GATE 2022 ME (ME2) – Question 61
At steady state, 500 kg/s of steam enters a turbine with specific enthalpy equal to 3500 kJ/kg and specific entropy equal to 6.5 kJ kg$^{-1}$ K$^{-1}$. It expands reversibly in the turbine to the condenser pressure. Heat loss occurs reversibly in the turbine at a temperature of 500 K. If the exit specific enthalpy and specific entropy are 2500 kJ/kg and 6.3 kJ kg$^{-1}$ K$^{-1}$, respectively, the work output from the turbine is ______ MW (in integer).
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Correct answer: 450
Explanation
**Step 1: heat transfer per kg.** The heat loss occurs reversibly at the constant temperature of 500 K, so
$$q=T\,(s_2-s_1)=500\,(6.3-6.5)=-100\text{ kJ/kg}$$
(negative: heat leaves the steam).
**Step 2: steady-flow energy equation** (kinetic and potential energy neglected):
$$w=(h_1-h_2)+q=(3500-2500)+(-100)=900\text{ kJ/kg}.$$
**Step 3: power for 500 kg/s.**
$$\dot W=500\times900=450\,000\text{ kW}=\mathbf{450\ MW}.$$