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Fluid Mechanics!

  1. Aug 22, 2007 #1
    hi, i've been searching around the site and found it pretty useful but i'm having a pretty tough time with a few questions i have

    its pretty much the same as This

    but the coefficient values are different, the problem i'm having is with the coefficients for the bends, could someone please explain what i'm meant to do with them? and where they go in the equation ?

    also i'm using the equation H = 4f *(l/d) * (v^2/19.6) is this right ?

    cheers for any help guys :smile:
     
  2. jcsd
  3. Aug 22, 2007 #2

    Astronuc

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    Staff: Mentor

    Fittings can be assigned an equivalent length for pipe of the same diameter.
     
  4. Aug 23, 2007 #3

    FredGarvin

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    Science Advisor

    Like Astronuc said, the coefficients for the bends and restrictions are a method of calculating an equivalent straight run length of pipe that would provide the same energy loss as the fitting. If you look at post #3 in the thread you referenced you will see that they are summed up. That same post also shows the end result of the equation used. It is a modified form of the Bernoulli equation to provide loss terms for frictional effects.
     
  5. Aug 23, 2007 #4
    Ahh right that makes more sense now, i tried it using the coefficient values in that equation and as you can probably guess got some stupid answer lol

    if i try and use the equation i quoted i won't get the right answer ?

    thanks for the help

    Edit: ok i've just tried it using the Eqn in the thread i referenced and got a head of 12.35m if i use L=50m, if i use L=80m (50m plus the equivalent lengths) i get 17.8m neither of those sound right considering the other guy got a 5m head and the only difference in his Q is he has a flow rate of 6litres instead of 5?

    Edit: screwed it up, i now have a head of 4.1m
     
    Last edited: Aug 23, 2007
  6. Aug 24, 2007 #5
    Solution

    This one's easy....

    Use the energy equation:

    (p($1)/rho*g)+((v($1)^2)/2g)+z($1)=(p($2)/rho*g)+((v($2)^2)/2g)+z($2)+h($f)+sum(h($m))-h($p)

    A $ sign within brackets mean that is a subscript.

    h($f)+sum(h($m))=v^2((f*L/d)+sum(k))/2g

    where sum(k) is the sum of all your minor loss coefficients due to bends, etc.

    If you have the book Fluid Mechanics by Frank M. White (5th Ed), you can find an example on page 392. If you don't and need clarification, just ask.
     
  7. Aug 24, 2007 #6
    Oh yeah, and h(p) is the losses due to a pump, if there is one in the system. Neglect if there is not.
     
  8. Aug 24, 2007 #7
    that's the equation i ended up using and got a value of 4.1 meters which sounded about right to me
     
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