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Hydraulics - Problem finding pressure

  1. Jan 13, 2012 #1

    Femme_physics

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    1. The problem statement, all variables and given/known data


    http://img37.imageshack.us/img37/638/pumpyu.jpg [Broken]
    (the pump is between 1 and 2)

    D1 = 300mm [diameter of pipe 1]
    D2 = 250mm [diameter of pipe 2]
    k(pressure lost due to valve) = 7
    k(pressure lost due to 90 degrees pipe angle change) = 0.9
    K (pressure lost due to entrance to a container) = 1
    L1 = 12m [length of pipe 1]
    L2 = 60m [length of pipe 2]
    Energy conversation efficency of pipe = 75%
    Q = 0.22 m^3/s [volumetric flow]
    f1 = f2 = 0.02 [friction coeffecient in both pipe]

    Fluid is water

    Find static pressure at the entrace to the pipe
    Find static pressure at the exit of the pump

    Answers:

    P1/lamda = 1.89m water
    P2/lamda = 49.7m water



    3. The attempt at a solution
    I can't get p2/lamda as the right result :(


    http://img810.imageshack.us/img810/1269/scan0007o.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 13, 2012 #2

    I like Serena

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    Just a few observations, since I do not understand everything you're doing yet.


    I see you have the kinetic pressure correctly now! :)

    And I presume the static pressure is ##p \over \lambda##?


    You appear to multiply the pressure-losses with the kinetic pressure, but I don't think you should do that.

    When you write ##p \over y##, can it be that you mean ##p \over \lambda##?

    You multiply the friction coefficient by the length and divide by the squared diameter.
    I'm not sure which formula you're using, but I'd expect that you should multiply by the surface area of the pipe, which is the length times pi times the diameter.
     
  4. Jan 13, 2012 #3

    Femme_physics

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    Yes! DOn't confuse my y and lamda. It's just how it looks in the book.
    That's just what the formula says last I checked.

    Again "Yes! DOn't confuse my y and lamda. It's just how it looks in the book.
    " :)
    OK. I will try this tomorrow morning. thank u.
     
  5. Jan 13, 2012 #4

    I like Serena

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    Isn't your formula the following?
    $$H_{S_A} + H_{V_A} + Z_A = H_{S_B} + H_{V_B} + Z_B + \sum Y_{A-B}$$
    I don't see that you should multiply ##H_{V_B}## with ##\sum Y_{A-B}##.


    Well, they confuse me, but I guess the book must be right. :wink:
     
    Last edited: Jan 13, 2012
  6. Jan 14, 2012 #5

    Femme_physics

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    Well that's what our teacher said
     
  7. Jan 14, 2012 #6

    I like Serena

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    Well, in your scan you appear to have left out a plus sign in the formula, which I assume is just a typo.

    But when you filled in the numbers, you appear to have introduced an extra factor that's not in the formula.


    EDIT: Wait! I think I'm as yet missing something about the pressure-loss terms that you didn't define yet.
    Can you say what the formulas are for the Y terms?
    That is, what is the formula due to pressure-loss caused by friction, etcetera?
     
    Last edited: Jan 14, 2012
  8. Jan 21, 2012 #7

    Femme_physics

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    I just wanna apologize for cutting off from this topic and the other hydraulic topic. It appears there is just a big mixture in formulas and definitions in that topic that I just decided to harass my classmates. I decided to leave hydraulics out of physicsforums :)
     
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