Page 137 - Advanced Thermodynamics for Engineers, Second Edition

124 CHAPTER 6 FINITE TIME (OR ENDOREVERSIBLE) THERMODYNAMICS

Thus
( )
1 Q _ C
T 1 ¼ þ T C ¼ 2 ð200 þ 400Þ ¼ 1200 K;
s U C A C
and

1200
T 2 ¼ ¼ 600 K:
2
This results in the temperature values shown in Fig. 6.3 for the engine and reservoirs.
The efficiency of the Carnot cycle operating between the reservoirs would have been h th ¼ 1 400/
1600 ¼ 0.75 but the efficiency of this engine is h th ¼ 1 600/1200 ¼ 0.50. Thus the engine which
delivers maximum power is significantly less efficient than the Carnot engine.
It is possible to derive relationships for the intermediate temperatures. Equation (6.9) gives
( )
1 Q _ C
T 1 ¼ þ T C ; ð6:9Þ
s U C A C
and Eqn (6.8) defines T 2 as
Q _ C
T 2 ¼ þ T C : ð6:8Þ
U C A C
_
Also Q _ C ¼ Q _ H W; and then,
Q _ C Q _ H W _
¼ :
U H A H T H U H A H T H U H A H T H
Substituting from Eqns (6.10) and (6.11) gives
Q _ C ðs T C =T H Þ
¼ : (6.15)
U H A H T H U H A H
1 þ
U C A C

T = 1600 K
.
Q = 400 units
T = 1200K .
W = 200 units
E
T = 600 K
.
Q = 200 units

T = 400 K

FIGURE 6.3
Example of internally reversible heat engine operating between reservoirs at T H ¼ 1600 K and T C ¼ 400 K
with U H A H /U C A C ¼ 1.