--- title: "Adiabatic air turbine: mass flow and areas" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 4 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:turbine-1000k-to-400k-areas" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/turbine-1000k-to-400k-areas site: BCT Engineering --- # Adiabatic air turbine: mass flow and areas Fundamental of Thermodynamics & Heat Transfer · Chapter 4: First Law of Thermodynamics - Asked in: 2073 Bhadra, 8 marks; 2071 Bhadra, 8 marks (asked 2 times) - Video: [Air flows steadily through an adiabatic turbine, entering at 1 MPa, 500C, and 120 m/s.](https://www.youtube.com/watch?v=jxIsdEFuoB0) — Book Knowledge - 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/turbine-1000k-to-400k-areas - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4 ## Question Air expands through an adiabatic turbine from 1000 KPa, 1000 K to 100 KPa, 400K. The inlet velocity is 10 m/s whereas exit velocity is 100 m/s. The power output of the turbine is 3600 KW. Determine the mass flow rate of air, the inlet and the exit area. [Take R = 287 J/KgK and Cp = 1005 J/KgK] ## Short answer Mass flow rate is found from steady-flow energy: ṁ = 3600 / [1.005(1000-400) + (10² - 100²)/2000] = 6.02 kg/s. Using ideal gas specific volumes, inlet area A₁ = 0.173 m² and exit area A₂ = 0.0691 m². ## Answer **Given:** adiabatic turbine, air: P₁ = 1000 kPa, T₁ = 1000 K, V₁ = 10 m/s; P₂ = 100 kPa, T₂ = 400 K, V₂ = 100 m/s; Ẇ = 3600 kW; R = 0.287, c_p = 1.005 kJ/kg·K. **Steady-flow energy equation** (Q = 0, ΔPE = 0): Ẇ = ṁ [ c_p (T₁ − T₂) + (V₁² − V₂²)/2 ] **Step 1: Mass flow rate** c_p (T₁ − T₂) = 1.005 × 600 = 603.0 kJ/kg (V₁² − V₂²)/2 = (100 − 10 000)/2 = −4950 J/kg = −4.95 kJ/kg Work per kg = 603.0 − 4.95 = 598.05 kJ/kg ṁ = 3600/598.05 = **6.02 kg/s** **Step 2: Specific volumes (ideal gas)** v₁ = RT₁/P₁ = 0.287 × 1000/1000 = 0.287 m³/kg v₂ = RT₂/P₂ = 0.287 × 400/100 = 1.148 m³/kg **Step 3: Areas** (ṁ = A V/v) A₁ = ṁ v₁/V₁ = 6.02 × 0.287/10 = **0.1728 m²** A₂ = ṁ v₂/V₂ = 6.02 × 1.148/100 = **0.0691 m²** **Answer: ṁ = 6.02 kg/s; A₁ = 0.173 m²; A₂ = 0.0691 m².** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the adiabatic turbine, the power is: A. Ẇ = ṁ[c_p(T1 - T2) + (V1² - V2²)/2] B. Ẇ = ṁ c_p (T2 - T1) only C. Ẇ = ṁ(V2² - V1²)/2 only D. Ẇ = ṁ R (T1 - T2) Answer: A. Q = 0 and ΔPE = 0, so work comes from enthalpy and kinetic-energy changes. (Key point: Energy equation) 2. The mass flow rate is: A. ≈ 5.97 kg/s ignoring KE B. ≈ 6.02 kg/s, from 3600/598.05 C. ≈ 0.6 kg/s D. ≈ 60 kg/s Answer: B. Work per kg = 603.0 - 4.95 = 598.05 kJ/kg. (Key point: Mass flow) 3. The inlet specific volume is found by: A. v1 = P1/(RT1) B. v1 = c_p T1/P1 C. v1 = RT1/P1 = 0.287 m³/kg D. v1 = RT1/P2 Answer: C. Ideal gas law per unit mass. (Key point: Specific volume) 4. Flow area is calculated from: A. A = ṁ V/v B. A = ṁ/(v V) C. A = V v/ṁ D. A = ṁ v/V Answer: D. From ṁ = A V/v; A1 = 0.173 m², A2 = 0.0691 m². 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