa convection
environment at 15°C with convection coefficient 25 W/mÄ.deg. Each fin has cross-sectional area
25 can’, perimeter 5 em and is made of a material having thermal conductivity 250 W/im-deg,
Neglect the heat loss from the tip of fin.
Solution : The heat loss from mins is given by
Qa mk Am (= 4) tanh mi

FF, CES À
lire cn an Eur
ere me Via, Tree

Casel n=6and [= 10cm = 01m
ml = 44772 x OA = 04472

case Mani 5 em = gon 15) anh (04472) = 13038 w
mi = AAT2 » 005 = 02036
Qs = 12 1250 25 a 10% 477 2
‘The arrangement 115 0 be preferred ast gives ny

EXAMPLE 57

= 15) tanh (02256) = 138.68 W
Ier ate of heat traer.

nn ne En
Sand na ta De at °C. Deena
sai sR gad fr he ih ma

P=26+5
226 + 04 = 68 mm 4
ABB 23 x 04 € 12 mn os tre
fir < i

moda
ETE)

free) amt ras

The arrangement cortesponds to afin with tip insulated and for that
Q=kAm (fy 1) tanh ml
= 180 x (12 10%) « 1944 x (60 35) tam (19.4 x 005)
= 0.0786 W
À ent los from the array of 10 fins,
= 0.0786 x 10 = 0.786 W
EXAMPLE 58
A heating, unit is made in the form of a vertical tube of 50 mm outside diameter and
12m height, The tube is ited with 20 steel fins of rectangular section with height 0 mm and
thickness 25 mm. The temperature at the base o fin is 75°, he surrounding air temperature
ls 20€ and the heat transfer coefficient between the fin as well asthe tube surface and the
fuounding ar is 9.5 WntK. If thermal conductivity ofthe fin material is 55 Wink, make
Salculatios forthe amount of heat transferred from the tube with and without fin
Solution : Heat flow rate from the tube surface without in
QuehA a= hx dH x 1)
5 x (x x 05 x 1.2) « 75-20) = 98.44 W
(0) Heat flow rate convected from the base
DEEE) à
where A, = (6x 0.05 x 1.2) - 20 (L2 * 0025) = 01284 mi
: Qu 95 x 01284 x (75 - 20)» 6709 W

438 7 Gasca of Hout and Moss Transfer
eut flow tate sonwecte from the fin,
Qn kA, (t,t) tanh mt
whore A, = erection aes of fi = 12 x 0.0025 = 0000 mi
P= perimeter of fin = 2 (12 + 00025) = 2405 m

FE.
EEE „un
YA" V8" 003,
Then = 20 = 55 » 0003 x 11.77 « (75 = 20) tan (1177 x 0108)
2987.75 W
Heat flow rate from the take surface when fins are fitted,
Qi = Qi + Q,= 67.09 + 937.75 = 100434 W

EXAMPLE 59
[An electronic semiconductor device generates 0.16 hr of heat. To keep the surface temperature
tthe upper safe limit of 75°C, i ls desired thatthe generated heat should be disipated to
the surrounding environment which i at 30°C. The task is accomplished by attaching aluminium
fins, 05 tm! square and 10 mm to the surface, Work out the number of fins if thermal
‘Conductivity of fin material s 690 rdc andthe heat wane ccficen is 45 aude,
[Neglect the heat loss from the tip of the fin.
Solution: Fora fin of rectangular cross-section,
Pz (b4 8) =2(05 + 03) = 2 mm » 2 x 10m
Abr B= 05 » 03 = 025 mm’ 025 » 10% m?
En, Cao
Via, CUT ETS
For a fin with insulated Hp,
Th Ae (l= 4) tanh m
2 690 035 «10% »
| £3977 2 10 K/h per fin

4 (25 - 30) tank (2284 x O01)

016
umber ot fins = O6 402
NPE SN ra

Thus, 4 fins are needed to dissipate the required amount oF heat.

EXAMPLE 5.10

is used for two fins of the same length ? Assume short fin with end insulated,
Solution + For a short fin with insulated tip,
Q uk A, m (ty 4) tanh ml

A rod of 10 mm diameter and 80 mm length with thermal conductivity 16 W/m-deg protrudes
from a surface at 160°C. The rod is exposed to air at 30°C with a convection coefficient of
25 Win?.deg, How does the heat flow from this rod get affected if the same material volume

ester from Extended Surfoces 11138

"23m

diana

a

Case

=F oom x 25 xa
4 02 x 25 (160 20) tanh 2) = 3995 w
Case I; Let Bethe new dameer Them forthe ame mat

4090097 m

hes

my! =29.73 x 008 = 2378

Q = 16 À (oon? «2975 « (40 30) nr

= 2385. w
For two fins Qy 22 x 2385 » 4770 w
Percentage increase in heat Now
477-398
o = ana

Comments : A thinner ór a low sectional area fin is a beter choice

EXAMPLE 5:11

A plate fin of 10 mm thickness and 60 mm length is dissipating heat from a surface at 197°C.
‘The ins exposed to air at 25°C with a convection coefficient of 2 War dey I thermal couv
ofthe fin material is 200 Wm-deg, determine the heat dissipation. Consider 1 m width of fin.

(0) To increase the heat dissipation, the following tw alternative ave Decn suggested
with the same material volume. =

(9 Split the fin into two fins of 5 man thickness each

i) Single fin 5 mm thick and 160 mm long,
Which will be the better choice ?

The ine may be tonic short th ip inate
Ssitin: Fo à shor fn wih y estad
=k Ava o 1 tanh
Hee: Pet 0220 sag) = 2am
Ans none
m= Veg "oo
I 4714 «008 «0377
Meat dspation, Q 2 20 «O01 «A714 (190 25) tanh (0379
2 56018 por m width
1) Case 1 Two fin 5 mm thik ad 50 mn eth
pears) 20 + 008

474 mt

|
|

:

140 7 Basics of Hoot and Mass Transfer
ee x 8 2 2 = 0005 © 0005 m}
FE. á
me Via Vaoxans 7 668 m
2665 «008 = 0502
{Q = 200 x 0005 » 665 (190 - 25) tanh (0882)
£504.26 W for one in
Heat dissipation from two fis = 106852 W per m width
Percentage incense in het cisiption
„ 105852 —56018
En
Case I: One fin 5 mm thick and 160 mm length
P-20>9=2( + 0008) = 201m
A, #081 x 0005 = 0.005

100 = 90.65%

FF. EEE sn
me fe = avs CES

mi = 665 x 0.16 = 1.064
Q = 200 x 0.005 x 6.65 (190 - 25) anh 1.064
263744 W
Percentage increase in heat dissipation
63,744 - 560.18
sons
‘There is an improvement of only 54.19% against 90.65% in case of two fins ofthe original
length. Obviously increase in length is not effective

x 100 = 54.19%

EXAMPLE 5:12
Two rods A and B ofthe same length and diameter protrude from a surface a 120°C and ae
spas to ir at 25°C. The temperatures measured at the end ofthe rods are SOC and 75°C
thermal conductivity of material Ais 20 Wind, acute the thermal conductivity of
‘material B. Adopt the condition of a fin inulated at the tp.
Solution : The temperature distribution fr a short fin with tp insulated is given by
1

Wet ch

1
10-3 "mi

For rod A:

8; ml = 201
75-35 4
120-25 ~ cosh mgl

9
5

For rod B

cosh mt = 21.9; m

257

Le

ER

"o ARA

since the rods are of some length and diameter,
Ay Ay Py Pe

sn

and are exposed 1 the same environment

and eh,
E
that gives: mao
“Thermal conductivity of rod 8,
DEE

un 20 * (1599) = 5113 Winde,
EXAMPLE

A cebifugal pump which csculats a hot Liquid metal a YC a Given Ey 4360 rpm
tlecric motor. The motor is coupled tothe pump impeller by à born ney stan Dan
in diameter If the temperature ofthe motor is limited to a mardmmum sane ot See

Ambient air at 25°C, what length ofthe shaft should be specified polen Ue coos al ne
pp, It may be presumed that the thermal conductivity o the shaft material 35 Wake ed
that the convective film coefficient between the steel shaft and the saben cee
157 Wink.
Solution ; The shat conducts heat from the pump (t=
nd also loses energy from its surface tothe suroundin

SOC) towards the motor (+ = 60°C)
88 (, = 25°) by convection Treating

the shat as a fin insulated at he tip, the temperature distribution would be prescribed by the
ration

% t-te | cosh)

Cr TE

Lun ly at x= length ofthe shat)
60-25 con 1
500-235 ° cosh ml cost

Si
1337 jt

cosh mt = LE
For a circular shaft of diameter 4,
Pd

142 7 Bases of Heat ond Mass Treaster

EXAMPLE 5:14

nectar 014 gan stream 1 19 be measured by Using wo thermocouple atached fo y
ENT So m and wall crsesetfonal arca 15 am?, The tube ls 250 mm Long and
iStrocnted norma to ine duct wall Y thermocouples are attached o he tube at 125 men and
me Ron he ct wall and indicate tbe wall temperatures of 80°C and 39°C respectively,

{me the gas temperature and the duct wall temperature. The heat transfer coefficient
the gas stream is 12 Wn?K and the thermal conductivity ofthe tube

eme the tube a
material is 25 Wink. Neglect any heat transfer into the exposed end of the tube.
Solution : The temperature distribution fora fin with insulated end (no heat transfer at the

‘exposes end) is given by relation
fot, ¿Osho

A
CT
far, [a

mm Vea EST
Inserting data appropriate tothe two locations of the thermocouples:
350-4,
vn
390-4 |

“

and

‘Simultaneous solution of expressions () and (i) would give
‘Temperature of the gas stream 4, = 417.8°C
Temperature of the duct wall ty = 11883°C

1e heated wall
at 20°C with

EXAMPLE 5.15
A very long copper rod 20 mm in diameter extends horizontally from a pl
maintained at 100°C. The surface of the rod is exposed to an air environım
int of 85 W/intdeg. Workout the heat loss if the thermal
order to be

convective heat transfer coeffi
conductivity of copper is 400 Wm-deg. Further estimate how long the rod be

considered infinite
Solution : For a circular shaft of diameter d

Bund‘
A sg a
já
[iP far
me eee 2061 m
FA “Ved ~ Yioox0.03 ~ >

Heat dissipation from an infinitely long fin is prescribed by the relation,
Qn =A, m (ty) = 400 x i (0.02)? x 2.061 (100 - 20) = 20.71 W

de

There is no heat loss from the tip ofan infinitely long rod and such it behaves as
tip were insulated. Therefore, an estimate of the validity ofthis approximation can be mad?

by comparing the following two expressions for the fin,

Mer Tra
Transfer rom Extended Surtoces 11143

Infinitely tong fin
in with ip input
{30 mt 209 or ml 2 265

IN
Oo = À Amo 1) an
These expressions provide equivalen ut y
Hence the rod can be considered iti 4
¿26
26
ot
54. HEAT DISSIPATION FROM A FIN LOSING HEAT Ar THE re
Te fin sof an Finite Length with te tp exposed for heat diipation. The relevant boundary

conditions are
(0-0 at x=0
(i The fin is losing heat a the Spe te heat conducted tothe a at x= equals the
heat convected from the end tothe surroundings a
a)
a,
[At the tip of fin, the cross-sectional are for heat conduction A, equals the surface area A,
from which the convective heat transport occur. Thus

HA)

“m
wre
ying these boundary conditions to equation 56,
CECEN ©
Further
e
ne etemG ete
19 Ee
- Cet eme Age o
At Ro ni

This is because 8 at x =I equals (C, e + Ce)
Solving expressions (a) and (0), the constants are determined as follows

aarp hee
je Ee
£
aleje
ere
Substituting these values ol constants C, and Gin equation, 56, one obi
expression for temperature distribution along the length ofthe fin

en)

the following

|
|
|
|

mita) 4 À [m0
ler

% re

Expressing in terms of hyperbolic functions

conta) Est

61m

The rate of heat flow from the fin is given by

From the expression for temperature distribution,

Elo)

deagos Eta]

E
cosh + si
eo)

a sinh ml + cosy mi
CRETE
cosh mis À à

sm

h
Sem + cosh mi
CARTER Er nes

oh + sh mt

x
me Josh nt

= PA (tot)

E
cost nl + sinh mt
D

tanh
en (6.12)

EXAMPLE 5.16
A horizontal steel shaft, 30 mm diameter and 600 mm long, has its first bearing located
100 mm from the end connected to the impeller of a centrifugal pump. If the impeller is

mersed in a hot liquid metal at 500°C, work out the temperature atthe bearings under the
conditions: (a) the shaft is very long (6) the heat flow through the end ofthe shaft is negligible
and (c) the heat is transferred to the surroundings from the end.

The temperature and convection coefficient associated with the fluid adjoining the shaft
are 35°C and 68 KJ/ahr-deg, For steel shaft, thermal conductivity k = 72 KJim-hr-deg.
Solution : For the circular shaft

ed
a

[RE (EE gt
Fees

(0) For an infinitely long fin

a

446 Basics of Heat and Mass Transfer
Therefore, temperature, at the bearing (x = 100 mm = Dm) 1
nero de
m Lay naa
35 + 406 © 0.3256 = 19642%C
(0) For à fn wich o heat los from the tip end

Wr) cosh (t=)
a PL
Tree, eat fat he bearing (= 100 mm 0 1) à
ante
a ar
2 son 35) SAL (OS 0.)
= 38 + 500 35) on (01.220. 5)
SB e
8 465 ar
(0 Fora fn dispating heat 6 the surroundings fom stp end
o ton) men) Ln)
8 oda) coshants sinh mt

Therefore, temperature £, at the bearing (x = 100 mm = 0.1 m) is

cosh (= x)+ LL sinh m (1-2)
+ y=) A Á

i
iS

cosh 11.22(06 04) +5 sinh 22.22 (06-01)

x
235 + (500 x — —
cosh (122x065) +; E sin (01.22% 06)

(LL OS) 7 2277 SC )

136.57 + 0.084 x 136.57

235 + 465 x 032 = 1838°C
EXAMPLE 5:17

A turbine blade 5 cm long, 45 cm? cross-sectional area and 10 em perimeter is made al
Stainless tel of thermal conductivity 100 kJm-hr-dez, The temperature atthe root of loe
le 500°C and itis exposed to product of combustion passing through the turbine at 850°C
Determine the temperature at Ihe middle of blade and the rate of heat flow from i TH
film coefficient between the blade and the combustion gases is 1100 kj/m?-hr-deg. Tht
blade may be treated a à fin losing heat a he tip,

Slatin ; Fora fin losing heat atthe ip, the temperature distribution is prescribed by UE

Host Tranat
Transfer trom Extended Surtaces // 147

O e mms

| =)

ooh ns er,
Le

where

mi = 49.48 x 005 = 2479

nue

Em "Tan 7 022

sink « 5881
850

2 cosh ml = 5965 ; tan
2 sh (122) 60.22 ihn)
[SSSR
From this expression, temperature at any distance x from the root
worked out, At the mid of turbine blade = 0038 m
cos m (= 3) "cosh 49.4 (005 = 0025) «1.236
sinh m (= x) = sinh 4944 (005 - 0.925) = 1576
1,2850 _ 1.23640.222%1.576
500-850 ~ 5.965+0.222%5.881 "0218

mt = 0.986

500

of the blade can be

La 850 + 0218 (800 - 850) = 77A7C
(9) The heat flow through the blade is given by

tanh ml
Orkan)
re Eon
09860222
7100 x (45 » 10- (4944) (00 - 85) 70.23 20.586

“man
The «ve sign indicates that the heat flows from the combustion gases to the blade
EXAMPLE 5.18

cylinders em in diameter and 1 mlongis provided with 10 longitudinal straight fins of material
having thermal conductivity 120 Wm-deg. The fins are 075 mm thick and protrude 125 mm
from the cylinder surface. The system is placed in an atmosphere at 40°C and the heat transfer
‘oefficient from the cylinder and ins tothe ambient airs 20 Wm'-deg, Ifthe surface temperature
of the eylinder is 150°C, calculate the rate of heat transfer and the temperature atthe end of fins.
Consider the fin to be of finite length
Solution : For a fin of rectangular cross-section
A, = bx Be 1x (075 x 10°) = 075 10% m

P=2#+8-26=2x1=2m

148 4 Benica of Maat and Mess Transfer

pe EI a
A VS O7 © 22.08 we

For a in of finite length the heat conducted to the end is convected
air and is prescrited by the relation:

=k A, m ty

mi = 2108 x (125% 109) = 02605
h 20
Tos TOO "0007006

tanh mí = tanh (02635) = 0.2576

120 075 x 109 «21.08 x (30 ao) « 2760-0706

Tr 0.007916 «012576
= 5529 W per fin

Heat transfer from unfinned (base surface)
=hlnd=nAJ Dx (ot)
= 20 [nx 0.05 10 x (075 x 109)] x 1 (150 - 40) = 3289 W
28.9 +10 x 55.29 = 8818 W
(2) For a fin dissipating heat to the surroundings from its tip end

pus or 10 fins

coto Le (t=)
Carr res

cosh mie xsinn
At the tp (x = D, the above identity reduces to

1 Cosh 0.2035 + 0007806 x sin 09
1, = 40 + (150 - 40) x 0964 = 146.04°C
5:5. FIN PERFORMANCE

The utility of a fin in dissipating a given quantity of heat is generally assessed on the basis of
the following parameters

+ Efficiency of fin relates the performance of an actual fin to that of an ideal or fully
effective fin. A fin will be most effective ie, it would dissipate heat at maximum rate ifthe
entire fin surface area is maintained at the base temperature

Actual eat transfer rate from the fin
"Vet ht would the whole surface of he fn were mu de ase emperors

away to the ambient

“The parameter
«implican,

above
For a very long fin
tank’ à
CT
Obviously the fin affine} drops with an increase ine length
(6) For small values of ml he fin efcency increases When the length reduced o zero,
uN tank! mi
m © ni
Thus the fin efficiency reaches its maximum value of 100% fo a tiva alu f = 0, ie,
no fi at all. Naturally maximizaion of fin performance with respect ls Inga does sot
Constitute the design criterion for a fi, The efficiency of fin, however loma en ee
judging the relative merits of fins of different geometries or meta
» Effetiveness of fin (e) represents the ratio of the fn heat transfer rate (hat dissipation
with afin) to the heat transfer rate hat would exist without a in.

Fig 58 shows the base heat transfer surface before and after the fin hasbeen attached. The
eat transfer through the root area A, before fin attachment i

Qe RAL)
(2)

Fass teins
After the attachment of an infinitely long fin, the heat transfer rate through
area becomes

ION)
SPA lt) | [PE 515)

Fin effectiveness, € aloe) "RA

AS
> 180.7 Bases of Mast and Mass Transfer

ectangolr fin of thickness 8 and width B
po 2028) 2

AT 6
DER: 6.15)

EXAMPLE 5.19
A steel rod (k = 30 W/m-deg) 1 cm in diameter and 5 cm long protrudes from à wall
Which is maintained a 20°C. The rod is insulated at its tip and is exposed to an environment
with = 50 W/mideg and f, = 30°C. Caleulate the fin efficiency, temperature atthe tip of,
fin and the rate of heat dissipation.

Solution : The fin efficiency is given by

tah mi

Me at
| FF. feed. [th os
where A EN E
| ape PAIDOS 96689 o 657%

(5220.05)

(0) The temperature distribution for a fin with insulated tip (no heat transfer a
end) is given By the el

the exposed

9-50 ERBEN
ST

70 eee
O 2 30 + 2579 = 65.79 °C
ren

Q RA, m (ly = 1) tanh mi

|
e

al
| (9 Te et oe cm af with inst pi
305 À (D à 2882 » (10-30) «anh (562 + 005)

3658 W

i 5.6. THERMOMETRIC WELL
Fig. 59 shows an arrangement which is used to measure the temperature of gas flowing
through a pipeline, À small tube called thermometrie well is welded radially into the pipeline
‘The wel is partially filled with some liquid and the thermometer i immersed into this liquid
‘When the temperature of the gas flowing through the pipeline is higher than the ambient

im aid surfaces

Fig. 59, Thermometric wet

rhe rte ube el canbe cased a low tn el ar ns
and ni and the emp on ca

À in with tip insulated

where fg is the temperature of the pipe wal, , is the temperature of hot gas or ar flowing

through the pipeline, and 1, is the temperatura any dance mesures Wom pipe wall
along the thermometric well. dl pe

Mx 1 then,

bate cosh
lt chen em
where fs the temperature recorded by the thermometer atthe bottom ofthe well

‘The perimeter of the protective well P= R(t +28) = xd, and its cros-sectonal area A,
Therefore,

u
mas 78
WE

then, Ge.
+ TO

Apparently, diameter of the well does not have any effect on temperature measurement
by the thermometer,
The error can be minimized by

BR 7 Gan of Heat ond Mass Transfer.

A aci vs sng
net
EEE nme eg di
En lan ae
SE Een

EXAMPLE 520
ueanurements of temperature of gas flowing through a pipe has been made by mercury-in
NLT eermometer dipped into an oililed steel tube (protective well) welded radially to
Re he thermometer indicates a temperature atthe end ofthe steel tube Which
ae ee Han the gas temperature due to transfer of heat by conduction along the protective
Sa How large is the measurement error if the thermometer reads E, = 85°C and the
tarea be base of the protective well (pipe wall) is fy 40°C. The protective tube
[STW tat tong and has 1.5 mm thick wall. It may be presumed thatthe thermal conductivity
ste tabe material is 56 W/mK and the local coefficient of heat transfer from gas to the
protective tube is 235 Wk.
Zotation : The temperature distribution along the length of the pocket is given by
SER

Per
here 1 i the temperature at the bottom of the pocket, is the temperature of gas flowing
And o à the temperature of the pipe wall,

th is arrested

For a rectangular fin m = Y

Lames
PRET DE
wm. . 0
then, nn = Pr

mi =1673 x 0125 = 20912
cosh mi = cosh (20912) = 4.109

4.109 (85 = 14) 00-4,
3109 1, = 85 x 4109 - 40 = 309.265

Therefore, the temperature of gas,
ME
ET]
Error in measurement of temperature is equal to

WAT - 85 = 447°C

14.47
Erz

= arc

Percentage of error

2100 = 14.55%

Host Teanater
EXAMPLE 521 from Extended Surteces 7153

Tae tenets A 0 Gn ps e
Filed steel well. The well is of 1.28 cm diamerer nea a mometer imaersed in an
ressure gauge fitted to the pipe line reads 105 bar ana a rm all thickens. A reliable
een ea ar
ee
Sense

i

M treating id as fin wit osetia
Dt), comm
lo)" comal

The perimeter of protective well P = x (4 +28), and the c
ame (4 +28), and the cross-sectional ares A, = #48

P
=
Then,
fm imc;
1, = 0978 x 1BLIC = 17657°C at x =
17657-1811 com)
120-181" cost Tn

cosh mil = 13488; mi = 320

length (depth) of the well = 2283 = 01113 m= 1133 em
wth (depth) of the wel, 1 = FE = 01013 em + 1133

a y te length of protective well equine is mare than the diameter ofthe sel pipe, it
cessary 10 locate the well obliquely (nclined) tothe pipe as shown in Fig, 510.

184 4 Basics of Hoot and Mass Transfer
REVIEW QUESTIONS

SE. Wei the uty of extended surfaces ?
52 Give a fem practical and speii examples of use of fn in het transfer,

53 How does afin enhunce heat transfer at a surface?

154 Under what conditions does the fn efficiency become nesiy 109% 7

SS. What type of boundary condition ix used a the fin edge 7

5. Discuss some of the important applications of Fins.

57. What ae the various types of fins?

SA Mention the most common types of fins and sketch them

59. List the assumptions made wale analysing the heat flow from a finned surface

50. Enomerate the various assumptions made in the formation of energy equation or one dimensional
eut cisipaion rom an extended surface

511. Derive the goverining differential equation for temperature distribution of constant area extended
surface inthe folowing form

ae
awe

where is the temperature excess above ambient air ofthe fin tempera
{oot Ps the Perimetr; A, the coveastional area ofthe
A the thermal conductivity of the material
Proceed to develop expressions for temperan
state conditions for an infinitely long fi
512. Fora constant cross-sectional are fin, obtain the temperature distribution and total heat flow rate
under steady state conditions when one end ofthe fin is atached to à Body at hightemperature
and other end of the fin is insulated,
A fin 30 em long and 10 mm diameter throughout is made of steel alloy of thermal
conductivity 43 W/m-deg. The fin attached to a plane heated wall at 200°C temperature,
extends into surroundings at 25°C and unit surface conductance of 120 W/mi-deg, Work
out the heat low rate from the fin to the surroundings. Presume that the ip ofthe fin
is insulated and thermal radiation effects are negligible. (Ans. 151.9),

513. Two long pieces of copper (k = 400 W/medeg) wire 15 mm diameter are to be soldered together
end to end. The surrounding ir temperature 30°C and the melting point ofthe solder is 230°C
If the convective het transfer coefficient between the copper wir and air 1 1S/r deg, ná
(he minimum energy input in waits to keep the soldered surface at 230°C, (Ans, 2096 watt)
514. À rectangular fin measuring 4 cm length, DOS em thickness and 30 cm width is made of cast ren
having thermal conductivity k= 180 Kj/m-he-deg. The base temperature of fin I 130°C, the fn
is exposed to surrounding air at 30°C and unit surface conductance 5 120 ki/m-he-deg. Make
«alulatons forthe rate of heat flow through the crosesectional area at Ihe root of the fi
Assume uniform temperature distribution at any cross-section perpendicular tothe length of in
and neglect heat low in the direction perpendicular to the fin profile aces. (Ans. 392 KM)

5.5. Show that the heat flow rate per unit width from a straight fin of rectangular cross section I
soverned by the relation

re a distance x from the
1 his he eat transfer colin and

distribution and total het flow rate under steady

Q =km80 tant
where ls thermal conductivity of fin material, m = JER7EB, his the convective coefficient &

is he fin thickness, Oy i the temperature excess at the base and denots the length ofthe fi.
Neglect th heat low through the tip of the fn

su

sm.

su.

Hoot Transfer
EEE rue from Extended Surtaces 11155
382 m ong. Assume that the sa ont 02 6m ter

Fd the rato of eat ot Kom he wat or hth

epee oe fod memo tre pe

tnd eal onc u 9 M

ass alsa’ BC whit the sunounding ar E a ees m ne der, hi
20 equally spaced longitudinal fins of rectangular sections ec an Pub 1 provided with
thermal conductivity of fin material k= 357 W/mdeg any go nu 10 the surroundings. Take
en ee cc en
A125 mm diameter rod of iron (k= 45 W/m. (hoa 1356 4)
soe recat lan
equations? What error in heat dissipation ie sage sn
=

nd 2 45 {me 0 05 mn dane wd Sen og mm
Pen le

‘spot a DA LV

nn ne MS

°C er. by the ue nie Fo
dis length by the use ofthe int rd

ono

Transient (Unsteady State)
Heat Conduction

‘The term transient or unsteady state designates a phenomenon which is time dependent,
‘Conduction of heat in unsteady state refers to the transient conditions wherein the heat flow
and the temperature distribution at any point of the system vary continuously with time. The
temperature and rate of heat conduction ae then undoubtedly dependent both onthe time and
space coordinates, Le, = f(y. 2 1, Transient conduction oecurs in
(0 heating oF cooling of metal Bellet,

( cooling of LE. engine cylinder;

(i) cooling and freezing of food

(2) brick burning and vulcanisaion of rubber

(0) starting and stopping of various heat exchange units in power installation.

Change in temperature during unsteady state may follow a periodic or a non-petiodic

(0 Periodic Variation ; The temperature changes in repeated cycles and the conditions get
repeated after some fixed time interval. Temperature variations in the cylinder of an LC
‘engine are considered periodic. During each eye, a definite variation af temperature occurs
‘with respect to the crank angle and this change continues as long as the engine continues io
“perte. The profil of temperature variation with crank angle for one cycle is called temperature

‘wave and the duration of each temperature wave is called period. Other notable examples of
periodic variation are

(6) variation of temperature of a building during a full day period of 24chours
(0) temperature variation in surface of earth during a period of 24 hours
(6) heat processing of regenerators whose packings are alternately heated by fuel gases
and cooled by air
(49 Non-periodic Variation : The temperature changes as some non-linear function of time
‘his variation is nether according to any definite pattern nor is in repeated cycles. Such a

variation includes the processes where the medium is heated or cooled by exposing it t0
another medium of given thermal state. Examples are

(4) heating of an ingot in a furnace

(©) cooling of bars, blanks and metal billets in steel works

Undoubtedly the time-dependent effects occur in many industrial heating, cooling and
deying processes. An increase or decrease in temperature at any instant continues until steady
temperature distribution is attained. For example during quenching of steel, there occurs 2

ral decrease inthe temperature of hot steel rod until the rod and the quenching medium
ain the same temperature

TRANSIENT CONDUCTIONWNSotios wen
CONDUCTA nage MT
ions San
à which presumes at the sold possesses ini anes DY Oe lumped parameter
resins nso at ath eee enc ne
ease othe Pen en seed ima
athe solid is SP: ay thc cl (v2) with, temperature. Consequently
0 Temperature though changing with time i ner nin oly oth te me
any time: Typical examples of this type of heat flow ares fo throughout the slid
21 cooling of a small metal casting ora Bilt non
ogee AS an at oran to

sn

sal

heating or cooling of à fine thermocouple wire
(ay heating , ple wire due to change in ambient temperature

= À — a meus gg

o né

ts
Lotto

Fig, 6:1. General system fr unsteady st conduction: Lopes parameter ana

Fig. 61. shows a general lump of material comprising the system of interet. A body of
sure area A, volume V, density p thermal conductivity Spec hate and inital tempora
(¿has been exposed to the surroundings maintained a temperature 1, The wransent opens

af he solid can be determined by relating is rate of change of internal energy with can ve
feat exchange at the surface, That is

a
= pve MA Ge) 463)
‘Ths expression canbe rearranged and integrated temperature an theme ae the two
EN)
et
ma
or tog, (e) sec
The integration constant C, is evaluated from the initial conditions: + = £, at € = 0; £,
‘gmbolizes the body temperature at the commencement of the cooling or hesting process
Therefore C= log, (t,t) and hence

aa, »
tog, (= 4) rm e og =

Y 158 1 Basics of Hoar and Mass Transfer

ba (ua
av alone‘)

ae)

Gh?)
a = (K/ pe] isthe thermal diffusivity of the solid, and Iisa characteristic length equal
to the ratio of the volume of the solid to its surface ares.

4, For simple geometrical shapes, the values of characteristic length 1 are

(63)

Au
MEA
Sphere 5
LR
Cyinder + 1e BE - À
et Jt
Cube ci

Fora at plate (thickness , breadth b, height A) the heat exchange occurs from both the

sides; area exposed for heat transfer is 2h, The characteristics length then equals

(6b /2bh) = 8/2, ie, half the plate thickness.

+ The non-dimensional factor (1/2) is called the Fourier number, Fy. It signifies the
degree of penetration of heating or cooling effect through a solid. For instance, a large
time + would be required to obtain a significant temperature change for smal vales
ot(a/1)

+ The non-dimensional factor (M/A) is called the Biot number, B, 1 gives an indication
Of the ratio of internal (conduction) resistance tothe surface (convection) resistance. À
small value of & implies that the system has a small conduction (internal) resistant
‘icy relatively small temperature gradient or the existence of a practically uniform
temperature within the system. The convective resistance then predominates and th
‘convective heat exchange controls the transient phenomenon, Essentially this has ben

> the basic assumption in the lumped parameter analysis made above.

Troe
"nt Cuneteady Stata) Hest Conduction 11159

e aplication ofthe lumped parameter aps

RE spares os indi Dl dono dvi shape sar o pate.

Ay time for a value of B, <01. Small Bot numbers can py coo) der by les than 5% at

any mal conductivity k and smal hat transfer cociente. in plates, and with large
me fomped parameter solution or tan oe

> ton can be conveniently sad 38
FEE TOPO B Ry a
rstantancous aná total het fw ate: The instant
Ma artaneous heat fow rte Q, may be computed
det tl ae
> pve B= pve E (ire HA |
ra za
= pve | -ta( 24 4 0
(e Tove)" ove
HA (6-1) exp(-#A.) rey

are‘)
is obtained by integrating equation 65 over the time

a+ fas

aná th ttl het flow Qu or ga)
= [ona -tp0p/- 22 ele
a re}

-[ aaa

“A Jove

466)
In terms of non-dimensionl Bio and Fourier numbers, we may we
Gn WAG eee ar
and Qe (pts) en

EXAMPLE 6.4
‘An iron (k = 65 W/mK) billet measuring 20 x 15 x 80 cm is exposed to a convective flow
resulting in convection coefficient h = 115 W/mK. Determine the Biot number and the
tality of a humped analysis to represent the coating rat the Mlle tally hoter

than the environment.

Solution: The characteristic linear dimension defined asthe ratio of the volume ofthe billet to
it surface area works out to be

15x80

rem
O]

200%:

160 7 Basics of Heat and Mass Transfer

m
Bot number 8, = At

„USB y,

© ee,

Since the Biot number is less than 0, the internal temperature gradients are smal
Consideration of the billet asa lumped system would be quite accurate, it will introduce an
error of no more than 5%,

EXAMPLE 62

AZ cm thick steel slab heated to 525°C is held in air stream having a mean temperatore of
25°C. Estimate the time interval when the slab temperature would not depart from the mesa

value of 25°C by more than 03°C at any point in the slab. The sec plate haste following
thermo-physical properties:

D = 7950 kg/m? : 6, = 455 Vkp-deg : k = 46 Wmdeg
À (heat transfer cotfficient on plate surface) = 36 WY/mf-deg

Solution : For a flat plate (thickness 8, breadth 5, height), the heat exchange occurs from both
the sides the area exposed for heat transfer is 2h The characteristic length then equate

volume ofplate | 58h 8 002

surface area" 2a“ 2 72 “0m
ET
Bit number 8, = AE O 60078 < 01

Since the Biot number is less than 0.1, the internal temperature gradients are small
Consideration of the steel slab as a lumped system would be quite accurate; it will Incoduce
an error of no more than 5 percent. The lumped. parameter solution for transient conduction

Now

AfA), 36
CURE TS
os
5-5

= 100 = 9.952 x 104

= exp[-9.952.¢10"*¢]

or plan“

or 9952 x 104 € = Jog, (1000) = 6.9077

EXAMPLE 6.3

An average comective heat transfer coefficient for flow of ar over a sphere has been measured by
observing the temperature-time history of a 12 mm diameter copper sphere (p = 9000 kg/m’

tof
@ week

= 2513 W/a-deg
EXAMPLE 6.4

A cylindrical states sec (= 25 Wak) ingot, 10 in ameter nd 25 am long pus
though heat treatment furnace which 3 wee Ile Te il ne Le Pas
REG the farma as at 127€ und ie combine adn nd cae re
cocción is 100 Wntk. Determine the maxima sped sobs ahh eee
Beni mue 60°C tempera 7 A io more reg
Take thermal difesivity & = 048" 0 wi

Solution: The characteristic nea imenaln defined asthe at fhe vole ong os
Siria le worked out to be

A Sx
Berle” Fev * 26425)

M 10030.0208
Biot number 8, „At MODA

Since the Biot number is less than 0.1, the internal temperature gradients are small
Consideration of the ignot as a humped system would be quite accurate, it will introduce an
tor of no more than 5%,

208 em

oa < 02

‘The lumped-parameter solution for transient conduction is stated as

la exo #4
Rave‘)

as Nn a en
à angaben Farin
wu: deu
ee ee (2) «ous #309 au 18
sois

ER ex 864 10d

90-1260
or expls6t x 104 9 = ES
for 864 » 104 + » log, (27209) = 100097

1.00097
Rs = 1158.53 s

The regired ingot velocity then becomes

famacelength 025 gig x 10%
Ara + 0216 107 mys

“an

EXANPLE 65
{Glas spheres of 2 mm dias and a 500°C ae fo De cooled by exposing them 10 an i
Seam at 25°C. Make calelatons forthe maximum value of convection coefiien that
fetmitetble, and the minimum time required for cooing toa temperature of 60°C. Assume
Ee tollowing propery values:
density 250 kg/m sp. heat 850 J/kgK and conductivity 15 Warde.
Solution + Surface cracks are caused by temperature difference within a sli No crocs wi
paar during cooling ifthe temperature gradient are small For that, Bit number shold be
AE than 01. Im the limit 8, = AI» 0
he character Tinea dimension defined as the ratio of volume of sphere 10 is surface
area's worked ot o be
LS
_volume_ _ 3"
surface area — 4nr*
PCL oy
Maximum permisible value of convective cocfcient
PROTA
USE
(0) The temperature variation with time is given by

25 W/nédeg

25 (3
7250850 01002,

Transient Una
a "any State) Heat Conduction) 163
See Pl 017647 q

je 0-5
En
or 017617 + = log, 1357 = 2008

a tego PA 5
Er 7647” 78 sec
faced in a furnace at 750°C with convection coefficient 80 WI an end at 30°C are
fe ie an u Ct eds
traen out from the fucnace after 280 seconds? pies Y ihe pieces:

rege tet pea en pay

solution : For a cylindrical piece, the characteristic linear dimension we

volume ak oh
surface area " Zarek)” Arch)
0.0125%0.03,

= aor rm

804.4110
ELIO an < 02

Since B, < 0.1, the lumped parameter model can be adopted. Therefore

m
Biot number 8, = À

so

Ted aaa * 0.008814

50-750

expl- 0.008818 Y Fee

0-750
0-76

150
» oossis
à be the temperature attained when the peces ar take out fom the funace ater
280 seconds. Then E

Time required for heating, + + 326 sec

1
= oxpl- 0.004814 x 280] = expl= 1348] = 5355

E
a ram ER we
Storia tempera = 00-50 = 37

164 7 Basics of Heat ond Mass Transfer

eur
Re wit ean dm of fc and 22°C place In a Bali water pan
ASSET TATUNG ah mere ae For bow long hos ae
Ser Tame Nesom be ales wien ken fm à setigeato a CT Use lumped
RS es and presume tne following proper fr de

Th Weg MIS mes sen File and p= 1250 km
Solution: Carter length

‘
„volume 3
Grace area" x
„OR 000666

ML 125%0,00666
Biot number B, = MZ 15x00066e

Since Biot number is less than 0.1, the solution can be worked out by applying lumped
parameter theory which states that

= 00094

HE = 00075 x (4 60) = 18

pve

temperature of boiling water in the pan = 100°C. Then
¿100 me Ea
O © 80

00
6.05

(0) Now. we have to find time + for the temperature valves:
1,= SC t, = 100°C and t = 87.60

' 100 = 876°C

or expl0.0075 1) =

00075 + » log, 7.65 = 2.036

2.036
anny = 27147 5 = 4524 minutes

EXAMPLE 6.8

Daring a eat treatment process alloy see spherical balls of 12 mm diameter are iil)
heated to 900°C in a furnace Subsequently these are cooled to 100°C by keeping the
immersed in an oil bath a 35°C with convection coefficient 20 W/ma deg, Determine tht
time required fr the cooling proces. Proceed to calculate the value of convection coefficient

Trono netas
à 1 dence 10 compete the colin pr
AA papel sl el Bl qe Pb à

Y State) Heat Conduction 1 188
Period oF 10 minster. The thermo-
& K : come

mean 59 ede

lume ger
ET y

ES

caros a = IE am
M. 20008

ET 7 00008 < 03

Since 8, < 02 the lumped parameter model can be ape

Biot number B, =

ed. Therefore

TE = 00008 y

800-35,

or expl0.00248 41 = =:
POS x} = RER = 11769
or 000248 + = log, 11.769 = 2465

Time required or cooing = 2465

TRS 99885 see

(0) If cooling isto be achieved in 10 minutes, then

hyd Aa

TEN A
s
un: un

80-35

2465
Convective costficiont h = 245 « 3333 wrmtceg
Duras 7 15 W/m des,

Biot number may be calculated to check the validity of lump parameter model
MM _ 31x00
30
= 0.00132 <01

Biot number 8, =

EXAMPLE 6.9

Acylindrical 5 ly a 800°C is dipped in wat
cal ingot, 25 mum radius and 250 men height, initially PP

25°C with convective heat transfer coefficient of 25 Wm'-deg and dipping continues tll

the temperature drops to 400°C. Subsequently the ingot is kept exposed to air at 25°C with

Convective coefficient of 27.5 W/m'-deg til it attains a temperature of 80°C. If the ingot

168 7 Basics of Most and Mass Transfer

moa Da thera conductivity 65 Wim, specific hat 250 kg XK and density 820 Kg,
ATMA TTT ute regula forthe ingot to eich the temperature from

Solution : For à cylindrical ingot the characteristic length is
volume. rl
" Gurfacearea "Qari ” 2
m msx0.0125
Biot number 8, = ML = 2150:
[As Biot number ie less than 0.1 the internal thermal resistance can be neglected and the
lump theory can be adopted
(0 The temperature variation with respect
to time when cool in water (Fig. 62) given
by

0.005
22 00125 m

= 008135 < 02

tte wale)

Le (ès)

Haha). le

ave * pel) "od
as
BAR os
=00

where

400-25

pl

or expl 0.0839 4]
or 0.0839 +

(6) The temperature variation with respect
to time when cooled in air (Fig, 63) is given

wore

001073

Tran
rt Unétaady Stee) Hest Conduction 187

a
pee à

EXAMPLE 6.10

co of convective heat taser fie upto which tamped parameter aaa

( value of volumeacea ratio upto which the lumped parameter model canbe used.
For the material of the component take,
(ermal conductivity = 100 Wm-deg and specific heat = 00 Jg K.
Sottion ; The characteristic linear dimension defined asthe ao ol he volume of component
lo its surface area is worked out to be Pica Somgee

Surface area”
= EB xk 00mm
7350 0.08 en
og oH GOOD
Biot number 8, = Ll 5 a
er 8, = = RARE» oo1a96 < 01

Since B, < 01, the lumped parameter model can be adopted, Therefore,

a 1
er

Now (BED) = 28735

30-3
Ex]

pa. = Bc

mor en Mens Patt

me and is called time constant

(6 The parameter BYE has units of

ha
pve , pe(V). #6
ne

2 Time constant 1° =

250x300 000216 = 1044 sec
a

(a For e vay of humped parameter analyse Rad good Biot mabe should ee
an 1 in he limit
ork _ 02x10
oe
he humped mode can be adopted for any value of les han 4690 W/m,
V) _ oak _ 02100
(=) = er ms

motes + = 4630 W/m?-deg

@
‘The lumped model can be adopted for any value of voluma/aren ratio lest than 0267 m.

