--- title: "Simple Rankine plant 2 MPa / 20 kPa: efficiency, power" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 6 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:rankine-2mpa-20kpa-50kgs" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/rankine-2mpa-20kpa-50kgs site: BCT Engineering --- # Simple Rankine plant 2 MPa / 20 kPa: efficiency, power Fundamental of Thermodynamics & Heat Transfer · Chapter 6: Thermodynamic Cycles - Asked in: 2068 Bhadra, 8 marks - Video: [Solving a simple Rankine Cycle Problem Engr 251](https://www.youtube.com/watch?v=egiIppF5Eyw) — asianswagstorm - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/rankine-2mpa-20kpa-50kgs - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-6 ## Question 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] ## 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. ## Answer **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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the pump work calculated? A. w_p = v_g (P2 - P1) B. w_p = v_f (P2 - P1) C. w_p = cp (T2 - T1) D. w_p = h3 - h4 Answer: B. The pump handles saturated liquid, so w_p = v_f (P2 - P1) = 0.001017 x 1980 = 2.01 kJ/kg. (Key point: Pump work) 2. How is the turbine exit state 4 found? A. Use T4 = T3 B. Use h4 = h1 C. Use s4 = s3 and find the quality x4 from s_f and s_fg D. Use P4 = P3 and h4 = h3 Answer: C. The ideal turbine is isentropic, so x4 = (s3 - s_f)/s_fg = 0.8900. (Key point: Turbine exit state) 3. What is the thermal efficiency? A. w_net / q_in = 896.6/2994.8 = 29.9 % B. (h3 - h4)/(h4 - h1) C. w_t / w_p D. q_in / w_net Answer: A. η = (w_t - w_p)/(h3 - h2) = 29.9 %. (Key point: Thermal efficiency) 4. How is the net power obtained? A. Mass flow rate x net specific work = 50 x 896.6 B. Pump work x mass flow C. Mass flow rate x heat input D. Turbine work alone divided by 50 Answer: A. W_net = m x w_net = 50 x 896.6 = 44830 kW, about 44.8 MW. 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