Page 87 - Advanced thermodynamics for engineers

72 CHAPTER 4 AVAILABILITY AND EXERGY

(b) Non-isentropic expansion
If the isentropic efficiency of the turbine is 85%, i.e. h T ¼ 0.85, then the turbine work is

w T ¼ h w T j isen ¼ 0:85 629:5 ¼ 535:0kJ=kg
T
The temperature at the end of the expansion is T 2 ¼ T 1 w T /c p ¼ 1273 535.0/1.100 ¼ 786 K.
The change of availability is given by Eqn (4.13b) as

Da ¼ a f2 a f1
¼ h 2 T 0 s 2 ðh 1 T 0 s 1 Þ
¼ h 2 h 1 T 0 ðs 2 s 1 Þ¼ w T T 0 ðs 2 s 1 Þ
The change of entropy for a perfect gas is given by

T 2 p 2
Ds 12 ¼ s 2 s 1 ¼ c p ln R ln
T 1 p 1
where, from the perfect gas law,

ð1:35 1Þ 1:100
k 1 c p
R ¼ ¼ ¼ 0:285 kJ=kg K
k 1:35
Thus the change of entropy during the expansion process is
786 1
s 2 s 1 ¼ 1:1 ln 0:285 ln ¼ 0:5304 þ 0:6562 ¼ 0:1258 kJ=kg K
1273 10
Thus Da ¼ 535 T 0 (s 2 s 1 ) ¼ 535 293 0.1258 ¼ 571.9 kJ/kg
(c) The change of availability of the surroundings is equal to the work done, hence
Da surroundings ¼ 535 kJ=kg
(d) The irreversibility is the change of availability of the universe, which is the sum of the changes of
availability of the system and its surroundings, i.e.

Da universe ¼ Da system þ Da surroundings ¼ 571:9 þ 535:0 ¼ 36:9kJ=kg
This can also be calculated directly from an expression for irreversibility
I ¼ T 0 DS
:
¼ 293 0:1258 ¼ 36:86 kJ=kg
Note that the irreversibility is positive because it is defined as the loss of availability.

Example 4.5.2: an air compressor
A steady flow compressor for a gas turbine receives air at 1 bar and 15 C, which it compresses to

7 bar with an efficiency of 83%. Based on surroundings at 5 C, determine (a) the change of avail-

ability and the work for isentropic compression. For the actual process, evaluate (b) the change of
availability and work done, (c) the change of availability of the surroundings and (d) the irreversibility.
Treat the gas as an ideal one, with the specific heat at constant pressure, c p ¼ 1.004 kJ/kg K and the
ratio of specific heats, k ¼ 1.4.