Modern Engineering Thermodynamics

292 CHAPTER 9: Second Law Open System Applications
EXAMPLE 9.5 (Continued )
Solution
First, draw a sketch of the system (Figure 9.9).
Data:
m air = 0.200 kg/s
(T in ) air = 90.0°C
(T out ) air = 75.0°C
Air
(T in ) water = 20.0°C
(T out ) water = 40.0°C
Air
Water
Area = ?, S P = ?
(a) Parallel flow
Water
Area = ?, S P = ?
(b) Counterflow
FIGURE 9.9
Example 9.5.
The unknowns are the heat exchanger area and the entropy production rate for parallel flow and counterflow conditions.
The heat exchanger area, from Eq. (9.21), is
_Q
A =
UðΔTÞ LMTD
where the log mean temperature difference has different values for the parallel and counterflow arrangements. From
Eq. (9.22), we have
a. Parallel flow:
b. Counterflow:
ðΔTÞ LMTD
=
ð75:0 − 40:0Þ− ð 90:0 − 20:0 Þ = 50:5K
75:0 − 40:0
ln
90:0 − 20:0
ðΔTÞ LMTD =
ð75:0 − 20:0Þ− ð 90:0 − 40:0 Þ = 52:5K
ln
75:0 − 20:0
90:0 − 40:0
For each case, the heat transfer rate _Q is obtained by applying the modified energy rate balance equation to only the air
flow stream. Then, using the ideal gas assumption and the fact that the value of _Q must be positive for use in Eq. (9.21),
then, the corresponding heat exchanger areas are
_Q = j _m air ðh out − h in Þ air
j = j _m air c p ð
air T out − T in Þ air
j
= jð0:200 kg/sÞ½1004 J/ ðkg.KÞŠð75:0 − 90:0KÞj = j −3010 J/s j = 3010 J/s
A parallel
flow
3010 J/s
=
140:W/ m 2 . = 0:426 m2
½ ð KÞŠð50:5KÞ and
3010 J/s
A counterflow =
140:W/ m 2 . = 0:410 m2
½ ð KÞŠð52:5KÞ Also, in this case, one of the heat transfer fluids is an ideal gas and one is an incompressible liquid. Combining Eqs. (7.33)
and (7.37) into Eq. (9.24) with the condition (p in ) air =(p out ) air (i.e., a negligible pressure drop) gives
_S P = _m air c p
air ln T
out
+ _m water c water ln T
out
T in air
T in water
(a)