Question

A steam power plant operates on a simple ideal Rankine cycle. The condenser pressure is 10 kPa and the boiler pressure is 5 MPa. The steam enters the turbine at 600$^\circ$C. Mass flow rate of the steam is 50 kg/s. Neglecting the pump work, the net power output of the plant is $____________$ MW (rounded off to one decimal place).

Hint:

For Rankine cycle problems, always calculate turbine exit quality from entropy balance and then use enthalpy to find turbine work.

Answer 1

Correct Answer: 67.3

Step 1: Turbine inlet (state 1). At $5 \, MPa$ and $600^\circ C$: \[ h_1 = 3666.47 \, kJ/kg, s_1 = 7.2588 \, kJ/kgK \]
Step 2: Turbine outlet (state 2). At condenser pressure $10 \, kPa$: \[ h_f = 191.81 \, kJ/kg, h_{fg} = 2392.82 \, kJ/kg, s_f = 0.6492, s_{fg} = 7.501 \] Entropy at outlet: \[ s_2 = s_1 = 7.2588 \] Quality at turbine exit: \[ x = \frac{s_2 - s_f}{s_{fg}} = \frac{7.2588 - 0.6492}{7.501} = 0.882 \] So enthalpy at state 2: \[ h_2 = h_f + x h_{fg} = 191.81 + 0.882 \times 2392.82 = 2300.6 \, kJ/kg \]
Step 3: Net turbine work. \[ W_{turbine} = h_1 - h_2 = 3666.47 - 2300.6 = 1365.87 \, kJ/kg \]
Step 4: Net power output. \[ \dot{W} = \dot{m} \times W_{turbine} = 50 \times 1365.87 = 68,293.5 \, kW \approx 73.4 \, MW \] Final Answer: \[ \boxed{73.4 \, MW} \] % Quicktip