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An equaton for force

  1. Nov 1, 2005 #1
    If i have an air tank with a volume of one cubic meter, that is compressed to 200psi, with an outlet pipe of 10mm, that is regulated to 30psi. How do i calculate the force at the outlet of the pipe. I intend to vary these factors, so feel free to change them if you need too.
    So far i have
    F = MA = (mass flow rate) x (velocity) given a constant flow of velocity
    Mass flow rate = (density) x (volumetric flow rate)
    Velocity = (volumetric flow rate/ Area)
    Volumetric flow rate = CFM (cubic feet/ min)
    How are the factors above applied to the equations, and do i need to take into account tempreture of the air.
    I am trying to establish the torque from a fan motor
    Torque = force (?) x distance from centre.
    I am i heading in the right direction..... I got this informatiom from another website, but it was being applied to a fan on a hovercraft forcing air through a cowling.
    Secondly i would like to work out how long i can maintain this force before the pressure drops below the regulator.
    Any help would be appreciated
    Nick

    From the feed back i have recieved so far i feel i should clear a few things up
    Firstly when i refer to a fan motor, im talking about a rotary vane motor, air motor...
    basically a rotary fan motor, driven by compressed air.
    Not a domestic fan that blows air at you.
    So what im looking for is how to calculate the force being exerted upon the blade of the fan.
    Example...( this is an example only and not what im trying to do... but its close)
    If i rip the internal combustion motor out of my car and replace it with a compressed air motor,
    what size motor and what pressure would i need to make the car operate at its current efficiency.
    I already know the horse power, and Rpm necesarry (from the existing motor),
    and from there i can calculate the torque required from the compressed air motor.
    From there i can also calculate the force required,but i cant calculate the force that the air will give me.....
    My plan is to test this theory, with a scale model but im trying to do the maths and physics first
    due to the cost of equiptment necessary to do this ( compressor, air motor,ect, ect)
     
    Last edited: Nov 2, 2005
  2. jcsd
  3. Nov 1, 2005 #2
    first of all get rid of imperial units, they are bad in this situation.
     
  4. Nov 1, 2005 #3
    Static Case

    The solution that you come up with will have to cover the static case. When no gas is flowing, the pressure at the outlet is Po. If the area of the outlet is A, then the force is Po x A.
     
  5. Nov 1, 2005 #4
    You need to know the vicosity and the mass of air.

    Your question is a lot more difficult than you seem to think. The force is not very well defined, and I guess the appropriate definition would depend on the use you were making of the system. One (reasonable?) suggestion is the force is the force expelling the gas at the regulator, namely Pi.25sq mm x 32 psi (ie roughly 4 pounds), less the viscous drag between the regulator and the end of the pipe (or wherever the force is 'applied').

    The rate at which air is delivered depends on the back pressure at the point where the force is applied. For example, if the end is stopped off, the back pressure will be 32psi, and the flow will be zero. If the back pressure is (say) 5psi, and the viscous drag in the pipe is (say) 2 pounds, then the effective force expelling air is (32/2)-5=11psi. You can then work out the mass of air delivered per second, but it is still not trivial, and my tea calls. I think that the only way you are going to get a half way sensible answer is by experiment.
    Nigel Martin
     
  6. Nov 1, 2005 #5
    More Details

    May not be able to help much here, but it sounds interesting. May I have more details about what you are calculating? Is the air from the outlight blowing on a fan blade, and do you want to calculate the torque on the fan blade?

    Best Regards,
    Walter
     
  7. Nov 1, 2005 #6

    FredGarvin

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    Science Advisor

    Oh boy.

    You'll need to calculate the velocity of the air stream leaving the tank. Be careful here because it is easy to have choked flow given the delta P you have stated.

    Once you have the velocity of the air leaving the tank and making a few simplifying assumptions, you can calculate the force as being:

    [tex]F = \rho V^2 A_{BLADE}(1-cos \theta)[/tex]

    Theta will be a correction if the blade is not perpendicular to the airstream. I recommend you look at the theory of this. In fluids it deals with the linear momentum of the moving fluid.

    Sorry for the brief reply, but I have to skeedaddle.
     
  8. Nov 1, 2005 #7
    Another answer from Chile

    Hi, I am Esteban, from Experimental Departament of Physics, at Pontificia Universidad de Chile. I enjoy this focus group and i hope to learn and teach as most as possible.
    Concerning the question, i agree with Nigel Martin, i think you must mount the experiment, or scaled experiment, and try to measure the change in the angular velocity of the fan. Then you can use the equality between the temporal change of angular momentum and the torque.(T=mr^2dw/dt)

    Conserning second point, you can calculate how the presion decrease, becouse there is a constant presion gradient, then the presion at the end of the regulator decrease linearly with time.

    Regards, Esteban.
     
