Project ForgeAt 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).
Explanation
Since the heat loss occurs reversibly at a single boundary temperature T=500K, the entropy change of the steam equals q/T (no entropy generation): Δs = s2−s1 = 6.3−6.5 = −0.2 kJ/kg.K, so q = TΔs = 500×(−0.2) = −100 kJ/kg (heat leaving the steam). Energy balance per unit mass: h1+q = h2+w, so w = h1−h2+q = 3500−2500+(−100) = 900 kJ/kg. Total power = ṁw = 500×900 = 450,000 kW = 450 MW.
Source: GATE ME 2022 Session 2 (ME-2) (2022)
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