EXAMPLE 611 —
ca bal bearings of 40 mm diameter and initially at uniform temperature ol SOC ae
Toe ano ba maintain a 50° temperature The heat anse coin! Ben
o and oll 325 W/m’K and th thermodynamic properties of bearings can be
taken as
cal conductivity k= 45 Wank and thermal diffusivity & = 125 10% m/s
Determine :
“te view duration for which bearings must remain in llo attain 225°C temperate,
(9 the amount of eat removed from bearings during this time, and
(O e instantaneous heat transfer fate from the Beasings when they are fist immer.
fll and when they reach 225°C.
Solution: The characteristic linear dimension defined as the ratio of volume of bearing to
surface area is worked out to be

4

volume 3 0.04 /2 >
some. M 5. MR „go x 109 m
Surfacearea an 3 3 i
Bot number 3, = Hf EGO „oo

pas
Sine 8 € 0, inma ete of the beings i negligible andthe lumped parame
‘model can be adopted. Therefore =

mah
Now, ME 24 (2
pe "pe 0)
120° aps à _ wos
5 “am 0

ent (Unsteedy State) Hast Conduction / 189
expt 001354 yy

POS De

a ES

hat gives

(m The total energy transfered is

1
Q = pve ut, ~t) lexpl =
P| pve

36 10 and

ha
pl he] = expt 01354 « 8458

03182

| * (20-50) (03182 - 1,
= 45281 J/s or 45.28 KW
flow rate is computes from the expression

CERTA ENS

(e) Instantaneous hes

A any
= 225 « ax [22 oo so (03182)
= 28576 J/s = 285.76 W

6.2. TIME CONSTANT AND RESPONSE OF A THERMOCOUPLE

‘tn we dei mesa as foo ta to pi faded bep ed»
ture difference exists between the punctins, an electrical potential is set up

the junctions. Such an arrangement is known as then
suman Sach a rang owen as thermocouple and is frequently used for the

ment of temperature by à thermocouple is an important
application of the lumped parameter anaisis ” ™

the

he time required for the thermocouple 10

Referring to the fumyped-parameter solution for transient heat conduction:

(68)

Wis evident that larger the parameter 11 / Ve. the faster the exponential erm will reach

zero or more rapid wi e the response ofthe thermocouple, larger valio of NA pe cen
be obtained either by in

be obtained asin the value of convective ofen, or by deeteasing the wire
ster, density and specific Neat

170 / Basics of Hoot and Mass Transfer

The sosie ofthe thermocouple e defined 3 he time required by te thermocoupy
tech 052% oft seedy sate value When such a condone lame equation 62 ane

ove
m

‘The parameter pVe/hA has units of time and is called time constant of the system ang
is denoted by +". Thus &

“ME bY
MA” ana
Using time constant, he temperature distribution in the solids can be expressed as

tends o reach the neady tie value À large me Corso compo te see Da
response, and a small constant represents a fast response. A low value of time constant can be
achieved for a thermocouple by
D Ame ta oe dia
ed Hate an si sol wlio pu
(ih roving as masher cin
Pepin cp te ec ald cad ma for dt
sf hates ies ese le basen GON GER sa
EXAMPLE 812
A ms coco Wid bald e nahen of m mdi bed o
Pepe rar tet ams oa et fis
O = DONS más and. he 10 Wark
wis to temps ange of he a a, ar vt
a tapes Casey tu no el walt ve in ooo
se O cl the pe
ere
Stein à Tie chen on tnd n'ai lt
na

nt sizes and materials

ar
‘The time constant fr the thermometer is
k 10__},, 0001
SE an '* (a) a ns
Since the time constant (6.67 s) is more than the time forthe temperature change ofthe fluid
(8) the thermometer will not give a faithful record ofthe time varying temperature ofthe fut
(6) The diameter of the thermocouple with the given properties can be worked out from the
correlation for time constant, Tha is, ilies

Trensien (Unsteady 9,
ee

pe kyr

añ” ail}

— "8646104 m à 0868 ma

ow
E (thermocouple junction) = 20 Win eg à

€ = 04 KIKE K and p = 8000 Ryan 5777 50 Meg

seltion: The time constant for a thermocouple

given by
ea OME MR) pre
cm)

ie, 3x350x1
‘0% a9) © 2000328 m = 0328 m

Junction diameter of the thermocouple, D = 2 x 0328 « 0456 mm
(@ The temperature variacion with respect o time is

8000 «0.000328 400
cc tl

xp Let

20-20 180
200-197 3

+ = log, 60 = 4094 ec

EXAMPLE 6.14
The following data pertains to the junction of a thermocouple wire used fo measure the
temperature of a gas stream

Density p = 8500 kg/m } specific heat € = 325 /AgK; thermal conductivity k= 40 Wak
athe heat transfer coeiien between the junction and ase 25 Wak.

1 thermocouple Junction can be appronimated as 1 mm diameter sphere, determine how
Jong it will ake for the thermocouple to ead 99 percent of the nal temperature difference?
Solution : The characteristic linear dimension defined as the rao of volume of sphere 1 is
sora area is worked out to be

volume

78 s Bancs of Heat ond Mass Transfer

MA HY AY, Wid), 28 (3
ve pl) ir)” Bo

001 exp ( 0046 2)

1
or exp (0867 1) = GE = 100

or 0467 € = log, 100 = 4605
$005,
To
‘The solution indicates that the operator must wait at least 10 for the temperature uf
thermocouple function to read within 1 percent ofthe inital temperature difference bein te
Junction and gas temperature

EXAMPLE 6.15,
A thermocouple junction in the form of 4 mm radius sphere is to be used lo measure the
temperature alu as stream. The Junction is initally at 35°C and in placed ina as stream
‘whichis at 300°C: The thernocouple is removed from he hot as srcam after 10 seconds
And Kept in ail air at 25°C with convective coefficient 10 We deg, Make calculations for
de) time constant of the thermocouple and (0) temperature attained by the thermocoupie
fonction 20 second after removal Irom the hot gar stream. Assume Ihe thermo-physed
properties as given below
r= 375 Wint-deg ; p = 7500 kg; ¢ = 400 Yg-deg

Solution: The time constant fora thermocouple is given by
ove _ Ark or
MA” haar)" 3h
700x000;

HS

(©) The temperature variation with respect to time during heating (schen placed in gs
stream) is

That gives x = 9865

o
2 10667 see

tet (ba
it 7)

Given it," 30°C ; = 35°C and t = 10 seconds

106.75)

‘Transion (Unsteady Stove Heat Conduccon / 173

= expl- 0.0887]

Tim
+00. 320
E ON
EN ssc
The Le th respe time during cooling (when exposed to ait) is
Lt ep LA |
et]

Given: = 8870 51,» 250
and + = 20 see

MA Ore | 7500%x0.00 400
ove 0

Ze

reat

h + 10 /mdeg

vos ©

63. TRANSIENT HEAT CONDUCTION IN SOLIDS WITH FINITE

CONDUCTION AND CONVECTIVE RESISTANCE (0 < B, < 100)
Consider the heating or cooling of a plane wall of thickness 2
ine y and = directions. Initially the wall is at uniform temperature 1 and ceddently both
surfaces (1 = + 6) are exposed to and maintained at the ambien! temperature (Fig 64). The
‘controlling differential equation for the transient heat conulecton 1

Y La
at "wae 169)

and the appropriate boundary conditions are
() t= 4, at = 0 initially the wall is at uniform temperature ,

(stas = 0 at x= 0; symmerial nature ofthe temperature profil within the plane wall;
symmetry in conduction occues at the mid plane (e = 0) o the wall

(i) KA (fds) = HA (t= 1, at x = à 8. Tis condition stems from the fact that conduction
heat transfer equals the convective heat transfer at the wall surface

25 and extending to infinity in

sore es anon
soni pls
ward À Les
—|
EEE

174 1 Basics of Heat and Mass Transfer

an 00 m wo

Cr

atthe conve of lane

Fig. 65. Hester nat ore

A RE
nn

176 1 Basics of Heat and Mass Transfer

al

a torcer! temperatura history na sphere

me

function of Blot number M/k, Fourier number aren, un lemperature her bec
i Si the baa aan
ste md bere a Pe

onthe
st

Transen Un

"dy Staves man. Comunion

pe ot ol the cor tern ta NN.
dione would BNE an expression for Vomp

Cons obtained aller rigorous matters wm bah
sol OU mathematal anal il on
El fa Mae

TER jon
opwiously when conduction resistance is
er act and the

Pe ya he menor parar 7

erature sto be abla In

Graphical charts have oe prepared forthe equation 610 à

in Fig 65 167 depict the dimen ne |

The values of Bot number and Fourier number, as used inthe Heiler chart, ae evaluated
asis of a characteristic parameters which i the semi-thicknes n ace of pote and hs
face radius in case of eylinders and Spheres in

extensively used to determine the temperature distribution and heat
th conduction and convection resistances ave almost of equal impor

,
CE LH

i ii
‘soi ome on 01 02 05 1) 235 0 HT

it
9.68 Cone nm mas

478 7 Ensies of Hoot end Mass Transfer

“7

temperature history in a cyinder

in IT ]
DO o tea

Bora 005 OF 02 05 19 2 9 5 m m 100

os

Fig. 610. Conecton factor chan Lor temperature history in a sphere

EXAMPLE 6.16

Alca en both sides to an environment nun lios 283

vrs 65°C, Determin the cen line tempera, ¿sico 245 We and
100125 min from the mid plane after 3 minanc, "and the Iemperatre

pute

for Steel
Thermal conductivity k = 425 Wink
‘Thermal diffusivity a = 0.083 mr

solution : The characteristic linear dimension for at late equals hal the pate thickness,
1

2 B
À 5 mm
ue Fy = SE = BREI)
Fourier number Fy = SE 202 «5g
LH Sgen)
Bot number mA AGIT) ny
Since Bot number greater han 01, the internal emperatre gradients ar ot small and
so he internal resistance Cannot be nego, Core of pl so and
seu then be appropriate Furr, the Bot umber its han i and ens te
Tian valasen ova Ve obtained by employing Haie one
{The following parametre values apply

Be à Less à eo qu pu

Eh
5

1

Eto 00
Pot, + 096 (fy = 1) = 65 + 096 (281 65) = 27236%C.
EXAMPLE 6.17
The nose section of a missle is formed of a 6 mm thick stainless steel plate and is held
info a à uniform temperature of 88°C. The mine enters the denser layers ofthe
atmosphere ata very high velocity. The effective temperature of air surrounding the nose
ery high velocity.
Sign stains the value 2200°C and the surface convective coefficient is estimated a
305 Wmèk. Make calculations for the maximum permissile time in these surroundings
the maximum metal temperature is nat to exceed 1095°C Also work out the inside surface
temperature under these conditions.

480 / Bees of Heat and Mass Transfer
me constant values for steel properties are : density p = 7800 kg/m"; specific hey
Le 48 ARR and thermal conductivity k= 54 W/mK.

Solution Th ive: section mar: folie as à Mat plate for which the characteristic ligar
po Re 3 mm

Arco equa hall ano plate thickest
I SAO
ot number 8 À SONY

mo Biot mme gastos han 01, the Rampe analysis would be inappropriate, Fur
#2 100 amd ancora the trarsicnt solution van be était by employing Heisler chante

us

en she plate folhawing parametric values apply
LR 2 829 and 4/1 = | (outside surface of now section)
tung from the chat or am infinite pan bag 08)

e

Now. from Fig. 65, for the above dimensionles
value of Fourier number Fy * 30.

ge Pee ee
ro E

e (amas loner

1 = 1813 seconds
The temperature fy a the inside surface (x = 0) is given by

=0508

1, + 0508 (,~ 1)
2200 + 0.508 (38 - 2200) = 1101.7°C

EXAMPLE 6:18
During the manufacture of plastic sheets 10 cm thick, the sheets are brought to a uniform
temperature of 175°C and then allowed to cool to a surface temperature of 52°C in ai
38°C before Farther processing, How long a cooling period will be required if natural conveot
cooling is employed with average surface coefficient of 39.2 K]/mv-hr-1eg? Also detraig
ths temperature a the centre of pantie sheet when the surface temperature has ech

"rani neta Sa
properties of plastic material ate Steel os Canguctión 1 181

density 9 1280 kyo, speci nag
thermal conductivity À = 098 kymneg. KNBR. and

gion +The characters Hncar dungen €
wi for «tat pane

sol

RES

Biot number 8,24). 92205.

Since Biot number is greater han 01, the lumped anaes wen

2100 and accordingly’ the tzansent latin can be anc be ee rept Fuster
[257 be obama by using Herr cha

+ or the Mat plate, (he following pare À

YA» In apy

À outside cure ofthe sheet),
Reading from the chart for an ine plane fag. a

2-38 fut)
ARM un
non, 7 asl izs-a8 }

Now from Fig, AS, ov thy above dimensionless
value of Fourier number F, = 18

mA wove th.

leer

or 2155 127870

-0204

1, + 0208 (1)
= 38 + 0204 (195 ~ 39) = 6595°C

EXAMPLE 6.19
Along cylindrical shaft of radius 7.5 cm comes out of an oven a 815°C throughout and is
‘soled by quenching it in à large bath of 38°C coolant, Ifthe surface costicient of Beat
transfer between the bar surface andthe coolant i173 Wang, calculate the Have takes
forthe shaft centre to reach 116°C.

|
1

482 7 Basics of Heat and Mass Transfer

“Assume that K = 17.5 W/m-deg and «= 0.0185 mr

{an What would be the surface temperature ofthe shaft when is centre temperature ja

16°C. Alo calculate the temperature gradient at Ihe outside surface at the same instant of

‘Solution: The characteristic linear dimension defined asthe ratio of the volume of
Shaft 10 le surface area works out as

ecylindsica,

m
Biot number 8, = À =
Since the Biot number is greater than O1, the lumped parameter solution s invalid Further,
Biot number s les than 100 and accordingly the transient Solution can be obtained by using
Heiler cho

ws.
Bar " 176005810

13

116-38
a O

=0 (centre ofthe bar)
From the chart for an infinite cylinder (Fig. 66), we rend the Fourier number Fy = 182

ar 018s
Cy

-192

1920.50
pu
(D At the cylinder surface : r/R = 1 and 1/8, = 133
an infinite cylinder, we read

= 0584 hu = 2102 see

‘The temperature {atthe surface then works out to be
121,407 (ct
38 +07 (116 38) = 926
“The temperature gradient at the outside surface is determined by the boundary condition
at r= R which equates the rate at which energy is conducted to the fluidsolid interface from
‘within the solid fo the rate at which i is convected away into Une fluid. That is

a
Kart) À eh GARE) (1)

7 175 .
yey = BE 026 - 28) « 546*C/m

“Transient
* Unsceady State) Heat Conduction y 183

PLE 620
DO diameter india bar, nal a nom Ieocrgar a <a
Az EM M 6SOC with a convective coefficient ot 5a Tape atte OF 40°C, is plac
cn 122 VI Determine the ine

um de to reach 255°C. Also workout Ihe temper
fo th RE of de bar erature of the surface a this instant.
or the ms

Thermal conductivity X = 20 Wimk; density p =
specific heat € = 1050 J/kgk D + 580 kg/m

‘The characteristic linear dimension defined a the ratio oft
Inc area works out to be the ratio of the volume of the cylinder

solution +

RL LR
zu 202 dm
Haro?
pot number 8, = E 20 „
Biot number 8, = P= PGA = 53

Since Biot number is greater than 01, a lumped-parametersoltion i invalid, Further Biot
number i ess than 100 and accordingly the Lransient solution canbe obtained by using Hessler
For a cylindrical soli, the following parameter values apply
Lok, om
CETTE]

= oasis

% 0 (contre ofthe br)

From the chart for an infinite cylinder (Fig. 66) we read Fourier number Fy = 0.18

Sn so

on

or (mass oan
di) At he olinder surface #/R = Land 1/8, = 03518
From the chart (Fig, 69) lor an into in, we rend

-018 += 1973.16 seconds

Lt o
The temperature a he src en wor at be
Ons toi) m ans en me

EXAMPLE 6.21

A10 cm diameter apple, approximately spheric
And placed in a refrigerator where temperatur
Coefficient over the surface of apple is 6 Wa
the apple after a period of 1 hour
Th >
'ermo-physical properties of apple are
1D = 998 Km; c= 4160 ASK; À = 06 Wk

al in shape is taken from 20°C environment
Wa seC and average convective het transfer
Se calculate the temperature atthe centre of

_

+ 194 Basics of Hest ond Mass Transter

sation: The charente Hear dimension defined nc a he volum of the app
Bee tone aros works aut to Le
Serien
Biot number 8, = ML = BROT) png

G 06
Since the Biot mumber is greater than 01, à Iumpat apseity approach is not suitable,
the 8.108 and according the transient solution can be obtained by using Heisler charts,
Or à spiel solid. the dimensionless parameters for Heisler charts are
k 06
SO

rf As. exc)

E ES

-0207

E 0 (mid plane or centre of the apple)
E 0 (aid
Reading rom he chat for sphere (F8. 67)
lot wos

Therefore, the temperature a the mid plane (centre) of the apple is
y=, 085 (= 1) = 5 + 085 (20-5) © 1775°C

EXAMPLE 6.22
A 36 cm diameter egg, approximately spherical in shape, is intially at 25°C temperature
To boll i to the consumers taste, it needs to be placed for 225 seconds in a saucepan of
boiling water at 100°C. For how long should 3 similar eng for the same consumer be boiled
‘then taken from a refrigerator at a temperature of 5°C? Thermo-physical properties of egg

5 Wimk à p= 1250 kg/m"; € = 2200 REX

and the nea: transfer coefficient for the shell and shell-water interface may be taken at

230 Win? K (6) Compare the centre temperature attained with that computed by teating the

‘gg an à lumped-heat-capacity system.

Solution : For a spherical solid, the dimensionless parameters for Heisler charts are
1k 25

BTE "20008

ar (4)e

me © (pc)

= 04%

252). 5
(ri Guise

LE =0 (aid plone or centre of 638)
Reading from charts for sphere (Fig. 67)

‘Transient (Uns
(Unsteacy Stare) Heat Conduction 1 488
erefore the temperature at the mid plane (entre
e A TE ene) el e eg
100 + 022 (35-109) = sc
hen the es taken from th refrigerator, he dimen
othe values

Wionles parameter for Heiser charts

i 1

8

00
Too nis

- 0496

“iR

=
i
8
As before, from the Heisler charts (Fig. 6), we read

ar (2
wen or (Eisen

= Fourier number Fy = 071

sl
m0 Jour = 071
Nev value of ine 25 seconds

(0) The characteristic length defined as the ratio of the volume of egg to is surface area
works out a

Biot number B, =

25 1 ms
ma Gonr
The lumped parameter solution fr transient conduction is stated as

h

Fourier number Fy = EE

la expl-8 Fi]

LE = exp (0672 » 548) = 00219

fo * 100 - 0.0219 (100 - 25) = 983°C

The lumped-heat capacity method is based on the assumption of uniform temperature
“hroughout the egg” The difference between the values of temperature at the centre ofthe es
(635€ evaluated by Heisler charts and 963°C obtained by the application of lumped heat
apacty method) is mainly due to this assumption.

64. TRANSIENT CONDUCTION WITH GIVEN TEMPERATURE DISTRIBUTION

Quite fen th instant of me is known fo the one-dimensional
y th temperature distribution at some instant of tine

steady heat conduction through a sold. The precited temperature iiribation is combined

‘ith the relevant conduction equation fo evaaat eer the rate of eat fw or the ate of

temperature variation at feront point win the i

186 / Basics of Heat and Moss Transfer

EXAMPLE 6.23

“The temperature distribution at à certain time instant Ihrough a 50 cm thick wall pres
ty the relation

= 300 ~ 500% + 10032 + 140
phere temperature is in degree celsius and the distance x in metres has been measungg

fromthe hot sut. I thermal conductivity of the wall materia 120 kU/mehrdeg coca
{he heat energy stored per unit aren of the wall,
Solutia : The temperature distribution is given as
12300 = 5005» 100? 1007?
SE = 500 + 200% + 42032
Het entering the wall from the (ace beng heated (x = 0) :
Qu 2 ta fi) 2 20 « 1 € 500) = 10000 khr
Heat leaving the wall face at = 50 cm = 05 m
a
CS) a,

= 20 x 1[- 500 + 200 x 05 + 420 x (OS)?
= 5900 4)/hr
+: Heat storage rat

= Qu = Quay = 10000 5900 = 4100 Kr

EXAMPLE 624

A large plane wal, 40 em thick and 8 m? area, is heated from one side

distribution ata certain time instant is approximately prescribed by the relation
f= 80 000 + 124? + 254) 20

in degree celsius and the distance is in metres, Make caleulaions

where temperature +
for the:

(O bent energy stored inthe wall in unit ime
(i) ate o temperature change a 20 cm distance com these ein heated and
GG cation where the ae of being roo maximum
For the wall mater’
thera conductivity £ = 6 Wk and thermal das a =
Solution: Te temperate eistbuion sven as
te re Te De at

02 mr.

a ;
Lu co à ax à an 0
drame
LE 2 + 1500 200
a

Le o an

Transient tra
(yest eri he Wl om te ce to
(a)
a Va hs 7 * 866)» so
ies ang Whe face str © Bene bd

4} State) Hest Conduction 11187
ro)

cu

ale)
Va

088.96 W

CREER ST

2x0
Heat storage rate

“O = Quy * 2980 - 208896 = 791.0 w
ia Rate of temperature change is given by

ad

002 [24 + 150 x 02 - 240 x 027) = O888°CHhr
(ús The rate of heating or cooling would be maximum at a location where

or 160 - ABl

EXAMPLE 625
Ata certain time instant, the temperature distribution in a long cylindrical fre tube can be
represented approximately by the relation
12 650 + 500r = 42502
here temperature is in degree celsius and radius ris in metre,
Make calculations for rate of heat flow and rate of change of temperature at the inside
and outside surface of the tube, and the heat energy stored inside. The tube measures

inside radius 25 cm, outside radius 40 cm and length 15 m For the tube material k= 33 Wn,
a = 0008 my

Solution

The temperature distribution i given as
+ SOU - 4230

= 800 - 85007

sw

(6) Heat Now rates at the two surlaces can be obtained by invoking Fourier law of heat

488 7 Besos of Hest and Mass Transfer

At the outside surface (r = 040 m)

un
. = = 55 x (2x x 0.40 x 1.5) (800 - 8500 * 0.4) = 5.388 x 10° W

At the inside surface (= 025 m)

== 55 x (2e x 025 x 15) (800 - 8500 x 025) = L716 105 W
Rate of heat storage = Qu, - Q.
#1716 + 108 5388 x 109 = - 3672 x 10° W
‘The negative sign indicates that here is depletion of heat energy, i, the heat is being ot

by the fire tube
(9) For a cylindrical object, the rate of te

ature change is

At the inside surface, ( = 0.025 m)
a 1 5

= 0004 [- 8500+ {800-a500%0.25)| = - 552° Cue

0004 [-s500+ ¿Ft »] m

AL the outside surface (7 = 0.040 m)

a

: A ,
= 8500 + 1 (600 -8500%0.40)] = - 60°C/he
Fr = 0008 [-ss00+ e )] hu

REVIEW QUESTIONS

62. What is meant by à Jumpod.copacity ? What are the physical dimensions necessary for à lunes
Unsteady ate analyse lo app)? M ii
À 75 cm diameter orange, subjected to a old ar environment, may be idealised at a sphere
Determine he sity of a amped analyte for predicting the temperature ofthe orange damn
colin The orange prépas are + thermal conductivity À 2 0597 W/m-dep and connecter
Stine h = 1135 N/mider
(Ans. 8, = 02376; lumped parametric analysis will not be acuta
62. Prov that the temperature of a body at any timo during newtonian heating or colin gen
by the mation

1m,
open
jeher Band Fae ih Biot and Fourier modulos respectively, he ambient tempera and
[isthe inal temperature of the Day
À oa aluminium (= 210 W/mK) cab at 100 has been exposed to a convective low renal
in convestve coefficient = 25 W/mK, Calclae the mai sige dimension u he cube IE
Jumped analysis to be accurate within 5%, de, ing valve of Mor number 6 01

(ans. 5010)

po

[2

aes ion arpa a 2
DE ne ne Ed
pea ade om the surface 1 ba.
Iv plo oem ene ama i
eh the te pate acquires a temperature lemas on a ma TE, by
= 485 W/mK. p 7 7900 Kg/m, = 046 3J/KgK andthe local cote of hea ander
te St it E se nese es "sae tom
Yet ae Hier cure? How the harried ai pete de
nd cn re a Si enger

Toe aa 1h dan tity a er
Ken ily ta empre 32 Dame ie ame
require (or he trier atthe aa The lt orate BO aha Ge eh me
‘Miter Aare meri Mi ese OC eal er
ene mete ae
"hems one E 29 W/o emo dey a= EE end ent
of heat transfer at Ihe surface of shaft in the furnace ke 198 Dr aut ii

A lange dis ol 15 cm thickness i nly ad at 200 and then suddenly expaed to ambient

rations at AP temperature. What would be the temperate a he conse eee Dank
ster hs change? The following propor ae gen

K= 48,7 W/mK: p = 1600 kg/m €, = 16 J/KgK ; h = 235 W/anik

(Ans. 173°C)

À 10 cm diameter cylindrical br, heated inthe furace to uniform temperature of 200% is

allowed to cool in an environment with convective cote 230 Wa and impartan ¿O

Determine () temperature required to coal the centre of Dr to SC. (1) tempered the rose

this instant For the material of the bar = thermal Conduct £ > 50 W/m K and thr

‘ihusivty a = 20 2 10° m/s (as 1752 minuten, 48300)

5 {8 the ita temperature of the body 1,
es ard Wi the weight othe dy Ng

te oc

ano

ll

Radiation: Processes and
Properties

Thermal radiation isthe transmission of thermal energy without any physical contact between
the bodies involved, Unlike heat transfer hy conduction and convection, transport of thermal
diation does not necessarily affe the material medhum between the heat source and the

receiver. An intervening medium is aot even necessary and the radiation can be affected
rough vacuum or a space devoid of any mater. Radiation exchange, in oct, occurs most
effectively in vacuum. A material present between Ihe heat source and the receiver would
‘ether reduce or eliminate entirely the propagation of radiant energy

Energy released by a radiating surface isnot continuous but is in the form of successive
“and separate (discrete) packets or quanta of energy called photons. The photons are propagated
through space as rays; the movement of swarm of photons is described as the electromagnetic
waves. The photons (as carriers of energy) travel with unchanged frequency in straight paths
and with speed equal to that of ight, For propagation in vacuum € = 32 10% m/s. When the
photons approach the receving surface, there occurs reconversión of wave mation into thermal
energy which is party absorbed, reflected or trarsmited through the receiving Surface The
"magnitude of each fraction depends upon the nature of the surface that receives the thermal
radiation,

‘Attention would be restricted in this chapter to the radiation processes which occur ata
single surface,

7.1. SALIENT FEATURES AND CHARACTERISTICS OF RADIATION
Some saliont features and characteristics of radiation are enumerated below

(0 Te olctromagnetc waves are mitts as a result of vibrational and rotational movements
of the molecular, atomic or sub atomic particles comprising the matter, The emission occur
‘when the body is excited by an oscillating electrical signal, ceetronic or neutronie bombardment,
chemical reactions ete. The emission of thermal radiations is associated with thermally exited
conditions which depend upon the nature of surface and its absolute tem

i) The distinction between one form of rad
wavelength which are related by
«(peed of light) = à (wavelength) x f (frequency)

Consequently longer wavelengths correspond to lower frequencies and a shorter wavelengths
to higher frequencies. Again, a high temperature body will have a high frequency quantum
and so shorter wavelengths,

(i) The general phenomenon of radiation covers the propagation of electromagnetic Wave®
of all she wavelengths from short wavelength gamma rays, X-rays and ultravoilet radiation 0
the long wavelength microwaves and radio waves, Thermal radiation is limited to range &

Uavelength between 0.1 and 100 gamit thus includes the entire visible and infrared, and a at
ofthe ulraviolet spectrum.

other lies ony in its frequency and

on and

190

in dependen is slo upon the thickessofemiting ayer and the eas mess O

cor fr is bsp. An bolted body whch reais co a e e

sends primary on the temperature
eve. With ater factors (hand A)
transfer due to conduction and convection roms fot sour a
AE to the surroundings a 200°C would practeally remain ane fhe pete
sumoundings take up the temperature values as PC and 100C ropa gen ns
the same temperature difference, the brat exchange by rodar pa mn la aan
lemperntre of the source and the surrounding.

The most vivid evidence of radiation transfer is hat represento by the solar ner which
as through interstellar space (conditions lose 10 thal or pera aca où ee ee
{oth surface Slat radiation plays an important part in the eng of heating and venting

systems. Heat transfer by radiation is encountered in boiler furnaces, bile rehesing furnaces,
and other types of heat exchangers. The design end construction of engines, ar turbines,
rer reactors and solar collector

is alo significantly influenced bythe ration hea transfer.

72. ABSORPTIVITY, REFLECTIVITY AND TRANSMISSIVITY
‘The foul radiant energy (Q,) impinging upon a be
would be partially or totally absorbed bs it (0).
‘flected from it surface (Q,) or transmitted through
À (Q) in accordance with the characteristics of Ihe
body (Fig, 7.) By the conservation of energy principle,

{he total sum must be equal to the incident radiation
cs be equal tothe incident rah

Q-0+0=0
Dividing throughout by Qy we have
2,2 o

ua"
A an

9.74. Abren tacon and
a transmission of radiation

192 1 Basics of Heat and Mass Transfer

where
“a = absorptivity oF faction of total energy absorbed by the body
D = reflectivity or fraction of total energy reflected from the body
= transmissivity or fraction of total energy transmitted through the body

The factors a p and + are dimensionless and vary from 0 to 1. The value depends upon
the nature of the surface of the body, is temperature and wavelength of incident rays, The
response ofthe body t incident radiations is, however, completely independent of and unaffete,
by the simulatneous emission from the body.

Black surfaces are effective absorbers of radiation inthe wavelengths that are encountereg
in heat transfer. Accordingly the name black body is assigned toa perfect absorber of radiate,
‘The thermal radiations impinging upon a black body are totally absorbed by it the radiations
are neither reflected from the surface nor tranamitted through it. For a black body a = 1 and
P = 1 = 0. Incidently this implies that a black body is a Perfectly non-reflcting and nos
transmitting surface. Snow, with its absorptvily 0985, is nearly black to thermal radiations
The absorptivity of surfaces can be increased 10 99-95% by coating their surfaces with lam
black or a dark rought paint. In actual practice there does not ext à perfety black had
‘which will absorb all the incident radiations. The absorptivity of a surface depends upon tre
direction of incident radiation, temperature df the surface, composition and structure of the
irradiated surface and the spectral distribution of incident radiation, When a surface aborts
a certain fixed percentage of impinging radiations, the surface i called the gray body. The
absorptivity of a gray body is necessarily below unity, but remains constant over the ene
range of temperature and wavelength of inciden radiation, This condition of constant absorptivity
100 is not satisfied by the real materials and as such even a gray body remains a hypothe
concept like the black body.

‘A body that reflects all the incident thermal radiations is called a specular body (the
flection is regular) or an absolutely white body (the reflection i diffuse). For such bodes
pal. anda t= 0, The specular and diffused type of reflections have been indicated in
Fig. 72.

Regular (specular) reflection implies that angle between the reflected beam and the normal
to the surface equals the angle made by the incident radiation with the same normal, Reflection

rom highly polished and smooth surfaces approaches specular characteristics. In a diffused
reflection, the incident beam is reflected in all directions, £e, there is directional independence
fof the reflected beam. Most of the engincering materials have rough surfaces, and these rough
surfaces give difused reflections Diffused reflection is sometimes likened 40 the situation it
which incident energy is absorbed near the surface and then remitted.

EAS in al

ig Y

‘Spor etc

mr Ar sure)

Fig, 72. Spocular and dilosos rections

ay tha allows ll he inciden adn
codicia For such bocce n= ado
der aon Asal ny bet a
of ne or an À tin as pa mn Tors ern rr aer
ator of a building,
= tion of a radiation is

The abso à à surface pheromone,
ps m Ic) material eo ero DOS cam
rrr ering ae tick enough to cover Re one
‘pal mano ema ud
(e, yack salt and most liquids inthe viste and no ar
sce a ard fae ee nd
20 sever pases have relatively gn Se gd
tice de du
med Mawes ict ESS E

aider age hollow sphere or linde provided with ony oo snl oe
NS Te ea oe el en nd
tap Eck whch sorb stat PS of he id lo ae
entering the hole strikes the inner surface Since the surface has a high share a
froin ofthe radiation is abvorbed and any small canon beeen nee ee
fy dos vot find any way out and gan este ie sate a ete ed
¿irte and party reflected. Likewise the related raten à done a en ed
Bea when if escapes ou it has only a gigi nal coca ed

Las Q represent the radlaı energy tha eter e ol Toe nics a ag
(1-0) = pO afer f internal sion pS ser eo e lacs fo
Shem rection and this approaches rot ter Te eh er
have suffered many rections the emerges ax hee re ae one ble
ve mp a = or the hole. A spall oe O a and al
fey nes sa lack body Because al the radian ney seeing Ree nae eee
Te smaller the opening” beter the approinatcn E Siar tea Cee pe
exrimetaon, a hole of 23 cm Garter nthe ed fs ow ¿indi 2 om Ig Da
‘Sem dameter would slice Iciermal maces nal ante specs Be
Brand ar frequent used 1 cal a aap emotes and er alone

15 in a very thin ayer.
Solis nd liquid encountered

they can be considered por
Peete, <a Be considered non-transparent

“raheem uy at
et

een

Fri

Para ,
‘The values pertaining to black body are commonly designated by the sufi

108 7 Bases of Heat and Mass Transfer
7.3. SPECTRAL AND SPATIAL ENERGY DISTRIBUTION

butin of rahont energy 5 nos uniform with respect L both Wa

5 Speer distribution: The rat eit y a suce coma of comet
a ame wavetength, and the term spectral refer tothe variation in thermal radiations
net. Magnitnde of the radiation at any wavelength (monochromatic) and the
Sal ration ane found o vary with the nature and temperature of the ermting surface

velength and direction

ail 2 u
Ela

ar ®
Fig, TA. Spacral and sail (recon) energy distribution

{ Spatial or directional distribution : A surface element emits radiation in al directions
the intensity of radiation w however different in different directions, The surface may emi
preferentially in certain directions eresting a directional distribution ofthe emitting radiations

EXAMPLE 7.4
What happens to the radiant energy when it is incident upon a surface? Give the generat
equation that relates absorptivity, reflectivity and transmissivity

‘Of the radiant energy 355 W/m incident upon a surface 250 W/m? is absorbed, 60 Wat
i reflected and the remainder is transmitted through the surface. Workout the values for
Sbsorplivity,reflectivity and transmissivity for the surface material,
Solution Absorptivity «t= fractional of total energy absorbed by the surface

250
on

Reflechvity p = fraction of total energy reflcted from the surface

o
ve em

350
Trammissivity € = fraction of total energy transmitted through the surface

350-2507 60) _ 9195

EXAMPLE 7.2

Why does a cavity with a small hole behave as a black body?
Thermal radiation strikes a surface which has a reflectivity of 035 and a transmissivity

of 0032 The sbsorbed flux as measured indirectly by heating effect works out to be 95 W

Determine the rate of incident flux.

