11–21
The Log Mean Temperature Difference Method
11-35C Tlm is called the log mean temperature difference, and is expressed as
21
TT
−
11–22
11–43 Water is heated in a double-pipe, parallel-flow uninsulated heat exchanger by geothermal water. The rate of heat
transfer to the cold water and the log mean temperature difference for this heat exchanger are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Changes in the kinetic and potential energies of fluid streams are
kW 208.3=
C)50CC)(85kJ/kg. kg/s)(4.25 4.1(
)]([ hot water
−=
−= outinph TTcmQ
The rate of heat picked up by the cold water is
kW 202.0=−=−= kW) 3.208)(03.01()03.01( hc QQ
The log mean temperature difference is
)m 4)(CkW/m 15.1(
kW 0.202
22
UA
Q
11-44 A stream of hydrocarbon is cooled by water in a double-pipe counterflow heat exchanger. The overall heat transfer
coefficient is to be determined.
kW 48.4=C)40CC)(150kJ/kg. kg/s)(2.2 3600/720()]([ HC −=−= inoutp TTcmQ
The outlet temperature of water is
C 87.2=
)]([
outw,
w
−=
T
TTcmQ inoutp
The logarithmic mean temperature difference is
C30=C10C40
C62.8=C2.87C150
,,2
,,1
−=−=
−=−=
incouth
outcinh
TTT
TTT
and
C4.44
)30/8.62ln(
308.62
)/ln( 21
21 =
−
=
−
= TT
TT
Tlm
The overall heat transfer coefficient is determined from
KkW/m 2.31 2=
=
=
U
U
TUAQlm
C))(44.40.60.025(kW 4.48
Water
10C
HC
150C
40C
Cold
water
Hot
water
85C
50C
11–23
11-45 Water is heated in a double-pipe parallel-flow heat exchanger by geothermal water. The required length of tube is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 The heat
60C
11–24
11-46 Prob. 11-45 is reconsidered. The effects of temperature and mass flow rate of geothermal water on the length of
the tube are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
U=0.55 [kW/m^2-C]
“ANALYSIS”
Q_dot=m_dot_w*c_p_w*(T_w_out-T_w_in)
11–25
geo
m
[kg/s]
L
[m]
0.1
46.31
0.125
35.52
0.15
31.57
0.175
29.44
0.2
28.1
0.225
27.16
0.25
26.48
0.275
25.96
0.3
25.54
0.325
25.21
0.35
24.93
0.375
24.69
0.4
24.49
0.425
24.32
0.45
24.17
0.475
24.04
0.5
23.92
0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
20
25
30
35
40
45
50
mgeo [kg/s]
L [m]
11–26
11–47 Glycerin is heated by ethylene glycol in a thin-walled double-pipe parallel-flow heat exchanger. The rate of heat
transfer, the outlet temperature of the glycerin, and the mass flow rate of the ethylene glycol 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
C38.6=
+=+=⎯→⎯−= C)kJ/kg. kg/s)(2.4 5.0(
kW 33.22
C20)]([ glycerin
p
inoutinoutp cm
Q
TTTTcmQ
kg/s 0.545=
−
=
−
=
−=
C]15)+(38.6C70C)[kJ/kg. (2.5
kJ/s 33.22
)(
)]([
glycol ethylene
glycol ethylene
outinp
outinp
TTc
Q
m
TTcmQ
11–28
11–49E The required number of tubes and length of tubes for a single pass heat exchanger to heat 100,000 lbm of water in an
hour from 60°F to 100°F are to be determined.
Assumptions 1 Steady operating condition exists. 2 The heat exchanger is well-insulated so that heat loss to the surroundings
AV
19.14
)s/hr 3600)(ft/s 4()ft 12/2.1)(4/)(lbm/ft 3.62(
lbm/hr 000,100
22 ===
AV
m
nc
Hence, the number of tubes required to heat 100,000 lbm of water in an hour is
11–30
11–51 The hot and cold fluid streams at specified temperature and mass flow rate enter a parallel heat exchanger. For the
known values of convection heat transfer coefficients and fouling factors, the overall heat transfer coefficient, the exit
temperature of the hot fluid and the surface area of the heat exchanger are to be determined.
