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Applied Integral Calculus What did I do Wrong?

  1. Mar 28, 2013 #1
    This is a problem I've been working on... I did some things wrong, and syntactically incorrectly, and just messy. Could you help me out? Tell me how I can write this more syntactically correctly, and cleanly, please?

    Also, I realize that some might believe this would be more appropriately put in a physics category, however, understand that I'm mainly concerned with the mathematical syntax, at the moment. Thanks.

    This is to find Lift or Thrust Force as a function of length and width of propeller or rotor, pitch angle of the blades (assumed to be a constant), and RPM.

    At the end, I get Force as a function of Area, which I'm pretty sure must be wrong, since, it seems as length of propeller increases, even with a constant area of propeller blade, force should increase, because average linear velocity along the propeller should increase.

    Also, I think the units don't match up at the end,
    F = A [RPM π r / (1 min)]^2 D sin(a)
    where A=Area, RPM = n revolutions, D = Density of Air, a = pitch angle (assumed to be constant -- I know that they wouldn't be, that will be handled later).



    Force of Lift or Thrust on a Spinning Rotor or Propeller:

    We start with a "paddle in a river" analogy. A paddle of area, A, is held in a river, with water moving at velocity, v, the paddle held at an angle, a, to the direction of the motion of the water. The water has density, D.

    What is the force exerted on the paddle by the water?

    m=mass, p=momentum, t=time, F=force, D=density, V=volume, v=velocity, x=distance(in the direction of the water's velocity), A=Area, and a=acceleration in "F=ma" and a=angle in "sin(a)," π=pi

    (Area)*(velocity)*(Density) = A(x/t)D = (Volume)D/t = m/t
    ∴ Av^2D = mv/t = Δp/Δt = F
    Note momentum (p) = mv, so Δp/Δt = mΔv/Δt = ma = F

    So (adding in sin(a), using the principle of vector components),
    F_Lift = A v^2 D sin(a)
    Note: this equation does not include a coefficient of friction, drag, or any other losses.

    ===

    Now we need the "average velocity" of the rotor:

    v(r) = RPM C(r) / t ; C(r) = 2πr; t=(1 min)
    v(r) = RPM 2πr / (1 min)

    lim(n→∞) [Ʃ(from r=0 to r_f) RPM 2πdr / (1 min)] / n

    v_avg = 1/(r_f - 0) (integral from r=0 to r_f): v(r)dr
    = 1/r_f (integral from 0 to r_f): RPM 2 π r dr / (1 min)
    = RPM 2 π / [r_f (1 min)] (integral from 0 to r_f): r dr
    = RPM π r / (1 min)
    substituting "v = RPM π r / (1 min)" into the equation for force,

    F = A [RPM π r / (1 min)]^2 D sin(a)

    ===

    F = 4π^2(RPM)^2D/min^2 ∫(0 to r): r^2 w(r) sin[a(r)] dr
    if width and angle are not constant


    Here are some numbers to use for things like Lift Force, and Air Density. 2000lbs of lift force. And 0.0509 lb/ft^3 for air density.

    I also know that it's not going to be 100% efficient. If you want you can put a .7 or .6 or whatever. I haven't looked into that, yet, so I'm not including it, yet.

    Thanks so much.
     
  2. jcsd
  3. Mar 29, 2013 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    If find it hard to separate the math from the physics. In analyzing a problem in mechanics, where there is no acceleration of a body, the net forces on the body will be zero. Mathematically this means that you must be careful doing any manipulations that imply dividing by zero.

    At any rate, you'll have to be careful when speaking of things like force, acceleration, etc. The force on one body doesn't have to have any relation to the acceleration of another body.

    If I imagine the paddle held in a trough and fluid flowing by it, the natural "frame of rest" gives the paddle zero velocity. The net force on the paddle is zero because it includes the force needed to hold the paddle in the trough. In this scenario, it seems that if you want to analyze anything that has a non-zero change in momentum, you will have to look at the particles of the fluid.

    In case you haven't searched for "dimensional analysis propeller", those keywords produce lots of hits, such as http://www.google.com/url?sa=t&rct=...l4HQBw&usg=AFQjCNGvtc0IbZUosn3-dhjfpHmNaiP-iw
     
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