Solution : From an energy balance,

Radio: Pro
085 and Properties 1 195

on & +0002 + 085 «1
incident Max Qy = 2,
Toa

Dal 272 W/m

onion : From an energy balance,
Se arpri=1

orar a= 0583s
Absorption. Q, = & Qu = 05155 x 800 = 43640 W/m?
05455
800 = 145.47 W/m

Recon 0, =0 AN

Transmission Q, = + 0,

irom a eat source at he fate of ZW. heat tano coi on bok teen ee
is stated to be 87.5 W/m*-deg, workout the reflectivity of the ne der

Solution : Heat lost by convection from both sides of he plate
“as
The factors 2 accounts for two sides of the

ss (S007) onen

For most of solid, the transmissivity is zero
Energy lost by reflection = 2.0 - 11 » 09 W

09
Reflectivity p = Q a A
Pos
EXAMPLE 7.5
on

Se nhs here is ation from eats surface othe space, On such a ig he
‘ale particles on the plant leaves radiate to he shy whose temperature may be taken as
BK. The water pacs receive het by convection fom the nrrunding ay he convective
al ane cciiiet has val of 90 Noe tbe water shoal not eee ma
¿lclaons forthe sir temperature
Solution : For water just to ra, ts temperature hast be € or 273K

‘Assuming water surface to be Back, the Best Bee gres

eat radiated to sky = hes cuite by conven

5.67 x 10 x À x (2734 ~ 2008) = 30 x À x (T - 273)

486 7 Basics of Host ond Mass Transfer

22422 = 30 (T - 273)
Temperature of air T= 72422 4279 = 280474 K or 747
Amy temperature of sir lower than this value will cause frost or freezing on the leave
surtaces

7.4. WAVELENGTH DISTRIBUTION OF BLACK BODY RADIATION : PLANCK'S LAW
‘The energy emitted by a black surface varies in accordance with wavelength, temperature ang
surface characteristics of the body. For a presribed wavelength the body radiates much mare
energy at elevated temperatures. Likewise the amount of emitted radiation i strongly influence
by the wavelength even if temperature of the body remains at a constant fixed value,

The laws governing the distribution of radiant energy over wavelength for a Black body a
à fixed temperature were formulated by Planck. Based upon extensive experimental evidence,
Planck suggested the following law for the spectral distribution of emissive power
ee

Ed CCS 02
The symbols used have the following meanings
I = Planck constant, 66236 x 10 Js
© = Velocity of light in vacuum 2998 x 10" m/s
À = Boltaman constant, 13802 x 10° J/K
À = Wavelength of radiation waves, m
T = Absolute temperature ofthe black body, K
Quite often the above expresion is written as
à cat
EX RAT oy

ee
aaa
rs «ir
Th ns Leo ner gt wig ym in
Pe as easton ns are ss ee
a dd
a
cet Salles al Es nile min rae wt win
cad the peal energy evan an sar nen oped and
let
Te rm a en mpl
Rennen foe nennen de motas Pital
Pra rn e al tb
errada ne me
nn a Wi hey era cert me po nate où
en)
To dl ny arr guiada De and) rain mn
tose

Pariacion: pr
Processes
ii (0 Properties 1187

ee mec (men tm
Fig. 7.5 Raten of Hack aly Bun enormer

(69 Te wavelength at which the monochromatic emisve powers maximum his in he
discon of shorter wavelengths as he temperature rate Theo spe han eee
tenor, mucho the energy ls emited ame band rangingon baten mo
which he monechromatic emisive powers maximum Fr Sean eso a
temperature ol about SHAT emita 0% ofits ratos between bt sd Se

(tan temperature, here under the moechromat

© rome mie power versus wave

rth rate of raat energy emied wit he waveangih menu Tse (ero
Upon integration over the entire range of wavelengths

Pees es
ie tg mesures the ttl ato unde the menchromae eme power vss
eng curve forthe Mack body, an repens heal emir powe per ont et
fant energy fx density radiate from Del
(8 or sorter wavelengths, the factor CAT become ver rg, at ast
eo [£]
ol
(pri the term (- 1) appearing inthe denominator ete Pacs distribution aw con
et compared to this large vale. The Plis aw then reds to
ox
ED *
Eh exp[Ga/ AT]

es

488 17 Basics of Heat and Mass Transfer

Equation 75 is called Wien's law, and itis accurate within 1 percent for AT Tess thay

3000 uk.
1) For longer wavelengths, the factor
expands in series to give

onda EE.

‘The Planck's distribution law then becomes

The ados

T is small. In that case exp (C/A) can by

ga

identity is known as Rayleigh-Jean’s Lave It i accurate within 1 percent fgg

AT > 8" 10° uk, As black body emits over 99 percent of its energy at AT values below the
limit: the Rayleigh-Jean’s formula is apparently well ouside the range of thermal radiation

‘The relation, however, is quite useful for analysing long wave radiations such as radio

7.5. TOTAL EMISSIVE POWER; STEFAN-BOLTZMAN LAW

The total emissive power E of à surface is defined asthe total radiant energy emitted by the
<urface in all directions over the entire wavelength range per unit surface arca per unit tine

‘The basic rate equation for rad

proportional to the fourth power ofits absolute temperature
En an
where 0, is the radiation coefficient of a black body, This rate equation 6

jm transfer is based on Stefan-Boltzman law which sate
that the amount of radiant energy emitted per unit time from unit area of black sur

be seta by the

integration of monochromatic emissive power over the entire band width of wavelength fr

md»
Pe ama

La nye à

With this substitution, the new integration limits a
AA = 0,1 = and -0

0 fra
sano

E Pte

Expanding [exp (9) = 1}? by a serie we obtain

srr

Prepa) replay. Já

‘The imegral is of the form

Lr leecan}iy E

es

Bacon;
Processes ond Properties 1188
7
alsttting the value for constant C, and Cx eh
CET OT LS

cr
Tem

we get me

0.37810"
sa

Iren

Be ena on
er 507 310° W/m

Is us to determine Ihe amount of radiations

‘wavelength spectrum fm simple ke coe

Nats mnt eat from other bea
Vat temperature Y, is completely oe
The net radiant heat tas ey

Q.

Solution A black body isan ideal or hypothetical Sanne
temperature of a a

lace having the following radiation heat

radiacion emite by a black
À independent of diction,
From Stefan-Boltzman law, the rate of energy transmission from a black boy is

Em GT = 567 » 10% x (400 + 27 = 116807 W/m?
solute temperature at which the radian Mos gete doubled. Then
2 (1631.7) = 567 x 0" 7

16.7

Let T be the

ES

EXAMPLE 7.7
A furnace having inside temperature of 250 K has a gas circular viewing cf 6 em diameter
1 the transmissivity of glass is 0.08, make calculations for the heat loss from the glass
Window due to radiation.
Solution : The radiation heat loss from the gl
QA Txt
‘vere + fs the transmissivity of glass

window is given by

200 / Basics of Heat and Mass Transfer

Q es nav 0 200 008 = 32853 w

EXAMPLE 78
“Measurements were made of the monochromatic absorptivity and monochromatic hemispherical
inadiation incident on an opaque surface, and the variation ofthese parameters with wavelength

maybe appeoimated by the esas shown below, Determine the absorbed radiat ac
{ot eatphericlsbuorpivity andthe fla eel a the surface

Fi so

5 asl

“ Fig. 7.6.

Incident fan 00 (8 = 2) = 480 4/08
‘Atenas fa = [apa

Solution

= fs (josue

= 800 (4 - 2) + 400 (8 - 4) = 3200 W/m?
Absorptivity a = 3200/4800 = 0.667
‘The requirement that all the radiant energy striking any surface may be accounted for is

otttpal
Here, 1 = 0 as the surface is opaque and therefore reflectivity ofthe surface is
pri-a
=1~ 0667 = 0333
EXAMPLE 7.9 _

À black body of total area 0.045 m? is completely enclosed in a space bounded by 5 cm thick
walls, The walls have a surface area 05 m? and thermal conductivity 1.07 Wav-deg. Ifthe
inner surface of the enveloping wall is to be maintained at 215°C and the outer wall surface
at 30°C, ealeulate the temperature of (he black body. Neglect the difference between inner
and outer surface areas of enveloping material.
Solution : Net heat radiated by the black body to the enclosing wall,
0,70 4 if -TS)
= 567 x 10% x 0.045 (488)
where T, is the temperature of the black body in degree kelvin
Heat conducted through the wall,
FAME | 107x0.3x@1
. 05

= 19795 w

1567 x 10° x 0045 (14880)

afi

Rep
ÜBEL EINST ae 9947

ature ofthe blac REM
Temperature ofthe back body. 7, = 9859 y

6. WIENS DISPLACEMENT LAW

From the spectral distribution of black body emi
Fried with maximum rate of emision de
eng surface. The nature of this dependen

Isive power, itis,
Pends upon the
«an be obtained

Parent that the waveen
pa length
absolute temperature of the

pression by diterentiating the Planck's

ih esp to and sting the derivative equal 1 ero That

Since the denominator # 0, we have
Tefal, SGEN
36, [o9{ ho + af (Ge
ar) ler) ao

Simplification gives the following transcendental equation

a, &\
& + [er(-&)1]-0

‘whose solution by hit and tial method gives

CES
Bo T= Gg E 298 «10 = 0029 mK (719
tes den aS
Tang dents the wavelength at which emisivepoweris maximum. The Wiens displacement
mann) À sat as “the product of absolute temperature and the wavelength at witch the
UE Power is maximum, is constant”. The lave suggests that, is inversely proportional
tomada alte temperature and accordingly the maximum spectral intensity of radiation shits
shorter wavelength with rising temperature, The locus of points described by

|

202 7 Basics of Heat and Mass Transfer

Wien's law has been plotted as the dashed curve in Fig. 76.
‘A combination of Planck's law and the Wier's displacement law yields the corre

maximum monochromatic emisive power for a black body. ion fy

= SDAIN ans m 10 76 Wen?
ze 12852 10% 15 W/m per metre warnen,

Thus the magnitude ofthe maximum monochromatic emisive
with the Rh power of the absolute temperature ofthe bick sure

Wiens displacement nw hols true for more real substances theres however some dvi
in the case of a metallic radiator where the product (hy, D is found to vary with ala
temperature. The law finds application inthe prediction of avery high temperature tne
measurement of wavelength
EXAMPLE 7.10
What are the ranges of wavelength of electromagnetic waves covering ul
infrarted and thermal radiation.

A small black body has a total emissive power of 45 KM}. Determins lts sur
temperature and the wavelength of emission maximum. In which range of the sp
does this wavelength fall?

Solution : From Stefan Boltzman law, the rate of energy transmission from a Back bay
Emo, T5 45 « 1000 = 567 x 108 75

er varie proportion,

oles visti,

ria eau

EH]

‘The wavelength of emission maximum is given by Wiens law, That

eas T= 28908 x 10
2.998310
Dae = 530.77

‘This wavelength falls in he infrared region ofthe spectrum,
EXAMPLE 7.11
Making use of Planck's law of distribution, establish the relation forthe Wien displace
aw.

“The sun emits maximum radiation at = 0.52, Assuming the sun to bea Back
«coat the surface temperature ofthe sun and the emúsive ability ofthe sus si
that temperature. Also determine the maximum monochromatic emissive power
surface

Solution : From Wien’s displacement law

= 546 x 10° m = 546 um

7 = 28%
from Stlan’sBotzman lm,” OSEA "STK
B98

~367 x 10% sra
matic emis

ve poner

CORRE ENS

= 125 à 100 gs
6908 » 105 we (6573

Per metre wave
EXAMPLE 7.12 length

Slee cot nd A TLS
pren sated tod ae
monnaie dat fh deny a 8 Min i

{i wavelength at which emission is maximum nd,
4 the corespondng radiata density

{i total emissive power, and

(9) wavelength À such that emission rom 0 toh
Solution: () From Planck’ law of distribu, 591 Ihe emision frm tom
os

ey, =
(OSG ATTA

Da

= 28110" Wa? per meter wavelength
(0) From Wien’s displacement law ; AuyT = 2898 x 192
2.89810"

a E910 m
Maximum radiant flux density,

1285: 10575 © 1.285 x 105 (200095
= 411 x 10" Win? per metre wavelength

(e) From Stefan-Rolteman law, :
ar
5 67 x 10 x (2000)! 907200 Wim?
Ane THE emission in he and with O 0.5 af the al emision om Oto. Theres
factional emissive power is 03. Corresponding to Fy, = 05 the wavelength temperature
Product as read from Table 72 is approximately 4100 um-K. Thus AT = 4100 and so

ER 6
20 7 205 m

72, KIRCHOFF'S LAW

Consider two surfaces, one absolutely black at temperature T, and he other nonback at
lemperatureT (Fig. 77), The surfaces are arranged pale to each other and s clas that the
edition of one falls totally om the ather. The radiant energy E ented by the nomblck
Surface impinges on the black surface and ges fully absorbed. Likewise the radiant energy Ey

804 1 Basics of Heat and Mass Transfer

“its by the Hack surface strikes the non-black surface I the non-black surface has absorp

{i ill absorb a, radiations and the remainder (1 Fy. pace nee

‘will be reflected Yack for fll absorption tthe Black surface

Radiant interchange for the non-black surface equals (E

AE. 1 both the surfaces are atthe Same temperature, T =

Ty’ then the conditions correspond to mobile thermal

‘equilibrium for which the resultant interchange of heat is A
Under these conditions

> “vi
ok taco

S
Era m Een a

‚This relationship can be extended by considering different
surfaces in turn, Thus in general:
Fig. 77. Exchange of heat between

E memes

1 This is because the absorptivity @ of a black body equals unity

Equations 7.14 shows that he ratio of the emissive power E to absorptivity & is same for
all bodies, and is equal to the emissive power of black body at the same temperature The
relationship is known as the Kirchoff's Law

The ratio of the emissive power of a certain non-black body E tothe emissive power of a
lack body £j, both bodies being atthe same temperature, is alle the emissivity ofthe bad
Emissivity of a body is à function of is physical and chemical properties and the state of za
surface-whether rough or smooth,

From equation 7.14

E/E,
where, € = EJE, is emissivity.

Kirchoff's law can also be stated as: “The emissivity € and absorptivity a of a real surface
are equal for radiation with identical temperature and wavelength. The equivalence of € and
‘a does suggest that a perfect absorber (the perfect black body) is also a perfect radiator.

a or ca 7.35,

7.8. GRAY BODY AND SELECTIVE EMITTERS
Consider two bodies, one absolutely black and the other non-black and let these be at the same
temperature. The monochromatic emissive power ofa non-black body varies significantly from
the black body monochromatic emissive power as illustrated in Fig. 78,

‘These curves indicate that a non-black body radiates less intensively than a black body
Further the radiation spectrum for a non-black body may be similar or radically different from
that of a black body.

When the emissivity of non-black surface is constant at all temperatures and through aut
the entire range of wavelength, the surface is called a gray body. The radiation spectrum for
a gray body, though reduced in vertical scale, is continuous and identical to the corresponding
‘carve for a perfectly black surface; there is no shit in the peak of the curves. However für
many materials the emissivity is differen for he various wavelengths of the emitted energy
The radiating, bodies exhibiting this behaviour are called selective emitters, The radar
spectrum fora selective emitter does not follow any definite pattern, and it varies entirely from
that of a black body.

Stefan-Boltzman law when applied to a gray body takes the form

Lae 016)
-Teconsiant rs diferen or fer
de aná its value depends upon ut
tots af the body the state ft.
Temperature, I is always lees thee
aad gation cocticient 0 for a Week
Dt value ange rom 00 1 6 à
10 W/m K

pe emsiviy ofthe gray suce may
se expressed as

Monochromatic missive Power
in Wn pm

E
TE
az
ie ean

Values of emissivities range from 0.0
win.

The emitted radiant energy flux
density for nop-black body, as prescribed

716 be Fig. 71 en
on 716 may de renin PF Ensen onda gro each nan
De eae
Ends fra bodies Envy of à
a er pere
TE ere power dde aes

sure ints is by o et ration
same empese eve repens dic

1) Monochromatic emissivity ato fm on
to e monociuomatic emisive poet aback sun othe 0
: atthe sue wavelength nd temperate
jag tet te menace ems gordos en ate
Pan, ie, the monochromatc emisiv poner er de ee ed
porto ak srl have the same rat forall wong e ml la oe
Se temperature
(i) Total emissivity € : Ratio ofthe total emi
ower of black surface atthe same lemperatue
(#9 Normal total emisivity e Rati of he oral component af he tensive power
taste othe normal component of he lll eme Back body a te sane tampa
(9) Mean and equilibrium emissioty Emi of mast he cinco mater
inlooced by temperature as well as wavelength, Fr à particular tempt the arcing of
‚nschromane emissivity at various wavelengths (within range of wavelegi) ele he
den mean emissivity. Likewise for porte wavelength anta sal ll de
enschrömat eis at various temperatures (ui à range of empero) seed
he lenporture.mean emissivity
Theemisivity of material though varying with temperature andthe nae oft surface,
ist ft in any way bythe nature of surface surrounding The toa emisivi rra
{onsant whether the material is in equilibrium with the suroundings or nl, The total emisivity
‘here, sometimes called the quiro emission
In general, the emissivity of a material is dependent upon its ature (colour, texture and

re power ofa Surface to the total emisive

206 1 Bosıca of Heat and Mess Transfer

tise, Nem-sonductors have a comparatively lange (generally exceeding

ous and ith the
À a haghly polished sort
Secret with temperature
Sa) emisavity. Depending upon specific mater
Anse oF decrease With temperature rise

EXAMPLE 7.13
What physical ratio determines whether a real surface is an almost specular reflector o
Almost diffuse reflector ?

(0) Define emissivity, How does it vary with temperature for conductors and non,
conductors?

‘The radiant heat transfer from a plate of 25 cm? area at 1250 K 10 a very cold enclosure
is 50 W. Determine the emissivity of the plate at this temperature
Solution : If the roughness dimension for a real futace is lage with respect 10 wavelength of
incident radiation the surface behaves as a diffuse reflector Ifthe roughnese dimension jy
onsideraly smaller than the wavelength, the surface reflects specularly

(9) A real surface has a total emissive power E les than that of à black surface E. The rtp
of the total emissive power ofa forface to thal of à back surface atthe same temperature
Called the total emissivity ze = E/E, Some common observatiors about the emissivity of à body

(@ The emissivity ofthe metallic surfaces is very small having the values as low as 012
for highly polished gold and silver
y improves the emissivity of metalic surfaces

(3) The presence of oxide layers general
(ri The non-conductors have large value of emissivity, generally exceeding 0.6
(o) The emissivity of a conducting material increases with increase in temperatur, but
‘emissivity of non-conducting materials decreases with increase in temperature
ar
A ATE
50
AR

Emissivity € =

“ons

Emissivity € = =
EXAMPLE 7.14
‘Can a surface behave essentially like a black surface and stil Look practically white tothe
oe

(6) A 100 watt ight bulb has a tungsten filament (emissivity € = 030) which is required
to operate at 2780 K. Ifthe bulb is completely evacuated, calculate the minimum surface
area of the lungsten filament, Assume that no radiant energy strikes the bulb surface and
that steady state conditions preval
Solution; The operating conditions stipulate that

(@) The bulb is completely evacuated, and the vacuum precludes any heat transfer by

(i) No radiant energy strikes the bulb surface

(it) Steady state conditions prevail Le, there is no storage of energy

pccording!

A That is

300 2035 (547 « 10% » 4 à zt
face area of the bulb fiament, "À AE

su

A GORGE * 098 10% me

une 745
Pad (€ = 08) emits the same amount
Att pate of he ay ey

at ct body 100 K an 3 a boty mn Kan
vad ha shou be he emi he py Reg © Sit the same amount of
mn Since the black and gray body are emiting the Lame amount

Of heat as a black body at 1075 K. Find out

Soto

presuming the two bodies 10 he of same arca! we have

Thee rs
E ws
or ner
we E) = 04096
come 7.16

ndiation and natural convection per metre length of the pipe. Take thermal radiation constant

O, (radiation) = e 0, À (TT)
0.2 x 567 x 10% x 0.157 (O - 295) = 1988 W
O, (convection) =hA at

= 1135 x 0187 (70 - 299)
13366 W

Q (otal) = 1988 + 13364 = 15352 W

Tre total heat exchange can be expressed as
Q, = WA AT where U ls the overall coefficient of het transfer

21391 W/m,

208 Basics of Hest and Mass Transfer

EXAMPLE 7.17
how that no surface at any temperature can have emissi
EEE et materials, the emissivity € is à function of temperature and for à cer

material has been specified as

greater than unit.

Aes.
where Tis in degree kelvin, over the temperature range 0 o 1500 K. Is there any reason ig
spot the validity of this Information?

"fy A meta sphere of surface area 00225 mi is in an evaculated enclosure whose way,
are held as very low temperature Electric current passed through resistors imbedded
ihe sphere using eecrical energy fo be dissipated at the rate of 75 wat Ifthe sia
Serie temperature is measured to be 560 K, while in steady sat, colcolate emiasivi
ihe sphere surface and slate the assumptions made im the estimation,

Solution: The requirement that all the radiant energy striking any surface be accounted or y
Presid by the relationship
apte

‘Apparently the largest faction of incident radiation that can be absorbed is uni
with pe t= Op

From Kirchof's law, the emissivity equals the absorptivity under the some temperate
conditions. That is = &

“These conditions clar indicate that emissivity of any surface cannot be
uni.

At 1500 K, the emis

ty works out as

E is
Homer he mini fay mae la number been and Obi er
te e by he dns crn. Fa
ee

75 = © x 0.0225 x (5.67 x 10%) x 560 = 125.46 €

y of phere,
"8

iia
{) the wall ae hl ats ow epoca TN pls ht he pre es, bd
(0 Fue naa copy evacuado pres any ant y

(ii) Steady state conditons prevail, ie, there is no storage of energy,

EXAMPLE 7.18
A gray surface has an emissivity e = 035 at a temperature of 550 K source. If the sus
is opaque, calculate its reflectivity for à black body radiation coming from à 550 K sou

(0) A small 25 mm square hole is made in the thin-walled door of a furnace whet
inside walls are at 920 K. Ifthe emissivity of the walls is 072, caleulate the rate at whi
radiant energy escapes from the furnace though the hole to the room.

the resistors moy

Rotts
on: Processes
tion he eglrement that allo he adi ‘nd Proportion 4 209

ies Mey striking any surface maybe accounted
ar pera
Here
Me 0 as the surface is opaque
pare =035

in accordance with Kirehof a
Hin which state that a
ander the same temperature conditions i ean ey
populist APR o cee
‘Thus the surface reflects 65 percent of incident ener
(e) Te small hole acts as a black body and secon
leaves the hoe is

coming rom à sure 550K.
ly the rate lc adam cy
eqn

2567 105 008 0085) x

= 25.38 watts po
Note The data about the eich ofthe mi wa ant nde,

EXAMPLE 7.19
Sut stefan oltzman lw of total radiation fom aback body How ia wean bows
See ite account radiation from a nomina beat be modified

(a) Oxidised with emissivity € = 080, and
(0) Polished with emissivity € = 007. Assumption may be made ofthe gay body behaviour
Take 0, = 5.67 x 10% Wm? KS. Comment on the results y

Solution : Energy emitted per unit area

9h € Ti = (567 x 10% x 08 x 7
From Kischolf' law, absorpuivity & equals the emissivity © ofthe surface.
Energy absorbed per unit area
» incident radiation = 08 x 750
er equilibrium conditions
6.67 x 10%) x 08 x T = 08 x 730, T= 339K
Likewise when the surface is polished « = € = 0.07.
(567 x 10) » 007 x TE =007 x 750; T= 339K
if the absorptivity and emissivity remain constant over the range of temperature
th, the equilibirium temperature remains unaffected by the nature of surface

73. INTENSITY OF RADIATION ANO LAMBERT'S COSINELAW |

Sabtended plane and solid angles : The plane angle is defined by a region by the rays of a
le and is measured as the rato ofthe element of arc of length! on the cie tothe radius
"of the cree: = tr

The solid angle wo is defined by a region by the rays :
„At a,

sphere, and is measured as

210 4 Basics of Heat and Mess Transfer

whore
A, = projestion of the incident surface normal to the line of propagation
A = aren of inciden surface
8 angle between the normal to the incident Surface and the line of propagation
+ * length of the line of propagation between the radiating and the inciden surface,

Fig. 79. Plane and soli angles

‘The relationship between A, A, and 8 has been ilysteated in Fig. 7.10. When the incident
surface isa sphere, the projection of surface normal to the line of propagation is the shou
disk ofthe sphere, which isa circle of the diameter ofthe sphere. The unt of measure
angle is the stcradian (er).

of soit

Soros A

Den

Br aod spheres! taco Araımal
pon ezo {ethene of poser
naked

Avent
Fig, 7.0, Projection ofan incite surface normal tthe ine of propagation

Intensity of radiation : Considera small black surface dA (emitter) arbitrarily located ata
pint nthe pace under coter and ening radon in dent econ À sh
Body radiation collector rough which Whe radiations pass is located at an angular posto
characterized by zenith angle 0 towards the surface normal and the azimuth angle 9 ola
spherical coordinate system. Further, the collector subtends a solid angle do when viewed
from à point on the emite.
The intensity of radiation isthe energy emitted (of all wavelengths) in a particular direcion
per unit surface area and through a uni solid angle. The area ss the projected are of tit
Surface on a plane perpendicular tothe direction of radiation, The collctor or the inciden
surface measures a variation of the emitted radiations depending upon its angular posó
Maximum amount of radiation is measured received) by the collector when it iat he post
4 normal to the emitter, and the intensity in a direction 0 rom normal tothe emitter follows
cine lw

Fig. 7.41. Spat tibio of ran ens om a ataco
“The intesty of radiation in à direction 8 from the normal u as
Muy ack enter is proportional to
1, denotes the normal intensity and J represents the ine
u represents the intensity at angle from Ihe normal,
la =1,c050 =

Apparently
8%.

When the collector is oriented at an angl
radiations striking and being absorb

1 energy radiated out decreses with crease in 9 and becmes

fam the orm 1 he emit, then he
sd by the collector can be pd oan

(Eo, = le doy dA
1, cos, du, dA
here day is the solid angle subtended bythe
The collector could be located at diffe
radal distance from the emitter, Let i
located in a direction 8,

(ar,

02)

collector at the surface of th eme,

different angular positions and sil maintain the same

it subtend a slid angle de at the emuter surface when

from the normal. Then the rate of low of energy through it will be
y * Ido dA

050, de, JA ca
quations 721 and 7:2 that for any surfce located at an angle @ from the
ing à solid angle do at the emiter dA,
GE =1, cos 8 de a 723)
tne Relation between the normal intensity and emissive power To establish aration betwen
era intensity and the emissive power, we rote the differential soba angle do lo the
si and azimuth angles by noting that for a spheical surface
réa of collector = (y 48) x (rind de) = F sin 8 48 dp

I follows from oq
oral and subtend

ino 404
solid angle do ae

212 4 Boss of Heat and Mass Transfer

‘Then the radiations leaving the emitter and striking the collector is
dE, = 1, cos 8 sino 48 49 4A

o
Ths total energy E radar bythe emir and passing through a sphere egos ¿729
worked out be integrating equatign 725 otr de Kens, ak
: 90160 À and 4 010$ 2x
Thus,
Jen +10 Pond tef
1
By dh) 1, nda 0
Bu te tl emiiv power ofthe emite with ars dA and temperature Ts
wy x
Bro, Maa
Combining this with equation 7.25, we get
ot ott
nan * A 25

arena

Fig. 7.12, irn ni ong in terms of ani and azur! sois
Thus fora unit surface, the intensity of normal radiation I, is the 1 toes the en
power E,
EXAMPLE 720
Define intensity of radiation and prove thatthe intensity of normal radiation is Y ire
the total emissive power.
A black body of 02 m area has am effective temperature of 800 K. Caeualt a) the In
rate of energy emission () the intensity of normal radiation () the intensity OÙ

ie
along a direction 60° to the normal, and (4) the wavelength of maximum monochro®!
emissive power

Radio
sing Fo Samar, eu gr een MPa vata |
Fy OATS OÙ ner ein

CET i |
i The intensity of normal da * 0!» 0426 Ww H
ST SED os i

(a) rom Lamberts cine EI traia,

2 739528 co 6 =
{ey From Wien's displacement a ced W/m? sean

the wavelengy
gees ' Ans for maximum monochromatic
2.896102

T

= 286010

3625 «104 q

EXAMPLE 7.2

he emitting surface, and

by the three surfaces Ay Ay and A,

D

hs hate

Mess inees Sn tig sty

Pozas

Solution: () For a diffused emitter,

Abe intensity fte ented adan i independent el
‘diection, Theretor id i

1 = 6500 W/mèar
or ech ofthe thece directions

{Pi terms of diferent surface and the distance radiation travels the slid ange is
AL face and the di

7
¿tre © the angle between the surface normal and the deco lados. Therefore
angle subtended by surace Ay with resp o sutace Ay 8

214 / Basics of Heat and Mass Transfer

(6) The rate at which radiation is intercepted by &
de abtaines from the expression.
Aa.

where 0, is ange bon noe the cing rl
of radiation. Therefore =

Q .2 = 6500 x 0.0015 cos $5 x 2275 x 10% = 1272 x 109 w
Qu 3 = 6500 x DOS cos 0 x 347 x 10% = 3383 x 109 W

ch of three surfaces Ay Ay and A, ey

A and the direction of propagation

Qi à = 6500 # 0.0015 cos 40 = 278 » 10° = 20.76 5 10% W
7.10. SOLAR RADIATIONS,

The sun is a source of heat radiations and it emits radiations in all directions The atmosphere
absorbs part of the heat radiations and air, clouds, dust particles etc inthe atmospheee san
the heat and light radiations falling on them. Obviously, the earth receives only à facon et
the energy emitted by the sun. The solar radiation that is felt atthe earths surface inched
direct radiation that has passed through the atmosphere; diffuse radiation from the sky ein
radiation from water, snow and other such materials on the surface, The appronimate distribu
Of the flow of sun's energy to the earth's surface is

= 9% is scattered

+ 15% is absorbed in the atmosphere and out of it 4% reaches the earth's surface by

+ 43% is transmitted to the earth directly and by diffuse radiation

33% is reflected back to space

When the sum lies at a won distance from the esrth, the heat flux from the sun o

the outer edge of the atmosphere has been found to be about 1350 W/m? and 47% of ths
(or 635 W/m ) would reach the earth's surface. A lorge fraction ofthe sun’s energy reaches th
earth's surface in ultea-voilet and visible wavelength; and re-radiation from the cool surface
the earth would be in wavelength that are generally far larger

Green House Effect

by the objects within the enclosure. As their temperature rises, they too radiate energy ie
o do A reno

ids re
Soliva absorbing surface ave en dep 9628888 and Properties y 219

¿poto much of the incident ran witha ¿a colo

ler extent, ozone iy loaded with

fi a parameter called solar constant is inten CONS

CO} 160 and o

by he earth its pon
een incon

à xed time
sorption by

is determined with the help of
Bt various altudes of

the stmosphr ae ni
he ana eme da ans

‘lt constant Sand the angular elevation (lie)

or log Sp = log $+ see Jog a u

From he straight Tine graph between log Syalong the

inercpt log $ on the vaxis is found and the ue sens

sole constant varies between 1335 and 1815 Ware
lst R

axis and se Zalong the rani the
onstant i evaluated thee rom. The

an distance of earth from the san
radius of the sun
Then the total amount of heat energy received by the
amount of hea energy radiated by unit surface
EN]
MUR
5 » 108 ms = 7 x 10° m and $ = 1650 9/0
MESÓN j y Ei
(or) * 26502 7876 x w wine
Eo, T= 567 1087
5.76% 10°"
ETS

sphere of adios Ris ARS, and he
‘rea of he sun inthe same ime works out as

Taking R

= 6016 K

‘This value represents the effective temperature ofthe sun ating as a Back body
Temperature ofthe sun can also be worked out from Wie displacement lave

Au, T2 289 x 10° mK
The wavelength of radiation for which the energy i maximum 049 micro. The temperature
the sun then works out as

«

289810
‘049x107
The sun A nde by the photospher. The pra
consists ola central hot portion sure pue lei
‘the photosphore, referred toas effective temperature 0 re sun, al taken

“sk

216 / Bssics of Hest and Mass Transfer

comiera?
higher than the ambinet temperature outside. y
Estat equim ampere tht prb es Lot lc EN

300 Wim’ For solar radiation absorptivity of aluminium is 03 and the average o

appropriate for aluminium at low temperature is 0.04. 22
Solution The heat input from the sun i determine bythe projected ares ofthe sun
Qn 094,
03300 Fy? = 707 W
Tre outgoing radiation gosto space whee the temperatures ow enough ob gis

Qu = a AN
where A is the total area of the sphere.
= 004 x (5.67 x 10°) x An (05)? T = 0712 x 108 18
Energy balance gives
0712 x 10% 1 = 707

A
Parar ide

The probe will attain an equilibrium temperature of 31547 K
EXAMPLE 7.23

Give a brief account of the effects of temperature and surface condition on emissivity
‘An artifical satellite flying round the earth has absorptivity a with respect to incident solar
radiation and surface emissivity ©. The satellite has no inner heal source and has uniform
absolute temperature T al over its surface. Set up a relation between T, and e. You may
ignore the effects due to solar radiation reflected from the earth and the radiations emitted
by the earth.

Presuming thai & = 0.18, € = 0.12 and incident soalr radiation E,, = 1560 W/o’, calculate
the surface temperature ot satellite. How this temperature will change ifthe satelite surface
is considered as gray?

Solution : Under steady state conditions of heat transfer from the satelite surface
radiant energy absorbed by the satellite = energy emitted by the satellite into the pace
@E,, A, =6 6,4, 7 0
where A, is projected area of the satelite on the plane normal to incident radiation and A, 5
the surface area of satellite. fr isthe radios of satellite, then,
Ayame and À, = An?
From expression ()

eh isthe required relation between 7, a and ¢

far given values of = 018 and e = 947
= 299 (0181

7-266 (555) = aan x.

or guy surface a = €, Ln that ease: T= 288 x

mes
O kn

recy oes a

pooh tp dent ert

Near the earth

shitea
its temperature

towards mas. What shall be the

rom the satelite uc,
cited byte slit mite space
0
w
ation na he car an man, À à
dent don nd An eta

CORRE
Near the Mars: a ESA, =e 04 4 78
were £ and Ea are the intensities of solar
rscted area of the satelite onthe plane near te
asl

‘Upon division, these identities give
4 (8
aS,
where $, and S, are distances of ear

(m
icy)

th and mars fom the sun,

= 25521 K = - 1779
EXAMPLE 7,25

Die observed that when the sun i ovehead the earls surface on a dew dy, he
ha ceived by the carth’s surface is 1 KW/m? and an additional 0.3 AW Labbe

Tee cas atmosphere. Assuming the sun t be blak body, determine the epee

Given: dia of sun =1.39 x 199 m; dia of earth = 126 x 10m
Soliton 1, Stance between the sun and earth = 135% 100 m
ha Let 7, be the temperature ofthe sun considered as back body. Then energy radiated

Qu + AG, TY = 1 2 4 (139 IOP (67 104) TE,

344x109 row

£218 / Basics of Heat end Mass Transfer

Considering sun to be point source, the mean area over which the radiation is dist

Then sla fix Q,,/A =

buted ig
4% 3 where sis the distance of sun from the earth,

= 4 (LS X LOU » 28.26 x 102 mi
ao
2826x107

1217 «10 Th,

Total radiation from the sun = 1 + 03 kW/m? = 13 x 10° W/m?

A7 10 Th, «13 x 10°

‘That gives: approximate temperature of sun,

a.

72

2.

24

75.

13x10 | = 51667 K

za?

REVIEW QUESTIONS

Enumerate some salient features of thermal radiation, Woat position does thermal a
in the electromagnetic spectrum ? What limits Uns band width on e short and fo
sides?

ition caps
ng valen

Explain the terms absorptivity. reflectivity and transmissivity of radlant energy. How
related to each other for a black body and an opaque body?
(6) By qualitative reasoaing establish the expression

here a = absorptivity 9 = rei and € à rnsmissivity ofthe body, Comment upon y
‘aly of the statement that this relation à valid ely fora pay de aa
(0) Based upon the reradiatng properties of absoepivity,ceflectvity and tonic, tow
would you distinguish between the flowing,
‘ack Body, white Body raneparent body and opaque body

(0) Two pees of wood ae placed in sa ight ane plete ped white and the ee ck
Which piece wil absorb more heat I te same two pies Bla oo tempera te
on he ground inthe mia sinter and a mg which psc wil ool asters Civ sgt
in support your ner

(© What do you understand by aback body anda gray body” as applied to raciaon potes
Explain how rason streaming ost oa small hal in a ige how boy cn o Coni
‘bck body radiación. Ñ

(O) A black body may be approximted by means of a large hollow container with se

"eng, Gacy VAL e code ota ey?

(@) Define monactromatic and total emisive power, Hove i the Inter relat to the aba
temperature? Describe how the meoochromaicenussve power vais withthe mane
for emissions ram à Hack body? At what wavelength the black body sonora
ceive power the maxima? °

(© Ste and explain the folowing laws ring o thermal radiation and tempera #3
rire body

Farc nw : San Bolteman law and Wins displace .

(0 State and prove Krebs law of ration, What restive condition ae inet #
servation of Richa le? i"

(0 Define Lamberts cosine fw of radiation and prove ha the intensity of ration a
Constant at any ange of emision fort use sur

m.

Processes,
PASSES ond Properties y 219
ante

Define the following plie 1
LS ad ised seein °°

such a surface 10 emit as black body, and calculate Duran Prius SOC. Consider
5 de conte

1 radiation heu

pres

ak lations for he vation thermal y pr so

A Block body, in the form of cube 1 m

(Ans Sa ki)
‘every flux density and the ttle

ho à tn
pen Kn
the black body cube, “
CE u)
dy ang a o
Con ene

Tong on the side
TY eed by

The tent of à 75 W MEN bub may
‘core 70€ The Mn! dame ONO ES à 6
Fada, make calculations for the lamen impune Pen
Wale rat er a rom a eat nee

pu bly wa conta em à Pa M

be considered as à bc

100K? The peut may be tated as
he aver copain tae

eee power onen
‘The mime spectral intensity of rain for a gray stc a 100K i
1372100 W msm of wavelength Determine he eee me

sity ofthe body sure and ne mate
en ours (Ans OM 26M 107 0)

000

ch the mass pl te ol Indie

Radiation Exchange Between

Surfaces

Engineering problems of practical interest are involved with heat exchange between top
soe rs and this exchange strongly dependen upon die prop emp
ei surface prometi (te and shape) and creation relative 10 caco ater For bee
rice necessary to deermine what potion a ration ete by ame nto
by the other srfacs are non-tlac, considerations have 1 be given 10 the ema
récit and transmissivity el he raiatingsurlacs The problem becomes more cong
After an absorbing or radating medhum beeen the auras Anton would be ds
in ths chapter on the geometn features of radiation exchange, and the rations nous ye
setup by presuming tht the surfaces are separated by non parta mesi

8.1, HEAT EXCHANCE BETWEEN BLACK BODIES : CONFIGURATION FACTOR
Consider heat exchange between elementary areas dA, and dA, of two black radiating bodies
having areas Ay and A, respectively, The elementary’ areas act at à distance 7 apart and te

normale to these areas make angles 0, and 8, with the line joining them. The surface dA, ira
temperature T, and the surface dA, dat temperature Ty

M the surface dA, subtendk a solid angle da, atthe entre of the surface dA, then radiant
‘energy emitted by dA, and impinging on (and absorbed by) the surface dA is

Qa = lo dan dA,
hy cos0, de, dA,
Projected area of dA, normal tothe line joining dA, and dA, = dA, cos 8,
Ay £050)

solid angle da, =

and therefore

AT! £050) cos0,dAy dá, my
Integration of equation 81 over finite areas Ay and A, gives

a

CA

Fig. 8.1. Radiant hat exchange baten two Back suecas

The solution to this equation is simplified by introducing a term called radiation shape
facto geometrical factor, configuration factor or view factor Te coniguraton actor deere
‘only on the specific geometry of the emiter and collection surfaces, and is defined as

“The fraction of the radiative energy that is difosed from one surface element and strikes
the other surface directly with no intervening relschens“

The radiation shape factor is represented by the symbol F, which means the shape factor

form surface, A, to another surface, A, Thus the radiation shape factor Fy of sec An to
suce Ay is

leet radiation from surface ncidnt upon surface 2.

y Total radiation from emitting surface 1

SAT
al Jota >
‘Thus the amount of radiation leaving A, and striking Ay may be written as
Qu = A ñas, TE > 64

Sarl the energy leaving A, and ariving A, 18
Qn = Aine 2
à energy exchange from A to Ay is
(Qu) ner = Ar Bor = Aa

‚When the surfaces are maintained at he some temperatures, T, * Ty there can be no heat
exchange

and the

0 (Aa

= Aa) Tt rauen

!
|
E
i
|

222 1 Basios of Hoot ond Mass Transfer

Seo, a T an th nome qui,
Ar Ay "0 or AVF An 86,
o ES
mire: =
Ki PR
os!
7]
os
os
o z
cS :
o
82 Ra po coo es pat os
so =7
7 me
MET To
val —
al Py,
4 y Lo +
04! A, 1
i 17 =;
LE
= “|
> Ti
CC CE TE
a
83 Rain sae aca u pts
‘The above result is known as reciprocity theorem. It indicates that the net radiant interchange

TRY evaluated by computing one way configuration factor from either surface to the ot
‘Thus the net heat exchange between surfaces A, and Ay is

Radiation Exchange Between | Surfaces 1 223
ida = Avia OME 18) A o tre ae
uation 87 applies only to back surfaces ang nn 127) as
ae ey rent from unity

ne Ag) o Serge emer te
cal form for he geometis normally encountered er And presente in

CON
ET ET TS
= 5
A te
Erg)
ad en
mi 15 er] TI
5
== CET

Der 7 +

Fig. BA Rasaton shape ac or perpen eng wih canon gp

32. SHAPE FACTOR ALGEBRA AND SALIENT FEATURES OF THE SHAPE FACTOR

Te shape factors or complex geometis (fr which shape ator char o equations ae not
tall) can be dived in term of known shape factor for ober roman Fo tat he
complex shape i divided into sections for which the shape actor eer Know ft can be
ray eshte

The unknown configuration factor i woke out by adding and subit known actors
ai ltd geometries. The method is based onthe dehnen of supe cr Le pret
Pipe and the energy conservation law

The interrelation between various

The folowing
mi

shape factors is called shape factor algebra

Y fact and properties wil be useful forthe calculation of shape actors of

y roms and forthe ra of radian hs eng ben us

{9 The value of shape factor depends only on the geometry and orientation of surfaces
with eapet each ter. On the shape co eee uy sre an
used for calculating the radiant heat exchange between the surfaces at any temperate.

