Topic: Thermodynamics
A jet engine consumes 1 kg of fuel for each 40 kg of air. Fuel consumption is 1 kg/sec. When aircraft travels at 200 m/sec, velocity of discharge gases is 700 m/sec. The power developed by the engine is
- A. 7200 kW
- B. 5600 kW
- C. 2070 kW
- D. 4140 kW
Answer: D. 4140 kW
Explanation: Thrust F = (ṁ_a + ṁ_f)V_e - ṁ_a*V_a = (40+1)*700 - 40*200 = 28700 - 8000 = 20700 N. Power = F * V_a = 20700 N * 200 m/s = 4,140,000 W = 4140 kW.