Question

A steam power plant produces 50MW of net work while burning fuel to produce 150MW of heat energy at the high temperature. Determine the cycle thermal efficiency and the heat rejected by the cycle to the surroundings.

Answer 1

$Nth$ =$\frac{W net, out}{Qh}$

50MW is our net output.

Then, we plug it in.

= $\frac{50MW}{150MW}$ = 0.333 or 33.33%

W net, out = Qh - Ql

Ql = Qh - W net,out

Plug the values in.

Then, it becomes:

= 150MW - 50 MW

= 100MW

Thus, the cycle thermal efficiency is 33.33% and the heat rejected is 100MW.