(9 The net heat exchange between surfaces Ay Ay is

| 4

EEE

|
|

—

224 1 Bssics of Hest and Mass Transfer

Od ne Ara TÍ = Aa Fn 09 TE
when the surfaces are thought to be black 0) = = 0) and are mining
same temperature (T, = Ta = D, there is no heat exchange and un au te
Om (A ña- Aula
Since 6, and T are both non-zero ente,
Ara = AP
‘This reciprocal relation is particularly useful when one ofthe shape factor
9 A dation reaming out rom a conter surface intra y
apres 2 As uch the hope face of convex surface with ep tthe een
3 nity Then in conformity with reciprocity theorem, the othe: shape Kan AR

merely the ratio of areas

(o) The radiant energy emitd by one part of concave surface ie intercepted by ais
DR Me same sure cordingly a concave suce hs shape fate ie
Fo sel. The shape factor fa surface with respect ot sl is denoted by I
Ea ar conte surface, shape factor with esparto si I zero Ts apg
stems from the fat tha for any pat of ator convex surface, one can! se anh
Part of the same surface.
(0) Hone of the two surfaces (say A) is divided ito sub areas Ay, Ay,
Ay Fy ZA Figs 89
Thus with respect 0 Fig. 85 () wherein the radiating surface A has been sl u
areas Ay and Ay,

Ay then

Ai Fra Ay Fin + Ay Faz

Obviously Fy + Es + Fg
Thus if the transmitting surface is sub
respect to the receiving surface is not equal to

vided, the shape factor for that surface wis
the sum of the individual shape factors,

Receive saco
*

usage Lu tc
Baas tly > m
a

Fig. 85. Rolain bounon shape Loos
I the receiving surface A, (Fg. 850] is divided into subareas Ay and Ay, we have
Ar Fin As Fis As Fg oF Fi = Fis + Fig
surfaces
Apparently the shape factor from a radiating surface to a subdivided receiving
simply the sum of the individual shape factors,
(oi) Any radiating surface will have fin

area and therefore will be enclosed by mar)
+ surfaces. The total unit radiation being emitted

El
by the radiating surface will be recited

Radiation Exena
Inge Between
y ah of he canin succes Sg hp ne
vingt rating surface and ling upon er te

fron of tt radiation
sos give

eng surface, he nergy balones

089)
ce has ben subdivided in parch pat
thet
sel
Feat

ith a numberof lack surfaces
aditig surface willbe

AR TT) A Bro (Tf 74)

(ral * Ar Bo (Tf ~

EXAMPLE 8.1
‘The sun is neatly spherical radiation source th
and located at 150 x 10" m from the earth: On a
solar radiation incident on the outer surface of
known to be 1135 War? with an additional 235
Assaming thatthe sun emits as black body,
Solution : The radiant heat exchange between

is approximately 1365 x 10° m diameter
clear day, the energy flux associated wth
cath has been accurately measured and
Wav" absorbed by the earth atmosphere.
estimate its surface temperature

the sun and the earth may be written as

4 $050 £080 day
Qu = (mA AA

here the suffe 1 and 2 pera to sun and eth respectively. From the given data

(9.8) = 8, 90"; hence cos cond = 1

(6) Temperatur T, ofthe ar canbe neglected in comparison 1 temperature T, ofthe

i) The sun, emiting difusly, appears as a disk of area 44, = (1/4) and
(} The heat interchange corresponds to unit are ofthe earth surface and so 44,» 1
CANINA
OR tg ol g
Qn Sade

ST 95 10
4nx(L50x10")
= 58548 K a
The sur temperature is generally accepted 1 be ofthis approximate magnitude
EXAMPLE 6 =F
(ileal the shape factors forthe configurations shown in the figure given below

(2) long tube with cross-section of an equilateral tangle

(0) black body inside a black encosute

(6) diagonal partition within a long square duet.

(135 + 285) =

Solution gives

and Mass Transfer

@ o
F 8.

Solution : The desired shape factors can be worked

reciprocity theorem hs 5%

and from inspection of the geomet
(0) By summation rule for radiation from surface 1.
Fut Ft Fy
The Mat surface 1 cannot see itself and so F,, = 0, Th
Eg + Fy 21
Duc to syonmetry, the radiation from surface 1 is equally
3, and therefore =

at gives

Avid between surfaces 2 ang
2“ Fy

Likewise, considering radiation rom surface 2
Fy thy

(because Fa, = 0)
Fay tl Fy

By reciprocity
A,
Fat Ay Ft = Fp (because Ay = A)
Ey 81-1 -05 = 05
(0) The concave surface 1 can see it andthe rest of radiation falls on the enclosed
surface 2 Invoking the conservation principle (summation rule) fr surface À
By reciprocity

As Fa

Fuso Fi
Further, all the radiations coming out from the black surface 2 are intercepted by the
enclosing surface, Therefore Fy, = 1, and so
4
a-4
mich
(6) From summation rule F + F + Fy =1
By inspection F = Oas the flat surface cannot see itself
Fat Fy
Due to symmetry, the radiation form surface 1 is equally divided between the surfaces
2 and 3 and therefore

sy reciroci

eater ts ths fa ee
Sin: Invoking the conservation principe (umo
sal Reet ule) for surface 1 ofthe cavity,

À à cavity with respecte lose

for the closing surface 2

‘Toe above relation vail fl all ype of cavities a shown in ig. 7

A
Koi
@

ci m
a 0
(9 Fora cylindrical cavity of depth of h and diameter d

228 Basics of Heat and Mass Transfer

(For a conical cavity of diameter d and depth hi

4 e1-2sna

4
Fy ets Ate
E

4

here [is he slant height of the cone and a is half vertex angle.
In terms of depth h

4 4
apr ae
vee

(i) For a hemispherical bow of diameter d

‘Thas, half of the radiation from hemispherical bow! falls on itself and
is intercepted by the plane closing surface
EXAMPLE 8.4
Calculate the shape factors for the configurations shown in the figure given below:
(a) sphere of diameter d inside a cubical box of length I= d
(6) hemispherical surface closed by a plane surface
(0 end and side of circular tube of equal length and diameter

Dd:

Posa,
Solution : The desired shape factors can be worked out by invoking the summation rule, Be
reciprocity theorem and from inspection of the geometry

{@) The convex spherical surface cannot see itself and the entre radiation emited by tis
surface falls on the enclosing box. Invoking the conservation principle (summation rule) I

the remaining hay

==
©

surface Y,
Fyt yet or Fast (because Fi = 0)
By reciprocity
Aporta
AE cause! = 4)
Fa ee ead à e o

(6) The hemispherical surface 1 can see itself and the rest of radiation falls on Ihe at ph
sürface

Radiation E

chan

in Beton Born Sutaces 1229
al the radiations emited f

pure al he ration rom the plane su
peated by e REMC roce yg ai ice

2 (whic
ret hich cannot see ii) are
py secrecy

DE
par 8S
E
Fy == Fy 2 1-05 205
(a rom the given geometry,
! a,
Lez and Pros
Wh reference 10 Fig. 83, the shape factor Fis hen read as =
ping summation rule: Fy + By + Few Y ay
The at surface 1 cannot see Il and so fy
Fg "1 = Fy = 1017 089
ay pro

O. That gives

cause 1 9)
EUWPLE 8.5

Coder atm of once Pa ara TT
Bera der to Have he vie a ha Se eine
Son For e coniguration comen ne any

(Ao
©

From reciprocity theorem : Ay Fig = Ay Fy
Sabstituting the relevent data,

oa Eos

= 00645 m = 645 cm

EXAMPLE 8.6
ser à very lo
en dimensions :

8 isocelles triangular duct shown in the aomponying igure For the

ab = ac = x and be =
Show that Fy = 0.75,
ion : The duct is very long and as such the radiation lost out ofthe end ofthe uct can
grrcd This implis thatthe shape factors ofthe Large surfaces with espect tothe tragar
fd suaces are vanishingly small ñ
From the summation rule for flat surfaces (1) and Q)
En? hy
Fy + Fy tl
Fy En

o
@

By symmetry

230 Basios of Heat and Mass Transfer

‘Suto tebe 0)
Sarton Sate)

‘Sule 2a)

“Then from identity (i), we obtain
Der
By reciprocity Uheorem ; Ay F3 = Ay Foy

or = 2 «05 = 025

Therefore from identity () Beer

Pa
EXAMPLE 87

Show that the ra
the centre of à small sphere is given by
Solution : With reference to Fig. 811, consider an
elemental area dA, of included angle d8 at an angle 8
4A, © (2x8 sin 0) R 48
‘Then area of the surface of the sector ab of the
sphere

à = [@xksineyR0

= 2284 [no do

ZAR (1-cosa)
‘Area of the sphere of radius R = 4 aR?
Zur’ (1-co50)

A ae

Fg. 814
1
= Ha-cosa)
F-60506)

tion shape factor between a dise of radius r located at a distance Fin
a

Pasai Exch,

19e Between Surfaces 1231
exami 88
Kee aha Blow ide fom aaa
sorte hr sra of ide O cn ng ape
tat rved surface of the einge? À he ane forte ane eee

eer Ais

with respect

ac Of the thin hollow cylinder

ren a
For the surfaces 1 and 2,
Ay Esa "As
anos (Resiprcity theory
Since A, = Ay we get N
Fp = Fy = O72
que fas Fa" © fd Ex Fy ec foam

By summation rule, the shape acto relations among ihe hee ras ae given by
soe int Pat Fy el oe hone, 2 oor
Ps 212 ae l= 0172 © 050
Aso, + Fat amt oF Fy thy o Ey et
Fy 81-2 Fy)
Invoking reciprocity theorm :Ay Fry = Ay Fy

ocr)

4
BESTER = 01656,

Thos
EXAMPLE 8.9

Arruncated cone of height 10 em has top and bottom diameter
#3 cm and 16 em respectively. The bottom surface is stated
o intercept 15 percent of radiation leaving the top surface
Determine the shape factor between the (i) top surface andthe
sonlcal side surface, and (ii) the side surface and ise
Solution : Area of the curved surfaces

Aamir t e
un (eat
= Tr TS = 405.83 cm? Kom

ARR pin rer TEE

+

232 7 Basics of Heat end Mess Transfer,

Fy 045
By reciprocity theorem : Ay F = Ay Fp
fap e
= Mg, EAE cos «0
Fao Athy os «06
Further, yt Fy + Ay 2
Fat Font © Rao)
o Fy t1-Fy 2 1-06=04
Asin Fat Fy ty
CRC ( Fao)
Fy 1 2 Fu = 1-085 = 085
i) By reciproci m Ar 08 «
{By reciprocity theorem: Fa + hy = es XO8S = 0421
oA a en
En Sify REE 104 « 0085

From the identity, F + Fa + Es =1
Es +1 FF
= 1 - 0.0495 - 0421 = 0.5295
EXAMPLE 8.10
‘Two areas 1 and 2 are inthe form of circular rings, coaxial and are in two parallel pany
ata distance of 10 cm. For area 1 the inner radius is 5 cm, the outer radius is 1 cm spain.

Somesponding values for area 2 are 8 cm and 20 cm respectively. Make calculations fore
shape factor between these two areas,

‘The following relation may be used for caleulating the shape factor between two circu
areas located coaxially in two parallel planes :

= hl ere]

E

¡14 BC and His

the distance between the two areas.
Solution : Refer Fig. 8.15 for the configuration and
nomenclature
Surface 1 = Ay Ay
Surface 2 = Aj 44
Fa Fas Fas 40
For surfaces 3 and 5

Radiation 6
Change Ba
men Surfaces,
, | 123
fee rl à
36-44) «0745
or surfaces 4 and 6
Rn Sem à Re Bem and
ER Bom and Ha 10cm
BaF 05; có,
an Kat ange out

= o ee]

1
= 7g (89-171) «036

‘Then from expression (),
E Fas

= 0765 - 036 = 0405
EXAMPLE 8.11
A large black enclosure consists of a box as show

ra in the adjoining figure. The bottom
suface 1 sa 530 K the op surface 2 450 K and al weed su
ack wall) are at 475 K. Find the net heat rones ia anne

rates Q and Qu
Slalom The net interchange of hal between wo sure nu
(doa + FA 0 TT)
(0) Net heat transfer ae from the bottom surface 1 ap ae 2:
From Fig. 82 with XL = 20/25 = 08 and YA = 23/2391,
Shape factor Fig = = 0.168
(des

68 * (025 « 020) x 547 « 10 (5801 — 450% = 18.03 W

BES

‘294 Bosics of Heat and Mass Transfer

to vertical surface 3
() Net heat transfer rate from the bottom surface 1 o vert
From Fig. 84 with VX = 25/20 » 125 and Z/X = 25/20 = 125,

Shape factor F,y > 0185 :
= (Qi "0185 x (025 » 020) x 567 ¥ 10° (30° - 475% = 1459

cree

Aa Ba 488 Fy

res
a

En ña

‘Thus 4% of the emission from the inn
ere and absorbed by it

From energy balance for the large sph

Fy * En "1 or Fug = 1 ~ Fy, "1-008 = 096

“Thus 96% of emission from the large sphere is absorbed by the inner surface ofthe sh
itself. The net interchange of heat between the two sphere i

(Qube = Fig Ay © (Tt 78

1» (dm x 0.0252) x 5.67 » 10° (50 - 280) = 3

surface of the large sphere is incident upon te

799 W
EXAMPLE 8.13
A25cm holc has been drilled completely trough a Sem thick metal plate that is maintained
2 uniform temperature of 500 K. Workout the loss of energy to the surroundings at 300
temperature. Both the metallic surfaces and the surroundings have black body characteris
Two ends of the hole may be teated as disks,
Solution: The arrangement ofthe system i shown

in Fig. 8.16. Let sufix 1 designate the cavity and o J
the suffices 2 and 3 denote the two ends of the
25 em diameter hole,
With both ends of the hole treated as disks
beth ends of the hole tested as disks, y, EAP o

1
2-05

With reference to Fig. 83, the configuration
factor Fa, is then read as 0.085,

Ratistion Exchan

198 Between
Considering radiations from end 2 ofthe hoj, Sortaces y 235,
Fu * Fat Fy 2
Now Fy 0 s the end conforms 1 à at surface and
yD Pn nest se ini

From reciprocity theorem

FOR * 995 = On
adopt same pecdur o my fs

RATTE + A Ares)
“AT era)

(because F3 » F1)
“Orr (x : eo

0025 x 0.05) (567 « 10%) (004 - 300% = 282 W

EXAMPLE 8.14

Solution: The evaluation of shape factor for such cases

ee
a aia T
Fee 7

ATA +A, R

El
|

+ AF
= As Fay Ay Fig + Ay Fr + Ay Fy
(As Fog + AL Fad (As Fag + Ay Fa) D a

a +
Each of the configuration factor on the righthand side N >

ofthis expression can be read from Fig, 84. as they.

Aa

correspond to perpendicular surfaces having a common u
inerecion line. Ihe values are tabulated below
2 y
«es (A :
+ x A
2
ha ons
1
A E o
e 1 020
2
2 on
As 1 va
1

i

1236 / Basics of Hest and Mass Transfer

Substituting these values in expression () we get
Ay Fig #035 + 1 ¥ 026) - « ON + 1 x 024) = 004

os
Fan 00

1 83. HEAT EXCHANGE BETWEEN NON-BLACK BODIES

83.1. Infinite Parallel Planes.

‘he analysis of radiant het exchange between two non-back parallel surfaces sha y
based on the following assumptions

(0 The surfaces are arranged at small distance from each other and are of Squal an,

{hat practically all radiation emitted by one surface als onthe other The cour
factor of either surface is therefore unity

(5) The surfaces ae diffuse and uniform in temperature, and that the reflective and emis

properties are constant over al the surface "

(it) The surfaces are separated by a norvabsorbing medium such as air,

Tre surface 1 emits radiant energy Es which strikes the surface 2 From ta par @ à
absorbed by the surface 2 and the remainder (1 - a,) E, is reflected back to surface 4 La
reaching surface 1, a part a (1 - 0 E, is absorbed and the remainder (1 — a) (L= 092.3
reflected and so on. The amount of radiant energy which left surface 1 per unit time

a Bes +05 1a) 0c) 6,44 (1-0) aa + ]

=E,-0, 0-09 E, [1+(1-01)(1-0:)+(1-0,
=E 0 =a) EE PAP ad

where P = (L- a) (0)
ain E een,

(agen:

(=a 0g",

DR met

Fig. 848. Radiant het exchange between two nor-Back prall surfaces

since P is less than unit

the series 1 + ps
en Pers

(06 FH absorbed andthe eit = a9 (ay Ee ena lc

et oy
‘Te net heat flow from surface 1 to surface 2 per uni
À 2 per uni imei then given by
area Area “m
From Stefan-Blotzman law or non-lack sure
Beat and Ez = ot
Te,
2)
on
where,
—s8 1

A erg are
and is called the interchange factor for the radiation from surface 110 surface 2

83.2. Infinite Long Concentric Cylinders

Consider two large concentric cylinders of areas A, and Ay emissvites and e and
‘hei surfaces maintained at temperatures 7, and Tz respectively

From reciprocity thvorem : Ay Fy = Ay Fy

The inner cylinder is completely enclosed by the our cylinder and as such the entire heat
{ieions emi by the inner cylinder ar intercepted by the outer ender be Fy = 1 and

1238 1 Basics of Host and Mass Transfer

Inn Ba nu ct fn

Surface 1 absorbs %

EXPERTEN BER

A fi 4
Surface 1 reflets ELE DE Ed ge = Eye) [1-1

It can be shown that the energy absorbed by surface 1 (inner cylinder) on the se
reflection would be
2 Alı-

sa,
Continuation of this process would show that total energy Jost by inner cylinder per uit

ne)

efes

Le q-eli-a-elr

sw

Radition Exch
ange
similarly the heat energy lost by the ou Botumen Surfaces 1 239)

pe a woul nt
ml)

q eu
nee
heat flow from inner el
er cylinder to Outer cylinder i
‘ inder is hen given by

From Stefan-Boltazman law for non-black bodies,

Bear and Feat

613)

change factor or equiv

yünders
Exuation 8.15 is equa
of equal length

isthe in
Jorg concent

alent emissivity for radian heat exchange between infinite

applicable to concentric spheres excep that for concentric cylinders

ERE

A ni dr
AL der Y
ale)

433. Small Gray Bodies
Consider two soa
repaire 7
the distance be

11 gray bodies having emissivities €, and € and absorptivities a and
he small size af the Bodies does signty that their Size is very small compared
teen them. The radiant energy emutted by surface 1 would be party absorbed

BAD / Basics of Host ond Mass Transfer

by surtace 2, and the umatsorbes reflects portion would be lost in space. It wi
reflected Rach to surtace 2 because of is small sie and large distance between the ti

Energy emited by body 1 = 4 6, 0, Tf

Energy incident upon body 2 = Fyy Ay 6,0, TE

Energy absorbe by body 2= a, Fiz Ayo Ti

Since a, = € he radiant energy transfer from body 1 10 body 2 is

Q = 6A, Bo

Likewise the energy transfer fom body 2 to body 1 would be
Qs = 61 62 Az Fx TS

‘Therefore, the net radiant Neat exchange between Ih two bodes is
ash ana

From reciprocity theorem Ay Fy = Ay Fy,

A no
sur

Qu akut)

= fa A Fo (Tt TE)
where fi, €, € respresent the equivale
heat exchange between two small gray bodies
83.4. Small Body in a Large Enclosure
The large gray enclosure acts like a Black body; it absorbs
incident upon i and reflects negligibly small energy back tothe

entire radiations emitted by the small body would be intercep
and as such Fu = 1

Therefore,

Energy emitted by small body 1
and absorbed by large enclos
En

815
À emissivity or interchange factor for radin,

Practically all the radiation
all gray body, Further
nd by the outer large encon

2 AT
yy emitted by enclsoure
Energy incident upon small body
Energy absorbed by small body

= AGO Tf
Ea 007

“m AGO
“Aa

Gecaune a, #6
Net exchange of energy
Qn = 51 41 0, Tee: Ar y 0, TÍ
Considering the trivial case in which the small body and its surrounding large ent
are atthe same temperature, be, T, = Tz and Oy = 0, we get
Aj = Az

and so Qn = A, (rt -

ha Ao TÍ

-1!) an

¿here fi, = «y represents the equivalent emissivity or interchange factor for radiation fat
exchange between a small body and a large enclosure
While calculating the

dant interchange between two gray surfaces, both the intescaret

and the geometric fa
tt i eT Oy ar cose ed ng
tom ie ang
uch Pac
FO Ar)

eu
EXAMPLE 815 »

md, parao, infinite pi

e oppore, pu plane ae main
TS tus between thee plane nes
A Doce tater which pat has uc a9

st 420K and 40 re
emissivity of 0

(9 the temperature difference ls doubled by ri y
te planes are assumed tobe Back? 7 NEM empertune 430

sein The ate of het erchange Dee o pass gen by

Qu Fon Ao (1 11)
‘The gray body factor (Fy is

A
For infinite long parallel planes which se each eier and

nothing else,
Fel and Ay= Ay ne

Eee

as ww

059 » 15 567 x 10% (480 = 420% = 73486 Wim?
Iisimmaterial which plane has which emissivity sin
temperature,

TET, = 510 K and the su
Qı 7 » 10%) x (540 = 420) = 190955 Wa?
u Dit the heat u increases by factor 0180958/754.6 = 243 when the temperatare
Ihe surfaces are black,
Qu" Ay 0 (T= 7)
= 1 (567 x 10% x (4808 — 429 = 124552 What

ce the emissiviies are independent of

ces he
059 «1%

© the given emissiviis, then

EXAMPLE 8.16

Distinguish between the configuration factor and the interchange factor

Determine the radiation heat flux between two closely spaced, black parallel plates
Bene 0 sch te their empresa 5 Kad respecta Rea
{he heat flux presuming that each ofthe parallel plates has an emissivity of O3 in each ese,
(he plates have an aren of a ii
Fito : The configuration factor F, considers the orientation and geometry of the Mack
ME surfaces; how the two surfaces view each other and to what extent the two surfaces
Kanals Solty to each other. The interchange factor f allows for the departure of the two
‘urlaces from complete blackness a function of the emits af dhe Po surfaces.

242 y Basics of Heot end Mass Transfer

For the black parallel plates ra
eshange i

ing only to each other Fa = Hand then the ada hey

Que Fa ies (1-7)
an ax (47 10 7 000-2
For the gray surfaces, the het exchange is
Qu = (Fhe Ao (M1)

For the given configuration of parallel plates which see ea
Ey, 2 and A nA,

no x 107 w

y other and nothing else,

Qu = 0.333 « 4 (567 + 10% (850 - 425)
lt may be noted that ifthe emissivity ofeach plate is one-half of a black body, heat fx
is reduced by a factor of 3.

Dawson —

Solution : From Stefan-Bolteman law E = 60, T*. That gives

nat and road
Ae
Tat gives Ten

“The net radiant het transfer between two infinite parallel planes is given by

Oj En AT)
Since T, = Ty we get:
O
M0
“Tat, the ne et flux Between the planes is zero,
EXAMPLE 8:18

A thermos flask has a double walled bottle and the space between the walls is evaconed
#0 as to reduce the heat flow. The bottle surfaces are silver plated and the emissivity of

Pasan Ergo Sem tcs 23
a ha ea tomos
be required if the same insulating effect isto be achieved by the una oy rest) moulé

I use of cork ?
jation : The rate of heat interchange between the two bot
bret jr the two bottle surfaces is given by

Qu = In), A OT 14)
he gray body factor (Fu i equal to

cesT ern cers
for finite Long paral planes which cach ther and nathing eee Fy = and A, =,

1
ue
ha EEE

ah
ne 1
Vernet 7 TOUS yaoi 7 001266

ip = 001266 x 1 x (567 » 104) (375 3004 = 838 W
(0) Let 8 be the required thickness of cork, Then

wrath)
5
ora = MAL a)
F
ma
be = © 0.268 m = 26.8 cm

5
EXAMPLE 6.19

A 250 x 250 mm ingot casting 1.5 m high and at 1225 K temperature, is stripped from it
sold. The casting is made to stand on end on the flor ofa large foundry whose wall, floue
nd roof can be assumed to be at 300 K temperature, Make calculation for Ihe rate of radiant
heat interchange between the casting and the room. The casting material has an emissivity
of O85. Take Stefan Boltzman constant 9, = 567 = 10% Want Ke

Solution: The rate interchange between the ingot and the oom given by
On = der AG (RH)

The configuration corresponds to a completely enclosed body, and small compared with
the enclosing body, he, Ay <<<A, and Fy, = 1 Hence

(ae + Win + (AE x Aga

1 =
ar 08

a"

Arca Ay ofthe ingot radiating energy
Ay = (025 = 025) + x 025 % 13) = 00625 + 15 = 15625 m?
yy = 085 x 15625 x 547 x 10% x (12258 - 30)
a2 10 Ww

244 1 Besics of Host end Mass Transfer.

en

nl), M0 (1-7)
where the suffix 1 and 2 denote the conditions at the tank and the enclosure respective

The configuration corresponds 10 a completely enclosed body, and small compare
the enclosing body, ie, Ay <<< Ay and Fig = 1. Hence ih
1

Cadi” Ge fers We ea AVA
a pate
(ayers
‘Area A ofthe tank radiating energy,

O =geradens2 (S302) «rot

Oye = 08 x 1.96 x (567 x 10%)» (350° - 2954) = 660.86 W
Some aluminium paints have an emissivity of about 0.3 and a coating of this paint on he
tank would reduce the heat loss to

03
ae x 23. (ll other parameters remain the sam
660 x QE = 24782 (al ther ps he same
Percentage reduction in heat loss
660.86 ~ 247.82
7 682

= 0625 or 625%

EXAMPLE 021
An enclosure measures 1.5 m X 1.75 m with a height of 2 m. Under steady tal
conditions the walls and
tel radiation to floor.
+, femssiity of ceiling and wall) = 085
€, (caisiviy of float) 075
Solution The at of radiant heat exchange between the eiling and walls (su 1) and de
Shor (sfx 2) is given by
Qn = (4), A 00 (Ti = 7)
‘The gray body factor (Fs is equal to,
Four

AS per given data

egulliiun
Ing are maintained at 525 K and floor at 400 K. Determine be

Tralee What Neo

1 total area of four walls and ceiling. R
220175: 2) + 2 (15 ¥ 2) + (15% 175) = 15.625 m

Padaton Exchange Ge

Ad een Surface y
A À oo = 15% 175 «2695 ap 248

‘The floor is completely enclosed by the ws
FR by the walls ad eng
From reciprocity theorem: Ay fi = Ay Fy

EXAMPLE 8.22

„a
Bastei
Eee 1

(SST ne ET Er

1
"0176659524 Tan = 023

Oz = 0323» 1568 » (547 x 104
ren
on 67 10% «cas

= 400)

<oresponding values for the hemispherical rol me Ta gy ey OF 05. The
radiation heat transfer from the roof to floor, 25, Determine the net

Solution: Refer Fig. 831 forthe conf

The net

ao
a= (Ey Ae ont 14)

where the suffix 1 and 2 denote the conditions a the Naor and root respectively, and

au
That

For

The

the radiations from the floor

Be (softs 1) each he zoo (fix 2) and hence Fy = 1

Ga

the given configuration a
ra tm

Opa = 0286 x 1 » 587 x 10° (800° 12001) = - 26981 War?
egalive sign indicates that heat flow is from roof to Moor Le. the far gains the heat

245 7 asics of Hoot end Moss Transfer
Radiation
ion Exchange Between

EXAMPLE 8.23 ds
; IR ephere of 50 mm ouside diameter and witha surface temperature of 600 K is tocatgg ute
Acteeometrc centre of another sphere of 300 mm inside diameter and an inner sup Oa ey Aa(rt 79)
sour of 300 K. How much of emission from the inner surface ofthe large Spheres? DH
rot {G9 The situation corresponds ny oes 2 D * 567 = 1108 (5
pond to relatively 5002008 uw

incident upon the outer surface of the small sphere? Also calculate the net intere
i af ange
Amal ag

eat between the two shperes.
‘Assume black body behaviour for both sides of the two spheres Am,
ere ar incident upon and absorb y ® 3

22005

0196

Solution : All the radiations emitted by the small pt

BE body completely ende

Surtacas 1 247

inca and be meri o bo
dei dla! het tw Utne che
Ona (r-T) Callate the surface emissivity of ua À

the sphere sel
[Net interchange of heat between the two spheres is

EXAMPLE 8.24 —— _ _ _ _>»E->E>A=AXA o
A steel tube, 5 cm outside diameter and 2 m long, is at 500 K temperature. This tube is rz = ia Alt 19)
recited central in () à large brick room having wall temperature 300 K and ( a quer where the sufi 2 and 2 refer tn
Tick conduit of 20 em side and at 300 K I the emissviies of steel and brick are 08 al
255 scopectively, make ealeulatons for the rate of heat loss by radiation from the tube in

teach case and comment on the results
ution à The rate of radiant interchange between the steel tube and brick enclosure i gie

For the given configuration,

a
Accordingly

38

& Oj = 0.229 x 0.2526 = 5.67 10% Gt 3009
Te ve sin implie thatthe low of heat i from ouside 1 nie
nce the heat flow isto be reduced by 50%,
Qu = (- 08 » ( 2948) = = 5.95 W

(The configuration coresponds toa completely enclosed body and much smaller compet

to the enclosing body, ¿e, Ay <<< A, Hence

the inner surface of he large spre. Accordingly A = I AN
From repre theory Fig Ay Fy a e
Fy o Ea = 00078 du
a gag Pg Oe a 7 0298 x (x x 0.05 » 2) x 567 x 104 « (5
Obviously 2.78 percent of emission from the inner surface of the large sphere is incite, Comment ; Radiation heat transfer depends GR - 300% = 768.08 w
upon ike ml sphere and absorb byt Pena ol on the tbe ac and he cond
Further. rom energy lance ofthe org sphere EXAMPLE 025 it en
Testy el oe Fe 2 fa 2 1 2 0078 + 0977 20 mn dlameter serial coin weh ,
Obviously 97.22 pereent of emission from the large sphere is absorbed by inner sure y paint = 90 K) is imvulated yy 88 Qui nitrogen under atmos
"Sphere of 5.0m dameter The nerseing sans a. Onkel wähle sate

The radiant heat flow between
(wo concentric spheres is given by

and outer container
respectively andthe gray factor

by
la A (Tt =) a
where the suffix 1 and 2 denote the conditions at Ihe tube and enclosure respect m Arm and an 202006 m
fas y
Was) *045

029

248 W

1240 / Besice of Heat and Mass Tronster

aut)
Edda
0.2826. (56710) «(50 = 308)
— Us

0282656710" (tu)

-28

\xaus =
)

8 = 2183 + 0445 = 22275

EAS 00689

8.4. RADIATION SHIELDS
Many situations are encountered where it is desired to reduce the overall heat transfer bet
eo radiating surfaces. The task is accomplished by placing radiation shields between te
‘emitting surfaces.

E

N

| am”
& 4 # &
EAN me Ad AA ANNE
ce Le =

Fg. Bat. Heat exchange been 10 para panes went aston shin

‘The shields are thin opaque partitions arranged in the direction perpendicular to
propagation of radiated heat, and made of materials of very low absorptivity and hi

{thin sheets of aluminium, copper etc) The shields introduce a sort of additional resistance

the heat flow path and accordingly the net heat flux is reduced

(@ With no radiation shields, the net heat exchange between the infinite parallel planes

given by
Oa (ly hair)

Both faces ofthe radiation shield have

ineciectity expression B29 vas

Aov (Tt
Year
62

Fig 822. He a E x E
oh na pps o

been assumed to have the same emissivity. implication

+ 1830)

{B50 4 Basics of Mast and Mess Transfer
ach side of expresion £29 represents the heat flow through the system. Substiut
Salve of Ty in lt hand side of expresion 829, we obtain
Ao, (1-18)
(22 bl 4
IO
“The ratio of radian energy transfer with ome shield to that without any shield is o
from expressions 8.28 and 8:31, That gives, ne
Radiant energy transfer

8 the

dee (63)

Mia 7
Wiboat shied E 1 ex

When €, =e € y Ihe above fraction takes the value 1/2, Thus by inserting one shied
eut transfer rat is reduced to one-half of the original value. The corresponding tempera
T, of the shield attains the value =

Less sry
me;(m+t) 83,
The electrical net works for the system with and without a radiation shild have ten
indicated in Fig. 834 and 835
Heat exchange without any shield

a

a

to with a rion

‘Comparison of expressions a) and (b) also shows thatthe heat flow r
shield becomes just half of what it would have been without the'radiation shield
If mradiation shields are inserted between the two planes, then
(0) there willbe two surface resistances for each radiation shield, and one for ec
plane. When the emissivity of all the surfaces are equal, then all the (27 +
resistances will have the same value (1 - €)/€

rite
y suit

viously the tla resistance ofthe phys

Sem wi be

R (w= shields) = Qn + 2)

iia :
od erre the eat exchange gr AT)

“

comparison of expresion (and (dc ind i
sue ant heat taste By a factor of (as OA UM te Prac ai reduces 1

EsaNPLE 826

Pa parallel square plate, each 4m aes re gr cope
er Ove late as a temperate of St K and ta on A
dis temperature of 300 K and a surface emissivity of 99, reg 0 06 Ml he abet
“ation between the plates. Hy 0103, Find the net energy exchange Cy

ita thin plished metal sheet of surface en

cena between the fo plates what wil bet
‘er would be altered? Neglect he conver,
esignicanee of thin exerce,

Satan The ae 0a Iterchange beeen he to pat ven ty

Qn = (6), A 04 (77 73)

For infinite long parallel planes which see each
hee

{04 gap of mm separaten

E

ity 08 on eth side ow oe
steady sate temper? oe ee,
16 eget any. Come at

‘other and nothing ee, Fy 1 and
1
Bee
1 1
Ves Ye 1% Yoryoo-t ~
p= 0562 x 6» 567 x 108 (00-209
= 511758 Watts or SLATS KW

{i Let sufix 3 designate the sheot which has been inserted between the two plates.
Heat flow from plate 1 to sheet,

Eu =

RAR

On = (5), Ao (rt)

ne Dom
Op = 0.09374 A, 0 (800-74) #
Heat fow fom sheet to plate 2,
Osa EA (18-18)

nn

252 # Basics of Heat and Mass Transfer

1
a We
Qs = 0.0989 A, a (13 004)

Under steady state conditions,
009374 A, (800 -T$)
Recognising that Ay = Ay we get

6108374 (00° -74) = 00089 (13 00°)

9.09374 x800° + 0,0989 «300°
KT}

Ty = 6765°K
‘Therefore equilibrium temperature Ty ofthe shield is 671.65°K

‚Any ane of the expressions () and (@) can now be used o workout he het itr
between the plates: a

= Heat flow through the system,
209378 x 4 x (567 107%) x (800 = 671.654)
= 438166 Watts or 4382 KW
epa ant 3 radiation shild reduces the radian heat transfer by à factor y
Tre = 11675 times Screens ae thus paced between surfaces tol down he eg

| ermometer ora thermocouple

Now be

1
rs * 00

89 4 ant sn)

EXAMPLE 8.27

Two large parallel planes with emissivity O are maintained at different temperatures od
change heat only by radiation. What percentage change in net radiative heat transfer monta

ceux if two equally large radiation shields with surface emissivity 0.04 are infrodu ea it
Parallel to the plates ?

Solution; Case () When shields are not used,
O (rar (7)

1)
AE
7)

Ant

as A, Ajand Fy = 1

02C where C= A (7-7)

Case (i) When two radiation shields are used

man.
Fn Ein Fat, we ge

0088
= PC 0.0098 on 5,
a 1%

EXAMPLE 8.28
Determine the net radiant heat exchange per m? arca fortwo infinite parallel plates held at

Kapeature of 800 K and 500 K respectively. Take emissivity as 06 forthe hot plate and 04
forthe cold plate

laced been them if
Wat should be the emissivity of a potished aluminium shield pl eh
et ito Ve reduce to 4 per ct of fn vale? Red eee De
Selbrium temperature ofthe shield. m

Selon: Refer Fig, 823. The rate of het inerchage between he two aes

CRAQUE)
ee the ray by factor (is

254 7 Basics of Heat end Moss Transfer

1 ET (ag

Y)
Y RL à 0 oy
fa lq Ja x BOOTY = 40, 2003.55, igo
Fo ft o ua plats wh ee chr a in an . RES o
Au 4, Pe

That gives: Ey = I
e
Qu = 318% 1 x (367 x 10% x (800% 5009 6200 W/m?
(Refer Fig 836. When a radiation shield with emissitivity e, on both sides ja
between the two plates, then ia Met pao

Fa

ADR 101580 ow yg

1
ons

TS
TS o y

«10

iin oe tin ten on et
jon: The net radiation heat transfer between tac
BE EEE a ane

Surface emimivinos
ween the surfaces fo

Ao,(tt-1$
For the given arrangement (Qn) one té = Na]
A, and Fa * Fy Cane re
That gees oa
octyl? ee Be
mE as A TE

lates separated by radiation
Since the radiant heat interchange is limited to 40% of original value, ¿cis of emissivity En Emm, we get by

Loco an 1 «(867104 «0m - 0 Aout)

Vesve,+2Ÿ Ve,

$ Oi) ais

for = D (no shield)

aa (ttn
Accordingly POE
| Asper the given condition, *
Qu) with no shield
(Qa) withn=stveias "75
or 2796 Ye, + Ves +2nfe, =(n+1),
A 2 Verve a "7
& E none 2167-25 «5706 um
2 | e CET
Desired emitsivity of seen e, = <2 = 0347 | 1354128440
(ii) Under steady state conditions. = 154181
Qu Qu" Qn | e 27
TE) 1x5.67x 10% (800* - 73)
e agg ME See ne ASS ag
CE El

Hence tree serens ae required to reduce the hat exchange By a face 7.

