heat exchanger thermal engineering

1. Heat Exchanger 1

2. Device that facilitates the exchange of heat between two
fluids that are at different temperatures without mixing
each other
Heat transfer in a heat exchanger involves
 Convection in each fluid
 Conduction through wall separating each fluid
Overall heat transfer co efficient accounts for the above
Conduction and convection effects
HEAT EXCHANGERS
Hot Fluid
Hot Fluid
Cold Fluid Convection
Conduction

3. • LMTD-F approach :
• known – Temperatures (ΔTlmtd ), mass flow rates,
overall heat transfer coefficient, Calculate the size
of the heat exchanger (heat transfer area)
• ϵ-NTU approach :
• Known: inlet temperatures on cold side and hot side,
type and size of the heat exchanger, calculate the
outlet temperatures, load of the heat exchanger

4. TYPES OF HEAT EXCHANGERS
• COUNTER FLOW HE
• PARALLEL FLOW HE
• DOUBLE PIPE HE

5. Double pipe heat exchanger:
Parallel Counter flow
T T
Hot fluid in Hot fluid out
Cold fluid in
Cold fluid out
Hot fluid in Hot fluid out
Cold fluid in
Cold fluid out

6. COMPACT HEAT EXCHANGERS
β >700 m2/m3 –Compact
Car radiator β =1000 m2/m3
Gas turbine HE β =1000 m2/m3
Human lungs β =20000 m2/m3
Achieve high heat transfer rates between two fluids in a
small Volume



volume
exchanger
Heat
area
surface
transfer
Heat
density
Area

7. CROSS FLOW HEAT EXCHANGER
Cross-flow
(unmixed)
Tube flow
(unmixed)
a) Both fluids unmixed
Cross-flow
(mixed)
Tube flow
(unmixed)
b) One fluids mixed, one fluid unmixed

8. One shell two tube pass
Two shell four tube pass
Shell-side fluid
In
Tube-side fluid
Out
In
Out
Tube-side fluid
Shell-side fluid
In
In
Out
Out

9. Overall heat transfer coefficient
Ai – Inside area of the tube
Ao – Outside area of the tube
hi – inside heat transfer coefficient
ho – Outside heat transfer coefficient
k – thermal conductivity of tube
o
o
i
o
i
i
o
wall
i
Total
A
h
1
kL
2
r
r
ln
A
h
1
R
R
R
R 












o
o
i
o
i
i
o
o
i
i A
h
1
kL
2
r
r
ln
A
h
1
A
U
1
A
U
1











U is meaningless unless area is specified
Cold fluid
Hot fluid
1
i
i i
R
h A
 1
o
o o
R
h A

wall
R
Hot fluid
Cold fluid
Wall
i
A
o
A
i
h
o
h
Heat transfer
i
T
i
T
o
T

10. If Rwall = 0 and Ai =Ao then
U is dominated by smaller convection co efficient
• If one side fluid is gas and other side it is liquid, then to
improve h.t.c. on gas side fin is introduced
Fin- Gas side –UAs
Tube is finned
η = Fin efficiency
o
i h
1
h
1
U
1


o
o
o
i
o
i
i
i
o
o
i
i A
h
1
kL
2
r
r
ln
A
h
1
A
U
1
A
U
1














11. Typical over all heat transfer co efficient
Shell and tube type heat exchangers
Tube side fluid shell side fluid
U
(W/m2
-K)
1 Gas (1 bar) Gas (1 bar) 5-35
2 Gas (200-300 bar) Gas (200-300 bar) 150-500
3 Liquid (1 bar) Gas ( 1 bar) 15- 70
4 Gas (200-300 bar) Liquid 200-400
5 Liquid Liquid 150-2000
6 Liquid Superheated vapor 300-2000

12. Evaporators
Steam on Tube side fluid
U
(W/m2
-K)
1 Natural convection 300-900
2 Forced convection 600- 1700
Gas heaters
Tube side fluid shell side fluid
U
(W/m2
-K)
Steam /Hot water Gas
a free convection 5-12
b forced flow 15-20
Condenser
Tube side fluid shell side fluid
U
(W/m2
-K)
1 Cooling water Organic vapour 300-1200
2 Steam- turbine condenser 1500-4000

