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Fluid Velocity Problem

  1. Oct 16, 2008 #1
    Problem- I do not have a picture of the example so I will do my best to describe it.

    Water is in a large tank of depth H. On the side of tank, on the very bottom, an open ended pipe of length L leaves the tank at angle theta. The question asks to determine the maximum height of the water (h) leaving the pipe as a function of angle theta.

    I know that velocity = the square root of (2gH), so using SI units, my velocity at the very bottom of the tank would be the square root of 19.62H. However the exit to the pipe is higher than the bottom of the tank, so would my velocity leaving the pipe be sqrt (19.62(H-L sin theta))? If this is the case, would the velocity of water leaving the pipe be (sqrt(19.62(H-L sin theta))) sin theta + L sin theta?

    I hope my description of the example is clear enough. Any suggestions/help is always appreciated.
  2. jcsd
  3. Oct 16, 2008 #2
    That seems to be correct to me for the velocity leaving the pipe at the angle theta.

    I'm not sure I understand what you are asking in this part. If you mean the vertical component of the velocity leaving the pipe would be

    [sqrt (19.62(H-L*sin theta))] * sin theta

    I would agree.

    I don't see where the extra term " + L sin theta " would come into play for the velocity, as that is a measure of the vertical height of the end of the pipe. It should be added to the vertical distance the water would travel after leaving the end of the pipe, not the velocity.
  4. Oct 16, 2008 #3
    I understand what you are saying. I am trying to determine distance h (max height between stream of water and the ground). To do this would I add the vertical component of the water velocity ([sqrt (19.62(H-L*sin theta))] * sin theta) and the vertical component of the pipe L (L sin theta)? Am I missing something?
  5. Oct 16, 2008 #4
    You cannot add a velocity to a distance. Use the velocity to determine the maximum vertical distance the water will travel, then add the height of the pipe.
  6. Oct 16, 2008 #5
    So I will take my initial height, L sin theta, and add it to (([sqrt (19.62(H-L*sin theta))] * sin theta) T), and subtract acceleration due to gravity (-4.9T^2). This I hope will give me h.
  7. Oct 16, 2008 #6
    I don't think you have a value for T. But you do have Vfinal (at least in terms of a vertical component), Vinitial, and acceleration due to gravity. You can calculate the maximum height from that information (refer to equations useful in projectile motion).
  8. Oct 16, 2008 #7
    Would it work if I substituted in (vy - vyo)/g for t? If I did that it would be ((([sqrt (19.62(H-L*sin theta))] * sin theta) T) - 0)/9.81 ?
  9. Oct 16, 2008 #8
    That will work fine. :cool:

    I was actually thinking of avoiding the calculation of time altogether by using the formula vf2 = vi2 + 2ad, but either way will give you the same answer if you crunch the numbers properly.
  10. Oct 17, 2008 #9
    Alright well i'm glad to see i'm making some progress. The equation you mentioned I had forgot about, and certainly makes things a bit simpler when not having to use time. If I used that equation I would set the formula up as d=(vf^2-vo^2)/2a ,with vf^2 being 0, and vo^ being ([sqrt (19.62(H-L*sin theta))] * sin theta)^2. Then divide by 2(9.81). Once that is done I would need to add my initial height L sin theta, to give me my total displacement h. Does this make sense?
  11. Oct 17, 2008 #10
    That makes sense to me. I think you're all set now.
  12. Oct 17, 2008 #11
    Thanks alot for the help
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