Page 115 - Advanced Thermodynamics for Engineers, Second Edition

102 CHAPTER 5 RATIONAL EFFICIENCY OF POWER PLANT

same two temperature limits (T 3s and T 1 ). If T 0 is not equal to T 3 , but is equal to a lower temperature,
T L , then the rational efficiency will be less than unity because energy which has the capacity to
0
do work is being rejected. This is depicted by area C in Fig 5.1(b), and the rational efficiency
becomes

Area A
Areas ðA þ B þ CÞ Areas ðB þ CÞ
R < 1 (5.9)
h ¼ ¼
Areas ðA þ B þ CÞ Area ðBÞ Areas ðA þ CÞ
Consideration of Eqn (5.8) shows that it is made up of a number of different components which
can be categorised as available energy and unavailable energy. This is similar to exergy, as shown
below
E F (5.10)
b ¼ a a 0 ¼ h h 0 T 0 ðs s 0 Þ ¼
|fflfflffl{zfflfflffl} |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} |{z} |{z}
available unavailable available unavailable
energy energy energy energy

Thus Eqn (5.7) may be rewritten
b w net ¼ b 2 b 1 ¼ h 2 h 1 T 0 ðs 2 s 1 Þ¼ dE 12 dF 12 ; (5.11)

giving Eqn (5.8) as
w net h 2 h 1 T 3s ðs 2 s 1 Þ dE 12 dF 12 ðT 3s T 0 Þðs 2 s 1 Þ
h ¼ h 2;th ¼ ¼ ¼ (5.12)
R
b w net h 2 h 1 T 0 ðs 2 s 1 Þ dE 12 dF 12
which can be written as

h ¼ h 2;th ¼ 1 ðT 3s TÞðs 2 s 1 Þ : (5.13)
R
dE 12 dF 12
If T 0 ¼ T 3s then the rational efficiency, h R ¼ 1.0. If T 0 < T 3s , as would be the case if there were
external irreversibilities, then h R < 1.0 because the working fluid leaving the engine still contains the
capacity to do work.
If the engine is not internally reversible then the T–s diagram becomes 1-2-3-4-1, as depicted in Fig
5.1(b). The effect of the internal irreversibility is to cause the entropy at 3 to be bigger than that at 3s,
and hence the entropy difference, dS 34 > dS 12 . The effect of this is that the net work becomes

w net ¼ðh 2 h 1 Þ ðh 3 h 4 Þ¼ T 1 ðs 2 s 1 Þ T 3 ðs 3 s 4 Þ
¼ T 1 ðs 2 s 1 Þ T 3 ðs 3s s 4 Þ T 3 ðs 3 s 3s Þ
¼ h 2 h 1 T 0 ðs 2 s 1 Þ ðT 3 T 0 Þðs 2 s 1 Þ T 3 ðs 3 s 3s Þ
(5.14)
¼ dE 12 dF 12 ðT 3 T 0 Þðs 2 s 1 Þ T 3 ðs 3 s 3s Þ
Hence, from Eqn (5.14)

h ¼ h 2;th ¼ 1 ðT 3 T 0 Þðs 2 s 1 Þþ T 3 ðs 3 s 3s Þ : (5.15)
R
dE 12 dF 12