11-121
11-140 Water is to be heated by steam in a shell-and-tube process heater. The number of tube passes need to be used 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
kg/s 046.2
C)20CC)(90kJ/kg. (4.19
kW 600
)(
)(
,,
,,
=
−
=
−
=
−=
incoutcpc
incoutcpcc
TTc
Q
m
TTcmQ
The total cross-section area of the tubes corresponding to this mass
flow rate is
24
3m 1082.6
m/s) 3)(kg/m 1000(
kg/s 046.2 −
===→= V
m
AVAmcc
Steam
20C
Water
90C
11-123
11-142 Cooling water is used to condense the steam in a power plant. The total length of the tubes required in the condenser
is to be determined and a suitable HX type is to be proposed.
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 The overall heat transfer coefficient is constant and uniform.
C12=C18C30
,,2
−=−=
incouth
TTT
and the logarithmic mean temperature difference is
( )
C28.7
/124ln
124
)/ln( 21
21 =
−
=
−
= TT
TT
Tlm
18C
11-125
11-144 Petroleum based organic vapor is to be cooled inside a shell and tube heat exchanger. For the specified flow rates of
each fluid, the number of tubes is to be found.
Assumptions 1 Steady state conditions exist. 2 Heat exchanger is insulated. 3 Negligible thermal resistance due to pipe wall
C 70.42 o
,=
outc
T
Now the logarithmic temperature difference is calculated as
( )
21
21
/ln TT
TT
Tlm
−
=
CTTT o
outcinh3.327.4275
,,1 =−=−=
and
CTTT o
incouth601575
,,2 =−=−=
( )
C 73.44
60/3.32ln
603.32 o
=
−
= lm
T
tubes 417=
n
11-128
11-146 Saturated liquid benzene is to be cooled using cold water at 15C. For the given value of overall heat transfer
coefficient, the surface area of the heat exchanger is to be determined for four different configurations.
Assumptions 1 Steady state conditions exist. 2 Heat exchanger is well insulated. 3 Fluid properties remain constant. 4
C33.82C15
K)4187(J/kg3.5(kg/s)
C45)–(75K)J/kg(1839 g/s)5(
)(
)()(
oo
o
,
,,
,
,,,,
=+
=+
−
=
−=−
k
T
cm
TTcm
T
TTcmTTcm
inc
pcc
outhinhphh
outc
incoutcpccouthinhphh
In order to use effectiveness-NTU method we first need to determine the heat capacity rates, capacity ratio and effectiveness
of the heat exchanger.
The heat capacity rate of cold fluid (water) is
W/K14654.5K)J/kgkg/s)(41875.3( === pccc cmC
The heat capacity rate of the hot fluid (liquid benzene) is
W/K9195K)J/kg1839(kg/s)5( === phhh cmC
Thus the capacity ratio is
627.0
5.14654
9195
max
min === C
C
c
(a) For parallel flow arrangement,
5.0
1575
4575
)(
)(
,,min
,,
max
=
−
−
=
−
−
==
incinh
outhinhh
TTC
TTC
Q
Q
Therefore, for the known values of effectiveness and capacity ratio we can find the number of transfer units (NTU) from the
relation given in Table 11–5.
032.1
627.01
)627.01(5.01ln
1
)1(1ln =
+
+−
−=
+
+−
−= c
c
NTU
From the definition of NTU we find the surface area as
2
m 12.65=
==
KW/m750
W/K)9195(032.1
2
min
U
CNTU
As
m 33.201
0.02(m)
)m(65.12 2
=
==
D
A
Ls
85.0
15.0627.0
15.0
ln
1627.0
1
1
1
ln
1
1=
−
−
−
=
−
−
−
=
cc
NTU
0.02(m)
)m(42.10 2
=
==
D
A
11-129
172.1
15.0
15.0627.0
1
15.0
2/1
2
2=
−
−
=
−
−
=
c
F
315.0
627.0172.1
1172.11
1=
−
−
=
−
−
=cF
F
Based on this effectiveness for 1 shell pass we calculate the NTU1 for 1 shell and 40 tube passes as follows.
