GATE_AE_2027_MOCK_03_Q53 GATE 2027 Aerospace Engineering All-India Master Mock 03 (Flight Dynamics, Controls & Space Trajectories) NAT · 2 Marks
Propulsion • Thermodynamic Cycles of Gas Turbines • Brayton Cycle & Component Efficiencies

In an axial compressor, air enters at stagnation temperature T_{01} = 300 K. The actual stagnation temperature at compressor exit is T_{02} = 650 K. If the isentropic efficiency of the compressor is \eta_c = 0.875, what is the ideal isentropic stagnation temperature T_{02s} in Kelvin?

Answer Key:
606.25

Step-by-Step Verified Solution

### Solution: \eta_c = (T_{02s} - T_{01}) / (T_{02} - T_{01}) ==> T_{02s} - 300 = 0.875 * (650 - 300) = 0.875 * 350 = 306.25. T_{02s} = 300 + 306.25 = 606.25 K. Correct: 606.25 (range 605.0 to 608.0).
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