H.T through the Tubes with parameters.ppt

heat transfer in tubes

2. Fluid Flow and Heat
Transfer in Tubes

3. The pipe, duct, tube, and conduit are usually used interchangeably
for flow sections. Most liquids are transported in circular pipes.
This is because pipes with a circular cross section can withstand
large pressure differences between the inside and the outside
without undergoing any distortion.
For a fixed surface area, the circular tube gives the most heat
transfer for the least pressure drop,
 Therefore, circular tubes have devastating popularity in heat
transfer equipment.
Flow in a tube can be laminar or turbulent, depending on the flow
conditions.
Fluid Flow and Heat Transfer in Tubes

4. Boundary layers in Tubes

5. Boundary layers in Tubes

6. Velocity Boundary layer in Tube
Hydrodynamic Entry Length
In case of pipe/tube flow , the development of boundary layer in
the same fashion as flat plate boundary layer.
Fluid velocity in a tube changes from zero at surface to maximum
at the centre line.
Formed parabolic region known as hydro dynamically developed
region
The region from tube inlet to the centre at which the boundary
layers merged is called Hydrodynamic entry region/length.

7. Velocity Boundary layer in Tube
 For calculation we use average/mean velocity (U
ave
).
In actual practice of heating and cooling, the mean velocity may
change due to change in density because of temperature change.
The thickness of boundary layer is depend on the radius of the
 tube.

8. Velocity Boundary layer in Tube
Boundary layers from the tube walls meet at the centre of the
tube and entire flow acquires the characteristics of the B.L.
Once boundary layer thickness become equal to the radius of
the tube, there will not be any further change in the velocity
distribution
This is known as fully developed velocity profile flow
conditions

9. Thermal Boundary layer in Tube
 Thermal Entry Length
The region of flow in which thermal boundary layer develops and
merged in the tube centre is called thermal entry region
Its length is known as thermal entry length.
The temperature profile in thermally developed region may vary
with x-direction of the flow.
Therefore, temperature profile can be different at x-direction of
the tube in the developed region.
The dimensional less parameter Prandtl Number (Pr) shows effect
of both the boundary layers.

10. Prandtl number and boundary layers
(Thermal and Velocity)
For laminar flow in a tube, the magnitude of the Prandtl number
is measure of the relative growth of the velocity and thermal B.L
 If Pr=1
Both boundary layers are coincide to each other.
If Pr>1
The velocity boundary layer out grows the thermal boundary layer
If Pr<1
The thermal boundary layer is more dominant than the velocity
boundary layer.

11. Reynolds Number Re and flow in tubes

12. Reynolds Number Re and flow in tubes
It is certainly desirable to have precise values of Reynolds
numbers for laminar, transitional, and turbulent flows.
The transition from laminar to turbulent flow also depends on the
degree of disturbance of the flow such as,
surface roughness,
 pipe vibrations,
fluctuations in the flow.
Under most practical conditions,
 If Re<2300
Flow is laminar
If 2300  Re 10000
Flow is transitional
If Re>10000
Flow is turbulent

13. Hydrodynamic and Thermal
entry length of Tubes
Calculation formulae

14. Hydrodynamic and thermal entry length
For laminar flow:
For Turbulent flow:
Hydraulic diameter (D
h
) of the tube
Friction Factor (f)
Fluid Flow and Heat Transfer in Tubes
DL
DL
th
h
.Pr.Re.05.0
.Re.05.0


DLL
thh
10
PAD
ch
/4
2.0
Re/184.0
Re/64



f
f Laminar Flow
Turbulent Flow

15. Fluid Flow and Heat Transfer in Tubes
The conservation of energy equation for steady flow of fluid in tube:
For average Temperature:
Conservation of heat flux at any location on the tube
For constant surface heat flux
For constant surface temperature
)(
iep TTCmQ 
cmean
AUm
mpTCmQ

)(
mss TThq 
)(
ieps TTCmAqQ  
)(
msave TThAThAQ 

2
TT
T
ei
m



16. Fluid Flow and Heat Transfer in Tubes
2
TT
TT
2
TTTT
T
2
TT
TT
ei
sam
esis
am
ei
aveam
)(
)()(






2
TT
T
ei
b


bsam
TTT 

17. Fluid Flow and Heat Transfer in Tubes
Logarithmic Mean Temperature Difference (LMTD)
For heating of fluid:
T
s
>T
i
and T
e
For cooling of fluid:
T
s
<T
i
and T
e












is
es
ises
m
TT
TT
TTTT
TT
ln
)()(
ln












i
e
ie
m
T
T
TT
T
ln

18. Problem
Water flows through a tube of ¾-inch internal diameter at the rate
of 0.4 gallon per minute. The viscosity of water may be taken as
2.36 lb /hr ft. Determine either the flow is laminar or Turbulent.
If it is laminar determine the entrance length.
Data
D=3/4=0.0625ft
V=0.4 gallon/min
=62.4 lb/hr ft.