13. Double pipe heat exchanger
Tube side
fluid
shell side fluid
U
(W/m2
-k)
1 Gas (1 bar) Gas (1 bar) 10-35
2 H.P gas L.P gas 20-60
3 H.P gas H.P gas 15-500
4 H.P gas Liquid 200-600
5 Liquid Liquid 300-1400

14. Fouling
Phenomenon of deposition of material on surfaces from the
fluids in form of scales, layered sediments or biological
agents
Fouling increases resistance to heat transfer
Chemical treatment plant – PH
Periodic leaning – Downtime – Penalties
Ai= πDiL Ao= πDoL
o
o
o
o
,
f
i
o
i
i
,
f
i
i
o
o
i
i A
h
1
A
R
kL
2
r
r
ln
A
R
A
h
1
A
U
1
A
U
1














15. Fluid Rf (m2
.K/W)
1
Distilled water
sea water, River Water
a) Below 50° C 0.0001
b) Above 50° C 0.0002
2 Fuel oil 0.0009
3 Steam (oil free) 0.0001
4 Refrigerants (Liquid) 0.0002
5 Refrigerants (Vapour) 0.0004
6 Alcohol Vapour 0.0001
7 Air 0.0004
Typical fouling factors for different fluids
Fouling – 0.0001 m2.K/W ̴ 0.3 mm thick times limestone
(k = 2.9 W/m.K ) per unit surface area

16. Example: ksteel = 15 W/m oC
kcu= 400 W/m oC
Do = 1.9 cm
Di = 1.5 cm
h0 =1200 W/m2 oC
hi= 800 W/m2oC
Rf,o=0.0001 m2oC/W
Rf,i= 0.0004 m2oC/W
Hot fluid in Hot fluid out
Cold fluid in
Cold fluid out

17. Solution:
Ai= πDiL = 0.0471 m2 Ao= πDoL = 0.0597 m2
Ui = 399.4 W/m2 oC; Uo = 315 W/m2 oC
Conduction resistance of copper tube
=
W
/
C
10
159
.
53
10
96
.
13
10
675
.
1
10
491
.
2
10
493
.
8
10
54
.
26
A
h
1
A
R
kL
2
r
r
ln
A
R
A
h
1
A
U
1
A
U
1
0
3
3
3
3
3
3
o
o
o
o
,
f
i
o
i
i
,
f
i
i
o
o
i
i































 
 
3
i
o
10
094
.
0
1
400
2
5
.
1
9
.
1
ln
kL
2
r
r
ln













(negligibly small)

18. ANALYSIS OF HEAT EXCHANGER
LMTD-F approach : Select HE that will achieve specified
temperature range in a fluid stream of known mass flow
rate
ϵ-NTU approach :Predict outlet temperature of hot and
cold stream in a specified HE
Assumptions:
•Steady flow
•KE and PE changes neglected
•Thermophysical properties contant over entire length
•No heat loss to surrounding
•H.T.C is constant over entire length

19. Applying energy balance
Heat capacity rate,
FLUID- Large heat capacity rate – small ΔT
Doubling - 1/2 ΔT
ΔTHot = ΔTcold when Cc= Ch
   
i
h
o
h
h
h
i
c
o
c
c
c T
T
Cp
m
T
T
Cp
m
Q ,
,
,
, 


 


h
h
h
c
c
c Cp
m
C
Cp
m
C 
 

m

   
i
h
o
h
h
i
c
o
c
c T
T
C
T
T
C ,
,
,
, 



20. Hot fluid in Hot fluid out
Cold fluid out
Cold fluid in
T
h
dT
c
dT
h,in
T
c ,in
T
h,out
T
c ,out
T
1
T

T
 2
T

 
.
h c s
Q U T T dA
  
1 h,in c ,in
T T T
  
2 h,out c ,out
T T T
  
.
Q

1 2
s
dA
h,in
T
h,out
T
c ,out
T
c ,in
T
s
A
PARALLEL FLOW HEAT EXCHANGER
From Energy balance,
,
h
h
h
c
c
c dT
Cp
m
T
d
Cp
m
Q 

 



Re writing
)
2
(
Cp
m
Q
dT
)
1
(
Cp
m
Q
T
d
h
h
,
h
c
c
c









in alternate form
  )
3
(
dA
T
T
U
Q s
c
h 




21. ΔT1=Th,i-Tc,i ; ΔT2 = Th,o-Tc,o
Equation (2) – (1)
  









c
c
h
h
c
,
h
Cp
m
1
Cp
m
1
Q
T
T
d




Substitute from (3)
Q


 

