College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 375
Heat Transfer
Spring 2007 Number 17629 Instructor: Larry Caretto
April 25 Homework Solutions
11-42
Ethylene glycol is heated from 20°C to 40°C at a rate of 1.0 kg/s in a horizontal copper tube
(k = 386 W/mK) with an inner diameter of 2.0 cm and an outer diameter of 2.5 cm. A
saturated vapor (Tg = 110°C) condenses on the outside-tube surface with the heat transfer
coefficient (in kW/m2 K) given by 9.2/(Tg - Tw)0.25, where Tw is the average outside-tube wall
temperature. What tube length must be used? Take the properties of ethylene glycol to be
 =1109 kg/m3, cp = 2428 J/kgK, k = 0.253 W/m°C,  = 0.01545 kg/ms, and Pr = 148.5.
In this problem we have to evaluate the overall heat transfer coefficient, U, before we can do the
heat exchanger analysis. Here we choose to base the overall coefficient on the outside tube
area, so we label this Uo,.
ln Do Di 
1
1
1



U o Ao h0 Ao
2kL
hi Ai

ln Do Di 
1
1
1



U oDo L h0Do L
2kL
hiDi L
We can multiply the second equation by DoL to obtain the following result.
1
1 D L ln Do Di  Do L
1 D ln Do Di  Do
  o

  o

U o h0
2kL
hiDi L h0
2k
hi Di
We know that the outside heat transfer coefficient is given by the equation A/(Tg - Tw)0.25, where A
= 9.2 kW/m2/K0.75. = 9200 WK0.75. We know that Tg = 110oC, but we do not know the average wall
temperature Tw. We do know, however, that the transfer to the condensing fluid must be the
same as the heat transfer from the ethylene glycol to the outer tube wall.
Q  ho Ao Tg  Tw  
Tw  Tb,avg
ln Do Di 
1

2kL
hi Ai
 hoDo LTg  Tw  
Tw  Tb ,avg
ln Do Di 
1

2kL
hiDi L
In this equation Tb,acg is the average temperature of the glycol in the tube which 30 oC, the mean of
the inlet and outlet temperature. Multiplying the second equation through by DoL gives
ho Tg  Tw  
Tw  Tb ,avg
Tw  Tb ,avg

Do L ln Do Di  Do L
Do ln Do Di  Do


2kL
hiDi L
2k
hi Di
Substituting the value for ho gives
ho Tg  Tw  
Tw  Tb ,avg
A
0.75




T

T

A
T

T

g
w
g
w
Do ln Do Di  Do
Tg  Tw 0.25

2k
hi Di
The thermal conductivity of copper is found from Table A-3 to be 385 W/mK. If we can compute
the inside heat transfer coefficient for the ethylene glycol, can find all terms in this equation
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
April 25 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 2
except for Tw. That will give us an equation to solve for T w. We have to compute the Reynolds
number to see if the flowinside the tunbe is laminar or turbulent. To do this we need the velocity.
1 kg
m
m
2.870 m
s
V



2
3
1109 kg 
A  Di 4 m
s
0.02 m 2
3
4
m


VD
Re 


1109 kg 2.870 m
0.02 m 
s
m3
 4121
0.01545 kg
ms
This is a transition flow; we will use the equation for fully developed turbulent flow in this case.
Here we use the Dittius-Boelter equation with a Prandtl number exponent of n = 0.4 because we
are heating the fluid, to find the Nusselt number and the heat transfer coefficient, hi.
Nu  0.023 Re 0.8 Pr 0.4  0.0234121
0.8
hi 
148.50.4  132.5
kNu 132.5 0.253 W 1677 W

 2
Di
0.02 m m  K
m K
Substituting this value for hi and the other data for diameters and thermal conductivity into our
equation for Tw gives the following computational result.

