Is it possible to сalculate discharge / flow rate on the pipe diameter and the pressure (for constant pressure)?

For example:

enter image description here

Pressure_internal = 5 bar, Diameter_pipe = 36 mm, Density_water = 1000 kg/m^3

Bernoulli's Formula:

$$\frac{\rho v^2}{2} + \rho g h + p = const$$

$$v \approx \sqrt{(p_i-p_a)\frac{2}{\rho_w}} = \sqrt{(5 \mathrm{bar}-1 \mathrm{bar})\frac{2}{1000 \frac{\mathrm{kg}}{\mathrm{m}^3} }}=28.284 \frac{\mathrm{m}}{\mathrm{s}}$$

Then the discharge should be:

$$Q = Sv = (\pi r^2) v = \pi * (18\mathrm{mm})^2 *28.284\frac{\mathrm{m}}{\mathrm{s}} = 1717.81 \frac{\mathrm{l}}{\mathrm{min}}$$

But these values are very far from reality. How can I perform a calculation correct?

  • $\begingroup$ If you are asking for help in doing a calculation, first you need to state the conditions of the problem. It is difficult to comment without understanding what the situation is. $\endgroup$ – sammy gerbil Mar 1 '17 at 8:15

It's called a Poiseuille flow, if the flow is laminar its given here.

If the flow is turbulent then your approach with Bernoulli should be correct.

If its in between then its more complicated, you can use a semi empirical Darcy type law.


There are many online calculators available to do this kind of thing. For example I found this one that gives the pressure drop when the flow rate is given. Playing around a bit until it gave a pressure drop of (almost) 4 bar gave the following:

enter image description here enter image description here

As you can see, taking account of real world effects (in particular the viscous drag from the wall) gives us a flow rate of ~1000 liter per minute. That is fast - but then 36 mm is a big pipe, and the flow velocity of 16 m/s is not unreasonable. Pointing the outlet of this pipe up in the air, such a jet of water could reach about 12 m high - considerably less than the 40 m head of water needed to generate the 4 bar of excess pressure. This makes sense as an order of magnitude.


If you are approximating the fluid as being inviscid (which apparently is the case if you are using Bernoulli), the pressure drop in a section of pipe of constant diameter must be zero, irrespective of the flow rate. So the upstream pressure in the pipe could not be 5 bars if the downstream pressure is 1 bar. Either both pressures are 5 bars or both pressures are 1 bar. Your equation implicitly assumes that the upstream velocity is zero (which is not the case).


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