Question

Air enters an aircraft engine at a velocity of 180 m/s with a flow rate of 94 kg/s. The engine combustor requires 9.2 kg/s of air to burn 1 kg/s of fuel. The velocity of gas exiting from the engine is 640 m/s. The momentum thrust (in N) developed by the engine is

• A. 47540
• B. 45660
• C. 49779
• D. 42400

Hint:

Include fuel mass when computing exit momentum for thrust.

Answer 1

The Correct Option is C

Momentum thrust = \(\dot{m} (V_{\text{exit}} - V_{\text{inlet}})\) Total mass flow rate = 94 + 1 = 95 kg/s \[ F = 95 \times (640 - 180) = 49779 \, \text{N} \]