A Tg  Tw

0.75 
T  Tw 0.75 
0.25 g
9200 W
m2  K
Tw  30 0 C
0.025 m ln 0.025 m  0.02 m  
2
385 W
mK
m 2  K 0.025 m
1677 W 0.02 m

Tw  Tb, avg
Do ln Do Di  Do

2k
hi Di
Tw  30 0 C
7.24 x10  6 m 2  K 745.4 x10  6 m 2  K

W
W
We see that the resistance of the copper tubing is small compared to the inside convection
resistance and we could have neglected it. Rearranging this equation and setting Tg to its given
value of 110oC gives

9200 W
W
Tg  Tw 0.75  1329
Tw  30 0 C
2
0.75
2
m K
m K


110 C  T 
T  30 C 
0.75
o
w
0
 0.06920 K 0.25
w
Note that the units are consistent because we are dealing with temperature differences. Thus
we can use a constant with units of kelvins to solve for a temperature difference in oC. Solving
this equation by calculator or spreadsheet software for numerical solution of an equation gives Tw
= 91.58oC. This gives the outside heat transfer coefficient, ho = 9200/(110oC – 91.58oC)0.25 =
4441 W/m2oC. We now have all the information we need to compute the overall heat transfer
coefficient.
April 25 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 3
1
1 D ln Do Di  Do
  o


U o h0
2k
hi Di
m 2  K 0.025 m ln 0.025 m  0.02 m  m 2  K 0.025 m 9.706 x10 4 m 2  K


385 W
4441 W
1677 W 0.02 m
W
2
m K
Taking the reciprocal gives Uo = 1030 W/m2oC. We can find the length by finding the heat
  U A T . First, we can compute the heat transfer by
transfer area using the equation that Q
o o
lm
applying the first law energy balance to the glycol flow.
1 kg 2428 J
W s
Q  m c p T2  T1  
40o C  20o C
 48560 W
s kg  K
J


Next we compute the log-mean temperature difference for this heat exchanger. Although we are
not told if it is parallel flow or counter flow, that does not matter since the hot side temperature is
a constant.
Tln 
T
h
 Tc ,in   Th  Tc ,out 
 T  Tc ,in 

ln  h

T

T
h
c
,
out



110 C  20 C   110 C  40 C   79.58 C
o
o
o
o
o
 110 C  20 C 

ln 
o
o
 110 C  40 C 
o
o
We can now find the desired length.
Q  U o Ao Tlm  U oDo LTlm

L
Q
48560 W

1030
W
U oDo Tlm
 0.025 m  79.58 o C
2
m K
L = 7.54 m
11-46
Steam in the condenser of a steam power plant is
to be condensed at a temperature of 50°C (hfg =
2383 kJ/kg) with cooling water (cp = 4180 J/kg°C)
from a nearby lake, which enters the tubes of the
condenser at 18°C and leaves at 27°C. The surface
area of the tubes is 42 m2, and the overall heat
transfer coefficient is 2400 W/m2°C. Determine the
mass flow rate of the cooling water needed and the
rate of condensation of the steam in the condenser.
From the information given on temperatures and area,
we can compute the heat transfer using the usual
  UAT . Once we know the heat
equation Q
lm
transfer we can compute the required mass flow rates
by first law energy balances. Since the temperature of
the steam does not change we can write our log-mean
temperature difference equation as follows.


April 25 homework solutions
Tln 
T
h
ME 375, L. S. Caretto, Spring 2007
 Tc ,in   Th  Tc ,out 
 T T 
ln  h c ,in 
 Th  Tc ,out 

Page 4
50 C  18 C   50 C  27 C   27.3 C
o
o
o
o
o
 50 C  18 C 

ln  o
o
 50 C  27 C 
o
o
We can now compute the heat transfer.



2400 W
Q  UATlm  2 o
42 m 2 27.3o C  3.752 x10 6 W
m  C
We can use this heat transfer to compute the mass flow rates of the cooling water and the
condensing steam.
Q  m coolingc p T2  T1 
water
Q  m steamh fg


