--- title: "Gas turbine 60 MW: heat loss and areas" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 4 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:gas-turbine-60mw-heat-loss-areas" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/gas-turbine-60mw-heat-loss-areas site: BCT Engineering --- # Gas turbine 60 MW: heat loss and areas Fundamental of Thermodynamics & Heat Transfer · Chapter 4: First Law of Thermodynamics - Asked in: 2072 Asoj, 8 marks - 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/gas-turbine-60mw-heat-loss-areas - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4 ## Question A gas turbine develops 60 MW of power output. Mass flow rate of gas is found to be 80 kg/s. Properties of gas at inlet and exit of the turbine are as follows: | Properties | Inlet | Outlet | |---|---|---| | Pressure | 8 MPa | 0.1 MPa | | Temperature | 500°C | 50°C | | Velocity | 50 m/s | 150 m/s | | Elevation above the reference level | 10 m | 5 m | a) Determine the rate at which heat is lost from the turbine surface. b) Determine the inlet and outlet areas. [Take R = 287 J/kg.k Cp = 1005 J/kgk] ## Short answer Using steady-flow energy equation, the net heat transfer rate is Q̇ = +24.6 MW (net heat supplied, not lost). Inlet area A₁ = 0.0444 m², outlet area A₂ = 0.495 m². ## Answer **Given:** Ẇ = 60 MW = 60 000 kW; ṁ = 80 kg/s; inlet: 8 MPa, 500 °C (773.15 K), 50 m/s, 10 m; outlet: 0.1 MPa, 50 °C (323.15 K), 150 m/s, 5 m. R = 0.287, c_p = 1.005 kJ/kg·K (gas treated as ideal gas). **Steady-flow energy equation:** Q̇ = Ẇ + ṁ [ c_p (T₂ − T₁) + (V₂² − V₁²)/2 + g (z₂ − z₁) ] **Step 1: Terms per kg** Δh = 1.005 × (50 − 500) = −452.25 kJ/kg ΔKE = (150² − 50²)/2 = 10 000 J/kg = 10.0 kJ/kg ΔPE = 9.81 × (5 − 10) = −49 J/kg = −0.049 kJ/kg Sum = −452.25 + 10.0 − 0.049 = −442.30 kJ/kg **Step 2: Heat transfer rate** Q̇ = 60 000 + 80 × (−442.30) = 60 000 − 35 384 = **+24 616 kW ≈ +24.6 MW** The sign is positive, which means that, with the data as given, heat has to be **supplied** to the gas (about 24.6 MW); the enthalpy drop of the gas (35.4 MW) is smaller than the 60 MW power output, so no heat can be lost. A heat loss would appear only if the power were below about 35.4 MW. The calculation is shown as per the given data. **Step 3: Areas** (ideal gas, ṁ = A V/v) v₁ = RT₁/P₁ = 0.287 × 773.15/8000 = 0.02774 m³/kg v₂ = RT₂/P₂ = 0.287 × 323.15/100 = 0.9274 m³/kg A₁ = ṁ v₁/V₁ = 80 × 0.02774/50 = **0.0444 m²** A₂ = ṁ v₂/V₂ = 80 × 0.9274/150 = **0.4946 m²** **Answer: Q̇ = +24.6 MW (net heat input, not a loss, for the given data); A₁ = 0.0444 m², A₂ = 0.495 m².** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Heat transfer rate is: A. Q̇ = Ẇ + ṁ[Δh + ΔKE + ΔPE] B. Q̇ = Ẇ - ṁΔh C. Q̇ = ṁΔh D. Q̇ = Ẇ Answer: A. Steady flow energy equation rearranged. (Key point: Energy equation) 2. The enthalpy change per kg is: A. +452.25 kJ/kg B. c_p(T2 - T1) = -452.25 kJ/kg C. -49 J/kg D. 10 kJ/kg Answer: B. KE and PE terms are small. (Key point: Dominant term) 3. The result Q̇ ≈ +24.6 MW means: A. Heat is lost at 24.6 MW B. No heat transfer C. Heat must be supplied, since the enthalpy drop (35.4 MW) is below 60 MW D. The turbine is adiabatic Answer: C. Positive Q is into the gas. (Key point: Sign of Q) 4. Areas use: A. A = ṁ V/v B. v = P/(RT) C. A = ṁ/V only D. v = RT/P and A = ṁ v/V Answer: D. A1 = 0.0444 m², A2 = 0.495 m². 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