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Turbine power

  1. Aug 19, 2008 #1
    Im trying to calculate energy of the turbines to lift an airplane , by this way, calculating potential energy necessary to lift mass of the plane to an height of, lets say, 2000 meters , ,supose it spend 1 minute in reach this altitude, it gives me a result of the order of 1 megawatts of power , is this correct?
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  3. Aug 19, 2008 #2


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    You'd have to base this on a jet that climbs while hovering, like a Harrier. The time it takes doesn't matter. Work done on the jet is the force times the distance the jet is moved.

    Work done on the air is a different matter though. Even in a steady hover, a huge amount of work is done on the air, accelerating it downwards and increasing the total energy of the air, and a huge amount of energy is consumed while hovering, even though there is no work done on the jet itself.
  4. Aug 19, 2008 #3


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    In any case, a megawatt could be a reasonable figure for a real airplane. But you didn't give us much information...
  5. Aug 20, 2008 #4
    I disagree....well sort of. I believe he is trying to calculate the power required to move a jet forward, not strait up like a harrier. If he was referring to a VTOL then you would be correct. I however will assume that he is talking about a conventionally powered airplane.

    In order to even remotely accurately find this required power you need to know the amount of drag on the aircraft and its climb rate and assume the coefficient of lift is enough to reach this climb rate. If you want to know the power for take off you will require the mass of the craft as well.
  6. Aug 20, 2008 #5


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    I would say that if you are interested in energy, you take the energy content of the fuel, which for JET-A is about 18505 BTU/Lbm and measure the fuel mass flow, exhaust temp and inlet temp to get a rough idea of how much energy is being used by the engine and what is being spent out the exhaust duct. You would have to make a swag at the mechanical losses.
  7. Aug 20, 2008 #6
    my preliminary calculation was, suposing the plane was already in the air, and a watt= 1 joule per second,then you calculate how much falls a mass in 1 second , its aprox 5 m,hence the energy to maintain plane is equal to potential energy to lift to 5 meters, ; supposing the mass of the plane be 40 tons, then take E= m*g*h= 40000k*g*5m = aprox 2 Megajoules /sec = 2 Mwatts
  8. Aug 20, 2008 #7


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    If the reference was regarding horizontal flight, then power consumed equals aerodynamic drag force times speed. If flight isn't horizontal, then add weight of the aircraft times sin of angle from horizontal to aerodynamic drag force, and still multiply by speed.
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