258 //Bosics of Heat and Mass Transfer

EXAMPLE 8.20
A 19 mm outside diameter pipe cares ryogeie lid at 100 K temperature Ano
of 13 mm outside diameter and at 280 K surrounds it coaxially and the space mi Pipe
ows completely eracused. Determine the aan het low fr 3 m length af see
Surface emissivity for both surfaces is 0.2 Proceed to calculate percentage redue ft
flow if shield of 115 mm diameter and 008 surface emisci

Solation : The rate of heat interchange between the inner sls 1) and outer su y
is given by Pipes
rp = 4 0 (1-7)
The gray body factor (Ea is
1
A
E-) ale dz
For the given configuration : Ey = 1. and therefore
Ao (1-18)
ES ( 14
€ Le JA
For 3 m length of pipe, the pipe arcas are
Ay = Rd = x x 001 «3 = 00942 m?
Az = = 0.013 x 3 = 0.1225 m?

Ai om

Eon“

(0.0942 (5.67 10) x (100 - 200°)

a
CTS

The negative sign implies that heat flows from outer to inner pipe.

di) When a radiation shield (sufi ) 1 placed between the two pipes

Pa ETES MEN SE TA AT
Ez a
Her F «Fa = 1, and trio

Eon“ Ze
ale

A, (hild pipe ares) = x x OO115 x 3 = 01084 m?

SA) fen

(15-0.

051-2)
TEE TETE TT
Hence percentage raduction in heat flow

3388-076
BO a
COEFFICIENT OF RADIANT HEAT TRANS e
© rh convection RAND RADIATION COMBINED
alte bet exchange between br syste ets
pied equation Hen (sucess ener calcd fm te
Oh AT)
tater han rom the elation 0
= fran (rt rt) o
Comparing thes denis

hatin où Tue) am
Th actor I called the coefficient of radiant hat tanfer rom sd sod an
‘mest in W/m de temperature deren been te enced oa See 8
Te value of h, can be calculated from the hat Aux eqeton for any contigs fey
ta, the val of forthe case of two large prall plat wold Le
CIEL

A

Teme

Small)
Dares

‘There occurs simultaneous heat exchange due to radiation and convection in many situations.

ig imports cha

(9 he heat loss from a hot steam pipe passing through a room.
(9 the heat loss from hot combustion producks when they pus through a cooled duct

W849)

k

258 7 Basics of Heat and Mass Transfer

‘The total heat transfer by both convection and radiation is then obtai

ined by the
superposition of heat fluxes due to these modes, That is nee
95004
Fora hot gas at temperature 4 Pasing through a dut with wall temperature ye
write + A
Th y= ty) + I TR}
meth) (fe)
tet) y= hy (aay
‘Te radiation heat transfer coefficient is a strong function of temperature in con)
convective heat transfer coefficient . wa
EXAMPLE 8.31,
EUR Liha race temperature of 480K i kept within a are ncorure whore a
Mantes 1 min he pipe surface to be black, aaa the coeticent ol arg
Sea rgd he het wansfer coefficient including the effect of radiation and concen Ma
349 Wa deg, find the convective heat transfer coefficient ne

Solution : The rate of radiant interchange between the pipe and the walls of the enclosure ig
Quo (m1)
=567 x 10% = 1 x (480% ~ 3804) = 1827.6 W
h, denotes the coefficient of radiant heat transfer, then
Q“HA (T,- To
1827.6 =, x 1 x (480 - 380)

1827.6 A
or BES. 1828 mee

Further,

Pratt Mg + Hn

Mn "349 + 1828 = 1662 Win?-deg
EXAMPLE 8.32
A hot water radiator of overall dimensions 25 m x 125 m x 025 m and having sur
temperature of 335 K is being used to maintain the room temperature at 290 K Calcule le

coefficient of radiant heat transfer, convective heat transfer coefficient and the heat ls
from the radiator due to both convection and radiation.

‘The radiations are considered close to that of a black body and the convective bet
transfer coefficient is prescribed by the relation
hogy 2132 (AT deg
For convection, the actual surface area of the radiator is twice thea
Solution: Area ofthe radiator

2 (ab + be + ca)

= 2 (25 x 1,25 4 125 x 0.25 + 025 » 25) = 8105 m?
Radiant heat loss from the radiator to the room,

Que = OATS TE)
In terms of the coefficient of radiant heat transer,
Mas A Ty = Ty) = ACHE TE)

ofits envelope

FE Exchange Bar
CE FEN SET en
#567 2 108 use

een Succes 289

un.
ant heat loss from the radiator D GRR à 230) « 666 +0
asian

En

Ey ATT

= 696 » 9305 »
active heat transfer cocfient 8205 » 235-290) = 2538 we
Pa "132 (SP = 132 295 «rags

Lacie het loss rom the radiator,
° gg ABI = 863 3 (23 8105)» 635 299 = 2057 y

toss due to both convection and radiacion.

dd = 3977 + 2538 = 5918 w

Com
468 Wimék

Array in à brick duct (€ 09) of 25 cm id square sein and a M À namen
‘Serine the radiation heat loss from each men f the Resting sip
® ‘Assuming the system to be in steady state condition, calculate the surface heat transfer
E

Op = Ene AG (MTI)
vier the sufixes 1 and 2 denote the conditions atthe heat pipe and the bik duc respectively
“The configuration corresponds to a relatively large body completely enclosed; the inner
Andy cannot see itself, Ez = 1 and hence

1
1/R+0-e)/ex A 74
1
LESNZIEZET ER ZEr A
ib
“Tele - DXA
Suse area per metre length of he pipe, ,
And] = an or 1r 056m
Sure ara per mete length ol du,
Ata
. 1 ==
Cda" TOO ROSS TEO

war

07616 « 0565 x 567 10% (54 - 30
ee endy sat conditions

LA T, denote the atmospheric temperature, Then under tad)

QA)

and Op

i

260 // Basics of Heat and Mass Transter
10444 =1 G00 ~ 285) or h=6927

Heat transfer coefficient h = 69.627 Wa? K
EXAMPLE 8.34

ata RP

vin

so arranged that heat

by conduction slong the wives oF the thermocoupie mt

0.85, make calculations for the tr vera,
Take radiation constant for

black body 0, = 547 x 10% Wimikt
Solution ; The arrangement of thermocouple

for measuring the temperature of gas flowing
2 pipe line is indicated in Fig. 824 There à

such, the tmperature indicated
by a thermocouple is laways less than the
actual gas temperature,

Convective heat flow from gas to
thermocouple, ee

AA Ta Ta) Fi. £24. Topert measure by ama

5465 AT, Frag)
Heat radiated by thermocouple to pipe wall
= FAO, (Teepe Ta)
The configuration factor E, corresponds to a completely enclosed body
and small compared with the Enclosing body (pipe lng
1
SS
- = ERBE.
(aero
Under steady state condit
flow from gas to thermocouple,

Chermocouy
ite Fa + Land A, ceca, 0

6-0
thr ss an uri ben e cometer
and deberás y moco ne ope
5465 A (Tao, ~ Tao) "0.85 AO, (The Ta)
Inserting he propia
Ty 75» BRITEN ea) «rr

Tho as temp 7, 274

Nate The hn ated by te ge oh muscle very small compr i

ee hat How nd eS
ELE ass
Tre sie ade uses of hallo bik Hing of race we wo be apt TK
Pepe bythe placement aan ssn een blow pee Ba

‚and radiation. The surface et fürnaee ï pres
BE rum

equ
on: Under steady tte conditions the a
ution eat transfer from the Lan.
ar ii equal the eat tano from the pa
e «cion and radiation.

ci

{Convective ent los Qu o surounding,
ns (TT peru area
45 (T= en,

2145 (80 30D 9957 y 98
(o) Radiation heat loss Q, o surroundings,

et los the
i Outside sure of the
1510 e songs by au

us is (TT) per ni arca

her factor

Oa Tea
forthe give artangemen Ay << 4 ad E
(Fin = = 086
Gus = 086 » [567 « 104 (800 - 00%
(a) With radiation as of equal emis
tet Alo tothe external Surface
1 alt)
ae
1 „567x100 30) soe
we ajo a Wot

“1 and therefore,

ies, the het lo rom intra surfe ofthe

Bon energy balance
as
N ums or
Baden shields required nen 13
OaupLe 835 =
Dry standard steam at 5 bar is carried within a steel pipe of 10 cm outside diam
Simic inten enla neni
S204 Wa and 9 vespa ihr seedings et nde
it acter ME ea dee iss Hee make
ion oor at enon tration en te hat
radiation and overall heat transfer coefficient and (ii)
Sesa from 49 mete length a he pe eng oe ie
Ion: (The heat oes by conduction and concn om one me gh
Ste worked ou by ung De

Q

métiers

dh
MELDE AC NOTE)

262 1 asics of Heat ond Mass Transfer 5

rere f is the temperature of dry saturated steam at 5 bar pressure. From st
isst

a 1518-

2 aire,
1518-25

= gg 7 13686 W/m length of pipe

TUE)

For finding the temperature on the outer
surface of the insulation

Qe
BETT]

una un

1518-1, i
0.788 Ml
ty 0950 \
„an bat os rm he ete N
surface ofthe pipe tothe surroundingsis given Suowehoy at
by the relation e
= (Fy), A00.(11-13)

where the factor

13686 =

Flo bas,

1
le” NARA

¡e in the atmosphere) corresponde tothe situation where

The arrangement (location of
Fa = 1 and Ay <<< Ay, Therefore,

9

1
a Taverne
Q,=09 x (nx 02 x 1) x 5.67 x 10° (316.958 - 298%)
68 Wim length
(0) Leth be the radiation es coco Then
Q,=h, As (= 4)
rsh NR Wan D (898 2)
h, = 5.939 W/m?-deg,
io colin.
2 hy + h, = 115 + 5939 = 17439 Wimi-deg
(The toss o heat rom 0 m length of pipe
Uran
= (9686 ©7068) 40 «ROLE» 8011/5
a bar pressure
2 2107 = 207 » 10 kg
Condensation rate of steam

94 9107 kgs = ir
Foro? 7394 * 10° kg/s = 1418 ky

Latent heat I, of st

the
aod nac a he &

ons for the (a) heat

E cseficient, and () convective he

gen: The geometrical

r= Sem

considering 1 m length
w

{toner steam film =

ds Pipe material The resistance offered 4
vie and the pipe thickness ig

cf sel 1 quite bi

(a) Asbestos insulation

(i) Outside air film =

‘The resitance of a ser

conductivity 0073 wet
ndings a

ini

composite system a >
the eet of MI

transferred

meen SE SPV nd

perme i
ter ene
of he age
CEE
Of Pipe the varus hama

ipe, quivalent. à
nits

dimensions
N76 Som
sistance to flow of Bent are offered

1
WA” Ra © Cons

small

C2) | tog tt1/6)
o 2

E, 1
WA” oa "0178

1 path is equal tothe sum of individual resistance,

LA 00516 + 1324010 = 150 dee
‘here heat os rough condition,

o-

fm

186.29 =

at 300-20

Sara wow

‘Thesame heat passes through each aye ofthe composite system, Then for he ide ir

2
0178

here gi the temperature on the outside surface ofasbests insulation

In terms of coefficient

= 18629 x 0178 + 20 = 586°C = 32616 K
of radiant heat transfer,

HALT, -T) =eA0,(19-7)

neo (T+7)(1 +72)
= 088 x (547 x 10% (2616 + 299) (SI +29)
= 5.938 wm,

Convective coufiient for outer surface,

814 - 5998 = 2202 Wades

264 1 Basics of Hest end Mass Transfer

£86, GASEOUS RADIATION
Some of the salient features of radiation from gases, vapours and flames are

1. Most solids and liquids have à continuous spectrum, i they emit and absorb yg
energy ofall wavelengths In contrast, gases and vapours emi and absorb radian att
eine ports ofthe spectrum called bands, and are commonly known as selective she
And emiers The number of bands and their widths, andthe monochromatic emisie na
‘of a gas at each wavelength within the bands vary for diferent gases and vapours PONE

Castor as vry wih temperatre, preto, tekst el fs volume a an
‘The total emisive power equals the Summation of he areas within the se OT
wavelength, nda

2 Only thin fayers (1 micron Lo 1 mm thick) participate in the process of thermal rad
through solids and liquids. Gases possess a much small radiating power and all tnt ween
participate in the radiation. ma

31 Monaatomic gases such as argon and helium and diatomic gases such as oxygen, ir
are extremely inert to thermal radiation. Their emissive power and absorptivity itso
they are considered practically diathermancous, Le, transparent to thermal fadiat
temperature ranges of common engineering interest

4. Polyatomic gases such ammonia, carbon dioxide, methane, sulphur dioxide, many o te
yatocarbons and water vapours re fay good absorbers and radiators over oran warn
ranges.

5. For very thick layers, gas radiation approaches black body radiation wikin the wave
‘of the band. The absorptivity of a gas is a fairy complicated Function of pressure, tempera
size and configuration of the gaseous region,

6. The following empirical relations have been suggested to workout the
of CO, and water vapours

to as”

mall ht
on ine

ceatssive pou

esse E)

where pis the partial pressure and I takes into consideration both the size and configuro
of the gaseous region; it is defined as

“Thus for to Infinite parallel plates a distance x apart, = 4 x Ax/2A = 2, The dation te
bean length for gaseous bodies of various geometrical shapes have been compiled by Hot
and the relevant graph are given in hand books on beat transfer.

The empirial relation sted above suggest that the emisiv power of gases and apt
deviates considerably from Stefan Bltzman's law which stipulates that Emi power
Proportional tothe fourth power of the absolute temperate.

CO, and H,O vapour are somewhat opaque to each other and as such when St at
present the total radiation becomes less than the sum oftheir separate individual ec

7. The radiant heat exchange between a gas at temperature T, and a black surface of
area A at temperature T i worked out fom the relation

Patio change Sete Surco 249,
2+ ACT a, rs)

‘The emissivity 6, of the gsi eva >
at te temperature of the surface MPA fe ga, and sor
ssid surface ofthe enclose i e

0 Back. ten com
tense rm he gas othe walls are made fom EN fe net a eat
ER am;
wher, lie mi apr ue en a
ofenand Lic, E, = (6, 4)/2 where, is the emis cca
' the emi le para al meras

* The average value accounts for the beams of ra
sated and fen sted by nie

and heaters. Most of the flames are luminous: and the eed Gel a ck
presence of glowing particles of carbon and ash, i u

‘oF rom pure chemical reactions wich occur
ein tempera, The et interchange o energy Demon fam and soa
given by

sation from the gas tothe walls which are

Qu «A Fu peut T5)
mer the subscript correspond lo
et fame enveope
The Mames produced inthe household stoves using eros or cis for be
purposes are non Juminows innate. The enn of energy Oy recia a

rain discrete wavelengths; obviously such flames cannot be treated either black uy pay
sas. o

coy
lames and to wall ofthe enclosure; A the area

REVIEW QUESTIONS

A1 Derie general relation forthe radiation shape factor incase of radio been mo sr.
“Two facing parallel plates radiating only rom their facing sides se y each oe Dt te eo
rectangular plates meeting at ight angles do not date solely 18 each eer How do You
{cout forthe variation in shape lator >

82 Show that
by here

1 quaniy of radian heat interchange beten two gray sulaces can be expe

e)

Tie notations ave te ua mening Ste al he assumptions made in dran is

How the relation gets moet when the surface ae parallel to each ober?

Using the definition of radios and irradiation, prove thatthe radian inter ange betwen two
say bodies ls given by elation

5

CCE
Ter Cee en Ver

CA

266 1 Susi of

se

ss

as

oat and Mass Transfer
re ws meanings
a ere were surface 1
ed im aha ae?

e at emperoure T

‘re notations Have t sos only another surface 2 Ho
How by

Give now examples of
mar es nod
en a small body oF emisiciy
entries € 0 (E = Tf). Express this result
Fa inte valo al radio het Water csi
Hof water (10 m x 10 m) at Com a clea

aan energy from a pond
tan eer elit nde of 20K TARE my eye

Toa rien constant
PATTES

a placed in à arg eneosute a
placed ina pe,

orm of

Neon

Workout he met ate cf
Deer wha th y ad
ae = 098 and the lc
(As. 17325
sma sage vs of exter ata 05 conti ar Can oc ie pl
Ae ea Oe. The tank puted Hck ca,
sri O8. How uc eatin m hat ss woul St ifthe outside sure og
ESS alu poi ol emisivity 05 7 N
Te reden constant 0, © 567 10 W/mK
Fe ian cm in mtr and wie soc perature LPG aed ie ah
AO cación 038 m = 035 anal walls a 435° Comparar a

at by ration

(05.2766 y

from one metre length of Ina tube

Determine the hourly Is of
Lars, 25

d+ 04) at 10°C (ce a brick wall 6 +07) a 300°C: Eine eg

A sirios sel plate (
a (ans 3573 Mi Waa

ee

À steam radiaor ih the
À elperung leí on the fr ola a

RAT omni 10% aluminium (6 Ó
ay. era he rate of est nerchange

Arge tan forming he ht we
Sere ses and sin spose

mm

A domestic hot water tank, 03 m diameter a
Ame uerommdings Ifthe tank surface oxidised copper
na eures ul estimation of eat hes x

tions for the adan heat
a ce temperature of BOC and an ambient temperate of 25°C

vest enredo nat on tb tank sure cen x coaing of luni
Mi wd 0327 Take radiation constant, = 567 = 10° W/mK
Kin: Area =2 (02) #287 (as, 086 DA

Dovngemerpney ater an acide the naked body of persons placed sn

aor ena a rn paves The og Strand: ebd rt

ai LDC and om msi 02, The skin temprano of he wc 23°
“ond bag are 11

ine ob emy is 08 and the elective radiating ares ofboth skin
ne erh rain? the Body can sna 2 200
temperature wl te bag give adequate protection ?

inc. Fed

À dend Hack cylinder of emisviy 095 is ep at 95°C in a arg enclosre at e
td a o pe nr er ls race: What woul te ation ls soe

an

Radiation Exch
Change Between
Surfaces 11267

tn onan oy mt Mac ray

(ns 56 and 245 Km

crs aan tee a
the engi can bat
ENS

slider ere surrounded by conce
‘al of emisivty B10? Take sot

two infinite plates mar
nine at cont em
plod por ach ter, Aether hn
Places par between them, Obtain on eres ee eth
Rested when ev "ey oad bo ernie à
17 Ep and 6) 2 0/2 0/2 What ml
Ath ci la e; on ah ide ziel
engen F
À te sid given by

peratures, and

8 placed between wo infin
and! Tree Way = tye oat A

reapers inte pars >

=)

ees 03 and 3 ar matar ot temperate SF
. $ LUIS spaced between Le emo

de) heat exchange per m? arc when the shield is presen

aaa

ras

Convection: Processes and

Properties

Thermal convection accurs when a temperature difference exists between a solid surface ang
a fluid Rowing past it Convection is essentially à process of energy transport affected bythe
itcolotion or ming ofa fluid medium which may be a gas, aliquid or a powdery substanee
‘The transport of heat energy during convection i directly linked with the transport of medi
ll, and as such convertion presents a combined problem of conduction, fluid flow ang
mining This chapter is concerned with brief introduction to free and forced convection: the
Fonction rate equation, the convection film coefficient and the different parameters affecting

the convection phenomenon,

9.1. FREE AND FORCED CONVECTION
With respect tothe cause of fluid circulation or low, two types of convection are distinguish
Free Convection Circulation of bulk fluid motion is caused by changes in uid density resulting
from temperature gradients between the solid surface and Ihe main mass of uid. The stagnant
layer of Hid inthe immediate vicinity of Ihe hot body gets thermal energy by conduction. The
energy thus transferred serves to nerense the temperature and internal energy of ud partes
Because of temperature rise, these particles become less dense and hence lier than the
Sarrounding uid particles, The lighter id particles move upssards to a region of low temperate
Are they mx wth and transfer a part of thse energy 10 the cold particles. Simultaneosly
The quo! Reavier particies descend downwards to fll the space vacated by the warm fla
Fast. The crculaion pattern, upward movement ofthe warn fluid and downward movement
SF cool fura, ls called convection currents. These current ae setup naturally due to gray
lune and are responsible for heat convection.

‘Designers of fornaces, house heating systems, architectural projets, roads and conce
structures wil be concemed with fre convection. Since there are no density forces (o gravitinl
Fila) in the orbiting satellites, Une space vehicles with a zero gravity tajecory, free convection
‘would be norvexistent in such vehicles,

Forced Convection : Flow of fluid is caused by a pump, a fan or by the atmosphere winds
Those mechanical devices provide a defini circa or the crulating current and tat spe
up the heat transfer rate. Examples of forced convection are cooling, of internal combustion
“engines, airconditioning installations and nuclear reactors, condenser tubes and ether hat

exchange equipment
9.2. LAMINAR AND TURBULENT FLOW

The convection heat is affected to an appreciable extent by the nature of fluid flow.
realms of fluid mechanics, essenislly two types of uid flow are characterised

Laminar flow ; The fluid particles move in flat or curved un-miing layers or steams au
follow a smooth continuous path. The paths of fluid movement ave w ved and he Di

inte

ticles retain their relative po even: Process ans races 269
to ebcily of individual uid elements fe

moet cca a
eg
te gn an
The turbulent flow resembles a crowd of commutersina ral rood ion au character
flow is governed by the following parameters à 7 o ees et

{characteristic dimension of the flow paca, for example the di
y, for example the diameter o 1e pipe

A grouping o tes variables results ino a dimensiones quantity, Vdp/y called ne
Reynolds number, This number represents he ata fiera 1 you ares, At fo Resol
pum the vous forces promise ná ts fw io At ih aes o Rema
number, the inertia forces overcome the viscous friction forces and consequently the fui
layers break up into a turbulent flow, = m

Tor Aid low through a pipe, tow Reynolds number upto 230 I indicativo laminar
fo. From R, = 2300 to 600, te-laminar lw begins ranita to turbulent low. Un
drow 1 completly turbulent at R= 6000

in many flow situations, the duct not cular bet i rectangular, trope

lar bu is rectangular, trapezoidal or even an

ulus formed by tube within ander tube In that case the characteristics dimension din
ihereation = Vdb/ isthe equivalent (hydraulic) diameter which deine a our ties
the comeacctional flow aca divided by the wetted perimeter.

cross-sectional low arca

Eauivaent dame , + au Sees om 493)
Thus for a duct of rectangular cross-section with length and breadth à
Pres
a Tine 2

a {IR anus (a low passage formed by a tube within ube) hasan inner meter (outer
ameter of inner tubo) of dy and an outer diameter inner diameter of cuter tube) of d then
the equivalent diameter is

4-4 e

33. NEWTON-RIKHMAN LAW : CONVECTION RATE EQUATION

Rares ofthe particular nature (ree or forced) the appropriate rate equation for he convective

It trastes between a surface and an adjacent ud is prescribed by Newton's law of cooling
armen 04)

270 1) Becics of Heat ond Mass Transfer

A i the sa ren exposed to Het aa,
TS: werte 0 {he nd, nd the ire temperate ofthe Reg 4%
Nadando pony hit fe hat er an me a tae De
Sperre deis Toe unt of Ware ng and tie eed oa a
Sr tient sois conductance or the Em een
"he va of cotcentr dependen upon
ati codon roughnet and eos
À geometry and orientan fe surta lat, abe and elinder
Some
«therm physical properties ofthe vi dent, vos
sn and Uma eoncucy
ature of ld Now laminar a but
2 toundar ayer configuration
2 real tera! cnrs

placed veria x

y. specifi heat, cocífici y

‘Typical values of convective coefficient are

f Free convection = Forced Convection

Oar” 3-7 Wark ”

(@ ar and
superheated steam 30-300 W/m’k.
60) gases 2-0 mE) 60 - 3000

(i) liquido 30 300 Gi) water 300 - 10000

Convection mechanisms involving, phase changes lead to important field of Boing
(evaporation) and condensation, The convection coefficients for boiling and condensation lee
the range 2500 - 100,000 W/miK.

EXAMPLE 9.1

A motor cycle cylinder consists of ten fins, each 150 mm outside diamter and 75 mm inside
diameter. The average fin temperature is 500°C and the surrounding airs at 20°C temperature
Make calculations for the rate of heat dissipation from the cylinder fins by convection when
(8) motor cycle is stationery and convection cocfficient h= 6 W/m?K (i) motor cycle
moving at 60 kayhr and h = 75 Win?K.

Solution : Since both sides ofeach fin ate exposed to the surroundings air, the surface area fo
convective heat transfer is
A= r0x[2 $ 0st -0.078)] = 0265 n°
‘The heat flow from the surface to the fluid is given by Newton's law of cooling
Q= la -t)
‘Therefore, when the motor cle is stationary

Q = 6 x 0.265 (500 - 20) = 7632 W
and when the motor cycle is moving

Q'= 75 x 0.265 (500 - 20) » 9540 W

EXAMPLE 9.2

Forced air flows over a convective heat exchanger fi

a room heater, resulting in a convecive
heat transfer coefficient 1.136 KW/miK. The surface

temperature of heat exchanger may be

Convection: Process [1
509 and Pr
«omidered constant at 65" €, and th peris 1 274

crite for 85 UW of hen #20. Determine he et tangy se
{atution : The convective heat flow from
Neston's law of cooling
TEA
Substituting the given data, we obtain
88 = 11364 (65 - 20)
Therefore heat exchanger surface area,
a8
Trees * 0072 mt

‘slid surface to the suroundig Aud ie given by

94. DIMENSIONLESS GROUP FOR FREE CONVECTION
The diferent variables specifying the system behav
represents the free convectiopn at fluid flow aver a

four have been indicated in Fig. 93. which
dimensions is M-LT-D-H system of units are

fat plate The physical quantités with their

[ Variable O Dimension]

[faa a an
Here : pa
Lee

5 Heo
Hid coefficient of thermal expansion Fa
Temperature difference wey f
Significant length i 4

| Heat transfer cocficient | ñ HT

; vs wee
o

nennen

urn

vun
Fig. 91. Dimensional analyse varies fo wo conection
‘The coefficient of thermal expansion, Bi prescribed by the relation:

1278 1/Sesies of Host. end Mass Transfer

pme + BU)
where py and pz are the Muid densities at temperature f; and 1, respectively,

Buoyant force = (py = pag = [pr {1 +B (t2-4)}-pa]s= px (Os at)

‘Tis suggests the inclusion of variables Bg. 87 into the ist of those important to y
natural convection situation, The parameter (By AT) represents the buoyant force and ha y
dimensions of (LT,

Buckingham s Method : 1 can be premised that the factional relationship is,

Fla Be cy Ba AT, = 0

There ate 8 physical quantities ( and AT are counted seperately) and 5 fundamen
anis: hence (85) or Steams, We choose Mud viscosity . hermal conduct I pe
(Bg 41) and the characteristic length 1 as the core group (repeated variables) with urhnos
exponents, and establish the xterm as follow

KW Be AD Ho
1 (MEA TPE x (HL TA Oy (LT (Ly MIA
Equating the exponents of fundamental dimensions on both sides
Mo: Oart
Loi On-a-bretd-3
To 0=-a-b-2%
9 : 0--0
Ho: ose
Solution gives se m=1 ; b=0 ; cn 1/2 and 43/2
Substituting these values in relation (i)
= (7 OP Bg A)? (HY? p

PP (Bg ar)” À 9° (Be aT)
ie y

The above step has been worked on the postulate that square of a term is alo nor
dimensional

Following the same procedure, one would obtain

mre Og aT He,

my tHe Bg AN
‘Thus the functional relationship becomes

[(PotfgaT ue a)
EM)
AE)

[Pos
o Eten
a = 9 (Gr m9)

17% a dimensones group called Nuss number
Gr "(PO RAT 14). à dimensiones group clad Grat number
Pr =i Gh, a dimensionless group cal Frand number |
it isa usual practice to rewrite the above correlation inthe fam
Na = C (Gr (ay |
The constant € ad the exponents sand bare evaluated through expert.
95. DIMENSIONALESS GROUPS FOR FORCED CONVECTION
‘The diferent variables specifying the system behavi
present the forced convection of Il flow over
Ih dimensions in MELTS system of unta a

jour have been indicated in Fig. 9.2 which
à Mat plate. The physical quantities with {

Ei a]
Passe | me
| nd am 4 a
Fi thermal oc 4 |e
| ad es cans | a i
epee sii ” |
| Flow vi me |
7 À
Het ner coles mess |

En

Fig, 92. Dimensional analysis vais o forced convection

Buckingham's x-Method. Wt can be premised that the functional relationship is
Fin Bo cy AL V,L

|

278 //Gasice of Heat and Mess Transfer
voy poe

+

De
panne
ID re

Tampere ro

|
= Von oty atic
CRETE

Fa pto

Omen ela a

Fig. 93. Velocity and tomporatur profes in convective hes! transfer

‘The Auid velocity decreases as it approaches the solid surface, reaching to zero (po,
condition) in the vid layer immediately next tothe surface. This thin ayer of strate fu
as been called the Aydradynamic boundary layer The quantity of heat traste sh
dependent upon the id motion within this boundary Layer, beng determined city D
thickness ofthe layer. The boundary layer thickness 8 is arbitrarily defino as the diac”

from the plate surface at which the velocity approaches 99% of free stream veloc, 0
Likewise a region of fluid motion near the plate in which temperature gradients exis sty
thermal boundary layer and its thickness 8, is defined as the value of transverse distan,

from the plate surface at which ,

-0%

4

p
At the plate surface, there is no fluid motion and the energy transport can occur on by

conduction. From energy balance, this heat transport must equal the heat transferred by conver
into the rest of the fluid. Thus

ef),

Heat flow rate is thus dependent upon temperature gradient atthe wall, and the temperature
gradient is influenced by the fluid velocity high temperature gradients are associated with de
higher velocities.

If temperature field of the fluid varies only in the di

plate surface, then
er es
BOY).

‘Thus, the convective coefficient /t can be evaluated from a knowledge of fluid temperate
distribution in the neighbourhood of the surface
Introducing a characteristic dimension J, the equation 9.6 can be recast as

Ali) 95,

ion ofthe coordinate normal tothe

[der +.) | on

EE]

‘The dimensionless parameter it
cay be interpreted sera aca Nee
temperature gradient, The paramere i

tame

ently the Nul number
"Can overall or rare.

+ Stanton Number St is the rato of hea rene

ee colin othe fy
et the vlc of te eis ow fat per unit
wo yt
st. Nw

9 asa] RR

Thus the

A1 should be noted that Stanton nuinber can
data. This ecos obvios when we oben ye nin correlating forced convection

gu Tie eee ‘we observe the velocity V contained in the expression lor
EXAMPLE 93

The temperature profil ata pancla Joan ma nam

by an expression of the form” ‘eemal boundary lye spreeibed
MISA Bye cr

where A, 8 and C are constants, S à

shee A et UP an expression for the corresponding het transfer
Solution : The heat transfer coefcen is given by th expression
{ar
ne E
GOW),
From the given boundary layer temperature profile

„dt
tease cg ; Spe acy

y
a
he Te

ho

nee

EXAMPLE 9.4

The temperature ie
nn eture profile at a particular location on the surface of plate

presribed by the

278 11 Basics of Hest ond Mass Transfer

TE thermal conductivity of ai is

tated to be 0.03 Wan
Sonvective heat transfer coefficient in

de,
each case. ©

determin the

va
Seton: Te cnc et rar à rl yt ey +, mis
\ Kia 2 flows over
ca coefficient i the we
mE tastes AP was at he a or aa
Dial, cae rn ad
sr incre nc ttre ed
à sin Ti hs owe M Nang p
® ten "ten '
CCD TEE" hein wes ¡
Mint |
Then from the relation: y= (dr) }
Tole],

‘Temperature gradient at the surta

(e) Ba

Ki.

De

DO = = em
Temperature a OS mm trom the ue is

= 804) 2 0.0005
4),
= 80 + (44636) 000005 = 60 - 22218 = ane
EXAMPLE 97

engen of at meaning 3 cm = Won wi rc ASS
Teeny Tess cs wing ar has hema comune DS ran eae
Seen 1635 10 a and ram epa conste he Nace eee nena
erate to be 425, Work ot the equivale dancer of We hay nee bem
$eyolds number and the convective hes fon odio

Ñ Salio: For a duct of rectangular rohen mb
cs + cum sr hydrlic diameters pre by as

2 2x04x0:

EXAMPLE 9.5 PR ones "OM

ng and breath the equivalent

Air at 20°C flows over a fat plate mal
at a distance of 05 mm from th
fair as 0.0266 W/m-deg, estimo

intained at 75°C. Measurements show that tempera Reynolds nante
1 surface of plate is 50°C. Presuming thermal condi

is equal to
ate the value of local heat transfer coefficient. a ee
Solution : The heat transfer cooficient is prescribed by Ihe relation nV "esse
Clearly the flow is turbulent in character

i) The Nuss
Assuming linear variation of temperature,

280 1 Basics of Heat and Mass Transfer

sa.

sx

34,

ss.

96,

97.

95

9.

920.

REVIEW QUESTIONS

Détente between machanims of heat transfer by free and forced convection. Men
(ofthe areas where these mechanisms are predominant
Exphin the phenomenon of hat traste by fee convection What frees control the id mo
{Can tee convection cur in space vehi wi a eros co ation
sin the phenomenon of ha transfer by forced convection. What fos conto the fy
maton ? Cie stable example to iste your acer Mig
ive general uation forthe rte feat amer y conveton and hence ein cof,
of heat transfer. List the various factors on which the value of ths coefcent depende TÜR
Tre forced convective het transfer coefficient for a hat ui lowing over a oo surface ha
ato 19 halle Te ld temperate parar! the ol surface a an
the surface i eld at 10°C. Deermine the eat taser, per uit surlac are, rom the fain
the sure, (Aas. 21450 kant,
ite em high and 30 em wide, having a surface temperature a 38°. contact wu)
AC. the observed convective heat transfer fate is 45 W for each side, compute the man
Sonvection soeliien, (Ans 1667 Wat fy
‘An let resistance heater at 125°C being cooled by rat °C What isthe average contra,
‘efficent the eat fl at theater surface i 5600 W/m? ? I the beat fox a ant
2800 W/m, what ve be he heat temperate? The convective colin remain at
(Ans. 86.15 Win" de soga
Define the Nusselt number How itis ated to temperature gradient in the fs immediate à
‘enact withthe solid surface? Menton the an apronches which fave suggest lr esmas
the value of Nussel number,
‘What tbermo-phyeicl variables ae involved in the equations describing the phenomenon e
‘Apply dinesional analysis to these varibles and develop a generalised corsation between cia
rondimensional parameters. Discuss the physical significance of Ihose parameters
List the vaiabls that affect the forced het transfer cocfficint
‘Using dimensional analysis, demonstrate that the following dimensionless parameters ae posite
combinations of the appropriate varibes desenbing forced convection

Vin som

Vee BE NE ang —t
eee GP

Name these groups and discuss their physical significance

ana

Empirical Correlations for Free and

Forced Convection

plicated due to
are, therefore, analysed through the

customary to evaluate the fluid propettes ene,
‘ea fl temperature, The mean bulk temperatur y denote th ae

Sat would result if the Quid at a erossscion was hood ana ea eae
For turbulent flow of fluids in duct, tht temper
peu ner the dit axis In hat exchange,
127 be heated or cooled during its low passages The bulk tempera a ee
the arthmetic mean ofthe temperatures lil and alex fom the ht hares ai
ie to 7 43/2 The mean fm temperate is de ore mea othe nee,
temperature 1, ofa solid and the undisturbed temperature 1.0 the Nard whch tone na
Beye et) = e

tre is very nearly equal to the uid
the fluid flowing tough the tubes

102. LOCAL AND AVERAGE CONVECTIVE COEFFICIENT

der the flow condition depicted in Fig. 101 where a fluid wit velocity U, and temperature

{we Past a stationary flat plate of length J and width B. If temperature 4 tthe plate

Iwan greater than the fre stream temperature then convective hat transfer occurs fom

the ate o the fluid. The flow conditions vary fom point fo point onthe surface, nd as such
Sonsective film coefficient and the heat flux would also voy along the surface. However

won lementary strip o length dx located ata distance x form the lading edge, the conceive

“ent can Be assumed to be practically constant. The local heat fx is then given by

N Qu =h, dA (4, = 1) eh (540 t= € dom
"pd the local convection coeficent

sarang, al heat transfer rate Q is obtained by ineprating the local flux over the entire

zu

282 Sesics of Heat and Mass Transfer

mati.

