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Force of a water flow through a U-shaped pipe

  1. May 12, 2013 #1
    1. The problem statement, all variables and given/known data
    Water flows at a speed of 0.75 m / s through a U-shaped tube whose cross-section is circular with a radius of 10 mm. How large is the force F to be to compensate for water power on the tube?




    2. Relevant equations
    I am not sure. Perhaps;

    Pressure=Force/Area
    Conservation of momentum
    Density = mass/volume

    3. The attempt at a solution

    I believe that the speed of water flow is the same through the whole pipe but the velocities have different signs in each side of the pipe. Of course the mass is constant.if I use conservation of momentum I will have:

    Fdt=m(Δv)=m(v_1-v_2)=m(v-(-v))=2mv

    Fdt=2mv. The mass is not given but if we knew the volume we could find it as m=ρV. And how can I separate F from Fdt?

    I tried this(wrong);

    Fdt=2mv=2m(dx/dt) --->F=2m(d^2x/dt^2)=2ma=0 since velocity is constant.
     
  2. jcsd
  3. May 12, 2013 #2
    Find how much momentum enters the pipe over dt, and how much leaves it. The differnce between these two momenta is the impulse.
     
  4. May 12, 2013 #3
    Momentum that enters the pipe: mv
    Momentum that leaves the pipe: -mv
    The difference is then mv-(mv)=2mv=Fdt. I just don't know what to do with dt and mass m.
     
  5. May 12, 2013 #4
    How do you interpret my phrase "how much momentum enters the pipe over dt", with emphasis on "over"?
     
  6. May 12, 2013 #5
    i believe it means during a very small time interval. Which is no different in my imagination. It enters p=mv at the top, whether it is a short or not so short interval. Is that wrong?
     
  7. May 12, 2013 #6
    The question is, how much mass enters over this very small time interval.
     
  8. May 12, 2013 #7
    I am not sure actually. During a very small time interval enters a very small amount of mass and if the time interval --->0 then the amount of mass that enters --->0. But I don't see how this helps....
     
  9. May 12, 2013 #8
    You are right, this does not help at all. You need to be more specific about what this very small amount really is.

    Imagine that at some time t you draw a plane perpendicular to the pipe at its entry, and then you let this plane move along the pipe with the flow. At time t + dt, you draw another such plane. These two planes contain your very small amount of mass. What is it in terms of what you are given?
     
  10. May 12, 2013 #9
    More specific? Well how about taking the derivative; (dp/dt)dt=dp=2mv which is the same thing as before, nothing new. I'm going in circles here. I think this is the same thing as the idea with the planes, although I do see that the difference will contain the mass.
     
  11. May 12, 2013 #10
    You completely ignored the second part of #8. No wonder you are going in circles.
     
  12. May 12, 2013 #11
    What is the rate at which mass is flowing through the pipe in kg/sec?
     
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