1J
3.752 x10 6 W
Q
W  S = 73.1 kg/s
 m cooling 

4184
J
c p T2  T1 
water
27 o C  18o C
o
kg  C

 m steam 



Q
3.752 x10 6 W
= 1.15 kg/s

h fg 2383 kJ 1000 W  s
kg
kJ
11-49E A 1-shell-pass and 8-tube-passes heat exchanger is used to heat glycerin (cp = 0.60
Btu/lbmoF) from 65oF to 140oF by hot water (cp = 1.0 Btu/ lbmoF) that enters the thin-walled
0.5-in-diameter tubes at 175oF and leaves at 120oF. The total length of the tubes in the heat
exchanger is 500 ft. The convection heat transfer coefficient is 4 Btu/h ft2oF on the glycerin
(shell) side and 50 Btu/hft2oF on the water (tube) side. Determine the rate of heat transfer
in the heat exchanger (a) before any fouling occurs and (b) after fouling with a fouling
factor of 0.002 Btu/hft2oF /Btu on the outer surfaces of the tubes.
From the data given we can compute the heat transfer coefficient by assuming that the thinwalled tubes do not add to the overall resistance composing the heat transfer coefficient.
ln Do Di 
1
1
1



U o Ao h0 Ao
2kL
hi Ai

1
1 1 h  ft 2 o F h  ft 2 o F 0.27 h  ft 2 o F
  


U h0 hi
4 Btu
50 Btu
Btu
Taking the reciprocal gives U = 3.704 Btu/hft2oF. From this U value and the other given data,
  UAFT where F is the correction
we can find the heat transfer by the usual equation, Q
lm
factor to account for the 8 tube passes in this heat exchanger. We compute the log-mean
temperature difference for a counter-flow heat exchanger (the basis for the correction factor
method) as follows.
Tln 
T
h ,out
 Tc ,in   Th ,in  Tc ,out 
T
T 
ln  h ,out c ,in 
 Th ,in  Tc ,out 

120 C  65 C   175 C  140 C   44.25 C
o
o
o
o
o
 120 C  65 C 

ln 
o
o
 175 C  140 C 
o
o
We have to compute the ratios R and P to determine the correction factor.
April 25 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 5
Ttube,out  Ttube,in
t 2  t1 120 o F  175 o F
P


 0.5
Tshell,in  Ttube,in T1  t1 65 o F  175 o F
T
T
T T
65 o F  140 o F
R  shell,in tube,in  1 2 
 1.36
Ttube,out  Ttube,in t 2  t1 120 o F  175 o F
From Figure 11.18(a) for one shell pass and any multiple of 2 tube passes we find F = 0.60 for
these values of R and P.
We can now find the heat transfer.




3.704 Btu
Q  UAFTlm 
523.6 ft 2 0.6 44.25o F  51.5x104 Btu/h
2 o
h  ft  F
With fouling we have to add the fouling factor to our calculation of the overall heat transfer
coefficient U.
2 o
1
1
1 h  ft 2 o F 0.002 h  ft  F h  ft 2 o F 0.272 h  ft 2 o F

 R f ,i  



U h0
hi
4 Btu
Btu
50 Btu
Btu
Taking the reciprocal gives U = 3.676 Btu/hft2oF. The other data (correction factor and logmean temperature difference) do not change and we find the heat transfer as follows.




3.676 Btu
Q  UAFTlm 
523.6 ft 2 0.6 44.25o F  51.2x104 Btu/h
2 o
h  ft  F
11-90
Cold water (cp = 4.18 =J/kg°C) enters a cross-flow heat exchanger at 14°C at a rate of 0.35
kg/s where it is heated by hot air (cp = 1.0 kJ/kg°C) that enters the heat exchanger at 65°C
at a rate of 0.8 kg/s and leaves at 25°C. Determine the maximum outlet temperature of the
cold water and the effectiveness of this heat exchanger.
There are two possible limits to the maximum water temperature. The first is simply the
temperature of the hot fluid entering, 65oC. However, it is possible that the maximum heat
transfer will give a lower limit for temperature. We compute the maximum heat transfer by
computing the products of mass flow rate times heat capacity and finding which is the smaller.
Ch  m h c p ,h 
Cc  m c c p ,c 
0.8 kg 1.0 kJ 1 kW  s 0.8 kW
 o
s kg o C kJ
C
0.35 kg 4.18 kJ 1 kW  s 1.463 kW

o
s
kg o C kJ
C
So Cmin = Ch = 0.8 kW/oC. We use this to compute the maximum heat transfer.


