Correct Answer - Option 2 : 34% and 590 kW

**Concept:**

The net work done by the power plant is the difference between turbine work and pump work i.e. W_{net} = W_{T} - W_{p}

And in other words, net work done by the power plant is the difference between heat absorbed and heat rejected, i.e. W_{net} = Q_{s }- Q_{R}

The thermal efficiency of the plant is given as, \(\eta_{th}=1-\frac{Q_R}{Q_S}\)

**Calculation:**

**Given:**

Q_{S} = 100000 kJ/min, Q_{R} = 66000 kJ/min, W_{p} = 1400 kJ/min

So, net work done is,

W_{net} = 100000 - 66000 = 34000 kJ/min

And Turbine work, WT = Wnet + Wp

W_{T} = 34000 + 1400 = 35400 kJ/min ⇒ \(\frac{35400}{60}=590~kW\)

**Thermal efficiency,**

\(\eta_{th}=1-\frac{66000}{100000}=0.34=34\)%