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Change in Flow of Momentum

  1. Apr 20, 2015 #1
    There is a pipe which is bent in the shape of a right angle as shown. The cross section area of the pipe is A, the velocity of the water inside is v, and the density of water is ρ. The momentum flowing per second towards the right corner is thus: Av2ρ towards right and momentum flowing per second downwards from the corner is Av2ρ downwards. Then why is the force acting on the liquid the vector difference of the flows of momentum per second? Isn't force equal to change in momentum per unit time? But we never calculated change in momentum, we calculated change in flow of momentum per unit time. This leaves me confused.
     

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    Last edited: Apr 20, 2015
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  3. Apr 20, 2015 #2

    Simon Bridge

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    Force is the rate of change of momentum.
    When you calculate the change in the flow of momentum, what do the units come out as?
     
  4. Apr 20, 2015 #3
    I know it is dimensionally correct but it doesn't make logical sense since it isn't equal to rate of change of momentum, it is equal to change in flow of momentum.
     
  5. Apr 20, 2015 #4

    Simon Bridge

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    How would you work out the rate of change of momentum?
     
  6. Apr 20, 2015 #5
    Change in momentum per unit time
     
  7. Apr 20, 2015 #6

    Simon Bridge

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    Yes - so do it: work it out ... do the maths.
     
  8. Apr 20, 2015 #7
    But we don't know what the momentum is. We know the flow of momentum ( that is already momentum per unit time). And we don't know that how long would it take for the change to take place.
     
  9. Apr 20, 2015 #8

    Simon Bridge

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    You are telling me that the change in momentum flow rate is not the same as the rate of change in momentum. Prove it!
    You are totally free to set up the problem however you like - but you do need to express the momentum in terms of a density, crossectional area, and an appropriately chosen length.
     
  10. Apr 20, 2015 #9
    Because, rate of change of means the the difference in momentum's with a slight increase in time (assuming momentum is a function of time, i.e. p(t)). But change in momentum flow rate simply means: momentum flow rate A - momentum flow rate B which does not give the difference of momentum's with respect to slight increase in time.
     
  11. Apr 20, 2015 #10

    Simon Bridge

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    The time won't increase because the time to go around the corner is the time to go around the corner and cannot be different.

    Also: the corner drawn is very sharp. This is key.
    The time to go around the corner is very small.
    Thus: time for a section of fluid length x to go around the corner is t=x/v.
    So work out the change in momentum for a mass ##m=\rho Ax## in going around the corner - divide it by the time it took to go around the corner.
     
  12. Apr 20, 2015 #11

    Delta²

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    Ok the math seem to work because velocity is constant in magnitude and also density is constant in space and time. What happens if velocity doesnt have constant magnitude and density also varies?
     
  13. Apr 20, 2015 #12
    OK. I'm going to give it my best show at explaining this. I hope it works for you.

    upload_2015-4-20_21-36-38.png

    The figure above shows the situation in your original figure. I have indicated a control volume in the figure (between the red dashed lines) into which fluid is flowing in, and out of which fluid is flowing out. The rate of flow in is equal to the rate of flow out.

    I have also shown a Before picture in which a small parcel of fluid is about to flow into the control volume at time t. I have also shown an After picture at time t + Δt in which the entering parcel has just fully flowed into the control volume, and another small parcel of equal size has just flowed out. So the net effect during the time interval Δt is that a mass of fluid equal to ρvAΔt has flowed in, and a mass of ρvAΔt has flowed out. The horizontal momentum of the mass flowing into the control volume during the time interval Δt is ρv2AΔt, and the horizontal momentum of the mass flowing out of the control volume during the time interval is zero. The vertical momentum of the mass flowing into the control volume during the time interval Δt is zero, and the vertical momentum of the mass flowing out of the control volume during the time interval is -ρv2AΔt. So the rate of change of momentum of the fluid passing through the control volume over the time interval is:

    Rate of change of horizontal momentum of fluid = ##\frac{0-ρv^2AΔt}{Δt}=-ρv^2A##

    Rate of change of vertical momentum of fluid = ##\frac{-ρv^2AΔt-0}{Δt}=-ρv^2A##

    These rates of change of momentum require forces to bring them about. These forces are applied to the fluid by the pipe wall. The pipe wall in the L pushes backwards against the fluid in the negative x direction with a force equal to the rate of change of horizontal momentum. The pipe wall in the L pushes downward on the fluid in the negative y direction with a force equal to the rate of change of vertical momentum.

    I hope this helps. If not, I've given it my best shot.

    Chet
     
  14. Apr 20, 2015 #13

    Simon Bridge

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    ... one of the assumptions in the theory is that the fluid is incompressible - so its density won't change. If it is compressible then you have extra thermodynamics to take into account like if the fluid were a gas.
    The derivation assumes laminar flow, a sharp corner, and that there are no forces besides those from the walls of the tube.

    For the velocity to change, other things must change too - like the crossectional area of the tube. These are related via Bernoulli's equation. You can do the math yourself to see what effect this has. You can also try the math for the case that the corner is not sharp.
     
  15. Apr 20, 2015 #14
    So Chestermiller, what would be the overall change in momentum per unit time?
     
  16. Apr 20, 2015 #15
    ##-\rho v^2A\vec{i}_x-\rho v^2A\vec{i}_y##

    Chet
     
  17. Apr 20, 2015 #16

    boneh3ad

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    @andyrk, are you familiar with the material derivative (also called the convective derivative or a total derivative)?
     
  18. Apr 21, 2015 #17
    Can you just add the x and y forces in quadrature, as in this example?

    Example - Resulting force on a bend due to mass flow and flow velocity
    The resulting force on a 45o bend with
    • diameter 114 mm = 0.114 m
    • water with density 1000 kg/m3
    • flow velocity 20 m/s
    can be calculated by as
    Resulting force in x-direction:
    Rx = (1000 kg/m3) π ((0.114 m) / 2)2 (20 m/s)2 (1 - cos(45))
    = 1196 (N)

    Resulting force in y-direction:
    Ry = (1000 kg/m3) π ((0.114 m) / 2)2 (20 m/s)2 sin(45)
    = 2887 (N)

    Resulting force on the bend
    R = (1196 (N)2 + 2887 (N)2)1/2
    = 3125 (N)

    Note - if β is 90othe resulting forces in x- and y-directions are the same.
     
  19. Apr 21, 2015 #18

    Delta²

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    Shouldnt we add ##pA## to the result for the x and y component?(p the pressure which we assume constant). I mean the rest of the fluid exerts pA force at the control volume along the dashed lines doesnt it?
     
  20. Apr 21, 2015 #19
    If you did the arithmetic right (which I haven't checked), this looks correct. (I don't know what you mean by adding the forces in quadrature.
     
  21. Apr 21, 2015 #20
    The momentum change calculation presented here using the control volume approach corresponds to the mass times acceleration side of the equation in Newton's 2nd law. This mass times acceleration is the result of the contact forces acting on the fluid in the control volume. The contact forces include the inlet and outlet pressure forces and the force exerted by the pipe wall on the fluid in the control volume. That's the other side of the equation.

    Chet
     
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