0.8 kW
Q max  C min Th ,in  Tc ,in   o
65 o C  14 o C  40.80 kW
C
This maximum rate of heat transfer gives the maximum outlet temperature of the water found as
follows form the usual first law energy balance.
Q
40.80 kW
Q max  Cc Tc ,max out  Tc ,in   Tc ,max out  Tc ,in  max  14 o C 
= 41.9oC
1.463 kW
Cc
o
C
April 25 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 6
The effectiveness is the actual heat transfer divided by the maximum heat transfer. The actual
heat transfer can be found from the inlet and outlet temperature of the air.

C T
T
Q
    air air,in air,outn 
Qmax
Qmax
11-93


0.8 kW
65o C  25 o C
o
C
= 0.784
40.80 kW
Hot oil (cp = 2200 J/kg°C) is to be cooled by water (cp = 4180 J/kg°C) in a 2-shell-passes
and 12-tube-passes heat exchanger. The tubes are thin-walled and are made of copper
with a diameter of 1.8 cm. The length of each tube pass in the heat exchanger is 3 m, and
the overall heat transfer coefficient is 340 W/m2°C. Water flows through the tubes at a
total rate of 0.1 kg/s, and the oil through the shell at a rate of 0.2 kg/s. The water and the oil
enter at temperatures 18°C and 160°C, respectively. Determine the rate of heat transfer in
the heat exchanger and the outlet temperatures of the water and the oil.
We compute the maximum heat transfer by first computing the products of mass flow rate times
heat capacity and finding which is the smaller.
C h  m h c p ,h 
0.2 kg 2200 kJ 1 kW  s 440 W
 o
s
kJ
kg o C
C
Cc  m c c p ,c 
0.1 kg 4180 kJ 1 kW  s 418 W
 o
s
kJ
kg o C
C
So Cmin = Cc = 418 W/oC. We use this to compute the maximum heat transfer.


418 W
Q max  C min Th,in  Tc ,in   o
160 o C  18o C  5.935 x10 4 W
C
In order to find the heat transfer we have to find the heat exchanger effectiveness. We do this by
computing the NTU and using the charts that give the effectiveness as a function of NTU and the
ratio of Cmin/Cmax. The surface area is required to compute the NTU. The heat exchanger in this
problem has 12 tube passes, each of which is 3 m long, with a diameter of 1.8 cm = 0.018 m.
Thus the total heat transfer surface area is found as follows.
As  NDL  12 0.018 m 2 m   2.04 m 2


340 W
2.04 m 2
UAs m 2 o C
NTU 

 1.659
418 W
Cmin
o
C
C min
C max
418 W
o
C  0.95

440 W
o
C
For these values of NTU and Cmin./Cmax, we find  = 0.061 from Figure 11-26(d) on page 637.
The actual heat transfer is the product of the maximum heat transfer and the heat exchanger
effectiveness.


Q  Q max  0.61 5.935 x10 4 W  = 3.62x104 W
11-118 The condenser of a large power plant is to remove 500 MW of heat from steam condensing
at 30°C (hfg = 2431 kJ/kg). The cooling is to be accomplished by cooling water (cp = 4180
J/kg°C) from a nearby river, which enters the tubes at 18°C and leaves at 26°C. The tubes
of the heat exchanger have an internal diameter of 2 cm, and the overall heat transfer
April 25 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 7
coefficient is 3500 W/m2°C. Determine the total length of the tubes required in the
condenser. What type of heat exchanger is suitable for this task?
From the data given we can compute the required area by computing the log mean temperature
difference.
Tln 
Th, out  Tc,in   Th,in  Tc, out 
 Th, out  Tc, in
ln 
T T
 h, in c, out





30 C 18 C  30 C  26 C   7.28 C
o
o
o
o
o
 30 C  18 C 

ln  o
 30 C  26 o C 


o
o
If we assume a simple configuration so that there is no correction factor, we can compute the
desired tube length.
Q  U o Ao Tlm  U o Do LTlm

Q
500 x10 6 W
L

U o Do Tlm 3500 W
0.02 m  7.28o C
2
m K

L = 3.123x105 m
This length of tubes will obviously require a heat exchanger with several tube passes. After
deciding on a final design we would have to recomputed the heat transfer to account for the
correction factor in a multi-pass heat exchanger.
