--- title: "Rankine cycle 2 MPa: efficiency vs Carnot" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 6 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:rankine-cycle-2mpa-efficiency-vs-carnot" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/rankine-cycle-2mpa-efficiency-vs-carnot site: BCT Engineering --- # Rankine cycle 2 MPa: efficiency vs Carnot Fundamental of Thermodynamics & Heat Transfer · Chapter 6: Thermodynamic Cycles - Asked in: 2071 Bhadra, 8 marks - Video: [Compare Carnot Ideal Rankine steam 10MPa 10kPa](https://www.youtube.com/watch?v=QRgjpCFmgR0) — Randall Manteufel - 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-cycle-2mpa-efficiency-vs-carnot - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-6 ## Question A Rankine cycle has a boiler working at a pressure of 2 MPa. The maximum and minimum temperatures during the cycle are 400°C and 50°C respectively. Determine the efficiency of the cycle and compare it with that of the Carnot cycle operating between the same temperature limits. [Refer attached table for the properties of steam] ## Answer **Steam table values (read from tables)** Boiler: 2 MPa, 400 °C (superheated): h₃ = 3248.2 kJ/kg, s₃ = 7.1290 kJ/kg·K. Condenser: saturated at 50 °C: P_sat = 12.35 kPa; h_f = 209.34, h_g = 2591.3 kJ/kg; v_f = 0.001012 m³/kg; s_f = 0.7038, s_g = 8.0749 kJ/kg·K. The cycle: 1 (saturated liquid at 50 °C) → 2 (pump) → 3 (turbine inlet) → 4 (condenser inlet). #### Pump w_p = v_f (P₂ − P₁) = 0.001012 × (2000 − 12.35) = 2.01 kJ/kg h₂ = h₁ + w_p = 209.34 + 2.01 = 211.35 kJ/kg #### Turbine (isentropic, s₄ = s₃) x₄ = (s₃ − s_f)/(s_g − s_f) = (7.1290 − 0.7038)/(8.0749 − 0.7038) = 0.8717 h₄ = 209.34 + 0.8717 × (2591.3 − 209.34) = 2285.6 kJ/kg w_t = h₃ − h₄ = 3248.2 − 2285.6 = 962.6 kJ/kg #### Heat and efficiency q_in = h₃ − h₂ = 3248.2 − 211.35 = 3036.9 kJ/kg w_net = w_t − w_p = 962.6 − 2.0 = 960.6 kJ/kg η_Rankine = 960.6/3036.9 = 0.316 → **31.6 %** #### Carnot cycle between same limits T_H = 400 + 273.15 = 673.15 K, T_L = 50 + 273.15 = 323.15 K η_Carnot = 1 − 323.15/673.15 = 0.520 → **52.0 %** #### Comparison **Answer: η_Rankine = 31.6 %, η_Carnot = 52.0 %.** The Rankine efficiency is about 61 % of the Carnot value. Heat in the Rankine cycle is added over a range of temperatures (liquid heating from about 50 °C up to the boiling point of 212 °C), not all at the highest temperature, so its mean heat addition temperature is lower. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is state 4 found? A. Constant volume B. Isentropic: s4 = s3 gives x4 from s_f and s_g at 50 C C. Constant enthalpy D. Constant pressure at 2 MPa Answer: B. x4 = 0.8717, h4 = 2285.6 kJ/kg. (Key point: Turbine exit) 2. Pump work is: A. Zero by definition B. h4 - h1 C. v_f (P2 - P1) = 2.01 kJ/kg D. h3 - h4 Answer: C. 0.001012 x (2000 - 12.35) = 2.01 kJ/kg. (Key point: Pump work) 3. Efficiency of the cycle is: A. 52.0% B. 31.6% C. 85% D. 61% Answer: B. w_net/q_in = 960.6/3036.9. (Key point: Rankine efficiency) 4. Why is Rankine lower than Carnot (52.0%)? A. Heat is added over a range of temperatures, lowering mean addition temperature B. Turbine is irreversible C. Condenser temperature is higher D. Pump work is large Answer: A. Liquid heating from about 50 C to 212 C occurs below T_max. 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