Page 82 - Advanced thermodynamics for engineers

4.3 EXAMPLES 67

The net useful work done by the system is
X
w u:net ¼ ðw use þ w R Þ¼ 26:46 þ 13:52 þ 62:18 ¼ 49:24 kJ=kg
The total displacement work done against the surroundings is

w surr ¼ 26:46 þ 33:63 ¼ 7:17 kJ=kg
The displacement work of the surroundings evaluated from
3
5
p 0 v 2 v 1 ¼ 1 10 1:0909 1 0:7896 10 ¼ 7:17 kJ=kg

Example 4.3.2: change of availability in reversible piston-cylinder arrangement
Calculate the change of availability for the process described in example 4.3.1. Compare the value
obtained with the work terms evaluated in example 4.3.1.
Solution:
The appropriate definition of availability for System A is the non-flow availability function,
defined in Eqn (4.12a), and the maximum useful work is given by Eqns (4.11) and (4.12b).Hence,
w use is given by
w use ¼ Da ¼ ða 2 a 1 Þ¼ a 1 a 2

¼ u 1 u 2 þ p 0 v 1 v 2 T 0 s 1 s 2
The change of entropy, s 1 s 2 can be evaluated from
dT dp
ds ¼ c p R
T p
which for a perfect gas becomes
T 1 p 1
s 1 s 2 ¼ c p ln R ln
T 2 p 2
550 2
¼ 1:005 ln 0:287 ln ¼ 0:41024 kJ=kg K
300 1
Substituting gives

w use ¼ c v T 1 T 2 þ p 0 v 1 v 2 T 0 s 1 s 2
1 10 5
¼ð1:005 0:287Þð550 300Þþ ð1 1:0909Þ 0:7896 300 0:41024
10 3
¼ 179:5 7:17 123:07 ¼ 49:25 kJ=kg:
This answer is identical to the net useful work done by the system evaluated in part (vi) above; this
is to be expected because the change of availability was described and defined as the maximum useful
work that could be obtained from the system. Hence, the change of availability can be evaluated
directly to give the maximum useful work output from a number of processes without having to
evaluate the components of work output separately.

Example 4.3.3: availability of water vapour
Evaluate the specific availability of water vapour at 30 bar, 450 C if the surroundings are at

p 0 ¼ 1 bar and t 0 ¼ 35 C. Evaluate the maximum useful work that can be obtained from this vapour if

it is expanded to (i) 20 bar, 250 C; (ii) the dead state.