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.
Step-by-step 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 =
![1 - (Tc)/(Th)](https://img.qammunity.org/2021/formulas/engineering/college/k93lij1j6o455k39h3n934hxx7o0aqwkkq.png)
==>
= 0.748 ≅ 75%
also recall that actual cycle efficiency =
![(W)/(Q)](https://img.qammunity.org/2021/formulas/engineering/college/56b3p287uxlk37jb3stahn7en9i68kosqm.png)
Where W is the energy output or work
a) for work of 28 kW,
eff =
=
= 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 =
=
= 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