Page 127 - Advanced thermodynamics for engineers

5.4 EXAMPLES 113

Hence the isentropic turbine work is

¼ c p T 3 T 4s ¼ 1:0045 1200 509:9 ¼ 693:2kJ kg:
w T isen
The net work from the cycle is w net ¼ w T þ w C ¼ 693.2 þ ( 407.9) ¼ 285.3 kJ/kg and the
w net 285:3
thermal efficiency is h ¼ ¼ ¼ 0:575:
th
q 2s3 496:1
This value is equal to the standard expression for the efficiency of a Joule cycle, defined as
1
h ¼ 1 k 1 , see Chapter 3.
th
k
r p
The maximum net work output is defined by
b w net ¼ b 3 b 2 :
Now, the values of exergy are defined, for a perfect gas as

T k 1 p
b ¼ðh T 0 sÞ ðh 0 T 0 s 0 Þ¼ðh h 0 Þ T 0 ðs s 0 Þ¼ c p ðT T 0 Þ T 0 ln ln :
T 0 k p 0
Thus, along an isobar,

T
b ¼ c p ðT T 0 Þ T 0 ln :
T 0
Since both 2s and 3 are at the same pressure

T 3 T 2 T 3
b 3 b 2s ¼ c p ðT 3 T 0 Þ T 0 ln ðT 2 T 0 Þ T 0 ln ¼ c p ðT 3 T 2 Þ T 0 ln
T 0 T 0 T 2
Thus

1200

b w net ¼ 1:0045 ð1200 706:1Þ 300 ln ¼ 336:3kJ=kg:
706:1
The rational efficiency of the cycle based on these dead state conditions is
w net 285:3
h ¼ ¼ ¼ 0:848:
R
b w net 336:3
This means that the ideal Joule cycle is only capable of extracting 84.8% of the maximum
net work from the working fluid, whereas under similar conditions (with the dead-state
temperature defined as the minimum cycle temperature) the Rankine cycle had a rational effi-
ciency of 100%. The reason for this is that the energy rejected by the Joule cycle, from 4s to 1, still
has the potential to do work. The maximum net work obtainable from the rejected energy is
b 4s – b 1 .Thisisequal to

T 4s
½ b w net Š rejected ¼ b 4s b 1 ¼ c p ðT 4s T 1 Þ T 0 ln
T 1

509:9
¼ 1:0045 ð509:9 300Þ 300 ln ¼ 50:99 kJ=kg:
300