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Using a Fan to move an object

  1. Mar 29, 2009 #1
    I'm trying to see how fast a fan that produces "x" cfm can move an object that weighs 150 lbs with a cof of 0.04. How would I set up an equation to find this?

    Thanks
     
  2. jcsd
  3. Mar 29, 2009 #2
    to over come friction you haft to produce a force=(.04)n times the normal force
    so f= 150(32)(.04) = 192ft lbs.
     
  4. Mar 29, 2009 #3
    ok thanks.

    one more thing, am I right in converting cfm to ft lbs?

    28 cfm = 1338.96 Nm/s = 987.5 ft lbs/s

    this doesn't seem right to me.
     
  5. Mar 29, 2009 #4
    I'm sure this is obvious to everybody but me, but... what does "cfm" stand for?
     
  6. Mar 29, 2009 #5
    cubic feet per minute , and im not sure how to convert the whole thing
    so u want to convert cfm to foot pounds per second.
     
  7. Mar 30, 2009 #6
    yes cfm is cubic feet per minute.

    I'm trying to figure out what size fan to use for a project of mine, and I am given the number of cfm the fan produces. I'm wanting to use that to figure out how fast it can propell an object. since it will take 192 ft lbs to move the object, I figured I needed to get cfm to ft lbs, but I am not sure if this would be the correct units for this.
     
  8. Mar 30, 2009 #7
    That doesn't sound like a direct conversion to me. The fan moves a given volume of air per time, but the force actually acting on the object would involve aerodynamics. How about some more details on the project?
     
  9. Mar 30, 2009 #8

    rcgldr

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    150 lbs normal force, .04 cof, so maximum static friction = 150 lb x .04 = 6 lbs.

    The cubic feet per minute rating isn't enough to determine the force.

    force = mass x acceleration = mass x (velocity change) / (unit time)
    force = mass / (unit time) x velocity change

    Ignoring compression effect, and assuming dry air at sea level,

    1 ft^3 / min ~= 0.075 lbmass / min = .00125 lbmass / sec = .000039 slug / sec

    So CFM x .000039 gives you mass flow.
    The change in velocity due to the fan times the mass flow will give you the force at the fan.

    Assuming no losses, and that the block captures and stops all the air, the initial force on the object (until it starts moving) will be the same as the force at the fan.

    6 lbforce = Velocity_from_fan (ft / sec) x CFM x .000039 slug / sec
     
  10. Mar 30, 2009 #9
    not sure what you mean by velocity_from_fan

    think of the object as a cart and the fan(s) are mounted onboard the cart pushing air backwards, thus propelling the cart forwards. The only information i have about the output of the fan is in cfm.
     
  11. Mar 30, 2009 #10
    just wondering... would hp be benificial in this calculation?
     
  12. Mar 30, 2009 #11

    mgb_phys

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    The force on an object from the airflow is
    F = 1/2 (density of air) * (speed of air)^2 * (cross section area) * (drag coefficient)

    The drag coefficient depends on the shape of the object and varies from around 1 for a flat plate, 0.5 for a sphere and perhaps <0.1 for an aerodynamic shape.
    From the CFM rating of the fan and it's area you can calculate the speed of the air flow (just consider a tube of air of that volume/second going through a fan of that area)

    Note that you don't care how much of the air from the fan actually hits the object assuming the air speed is constant.
     
  13. Mar 30, 2009 #12
    Let me just see if I'm doing this correctly.

    Fan specs: 95 cfm with outlet of 2-11/16 x 2-1/16 inches

    Cross section area = 0.038493 ft^2

    So the speed of the air would be:

    95 cfm / 0.38493 ft^2 = 2467.99 ft/min = 41.1332 ft/sec

    So assuming standard air density of 0.075 lb/ft^3 and drag coefficient is 1:

    F= 1/2(0.075 lb/ft^3)(41.1332 ft/sec)^2(0.038493 ft^2)(1)
    F= 2.4423 ft lbs/sec^2

    Is this correct?
     
  14. Mar 30, 2009 #13

    mgb_phys

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    0.1N from a typical CPU fan - sounds reasonable.
     
  15. Mar 30, 2009 #14
    Why is this? I also don't understand why the cross section area in your equation is for that of the fan and not the object. It seems to me that on a small object, a large fan would exert less force than a small fan displacing the same volume of air (at a faster velocity).
     
  16. Mar 30, 2009 #15

    mgb_phys

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    All you care about is the airspeed so the same CFM from a larger fan is a different air speed.
    Think about the wind - in a gale do you realy care if it a local gust or a weather front 100miles long?
     
  17. Mar 30, 2009 #16
    Ok, so then is the cross section area in the equation of the fan or of the object?
     
  18. Mar 30, 2009 #17

    mgb_phys

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    In the drag equation it is the cross section of the object.
    but in order to convert from cfm to ft/min in the fan you need the area of the fan.
     
  19. Mar 30, 2009 #18

    rcgldr

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    Fan specs: 95 cfm with outlet of 2-11/16 x 2-1/16 inches
    Cross section area = 0.038493 ft^2
    95 cfm / 0.038493 ft^2 = 2467.99 ft/min = 41.1332 ft/sec
    95 cfm = 1.5833 ft^3 / sec

    1 lb = 1 slug ft / sec^2.
    air density = .075 lbmass / ft^3 = .002331 slug / ft^3 (at about 65 degrees Farenheight)

    thrust equation: force = mass flow x exit velocity (assuming initial air velocity is zero)
    fan force = (.002331 slug / ft^3) (1.5833 ft^3 / sec) (41.1332 ft / sec) = .1518 lb


    Compare to 1755 watt (2.35 hp), 90 mm electric ducted fan:

    thrust = 6.06 lb
    rotor diameter 90 mm = 3.5433 in
    rotor swept area = 9.861 in ^2 = .06848 ft^2
    volume_flow = velocity (ft / sec) x .06848 (ft^2)

    fan force = (.002331 slug / ft^3) (volume_flow (ft^3 / sec)) (velocity (ft/sec)) = 6.06 lb
    fan force = (.002331 slug / ft^3) (.06848 velocity (ft^3 / sec)) (velocity (ft / sec) = 6.06 (slug ft / sec^2)

    (.002331) (.06848) (velocity^2) (slug / ft^3) (ft^3/sec) (ft / sec) = 6.06 (slug ft / sec^2)
    (.002331) (.06848) (velocity^2) = 6.06

    velocity^2 = 6.06 / (.002331 x .06848 ) = 37965
    velocity (ft/sec) = 194.85 ft / sec
    volume_flow (ft^3/sec) = 13.343 ft^3 / sec = 800.6 cfm
     
    Last edited: Mar 30, 2009
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