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Npsh problem

  1. Sep 6, 2009 #1
    At an oil refinery, a mobile centrifugal pump is used to pump warm crude oil from a storage tank. Liquid level in tank at elevation of 3 metres above pump. Suction side of pump is fed with a flexible hose that is 10 metres long.
    Crude oil density: 846 kg/m3
    Crude flow 4000 kg/h
    Internal diameter of flexible hose: 50mm
    Vapor Pressure of crude oil: 0.0654 bar (a)
    Friction head in suction line: 1.2 m

    NPSA Available = [P(atm) - Pv] / [density * grav] + hs - hf

    Velocity Head: H(v) = u^2/2g
    [where u = Q/pi*r^2]

    I have plugged in my values into the two above equations to find out NPSHa and Velocity head.
    However, the 10m value that is mentioned for the flexible hose i did not plug in.Does this 10m value need to be used in the solution or is it just for the purpose of a sketch?
    My solution is attached, any help will be kindly appreciated.

    Attached Files:

  2. jcsd
  3. Sep 7, 2009 #2


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    Re: Npsh

    Think about a term for the friction in the inlet piping...You already show the term for it.
  4. Sep 7, 2009 #3
    Re: Npsh

    So the Head loss in the suction line will be the frictional head loss which is the 1.2m value given in the question..

    Also The formula is H(v) = u^2/2g
    [where u = Q/pi*r^2]

    Therefore when calculating Q, i must divide by 3600 and also multiply by radius squared and Pi and After converting Q to m3/s,

    My final answer for the velocity head is 9 x 10-3
  5. Sep 8, 2009 #4


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    Re: Npsh

    Your NPSHa looks good. I get .022 for the velocity head though.
  6. Sep 8, 2009 #5
    Re: Npsh

    Thanks Fred,

    I reworked my solution to the velocity head and got an answer of 0.02281m also..

    Converting my u value to m3/hr involved 4000/846 then multiplying by 3600 and the area
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