Heat transfer course : physical origins and rate equations

HEAT TRANSFER COURSE

1. HEAT TRANSFER:
PHYSICAL ORIGINS
AND
RATE EQUATIONS
Chapter One
Sections 1.1 and 1.2

2. HEAT TRANSFER AND THERMAL ENERGY
•What is heat transfer?
Heat transfer is thermal energy in transit due to a temperature
difference.
•What is thermal energy?
Thermal energy is associated with the translation, rotation,
vibration and electronic states of the atoms and molecules
that comprise matter. It represents the cumulative effect of
microscopic activities and is directly linked to the temperature
of matter.

3. HEAT TRANSFER AND THERMAL ENERGY (CONT.)
Quantity Meaning Symbol Units
Thermal Energy
+
Energy associated with microscopic
behavior of matter
Temperature A means of indirectly assessing the
amount of thermal energy stored in
matter
Heat Transfer Thermal energy transport due to
temperature gradients
Heat Amount of thermal energy transferred
over a time interval  t  0
Heat Rate Thermal energy transfer per unit time
Heat Flux Thermal energy transfer per unit time
and surface area
or U u J or J/kg
T K or °C
Q J
q W
q
2
W/m
+
Thermal energy of system
Thermal energy per unit mass of system
U
u


DO NOT confuse or interchange the meanings of Thermal Energy,
Temperature and Heat Transfer

4. MODES OF HEAT TRANSFER
Conduction: Heat transfer in a solid or a stationary fluid (gas or liquid) due to
the random motion of its constituent atoms, molecules and /or
electrons.
Convection: Heat transfer due to the combined influence of bulk and
random motion for fluid flow over a surface.
Radiation: Energy that is emitted by matter due to changes in the electron
configurations of its atoms or molecules and is transported as
electromagnetic waves (or photons).
• Conduction and convection require the presence of temperature variations in a material
medium.
• Although radiation originates from matter, its transport does not require a material
medium and occurs most efficiently in a vacuum.
Modes of Heat Transfer

5. HEAT TRANSFER RATES: CONDUCTION
2 1
x
T TdT
q k k
dx L

 
1 2
x
T T
q k
L

 (1.2)
Heat rate (W): x x
q q A 
Application to one-dimensional, steady conduction across a
plane wall of constant thermal conductivity:
Conduction:
General (vector) form of Fourier’s Law:
Heat flux
q k T 
Thermal conductivityTemperature gradient
2
W/m W/m K °C/m or K/m
Heat Transfer Rates

6. HEAT TRANSFER RATES: CONVECTION
Convection
Relation of convection to flow over a surface and development
of velocity and thermal boundary layers:
Newton’s law of cooling:
 
s
q h T T

  (1.3a)
2
Convection heat transfer coeffici: (W/m n K e t )h 
Heat Transfer Rates

7. HEAT TRANSFER RATES: RADIATION
Radiation Heat transfer at a gas/surface interface involves radiation
emission from the surface and may also involve the
absorption of radiation incident from the surroundings
(irradiation, ), as well as convection  if .
s
T T


Energy outflow due to emission:
4
b s
E E T  
(1.5)
 emissi:Surf vityace 0 1   
blackbody:Emissive power of a (the perfect emit r ) te
b
E
 
2
Emissive powe: rW/mE
 
-8 2 4
: Stefan-Boltzmann constant 5.67 10 W/m K  
Energy absorption due to irradiation:
abs
G G
2
abs
: incidAbsorbed radiationent (W/m ) G
 absorpti: Surfa vityce 0 1   
 
2
Irradiation: W/mG
G
Heat Transfer Rates
(1.6)

8. HEAT TRANSFER RATES: RADIATION
(CONT.)
Irradiation: Special case of surface exposed to large
surroundings of uniform temperature,
sur
T
4
sur sur
G G T 
  
4 4
rad sur
If , the from the
surface due to exchange with the surroundings is
net radiation heat f
:
lux
b s s
q E T G T T
 
  

   
(1.7)
Heat Transfer Rates

9. HEAT TRANSFER RATES: RADIATION
(CONT.)
Alternatively,
 
 
  
rad sur
2
2 2
sur sur
Radiation heat transfe: W/m r coefficient K
r s
r
r s s
q h T T
h
h T T T T
 