Fo. 101. Loca and average comecton concent or ot pa a a pl

= forarr(.— 11 fa ae ae
Defining an average coeticient for he entire surface, he total eat rase rate may ay
be expressed as
QF ENG - 1) (03
From equations 102 and 103, the local and average convection cocficients are related y)
an expresion of the form,
1
i htt
“The average value of convection coefficient based upon th total surface arn used in y
Non kan relation to calculate the oa amount of heat hat stranored fo he ies
A ondions: The local convection coefficient becomes the controlling factor when thea
rate nthe boundary layer has a ers temperature mit which i not o be exceed

EXAMPLE 104
Experimental results indicate that the local heat transfer coefficient h, for flow overall
plate with an extremely rough surface is approximated by the relation:

hema xo
where isa constant coefficient and x Is distance from the lading ce ofthe plate. St
Ups relation between this local heat transfer coefficient and the average heal tars
Seficient for plate of length x i
Solution: The Ical and average convection coffcints are related by an expression ale
form

En] sess ma,

is cme sik es |
du a td
Pen
+ Grat sata cr» 2202) |

Me: pe Cox

form

Ped from dimensions analy for wen demining eva

ao nos
con with various Mid (bh gun
nd vera pats, the exponen an have

Fits the correlation 105 lo be rewriten in Ihe
Nu = C (Gr x pro

‘The constant € and exponent m depend upon the nature of fk
geometrical configuration (plate or cylinder) and its orientation (vere or homromal) For
flee convection, the fluid properties needed fr the determination of Gr and Pi re heran
evaluated at the mean flm temperature, The product (Gr + Pr) is often rtered oa RER
Somber, and its value sets the eriecion of laminar or turbulent charter l los

Thus

(006)
low (laminar or turbulent,

108 < Gr Pr <0?
Gps

for laminar ou
for turbulent dow
103.1. Horizontal Plates, Cylinders and Wires
(Plates: heated surface up or cooled curface down
+ laminar flow 2 x 10° < Gr Pr € 2 x 107
Nu = 084 (Gr PPS 08)
‘turbulent flow 2 x 107 < Gr Pr < 3 x 108
Nu = 0.18 (Gr PAP wos)
{i Plates : heated surface down or cooled surface up
+ laminar flow 3 * 105 < Gr Pr € 7 x 108

Nu 027 (Gr Ps) os
+ turbulent flow 7 x 10° € Gr Pr < 11 x 108
Nu =0107 (Gr Pr 10.20)

In these correlations, the characteristic length used in computing the Nusselt and Grashof
numbers isthe length of the side of a square, mean of the two dimensions ofa rectangular
Surface oF 09 times the diameter of a circular disk,

{ Long cylinder 1/0 > 60
+ laminar flow 10% < Gr Pr < 10°

Nu = 053 (Gr Pr ™® ao
+ turbulent flow 109 < Gr Pr < 102
Nu = 033 (Gr PAS Pur)

For cross flow over horizontal pipes,
Nu #037 (Gr en
The characteristic length isthe cylinder diame

033)

(1) Fine horizontal wires D < 0005 cm
Nu =04 (Gr PA? = 04 ats)
length is the wire diameter

284 7 Beeice of Host and Mass Transfer

10.3.2, Vertical Plates and Large Cylinders
+ laminar flaw 10% < Gr Pr € 10°

Nu = 0.59 (Gr Pr)? (00)
+ turbulent low 10°, Gr Pr < 10% tes
Nu = 023 (Gr Pr) ty

‘The characteristic length is the vertical dimension of the plate or cylinder,

1033. Incined Plates
Multiply Grash of number by cos, where @ isthe angle of inclnation from the yy
and use vertical plate constants id

10:34. Miscellaneous Solids (Spheres, Short Cylinders and Blocks)
‘The following correlation has been suggested for free convection for à sphere
Nu 2 4 043 (Gr PA «om
For higher values (3 x 10° < Gr Pr < 8 = 10%) 043 is replaced by 0.50
For short cylinders (D = H)
Nu 0775 (Gr Pro (0039

For ther sis
Nu = 082 (Gr pm as
where the characters length iste distance Waele by parle in bound
For rectangular alt the character Teng worked out rm the rela
LLE
ae
ebene 1, and rele 1 the sigan veia and hola dimensions. Fut when
Mon I onen hrosgh a none condal such ae quer rem duc ee
po pipes und NN aes

defined as;
cross-sectional area
PE ted perimeter

EXAMPLE 10.2

A horizontal heated plate at 200°C and facing upwards has been placed in still air at 2°C.
If the plate measures 1.25 m * 1 m, make calculations forthe heat loss by natural convection.
‘coefficient for free convection is given by the following empirkal

‘The convective
relation z
f= 0.32 (0)°% Wat
where 0 is the mean film temperature in degrees Kelvin.
Solution : Mean film temperature,
9 = 273 + (200 + 20)/2 = NC
Convective film coefficient,
h «302 x (383)°% = 1336 W/m?-deg,
Rate of heat loss, Q=h À At
1336 » (125 x 1) » (200 - 20) = 3006 W

Epica,
EXAMPLE 10.3 Ome Fe or au
ld ae an
that any mutual heat effect i preclndea treo 204 1S cm
wwhilat the ambient str The mains are at the ag 04 in à
tf the ha ne rom ne nage et he no i e
jon : The steam mains are located in wen ot IM meins, Pt anale condtiient,
Set an en copes ae ae hs ey
The 110 that of free conve 7 the ambient air ia
m. ann

Nu = C (Gr ps

xis he a popes nd ei
nad peo ve

Care
EM” or

RON

Heat loss Q = h A at

Qe head, M
DETTES

EXAMPLE 40.4
‘Aspherical heater of 20 cm diameter and a Ch ma =
Determine the value of convective heat anses cosa à MEP

Solution: At he mean film temperature t= (60 + 20)/2 =
Sein AI erature t= (60 + 20)/2 = 40°C, the thermo physica properties

Pr 9922 kg/m ; v= 0459 x 106 m/s, Brass
N 0683 W/mdeg and B = O41 x 10 per degree kelvin
y. Dog Pagar
Gre SOBER See Car)
(027 0.4110" 981 x (60

(00x10)
Gr x Dr = 2964 x 10 x 4.34 = 1286 » 10°
For a sphere, the general correlation is
Ad
= 26 043 (Gr Py om
2 + 043 (1286 » 10% =14681

ae = 458.32 Wintdeg

Nu

£
hanes to 10681

|
|
|

286 7 Besics of Most and Mass Transfer

exAMPLeE 105

Anar pipe #2 cam dates end 23 m long bua been peca Emi ane

Ba eee ar opens poker

mean temperature of 160°C, the thermo-physical properties of air are : the
K = 3.64 x 107 W/m-deg v 30.09 x 10% ms

A
P= 0682 and B= gps = 291 *10% perk
D° p° (Bg AT)
Soutien: Grshot number, Gr + PE GEST
DBs aT)

0.05" «2.314107 x 9.81 (295-25)
ETATS
Gr x Pr = (08447 = 10% x 0682 = 0.576 x 10°
For laminar flow over horizontal eylinders within the range 10° < Gr x Dr < 10?
Nu =053 (Gr Plt

8447 « 108

R
BD ass ose» 109 «do
ges the conectivo covet as

= 1063 W/me-deg

Heat lose Q = À at
= 10,63 x (x x 005 x 25) x (295 - 25) = 11265 W

EXAMPLE 106 ==
IR muher reactor with Is core constricted of parallel venia plates 225 m high ad
wide a Sten designed on fre convection heating of liquid bismuth Mel
a: lit the maximum suce temperature of the plat 1 975 C and he oe
en pelar of bnmuth 28 €. stout the maximum posite het disipa
irom both sides ofeach pate
“the appropiate conto or the convection coefficient
Nu = 0.13 (Gr PA
where the different parameter) av vaste an ihe mean film temperate;
075 à 525)/2 = 650, te nem

Solution : At the mean film temperature. !
properties of bismuth are
= 312 ka/mchr ;
6, 71507 )/kg des

patie tal

p= 10 kg/m?
= 1302 W/medeg

2 1 = 1.08 « 10% per deg kelvin
a 7108 * 10° por deg

008 (975-5)
RT 10 + 100
Gr Pr ©1085 10 «001 «149. 198
Using the given correlation ü
u

Na = aaa?

non rm À

= 019 (1043 x 198313 y DR „ a
A a D = 4866 W/m deg
This gives a heat transfer of Q = 2h 4 ar
“The factor 2 accounts fortwo sides ofthe plate
Q=2 x 34866 » (225 « 15) x (975 - 325)
= 153% 10" watt = 153 MW

EXAMPLE 10.7
thin walled duct of 0.5 m diameter has been laid in
and conveys a particular gas at 205°C, The boun

slficent of heat transfer is given bys 7

EPT
TO
are
war (2) was
st enh tthe o, ten a cee open

ith that computed from the following non-dimensional orrelation for laminas flow natural
Convection for à large vetical cylinder,

y 7057 (Gr PPS
ey ene a sake gh GT a ane
a) 1 (E
aan te so ines

5086 W/m deg,

€ At mean film im
of ee úl à

La 3194 «10 W/mdeg

1.1 261x107 per deg kelvin
PenO7 and Bo ba pag 260107 per deg

y= 2410 x 10 m/s

i
Î

288 Bosics of Heat and Mass Transfer

or Pia Pogat

261109 9813 (20515)
UND X91 4(208-15) iy
not)

Using the given correlation,

Msgs «10 sarap

This gives the convective coofficent as

15789

neun E uray MO

= SO W/m deg
Heat lost by convection

HA Bt = h (md) at
= 5043 x (m x 05 x 1) x (205 - 15) = 1501 W
EXAMPLE 10.8
Estimate the heat transfer from a 40 W incandescent bull at 125°C to 25°C in quiescent a,

Approximate the bulb as a 50 mm diameter sphere. What percent ofthe power slo by fy
‘The appropriate correlation for the convection coefficient is
Nu = 0.60 (Gr Prt
‘where the different parameters are evaluated atthe mean film temperature and the characterise
Length is the diameter of the sphere.
Solution : At the mean film temperature, t= (125 +25)/2 =75%
for air are M

the therma-physicl proper
Y "2055 «10% m/s; k= 003W/mdeg
Pr = 0693 and B= 335575 = 287 * 10° per deg kelvin
gar „ PBg at

w v
(0.05) 2.87 «10° x 9.81 (125-25)
OA 80 x 108
(poso

Using the given correlation,

m
Nu nd = 06080 x 105 x 00937025

h

222 040 80 18 2063 «9825 Wink
Ths girs fat tae

AI nn «ZW
Tree he prone a a ye cone

2
77h 100 = 19278 %

EXAMPLE 10.9 “ ve
ware Plate 40 cm x 40 e
gore TRE LE Tps
julations forthe heat los from both suas oc 0 AMO ay y
De ee faces ft plat 2 ns
ne following empcica arrlaons have been su
Ni ©0125 (Gr Pf or uy ed
Nu 2072 (Gr PA 93 for upper sufge PMY and
= 035 (Gr PAP for ower srt
wer hear propres ae evaluated a the mens cape
Solution At the mean temperature, t= (100 + 292 à te
oon DWZ GPC; the thermophy sis probertes
D 2106 kg/m, k= 0098 Winde
GMOS HAE K and v + 1897 #104 ts
1

Ic
Pa 7m 273460 "0.008 per degre kelvin

(1897 x10°) «(1.08 1000)
Nag 078

DRE
1 240.003 x9 (00-20)
(18:97:10
Gr Pr = (429 x 10) x 074 = 3089 «108
(2) when the plate oriented vertical,
Nu = 0125 x (208 » 109 99 = 7369

À u

N os 5

= Nu T= 7869» QU» 5505 W/m Ke
This gives a heat tronsfer of : Q = 2h A at
The factor 2 accounts for two sides ofthe plate

Q =2 x 5508 x (04 x 04) (100 - 20) = 181 We
(#) When the plate is positioned horizontally
(9 For upper surface

Nu = 072 (3003 x 104% = 95
k 0.008
= nu E 950 208 545 w,
Wena 5 = 95 x S08 0 665 W/o K

O, = h À at = 665 x (04 » 04) x (100 - 20) = 8512 W
For lower surface;

Nu = 035 (033 x 10808 = 4619
ous
04

323 W/m? K

k
mena «4619

|
H
|
|

290 1 Basics of Heat and Mass Transfer

Q,=hA at = 3.23 x (04 x 04) x (100 - 20)
D = Q + Q,= 8512 + 4135 = 12647 W

ausw
Comments: The above calculations show thatthe plate loses more heat when it is orien

vertically. Obviously natural cooling can be achieved more effectively by keeping the pla

vertical position. :

EXAMPLE 10.10
Calculate the rate of heat loss from a human body which may be considered as à vez
cylinder 30 cm in diameter and 175 em high in still air at 15°C. The skin temperature is As
And emissivity at the skin surface is Od, Neglect sweating and effect of clothing.
Solution At the mean film temperature 4," (35 +15)/2 = 25°C the thermo-physical properties
of air are

VS 10% m/s 5 k= 0.0268 W/medeg ; Pr = 07

iz de ll 5

and Babel o gg = 0008 per degree tevin

„oben „Bngar (PASADA 5095-15)
rs (05.53 «10°F
Gr = Pr = 146 x 10 07 = 1022 » 108

Since (Gr x Pr) is more than 10, the flow turbulent and accord

Gr 214610»,

f
ml an are
= 0.13 (1.022 x 101003

Convective heat loss, Q, = h A At
eh dl AL = 3924 » (x x 03 x 1.75) x QS - 15) = 12937 W

Radiation heat less, Q, = co A(TÍ ~ Tf)
=04 x (567 x 10%) {x x 03 » 1.75) (308% 2889 = 7924 W
‘Total loss from the human body, Q, = Q, + Q, = 129.37 + 79.24 = 208.01 W

EXAMPLE 10.11
A thin walled vertical duct of circular cross-section is 04 m in diameter. That duct cast
à gas at 470 K and the surrounding air may be considered still at 290 K. Determine the Bet
transfer rate from one metre length of the duct assuming that the boundary layer is lamina

‘The general non-dimensional correlation for laminar flow, natural convection Für
large vertical cylinders is.

a cera"
x 2

%

The fluid properties are to be evaluated at film temperature which is defn
average of the Buk fluid and wal temperature,

ed as de

(o) The heat transfer coefficient is to be prescribed Comecton y 294
e Oye rin
ve mx

where the length parameter Ls the metres, Cakctne y
ie ac ae the value of constant C which would

at he film temperature the thermo physical propeie of he as ae
POS kyle? 5 Ge LIEUX à nes 04
PSS «10 Wink’ amd Ba 24 0) pr dp er
22.016 x10 x(Lo2x 10)

Solution

8 81x(470-290)

amsn DE
= 056 (6.316 x 10° x 069003 = 15421
‘Thus the convective film cocfiient,
a £ 32510?
ne sank = sm = 4968 W/m? K

T
Heat ow rte = h À at
ADS (ex 04 2) (470299) = masas W
(9 Substiating the relevant data in the press create
(470-290 P5
(A cm c= 2%
( ) à Constant c= dat

4968

EXAMPLE 10.12
A am pipe 6<m in diameter covered with 2 cm ick lye ola which as à
Surface emissivity of 092, The insulation surface temperature 1 75°C and the pipe is placed
inatmospheric air at 25°C. Considering, heat loss both by radiation and natural convection,
‘timate the heat loss from 5 m length of the pipe. Also calculate the overall heat transfer
‘efficient and the heat transfer coefficient due to radiation alone.

Solution: At the mean film temperature, (75 + 25/2 = SPC, the thermo-physia properties

p= 1.092 kg/m? = 1007 gg,
= 1957 x 10%kg/ms and k= 2781 x 10° Wide
Corresponding to these properties, the relevant parameters are

[2 096 + 10° per degree Kevin
Ba 1957x10#x 100

pre le à Bs 7 070
ET

|

4

292 1 Bosics of Host and Mass Transfer

cow Pas
A

4=6+2%2=10cm=01m

1 x: x10°x9.81x(75- 25)
01 (0.05 3.096210 CE
so)

A quiescent atmosphere corresponds to natural free convection for which we can

following correlation for finding the convective heat transfer coefficient ehe

há

Mu = 22 = 053 (Gr Pros

ne $x 0.53(Gr Pr |

„Zapf san)”

R = 630 W/mdeg
Heat lost by convection, Q, + À af + hyn dl at
620 «(m 01 x 5) x (75-25) = 49455 W
Hest lost by radiation, Q, = € ATH)
#092 (67 10-4) x (x 01 x 5) x AE 28)
255528 w
Total heat loss, Q = 494.55 + 55528 = 104983 W
In terms of total (over al) het ranser offene
Qh Aat
=, es
Bai” (exo1%05)x(75-25)
Het transfer coefficient due to radiation
h, al, h = 1337 - 630 = 7.07 Wn?-deg
104. CORRELATIONS FOR FORCED CONVECTION
For te usual freed circumstances, the following dimensionless numbers apply

he 1337 W/mk deg

+ Nusselt number Nu

y
+ Reynolds number Re = IE |

+ Prandtl number

+ Stanton number

The conventional
‘coefficient ha

|
Baneralised basic equations fr use in determining the value of conectan |

A LAMINAR FLOW á
10.41. Plane Surfaces: Flow Past Fat Plates and Walls
(6) The local film coefficient for laminor fox past
ais Lona past à Dat plate may by obtained from the
N Nu, = 030280 AS pps
(Fluid properties are evaluated atthe mean film tempera
of the tempcatune ofthe Aid and the temperature of he acca ee avarage
(Reynolds number must not be less than 40 dir Es
(i) Pr must be more than 04
Fora plate of length 1, an avers
te obtained by integration.

"ias

(10.39)

8€ value of Nusselt number or convection coefficient may

a. OS

or Nu 0661 (Rep (ppt

(1029
(9 The following correlation has been proposed for liquid metas
Nu, =0565 (Re, x Pr) (0023,
and Pose qa iy of the above rain, the ud proper are evaluate en temperature

1942. Sytindrical Surfaces: Flow Inside Pipes and Tubes
(© For uniform heat ux: Nu = 436
(© For constant wall temperature
(0 slug flow : Nu = 378
(9 Full developed flow : Nu = 365

*Teoos[lasix ke Pep 0024)

(9 Pipe length is much greater than diameter.
(6) Fluid properties are evaluated at the bulk temperature

(ii) Re and Nu are calculated on the basis of pipe diameter as length parameter
(ic) Prandtl nuniber lies in the range 05-100

(©) Constant wall temperature and fully turbulent flow.

su
Bu LA pe à

2 met 16 Pre] al 00

which is esently valid or short tubes. Te ui properties are evaluated atthe

temperature exept, which calculated a the mean sua temperate The pipe dest

isthe signcant Feng parameter Further Re Vd and should have valu wiki the og,

1
ee ; L>2andp>t
Ben; À w

B.TURBULENT FLOW

10.4.3. Turbulent Flow Over Flat Plate
‘The general equations giving he local het transfer coficient for turbulent fw (Re, > 510)
past flat plate

ia, = ORI Re Pr

IS
and Na, 00292 (Re (PAPE co
‘where the properties are evaluated atthe mean film temperature
Fora plate of length average Nusselt number would be given by:

Nu = 0.086 (Re) (PAP om
When the low lies in the transition range,
on

Fu = 0036 (PAS (Rey? = Al
‘The constant A takes the value 18700 at transitional Reynolds number 4 10% and 230
at transitional Reynolds number 5 x 10
10.44. Turbulent Flow in Tubes
1. Me Adam has suggested the following general correlation for heating and coli ©

Ads in ben fw through long pipes
Nu = 0.023 (RS (Pr wa

ad

294 1 Basics of Heat and Mass Transfer

(e) The following correlations have been suggested for flow inside tubes
hd oy 0.0668 (2/9 Re Pr
|

where
9 n = 04 if the Avid is being heated
+ 03 if the fluid is being cooled

li) Fluid properties are evalued a
Dieter mn
(i) 07 Pr < 120 and
(o) a> 60
or gases such as ait, Pr is essentially constant a

Br NY const and an be pp by caging e
2. Colburn proposed a correlation in tems of Stanton sume

St = 002 (Ray 02 (pou E

bulk temperature

me
V3 vante a he man uk temp
(9 Bean Pac latte vas temperate
00 rant do ho eat ha
ye a ene
Tue Me Adams and Colum comatose a
pei omnes na ae pere
ie pores Do alone cc

St = 0023Re py w(t)

m 0931)

where
(9 all fluid properties except,
at the wall temperature
(i) Re> 10"; 07 < Pr € 7000 ; ya < 60
For flow in short passages, these correlations need tobe modified to account forthe variable

velocity and temperature profile along the axis
ig and temperature proble along the af Now. The flowing cations have ben

are calculated atthe bulk temperature, is evaluated

and ente óÍ tor mete
for Weich 11032)

For vr rg tempertr fee (4 ia Desman and Sansa
eng coat ns ga

Nu= ps (ene
Ses 0.026 (Res (Py 1038)

© 1/4 upto 355
6) Re 2 10
(i) Nu and Pr are evaluated at mean bulk temperature
(0) Reis evaluated at mean film temperature
945. Turbulent Flow Over Cylinders y :

nel à care os

‘The values of constants Cand depend on the flow Reynolds number Further, all theme

Physical properties of fluids are evaluated at the fi temperature

296 7 Basics of Host and Mass Transfer

10.46. Turbulent Flow Over Spheres
A. For flow of gases over spheres
Na = 037 (Re) 0%

for 25 < Re € 108 u)
oid properties are to be evaluated at the film temperature
“0:
for 1 < Re < 2000 de
Fluid properties are to be evaluated at the film temperature.

EXAMPLE 10.13
Air flows at 18 ays past a 15 em diameter pipe in a direction normal

surface is at 30°C and it receives 495 kKW/m? of heat from the air at 25
convective heat transfer coefficient is prescribed by the relation

'na] [vac,P?
AE
Under similar operating conditions, make calculations for the heat loss from carbon
‘ioxide at 200°C when it flows at 1.5 nys normal to a 25 cm diameter pipe at 25°C;

‘The relevant thermo-physical properties for air and carbon
For air at 50°C :

cra

oxide are :

1038 K/gK ; km 4270 » 10% Wink
For CO, at 200°C :
E "099 KVkgK ; k= 312 x 107 Wink
Solution : Let suffix 1 denote air and surffix 2 denote carbon dioxide, Then,
hé [Maen dl
ha Ut
Dal

“ef

Ra TR
m 015, 312x10? _
ig" 025" 427x107

hh fées]
Mb

P
h
ous 7

a en ge
ee
E

1
CET

1592

Empirical Correlations fo

and Focas
Heat loss from carbon dioxide, Comet 1287
495
Qs apy "2516 We
E10
xan
gue ine fect of following condon me eg,
ee ae ge ra
in flow

fp Two- fold increase in flow velocity
{iy Two-fold increase in the diameter
py ¿change in the rate of liquid flow.
Y may be presumed that there is no change

the temperatures ofthe
oc wall and thatthe flow through the tbe is ent a a Ud and the

by varying mass flow rate;

of tube, the flow velocity is maintained constar

Dee ee
i

Eh a tbe, the average value of heat transfer offen:

= ME 0023 (pe (pn
pl auf ee e
Vu a] Ved G

Further as there e no change inthe temperatures, th ud properties remain estat
{When the tube diameter and the tuk! properties remain concent
ho (Ma

That sa two-fold increase in the low velocity results into 74.1% increase in heat rate.
() When the flow velocity and the Auid properties remain unchanged

Tati, with two-fold increase in ube ame, the bat ae can by 1%
CXAMPLE 1045

Glestate the rate of heat loss from a human body which may be considered à à vertical
liner 0 em in dimeer, nd 375 cm high while sanding na 3 wind at SC. The
suas temperature ofthe human i 3°

Solution: At the m emperatare, t= (85 +
ten: At the mean film temperature = (35

\/2= 25°, the termes phil properties

Ve 1829 106 ms 3 ROO Wardeg 7 Pre 07
Be ,

Wind velocity = 30 km/hr = PE

33403. „or

„33 (mina)
y "amo?

298 / Basics of Heat and Mass Transfer

Nu > = 0664 (RAS x (Pr
= 0664 (Résa) x (008 = 23833

& 0.0263
= Nuk «238.3 9 Wm? de
ha Nu xk «m6 23 008 « 2049 W/m eg

Convective heat Toss,
QehAM=h AD ar
= 2089 x (1 x 03 x 175) x (35 - 15) = 688.86 W

EXAMPLE 10.16
Air moving at 03 ms Blows over the top of chestaype freeze. The top of the fay
measures 0.9 m by 15 m and is poorly insulated so that Ihe surface remains at IC Wt
temperature of sir is 30°C, make calculations for the maximum heat transfer by 4

convection from the top of the freezer. nel
Solaion : At the mean film temperature, 1 (10 +30)/2 = 20°, the thermo-physical property

of air are:
¥=1506 104 mi/s ; k= 00259 W/medeg ; Pre om

sno Ds mu average heat ranser ocur whe the ai wi he diction of

pvt vi _oax09

1 v 506107

Nu = 0.664 (Re) (Pr) 0 = 0.64 (17928775 x (070392 = 79214
0.0259

09

Re = 17928 (aminas)

- A ae wines

Convective heat flow, Q = # A at
#228 x (09 x 15) x (10- 30) = - 61.56 W

The negative sign indicates that heat transfer is towards the freezer.

EXAMPLE 10.17
The oil pan of an 1.6 engine approximates a flat plate 03 m wide by 045 m long and
protrudes below the framework ofthe automobile, The engine ail is at 95°C and the ambi!
air temperature is 35°C. Ifthe automobile runs at 36 km/hr, make calculations forthe ae
of heat transfer from the oil-pan surface. Assume negligible resistance to conduction through
the oil pan.

Schon ba man Min tempers + 39/2 °C he hemo yal rep
ee V=1846 x 10* m/s ; k= 0.0293 W/m-deg Pr = 0695
Vela of atanobie y = 36 tne = 228 10 wy
vi vt, 0208

= 2498 10° (laminar)

Ree

O

Nu = 0.664 (Re (PHO
= 0.664 (2438 = 1025 x (0695)0% = 290.81

convective eat transfer, Q = à A ay
= 1893 x (03 « 045) «95

de ma Ww
feel plate measures 3 m «Im,
lo 1 m edge at 9 km/h. Ifthe air
des of the plate.

I Atthe mean fl temperature, t= (0 + 10) «

is italy 30 Winco
MPC calla on ME
276 te emo physi ropes

VOTE k= 0038 W/mdeg Prag

91009
oi of aie V=9 km/h = 22200
pe Vi 250
HV gag 160 (mia)

AL
Na =F = 0664 (RES (pe

= 01664 (1660033 x (0703039 = 249 83

k

he Nux

20%
E ug

‘Convective heat transfer from both sides ofthe plate,
= Q=h@ayar
2 623 x (2x3 x 1) x (0-10) = 707.6 W

EXAMPLE 10.19
A metalic cylinder of 125 mm diameter and 95 ram length was heated imemally by an
retical heater, and was subjected to cross flow of air in «low speed wind tunnel. Under
a specific set of operating conditions, the following data were recorded

Velocity and temperature of free stream air = 10 m/ and 235° respectively

Average temperature of cylinder surface = 1285°C

Power disspation by heater = 45 W

115% of the power dissipation is lost through the insulated end pieces ofthe cylinder,
determine the experimental value of the convective heat transfer coefficient. How this value
‘Compares with the covection coefficient obtained by using the correlation =

fe
Ni = 026 (Rey (pops |
Pperies, except Pr, ae evaluated atthe mean bulk (re sra) temperature of ar.
K = 00264 Wink; v= 1585 «104 mis and Pr= 0706
Pr, is the Prandtl number of ai evaluated at the average temperature of tinder surface
bso

lon: Heat flow from the heater tothe ar flowing past ts given by

Qrhaar o
085 x 45 =i (x x 00125 » 0.095) x (1285 - 255)

XUNTA

300 1’ Basics of Heat end Mess Transfer

Convective film coefficient,
05% 5

AS
NS y

va 100.0125

a ESTO

= 7886

a
me

Convective film coficient,

26 (7886) (0.706) = 5023

0.0268

E
N ps

106 Wek

EXAMPLE 10.20
Air flows through a 10 cm internal diameter tube at the rate of 75 kg/hr. Measurement,
indicate that ata particular point in the tube, the pressure and temperature of air are 13 te
and 325 K respectively whilst the tube wall temperature is 375 K. Make calculations fort,
hat transfer ate from one metre length in the region of this point.
‘The general non-dimensional correlations for turbulent flow in the tube is
Nu = 0023 Re pas
‘where the fluid properties are evaluated at the bulk temperature,
Solution : For air at 4,= 325 K and

1 atm, the therme-phy

cal properties of aie are

1967 x 105 kg/ms ; = 00292 W/mK and Pre 0715
‘These properties may be assumed tobe independent of pressure to an excellent approximation
vd md _ am _ ax,
Re = PU nt Am EI). = 13492

“Au RAR TES
This is well in excess of the critical Reynolds number for flow in tubes (2500); the low is
turbulent and the given correlation applies
Nu = 0023 (133928 (0.713) = 4046.
Convective film coeticient,
je Neck _ 40.46 0.02792
ci

= 11.296 WmK

Heat flow rate, Q =f A At = 11.296 « (nn O1 x 1) x (375 - 325) = 177.35 W
EXAMPLE 10.21
‘A square channel with a side 10 mun and length 1.3 m caries water with a velodity of 5

feasurements indicate that lengthwise mean temperature of water is 30°C whilst the ins
surface of channel is at BOC, Caleulate the convective coefficient of heat imasfr for e
channel wall to the water. Use the correlation +

Nu = 02 RAS pa (5)

where the thermo- physical prope
‘water. Pr, corresponds to the value
and equivalent diameter is the refer

les pertain to those at the mean bulk temperature of
sf Prandtl number at the channel surface tempers
ence dimension,

operties of water

me physical prop 31 30°C are:

E 9 = 95407 kale? a 8174 gx
S28 bajado; 05 ra wi

at wall : temperature f, = 80°C and Pr_ > 22

san: Equivalent diameter of the chama,

EXAMPLE 10.22
Air at atmospheric pressure and 20°C flows with 6
sie conditioning system. The duct is recta
Determine heat loss per

The relevant thermo-physical properties of ar are

Ve 210% mje; Ge 772108 mts and k= 007$ Wider
Seton: For the rectangular dct, charter eng
PR
CL Tre
DÉCRET ABE
TES
TT

uv 15x10
ly the flow is turbulent and accordingly

al

nu.

0023 (Rae » PoP
0.023 (0.213 x 10898 x (0701704 = 36534
Convective coeticint, h = 36534 « À = 534 x BE
noir 135
Area of duct per metre length = 2 (a + #) x 1 = 2 (08 + 04) x 1 + 24m?

| Convective heat loss, Q = h À at 17823 24 1 = 42268 W
| EKAMPLE 1023

na cent for water flowing inside a colar
tape] Or affect the value of convection coefficient fr water lowing

= 1752 Wastes

302 1 Basics of Hoot and Mass Trensfer

Within a condenser shell, water flows through one hundred thin-walled ica py,
(diamerer = 225 mm and length 5 m) which have Been aranged in parallel The mar aro
rate of water i 65 kgs and its inlet and outlet temperatures ate known 10 be 22°C ang OR
respectively. Predict the average convection coefficient associated with water ya DC
Splaton: À De mean bth temperature (2 + 28/2 = 25 the thermorph slopes

99665 kg/m° : = 903,01 x 10-0 kg/ms
67 “41776 KJ/kgK ; km 2.1893 Kl/mihrk

(03.1 10° x 3600) 4.1776

Tiss
Mass flow of water through each tube,
65
L0651g/s
"Oh,
vap mi im
Ree an mn
4x0.65

“0.0225 x 903.06 x10 074978

This is well in excess of the critical Reynolds number for low through tubes (2509; de
flow is turbulent and the following correlation applies +
Nu = 0025 (Re) (pra
= 0.023 (40749.78)* x (62)04 = 23269
232.69 «2.1893,
0m

= 22641 Kymichrdeg
EXAMPLE 10.26

(0.90.15) <1 912199
= PAS) arret ons og y
(10.1104) ag

Gr Pr 1287 a 19 x 0677 «03h, yy
ni on a ange en

Nu = 054 (Ge Pa
i es the convection as
me NK, DI
7 E 09x015 "IB W/m? kK
since both cb ofeach in are expo tn see

10, towing coe
SA ATS à A PP

ares for convective heat

+210 2035-007) «0265 ns

Be dpaiôn. Q =h A At = 928 «0265 0 «2m = 13129
D Mr cle Le mag (red comet
arm

Speed of motor cycle= 60 km/hr = 1667 m/s

2 PALI 1667099009).
a ‘516

‘which indicates a turbulent flow for which the following correlation applies
Nu = 0036 (Re) (Px? = 0036 (5416) 0677 = 197.1

Nuxk _ 19711042 5
ES

Heat dissipation Q = À At 61.76 » 0256 (480 -20) = 5286 W

EXAMPLE 10.25

Re

A motor cycle cylinder consists of 10 fins cach 15cm outside diameter and 75 em ie
diameter, Calculate the rate of heat dissipation from the cylinder fins when () motor cx
is stationary and (i) motor cycle is running at 60 km/h.
‘The atmospheric air is at 20°C and the average fin temperature is 480°C. The rele

thermo-physical properties at the average temperature of 250°C are

9 0678 km” 7 cy = 1038 [gk

K= 0427 Wink ; 1 = 0677

¥=4061 x 10% mijo and

312 x 10% per degree kel

ate value of heat transfer coefficient may be evaluated by idealing the fit
as a single horizontal flat plate of the same area.
Solution : (a) Motor cycle stationary (free convection)

cr = PBEM! Pop

Ina rein glass making process,» square le of gas 08 m? ac nd 3 mm ck B
Sted uniformly to 30°C Subsqueny i scold Oy BFC template ht flow &
Fins over bth sides para 1 the plate agas tempe aint fn he Gas
pit and considering only forced conection, estimate he al ot to he Pate
How would thie cooling tate De alicia even i iced ES me
For glass take p= 2500 ky? and cy 06 Mrd
{At the mean a temperatur. 01» 20/2 SC e erp properties

PALO kg/mm’; #1008 KK à E 00286 W/O
and 9 198 106 No/m?
„DIV 2076x082
Ru 198% 10%
Bey DEI aus
Po
Nu = AE 6 (Ro? » (POP?

Lau mn» 217

ra ne

304 7 Basics of Heat ond Mase Transfer

oak 0.0286 .
par x À aan SES = 5562 W/medeg
i Heat flow from both sides of the plate is given by
Q = (24) At = 5562 = (2 » 08) x (90-20) = 623 w
The instantancous heat los from the plate is also given by
Q=m 6, At = (p AD 6, at
(2500 » 08 x 0,003)» (065 x 1000) at = 3900 ar
a
Rate of cootng, at « SE
ah, (tay ( Vs
© nn (ra) ln) +) sta
Percentage increase in cooing rate = 11.8%
EXAMPLE 10.26

0.1597°C/sec

A Copper bus bar of round cross-section wit
Of dry air. The air is at 20°C and it flows pa
«calculations for the coefficient of heat
What would be the permissible curren
| is not to exceed 75°C

'h 15 mm diameter is cooled wi
st the bus bar with a veloc
asfer from the surface of bus bs

tensity for the bus bar if its su
? Resistivity of copper = 0.0175 x 10% ohm in,
For a single cylinder placed in cross flow, the follow
been suggested:

Nu = 0.43 Res
and Nu = 0.22 Reds
‘The reference dimension is the diameter
at the temperature of free stream

10 < Re < 10?
10 < Re < 2 » 109

flow (# = 20°C) a
Y= 15.06 x 10% mys; k = 0.0259 W/mdeg
Solution : Reynolds number

Va _ 15x0.05 A
Re ER 1494 x 10
he given conditions, the second correlation applies
Nu 0.22 x (1.494 x 1099 = 17.66

Nuxk _ 17.66%0,0259

Apparently, for

he Ze mies

ne ne 049 W/mé.deg

With the limiting surface temperature of 75°C, the heat dissipation to air is
Q=h a at

= 3049 (x x 0015 x 1) (75 20) = 78.985 W/m length
M 16 the current intensity, then heat generated in the bus bar

peel. x10) «1

Then from energy balance,
99.08 x 10-6 7? = 78.985,

th à cross flow
ty of 1.5 mys. Make

tothe cooling ais
face temperature

18 empirical correlations have

ofthe cylinder and the relevant physical propone

9908 10%

+ MAS

Empricas
MORE Cris Fon sa

QuE 1027 men 308
id mercury flow through à copper Fone y 2
sgt ee my emer ee 8 on gaa

at
mine the tube Leng which would sait the conan Pao eg
uh ao ge ope ton, nn

uid metas, the following comen y re
ori Presumed to ge
seuls

eh pen
27 ats

(05+25/2 20. te der gh
ECK 8685 W/m

V2 LUS x 10 mija
Q= meat =1

Wa properies

2 and Pre ous
+1935 « 05-15) = 17418 Ww
Vdp_md_ im __im

An edu rap

sx12 :
TRADE * SNS
Then from the given correlation
Nu = 7 + 0.025 (Pr x Rojos
=7 + 0025 (51115 = 0.0249) = 14516
Nuxk | 161628. a
A = 6807 W/m
The heat gained by the fluid results from the convective process Ths
Q=HAAT

1741.87 6347 (x x 002 x 9 (25-15)
Therefore the requi

length of the pipe is
— 7487
A

10m

EXAMPLE 10.28

at a temperature of 25°C is blown across a flat platea mean velocity of 7S mi 1 the

Pate surface temperature is 575°C, make calculations for ie heat tansfemed per metre

“ih From both sides of the plat over distance of 20cm fom he Wading sg
For heat transfer from a plate with large temperature between the pate

Me local Nusselt number is given by.

antic
"here all the film temperature, T, and 7. dimension.
properties ae a the mea ae
eater ance eno sc) e

D dance tom the eng abe

306 / Basics of Heat ond Mass Transfer

Solution: At the mean fil temperature = (975 4 29/2 = 30°C, the royo

properties of air are
D = 0615 kg/m?
À «01659 K)/mhrk ; and

evi _ 06

“=

Gp 10465 K/kgK :
D = 29728 x 10% kg/ms

75x02
107

31035

„ (29.724 «10% x 3600) 1.0465 |

pr = BE

ry 01659 =>
Nu, = 0.392(0.675)" ra (a
a) 7
Local heat transfer coefficient,
58 0.1659 a
o A a KJ/matuk

‘The local and average convective cooficients are related by an expression of the form

i

Hna

ape na" (08) a
E Kon ir
em) (EE)

rey ÉS

kml
= 0.302 510) ¡e

La

ooo (EJ) ar

7,

crier (TE

2 x local heat transfer coefficient
2 x 4811 = 9622 KJ/mibrK
Heat loss from both sides of plate = 2 (4 AM) .

22 (9622 (02% 1) (675 - 29] = 211684 Kir

REVIEW QUESTIONS

103. Enumerate some ofthe empircal relations which are used to compute the convective oido

for free convection.

Inthe product of Grand Pr, what group of properties ls concerned only with Ihe propia

fpaviational feld ? What Us group of properties called? What are the di
this group of properties

m

ws

us

m,

Ws to ope MT nm arc Cae
pe relevant physical. properties of water a he conan

PM kein; ea fl tempo

DETTE PC
#805 Wak and bras

A restaurant pill LO m 08 mis mained 15 ne zum

Quite the eat load generate bythe gs" PC M the rom emp nase

Tre approslamte correlation e

wae À 2 028 (Ge x pn

eher the significant Fengt isthe average ofthe to e,
Se at the mean film temperature (0%) are
À = 00804 W/mk

an he étant pa proper
VAIO m/s and pre ou
‘aan 629

Wm and sre unpeatum ar
convective codon an ea of Rat

A horizontal cylindrical beat exchanger of she dane
{Eto be coke by the ambient ir ar PC oca he
ies from unit surface area ofthe het exchanger
Use the correlation.

nu = M os cor» rs
where ie pica rapie Bad and uk temp e
Mets RSR OS ond Peom
(Ans. 59 Wha deg, 1003 War)
A long horizontal pipe 15cm ouside mer ná wi oid aie pss og à
Beam The stra tempest ofthe pe à PC ale seen du DC
Washout the convective coin lr he ion Ue de cra

Nu = nass (rr

and take the air properties atthe mean in pert (00) as
SO ler ao |
Lite es md Dee nn Sata)
AB mm diameter electric wie covered ih 075 hk nto e ed eon
y capo a BC we dee me at de
evened ths het usury convene Seen fra
temperature. Asume fee convertion besa yer Bani nd we Decora

RU (TRS when his memured in WR rt dpe e and dm men.

10cm nie ameter tte Pot
Mtb For he ble om,

gto air at bl emp of 35K fo enh
‘the coefficient of beat transfer by convection between Ihr a
te he cc

Nu = 0.23 (Re) = (Pr) eK)
and take the following properties at he bulk tompertue OK) q
da en prope a a Ne a ne
O eA ote wel bese EY
pos

308 7 Basics of Mont and Mass Transfer

Use the correlation

N = 0459 (n° pa wo
The recat phys popes ir evant a the ln temperature (5°C) ae
en Ve 1795 x 10 m/s ; k= 00282W/mK and Pr = 0608
(8539 Wa
104. Water a 75% ou through a 5 mm diameter tbe wih a vts of m
Seminar à BC ma Caen fr I hea amr coe
Uae ie eration
r= ao na
and tke the following termo yt! ropeies of air
DRE à OT WAR à Ge ANI and 9 op
thos ae
1010. Esme he hat ae fm à 40 at cane lt RC 10 29°C ae
SS eve. bal may De appronnated sos 50m lama pee, Abo lo
of powes lau by ocean
She average tea rants cen gen a
tp
‘to
The ruled properties evaluated atthe mean fl temperature are
Vs lo mojo and DO Wink

ph
8 the tae oy

rer

(ns. 227 sm
1031. For a certain forced conection process,

he following correlation applies

Nu a 01 (Ref (en

Workout the percentage change i the rate of eat flow per degree temperature difccce nie

the orignal coolant replaced by another fluid having viscosity equal to ler dul aaa

‘original coolant Assume that other fluid Variables and configuration remain the sane
Karen

000

Hydrodynamic and

Thermal Boundary Layers

Li ne aes ar op a
When thin layer of fluid close to the solid ay
eence the velocity distribution. The Rud ue
SES nap yong id wath tinier Renee hod
12S ao made to thre fg eae st en
dpe Sa ont andar gent em Se
and explain the methods for their solution, - dtl

111. HYDRODYNAMIC BOUNDARY LAYER : FLAT PLATE

Considera continuous flow of Mid along the surface of
‘sige se parallel to flow direction (Fig LE,

CH Moves through» body of aid their

race within which 9
iy var ew see

a thin plate with ts sharp lei
The sen features of the flow situation are
() The free stream undisturbed flow has a uniform velocity U, in the rom Peles
suid adhere to the plate surface as they approach and the fed O domed dem ne
The fvid becomes stagnant or virtually 50 in the immediate vic of he pa pa
Generally it is presumed that there is no sip between the Quid and th slid torta, Thee,
there exists à region where the flow velocity changes from that of solid boundary te that of
mainstream fluid and in this region the velocity gradient exist in the fluid Concequenty the
Sow is rotational and shear stresses are present, This thin layer of changing veloc has been,
called the hydrodynamic boundary layer

(i) The condition du/2y # 0 is true for the zone within the boundary layer whilst the
conditions for flow beyond the boundary layer and ts outer edge ae

de
E

Thas all the variation in fluid velocity is concentrated in a comparatively thin Laver in
Immediate vicinity of the plate surface un

(1) The concepts of boundary layer thickness and outer edge of te boundary layer are
Qu tous steep spe ton our on do
de. Velocity within he boundary ver approaches ere secar elo amply
Pal the boundary layer thickness 8 is taken to be the distance from the pate surta
Front At Which the velocity is within 1 percent of the asympttic Him de, a = 099 UL The
Parameter 3 then becomes a nominal measure ofthe thckness of Boundary layer, ¿e 0

sm

0 md veu,

310 7Basıca of Moor and Mass Transfer

gon in which the major porton af velocity formation fakes place, The hicknes is mes,
Sa 0 he plate thiekness. The Boundary layer is normally very U
Gimensions of the body immersed in the flow.

