11–41
11-61E Steam is condensed by cooling water in a condenser. The rate of heat transfer, the rate of condensation of steam, and
the mass flow rate of cold water 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 There is no fouling. 5 Fluid properties are constant. 6 The thermal
F35=F55F90
,,2
−=−=
incouth
TTT
F8.26
)35/20ln(
3520
)/ln( 21
21
,=
−
=
−
= TT
TT
TCFlm
lbm/s 2.02===⎯→⎯= Btu/lbm 1043
Btu/s 2105
)(
fg
steamsteamfg h
Q
mhmQ
(c) Then the mass flow rate of cold water becomes
−=
)]([
watercold
inoutp
TTcmQ
70F
11–43
80 85 90 95 100 105 110 115 120
1000
1500
2000
2500
3000
3500
4000
4500
1
1.5
2
2.5
3
3.5
4
4.5
Tsteam [F]
Q [Btu/s]
msteam [lbm/s]
80 85 90 95 100 105 110 115 120
50
100
150
200
250
300
Tsteam [F]
mw [lbm/s]
11–46
300 350 400 450 500 550 600
20
30
40
50
60
70
80
90
100
110
202
204
206
208
210
212
214
Texhaust,in [C]
Q [kW]
Texhaust,out [C]
heat
temperature
300 350 400 450 500 550 600
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
0.05
0.055
Texhaust,in [C]
mw [kg/s]
11–47
11–65 The waste dyeing water is to be used to preheat fresh water. The outlet temperatures of each fluid and the mass flow
rate 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 There is no fouling. 5 Fluid properties are constant.
10
outh,c,inouth,2
−=−=
TTTT
and
)10()80(
outh,outc,
21
−−−
−
TT
TT
Dyeing
water
Th,out
11–48
11-66 Counterflow double pipe heat exchanger with a surface area of 7.5 m2 and U = 450 W/m2·K is used to heat the engine
oil using water at 100C. It is to be determined if fouling has occurred in the heat exchanger over a period of time.
Assumptions 1 Steady state conditions exist. 2 Heat exchanger is well insulated. 3 Fluid properties remain constant.
C 80
K)J/kg4206(kg/s)75.1(
kW 47.31
100 o
,, =
−=−= C
cm
Q
TT o
phh
inhouth
The heat transfer rate is calculated as
lmsTUAQ=
K W/m4.394
C))(49.8m(7.5
W47.31 2
o2 ==
=k
TA
Q
U
lms
11–50
temperature not exceeding 93.3°C (maximum temperature for CPCV pipes recommended by the ASME Code for Process
Piping). The flow configuration should be used, whether parallel flow or counter flow, is to be determined.
Assumptions 1 Steady state conditions. 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 Effects of fouling are
𝑄̇=𝑈𝐴𝑠∆𝑇lm
Parallel flow
The LMTD for a parallel flow configuration is
∆𝑇lm,PF =(𝑇ℎ,in −𝑇𝑐,in)−(𝑇ℎ,out −𝑇𝑐,out)
ln(𝑇ℎ,in −𝑇𝑐,in
ln(105−10
ln(𝑇ℎ,in −𝑇𝑐,out
𝑇ℎ,out −𝑇𝑐,in)=(105−80)−(93.3−10)
ln(105−80
93.3−10)=48.44℃
This gives
𝑈(𝜋𝐷𝐿CF)∆𝑇lm,CF =𝑚̇ℎ𝑐𝑝ℎ(𝑇ℎ,in −𝑇ℎ,out)
The counter flow heat exchanger length is
11–51
If it is a counter flow configuration, it would require 4.86 m of the heat exchanger length to cool Th,out below 93.3°C.
11–52
11–69 A double-pipe heat exchanger is used to cool a hot fluid such that the fluid flowing into a pipe system is below the
temperature limit for HNBR O-rings, 150°C. The fouling factors for a parallel flow and a counter flow configuration are to be
determined and compared.
Assumptions 1 Steady state conditions. 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 Properties of fluids are
Thus,
𝑅𝑓=(𝜋𝐷𝐿)∆𝑇lm
𝐶ℎ(𝑇ℎ,in −𝑇ℎ,out)−1
ℎ𝑖−1
ℎ𝑜
Parallel flow
The LMTD for a parallel flow configuration is
The LMTD for a counter flow configuration is
∆𝑇lm,CF =(𝑇ℎ,in −𝑇𝑐,out)−(𝑇ℎ,out −𝑇𝑐,in)
ln(𝑇ℎ,in −𝑇𝑐,out
𝑇ℎ,out −𝑇𝑐,in)=(180−30)−(150−10)
ln(180−30
150−10)=144.94℃
Thus,
11–55
11-72E Glycerin is heated by hot water in a 1–shell pass and 8-tube passes heat exchanger. 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
50.0
09.1
175120
14080
58.0
17580
175120
12
21
11
12
=
=
−
−
=
−
−
=
=
−
−
=
−
−
=
F
tt
TT
R
tT
tt
P
In case of no fouling, the overall heat transfer coefficient is determined from
1
12
F/Btuh.0006493.0
)ft 9.418(F).Btu/h.ft 4(
1
ft 9.418
F/Btu.h.ft 002.0
)ft 9.418(F).Btu/h.ft 50(
1
11
222
2
22
=
+
+
=
++=
ooi
fi
ii AhA
R
Ah
R
175F
Hot water
140F
11–58
11–75E A 1-shell and 2-tube heat exchanger has specified overall heat transfer coefficient, inlet and outlet temperatures, and
mass flow rates, (a) the log mean temperature difference and (b) the surface area of the heat exchanger are to be determined.
cpc = 1.0 Btu/lbm·°F.
Analysis (a) Using Fig. 11-19a, the correction factor can be
0.94
0.3
80100
120180
2.0
80180
80100
12
21
11
12
=
−
−
=
−
−
=
=
−
−
=
−
−
=
F
tt
TT
R
tT
tt
P
(Fig. 11-19a)
F54.2=== )F7.57(94.0
CF lm,lm, TFT
(b) The surface area of the heat exchanger can be determined using
in ,out ,
)(
TTcm
Q
−
