Gas turbine 60 MW: heat loss and areas

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.