Is there any equation that states the relation between pressure and water flow.

I.e. Let's say that in 1 hour with 8mca (water collum meters) pressure I obtain 50m3. What if (giving the same contditions) I execute the same test but now with 15mca?

Is that just proportional?


The water velocity into a region with atmospheric pressure goes as the square root of the pressure difference by Bernoulli's law. So if you quadruple the pressure difference, you get twice the speed.

This is not exact, because as the water fills up your container, unless it is coming from the very top, the building up water will produce a counterpressure.

The law is that when fluid is dynamically moving, the kinetic energy of the water coming out per unit mass, which is half the square of the velocity, is the loss in potential energy of the top of the imaginary water column of height h per unit mass, which is gh. So the velocity at the outlet is:

$$ v=\sqrt{2gh} = \sqrt{2P \over \rho}$$

Exactly the same as if you dropped the water from the top of the imaginary column to the point where you let out the water, and the answer doesn't depend on whether you have the pressure produced by a column as I imagined it above. This answer conserves energy, since water is disappearing from the top, and appearing at the bottom moving with the same speed as if you dropped it from the top of the column.

  • $\begingroup$ ...at high Reynolds numbers $\endgroup$ – Alan Rominger May 10 '12 at 18:18
  • $\begingroup$ @AlanSE: This is not really dependent on Reynold's number--- it's conservation of energy and the flow in a big pipe behind the nozzle is usually laminar. Any turbulence usually happens past the nozzle, but if you have viscous dissipation in a tube, you need to include the loss in energy in the turbulent flow, and the speed is the remaining energy. $\endgroup$ – Ron Maimon May 10 '12 at 18:21

All what Ron explained/wrote can be extracted directly from Bernoulli equation

$$\frac{1}{2} \rho v_1^2 + \rho g h_1 + p_1 = \frac{1}{2} \rho v_2^2 + \rho g h_2 + p_2.$$

Say, if you have the huge open container with small hole at the bottom, you have the same (atmospheric) pressure on the both ends $p_1 = p_2 = 1.03 \times 10^5$Pa and velocity of water at top of the container is negligible $v_1 \approx 0$, you get expression

$$v_2 = \sqrt{2 g (h_1-h_2)} = \sqrt{2 g \Delta h},$$

where $\Delta h$ is height of the water column.

You can obtain relation between water column height and pressure in container from Bernoulli equation too, taking $v_1 = v_2 = 0$, which gives

$$p_2 - p_1 = \rho g (h_1 - h_2),$$

$$v_2 = \sqrt{\frac{2}{\rho} (p_2-p_1)} = \sqrt{\frac{2}{\rho} \Delta p}.$$

  • 1
    $\begingroup$ What's the point of this answer if it's the same as Ron's ? $\endgroup$ – Physiks lover May 10 '12 at 18:03
  • 1
    $\begingroup$ To explain where it comes from. $\endgroup$ – Pygmalion May 10 '12 at 18:30
  • $\begingroup$ But doesn't Ron's answer show this? Maybe he edited in this later. $\endgroup$ – Physiks lover May 10 '12 at 18:35
  • $\begingroup$ Oops, he did mentioned Bernoulli's law - I'm lousy reader. Well, it is not completely in avail, as he forget $\rho$, which he added later. $\endgroup$ – Pygmalion May 10 '12 at 18:40

so basically velocity is proportional to the square root of differential pressure!!! and there is another equation which says Q = V x A where Q is the volume flow, V is the velocity of the flow and A is the cross sectional area!


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