Assumptions 1 Thermal resistance due to pipe thickness is negligible. 2 Thermal properties of the hot and cold fluids are
kW 83.6C)3070( K)J/kg 4180( kg/s) 0.5()( o
,, =−=−= incoutcpccTTcmQ
C 125.8 o
=
−= W/K3450
W106.83
150
3
,outh
T
(c) Using the concept of log mean temperature difference, the heat transfer rate from the heat exchanger can also be
and
C 8.55708.125 o
,,2 =−=−= outcouthTTT
( )
8.55/120ln
Thus the surface area of the heat exchanger is,
2
m 4.97=
=
=
C) (83.84K) W/m(200.67
W106.83
o2
3
lm
sTU
Q
A
11–32
11–53 A counter-flow heat exchanger has a specified overall heat transfer coefficient operating at design and clean
conditions. After a period of use built-up scale gives a fouling factor, (a) the rate of heat transfer in the heat exchanger and
(b) the mass flow rates of both hot and cold fluids are to be determined.
Assumptions 1 Steady operating condition exists. 2 The heat transfer coefficients and the fouling factors are constant and
uniform. 3 Fluid properties are constant.
11–34
11–55 Oil is cooled by water in a thin-walled double-pipe counter-flow heat exchanger. The overall heat transfer coefficient
of the heat exchanger is to be determined.
2.20 kJ/kg.C, respectively.
Hot oil
11–36
Tw,in
[C]
U
[kW/m2C]
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
22.93
23.39
23.88
24.38
24.9
25.45
26.02
26.61
27.24
27.89
28.58
29.31
30.07
30.87
31.72
32.61
33.56
34.56
35.63
36.77
37.98
5 9 13 17 21 25
22
24
26
28
30
32
34
36
38
Tw,in [C]
U [kW/m2-C]
11–37
11-57 Engine oil is heated by condensing steam in a condenser. The rate of heat transfer and the length of the tube required
are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings
kW 25.2=C)20CC)(60kJ/kg. kg/s)(2.1 3.0()]([ oil −=−= inoutp TTcmQ
The temperature differences at the two ends of the heat exchanger are
C70=C60C130
,,1
−=−=
outcinh
TTT
Steam
130C
11–38
11-58E Water is heated by geothermal water in a double-pipe counter-flow heat exchanger. The mass flow rate of each fluid
and the total thermal resistance of the heat exchanger 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
lbm/s 0.667=
F)140FF)(200Btu/lbm. (1.0
Btu/s 40
)(
)]([
water
water
−
=
−
=
−=
inoutp
inoutp
TTc
Q
m
TTcmQ
lbm/s 0.431=
F)180FF)(270Btu/lbm. (1.03
Btu/s 40
)(
)]([
watergeo.
watergeo.
−
=
−
=
−=
inoutp
inoutp
TTc
Q
m
TTcmQ
The temperature differences at the two ends of the heat exchanger are
F40=F140F180
F70=F200F270
,,2
,,1
−=−=
−=−=
incouth
outcinh
TTT
TTT
and
F61.53
)40/70ln(
4070
)/ln( 21
21 =
−
=
−
= TT
TT
Tlm
Then
F/Btus 1.34 =
==⎯→⎯=
=
=
=⎯→⎯=
FBtu/s. 7462.0
111
FBtu/s. 7462.0
F53.61
Btu/s 40 o
ss
lm
slms
UA
R
RA
U
T
Q
UATUAQ
200F
11–40
11–60 A double-pipe counter-flow heat exchanger is used to cool a hot fluid such that the fluid flowing into a pipe
system is below the temperature limit for EPDM O-rings, 150°C. The hot fluid temperature at the heat exchanger outlet is to
be determined. If the outlet temperature is above 150°C, the length of the heat exchanger necessary to cool the hot fluid
below 150°C as it exits the heat exchanger is to be determined.
Assumptions 1 Steady state conditions. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is
𝑇ℎ,out −𝑇𝑐,in)= 𝐶ℎ(𝑇ℎ,in −𝑇ℎ,out)
The outlet temperature Th,out can be determined implicitly, or by trial-and-error,
(2000 W/m2⋅K)(𝜋)(0.025 m)(5 m)(210−80)−(𝑇ℎ,out −10)
ln(210 −80
𝑇ℎ,out −10)=(3000 W/K)(210−𝑇ℎ,out)K