A jet engine consumes 1 kg of fuel for each 40 kg of air intake. Fuel consumption is 1 kg/sec. When aircraft travels in still air at 200 m/sec, velocity of discharge gases with respect to engine is 700 m/sec. The power developed by engine is

Explanation: 
Given:

\( \dot{\mathrm{m}}_{\mathrm{f}}=1 \mathrm{~kg} \)

\( \dot{\mathrm{m}}_{\mathrm{a}}=40 \mathrm{~kg}\)

Velocity of aircraft V_a = 200 m/s

Velocity of discharge V_d = 700 m/s

F_net = \(\left(\dot{m}_a+\dot{m}_f\right) \times V_{\text {exhaust }(d)}\) - \(\dot{m}_a \times V_{\text {jet }(a)}\)

Power = \( \left[\left(\dot{\mathrm{m}}_{\mathrm{a}}+\dot{\mathrm{m}}_{\mathrm{f}}\right) \mathrm{V}_{\mathrm{d}}-\dot{\mathrm{m}}_{\mathrm{a}} \mathrm{V}_{\mathrm{a}}\right] \mathrm{V}_{\mathrm{a}} \)

= [(40 + 1) 700 - 40 × 200] × 200

= 4140000 W

= 4140 kW