- 2068 Bhadra · 8 marks
Simple Rankine plant 2 MPa / 20 kPa: efficiency, power
Short answer
Pump work 2.01 kJ/kg; turbine work 898.6 kJ/kg; net work 896.6 kJ/kg; heat input 2994.8 kJ/kg. Thermal efficiency = 29.9%. Net power = 50 kg/s × 896.6 kJ/kg = 44.8 MW.
A steam power plant operates on a simple Rankine cycle between the pressure limits of 2 MPa and 20 kPa. The temperature of the steam at the turbine inlet is 400°C, and the mass flow rate of steam is 50kg/s. Determine: a) The thermal efficiency of the cycle b) The net power output of the plant [Refer attached table for the properties of steam]
Steam table values (read from tables)
State 3, 2 MPa, 400 °C (superheated): h₃ = 3248.2 kJ/kg, s₃ = 7.1290 kJ/kg·K.
At 20 kPa (T_sat = 60.1 °C): h_f = 251.40, h_fg = 2357.5 (h_g = 2608.9) kJ/kg; v_f = 0.001017 m³/kg; s_f = 0.8320, s_fg = 7.0753 (s_g = 7.9072) kJ/kg·K.
The cycle is 1 (saturated liquid at 20 kPa) → 2 (pump) → 3 (turbine inlet) → 4 (turbine exit).
Pump
w_p = v_f (P₂ − P₁) = 0.001017 × (2000 − 20) = 2.01 kJ/kg h₂ = h₁ + w_p = 251.40 + 2.01 = 253.41 kJ/kg
Turbine (isentropic, s₄ = s₃)
x₄ = (s₃ − s_f)/s_fg = (7.1290 − 0.8320)/7.0753 = 0.8900 h₄ = h_f + x₄ h_fg = 251.40 + 0.8900 × 2357.5 = 2349.6 kJ/kg w_t = h₃ − h₄ = 3248.2 − 2349.6 = 898.6 kJ/kg
(a) Thermal efficiency
q_in = h₃ − h₂ = 3248.2 − 253.41 = 2994.8 kJ/kg w_net = w_t − w_p = 898.6 − 2.01 = 896.6 kJ/kg
η = w_net/q_in = 896.6/2994.8 = 29.9 %
(b) Net power output
Ẇ_net = ṁ w_net = 50 × 896.6 = 44830 kW
Answer: η ≈ 29.9 %; net power ≈ 44.8 MW (turbine 44.9 MW, pump 101 kW).
✓ Answer checked by Claude Opus 5.5 (high effort)
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