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Relationship between water flow and pressure drop

  1. Nov 2, 2015 #1
    A building has a water pipe inlet. I have found that at if I open the valve at this location fully, then I get 180 liters per minute at a residual pressure of 0,2 bars (static pressure is 5,4 bars). Is there a fixed relationship between the flow and pressure drop so that I can easily predict what the residual pressure will be at say 50 liters per minute?
     
  2. jcsd
  3. Nov 2, 2015 #2
    Yes, you want the Bernoulli equation for steady, incompressible flow.

    P1/ρ + V12/2 + gz1 = P2/ρ + V22/2 + gz2

    For your situation, gz1 = gz2 and can be ignored.

    So:

    (P2-P1)/ρ = (V22-V12)/2

    thus ΔP = ρ(V22-V12)/2

    where Vx = V°/A1 and A1 = πr2

    In your case A1 = A2

    So you will also need to know the inner diameter of your pipe.

    ΔP = ρ/2((V°2/A)2-(V°1/A)2)

    edit:
    I don't know what your background is so if you use this equation you have to make sure your units work out.

    If in doubt put pressure in Pascals (not kPa), volumetric flow rate in m3/sec, area in m2, density in kg/m3, answer will be in Pa
     
    Last edited: Nov 2, 2015
  4. Nov 2, 2015 #3
    My opinion is,
    as the Area is uniform, acc to continuity eqn Q=A.V,i.e velocity at both ends is same,which makes RHS ZERO..
     
  5. Nov 2, 2015 #4

    boneh3ad

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    That's because his original answer is incorrect. If you want to use Bernoulli's equation, you will need additional terms to account for the head loss in the pipes and fittings.
     
  6. Nov 3, 2015 #5
    The valve position is going to induce a pressure drop with a velocity increase in the fluid.

    Piping losses would be negligible for this purpose. If it was dealing with a section of pipe, then you would consider it. The only thing that isn't on there that could possibly effect the answer would be the kinetic energy correction factor.
     
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