  
(1.8)
(1.9)
For combined convection and radiation,
  
conv rad sur s r s
q q q h T T h T T

        (1.10)
Heat Transfer Rates

10. Schematic:
Problem 1.87(a): Process identification for single-and double-pane windows
PROCESS IDENTIFICATION
Convection from room air to inner surface of first paneconv,1
q
Net radiation exchange between room walls and inner surface of first pane rad,1
q
Conduction through first panecond,1
q
Convection across airspace between panes,s
q
conv
Net radiation exchange between outer surface of first pane and inner surface of second pane (across airspace)rad,s
q
Conduction through a second panecond,2
q
Convection from outer surface of single (or second) pane to ambient airconv,2
q
Net radiation exchange between outer surface of single (or second) pane and surroundings such as the groundrad,2
q
Incident solar radiation during day; fraction transmitted to room is smaller for double panes
q

11. Problem 1.40: Power dissipation from chips operating at a surface temperature
of 85C and in an enclosure whose walls and air are at 25C for
(a) free convection and (b) forced convection.
Schematic:
Assumptions: (1) Steady-state conditions, (2) Radiation exchange between a small surface and a large enclosure, (3)
Negligible heat transfer from sides of chip or from back of chip by conduction through the substrate.
Analysis:
elec conv rad
P q q 
(a) If heat transfer is by natural convection,
   
   
5/4 5/4
2 5/4 -4 2
conv
-4 2 -8 2 4 4 4 4
rad
elec
= 4.2W/m K 2.25 10 m 60K = 0.158W
0.60 2.25 10 m 5.67×10 W/m K 358 298 K = 0.065W
0.158W 0.065W = 0.223W
s
q CA T T
q
P

   
   
 
(b) If heat transfer is by forced convection,
   
2 4 -4 2
conv
elec
250W/m K 2.25 10 m 60K 3.375W
3.375W 0.065W 3.44W
s
q hA T T
P

     
  
   
4 4
surs s
hA T T A T T 

   
 
2
2 -4 2
0.015m 2.25×10 mA L  
PROBLEM: ELECTRONIC COOLING

12. RELATIONSHIP TO
THERMODYNAMICS
Chapter One
Section 1.3

13. ALTERNATIVE FORMULATIONS
• Alternative Formulations
Time Basis:
At an instant
or
Over a time interval
Type of System:
Control volume
Control surface
• An important tool in heat transfer analysis, often
providing the basis for determining the temperature
of a system.
CONSERVATION OF ENERGY
(FIRST LAW OF THERMODYNAMICS)

14. CV AT AN INSTANT AND OVER A TIME
INTERVAL
•At an Instant of Time:
Note representation of system by a
control surface (dashed line) at the boundaries.
Surface Phenomena
,
in out

: rate

energy transfer across the control

of thermal and/or mechanical
due to heat transfer, fluid flow and/or work interactions.surface


E E
 
Volumetric Phenomena
: rate of due to conversion from another energy form
(e.g., electrical, nuclear, or chemical); energy conversion proc
thermal ener
ess occurs w
gy generatio
ithin the sy e
n
m.

st
gE

st
energy: storate of rage in chang the se o ysf m. te
E

Conservation of Energy
in out st
st
g
dE
dtE E E E
      
(1.12c)
Each term has units of J/s or W.
APPLICATION TO A CONTROL VOLUME
• Over a Time Interval
Each term has units of J.
in out st gE E E E    (1.12b)

15. At an instant
t
dU
qW
dt

 
•Special Cases (Linkages to Thermodynamics)
(i)Transient Process for a Closed System of Mass (M) Assuming Heat Transfer
to the System (Inflow) and Work Done by the System (Outflow).
Over a time interval
tot
st
Q EW  (1.12a)
CLOSED SYSTEM
For negligible changes in potential or kinetic energy
t
Q W U 
Internal thermal energy

16. Example 1.4: A long conducting rod of diameter D and electrical resistance per unit length is
initially in thermal equilibrium with the ambient air and its surroundings. This equilibrium is
disturbed when an electrical current I is passed through the rod. Develop an equation that
could be used to compute the variation of the rod temperature with time during the passage
of the current.
• Involves change in thermal energy and for an incompressible substance.
t
dU dT
Mc
dt dt

• Heat transfer is from the conductor (negative )q
• Generation may be viewed as electrical work done on the system (negative ) W

EXAMPLE 1.4

18. Example 1.6: Application to isothermal solid-liquid phase change in a container:
Latent Heat
of Fusion
latt sf
U U Mh  
EXAMPLE 1.6