 









 0
i c
c
h
h
s
o
i c
,
h
c
,
h
Cp
m
1
Cp
m
1
UdA
T
T
T
T
d





















c
c
h
h
s
i
,
c
i
,
,
h
o
,
c
o
,
,
h
Cp
m
1
Cp
m
1
UA
T
T
T
T
ln


   
0
,
h
i
,
h
h
h
i
,
c
o
,
c
c
c T
T
Cp
m
Q
T
T
Cp
m
Q 


 




22. 




 














Q
T
T
Q
T
T
UA
T
T
T
T
ln i
,
h
o
,
h
i
,
c
o
,
c
s
i
,
c
i
,
,
h
o
,
c
o
,
,
h


Rearranging
For parallel flow HE where ΔT1=Th,o-Tc,o
ΔT2=Th,i-Tc,i





























i
,
c
i
,
,
h
o
,
c
o
,
,
h
i
,
c
i
,
h
o
,
c
o
,
h
s
i
,
c
i
,
,
h
o
,
c
o
,
,
h
o
,
c
o
,
h
i
,
c
i
,
h
s
T
T
T
T
ln
)
T
T
(
)
T
T
(
UA
T
T
T
T
ln
)
T
T
(
)
T
T
(
UA
Q











2
1
2
1
s
T
T
ln
T
T
UA
Q















2
1
2
1
LMTD
T
T
ln
T
T
T





LMTD
s T
UA
Q 



23. Parallel flow HE Counter flow HE
Hot fluid in Hot fluid out
Cold fluid out
Cold fluid in
1 h,in c ,in
T T T
  
2 h,out c ,out
T T T
  
Hot fluid in Hot fluid out
Cold fluid in
Cold fluid out
1 h,in c ,out
T T T
  
2 h,out c ,in
T T T
  

24. Counter flow heat exchanger
,
h
h
h
c
c
c
dT
Cp
m
T
d
Cp
m
Q








)
2
(
Cp
m
Q
dT
)
1
(
Cp
m
Q
T
d
h
h
,
h
c
c
c










  )
3
(
dA
T
T
U
Q s
c
h 



  









c
c
h
h
c
,
h
Cp
m
1
Cp
m
1
Q
T
T
d




Equation (2) – (1)
h
dT
c
dT
h,in
T
c ,in
T
h,out
T
c ,out
T

25.  

 












0
i c
c
h
h
s
T
T
T
T c
,
h
c
,
h
Cp
m
1
Cp
m
1
UdA
T
T
T
T
d
ci
ho
co
hi





















c
c
h
h
s
o
,
c
i
,
,
h
i
,
c
o
,
,
h
Cp
m
1
Cp
m
1
UA
T
T
T
T
ln


   
0
,
h
i
,
h
h
h
i
,
c
o
,
c
c
c T
T
Cp
m
Q
T
T
Cp
m
Q 


 



Since
Finally we get





 














Q
T
T
Q
T
T
UA
T
T
T
T
ln i
,
c
o
,
c
i
,
h
o
,
h
s
i
,
c
i
,
,
h
o
,
c
o
,
,
h
















i
,
c
o
,
,
h
o
,
c
i
,
,
h
i
,
c
o
,
h
o
,
c
i
,
h
s
T
T
T
T
ln
)
T
T
(
)
T
T
(
UA
Q


26. For counter flow HE where ΔT1=Th,i-Tc,o
ΔT2=Th,o-Tc,i










2
1
2
1
LMTD
T
T
ln
T
T
T





• Cold fluid is heated to inlet
Temperature of hot fluid
Tc,o can never exceed Th,i
• Violation of 2nd law of
thermodynamics
h,in
T
c ,in
T
h,out
T
c ,out
T
Hot fluid in Hot fluid out
Cold fluid in
Cold fluid out
Limiting case

27. Specified inlet temperature:
(LMTD)Counter flow > (LMTD)Parallel flow
• Counter flow HE Surface area smaller for same heat load
Thus more popular