‘Nominal ko boundary layer
Ron 55x10"
DS

Leasing | Transition Laminar
ES font sible

Fig 118 Deepen Bout ye na o

(ie The hiess of the boundary layers variable alo the Mow direction is ee at
leading edge ofthe plate and increases as the distance x from the leading edge incre
‘This aspect may be atibuted to the viscous Lores which dissipate more and more energy u
the fluid stream as the flow proceeds. Consequently, a large group of the fluid past à
slowed down.

“The boundary layer growth is also governed by other parameters such as the magriude
‘of the incoming velocity and the kinematic viscosity ofthe owing fui. For higher nconing
‘Velocities, there would be less time for viscous forces to act and accordingly there would be
less quantum of boundary layer thickness at a particular distance from the leading ee
Farther, the boundary layer thickness is greater fr the Muids with greater Kinematic vs

(0) For some distance from the leading edge the boundary layer s laminar and the ty
profile is parie im character. Flow within the laminar boundary layer is smooth and te
Streamline ae esentlly pora othe plate, Subsequently the lamina boundary ayer become
Unstable and the lamina Now undergoes change nits low Structure a certain point all
transition point, in the flow Bel. Within a transition” zone, the flow is unsable ad s
feferred 10 as transition flow, After going through a transititon zone of finite Teng e
‘boundary layer entirely changes to turbulent boundary layer

(6) The turbulent boundary layer does not extend to the solid surface. Underlying a
extremely thin layer, called laminar sublayer is formed herein the flow is senti d
laminar character, Outside the boundary aye, Ihe main lud maybe either laminar o tb

(ot) The pattern of flow in the boundary layer is judged by the Reynolds rune
Re = Ux/v where x e distance along the plate and measured from its leing edge. Me
transition from laminar to turbulent pattern of flow occurs at values of Reynolds numb
between 3 x 10 1 5» 10% Besides this critical Reynolds number, the co-ordinate pins 4
‘which deterioration of the lamina layer begins and stablizedturbuler low sets in is depende
on the surface roughaess plate curvature and the pressure gradient, and the intenso À
turbulence ofthe fee stream ow

(i) In laminar boundary layer, the velocity gradient becomes les steep as one por
along the flow. Is because now the change in velocity from no slip a he plat surface IR"

value in the potential core occurs over a greater

eam va distance, Nevertheless in à turca,

fe ee re ons a
Peon ray amen e pas €

mene (ne boundary layer Consequent

EEE undary ayer Pas alr vey pie
ttle coer velo rodent pucca Y
el wt loro mar ooo
2

he at the pie nun Fars mr
tosh rahe etc grade comes ae
ante pow don and e do es ut Le
angie fr turbulent Boundary pr te nr +
Howe rine ue ena Sh vts ww,
ona ne op odo pede UT PA Wey sane ew ne

nbsp of bunny ye fe ow ur

Lora ion similar at fr ow al aA ple However, thks a
JESS yer limited to the pipe rds tects fete ete oe
Dag. Boundary layers from the pipe walls met at he center ofthe pip and ie emo oc,
eee he characters of «boundary yer Beyond tas pete he a enr Bo
San 1 ad 1 const al pal ae Pre do ot
Sr sens ae reste te pipe oca and dep ana aa ee
End fly a uch tonite eae ns
BEE and when bear leer Gow bates pea Senet Se

oy umi

So ia
entrance angen —e
= 5010803) yen
Fig. 113, Boor is gow ap

ae Fly developed turbulent flow is
entrance length required for the flow o become nis
dependen on te surface fn ntl evel ables downs nis ro

‘nd is generally estimated to be 50-80 times the pipe diameter.

312 11 Basics of Meat and Mass Transfer

cient C,, refers to the ratio of the local wall she,
st ln ont Ge os
soe nai oars heer star i
>" Feet) oy

Depenting upon the natur of velo profile the values of Boundary layer Winey gy
skin fiction coeficien are +

+ For the velocity

NÓ)

+ For the velocity profile 77

58 1m
E as
+ The latas technique for an ect scltion showed that
3.5 13%
FU Ge as

In the above identities 2, = PU

is the Reynolds number based on distance fom be

leading edge of tie pia.

An estimate of the average value of kin friction coefficient T, can be made by integntg
‘the local skin fiction coefficient C, from x = 0 o x= (where {isthe length of plate) and ti
dividing the integrated recul by the plate length,

A

7

ur

06 a a
Titre Lee Type Pl
AM | 130818

JET True Vee

where Re is the Reynolds number based upon to

s number based upon total length {of the plate.

The average shin con otic Is quite ole refered 103 dag colic

‘Then : Friction (drag) force = Cy x pu
(drag) force » Cy pu xaren of plate

room ie an Ty
TETE TE

rep
An P nto estimate the Boundary lave at is 4 à velocity of 3
ration enge en Fre
Sta rom Ihe leading edge ofthe lt on har esha Icon cnt
A esponding values obtained Kom the prima vn an compare wih he
sary mel scopes
efor nt at 25C, v= 1533 x 10% 3 od her
ai sl
“ Au, Reynolds qumber
Emm mn
(9 Exact Blasio solution

Sx 3x1
oe ae
Re, az
0.664 0664
Vie, Tea

= 0001376 m = 1436 cm

= 1511 103
dí Approximate solution with assumption a cubic velo peo,

CRE
‘Vie, "Jona "000

10
for the local skin fiction cocfficient
EXAMPLE 11.2,
A plate 05 m x 0.2 m has been placed longitudinally in a steam of crude ol which flows
with undisturbed velocity of 6 ays. Given that oil has à specific gravity 03 and kinematic

vscosty 1 stoke, calculate the boundary layer thickness and shear stress at the middle of
plate. Also calculate friction drag on one side of the plat.

Solution: Kinematic viscosity v = 1 stoke = 1 x 10 m'/s. At the middle of plate,
1/2 x 03 = 02 m
al 5x6
v Tot
Since the Reynolds number is les than 5 * 10, he boundary layers of laminar character
And the Blasius solution gives.

“ir

0.25
ao
102 «10 m= 102 em.

(gage Genios ol Heat end Moss Transfer
„000,006
Gr Re Tarot

= 0592 x 102

By definition,

‘Shear stress ty, ypUEXC,

= (Laos ruonxet)sossaxi = 878 Nt
0) Regis somber ah ling ig of he ae,

05x6
vet

Re

3x10!

Even a the trling edge, the boundary layer is laminar and therefore the average dap

(fiction) coefficient is

=:
FN

Friction (drag) force

% = 0766 x 107

rt

Ts pUË are of plate on one ide
= By} pt! xare of pateon one sd
1 A
= 0766 x 102 x (1x09x1000%6?)x(08%02) «
02x(} ) naw

EXAMPLE 113,

Air at 25°C and 1 bar flows over a flat plate ata speed 1.25 ny, Calculate the boundary lye
distances of 15 em and 30 cm from the leading edge ofthe plate. What wald
"e the mass entrainment (mass flow entering the boundary layer) between these two section?

thicknesses

Assume parabolic velocity

220)

The viscosity of air at 25°C is stated to be 6.62 x 10°? kg/he m.

Solution : The density of ar is calculated from the charaerisie gas equation: p = 9 AT

Given 1 bar = 2 x 105 N/m?
T=250C = (05 +273) = 208 K
R = 287 J/kg K

e S
rana ~ 1169 kg/m
The flow Reynolds number is, Re, = (+ p Un
Dream Reo 219X116 (1.25 3600)
EN 919

rame nd hema mie La
CESEN

emo; Re,
aur 30 i ax

ar the given parabolic velocity distribution, the ou R
Mo Mary layer thickness prescribed by

sé
Re
Ti =
464x000

Voss

( At any postion, the mass flow inthe Boundary lye is given by he integral

axe dem ir 899: 10%m

where the velocity is given by

a CECI

Evaluating the integral with this velocity dis

‘Thus the mass entrainment between the two sections is

Su. (6,
. 8, "PU (6)

A)
= 2.393 x 10° kg/s = 8.61 kg/he

"82, TERMAL BOUNDARY LAYER

Mena us o as

indicated in Fig
Di, the nen

ented or ol su, tempera eld ise wp in he id next
Soc te ne pera ton le
N Una peau Rl ecmpacs» ery sal ein ei
the repo a dean hen oy te pine some à in yer ar he
gues. This zone oe thi layer hen Ih epee ls cas the thermal
rtndary layer The tempora ray heat exchange between the plate a
Re layer The temperature gradient results dueto eat chan

The
mue

hicks of thermal boundary layer is abia) defined asthe distan y from he
‘Surface at which

4
]
À
|
:
À

316 Bosics of Heat end Mass Transfer.

mn
to nea
un ee

u Free stam

Temperate
rata)

a)

or
Nm otectedy

x eating
man

eS
Lama

Torten
Fa. 114 Tema our je ung How Au over a am ae

The convection of energy reduces the outward conduction inthe fluid and
(he temperature gradient decreases away from the surface. Further, the tempor
is inte at he leading edge ofthe plate and approaches zero as he ayer develope don eat
Moreover in the turbulent boundary layer, the action of dios aten the temperature

Ifthe approaching re stream temperature is above the pate surface tomperatu es
{ermal boundary layer will have the shape as dpicte in Figure 114. The tempera si
Aid changes from a minimum at the plate surface 10 the temperature of the mal oe
tan distance from the surface, At point A, the temperature ofthe fluid is Ihe oe ee
surface temperature }, The fluid temperature increases gradually uni it acquies thet
stream temperature The distance AB, measured perpendicularly tothe plate serace dar
the thickness of thermal boundory ata distance x from the leading edge of the Plate

consequen
rate sade

Fi. 115, Thema bouncy layer in Hw of wa Mit ovr aca ae
The concept of thermal boundary layer is analogous to that of hyrodynamic bound

{ayer the parameters affecting their growth are, however, different The velocity profe of
Ridtedynamic boundary layer is dependent primarily upon the viscosity of the fui. laa
thermal boundary layer the temperature profile depends upon the flow velocity. specific hat
Niagara and thermal conductivity of the fluid, The thermo-physical properties ofthe fil
Gel the relative magnitude of 8 and 3, and the non-dimensional Prandi nua!
(Pr y 6,/4) constitutes the governing parameter

(When Pr=1 3-5

(i When Pr>1 <8

(i) When Pret 858

Hrérodmamie ang

dimensiones temperature distribution for

rma Boundary Layers y 317

Fd number has ben

‘WPA for air and several er gree

2 wing results are of practical in
"ie following, 4 nero
{hte cts of ermal ound pr kg

10 be the distance from the plate
rte for W the pl

099

this value of non-dimensional tem;

ji erature parameter, here
alle of Prat number Prat, ere ar sexe curves with
ss
se
40
532
Tea

16

os

0 “02 Os 66 a 10
BE

Fig. 1.8. Tampere un in m nr pr esla ai
(6) When Dr =

& „ns

The thickness of thermal boundary ayer 8 os proportional o Zr. Wil ees in
Sates rom the leading edge, the fe of hat tnt: rena ar ino de fe
AN and the thermal boundary layer grows. Comparison of equations 1L and 117 reveal
A en Pr= 1, the thermal and yjérodyrame Hours lanes ae of equal icknes and
Aie temperature and velocity surfaces are dental Since Prantl number for mon gases
Se Sfienly close to unity (08 < Pr 1.0), the two boundary layers would be very

he gases,

Myton ang

Therma Boundary Layers 1318)

averages over the Interval O< x « ,
"rk out the average heat vans

TRE nd Nusselt number to be;

cent

ue com

le

Apparently the thermal boundary
layer ls thicker than the byérody namic
boundary laver forthe flow situations
swith Pr < 1, This has to be so because
Pr <1 implies that thermal dtfusivity
fs more than the momentum diffusivity
A large thermal diffusivity means more
penctration of the temperature elfects
And consequently a large value of the
thermal boundary layer thickness.

© When Pr>1,

550 50
E]
Apparently 8, <8, when Pr> 1.
Pohlhausen has suggested the following general correlation between Prandtl numer
and the relative values of thermal and hydrodynamic boundary layer thicknesses.
5,75 re am
(i) The heat flux at the surface may be writen as

Fig. 11.7. yéroénamie and hema Boundary yes,
Yor arm Prandi numbers”

La oy

20 Maid mation

This expression is quite appropriate because at the plate surface, the
and the heat transfer can occur only through conduction.

“Through rigorous mathematics analysis, it has been suggested by Prandil that temperate
gradient may be prescribed as

332

2
TOS

Upon substitution in equation 1111,

Re, (er

2 2-1 rom É (e -
htt, t,) 20.332 (
Solving for hy, we obtain

1 nase a on

Equations 11.12 and 11.13 relate local values of the convective cocffii

number; these apply at a specific value of distance x from the leading edge of the pl“

22 op py
Re (pep ie

= 2% local film cetcient fora plate length 1

and (Ra np

within the boundary layers taken o be ofthe cub form

44)

aus

PQ ine velocity profi

Y

3
(9 The temperature distribution within he boundary layer sic the conditions
et at ys

0119)

mo yea,
a yaa
y st yen

When these boundary conditions are fitted toa cubic polynomial

the temperature distribution age he frm

u). ya)

104 au

28) ie) en

temperature distbatiens ae inserted
raten tebe à (heal

de identities for the velocity and

boundary layer thickness) and 6 (hydrodynamic boundary layer thickness)
ee] sun
sel tes]

LÉ heating of he plate stats from the Isdin
‘seated over the entire length, xy = 0

320 / Bosics of Heat end Mass Transfer

he local heat transfer coefficient h, can be worked out from the quality, focegoing analysis i valid only f

ne 5 minar conditions y
Go eased 4) gue he low may have transition from laminar 1 turn a TS in Leg ofthe
1 Sa Pu anda to De posed onthe plate lr along ra GUS Apr en a
= k(t 1d) o met
1 hen A Al therme-couple is positioned in a thermal boon
OT] ; A aer flows a 30°C and 0:15 as. The plate oa nt plate pat
From the expression for temperature distribution sich location ofthe probe, the thiknes of tesa Er entre SC
ind ai st as cared byte pea nay ae
(78)
O
eine) Het fo rom pate wie; and he et andr coi
sept At de rent ils imepertar y= Gl SOY + O Be a ay
water is 0.633 W/m-deg. id ds =
mm
um
For a cubic polynomial fr velocity and temperature distribution, we had ©) Het transfer coefficient
& _ 0376|, [x] daz
Ts
Substitution in equation 1120 yields
E zen 151 12008
p> E LD a
DÉTENTE) CET deg
IE 1 EXAMPLE 11.5
= 0392 (re) (PAY? x zu MEN Air at 25°C approaches a 09 m long by 06 m wide fat plate with an approsch ve
x peer] 45 my. The plate 19 heated 10 a seface temperature of 135°C, Make calculations for
it (local het transfer coefficient la distance of 05 m rom the leading edge à
(dia rte of hea anses from fhe ae 1 De ME abus proben
un He 0382 (Re) (Pr) om Solution: At the mean am temperature 4 25 39/27 80%. de Heme
. tí air are: in Pr = 0682
[-00/37"] VER 104 mips à OOH Nm,
When the plate heated over the entire ler o kl
plate is heated over the entire length, , o o 5, 25208 1070 es
In Re am ve the lading ge and acom
E The low is laminar at distance of 5 m from uud
a qu ingen a = 0502 (1.087 = 1095 7 DE 5
an Nu, = 0382) (exp 0 Nyy 0392 (eS (PHP =

Mass Transfer ins
y Basics of Heat end dynamic 80d Then
se 00004 A EXAMPLE 11.7 Boundary Lars 1 329
neuen = 583 6/08 K ir at 1 atm and 20% flown actors a fn pl
E gil to 30 em long * 10 cm wide and when the rw is tong neg 102°C. The plate mon 4

(0 For the entire plate length aber i 40000. Make calculations fr the heat mance geo eth lw Ropas

a se wi be aeted 0 ae om fy
el O 2 po «06 810 feat low will e affected he flow eich la ats eme athe au Hw thie
u ne te Cereal)
bc he Am iat for the enti plate length, nd according sation: Ath mea mener 4 N nn
20668 (RS (PH BE one P = 106 kam? à ee 1005 ja x pre
Lakes (1.923 109 x (0.6907 = 2579 "00289 men and ns
don 2257 49 «2000 « 870 /mdes Since the Reynolds number is 40000 Ihe boundary M/S 07 Nom
¡rt a er omar an acond
"hr = 0.664 (Re)? x (Pr
Heat os from one side of the pate, o x ; y
ree ON 9 x 06) x (03525) = 51678 a
EXAMPLE 11.6 aa
PA aC IOCtemperaar and fre seam velocity of 25 m/s flows slong te 102 nids
Aramis an dt 3 WC Te Luk Convective het tear oc oth ico te pn
eng of Fa ae Sd em, 50 em and 25 cn thermal conductivity ol he a ones
ath and ches ofthe Pas clation fo a) hat lost By the plate) tempera igre co I MT +0203)» D an = 56
ac of the Pe lor sen sate condition. iar es
a ae ern = (090,2 = PC he here py proper nu = one cmp
of ais are o =1061g/m0 à en J/g K à Pr 0686 eg;
2 2002 Nm and u 20 x 10° kg/m (5) HE) fotos that
vn 10n25x1 Ie Peony
pe A a le A
aa % te) i) Ae) Ue) Ue

than 5 x 108, hence the flow is laminar and accordingly

The Reynolds number is less Since €, J, w and & remain constant and

it aon Va=2V, and p,= 4p, (as temperature is constant)
A 2 064 (Ra Pr I = hy YP? (ZÉ 17.02 (JV QE = 48:14 W/mideg,
0664 (2.995 = 10903 x (0.696) = 21439 Percentage change in heat transfer

A E

va

and 2100 = 1828 (increase)

Convective heat loss rom the pate EXAMPLE 11.
Q =H A At = 6.196 x (1.0 x 05) x (90 - 0) ale Ambient air at 20°C flows past a flat plate with a sharp leading edge at 3 ays. The plate is
‘This convective het Joss must be conducted through the plate, Then from energy ated uniformly throughout its entire length and is maintained a a surface temperature of

Sais (0-4) $C: Calculate the distance from the leading edge at which the flow in the boundary layer
( = ges from laminar to turbulent conditions. Assume that transition occurs at a critical

e

à À Reynolds number of 5 x 10.
Bottom temperature of the plate, Make calculations for the following parameters at the location determined above
er {9 thickness of the hydrodynamic and thermal boundary layers,

LE = 9037 C

25x(L0%05) (lsat and average convective heat transfer coefficients
i) convective heat flow from plate to ambient ar, and

(6) mass entrainment in the boundary layer. Consider unit width of the plate.

324 1 Basics of Heat and Mass Transfer

Solution: At the mean film temperature f= (4, + 3/2 = (40 + 20)/2 = 30°C, the then
Pays properties of air are mo.
4 KO Ki/menrdeg 5 Pr=070
V=16% 10% m/s and p= 1165 kg/m?
The value of y = x. at which flow changes fom lamina to turbulent conditions

from the expression for rica Reynolds number LT

Rs 2
© Assuming cubic velocity profi,
IS 00178 m = 175 ram
Wee. re
09768 _0076x0.0175
a 001923 m = 1925 mm
| (i) The local Nusselt number atx = x, given by
Nu, = 0.332 (Re. (Pr
= 0.332 x (5 x 10°"? x (0.701) = 208.47
Local heat transfer coufcent at x = x,
REDEE
= 207
Average heat transfer oeficien is oblained as
À fa

h, © 2 » 7519 15,038 Ky/mhrd

"2
(ii) Convective heat flow from both sides of the plate to ambient
Q=f Qa) x ar

15.038 (2 x 267 x 1) x (40 - 20) = 160606 K/hr
(0) Mass flow through the boundary layer from the leading edge to the point whet
transition occurs is given by :
5
m = SOUL (5-51)
570 at x=0 and 8=00075m at zen 2670
5

m = 3 x1.165 «3 (0.0175 -0) = 0.0382 ky/s = 137.62 aha

8

EXAMPLE 11.9,
In a certain pharmaceutical process, castor oil at 35°C flows over a flat plate at 6<avs. 7%

~ plate is 6 m long, is heated uniformly and maintained at a surface temperature of
Make calculations forthe (a) hydrodynamic and thermal Boundary layer thicknesses

atte

ng edge ofthe pte, toa
transfer coefficient at the end of the ser Sit ith on on
st ‘of he plate and (tl an

[A the mean film empero =

CE

Thermal dif,
sit a= 72 ga
‘Thermal conductivity ka an ay À

Kinematic vis
scosity v=
Density p = 9568 iy
s/o
Soltion + The Reynolds number atthe endo plate
i
DS

Since the Reynolds number is less th,
is laminar in character.

(o) In the laminar ra
Blasus solution is

945 x 10 the boundary ayer over the entire plate
ee the ches of ya boundary Laye a recibes by

5
ee
Thickness of hydrodynamic boundary Iyer a the ling eg «i
5x6
55363 * 04082 m
When the plte is heated ove the etre Le

layer thicknesses are related 10 each Noa manic ad Werl today

Other by the expression.

where, Prandtl number Pr = Y.

0408 m

(0) The average skin friction coeficient (drag coefficient) for the entire plate is

1308 138

UB.» aca of plate for one side

= 0017s x (3958x008 } (6x1) = 0184 per metre width’

(6) The local Nusselt number at x = is given by

326 1 Bosics of Heat end Moss Trensfer

Loca tat rater content athe plate end
DRE
6
¿0 Toe average hes ner fin is obtained as
it
aif
yn 2 8488 «16916 W/m? K
Hat os rom one seo he pate
QnR Aa
LAS = 6+) « 5-35) = 9089 W por mete win

8458 W/mK

a

EXAMPLE 11.10
À bin at plate of length f= 1m and breadth b= 045 ms exposed toa flow far paa
do its surface, The velocity and temperature of te fre steam flow of at are resp
ee ee
‘neat loss From 50 dm length of plate measured from the trailing edge.
Solution + At the mean film temperature f= (4, + 1/2 = (35 + 25)/2
the thermo-physcal properties of air are
P= 1060 kg/m? à Ke 2894x107 W/mK; Y
(0 For the fist 50 cm of plate length (x= 05 m)
JUL, 05x25_
Tora?
Nu, + 0332 (Re RS x (PP
20332 x (65800) x (2.696)
26007
05
The average value of the heat transfer coeficient is twice this value
= 2h, = 2% 4376 = 8752 W/m? K !
‘The heat os from one side of plate i, ef
Q=ha al
= 8752 (05% 045) = (95
(i For the entire 1 m length of the plate

xu,
Ree

ore

18.97 x 10 m/s and Pr = ue

re, = ET = 65800

517

hn, = Nu, E=75.617% = 4376 W/m? K

) = 137.86 W

erie

cote sons » eu = 10686
k _ 106.86 x 2.894 «107
x A $
2h, 2 2x 3092 = 6188 w/m? K
Q, = FA at = 6184 x (1 x 045) x (95 - 25) = MDN

Au

= 3.092 W/m? K

hy” Nay

Prosa ang
ence Meat os rom 50cm of pa ang Dr cy a

E fom the
A ange NOR me

gunmen ees
a Plate at 25 ays. The pla

Sans ats an temita D IN mee
spade parallel to the 30 em side?
9 = 1060 kg/m k=2894 à 102 W/m
fra hom pani pec ee
Re, = El. 06325
TBA comm a
Nu, = 0392 (Re x (69933
A à pis «eae
À even
ee
SP» (08 203) 65-25 = sow
Macro pinta soe a

+ BR = rc, the there

Y* 1897 2 10% o/s and r= 0406

oa Egg 2
x 0°

7996 Wie? K

00006 x 19
Ni, #0382 (Re) x (PHP
= 0232 x (0.0396 x 10485 x (0656)
E 50505 288107
fh, = Ni, £58,538 BIO oso wre
oa Week
B= 2h, = 2 «564 = 1128 W/m k
O) = HA At = 1128 x (06 x 03) x (5 - 25) = 182328 W
ex semi more heat loss occurs when the ar lows parate the sorter side andthe
centage heat increase is,

=Q gg = 12128-10050

& m

338

2100 = 12

113. TURBULENT BOUNDARY LAYER FLOWS: FLAT PLATE
or turbulent boundary layer flows, the following corelatios have been Se for boundary

er thickness and skin fiction coefficients;
ox

me 125)

(9 Boundary layer thickness 8 =

328 // Basics of Heat end Mass Transfer

{D Local shin friction coefficient, *

0
Gam a
Tour (Rad a
3 y
(ii) Average value of skin fiction coefficient
1
B= Heras
1f_ 005% à
Dion ur us
576 (N ar (21,002
= 00576 [Lar 0072 (2) 00
Go (itz) ie mn
At may be remarked that equation 1.27 results when the velocity distention
one-seventh power lave which essentially is valid for 5 103 < Reve 10° Mow
Experiments indicate that for Reynolds number between 107 and 10 th vl di
deviates from the one-seventh power law and close ft with experimental dats atin
the following empirical relationship suggested by Prandtl and Schlichting et
„ass
(oso 86" cum

‘The correlations listed above (Equations 11.25 to 11.28) are valid on the
the boundary layer is turbulent from the leading edge onwards. However, respecte
laminar or turbulent characteristics of the mainestream flow, the boundary layer Jomar
commences with a laminary boundary layer at the leading edge of the plate Even io
laminar boundary layer gets transformed to turbulent boundary layer ater passing thug,

transition zone. (Fig. 11.8)
Laminar Turbulent
en, T

sn

Fig. 11.8, Dog de to lominor and ete boundary es ño
of turbulence al

‘The position of the point of transition depends upon the intensity of ir
mainstream flow, on the roughness of the plate surface and is prescribed by the critical Repos
‘umber which normally ranges from 3 * 10° to 10 x 1 e
Drag force for the region of turbulent boundary layer can be estimated from th eb

Fag = Fast wd = Fr
here Fat Presets hed which would cu I a urbuentpounday le Mi
along the entire length ofthe plate, and Ear, represent the dag due fii

boundary layer from the leading edge to distance x, (Fig. 11.8)

assumption à

es
layer en

à Ÿ
the plate i Mriroomamie
oun Bag poe be ogy Terra Bar
w Perot ht Rs 29
Fang = ss SP un 2, hen,

ogg Re PH PUR (3) Lam
per joe un

Within the length,
force 6

and a such the ttl ra fore equa

(Re,
Invoking the relation,

fe

and therefore
£ [er os
CT

Presuming that transition cccurs at e,

5010.

and hence

[oso A

Mie Reynolds aumbr A, remains below 10

“|

then the flowing colon can setup
0072 _ 1670)
eae | 0129

114, REYNOLDS ANALOGY
(inter relationship between fluid friction and Newton's lee of viscosity)

a

330 1 Basics of Heat and Moss Transfer

Heat flow along the y-direction follows from the Fourier equation

a
11
La 0]
Temperature and velocity profiles are identical when the dimensionless Prandtl ny
tunity which is approximately the case for most gases (06 < Pr < 10) here
ne
ei or

‘Combination of expression (1), (ii) and (ii) yields =

a
e--san de

Spain the ate an at hn se i
odie Oa sure
D Wet athe eg ft |

ay

Aue vun
the het tecnico Far rom he o

Luz
in rction coefficient, Gy = Ta / (PU). Making these substitutions in equation 1132 e |

1
hi = Cy x} pub x
Ie Cp x GPU

or in dimensionless form
Ge i
a um |
‘The dimensionless group of terms h,/(p, U.) is called the Stanton number St, and à

represents the Nusselt number divided by the products of the Reynolds and Prandtl num,

Nu Sa
u, SO u
Ro °F u

The physical significance of Stanton number is |

mat actual heat flux tothe fie

$ ¿UL At eat flux capacity of the id flow

Equation 11.34 is called the Reynolds analogy and is an excellent example ofthe simi
nature of energy and momentum transfer. This inter-relationship can be used direcuy to
heat transfer data from measurement of shear stress,

(Re)
ing both sides of expression

Diving Presion (2) by the product pa,
. Den,

Rem

‘The Ket hand ide ofthis equality cn be rc u)

BED = 6 par

0135)
‘The inter relationship between heat and moment
transfer then becomes

St, PA

the Colburn and Reynolds analogies are the ame" 9 7 10 be noted that for Pr = 1,
en

‚worked out by the application of Colour analog). The peep Pa f

“ncn keane eee eons rca la en

layer flow past à fat pate can be

fre

= faune

|
or Nu, = 0.0288 (Re WS (Pr |

o .
“ simon ACH a‘)
Taking averages over the interval D < x <1
|

com) wd fre

232 1 sie of Het end Mass Transfer
aser Oe)” py nee
alo
„eat 242) eo
iw

vena

as
ii us 1
and Boney" sie pi
Serien of mia ad bat: An expresion fr te average bat wae
Cosina ove à pt lng when ot tania and relent tour te
preset canbe worked out by evaluating the intra
ga (inca
Papua]
Ilan ette
al ose (a) (nde «fas (ne, ce |
loe Va a aconnal Ue) ff a
oral (mae coun)" f à a]
Econ” [se +0036 (Ro) (ne) Y ay

Presuming that transition occurs at a critical Reynolds numberof Re, = 5 x 108, we in:

Te zen” [nas (6x10)" +(e -006(500)"]

= en fe. 036 Ra)" 836] am
and A
Ra en Bas" -836] au
EXAMPLE 11.12.

‘fla ple was positions at zero incidence i a uniform low scam ofa Ami
boundary layer tobe turbulent over he er plat, workout the rato of shit iin
onthe font and ver half par of the plate,

ait
Solution For turbulent boundary, he average Ara (kin fiction) coefficients pret
bereiten

- um
Dar
Forte enti plate Ry =U and fr theft hal of he plate Ry = (72 UD

Mérosnome ang
Diag fore Per ui width, fr teenie pas y Terra! Boundary yes y 333

<4

Feu lo
AA « area per unit with
Mn

Qu japo run

aie, the dg force per eit width o u
u FE re tl Pn pt
A tl
u.a? ze 2

m tr.

pare bout decaf

[Tr

Drag force for he rear half ofthe plate is
ESPE,

Hence,

EXAMPLE 11.13,

For a particular engine, the underside of the crankcase can be idealnad as a flat plate
measuring 80 cm x 20 em. The engine runs at 80 kayhr and the crankcase is cooled by the
air flowing past it at the same speed. Make calculations for he loss of het from the tank
use surface (, = 75°C) to the ambient air (= 25°C). Due to road induced vibrations, the
boundary layer becomes turbulent from the leading edge itself

Solution : At the mean film temperature, = (+ 13/2 (15 + 29/2 = SOC
the thermo-physical properties of ai are
k= 2524 «102 W/mK ; v= 1795x104 m/s and Pr= 0698
Engine speed = 80 km/hr «(80 x 1000/3600 = 2222 m/s
i, We | 08x22
rar
For turbulent boundary layer ; Nu = 0036 (Re)®® (Pr
Sms «ss rp AA

lax

Heat transfer coefficient, Ñ =

Heat lost by tve crankcase Q = A AT .
ls Os ps og x 75-25) = 65K W

334 1 Basics of Heat and Mass Transfer

ae 1. . —_ _ _ ___
A Mat plate 1 m = 1 m in placed in a wind tunnel. The velocity and temperature y
Stream sr ace 80 Vs and 10°C respectively. The flow over the whole length ofthe nur“
‚made torbulentDy torbulizng grid placed upstream af the plate. Make calculan fare
following parameters +

(a) thickness of hydrodynamic boundary layer at trailing edge of the plate

(0) heat flow from the surface of the plate,

‘The plate is maintained at 50°C and use the following thermo- physical propetien ur,

9125 kg/mm? ; Km 0022 Wmdeg sv = 1415 * 10% mys; c= 1000 Jag,

; pui U, PC
Solution; Flow Reynolds number Rg = Pal a Met LRO sr
For tarbulnt boundary layer, the boundary layer thickness prescribed y the tn,
s_om
RE
5 21 = 0.01655 m = 1655 mm
Box

Hp Ve 1.25% 14.1510" 1000 |

(0) Prandtl number pr = Hr = 0022 u

Ru = À = 0026 (Rep0* (rH
= 0,036 (5.65 = 106998 (0.804,09 = 8437

k 0.022
Y

Fe Nu pasaros 18561 W/atteg

Heat flow from the plate (one side)
TTA Af = 18561 (LT) x (50 10) = 74245 W

EXAMPLE 11.15.
Air at atmospheric pressure and 20°C flows past a flat plate with a velocity of 4 nf. To
plate is 30 cm wide, is heated uniformly throughout its entire length and is maintained #
à surface temperature of 60°C. Make calculations for the following parameters at 40 «8
distance from the leading edge =

(a) thickness of hydrodynamic and thermal boundary layers

(9) local and average friction coefficient

(6) local and average heat transfer coefficient and

(4) total drag force on the plate

Take the following thermo-physical propertice of air at the mean film tempe
ac

P= 118 kg/m

rate

y v= 17 10% mys; = 1007 Yugedog and. = 0.0272 Windes

Ue 04%4

1 : Flow Reynolds number, Re, = “H+ = HDL

OHI amber Ré Tre 7 age

Since the Reynolds number is less than 5 x 108, The boundary layer is laminar in
‘Assuming cubie velocity profile

Sol

= 941 x 104
pa

ss Sorcery Lane an
Re
Beg +

PAI 62 mn

When the plates heated ov

boundary layer thicknesses are roce ef

tof he plate, he hy

Felted 1 cach ter by rm ná rm,
1.0768
TA
where Prandtl number pr = À BM 1.18417 10 1007
a [T7 7 0782
ae 25762805 Sexo
3 TE 657 mm
(6 Local friction cotcen atx = 04 is ive y

fea

gie si

no eee RU 0
Re pans me

2 000216 = 00042

number at == 04 m à given by

e friction cooficent Cy = 2C, = 2
(6) The local Nuss af

Nah = (op
"0332 (9.41 x 10995 (0.7429 = 9227
Local heat transfer coefficient, = Nyx 90.27 « ER ang wae
ice, = Nan» EEN
Average het transfer ceticent
Fe 2h, = 2 6274 = 12548 Wim K
(8 Friction (drag) force = Ty x pu » area of
A Lou late for one ide

son» (Seta) uarase arme

1
a
EXAMPLE 11.46,

nefigerated truck is traveling on the high way at a speed of 0 Kay ina desert area
Sigue Ihe temperature is 70°C. The body ofthe truck is considered to be arcana box
aaa Ride 2 m high x 45 m long, The boundary layer i turuleat ove the ente surface
{m geMPerature of the surface i uniform at 10°C. Estimate (heat los from four surface,
Sein age of the refrigeration required, and (i) power required to overcome resistance
font on the four surfaces. Neglect any heat transfer and the fictional resistance ros the
tne att Back end of the truck ; and presume that or every 3500 W of Beat loss we need
SA capacity of the rfrigertig unit
on A the mean file temperature

me 10/2 = wre

396 / Basics of Heat and Mass Transfer
the thermophysical properties of air are
PLAT kg/m; ge 1911 10% kg/me
671007 J/kgdeg and k= 271° 10° W/medeg
Using these properties he pertinent parameters are
ALADO? „
CIT
Velocity of track, V = (80 » 1000)/3600 = 2222 m/s
Vip 224450117
a
Apparently the boundary layer is turbulent for which
Nu = 0036 (Re) (Pp?

Pre on

Re 589 » 10

m ME 0006 039 x 1078 (ap = 8
Heat transfer co-efficient, h » 8381 * 27,1 x 10/45 = 5047 W/mi.deg
and heat ost from the four side ofthe tack
=A Al = 5047 [2 (LS x 2 + 45 HI] 010) 130 9
(The cooling capacity required in tons of retrigeraion
136269

= BS 5096 tons
(© Average skin fiction cosficen,
=. 007 07
SI oa
Re” [590]

Drag force = Cy x(1 py?) area
u ze)
1 E
"00008 » (4x1127x2222 )«[26.5x2+45x5)] = 3881 N
ower = drag force speed
= 3981 2 222 = 58458 Nm/s = 89458 W
EXAMPLE. 1147,

Air at 20°C flows over a flat plate, 15 m long and 1.2 m wide and the plate surface is
‘maintained at 80°C. Ifthe rate of energy dissipation from one side of the plate i 390%
calculate the velocity at which the air must flow over the plate along its length.
Solution ; Heat lst from the plate to air is given by
Q=hAaT
3500 = (15 x 12) x (80 - 20) = 108 4
heat transfer coofcient, h = 3800/108 = 35.18 W/m£ deg

Let it be presumend that the

€) boundary layer is turbulent, and

1) transition

The average h

‘occurs at critical Reynolds number of 5 » 10°
heat warsfer coefficient is then given by

RME we Ther Bory Lors y 337

4 N
ho HUE TON
{At the mean film temperature 1,= (0 + 20/2 = src
the thermo-physical properties of air are
P= 1D) kg/m? sc,» 107 )/kg-de
BT 1936 X 104 Kgs and À 2781 10 Warez,
3007
ORNE ETES
Inserting the appropriate values in expression ()

0]

HE 19.56.
eg 07m

a E PAPERS PES

can owas]

Flow Reynolds number,

(20
ME

Now. Ra = Vipfn
Raxu LITIO 1957010 |

vorn

Flow velocity
1p fer)

1671 ays

EXAMPLE 11.18

Air at 10°C flows past a fat plate 1 m wide x 15 m long, The plate is maintained at 90°C
temperature and dissipates 3.75 KW of energy. Determine the convective coeiicient and the
velocity at which air flows over the plate

Adopt the following correlations

Nu = 5 = 0664(Re)™ (Pr) for laminar flow

ry

= Hoax ne 836 ]e{Pr) for trbatet Sow
At the mean temperature of 50°C, the tbermo-physcal properties of ar are
D 210 gm? ke 0008 Wordeg ; Pr = O78
Ey =10075 YAg-deg and p= 2029 x 105 kgs
Solution : Convection heat flow, Q = h A a?
375“ Wenn x16 15) « (0-10
x 10)

a
Convective coeficient =

= 15625 W/m

FEE + 15025 Wades
Al 18x15
MISES 09705

er

j

II N

338 1 Basics of Hest and Mass Transfer

Assumung laminar flow along the plate
Nu = 0664 (RAP (PD

83705 = 0.664 (Ra x (070; Re= 201 x 196

‘The Reynolds number greater than the critical Reynolds number 2 108; the sssumpigg

made of laminar flow is wrong,
‘Ax such the Auid ow is turbulent, Then

Au = [one Ra spe
Considering ow parali to length tube
83705 » [0.006 RO" - 836)» 0.705"

A E
“Los Tame * 91
y

o Res

or Re = alas a 7 354 x 105

y (7.358 x10%) x
Airflow velocity, Y = (350010),

1) 9g 2.029108
(7.3540 10°) x MIO 126 m/s
( as © 246 mA

EXAMPLE 11.19.
A smooth, thin model airfoil is to be tested for lift and drag in a wind tunnel. To obtain the
desired conditions the model is heated to 35°C surface temperature with internal eet
heaters while the free stream air temperature is 15°C. The model chord length is 15 mand
the air stream velocity is 20 m/s. Make calculations for

(a) the local heat transfer coefficient 0. m from the leading edge

(6) the average heat transfer coefficient for Ihe entire chord length, assuming that rial

Reynolds number is 5 * 10° and
(0 the heat transfer rate per metre of air foil width
It may be assumed that flat-plate theory is applicable and take
v=156*10% mys ; k= 0026 Wmideg and Pr = 0207

Vex, 20%0,
Solution: Re, © ==. 1026 x 10 75 x 105
Be ga 7 1026 * 109 x 75 x 10

Obviously the boundary layer is turbulent and therefore
Nu, = 0.0288 (Re,)Y5 (Pr)
= 0.0288 x (1.026 x 104 (0707273 = 165435

54.35 0.026
654.35 x 2228 5376 wm?
x = 5376 W/m"xdeg,

D Rae end + 1.923 «106 (turbulent)

Hydrogineme, on Therm
ma Boundary Lore 398.