20. (ii)Steady State for Flow through an Open System without Phase Change or
Generation:
flow o w rkpv•
  enthalp y
t
u pv i  •
 
in out in out
ideal gas constant specific he For an w : atith
p
i i c T T  
•
 

in out in out
in out
For an :incompressible liqu
0
id
u u c T T
pv pv
  
 
•


2 2
in
in
out
0
2 2
0
out
For systems with significant heat transfer:
V V
gz gz
 
 
•
At an Instant of
Time:
2
out
0
2
t
m u pv V gz W
 
      
 

•2
in
2
t
m u pv V gz q
 
    
 

(1.12d)
OPEN SYSTEM

21. SURFACE ENERGY BALANCE
A special case for which no volume or mass is encompassed by the control surface.
Conservation of Energy (Instant in Time):
outin
0E E  
(1.13)
• Applies for steady-state and transient conditions.
Consider surface of wall with heat transfer by conduction, convection and radiation.
cond conv rad
0q q q    
   
4 41 2
2 2 2 sur
0
T T
k h T T T T
L
 


    
• With no mass and volume, energy storage and generation are not pertinent to the energy
balance, even if they occur in the medium bounded by the surface.
THE SURFACE ENERGY BALANCE

22. METHODOLOGY

• On a schematic of the system, represent the control surface by
dashed line(s).
• Choose the appropriate time basis.
• Identify relevant energy transport, generation and/or storage terms
by labeled arrows on the schematic.
• Write the governing form of the Conservation of Energy requirement.
• Substitute appropriate expressions for terms of the energy equation.
• Solve for the unknown quantity.
METHODOLOGY OF FIRST LAW ANALYSIS

23. Problem 1.57: Thermal processing of silicon wafers in a two-zone furnace.
Determine (a) the initial rate of change of the wafer
temperature and (b) the steady-state temperature.
PROBLEM: SILICON
WAFER
KNOWN: Silicon wafer positioned in furnace with top and bottom surfaces exposed to hot
and cool zones, respectively.
FIND: (a) Initial rate of change of the wafer temperature from a value of 300 K,
w,i
T and (b)
steady-state temperature. Is convection significant? Sketch the variation of wafer temperature
with vertical distance.
SCHEMATIC:
•

24. PROBLEM: SILICON WAFER
(CONT.)
ASSUMPTIONS: (1) Wafer temperature is uniform, (2) Hot and cool zones have uniform
temperatures, (3) Radiation exchange is between small surface (wafer) and large enclosure
(chamber, hot or cold zone), and (4) Negligible heat losses from wafer to pin holder.
ANALYSIS: The energy balance on the wafer includes convection to the upper (u) and lower
(l) surfaces from the ambient gas, radiation exchange with the hot- and cool-zones and an energy
storage term for the transient condition. Hence, from Eq. (1.12c),
in out stE E E 
  
or, per unit surface area
rad, rad, cv, cv,
w
h c u l
dT
q q q q cd
dt
      
    
4 4 4 4
sur,sur,
w
w c w u w l wh
d T
T T T T h T T h T T cd
dt
  
 
       
(a) For the initial condition, the time rate of change of the wafer temperature is determined
using the foregoing energy balance with
,
300 K,
w w i
T T 
   
8 2 4 4 4 8 2 4 4 4 44
0.65 5.67 10 W / m K 1500 300 K 0.65 5.67 10 W / m K 330 3 00 K
 
        
   
2 2
8W / m K 300 700 K 4W / m K 300 700 K      
3
2700kg / m 875J / kg K   0.00078 /
w i
m dT dt
 104 K/s
w
i
dT / dt <

25. PROBLEM: SILICON WAFER
(CONT.)
(b) For the steady-state condition, the energy storage term is zero, and the energy balance can
be solved for the steady-state wafer temperature,
,
.
w w ss
T T
   
4 4 4 4 4
,
0.65 1500 K 0.65 330 K
4
w,ss w ss
T T   
   
2 2
w,ss w,ss
8W / m K T 700 K 4W / m K T 700 K 0      
1251 K
w,ss
T
To assess the relative importance of convection, solve the energy balances assuming no
convection. With  
,
101 K/s and 1262 K
w w ssi
dT / dt T  , we conclude that the radiation
exchange processes control the initial rate of change and the steady-state temperature.
If the wafer were elevated above the present operating position, its temperature would
increase, since the lower surface would begin to experience radiant exchange with
progressively more of the hot zone. Conversely, by lowering the wafer, the upper surface
would experience less radiant exchange with the hot zone, and its temperature would decrease.
The temperature-distance relation might appear as shown in the sketch.
<

26. PROBLEM: COOLING OF SPHERICAL
CANISTER
Problem 1.64: Cooling of spherical canister used to store reacting chemicals.
Determine (a) the initial rate of change of the canister temperature,
(b) the steady-state temperature, and (c) the effect of convection
on the steady-state temperature.
KNOWN: Inner surface heating and new environmental conditions associated with a spherical
shell of prescribed dimensions and material.