28. Example:
Thi=110 oC Tci=30oC
Tho= 60oC Tc,o=55oC
C
24
.
41
30
60
55
110
ln
)
30
60
(
)
55
110
(
T
T
T
T
ln
)
T
T
(
)
T
T
(
T o
i
,
c
o
,
,
h
o
,
c
i
,
,
h
i
,
c
o
,
h
o
,
c
i
,
h
LMTD 



























C
05
.
27
30
110
55
60
ln
)
30
110
(
)
55
60
(
T
T
T
T
ln
)
T
T
(
)
T
T
(
T o
i
,
c
i
,
,
h
o
,
c
o
,
,
h
i
,
c
i
,
h
o
,
c
o
,
h
LMTD 



























For parallel flow Heat exchanger
For counter flow Heat exchanger

29. Example :
• Counter flow HE
• Water flows through the inner tube
• Oil Flows through annulus
Find Surface area
H2O
Oil
30 C

60
0 1
C
. kg s

40 2
0 2
. C
. kg s

100 C


30. Oil Water
Tmean= (100+80)/2=60oC Tmean ̴ 35oC
Cp=2.131 kJ/kgK Cp=4.187 kJ/kgK
μ = 3.25 × 10-2 Pa-s μ = 725 × 10-6 Pa-s
k = 0.138 W/m-K k = 0.625 W/m-K
Pr=501.87 Pr= 4.85
Solution:
  Water
30
T
187
.
4
2
.
0
8524 o
,
c 



  Oil
W
8524
)
60
100
(
131
.
2
1
.
0
T
T
Cp
m
Q 0
,
h
i
,
h
h
h 





 

Tc,o = 40.2 oC

31. ΔT1=Th,i-Tc,o =100-40.2 = 59.8 oC
ΔT2=Th,o-Tc,i= 60 - 30 = 30 oC
Water side
Oil side Dh=Do-Di = 45 - 25 = 20 mm
C
T o
LMTD 2
.
43
30
8
.
59
ln
30
8
.
59










14050
10
725
10
25
2
.
0
4
D
m
4
Re 6
3
i
D 






 





K
m
/
W
2250
h
k
D
h
90
Pr
Re
0023
.
0
Nu 2
i
i
i
4
.
0
8
.
0
D
D 




  56
10
25
.
3
)
025
.
0
045
.
0
(
02
.
0
1
.
0
4
D
D
D
m
4
VD
Re 2
2
2
2
i
2
0
h
h
D 









 




 

32. For Do/Di=0.56 NuD = 5.56 = hoDh/k
Hence ho= 38.4 W/m2K
Finally
Q=UAsΔTLMTD
8520 = 37.76 × π ×0.025 × L × 43.2
L= 66.54 meter
Comments:
•ho controls over all h.t.c
•hi >> ho Wall temp. follows coolant water temperature
•Hence uniform wall temperature assumption
K
m
W
U
h
h
U o
i
2
/
76
.
37
4
.
38
1
2250
1
1
1
1







33. Ch = Cc
ΔT1 = ΔT2 = Constant
h,in
T
c ,in
T
h,out
T
c ,out
T
Inlet Outlet
1 2
T T T
      Constant
1
T

2
T
  
 
.
h h ph
.
c c pc
.
c c ,out c ,in
.
h h,in h,out
C m C
C m C
Q C T T
Q C T T


 
 
Heat exchanger with same specific heats and mass flow rates

34. Condenser and Boiler
fg
h
m
Q 
 
on
condensati
of
heat
Latent
h
on
condensati
of
Rate
m
fg 


finite
T
Cp
m
Q h
h 

 
0
T
Cp
m
C h
h
h 


 

Condensing fluid
.
Q
T
Boiling fluid
.
Q
T
Inlet Outlet Inlet Outlet
Condenser h
C   Boiler c
C  

35. Rate of heat transfer in HE
U = Over all h.t.c
As = Heat transfer surface area
ΔTLMTD= Appropriate temperature difference
LMTD
s T
UA
Q 



36. MULTI-PASS AND CROSS FLOW HEAT EXCHANGERS
F is obtained by charts
Hot fluid in Hot fluid out
Cold fluid in
Cold fluid out
1 h,in c ,out
T T T
  
2 h,out c ,in
T T T
  
Cross tube or multi pass shell
and tube heat exchanger
Heat Transfer rate
.
s lm ,CF
Q UA F T
 