627.01627.01315.0/2
1
11/2
1
2
2
1
+−−−
+−−−
cc
11-131
11-151 Using the values provided in Prob. 11-150 and the calculated value of the total resistance, calculate the arterial and
venous overall heat transfer coefficient.
Assumptions 1 Steady state conditions exist. 2 Heat transfer coefficients of artery and vein are constant and uniform.
Analysis The overall heat transfer coefficient for the arterial and venous is calculated from the following equations
11-152 Reconsider Prob. 11-150 but with fouling resistance due to physiological inhomogeneities for both artery and vein.
Assumption 1 Steady state conditions exist. 2 Heat transfer coefficients of artery and vein are constant and uniform.
Analysis The total thermal resistance of the cardiovascular system can be formulated as,
f
farteryvein
R
R
DD
)/ln(
11
K/W 108.7=
+
+
m)0.05(m)0.0006(
K/Wm 0.0003
m) m)(0.05 0.0005(
K/Wm0.0005
2
2
11-132
11–153 The cardiovascular system used as a counter-current heat exchanger is used to warm venous blood. For the known
temperature and mass flow rate of arterial blood, the overall blood vessel length is to be determined.
11-134
Review Problems
11–155 The inlet and outlet temperatures of the cold and hot fluids in a double-pipe heat exchanger are given. It is to be
11-156 It is to be shown that when T1 = T2 for a heat exchanger, the Tlm relation reduces to Tlm = T1 = T2.
Analysis When T1 = T2, we obtain
0
21 =
−
TT
11-136
11-158 A shell-and-tube heat exchanger is used to heat water with geothermal steam condensing. The rate of heat transfer,
the rate of condensation of steam, and the overall heat transfer coefficient 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 Fluid properties are constant.
kg/s 0.659=⎯→⎯=
mm
)kJ/kg 2203(kW 1451
(c) The heat transfer area is
2
m 3.378=m) m)(3.2 024.0(14
== LDnA ii
)102/46ln(
)/ln( 21
1
0
1874
120120
55.0
18120
1874
12
21
11
12
=
=
−
−
=
−
−
=
=
−
−
=
−
−
=
F
tt
TT
R
tT
tt
P
Water
6.2 kg/s
14 tubes
120C
11-137
11-159 Hot water is cooled by cold water in a 1-shell pass and 2-tube passes heat exchanger. The mass flow rates of both
fluid streams 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 Fluid properties are constant. 5 There is no fouling.
36C
21 TT =
88.0
0.1
6036
317
45.0
607
6031
12
21
11
12
=
=
−
−
=
−
−
=
=
−
−
=
−
−
=
F
tt
TT
R
tT
tt
P
(Fig. 11-19)
Water
60C
1 shell pass
2 tube passes
31C
11-139
11-161 Water is used to cool a process stream in a shell and tube heat exchanger. The tube length is to be determined for one
tube pass and four tube pass cases.
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
C2.114C8.45C160
,,1 =−=−= outcinhTTT
C90C10C100
,,2 =−=−= incouthTTT
C6.101
90
2.114
ln
902.114
ln
2
1
21 =
−
=
−
=
T
T
TT
Tlm
The Reynolds number is
968,11
skg/m 0.002
)kg/m m)(950 m/s)(0.025 (1.008
Re
m/s 008.1
4/m) (0.025)kg/m (100)(950
kg/s) (47
4/
3
232
=
==
====
VD
DN
m
A
m
V
tube
which is greater than 10,000. Therefore, we have turbulent flow. We assume fully developed flow and evaluate the Nusselt
9.92)14()968,11(023.0PrRe023.0
14
C W/m0.50
C)J/kg s)(3500kg/m 002.0(
Pr
3.08.03.08.0 ====
=
==
k
hD
Nu
k
cp
Process
stream
160C
11-140
6.281)14()872,47(023.0PrRe023.0 3.08.03.08.0 ==== k
hD
Nu
C W/m.50.0 2=
k