8-105
8-150 Water is heated by hot oil in a heat exchanger. The outlet temperature of the oil and the rate of
entropy generation within 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 heat transfer to the
cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid
properties are constant.
Properties The specific heats of water and oil are
given to be 4.18 and 2.3 kJ/kg.qC, respectively.
Analysis (a) We take the cold water tubes as the
system, which is a control volume. The energy
balance for this steady-flow system can be expressed
in the rate form as
E E out
in
Rate of net energy transfer
by heat, work, and mass
E in
Q in m h1
Q
in
'E systemÊ0 (steady)
0
Oil
170qC
10 kg/s
70qC
Water
20qC
4.5 kg/s
Rate of change in internal, kinetic,
potential, etc. energies
E out
m h2 (since 'ke # 'pe # 0)
m c p (T2 T1 )
Then the rate of heat transfer to the cold water in this heat exchanger becomes
Q
[m c p (Tout Tin )]water
(4.5 kg/s)(4.18 kJ/kg.qC)(70qC 20qC) = 940.5 kW
Noting that heat gain by the water is equal to the heat loss by the oil, the outlet temperature of the hot oil is
determined from
Q
[m c p (Tin Tout )]oil o Tout
Tin Q
m c p
170qC 940.5 kW
(10 kg/s)(2.3 kJ/kg.qC)
129.1qC
(b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the
entropy balance on the entire heat exchanger:
S Sout
in
Rate of net entropy transfer
by heat and mass
Sgen
,
Rate of entropy
generation
m 1s1 m 3s3 m 2 s2 m 3s4 Sgen
m water s1 m oil s3 m water s2 m oil s4 Sgen
Sgen
'Ssystem©0 (steady)
Rate of change
of entropy
0 (since Q
0)
0
m water ( s2 s1 ) m oil ( s4 s3 )
Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is
determined to be
Sgen
m water c p ln
T
T2
m oilc p ln 4
T3
T1
(4.5 kg/s)(4.18 kJ/kg.K)ln
129.1 + 273
70 + 273
(10 kg/s)(2.3 kJ/kg.K)ln
170 + 273
20 + 273
0.736 kW/K
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