Modern Engineering Thermodynamics

348 CHAPTER 10: Availability Analysis
Internal
heat transfer
Cold
Hot
Internal
heat transfer
Hot
Mixed
Cold
Nonmixing heat exchanger
(a)
Mixing heat exchanger
(b)
FIGURE 10.20
Schematic examples of nonmixing and mixing heat exchangers.
EXAMPLE 10.15
The inlet air to a gas turbine engine is preheated with the engine’s exhaust gases. The preheater is insulated so that all
heat transfer is internal. The engine’s exhaust gas enters the preheater at 500.°C and 1.10 atmospheres pressure and
exits at 400.°C and 1.00 atm. The inlet air enters the
preheater at 20.0°C and 1.50 atm and exits at 1.40 atm. The
mass flow rates of the inlet air and the exiting exhaust
are approximately the same at 0.800 kg/s. Both gases can
be treated as constant specific heat ideal gases. The specific
heat and gas constant of the exhaust gas are (c p ) exh. =
0.990 kJ/(kg · K) and R exh. = 0.272 kJ/(kg·K). Neglecting all
kinetic and potential energies and taking local environment
(ground state) as p 0 = 1.00 atm and T 0 = 20.0°C,
determine
a. The exit temperature of the inlet air.
b. The second law availability efficiency of the preheater.
Solution
First, draw a sketch of the system (Figure 10.21).
The unknowns are the exit temperature of the inlet air and
the second law availability efficiency of the preheater.
To engine
1.40 atm
Engine exhaust
500.°C
1.10 atm
Ground state:
T 0 = 20.0°C
p 0 =1.00 atm
FIGURE 10.21
Example 10.15.
Internal
heat transfer
m air = m exhaust = 0.800 kg/s
Engine exhaust
400.°C
Inlet air 1.00 atm
20.0°C
1.50 atm
a. The air exit temperature can be obtained from an energy rate balance on the steady state preheater with _Q = _W = 0
and, with the kinetic and potential energy terms neglected, as
_m exh ðh in − h out Þ exh
= _m air ðh out − h in Þ air
Since both sides of the heat exchanger are ideal gases, this can be written as
_m exh ðc p Þ exh ðT in − T out Þ exh = _m air ðc p Þ air ðT out − T in Þ air
from which we can solve for the exit air temperature as
ðT out Þ air = ðT in Þ air +
_m
exhðc p Þ exh
ðT in − T out Þ
_m cold ðc p Þ exh
air
or
ðT out Þ air
= 20:0°C +
ð0:800 kg/sÞð0:990 kJ/kg
.KÞ
ð500: − 400:°CÞ = 119°C
ð0:800 kg/sÞð1:004 kJ/kg .KÞ