Chapter 1 Gas Power Cycle

Example 8.1 Show that the thermal efficiency of a Carnot
cycle operating between the temperature limits of T L AND
T H is solely a function of these two temperatures.
Solution
from
Wnet
QL
ηth
= = 1−
QH
QH
TH 1
2⎛
δQ
⎞
and S2
− S1
= ∫ ⎜ ⎟
1 ⎝ T ⎠rev
for rev. Isothermal process, T = const.
TL 4
1 2
1Q2
then S2
− S1
= Q
T ∫ δ =
1 rev T
or
1Q2
= T ( S2
− S1)
Therefore,
QH
= TH
( S2
− S1),
and QL
= TL
( S3
− S4
)
But S1
= S4,
and S2
= S3
QL
TL
( S3
− S4
) TL
ηth
= 1−
= 1−
= 1−
Q T ( S − S ) T
H
H
Actual Gas Cycle
2
1
H
T
Q H
Q L
s 1 =s 4 s2 =s 3
� Open Cycle (intake, discharge)
� Working fluid is not a pure
substance
� Heat input by COMBUSTION of
a fuel
� Involve friction
Source: Assoc.Prof.Dr.Sommai Prepream
Exhaust
gas
2
3
s
Fuel + air
mixture