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Find the speed of a fan

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data
    You are part of a team which hopes to build the worlds first flying car. Your proposed design has four fans each of radius 50cm. Each fan sucks in air from above the car and blows it downwards.The downward flow of air can be apporximated as a column of radius equal that of the fan, travelling at a uniform speed downwards.How fast must the air be blown down wards if the car plus the passengers weigh 1000kg?


    2. Relevant equations
    Assume the density of air is 1.2 kg/m^3


    3. The attempt at a solution
    im completely stumped. Ive considered using inertia but we are lacking the mass and shape of the fa.n
     
    Last edited: Apr 9, 2012
  2. jcsd
  3. Apr 9, 2012 #2

    cepheid

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    Welcome to PF,

    What is the momentum flux (rate of change of momentum per unit time and per unit area)? across the fan blades? This will tell you the downward force (per unit area) exerted by the fan blades on the air. Presumably the air, in turn, exerts the same upward force on the fan.
     
  4. Apr 9, 2012 #3
    that information is not provided. All provided information I have stated in the question
     
  5. Apr 9, 2012 #4

    cepheid

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    No, you misunderstood me. I was prompting you to compute it starting from basic principles. This is the direction in which you need to proceed in order to solve the problem.

    Let me rephrase what I'm asking. We want to figure out what sort of "downforce" the fan blades exert on their columns of air. We know from Newton's 2nd law that F = dp/dt. How would you compute dp/dt for this situation? Hint: the flow speed of the air is constant.
     
  6. Apr 9, 2012 #5
    well we could assume the air starts with zero velocity above the fan and hence zero momentum above the fan. We then calculate the velocity it has below the fan and hence the change in momentum. from the velocity we have calculated we can calculate the change in time?
     
  7. Apr 9, 2012 #6

    cepheid

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    It's more of a steady state situation where, by the time a "parcel" of air reaches the fan, it is travelling at the flow speed v, and moves across the fan at that speed. Don't even worry about how it got up to that speed. Start off even simpler than that. You want to know what dp/dt is so that you know F. Well, what is the definition of p? What happens when you differentiate that?
     
  8. Apr 9, 2012 #7
    I think the wording means that the air is not moving above the fans then comes out at a steady speed of v
     
  9. Apr 9, 2012 #8

    cepheid

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    Whatever, it doesn't matter. Just answer this question: what is the definition of p, and what do you get when you differentiate that?
     
  10. Apr 10, 2012 #9
    with respect to what?
     
  11. Apr 10, 2012 #10
    I got an answer by assuming the air had no speed on top of the fan and it speeded up going through the fan. I don't know what is meant by the flow speed is constant.
     
  12. Apr 10, 2012 #11

    cepheid

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    Time, obviously. You're trying to find dp/dt
     
  13. Apr 10, 2012 #12
    Do you know that force = change in momentum per second. (dp/dt)
    Assuming the air speeds up as it passes through the fan can you see how to calculate the momentum increase of the air?
     
  14. Apr 11, 2012 #13
    I got 51m/s. Did anybody else do this question and explain what they did?
     
  15. Apr 14, 2012 #14

    cepheid

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    Here's what I was trying to lead the OP towards all along:

    You want to find the lift force from the fans, which depends on the rate dp/dt at which they transfer momentum to the air. Since p = mv, we can differentiate, obtaining:[tex] \frac{dp}{dt} = \frac{d}{dt}(mv) = v\frac{dm}{dt} [/tex]where the last step comes from the fact that the air flows through the fans at a constant speed. The factor dm/dt is the mass flow rate of the air. It is the rate at which mass passes through the fans in kg/s. How can we compute this flow rate?

    Consider a small interval of time dt in which a sheet of air perpendicular to the fan moves downward by a distance dx. Then the total volume of air that passes across the fan surface in this time dt is a cylinder of thickness dx and cross-sectional area A.[tex] dV = Adx[/tex]The mass of this cylinder is given by[tex] dm = \rho A dx [/tex]Divide this by dt to get dm/dtt:[tex]\frac{dm}{dt} = \rho A \frac{dx}{dt} = \rho A v[/tex]Therefore[tex] \frac{dp}{dt} = v\frac{dm}{dt} = \rho A v^2[/tex]Now if you equate this rate of change of momentum to the weight of the car + passengers (actually 1/4 of this, since there are 4 fans total), then you get:[tex]mg = \rho A v^2[/tex][tex]v = \sqrt{\frac{mg}{\rho A}}[/tex]Plugging in numbers, I get 51 m/s as well.
     
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