• Ask a Question
  • Create a Poll
300
Insert Image Size must be less than < 5MB.
    Ask a Question
    Cancel
    300
    More answer You can create 5 answer(s).
      Ask a Poll
      Cancel

      Consider the design of turbojet engine intended to produce a thrust of 25,000 lb at a takeoff velocity of 220 ft/s at sea level. At takeoff, the gas velocity at the exit of the engine (relative to the engine) is 1,700 ft/s.

      Consider the design of turbojet engine intended to produce a thrust of 25,000 lb at a takeoff velocity of 220 ft/s at sea level. At takeoff, the gas velocity at the exit of the engine (relative to the engine) is 1,700 ft/s. The fuel-air ratio by mass is 0.03. The exit pressure is equal to the ambient pressure. Calculate the area of the inlet to the engine necessary to obtain this thrust.

      1 Answers

      \[ T = \left ( \dot{m}_{air} + \dot{m}_{fuel} \right )V_{e}-\dot{m}_{air}V_{\infty }\]\[\dot{m}_{fuel} = 0.03\dot{m}_{air}\]

      \[\Rightarrow T = \left ( \dot{m}_{air} + 0.03\dot{m}_{air} \right )V_{e}-\dot{m}_{air}V_{\infty }\]

      \[\Rightarrow T = \dot{m}_{air}\left ( 1.03V_{e}-V\infty  \right )\]

      \[\Rightarrow \dot{m}_{air} = \frac{T}{1.03V_{e} – V_{\infty}} = \frac{25000}{1.03\left ( 1700 \right )-220}=16.33\,slug/s\]

      \[\dot{m}_{air} = P_{\infty }V_{\infty}A_{inlet}\]

      \[A_{inlet} = \frac{\dot{m}_{air}}{ P_{\infty }V_{\infty}} = \frac{16.33}{0.002377\left ( 220 \right )} = 31.23\,ft^{2}\]

      Answered by techAir on 8th December 2023..

      Aero Tool