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Fan cfm output: reliant on speed or torque?

  1. May 9, 2005 #1
    basically i wanna know for a fan whether the air flow created is more reliant on speed or torque, if there's some sort of equation out there that'd be helpful too!
  2. jcsd
  3. May 9, 2005 #2
    Its funny, I would say that the output of the fan would depend most directly on Power (watts).

    Power = (Rotation Speed)* (Torque)

    This equation is a good first step for thinking about motors in general.
  4. May 9, 2005 #3
    The flowrate across a fan depends upon the fan speed and the primemover should have enough power to move the air. You can use affinity laws of fan.
    Q1/Q2 = N1/N2, H1/H2 = (N1/N2)^2 and P1/P2 = (N1/N2)^3, where Q is flowrate, H is head developed and P is power. When flowrate is high, the primemover develops power till it gets overloaded and trips off.

    http://www.tcf.com/TCFBlower/literature.htm#bul [Broken]
    The above link provides you basic and detailed knowledge on fans.
    Last edited by a moderator: May 2, 2017
  5. May 9, 2005 #4


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    If your in the same medium, isnt the flow rating a function of the speed which is a function of torque?
  6. May 10, 2005 #5


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    The volumetric output will be a function of speed. Saying the speed is a function of torque is a bit misleading since the torque is going to be very dependent on the fan design. Like Crosson and Quark said, one should really think in terms of power for the prime mover.
  7. May 10, 2005 #6
    ok thanks guys...but one slight question. What're N1 and N2? lol, because this is from air to air.
  8. May 10, 2005 #7


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    N refers to the rotational velocity of the motor/fan.
  9. May 10, 2005 #8


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    Thank you so much for providing that; it's exactly the sort of thing I've been trying to find to aid in my hovercraft design. :smile:
  10. May 10, 2005 #9
    if N refers to the rotational velocity (im assuming thats rad/s) why do i need N1 and N2 if i only want to find out some information for 1 speed? And shouldn't size of the fan be an issue?

    also another sort of random question. Since (according to part of Bernoulli's equation) the energy density of a moving fluid is 1/2pv^2 (where v is velocity in m/s and p is density in kg/m^3), then would this make sense:

    Say the fluid we have is water (p = 1000 kg/m^3), moving at about 1 m/s (just to make it simple. thus the kinetic energy density is 500 j/m^3. If the water is flowing at a rate of 2 m^3/s then would this mean the energy (or power rather) of the water is 1000 j/s (which is equal to 1000 W) because 500 j/m^3 * 2 m^3/s (the m^3 cancels and you get j/s) = 1000 j/s

    does that make sense?
    Last edited: May 10, 2005
  11. May 11, 2005 #10
    I didn't get what you exactly intended to ask. Flowrate through any fluid moving device, fundamentally, depends upon the speed of the moving device(or the linear velocity of the fluid) and its cross sectional area. For a fixed flowrate, if you increase cross sectional area, you can work with lower speeds. This is a trade off between initial investment and operational cost.

    For a fixed flowrate and head, power varies only with the gross efficiency of fluid moving device+prime mover assembly. Fluid kW(or Fluid HP), a term referred to indicate the theoretical power required to move a fluid at a certain flowrate and against a certain resistance, is always constant(if flowrate and head are constant) irrespective of the speed of the fluid moving device.

    Hope this clarifies your doubt.
  12. May 11, 2005 #11
    ok so the flowrate of a big slow moving fan can be the same as a small (very) quickly moving fan?

    and what about the question in my "edit" post?
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