Answer :
Answer:
a) efficiency is equal to 67.2%, and this is lesser than the maximum obtainable efficiency, so this power output is possible.
b) efficiency is 75%, this is approximately equal to the maximum obtainable efficiency, but not more that it. This power output is also possible.
Explanation:
Cold reservoir temperature Tc = 295 K
Hot reservoir temperature Th = 1175 K
Energy input Q = 150000 K/h
Converting to kJ/s, Q = 150000/3600 = 41.66 kJ/s
Maximum efficiency that can be obtained from this cycle = [tex]1 - \frac{Tc}{Th}[/tex]
==> [tex]1 - \frac{295}{1175}[/tex] = 0.748 ≅ 75%
also recall that actual cycle efficiency = [tex]\frac{W}{Q}[/tex]
Where W is the energy output or work
a) for work of 28 kW,
eff = [tex]\frac{W}{Q}[/tex] = [tex]\frac{28}{41.66}[/tex] = 0.672 ≅ 67.2%
this is lesser than the maximum obtainable efficiency, so this power output is possible.
b) for work of 31.2 kW
eff = [tex]\frac{W}{Q}[/tex] = [tex]\frac{31.2}{41.66}[/tex] = 0.748 ≅ 75%
this is approximately equal to the maximum obtainable efficiency, but not more that it. This power output is possible.
NB: kW is also equal to kJ/S