No, this has nothing to do with propellors. Carlton said this is about the thrust of a turbojet engine. The thrust is the force the jet engine produces in the direction of motion, so this simple equation to calculate power of the engine applies correctly. Sure, you could also define the power of the engine as the power turning the turbine, but the original poster didn't say anything about the power of the turbine and in this case there is no direct way to compute the power based on thrust.
I will admit that my knowledge of how the power of jets is defined is based on a different kind of jet, i.e. rocket motors. In this case, the power is quite unambiguously the equation stated below. On Tuesday 11 May 2004 08:46, Bill Hooper wrote: > On 2004 May 11 , at 12:18 AM, J. Ward wrote: > > If the force of thrust is in newtons, and the speed is in m/s, then > > the power > > in watts is simply > > Power = Force * Speed. > > While you equation is undoubtably correct, you have overlooked an > important point. The force must be acting on the object which is moving > at the speed. That will give the power of produced (or expended) by > that object. Since the original question, I believe, asked about the > thrust of the propellor on the air (equivalent to the thrust of the air > that pushes the plane) and the power output of the engine, the above > equation could not be used (not directly anyway). > > In a plane, one has to be concerned with the power generated by the > engine and the power expended by the propellor against the air. It is > the power of the propellor against the air which you get if you > multiply the thrust (force) of the propellor against the air times the > speed of the propellor through the air*. That power may not be the same > as the power generated by the engine. > > The power output of the engine would be the force of engine against the > shaft of the propellor multiplied times the speed of the propellor > shaft. (Since this is a rotational situation, the appropriate variation > of the above equation would be: power equals the "rotational force" > (called the torque, in newton-metres in SI) multiplied by the > rotational speed of the shaft (the angular speed, or angle turned per > unit time, measured in radians per second in SI). > > > Regards, > Bill Hooper > Fernandina Beach, Florida, USA > > *PS > I may not have been completely correct above when I referred to the > speed of the propellor through the air. The air arrives at the > propellor with one speed (the speed of the plane) and leaves with a > different speed, because it has been pushed toward the rear by the > propellor. This change in speed of the air as it goes past the > propellor would have to be taken into account for my description above > to be accurate. > > The science is easy; the engineering is where it gets difficult ... as > always. :-) > > ======================== > SIMPLIFICATION begins with SI > ========================