 
1 2
1 2
lm ,CF
T T
T
ln T T
  
 
 
1 h,in c ,out
T T T
  
2 h,out c ,in
T T T
  
where

38. One shell pass and 2,4,6 etc (any multiples of 2), tube passes
1
T
2
T
2
t
1
t
2 1
2 1
t t
P
T t



1 2
2 1
.
P
tube side
.
P
shell side
mC
T T
R
t t
mC
 
 
  
 
  
 
 
0 0.5 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0.5
0.6
0.7
0.8
0.9
1
R = 4 3 2 1.5 1 0.8 0.6 0.4 0.2
Correction
factor
F

39. 1
T
2
t
1
t
2
T
0 0.5 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0.5
0.6
0.7
0.8
0.9
1
R = 4 3 2 1.5 1 0.8 0.6 0.4 0.2
Correction
factor
F
Two-shell passes and 4, 8, 12, etc, (any multiple of 4), tube passes
2 1
2 1
t t
P
T t



1 2
2 1
T T
R
t t




40. Single pass cross-flow with both fluids unmixed
1
T
2
t
1
t
2
T
0 0.5 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0.5
0.6
0.7
0.8
0.9
1
R = 4 3 2 1.5 1 0.8 0.6 0.4 0.2
Correction
factor
F
2 1
2 1
t t
P
T t



1 2
2 1
T T
R
t t




41. 1
T
2
t
1
t
2
T
Single pass cross-flow with one fluid mixed and other unmixed
0 0.5 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0.5
0.6
0.7
0.8
0.9
1
R = 4 3 2 1.5 1 0.8 0.6 0.4 0.2
Correction
factor
F
2 1
2 1
t t
P
T t



1 2
2 1
T T
R
t t




42. Steam in the condenser of a power plant is to condensed at a temperature of
30 °C with cooling water from a near by lake, which enters the tube of the
condenser at 14 °C and leaves at 22 °C. The surface area of the tube is 45 m2,
and the overall heat transfer coefficient is 2100 w/m2. °C . Determine the mass
flow rate of the cooling water needed and the rate of condensation of the steam
in the condenser.
Assumptions: 1) Steady operating conditions exsist. 2) The heat exchanger is
well insulated so that heat loss to the surrounding is negligible and thus heat
transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3)
Changes in the kinetic and potential energies of fluid stream are negligible. 4)
There is no fouling. 5) Fluid properties are constant.
Solution: Steam is condensed by cooling water in the condenser of a power
plant. The mass flow rate of the cooling and the rate of condensation are to be
determined.
Properties: The heat of vaporization of water at 30 °C is hfg = 2431 kJ/kg and
the specific heat of the cold water at the average temperature of 18°C is Cp=
4184 J/kg. °C.
The Condensation of steam in a Condenser

43. Steam 30°C
Cooling water
14°C
22°C
30°C

44. Analysis: The condenser can be treated as a counter flow heat
exchanger since the temperature of one of the fluids ( the steam)
remains constant.
The temperature difference between the steam and the
cooling water at the two ends of the condenser is
That is, the temperature difference between the two fluids varies
from 8°C at one to 16°C at the other. The proper average
temperature difference between the two fluids is the logarithmic
mean temperature difference (not the arithmetic), which is
determined from
 
1 30 22 8
h,in c ,out
T T T C C
       
 
2 30 14 16
h,out c ,in
T T T C C
       
Steam
30°C
Cooling
water
14°C
22°C
30°C

45. 1 2
1
2
8 16
11 5
8
16
o
lm
T T
T . C
T
ln
ln
T
   
   
   

 
 
 

 
This is a little less than the arithmetic mean temperature difference of (8+16)/2 =
12C. Then the heat transfer rate in the condenser is determined from
kW
.
T
UA
Q lm
s 1087
5
11
45
2100 



 

Therefore, the steam will lose heat at a rate of 1087 kW as it flows through the
condenser, and the cooling water will gain practically all of it, since the condenser is
well insulated.
The mass flow rate of the cooling water and the rate of the condensation of the
steam are determined from
 
  fg
er
coolingwat
in
out
p h
m
T
T
C
m
Q 

 


   
s
/
kg
.
h
Q
m
;
s
/
kg
.
.
T
T
C
Q
m
fg
steam
in
out
p
45
0
2431
1087
6
32
14
22
184
4
1087