FIND: (a) Governing equation for variation of wall temperature with time and the initial rate of
change, (b) Steady-state wall temperature and, (c) Effect of convection coefficient on canister
temperature.
535 J/kg·K

27. PROBLEM: COOLING OF SPHERICAL CANISTER
(CONT.)
ASSUMPTIONS: (1) Negligible temperature gradients in wall, (2) Constant properties, (3) Uniform,
time-independent heat flux at inner surface.
PROPERTIES: Table A.1, Stainless Steel, AISI 302:  = 8055 kg/m
3
,
p
c = 535 J/kgK.
ANALYSIS: (a) Performing an energy balance on the shell at an instant of time,
in out st
E E E 
  
.
Identifying relevant processes and solving for dT/dt,
   
2 2 3 3 4
4 4
3
i i o o i p
dT
q r h r T T r r c
dt
   

    

 
 
2 2
3 3
3
i i o
p o i
dT
q r hr T T
dt
c r r

   
  

.
SCHEMATIC:
<

28. PROBLEM: COOLING OF SPHERICAL CANISTER
(CONT.)
(b) Under steady-state conditions with
st
E

= 0, it follows that
 
2 2
4 4
i i o
q r h r T T 

  
2 25 2
2
10 W/m 0.5m
300K 439K
0.6m500W/m K
i i
o
q r
T T
h r

   
       
  
(c) Parametric calculations show a sharp increase in temperature with decreasing values of h < 1000
W/m
2
K. For T > 380 K, boiling will occur at the canister surface, and for T > 410 K a condition known
as film boiling (Chapter 10) will occur. The condition corresponds to a precipitous reduction in h and
increase in T.
Substituting numerical values for the initial condition, find

  

2 25
2 2
3 3 3
3
W W
3 10 0.5m 500 0.6m 500 300 K
m m K
kg J
8055 535 0.6 0.5 m
kg Km
i
dT
dt
 
 
 
 

 

  

0.084K/s
i
dT
dt
 <
<

29. PROBLEM: COOLING OF SPHERICAL
CANISTER (CONT.)
Although the canister remains well below the melting point of stainless steel for h = 100 W/m
2
K, boiling
should be avoided, in which case the convection coefficient should be maintained at h > 1000 W/m
2
K.

COMMENTS: The governing equation of part (a) is a first order, nonhomogenous differential equation
with constant coefficients. Its solution is  
1
Rt Rt
i
S/R e e 
 
   , where T T

  ,
 
2 3 3
3 /
i i p o i
S q r c r r  ,  
2 3 3
3 /
o p o i
R hr c r r  . Note results for t   and for S = 0.

30. SECOND LAW
For a reversible heat engine neglecting heat transfer effects between the
heat engine and large reservoirs, the Carnot efficiency is
where and are the absolute temperatures of large cold and hot reservoirs,
respectively.
• An important tool to determine how heat transfer affects
the efficiency of energy conversion.
SECOND LAW OF THERMODYNAMICS
out
in in
11
C
c
h
QW
Q Q
T
T
   
c
T
h
T
For an internally reversible heat engine with heat transfer to and from the
large reservoirs properly accounted for, the modified Carnot efficiency is
out out
in in in
,
,
1 11
c i
h i
m
Q qW
Q
T
TQ q
     
where and are the absolute
temperatures seen by the internally reversible
heat engine. Note that and are heat
transfer rates (J/s or W).
,c i c
T T
,h i h
T T
out
q
in
q
(1.15, 1.16)
(1.17)

31. SECOND LAW (CONT.)
In reality, heat transfer resistances (K/W) must be non-zero since
according to the rate equations, for any temperature difference only a
finite amount of heat may be transferred.
• Heat transfer resistances associated with, for example, walls separating the
internally reversible heat engine from the hot and cold reservoirs relate the
heat transfer rates to temperature differences:
 , i ,nh h i t h
T q RT 
The modified Carnot efficiency may ultimately be expressed as
From Eq. 1.21,
  ,, outc i c t c
T T q R 
where
tot , ,
in tot
1
c
m t h t c
h
T
R R R
T q R
   

tot
only if could be made infinitely sma ll.
m C
n R 
(1.21)
(1.18 a,b)
tot
For situations (realistic 0), .
m C
R n  
Good heat transfer engineering is a key
to improve the efficiency of heat engines.