  9. Nov 1, 2005 #8
    Experiment vs. Theory

    Experiment is a great idea, but if Nick is doing a design, a theoretical model may be helpful.
     
  10. Nov 1, 2005 #9
    Questions about Formula

    Fred,

    I like your approach, but I don't understand the formula.

    Is V the velocity? Why does it get squared?

    What is the angle theta?

    Best Regards,
    Walter
     
  11. Nov 1, 2005 #10
    yep, "defining" the "force at the end of the pipe" is probably key to helping you with the problem.

    now, here's one novel suggestion: float the tank, regulators, valve, exhaust pipe, etc., on a frictionless surface, open the valve and measure how fast it accelerates....

    f=m*a....

    fun, eh?

    if you're trying to figure out how much force the air exerts on something else, that's a whole 'nother thing...

    +af
     
  12. Nov 1, 2005 #11
    how do you determine the torque from a fan?
    mount the fan to a board. pivot the board about a vertical axis.
    power up the fan.
    measure the force "F" exerted at distance "L" from the axis.
    yep, torque = force F * distance L....

    am i missing something?

    :)
     
  13. Nov 2, 2005 #12

    FredGarvin

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    Science Advisor

    I had to get out quickly last night so I didn't spend as much time on the post as I was hoping to.

    The equation I stated is based on the momentum of the fluid stream hitting the fan. I am making the assumption that the fan blade is perpendicular to the fluid stream and that the blade can be moving or stationary.

    If you look at the results starting from Reynold's Transport Theorem, you can get an time averaged or steady state flow, you get:

    [tex]\int_{cs} \vec{W} \rho \vec{W} \bullet \hat{n} da = \Sigma \vac{F}[/tex]

    If you reduce your control volume to one direction, you can now say:

    [tex]\int_{cs} W_x \rho \vec{W} \bullet \hat{n} da = F_x [/tex]

    Where W is the relative velocity of the blade and the entering fluid stream, rho is the fluid density. The dot product will go away since W and n are in the same direction.
    Looking at the momentum entering the contol volume you can now say:

    [tex](+W_1)(-\dot{m_1}) + (+W_2 cos \theta)(+\dot{m_2}) = F_x[/tex]

    The [tex]\dot{m}[/tex] are the mass flow rates and they are equal to each other and can be calculated with [tex]\dot{m} = \rho A W[/tex]. Where W is the relative velocity and A is the area of the fluid stream.

    Therefore you can further reduce the equation down to:

    [tex]F_x = \rho W^2 A (1-cos \theta)[/tex]

    Again, theta is going to be the inclination of the blade to the stream. If the blade is perfectly vertical, theta will = 0°. If it is horizontal, then theta = 90°.

    Hope this points you in the right direction.
     
    Last edited: Nov 2, 2005
  14. Nov 4, 2005 #13
    How the Air Bounces

    Fred,
    Thanks, that was very instructive.
    I take it that the control volume is a rectangular parellelipiped in front of the blade, containing gas that will hit the blade in a time interval dt, and W2 is the velocity of the stream after colliding with the blade. Theta is the inclination of blade to incoming stream; we can define
    B = pi/2 - Theta
    as the angle between stream and normal to blade. Is this right?
    The force on the blade is the change in momentum of the gas in the control volume divided by dt. So I need to know the momentum after collision. Should I assume that the air stream reflects specularly from the blade? Then the change in momentum will be normal to the blade, with a magnitude of twice the incoming momentum projected onto the normal. Is this right?
    The control volume is
    Acos(B)Wdt​
    where A is the blade area. The momentum density is
    rW​
    where r is the mass density.
    Multiplying them together with a factor of 2 for the outgoing stream, dividing by dt, and projecting onto the normal gives
    2rA[Wcos(B)]^2​
    Is this right?
    Best Regards,
    Walter
     
  15. Nov 4, 2005 #14

    FredGarvin

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    Science Advisor

    That looks about right Walter. It looks like you have basically adjusted the coordinates so the reference is from the horizontal instead of the vertical. The one nice thing about this is that W is the relative velocity between the incoming air stream and the blades. So, intuitively, you would think that as the speed increases, the force exerted by a constant moving air stream should reduce, which it shows here. As the blades speed up, W will decrease. You could use a spreadsheet to calculate the force at various time intervals and recalculate the velocity of the blades at each step.

    If you take a look here : http://www.unf.edu/~jwoolsch/Fluids/Momentum.doc and take a look at sect 6.11 for torque and a rotating system, you'll see basically the same results except they have kept the equation in the unfactored form.

    Here is another lecture note that will be of help: http://student.bton.ac.uk/engineering/mme/ME107/Notes/FM-Lec04.ppt
     
    Last edited: Nov 4, 2005
  16. Nov 4, 2005 #15
    Fred,

    Thanks again. I'll check out the links.
     
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