Answer
a) $\dot{W}_e=35.3\ MW$
b) $\eta=45.4\%$
Work Step by Step
Total heat rejection:
$\dot{Q}_L=145+8=153\ GJ/h$
Net power, with $\dot{Q}_H=280\ GJ/h$:
$\dot{W}_e=\dot{Q}_H-\dot{Q}_L$
$\dot{W}_e=127\ GJ/h=35.3\ MW$
Thus, the efficiency:
$\eta=\frac{\dot{W}_e}{\dot{Q}_H}=0.454=45.4\%$