When transition occurs at critical Reynolds numberof «10 she
piven by Ara Nr number

Fa «foutre Jo

D CONTE OC NE EE

O CAES Wim
(e) Heat transfer per metr of sito width (taking both des ite account

regias

1505-19 za i

mee |

the thermo-physical properties of air are
= ORM kg/m” ; à 2 284 = 10° kg/m
Ey "1017 hd and = 3459 = 109 Mmes
From these properties, Ihe relevant parameters are

DO ,

Pr on

2105 » 10

!

Presuming that transition cursa al Reyoids number ol 5 10) the average Noe
number

Nu = [om Ku en”

[oso =

wenn O ose ees
en

and heat ant rom one ide of he pi

a 75x08) x az nam

Further skin fiction or drag co-efficient is

(940 7 Basics of Heat and Mass Transfer,

0072 1670
RT = 1.0029
Esa tar

“and drag force on one side of the plate is

1
XL OV x ares of plate surface
Cox ov plate suf

= 00029 » (3 x0.894 x40") x(0.750.8) 038 N

(9) For the laminar portion it is necessary to locate the transition poi

The distaste y
om the leading edge at which the transition takes place is ‘

Re xu SMA „
02 Edo
For laminar portion, average value of Nusselt number is
Nu = 0664 » (Re,J » (PU
0664 x (5 x 1055 x 0799 = 417
Ak 41734.593010
x 0387
and the heat transfer from the laminar portion is
= 404 x (0357 x 04) » (275 - 25)
=1442W ; 308 % of total
Further the drag coefficient for the laminar portion is
1.328 1.328
wee,

0357

d= Nux. 404 W/mdeg

&

rome

snd hn dig foc om e ar porn e
«none» (¿osa (0.987204)

=0279N ; 308% of total
The calculations made above clearly indicate that fractions of the heat transfer and de
force from the laminar portion of the flow are equal.
(6) Mf boundary layer is considered to be turbulent from the leading edge, then
Na = 0036 (Re (POD
= 0036 x (1.05 x 10898 x (070/03 = 2100
2100 34.59 «10
ET = 9685 W/mdeg

Q = 9685 x (075 x 04) x (275 - 25)

h

264 W

‘which is (7264 - 4690)/4690 = 55% higher than that found in part (a)
Further,

007 m
p= ar = 0.0045
RIT nor)

< and Terre Sr) Lyra y 249

,
on fre = 8005 (Da mg à

whichis (030 -058)/058 «55% higher than tha
EXAMPLE 11.21.

md in par (a,

3:5
FE stare
the thermo-physical properties of ir are =

OO Wa à v0 10m} ; Bre
For the entire plate length, the Reynolds number is de ae

„DU. WW, 06x50
CRETE
© X, from the leading edge at which transition takes place is
30.0910
5
laminar boundary layer region, the average heat transfer coefficient is

os (PP

RCE

The dist

2 = 03009 m

(9 For the 1

OO a?
or

ro” 0600)

18005 KJ/mtrdeg

ae heat transfer from the laminar portion i

(03009 x 1) (295 - 25)» 14628 4J/hr
(H) The average heat transfer coofiient ofthe region of turbulent boundary layer is

DESSUS

And heat transfer from turbulent portion is

Q=h As \
37094 = (02991 « 1) x (95-25

42 Basics of Heat ond Moss Trnstr
Tou het fr rom he plat
701° Qu 12 + 2996 = us ar
estao cout sh te blind yang the dire formula fr veal my
cost
Ñ disc
= Hoeneß” eos

amero" cola
pora)" oan

275.06 Ki/mehe-deg
Q = 275.06 x (06 x 1) x (295 - 25) = 44560 KJ/hr
(0) 1 the boundary layer is presumed to be turbulent from the very beginning,

TS

2.1310 peu" x (08720
A210 0.682" x(9.97 x 10

7495.46 W)/m?-hesdeg
Q = 43586 » (06 x 1)» (295 - 25) = 70609 Ki/hr

= 0.036

% age error = = % 100 = 58.37 % on the higher side

4584
EXAMPLE 11.22.
In a gas turbine system, hot gases at 1000°C flow at 75 mys past the surface of a combustion
chamber which is at a uniform temperature of 300°C, What would be the heat Loss from the
{gases lo the combustion chamber which can be idealised as à flat plate measuring
100 cm » 50 con ? The flow is parallel to the 100 em side and the transition from laminar to.
turbulent conditions is anticipated to be abruptly at a critical Reynolds number of
Re, 5 x 10
Take the following properties of gas

70496 kg ; v= 935 x 10% mis

K = 2.0744 W/mK and Pr = 0.625
Solution : The average value of Nusselt number over a flat plate when both laminar an
turbulent boundary layers are present is given by

[otre + 0.06 {(Re)* (80)

When transition occurs at a critical Reynolds numberof Re, = 5% 10% the above expresion
simplifies to

Nu

Ne

12 (ray 36]
The Reynolds number at he plate en is
fou, WU, _1x |
TE Zr * 802 * 10
Apparently the boundary layer becomes turbulent and the above correlation apple

Re

Ru «(0425 [pasoo] m0] «9420

Heat loss Q=hA at
6802 (1005) (000 a0) =

23809 1

An Sad RE me
ipa cot Rye neat omar es de
calculate forthe plate and thickness of thermal bounty eye ee

= 1000 kgm ;

Physical properties of water are

‘00
£2058 Wines bod pri 7
Solution : For the entire plate length te Reynolds nerve
Ae = PU 1000055
j 2 1700210 * 14 * 10
The distance x, from he Leading edge at which tansitoneccursi¢
E Os
E ÓN

The average value of heat transfer ocfiiet can be worked out by using the relation

ut]

.- Fern [orgy 6 [rm

= 255 nf e fe tal
A 7]
"2609 (DGA = 61245 + 006 2585-314» 605 Win? K
U 05x08
ro Oss

Nu

= 5159
EXAMPLE 11,24

Pons flat plate is oriented parallel to atmospheric air which flows at 25 ays. The air is
aan nperature of f= 20°C while the plate surface is maintained, = 20°C. Measurements
om that the plate experiences a drag force of 1175 » 10 N. Use the interclatonship
een energy and momentum transfer o workout the hes taster cei 3 the pate
e. Assume laminar flow and take the following properties of ar
2 E AR Kg K and Pr= 684
tion à The drag force is

El
Fe Cy} pu? xarea
jot

ABE Duta) as be

344 // Bastos of Heat and Mass Transfer

os Ben

FE en

E x puË KP = 0.664 pt a b
ee] ES luz

Inserting the appropriate values

1175 «104 Pon

(mx

aa)
The Reyna ruse te pate end

UT Fre” fas
From Cotburn relationship between energy and momentum transfer
$ & h ps 7
pon EL o
ee oy 4

U pl
ne
393x107, 1.029%2.5% 1.009
TR

= 6497 Wim? K
497 (0.964 x 0.964) x (120 - 20) = 603.80 W

Heat joss, Q = À A Ai

= 001200

EXAMPLE 2 ——

Air at 1 atm pressure fiuces over a plate 0.5 m long with a free st
the drag coefficient for one side of Ihe plate is found to be 0.0037,
value of convective heat transter coefficient.
At the operating mean film température, take
(P= 1.06 kg/m? ; c= 1005 /kgK, k= 0029 Wmdeg and Pr
Solution : Invoking the relation for the average drag (friction) coefficient

2,138 1308) pa Pon
a ES

Fi of
os ps «359
Ge Murten ER 226 wat

ream velocity of 5 ms
make calculations for e

= 0596

(micas

Ther
Alternatively : From Colburn anaoiy ma Boa Lopes 7348 À
sr pau „Cr _ 0.0037
27 os
0.00185

= 0.00
Waseem * 9002356

FE 79002356
1h 01002356» 1.05 10055 = 1255 wher
EXAMPLE 11.26.
at 20°C and at atmospheric pressure is flow
Air at 20°C pheri pr flowing past a fat plate 3 ays velocity. The

ean een ape Fe: Ce

‘Also make calculations for the drag force exerted on the first
Use the analogy between fluid friction and heat transfer nn
‘The relevant thermo-physical properties of air are

i
i

P= 1128 gm? c= LO Agde à 1 = 00 md
zu B= 183 «10 Kal = races
Solo + At tance x 03 m form the leading ee he Reynolds nunder 6
ROOMS
TA "so

REZ:
{es io xs
a RS
he flow is laminar and the following colton applis for he focal Nusselt number
Na, = 0332 (Re? (299
332 x US 06)" 6789
at 0.059
Nu, 67.892002
The average value of het transfer oeficien eee this vale
B= 2h, = 2 2240 = 4480 N/a dedos
The heat loss from one side of plate is
Q-Bası
= 4.80 (03 » 1) «(0 - 29 = 76 KV
ionship between fu Fiction and het un

40 Ky / mde

(0) The interet

pan a EL
sipep «Y
ñ

Pu
Substituting the appropriate values,

— ___ 210,977"
Ta)

(erp? a YY

(348 Gasics of Hest and Mass Transfer

sonra wo
nz

ña
5

= 0.023 » (3,976 x 105° x (0.702) = 617.41
“and the beat transfer coefficient N

fw ALTO? 39,6 yor
——
“The length of tube is determined from the heat balance equation
Qandil (= 1) = me, (4-1)
The rate of water flow and the amount of est transfered to i is

Ru oro

m= EA vo E RON 105998 «1259 19/6

Q = inc, (4, = £) + 1253 x 4183 x (25 - 50) = 524130 W
524130

a OS™

Tobe length!

REVIEW QUESTIONS

111 What do you understand by the hydrodynamic and thermal houndary layers ? state wih
reference lo flow over Dal Bete plate

112 Callate the fiction drag ona plate 15 em wide and 45 clog placed longitudinally in asa
‘of od (epi gravity 0985 and Kinemate vacosity 09 stokes) Nowing with fre Srs vey
‘of 6 m/s Als find the thickness of toundary layer and shear stress atthe train edge

(Ans. 1764 kg 0013 m 65 kg

113 Air at 20°C and 1 atm flows with a velocity 75 m/s past a fat plate placed at zero age
incidence, The pate surfaces maintained a uniform temperature of TAC. Calculate () hd
‘boundary aper 08 m rom the leading edge, (i) postion ofthe point o transition 1 tebe
Tow ithe rial Reynolds number is 510, and (i) drag force on one side of he pao
he fest 0 m length. Ansume unit width of pate (ans 0731 m, 1385 m O08 N

114 Experimental results for heat transfer over a thin lat plate were found to be cor by #
presion ofthe form

Mu, = 0352 (Re) (PAS .
here M, st oca value of Nuselt number a a postion x measured from the leading ss

Se pe Otal an pra o teat the Serge het ae len Beh

115. Atmos ala PC fons ac ft pt a 28 m/s fhe plate is video AE
is maine at PC, make calculators for long params dae oF 075 ©
j iting sage
(9 yéréyname boundary ae hike and he ca ion offen
other fount er he andthe al heat Water ofen,
(Oo am and QE
(0 LS em and 350

anak

ns

m

ns

ns

A bin fat plate has been placed
Andisturded velocity of 73 m/s. eus
FAI. Calculate the heat ander soe
cult the rate of het ander fem
Same unit width of he plate

Water at velocity

Born eh

1 1.25 m. The tem rn
of the main water stam are en

he man wae rom Pe Coe et
hl coi tthe md po

Palo at HO' and the temperature
and hydrodynamic boundary layer tudo
ofthe pl

fee plate

ae ee pro te

pil param fh pte us he han ne pd

the trailing edge of the plat, () average cran ah TE fincas at
ee

nge Nusselt number far turbulent flow over a Mt plate u
Sur abrup when the local Reynolds number 4 0 PAN he the Karten

(Aes (oas(ReP* oto) >)

Determine the ratio af ha an cin. rum the al afte o mates A) and a
{ip The tubes are of equal mer and the funy ml nn OL Ros

‘umber and Prant number, What will be th america! value of hs abo Y watt imparta
127 200°C and sir temperature f= 25°C > ieee

Condensation and Boiling

Condensation and being are the convective heat transfer procesos that ae associated wi
change inthe hase of a oi Condensation refers 10 a change fom the vapour to ey
phase and boiling involves change from liquid to vapour phase of a fluid substance
presses an very common in power plans (boilers and condensers), refrigeration syste
{evaporators and condensers), process heating and cooling, melting of metals in furnaces an
heat exchangers used in refineries and sugar mill etc where heating of the process fluid ana
by the condensation of steam, Daring a phase change, there is ether liberation (condenser)
absorption (boilers and evaporators) of latent het, the rate of energy transfer is quite pd
even though the accompanying temperature differences may be quite smal. The phenome
are quite dificult to describe due to change in fluid properties (density, specific heat emma
conductivity and viscesty) and due to additional considerations of surface tension latent han
of vaporisaion, surface characteristics and other features of two phase flow.

This chapter will be devoted to examine the condensation and boiling, processes with
regard fo their fundamental mechanism and the means of describing and predit ting heat tater
rates.

12.1. CONDENSATION
Fluid in a gaseous or vapour phase changes toa liquid state with the liberation of hat fon
the vapour
When a vapour iin contact with a surface whose temperature, is lower than the sturatan
temperature. corresponding tothe vapour pressure Ihe condensation sets in andthe ap
Changes to liquid phase, The condensation of vapour liberate latent heal and there he os
to the surface, The liquid condensate may gotsomewhat sub-cooled by contact withthe coste
surface and thet may eventually result in more vapour to condense on the exposed surat
‘pon the previously formed condensate,
Depending upon the behaviour of condensate upon the cooled surface, the condensa
proces has been categorised into the following distinct modes
‚9 Film condensation : The liquid condensate wets the solid surface, spreads out nt
forms a continuous film over the entre surface. The liquid flows down the cooling suet
under the action of gravity and the layer continuously grows in thickness because e M
<endensing vapours. The continuous film offers thermal resistance and restrict athe ar
‘of eat between the vapour and the surface,
Film condensation usually ocurs when a vapour, relatively fre fom impurities
tocando a dan uae, AP re fe rom mp
(en Dropwise condensation : The liquid condensate collects in droplets and does noté
break away f P ‚ice
ar tom the surface, knock off other droplets and eventually run all the Sue!
‘tout forming a film. A part ofthe condensation surface is directly exposed o the OP

350

¡salon

NS O SS

shout ap insulting im of condensa hu fy
rot higher heat transfer rates are experi ee arn Deren

heh

° 7]

au

Le

Corson tm
10m eondensten

comes coran
Flo. 121. Fam and Gnome corte ana vec saco

Dropwise condensation has been observed to ce
on surfaces contaminated with impurities lke fat
condensation gives coefficient of heat transfer ger
condensation. Because of potential performance gi
by surface coatings, called promoters, hat inh
of waxes and fatty acids are often used for this purpose These substances are ether appt
to the heat transfer surface or intecduced into the vapour. However. the phenomenon ie ny
unstable as these coatings gradually lose thee efetiveness due 19 oxidation ula, o ou
removal and the surfaces become wetted when expused to coming vapour ever an extended
length of time, Consequently film condensation is generally encountered in usta applcanons
and is usually planned for condenser design cakulanons

122. LAMINAR FILM CONDENSATION ON A VERTICAL PLATE

Gonsider the process of film condensation ocuring on the surface of à Nat vertical plate as
Scpicted in Fig, 122. The coordinate axes of the system have been o chosen that the origin
& # a the upper end ofthe plate, the x-axis hes along the venal surface (poste dico
95 measured downward) and the van 5 perpendicular tit The theknew of quid lm,
ich i ero at the upper end of plate gradually increases as further condensanon cure at
Uh lguid-vapour interface and attains its maximum value a the lower eng he pate The

ur either on highly polished surfaces, or
acts and organ compounds, Drops
neral five to ten times larger than with Ilm
in. dropwise condensation 1 provoked arly
i wetting. Sihcons, mins and an amortment

tal plate has height width , and 8 denotes the thickness o im at à tance rom the
engin
capt timate of the heat transfer coetiient forthe liquid film cn te made by setting up

‘ for the velocity distribution, the mass flow rate and heat flux through the
maak The analysis is based on the findings of Nusselt (1916) and st makes the toikwang
sumptions
* The liquid film is in good thermal contact with the cooling surface and therefore
femperoture at the inside of the film is taken equal 10 the surace temperature f,
Further, the temperature at the outer surface ofthe il (nterace o qu and vapour)
“quals the saturation temperature 1, at the prevailing pressure,

i

352 7 Basics of Heat and Mass Transfer

Là The condensate film is o thin that a linear temperature variation exists between y
plate surface and the vapour conditions. the
The physical parameters (the thermal conductivity k, the dynamic viscosity. ang
en 9) of the condensate film are independent of temperature, the
+ Vapour density is small compared to that of the condensate

ME
Ys

on

Ran gown pie one mean

The vapour delivers only the atent that (Le, there is no under cooling of condensat),
and the heat flow across the plate is by conduction

© An element of fluid mass within the film is influenced only by the viscous shear and
the gravitational forces the forces of inertia appesring in the condensate film se
disregard

+ There is no velocity gradient at the liquid-vapour interface
shear at the phase interface is negligible.

+ The condensing vapour is entcely clean and free from gases, air and non-condensing
impurities.

+ Drainage of condensate film along the vertical surface is by the
is through a laminar motion.

+ Radiation between the vapour and the liquid film; horizontal component of veloc $
any point in the lui fm and curvature ofthe Alm aro considered negligibly sl

md obviously the vous

sn of gravity and

* Velocity Distribution
‘An equation for the velocity distribution u as a function of distance y from the wall sr
can be setup by considering the equilibrium between the gravity and viscous forces 9 #
@lementary volume ( x dv) of the liquid film
Gravitational force on the element = pg ( dx dy)
du

Viscous heating tes on the element ace at y = A

change in shearing stress in distance

quating the gravity force 0 Ihe net shat for

à a
Ps (e de dy) = na [de
Le eus (ad

upon simplication,

Imegroting twice a

ese conditions determine that

470 an ge

h
Therefore the velocity distribution through the fin spree roving en
reatonship prescribed be owing parole

Pal gy)

The mean flow velocity Y of e ld fin a à dia rom he op gt cn de
determined from the expression y “rss

velfuay

1pos(ay |, . 988

paf 28% 123)
CINE TNT e

* Mass flow rate
The downward flow of the liquid at any elevation x (be. over the layer of thickness )
mass flow rate = mean flow velocity * flow area * density

ES ay
mo tin

gl mass wis has. anton of hs Be ef tenes Di ey
dent upon à =“
An increase inthe mass flow rate of condensation during dotará Dow ON

rom vith respecto oF 8
fom x 10 x + ds can be worked out by diferentating equation 124 with espe

end Mass Transfer

354 / Bass of He

ale) | 3h! 5246
im m D 25,
+ Heat Fux

The heat over
a he surface, Tus,

e into the im, 4Q, equals the rate of energy release due 1 condensation

hs
A gdm = hg OE ie

veer i the tet eat of condensation
Nuwelrfresumed thatthe heat eased during condensation flows ony by conducto
through te fm

bade
iq = un) Linh

Combining equations 126 and 127, we get

auge
Bas = 0)

En

Integration yields an expression for the thickness of condensate layer
E pre
TS

Substitution of the boundary condition 8 0

fa

Evidently the film thickness increases as he fourih root ofthe distance down the sur
the increase is rather rapid at the upper end of the vertical surface and slow therealtee

o

as,

* Film Heat Transfer Coefficient
[Nusselt had presumed that heat low from the vapour to the surface is by conduction
through the quid film, Le,

re

3
‘The heat flow can algo be expreseed as
dQ = h, (0) (hay = 1
where isthe local heat tranefercoelicent. It follows from these expressions thal

à
et

má 029
Thus at a definite point on the heat transfer surface, the film coeffciont h, is direc
proportional to thermal conductivity K and inversely proportional o thickness of lm dat lat
point
Substiuting the value of fil thickness 5 from equation 128,

219)

Condensation aná Gong y 338.

jocal heat transfer cosfficient atthe
ë the lower edge at plate te, at x =

Léo, Vs

Tilt)
Undoubtedly the at of condensation host ose
spon atthe lower end Heri higher atthe upper end 0 e pie

| Ppp integrating the local value of conductance (e
“get the average heat transfer colis

[ve

ay[ een
“Tho! tux,

"ation 12.10) ver the entire length of the

plate, we

2

Tessa, PE

J PE af or, 7”
el far] ses) “54 a
were hi e local hea tante cociente ler edge al the late
Tho follows rom equation 129 tet
au
Bez aan

where &{) is he film thickness at the lower end of the plate Obviously the average heat
transfer coeticient à 3/3 times the local heat taster coefficient atthe ling edge of the
plate

Equation 12.12. is usually writen in the form,

| sas

The Nusselt solution derived above is an approximate one because of the assumptions
Admitted inthe statement of the problem. Experimental result have shown thatthe Nusselt
“squation 1 conservative yield results which are approximately 20% lower than the measured
values Accordingly, use ofa value of 113 in place ofthe cocffcint 043 hasbeen resocrmended

met

i as

The results concerning devel ondensaton along a vertical at plate indicate
concerning development of film conden ni r
thatthe thickness of film increases with increase in plate height, equation 123), Since the
Thermal resistance increases with fl thickness a decrase in het ranser coefficients expected

and that is evident from the relations 1210 and 1215. The dependence of these parameters on

355 Basics of Heat and Mass Transfer

aight of plate hos been shown graphically in
Fa 123 unter the convection coefficient
Anse with temperature diference (ng 1)
‘Tne may be tuted to an increase in he ln
cines which resul fom incense in condensate
fate at high temperature diferences

Wii computing the average count vide
uation 1215 all the id properties are evated
the film temperature f= (o, + 19/2 andthe
Napour latent Ret of condentition 1 evaluated
At. Having thas determined the average
Condensation’ coefficient, the. following
expressions are used o aba the oa et tater
and the total condensation rate

mia) 0226

fi cost

eae

Fig. 123. Voisin of fn tiens and ten
and

„2. Mami)
Mg My
For a vertical surface, the flow aren Ais DB

Fa. 124 Geomayo an ninos fat condensation saca
For inclined flat surfaces, the gravitational acceleration in the basic Nusselt equation
replaced by a projection of the gravity acceleration vector on the waxis:g, = 4 ind where
is the inclination angle with the horizontal, This yields the following expression for fin
condensation on a flat inclined surface;
PoFahy (seine) e219
Han) |
Evidently,
DATE am
Equation 1219 must be used with caution for small values of inclination 8; it acc
comes poor as the angle of inclination approaches the horizontal

123: TURBULENT FUN concen
comes undulating in the mide seen Pat: then
Insisance as it does in laminar fm, fe Pa
fusion which is a charscteshe of non Ca
aa encountered when the fl hicks eat
appreciable, when the condensation ate nn

lent a |

‘The parameter indicating he commencement of FB-124 Ron tm
nr Roms ei
Wend os Een Vi n/a where

That gives

and

Since m =p AV = p (BÖJV, we may aso write
sm

Expressing he mas flow rate m in terms of he est ar,
2 Alt)
Ir

iy
we get M

Teno thy «a

sox tion for lamina 1 urbulet Now accra tel Reynolds unir 190
or turbulent film condensation on vertical surfaces, Kirkbride has suggested the following
omtion for the average heat ranfer euffiient. “se

Four [E] man
which is valid for Re > 1800. 5
y distinguishing features between he lamina nd turbulent fin condensation as anveyed
Y ains Tara an IE ae
{Inthe laminae fl the average im colicin cress with distance This i due
10 gradual cree in he thks of aia o

sa asco nis te

12.4. CONVECTIVE COEFFICIENT FOR FILM CONDENSATION ON TUBES

(a eral te «When e tbe later s og compared 1 he fm cen,
coon sft ot oem cn vera abe Ey the Newent
iar condemn an a venal fat ura: That on

vi Single uvizutal be; The average coefficient for fm condensation of a pore saura
apour on the outside of a single horizontal tube is determined from the correlation

Kreg
HD Gar
where Ds the outside diameter of the tube
Ieinteresting to note the relative efestvenssof horizontal and vertical aes as comeing

ar

LE sal

E 7 _ gage LY"
SL)” oz E
“own 5)

(22

When À, = solution ofthe above relation yields 4/0 = 256. Thun fora cylindrica tbe
witha Tength-to-dlameter io of 286, equal amount of heat anse oss from bt horn
Ind serial erientions. As the ratio 1/0 ineccoses, Ihe greater heat anser and 5 igh
donation rate 1 possible wit a horizontal tube. For mon seam condensrs. the re

1/0 for à tube is in the range 50 to 200 or even mo

seo 6 0
O) Ö ©

Fg, 128. Cones of ecient tube amar fm oral bank tubo
(ii) Banks of horizontal tubes: For a vertical ie of horizontal tubes, the average comes
coefficient for film condensation is

h 0.735).
where the equivalent tube diameter :
nr ee

aging tbe inthe bank
à reduction the fim oie ir,
ae cn Cath success uber da ao ed a nr nthe
tubes. Obviously it i advantageous to stagger the tubes eee” of dep from the ope
fo the upper Fo i a at paral ore 127 a ac

sd by the drip as falls fom ane ue li ic

ft
a

1.12. Fn onto on tor on
(o) Contention inside tbe : Condereson of

lersation of pour ocur onthe inside suce of
cylindrical tubes used in condensers of referat and conne ese nd eae
atthe chemical and petöchemical industries Te owing apour Roue arde ones
te Shrce nd thicken ofthe conden fm att see hel ear Te
mognitude ofthis effet ts dependent upon the helena or vera ponian othe tes
and also upon the upward or dowaward Bow isn E

‘by the station

Following empirical correlations have been suggested to workout the hat tarser when
steam condenses on the inside surface of tubes,
For saturated steam

Qe Gas + 116) an) 225)
2+ son” 139

17 STE diste cooled length, A is the surface ares based on inside diameter ofthe tube and
Vis the entrance velocity.

360 1 Basics of Heat and Mass Transfer

EXAMPLE 12.1
A condensation experiment for steam on plate type vertical condenser has been setup for,
patticolar fluid with a given temperature difference. The same setup was subsequent
{sed with another fluid with thermo-physical properties given as :

Page à Beg, Men à
CT à
ic temperature difference treed to 80 percent, make calculations forthe percentage
change in the convection coefficient.
Solution: The average hea transfer coefficient for vapour condensation on a vera ple,
Fo}

CN T°
ove let]

# OUEN
Boa T°
an py = 0.949) FE
Substituting py 0459, à A : 120:
D 95 Mag Clay = tp © Ol = ty ary et

be woul PRA
IAN zu oh xO,

CE Pf
"os. T4

ie. there is an increase of 15.2% in the convection coefficient,

EXAMPLE 122
(A plate condender us designed to be hep vertical How would Ihe condensation coc
KR A duc tonite contain, as 10 De ept at 60 to the horizontal?

( A plate condenser of mensions 1b ha been designed toe kept with ie
the veal postion. However due to ovessight during eetion and (stalin, it as
‘Min side b verde’ How would tis aca the heat rnsfe ? Assume laminar coos
Ind same terme physical properties in both cases and tke b= V2

(6 Determin the length of 28 em outer diameter tube the condensate formed one
saute ofthe tbe isto De same whether its het venia or horizons

Solution () For a vertical fat plate

ee gg [7
tro = 05) PB
nn
For inclined fat surfaces, the gravity acceleration y srl
inciaton angle withthe horizontal, Then

sd by g sin where 08 be

s Condesa
À pridently lo Mo (sin à ion and By
‘This implies 3.53% réduction in oo Gin 608 « ogg baad

lnsation caticient

feet

(o With side vertical

comments: The condensation coefficient and
| aorer side is Kept vertical For better condensatgs
À Saver side vertical

according heat fo o
acord I increases when the
he condensers shouldbe istlled with

(e For laminae film condensation on a vertical tube

horas Ls FO
| | o
i rer]?

ee mn]

From expression (D and (i),

| 0
For equal amount of condensation, the heal transfer rate and accordingly condensation
(ocfiient should be same for the horizontal and vertical orientation. In tat case

Js

75 cm

| example 123
la For condensing conditions, compare the condensation rate when a 65 cm
125 m long pipe is kept () horizontally and (1) vertically. Assume that ot

(0) For condensing conditions, compare the values of convective heat transfer coefficients
era pipe of diameter with that of two pipes having the same total cscunference when
Ai pies are horizontal and parallel and (i the pipes He one over ie other, Assume
that other conditions remain same.

Solo: () For vertical position
he ooo) Peon] N

Ina)

And fr horizontal position

|
|

362 7 Basics of Heat and Mass Transfer

nf
" un i)
From iden () aná ()
Roms (1) 07s (135)
à ose (a) “ass” (ones,
Cbviounty horizontal postining provides 61% more he transfer. This may beat
solange nes wih incense engi. Accordingly condenses are general of hres
ore
(9 Te average heat transfer cefcent for vapour condensation on horizontal ue i

given by
Bag |”

HD Ga 1)
Case (i): With same total circumference KD = 2 x nd and therefore d = D/2. When diameter
is reduced to half the value, the convective coefficient becomes (2)° = 1.189,
EXAMPLE 124
A vertical cooling fi, approximating a flat plate 40 cm in height is exposed to steam y
Äimospheric pressures surface ofthe fin e held at 80°C, make calculations for the fllowing
parameters:

film thickness at the bottom edge ofthe fin,

(9 overall beat transfer coefficient,

(i) Neat transfer rate and the condensate mass flow rate.

Assure unit width of he fin and check the flow Reynolds number for the assump
of laminar flow conditions.

(6) Estimate the minimum height of the plate necessary for condensate to become just
turbulent
Solution: For saturated vapour at atmosphere presu he saturation temperature = 10°C
and the latent hat of vaporsaton by = 2257 * 10° J/kg,

For saturated water at the mean film temperature,

y= (lg + 19/2 = (10 + 80)/2 = 90" €
the relevant fluid properties are :
p= 9633 kg/m; Ke 67995 x 102 W/mK and = 3.153 x 104 59/08
(9 The film thickness at a distance x fom the top edge is,

ETES D
e Piel |

[4x67.355x10° 3.153% 10+ 10090
REN
214278 x 104 0% il
At the bottom edge ofthe fin x = 04 m,
Bo) = 14278 » 20+ « 0402

rar

135 x 10-4 m = 0.1135 mm

ds) Overall eat transfer oc,

; Es
pong ME Adams comio ra

BE, et
= 105792 wy pit
| qy Heat water tale Qh u OK
= 108792 ar
com condensation rate O41) * 00 0 « ass yy
mL. 558
Ig” Brea 000521570

| checking the film Reynolds number, we get 7
=. 4200193
Ab" 3255x107 <7 * 248 < 1600
mus the assumption of laminar fl as been comet
{h From the correlations

Re

ño. onl ty
=) -a2ın
Aal)
and re = A, tat)
HO ly us

itis note that Re «194 oF «Re in the laminar regime

EXAMPLE 12.5
Saturated steam at atmospheric pressure condenses on the outer surface ofa veal ni
stlength 1m and outer diameter 75 mm. The tbe wall retail a actos ee
Iemperalre of 40°C by the flow of cooling water inside the tube. Esimate the sica
condensation rate and the heat transfer ae tothe tube, What wae om tte will eal ie
48C temperature difference of water between the ont and ne of pipe? Alo cele
«low Reynolds number to check the assumption of laminar low conditions
Selon: For saturated vapour at atmosphere prasue,
tas "100 °C and tg 2 22587 Ug
For saturated water at the mean film temperature
ty llas + 49/2 = (100 + 40)/2 = 70°C
the relevant Auid properties are;
metan 7 han) 480 /mirdeg à = 406% 104 g/m
Him thickness at the bottom edge ofthe tbe i,

[simi en]
Peek |

364 7 Basics of Hest and Mass Trenster

Inserting the appropriate values in consistent units,

rage heat transfer coef
she averse cen fr sea, 6 Bin
45 2.409 (4,065 10* x 3600) (100 40) 1 “ „ee
ir NE ss
(77.8) «(981 x 3600 x 3600) x o > ee
Am > 258.76 Eo Lean,
= 2358 x 10% m © 02354 mam Dante
‘The average heat transfer coefficient i, |

ta ee?)
935° 3 Tags 13610 W/mardeg.

lying Me Adam's correction for steam condensing on vertical
À = 12 x 13610 = 16992 KJ/m2hrdeg

() Heat transfer, Q = h A (= +)

FIR > 007841 00-49 2071 We
and the steam condensation rate is ul

Fon (3007

= 0,725 [RA 295.9 9, .
ie

Plates oreytindes, | = 520534 Wyk |

aa surface aes of 400 tubes = 400 « xg

Appl

OR 0006 «1075
QT
Heat flow rate, Q = AA 4) = mi,

Tus,
TE a
Checking the low Reynolds number, we get 1096107 = 485195
3m 4210217 135
Rew im, be length of each tube, t= UE a
Hb” [8.065107 3600) a "117 < 1800 | a LS
‘Apparently the fil ow is laminar in character. | ns. Bowne
(0) he heat released during condensation is picked up by water flowing inside the tte, | Bol constitutes the convective heat wafer proces ut ins à ph charg kom
Therefore, = Ji to vapour state. This is achieved Uough Fe supped w te heed al ee ee,
Qe lly 5 23071 = me x 4186 x 5 hated surface is expose to the liquid and tuintained asters sore arg
Flow rate of water

s
ET, co vga

EXAMPLE 12.6

temperature of the liquid. The boiling process has widespread aplicas eg, m

{9 production of steam in nuclea and steam power plans for generation an frst
processes and space heating

(absorption of heat in refrigeration and airconditioning systems

Gi een, detyáraion nd ding lts ad e
A condense sto be designed ioconrse 23501 o Samper osa rem (ln dean an rg
A square ary of A tubes each of mm in lamer is seal ar te ath he reining of

surface temperature is to be maintained at 26°C, make calculations for the length ofeach eater importance has recently been given o boing heat as bcs desarme
tube.

Solution : For the saturated water at 015 bar pressure
yg "SPC and hy = 2373 x 10° J/kg
For saturated water at Ihe mean film tempeature,
the relevant fluid properties are

n = AG = 20

= HD = 20 x 006 = 012m

Cu + 192 =

PH ODD kg/m ; k= OM x 102 W/mK
= 285.602 x 102 kg/hr = 654 x 10% kg/se0
The number of tubes m in a vertical column of the square array of 400 tubes is,

and for these 20-tubes in a vertical column the equivalent tube diameter i

| dits

rade reactor spa and eins mi o ques es PS à
lined space and ar o De disiated a rts as igh a 18 W/o Su gh he te
ts can be well appreciated by comparing it with he maximum heat transe ate ina modem
aed ohare
TRetolng phenomenon owe to inte evn m
| opa de er
{9 Poot toting : The liquid above the hot surface is essential sagnan an ©
‘es rth arica shoe ol eee a ong nd es
‘detachment The poo! Boling occurs sen ole pera I NS

0 Foret conection ting +The uid moto inde y eel eas Te ES
ES ta ec acs the sae in cold mans De

| NéPlsinduced-mixing also comite toward id pc ence
i Subcooted or loca iting : The temperature of the lq Ren sel a shart

ing takes plac ong vc ee neds TEP

ewe A

366 // Basics of Heat and Mass Transfer:

and the vanish apparent hey condense inthe blk of he gd Which stg

des han the Boiling point pe
(e) Saturated ing: The temperature f the gd excceds the tro

The vapour bes generated a the solid srfae(igud-sols interact) see pe,

Ve id tuovancy effets and eventually scape rom he surface (iin N he

The actual evaporation process then sels in "cine

126. BOILING REGIMES

Whether the boing phenomenon correspond to poo! bel
thre are the elit regis of bling acct wih ron an hi
These regimes have ben identified in Fig 128 which shows Ihe rt PE ot
(om Solicien) and the temperature excess (= fa) 1 te nenn Meng
dai ten
is being evaporated Cie
Are inthe slope of th curve indicate an increasing heat Tux it
exces andthe boing epee Wen te sop dese eh ok
and must be avd Tae regimes cane tasca by come any
heated art we aber ina poof iq ss impar JAS
is easily controlled by voltage drop across a wire of fixed restan hy
9 Espora rcs wi ab oration te pr The beating
ina tn arf id whch aus the ated sure Te hd ne on
divas apenas emptor
Been pes TR spe ud eso the up inane
takes place The id motion is deveined primal by

convection effets Toe kee
transfer rate increases, but gradually, with growth in à temperature exe

ay

8 OF force circulation

easing te

E
E

Pot nose

marge Mig
iosoce ao
y 19 10 LE]

Temporis ent] ©
Fig. 128. Boing cae or or: arc Pes us funcion of exe era

saturation MP

2) Nace ig; When the iui is overtatd in elation o str PE
‘pour bites ar rm cri favourable pts led the mun de
these may be the wall surface irregularities, ai bubbles and the particles of dis Ta

grow to certain size influenced by pressure, temperature and surface tensión

vapour interface, De
the following stage
(e) Bubbles form and cot
0) Bubbles form onthe area PE
{tom the surface a
Elo bat ay fom eed
SE, Ge ho: ad mi eg
id sara and A

ate dire condense

‘evaporation Y pele the vapor They ee,
ace and tha eps rp

‘The mechanism and the me

(@ Liquid next o the

Top of he ube comes ito cone
arrest the bute guna
(©) nea o he lid nd he
pasivo ths it ea
the feted sure
Buble begins 1 op and 6
CO re pe and ec pa ely wf ne et
(0 Babe saltar
the hen sae
(i) Exel the enki gene ator the rat ame
{98 for bul fermion and cla De, er

th the cooker gud and at as tendency to

bubble hus caused the bubble to grow
‘more hato the cooler aid th i cca tua

anit guns by condo foe,

laps and inertia ofthe cooler iui brings tino contact with

nn
A

At xt

SS SSS ae

ab AM

RES It SS

Fi 129, Mein e Et bmn ed catan
‘nucleation boing is us characterised by the formation of bubbles a he mucin
‘and the resulting liquid agitation. The bubble agitation induces considerable ud ming
And hat promotes substantial increase in the het lex and he boing heat eae
(9) Fm boring: The bubbie Formation very rapid the bubble blanks the heating sce
nd prevent the incoming fresh liquid from taking thei place Eventually the babies coalesce
a erm a vapour fim wich cover he surce come sing ef fe vapuz
{im low thermal conductivity) oveshadons he Benin efit of gui again and
‘consequently the heat flux drops with growth in temperature exes. Within the tempera

The
sites

ia ann

Heat transfer by dr. d.s kumar

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