--- title: "Adiabatic steam turbine 50 MW: mass flow, isentropic efficiency" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 5 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:steam-turbine-5mpa-500c-50mw" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steam-turbine-5mpa-500c-50mw site: BCT Engineering --- # Adiabatic steam turbine 50 MW: mass flow, isentropic efficiency Fundamental of Thermodynamics & Heat Transfer · Chapter 5: Second Law of Thermodynamics - Asked in: 2068 Magh, 8 marks - Video: [Steam Turbine Isentropic Efficiency |Thermodynamics](https://www.youtube.com/watch?v=5uRyos0qdeA) — Engineering Hack - 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/steam-turbine-5mpa-500c-50mw - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5 ## Question Steam enters an adiabatic turbine at 5 MPa, 500°C and with a velocity of 50 m/s and exits at 50 kPa, 100°C and with a velocity of 150 m/s. If the power output of the turbine is 50 MW, determine (a) the mass flow rate of steam flowing through the turbine, and (b) the isentropic efficiency of the turbine. [Refer attached table for the properties of steam] ## Short answer (a) Mass flow rate from steady‑flow energy: w = (h₁−h₂)+(V₁²−V₂²)/2000 = 742.1 kJ/kg; ṁ = 50 MW / 742.1 kJ/kg = 67.4 kg/s. (b) Isentropic exit quality 0.9054, h₂s = 2427.2 kJ/kg; η_T = (h₁−h₂)/(h₁−h₂s) = 74.7%. ## Answer **Steam table values (read from tables)** State 1: 5 MPa, 500 °C: h₁ = 3434.5 kJ/kg, s₁ = 6.9778 kJ/kg·K. V₁ = 50 m/s. State 2 (actual): 50 kPa, 100 °C, superheated: h₂ = 2682.4 kJ/kg. V₂ = 150 m/s. At 50 kPa (saturated): h_f = 340.5, h_fg = 2304.7 (h_g = 2645.2); s_f = 1.0910, s_fg = 6.5019 (s_g = 7.5930). #### (a) Mass flow rate Steady flow, adiabatic (q = 0): w = (h₁ − h₂) + (V₁² − V₂²)/2 w = (3434.5 − 2682.4) + (50² − 150²)/(2 × 1000) w = 752.1 + (−10) = 742.1 kJ/kg ṁ = Ẇ/w = 50 000/742.1 **ṁ = 67.4 kg/s** #### (b) Isentropic efficiency Isentropic exit at 50 kPa has s₂s = s₁ = 6.9778 kJ/kg·K, which is less than s_g = 7.5930, so it is wet: x₂s = (6.9778 − 1.0910)/6.5019 = 0.9054 h₂s = 340.5 + 0.9054 × 2304.7 = 2427.2 kJ/kg η_T = (h₁ − h₂)/(h₁ − h₂s) = (3434.5 − 2682.4)/(3434.5 − 2427.2) = 752.1/1007.3 **η_T = 0.747 (about 74.7 %)** (Including the kinetic energy change in both the actual and isentropic work gives 74.4 %, so the result is about 74–75 %.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which energy equation gives the specific work of this adiabatic turbine? A. w = (h2 - h1) + (V2^2 - V1^2)/2 B. w = (h1 - h2) + (V1^2 - V2^2)/2 C. w = h1 - h2 only, ignoring velocity always D. w = cp (T1 - T2) - V1^2 Answer: B. Steady flow with q = 0 gives w = (h1 - h2) + (V1^2 - V2^2)/2. (Key point: Energy equation) 2. How is the mass flow rate obtained from the specific work? A. m = power/w B. m = w x power C. m = w/power D. m = power x (h1 - h2) Answer: A. m = W/w = 50 000 kW / 742.1 kJ/kg = 67.4 kg/s. (Key point: Mass flow rate) 3. Why is the isentropic exit state at 50 kPa wet? A. Pressure is below atmospheric B. T2s is below 100 C C. s2s = s1 = 6.9778 is less than sg = 7.5930 D. h1 is above hg Answer: C. An entropy below sg at that pressure lies inside the dome, so x2s = 0.9054. (Key point: Isentropic exit state) 4. Using enthalpy drops only, what is the isentropic efficiency? A. About 74.7%, from 752.1/1007.3 B. About 50% C. About 90.5% D. About 135% Answer: A. eta = (h1 - h2)/(h1 - h2s) = 752.1/1007.3 = 0.747. 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