46. HEATING WATER IN A COUNTER FLOW HEAT EXCHANGER
A counter flow double pipe heat exchanger is to heat water from 20C to 80C at a
rate of 1.2 kg/s. The heating is to be accomplished by geothermal water available at
160C at a mass flow rate of 2 kg/s. The inner tube is thin walled and has a diamater
of 1.5 cm. If the overall heat transfer coefficient of the heat exchanger is 640 W/m2,
determine the length of the heat exchanger required to achieve the desired heating.
Cold water 20C , 1.2 kg/s 80C
Hot geothermal water 2 kg/s , 160C
Solution: water is heated in a counter flow double pipe heat exchanger by
geothermal water. The required length of the heat exchanger is to be determined.
Assumptions: 1. steady operating conditions exist. 2. the heat exchanger is well
insulated so that heat loss to the surroundings is negligible and thus heat transfer
from the hot fluid is equal to the heat transfer to the cold fluid. 3. Changes in the
kinetic and potential energies of fluid streams are negligible. 4. there is no fouling.
5. Fluid properties are constant

47. Properties: we take specific heats of water and geothermal fluid to be 4.18 and 4.31
kJ/kgC, respectively.
Analysis: The rate of heat transfer in the heat exchanger can be determined from
 
    kW
.
.
T
T
C
m
Q
water
in
out
p 301
20
80
18
4
2
1 





 

Noting that all of this heat is supplied by the geothermal water, the outlet
temperature of the geothermal water is determined to be
 
 
C
T
.
C
m
Q
T
T
T
T
C
m
Q
o
out
p
in
out
geothermal
out
in
p
125
31
4
2
301
160













C
ln
T
T
ln
T
T
T o
lm 92
105
80
105
80
2
1
2
1 























C
T
T
T
C
T
T
T
o
in
,
c
out
,
h
o
out
,
c
in
,
h
105
20
125
80
80
160
2
1












 
2
11
5
92
640
301000
m
.
T
U
Q
A
T
UA
Q
lm
s
lm
s 









48.  
m
.
.
D
A
L
DL
A s
s 108
015
0
11
5








Discussion: The inner tube of this counter flow heat exchanger (and thus the heat exchanger
itself) needs to be over 100 m long to achieve the desired heat transfer, which is impractical.
In cases like this, we need to use a plate heat exchanger or a multipass shell and tube heat
exchanger with multiple passes of tube bundles.

49. Heating of Glycerin in a Multipass Heat Exchanger
A 2 shell passes and 4 tube passes heat exchanger is used to heat glycerin from 20C
to 50 C by hot water, which enters the thin walled 2 cm diameter tubes at 80 C and
leaves at 40 C. The total length of the tubes in the heat exchanger is 60 m. The
convection heat transfer coefficient is 25 W/m2. C on the glycering (shell) side and
160 W/m2. C on the wataer (tube) side. Determine the rate of heat transfer in the
heat exchanger (a) before any fouling occurs (b) after fouling occurs with a fouling
factor of 0.0006 m2. C/W occurs on the outer surfaces of the tubes.
Cold glycerin
20° C
40° C
80° C
50° C
Hot water

50. Solution: Glycerin is heated in a 2 shell passes and 4 tube passes heat exchanger by
hot water. The rate of heat transfer for the cases of fouling and no fouling are to be
determined.
Assumptions: 1. steady operating conditions exist. 2. the heat exchanger is well
insulated so that heat loss to the surroundings is negligible and thus heat transfer
from the hot fluid is equal to the heat transfer to the cold fluid. 3. Changes in the
kinetic and potential energies of fluid streams are negligible. 4. Heat transfer
coefficients and fouling factors are constant and uniform 5. Thermal resistance of the
inner tube is negligible since the tube is thin walled and highly conductive.
2
77
3
60
02
0 m
.
.
DL
As 



 

The rate of heat transfer in this heat exchanger can be determined from
CF
,
lm
s T
F
UA
Q 


where F is the correction factor and Tlm,cF is the log mean temperature difference
for the counter flow arrangement. These two quantities are determined from
C
T
T
T
C
T
T
T
o
in
,
c
out
,
h
o
out
,
c
in
,
h
20
20
40
30
50
80
2
1











