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Power and Thrust of a Motor/Propeller

  1. Nov 6, 2013 #1
    I'm having some issues finding the relationship between power and thrust of a motor of a standard propeller based aeroplae.
    I've found two simple relationships that both make sense, but both (seem to) contradict each other.

    Thrust is equivelant to change in momentum:
    T = Mdot*v
    Power is the integral of thrust with respect to velocity:
    P = int(Mdot*v)dv = 0.5*Mdot*v^2

    Mass flow rate is ρAv, hence:
    T = ρAv^2 and
    P = 0.5*ρAv^3
    Or, rearranging gives:
    P^2 = (T^3)/(4ρA)

    I can follow that fine and it makes sense to me. But there's also an alternative solution:
    Work = Force * distance, so:
    Power = Work/Time = Force *Velocity = Tv = ρAv^3
    Rearranging gives:
    P^2 = (T^3)/(ρA)

    This is the same as before but without the factor of 1/4. Both make sense to me but obviously can't both be correct. Could anyone provide some guidance on the matter?

    (If it helps, both formulas are contained on wikipedia: http://en.wikipedia.org/wiki/Thrust . My scenario concerns a motor spinning a propeller to provide thrust for an aeroplane, which seems to be more in line with the first equation. But given that the first equation gives a lower power requirement I'm really more inclined to go with the second just to be safe.)
     
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  3. Nov 7, 2013 #2

    rcgldr

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    Note that velocity changes from zero to v as the air interacts with the propeller (assuming a static situation as mentioned in the wiki article). In the immediate vicinity of the propeller, the velocity is about 1/2 v, so power = T .5 v = .5 ρAv^3, same as the first equation. Link to Nasa article:

    http://www.grc.nasa.gov/WWW/K-12/airplane/propanl.html
     
  4. Nov 7, 2013 #3

    CWatters

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    The velocity of the air through the prop (approach 1) is not the same as the velocity of the aircraft (approach 2).

    The second approach could also be written..

    Power = drag * velocity (of the aircraft)

    It's not accelerating so Thrust = drag and so

    Power = Thrust * velocity (of the aircraft)
     
  5. Nov 7, 2013 #4
    Thanks for the replies. So the consensus is that the second formula applies to the craft, not the air? I would assume that (neglecting efficiency/stray losses) the thrusts and powers should be equal, it would just be the velocities that differ. Is that right?
    And as a conclusion, for calculating the power required from a motor to exert a certain amount of thrust on the craft I should use the first formula, not the second?
     
  6. Nov 7, 2013 #5

    CWatters

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    I would suggest it depends which velocity you know.

    You are much more likely to know or be able to measure the velocity of the aircraft than the velocity of the air through the prop.
     
  7. Nov 9, 2013 #6
    I'm afraid you've lost me a little bit again. The point of these calculations (for me, at least) is so that I can estimate what power my motor will have to supply to achieve a desired thrust, which should allow me to pick the right motor first time and not waste budget/time. As I don't really know the velocities, I arranged the equations such that the velocities cancel out and I'm simply left with a relationship between power (unknown), thrust (known), air density (estimated) and propeller area (known).
    Although now that I think about it, the area for the second equation should really be the cross section of the craft that drag is acting upon, rather than the propeller area.

    Reading up on wind turbine theory reveals that the first equation is most widely used. As propellers are essentially wind turbines in reverse I think I'll just go with the first one and add in a safety margin to be safe.
     
  8. Nov 9, 2013 #7

    CWatters

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    Do you have info on the prop?

    You can achieve the same thrust with a fast spinning small prop (high velocity air) or a large slow prop (and low velocity air flow).

    PS if this is for a model aircraft application best ask at a club what combination they recommend.
     
  9. Nov 11, 2013 #8
    I'm free to specify the prop myself. I did a bit of research and various model plane people suggest a prop diameter of ~ 25% of the whole wingspan and a pitch/diameter ratio greater than 0.6. Also, it seems to be important to keep the tip of the prop below the speed of sound as this drastically effects drag.
    At the moment I've been working off an estimated wing span of 800mm (31.5 inches) so a prop diameter of 7 inches (177.8mm) and pitch of 5 inches, giving a pitch/diameter ratio of 0.714. This also gives a maximum rpm of ~36500. I don't anticipate getting anywhere near that kind of rpm so it shouldn't be an issue.

    The way that the project is going it's looking more like we're going to use several smaller motors/props rather than one big one, but I imagine that I can just scale the prop sizes and specify motors so that the total combined thrust of several small motors is the same as the total thrust from a single large motor.
     
  10. Nov 11, 2013 #9

    rcgldr

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  11. Nov 11, 2013 #10

    CWatters

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    Never heard that 25% rule. I started flying electric RC models in 1984 and in the past I've flown in several international competitions. Now into scale models.. here is one I made in 2001..

    http://www.proctor-enterprises.com/photo_gallery/nieuport11/images/colin_watters_n11-1w.jpg

    Do you have info on your model? If it's a kit I strongly recommend investigating what others have put in it. Find out what worked for them and copy it.

    A 7*5 diameter prop at 36000rpm is going to be fast. Pitch speed works out at 170mph (not accounting for slip so slower in practice). So it's probably not a scale WW1 biplane you're planning, probably something more streamlined like an F5D racer (they use 4.5" props at upto 45,000 rpm but then their whole kit is expensive).

    If you haven't flown a fast RC model before the first flight probably won't last long! Strongly recommend seeking local help. Some clubs have training models you can have a go on before wasting a lot of time and money on the wrong kit.
     
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