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I need an easy explanation for a flow formula

  1. May 24, 2006 #1
    In order to calculate flow rate when using a differential head pressure device, the simplified formula is Flow = (coefficient)(square root of the differential pressure)

    My question is why? Why the square root, can some explain this to me please? Thanks
  2. jcsd
  3. May 24, 2006 #2


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    Basically because flow is proportional to the velocity. In the Bernoulli Equation, the velocity components are squared terms.
  4. May 25, 2006 #3
    A basic flow equation is V=(2gh)^0.5
  5. May 26, 2006 #4


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    The pressure differential over a distance represents potential energy (like a voltage), and the kinetic energy of the flow is proportional to the square of the mean velocity of the flow.


  6. May 29, 2006 #5


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    While the above posts have answered the question "why ?", let me answer the question "why not ?"

    In many applications we actually use a linear relationship between fluid flow and pressure drop. This too is an approximation, but a reasonable one for a certain range of flow characteristics.

    The reason the square root dependence is not strictly obeyed in real fluids is because of viscous loss. The corrected expression for pressure drop is :

    [tex]\Delta p = \frac{\Lambda}{D} \cdot \frac{\rho v^2}{2} \cdot L~~~~(1) [/tex]

    where [itex]\Lambda[/itex] is the "friction factor". For laminar flow, the friction factor is given by

    [tex]\Lambda = \frac{64}{Re} ~~~~~~~(2)[/tex]

    Re is the Reynold's Number, which in the laminar regime goes like

    [tex] Re = \frac{vD}{\nu} ~~~~~~~(3)[/tex]

    where [itex]\nu[/itex] is the viscosity.

    Plugging these into (1) gives :

    [tex]\Delta p = 32 \rho \nu L v~~~~(4) [/tex]

    However, for this equation to hold, we must have L/D >>1, so that the fluid attains "terminal velocity" from viscous drag. Over short lengths, the behavior will look more like the square root dependence (which is only exactly true in the ideal limit L/D -> 0 ).

    In terms of the flow rate, Q = vA, this gives the more familiar (Ohm's Law like) equation

    [tex]\Delta p = const \cdot Q \cdot \frac{L}{D^2} [/tex]

    With turbulence, the relationship becomes more complex and the notable difference is that the dependence on the diameter takes on a higher number, typically somewhere between 3 and 4.
  7. May 29, 2006 #6


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    To expand on Gokul's comment, you might wish to check up on Darcy's law.
  8. Jun 6, 2006 #7
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