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Thread: Pressure Drop in Circular Pipe

  1. #1
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    Pressure Drop in Circular Pipe

    I am trying to find the pressure drop of air flowing through a circular pipe. The equation provided by the engineers edge website is (P_1^2 - P_2^2)/(2P_1) = Friction coefficient * L/D * density * w^2/2 * T_avg/T_1.

    The link is: http://www.engineersedge.com/fluid_f...ssure_drop.htm

    The problem with this equation is that the units don't match up. The flow speed is squared which gives something over seconds squared but there is nothing else in this equation that cancels out the seconds squared. The web page doesnt tell you what units you should be using. Can anyone provide a solution? Thank you.

    Ian

  2. #2
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    Quote Originally Posted by Ian View Post
    The web page doesnt tell you what units you should be using. Can anyone provide a solution?
    More than one way to skin a Sloth!
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    calculation pressure drop in pipe

  3. #3
    Administrator Kelly Bramble's Avatar
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    Quote Originally Posted by Ian View Post
    I am trying to find the pressure drop of air flowing through a circular pipe. The equation provided by the engineers edge website is (P_1^2 - P_2^2)/(2P_1) = Friction coefficient * L/D * density * w^2/2 * T_avg/T_1.

    The link is: http://www.engineersedge.com/fluid_f...ssure_drop.htm

    The problem with this equation is that the units don't match up. The flow speed is squared which gives something over seconds squared but there is nothing else in this equation that cancels out the seconds squared. The web page doesnt tell you what units you should be using. Can anyone provide a solution? Thank you.

    Ian
    Yes, I think the equations need units and clarification, as well as an associated calcuator - on the list..

  4. #4
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    Nothing wrong with the dimensions

    - P_2^2)/(2P_1) = Friction coefficient * L/D * density * w^2/2 * T_avg/T_1.\

    P_2^2)/(2P_1) dimension of pressure
    Friction coefficient * L/D dimensionless
    density * w^2/2 pressure lb/ft^3/Ft/sec^2*(ft/sec)^2 ....................... lb/ft^2 or pressure

    So the left half and right haalf of the equation have the dimension of pressure.

  5. #5
    Administrator Kelly Bramble's Avatar
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    Good see you posting around Zekeman? ...

    Quote Originally Posted by zeke View Post
    Nothing wrong with the dimensions

    - P_2^2)/(2P_1) = Friction coefficient * L/D * density * w^2/2 * T_avg/T_1.\

    P_2^2)/(2P_1) dimension of pressure
    Friction coefficient * L/D dimensionless
    density * w^2/2 pressure lb/ft^3/Ft/sec^2*(ft/sec)^2 ....................... lb/ft^2 or pressure

    So the left half and right haalf of the equation have the dimension of pressure.

  6. #6
    Associate Engineer
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    Thanks Zeke!

    I guess I did not realize that density for imperial units is in slugs/ft^3!

    Ian

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