Page 80 - Advanced thermodynamics for engineers

4.3 EXAMPLES 65

Then evaluate for the total process, 1-2, the following parameters.
(v) The gross work done by the system, S(dW sys þ dW R );
(vi) The net useful work output against the environment, S(dW use þ dW R );
(vii) The total displacement work against the environment, S(dW surr ) and
(viii) The work term p 0 (v 2 v 1 ).
Solution:
It is necessary to evaluate the specific volume at the initial condition, state 1.

p 1 v 1 ¼ RT 1 ; hence v 1 ¼ RT 1 p 1 :
k 1 0:4 1:005
The gas constant, R, is given by R ¼ c p ¼ ¼ 0:287 kJ=kg K, and hence
k 1:4
3
0:287 10 550 3
v 1 ¼ ¼ 0:7896 m =kg
2 10 5
The intermediate temperature, T a , should now be evaluated. Since the process a-2 is isentropic then
T a must be isentropically related to the final temperature, T 2 . Hence
ðk 1Þ=k 0:4=1:4

p a 2
T a ¼ T 2 ¼ 300 ¼ 365:7K
p 2 1
p 1 T a 365:7
Thus v a ¼ v 1 ¼ v 1 ¼ 0:6649v 1
p a T 1 550
Consider process 1-a:
Z a
w sys j 1 a ¼ pdv ¼ p a ðv a v 1 Þ; for an isobaric process
1
2 10 5
¼ ð0:6649 1:0Þ 0:7896 ¼ 52:96 kJ
10 3
Z a 5
1 10
w surr j 1 a ¼ p 0 dv ¼ 10 3 ð0:6649 1:0Þ 0:7896
1
¼ 26:46 kJ=kg
The useful work done by the system against the resisting force, F, is the difference between the two
work terms given above, and is

w use j 1 a ¼ w sys 1 a w surr j 1 a ¼ 26:46 kJ=kg
The work terms derived above relate to the mechanical work that can be obtained from the system
as it goes from state 1 to state a. However, in addition to this mechanical work the system could also do
thermodynamic work by transferring energy to the surroundings through a reversible heat engine. This
work can be evaluated in the following way. In going from state 1 to state a the system has had to
transfer energy to the surroundings because the temperature of the fluid has decreased from 550 to
365.7 K. This heat transfer could have been simply to a reservoir at a temperature below T a , in which
case no useful work output would have been achieved. It could have also been to the environment