DU
R
ave
e
DL
h
Re05.0
 
2
22
ave
ft0030670A
06250
4
143
D
4
Area
A
timeVolume
U
.
.
.
/





19. sec/./
).(.
/
3
3
ft00089130timeVolume
60
ft1337040
timeVolume

  
3
ft13370gal1 .
sec/.
.
./
ft290U
0030670
00089130
A
timeVolume
U
ave
ave


23001725R
3600362
06250290462DU
R
e
ave
e



/.
...


This value of Reynolds number is lower than the 2300
Therefore, the flow is Laminar .
0625.0172505.0Re05.0  DL
h
ftL
h39.5

20. Problem: Heating of Water by Resistance Heaters in a Tube
Water is to be heated from 15°C to 65°C as it flows through a 3-cm
internal diameter 5-m-long tube as shown in Fig. The tube is
equipped with an electric resistance heater that provides uniform
heating throughout the surface of the tube. The outer surface of the
heater is well insulated, so that in steady operation all the heat
generated in the heater is transferred to the water in the tube. If the
system is to provide hot water at a rate of 10 L/min.
Determine:
i.The power rating of the resistance heater.
ii.The inner surface temperature of the pipe at the exit.

21. Solution
Properties: The properties of water at the bulk mean temp:
@ Tb = 40°C (From table A-9).
CT
CT
TT
T
o
b
o
b
ei
b
40
40
2
6515
2






sm
CmWk
CkgJC
mkg
o
o
p
/10658.0/
32.4Pr
/631.0
/4179
/1.992
26
3









22. For Cross-sectional area of the tube
For total surface area of the tube
Mass flow rate

24
c
22
c
m100697A
030
4
143
D
4
A



.
.
.
2
s
s
m4710A
5030143DLPLA
.
..


sec/.
..
min/.
min/
kg16540m
0101992m
m010V
L10V
Vm
3











23. Heat supplied to water to increase the temperature from 15 to 65
o
C
Power required for electric resistance heater is 34.6 Kw, because
Heater is totally insulated.
Surface temperature of the tube
Surface Heat Flux
 
 
Kw634Q
KJ634Q
1565179416540Q
TTCmQ
iep
.
sec/.
..








2
s
s
s
ms
mss
mKw4673q
4710
634
A
Q
q
h
q
TT
TThq
/.
.
.
)(










24. For heat transfer coefficient (h)
Where Nusselt number (Nu)
For Reynolds number
Nu
D
k
h
k
hD
Nu


4080
0230Nu
..
Pr.Re.
sec/.
.
.
Re
m2360U
100697
010
A
V
U
DU
mean
4
c
mean
mean








25. 10760
106580
03023600
6





Re
.
..
Re
This Reynolds number is greater than the 10000, therefore,
The flow is turbulent in this case and the entry lengths for
hydrodynamic and thermal are,
Now, substitute values in equation
m30LL
03010D10LL
thh
thh
.
.


4080
0230Nu
..
Pr.Re.
CmW1462h
569
030
6310
h
o2
/
.
.
.


Nu
D
k
h

26. For surface temperature
Discussion:
Note that the inner surface temperature of the pipe will be 50°C
higher than the mean water temperature at the pipe exit.
This temperature difference of 50°C between the water and the
surface will remain constant throughout the fully developed flow
region.
C115T
1462
46073
65T
h
q
TT
o
S
S
s
mS



.

27. Problem
Water enters a 2.5-cm-internal-diameter thin copper tube of a
heat exchanger at 15°C at a rate of 0.3 kg/s, and is heated by
steam condensing outside at 120°C. If the average heat transfer
coefficient is 800 W/m
2o
C, determine the length of the tube
required in order to heat the water to 115°C.

28. SOLUTION:
Properties: The specific heat of water at the bulk mean temperature
CT
CT
TT
T
o
b
o
b
ei
b
65
65
2
11515
2






CkgJC
o
p /4187
The enthalpy of steam at 120°C is 2203 kJ/kg (Table A-9).
Heat transfer rate determine by
kWQ
Q
TTCmQ
iep
6.125
)15115(187.43.0
)(







29. The logarithmic mean temperature difference is
CT
T
TTT
CT
T
TTT
o
i
i
isi
o
e
e
ese
105
15120
)(
5
115120
)(


















i
e
ie
T
T
TT
T
ln
ln
CT
o
85.32
105
5
ln
1055
ln 









30. The heat transfer surface area is
Length of the pipe for heating
2
ln
ln
78.4
85.328.0
6.125
mA
Th
Q
A
ThAQ
S
S
S








mL
D
A
L
DLA
S
S
61
025.014.3
78.4





