Page 136 - Advanced Thermodynamics for Engineers, Second Edition

6.2 EFFICIENCY AT MAXIMUM POWER 123

vW _ 2 T C
Hence, ¼ 0 when s ¼ N or s ¼ .
vs T H
Considering only the nontrivial case, for maximum work output
1=2
T 2 T C
¼ : (6.13)
T 1 T H
This result has the effect of maximising the energy flow through the engine while maintaining the
thermal efficiency (h th ¼ 1 T 2 /T 1 ) at a reasonable level. It compromises between the high efficiency
(h th ¼ 1 T C /T H ) of the Carnot cycle (which produces zero energy flow rate) and the zero efficiency
engine in which T 2 ¼ T 1 (which produces high energy flow rates but no power).
Hence the efficiency of an internally reversible, ideal heat engine operating at maximum power
output is

1=2

T C
h ¼ 1 : (6.14)
th
T H
Applying this result to the example given in Fig. 6.2, we obtain h ¼ 0.5528, as shown above.
An example will be used to show the significance of this result.
EXAMPLE
Consider a heat engine connected to a hot reservoir at 1600 K and a cold one at 400 K. The heat
transfer conductances (UA) are the same on both the hot and cold sides. Evaluate the high and low
temperatures of the working fluid of the internally reversible heat engine for maximum power output;
also calculate the maximum power.
Equation (6.11) gives the work rate (power) as

T C
W _ s T H
ð1 sÞ:
¼
U H A H T H U H A H
s 1 þ
U C A C
1/2
For maximum power output, s ¼ (T C /T H ) 1/2 ¼ (400/1600) .
Then,
1
W _ 2 1 1 1600
4
¼ 1600 1 ¼ ¼ 200 units
U H A H 1 ð1 þ 1Þ 2 8
2
and, from Eqn (6.10),
_
Q _ H W
¼ ¼ 400 units
U H A H U H A H ð1 sÞ
Then,
Q _ C Q _ C 1
_
_
¼ ¼ Q W ¼ 200 units:
H
U H A H U C A C U H A H