 C
.
ln
T
T
ln
T
T
T o
lm 7
24
20
30
20
30
2
1
2
1 
























51. 91
0
75
0
80
40
50
20
67
0
80
20
80
40
1
2
2
1
1
2
1
2
.
F
.
t
t
T
T
R
.
t
T
t
t
P






















In case of no fouling, the overall heat transfer coefficient U is determined from
C
.
m
/
W
.
h
h
U o
o
i
2
6
21
25
1
160
1
1
1
1
1





W
.
.
.
.
T
F
UA
Q CF
,
lm
s 1830
7
24
91
0
77
3
6
21 




 

Net rate of heat transfer is given by
When there is fouling on one of the surfaces, the overall heat transfer coefficient U is
C
.
m
/
W
.
.
R
h
h
U o
f
o
i
2
3
21
0006
0
25
1
160
1
1
1
1
1







W
.
.
.
.
T
F
UA
Q CF
,
lm
s 1805
7
24
91
0
77
3
3
21 




 

Net rate of heat transfer is given by
The rate of heat transfer decreases as a result of fouling. The decrease is not dramatic,
however, because of the relatively low convection heat transfer coefficients involves.

52. Cooling of an Automotive Radiator
40°C
Air
(unmixed)
20°C
Water flow
(unmixed)
65°C
90°C
A test is conducted to determine the overall heat transfer in an automobile radiator
that is a compact cross flow water to air heat exchanger with both fluids (air and
water) unmixed. The radiator has 40 tubes of internal diameter 0.5 cm and length 65
cm in a closely spaced plate-finned matrix. Hot water enters the tubes at 90°C at a
rate of 0.6 kg/s and leaves at 65 °C. Air flows across the radiator through the interfin
spaces and is heated from 20°C to 40°C. Determine the overall heat transfer
coefficient Ui of this radiator based on the inner surface area of the tubes.

53. Solution: During an experiment involving an automotive radiator, the inlet and exit
tempertures of water and air and the mass flow rate of water are measured. The
overall heat transfer coefficient based on the inner surface area is to be determined.
Assumptions: 1. steady operating conditions exist. 2. the heat exchanger is well
insulated so that heat loss to the surroundings is negligible and thus heat transfer
from the hot fluid is equal to the heat transfer to the cold fluid. 3. Changes in the
kinetic and potential energies of fluid streams are negligible. 4. Heat transfer
coefficients are constant 5. Fluid properties are constant.
Properties : The specific heat of water at the average temperature of (90+65)/2 =
77.5C is 4.195 kJ/kg. C
Analysis: The rate of heat transfer in this radiator from the hot water to the air is
determined from an energy balance on water flow.
 
    kW
.
.
.
T
T
C
m
Q
water
in
out
p 93
62
65
90
195
4
6
0 





 

The tube side heat transfer area is the total surface area of the tubes, and is
determined from
2
408
0
65
0
005
0
40 m
.
.
.
L
D
n
A i
s 




 


54. Knowing the rate of heat transfer and the surface area, the overall heat transfer
coefficient can be determined from
CF
,
lm
i
CF
,
lm
i
i
T
F
A
Q
U
T
F
A
U
Q



 


C
T
T
T
C
T
T
T
o
in
,
c
out
,
h
o
out
,
c
in
,
h
45
20
65
50
40
90
2
1











 C
.
ln
T
T
ln
T
T
T o
lm 6
47
45
50
45
50
2
1
2
1 























97
0
80
0
90
65
40
20
36
0
90
20
90
65
1
2
2
1
1
2
1
2
.
F
.
t
t
T
T
R
.
t
T
t
t
P






















C
m
/
W
.
.
.
U o
2
3341
6
47
97
0
408
0
62930




Note that the overall heat transfer coefficient on the air side will be much lower
because of the large surface area involved on that side.

55. LMTD approach
LMTD-F approach : Select HE that will achieve specified temperature
range in a fluid stream of known mass flow rate
1. Select the type of heat exchanger suitable for the application
2. Determine any unknown inlet or outlet temperature and the heat
transfer rate using an energy balance
3. Calculate the log mean temperature difference and the correction
factor F, if necessary
4. obtain (select or calculate) the value of the overall heat transfer
coefficient
5. Calculate the heat transfer area