Page 81 - Advanced thermodynamics for engineers

66 CHAPTER 4 AVAILABILITY AND EXERGY

through a reversible heat engine, as shown in Fig. 4.2(b). In the latter case, useful work would have
been obtained, and this would be equal to

dw R ¼ h dQ;
R
where h R is the thermal efficiency of a reversible heat engine operating between T and T 0 , and dQ is
the heat transfer to the system,

T T 0
¼ dQ:
T
Hence, the work output obtainable from this reversible heat engine as the system changes state
from 1 to a is

Z a Z a
T T 0 T T 0
w R ¼ dQ ¼ c p dT
T T
1 1

T a
¼ c p ðT a T 1 Þ T 0 ln
T 1
365:7

¼ 1:005 ð365:7 550Þ 300 ln ¼ 62:18 kJ=kg
550
Now consider process a to 2.
First, evaluate the specific volume at 2, v 2 .
p 1 T 2 2 300
v 2 ¼ v 1 ¼ v 1 ¼ 1:0909v 1
p 2 T 1 1 550
k
The expansion from a to 2 is isentropic and hence obeys the law pv ¼ constant. Thus the system
work is

ð1 1:0909 2 0:6649Þ 0:7896 10 5

w sys ¼ ¼ 47:16 kJ=kg
a 2 1 1:4 10 3
Z 2
1 10 5
w surr j a 2 ¼ p 0 dv ¼ 3 ð1:0909 0:6649Þ 0:7896 ¼ 33:63 kJ=kg
10
a

w usej a 2 ¼ w sys a 2 w surr j a 2 ¼ 13:52 kJ=kg
The energy available to drive a reversible heat engine is zero for this process because it is adiabatic;
hence

w R j a 2 ¼ 0:
The work terms for the total process from 1 to 2 can be calculated by adding the terms for the two
subprocesses. This enables the solutions to questions (v) to (viii) to be obtained.
The gross work done by the system is

w gross ¼ X w sys þ w R ¼ 52:96 þ 47:16 þ 62:18 þ 0 ¼ 56:38 kJ=kg