I have to test some fire hydrants and need to get the dynamic pressure. For that I know I can use a Pitot tube but, I don't have one. After Googling a lot, I found two formulas allowing to compute the dynamic pressure: one is using speed of fluid, the second is used in fact to compute the flow using dynamic pressure which is the opposite and, reversing the formula, give same result as first one.

The problem is that I've some data from French fire-hydrants and the "real value" are completely different from the one I get from the formulas...

First formula is:

$$P = \frac12 \rho V^2$$ where $P$ is the pressure in Pascal, $\rho$ the volumetric mass of fluid in Kg/M$^3$ (so 1000 for water), and $V$ the speed of fluid in m/s. If I use, for test a 100mm diameter hydrant, with a flow rate of 2000 liters per minutes, I get:

$$2000 {\rm L/min} \to\frac{2000}{60} = 33,333 {\rm L/s}\to 33333 {\rm cm^3/s}$$

10cm diameter section of tube is: $\pi\cdot\left(\frac{10}{2}\right)^2 = \pi\cdot25 = 78.54 {\rm cm^2}$

In one second, the water "run": 33333/78,54 = 424 cm so 4.24m/s

My dynamic pressure is : $$ P = \frac12 \rho V^2 = \frac12\cdot1000\cdot(4.24)^2 = 8988 {\rm Pa} = 0.089 {\rm bar} $$

Problem: it's impossible. In France, we have many fire hydrant of 100mm, flowing 2000 Lpm. And the normative minimum dynamic pressure allowed is 1 bar. The result of the formula seems to be far from reality.

Second formula, coming from a Canadian book about fire hydrants! This formula is:

$$Q = 0.0666 \cdot c \cdot d^2 \cdot \sqrt{P}$$

where $Q$ is the flow rate in liter per minutes, $c$ a constant which is of $0.9$ in case of hydrant, $d$ is diameter in mm and $P$ the dynamic pressure. The goal is to calculate the flow rate, from the dynamic pressure.

When you reverse the formula, you get the same as the first one. And when I apply to this formula, data from real French hydrant, the result is also wrong: using a 100mm hydrant with a dynamic pressure of about 1.2bar, I get about 7000 liters per minute, when in fact such and hydrant flows about 4 time less...

Actually, I've three options:

  1. the formulas are right but I've made some mistake while calculating

  2. formulas are wrong

  3. what we call dynamic pressure in France is not te same "dynamic pressure" used in these formulas

Any ideas?


3 Answers 3


I answer my own question and give a good thanks to DavidPh, who has not really gave the answer, but in fact, it was impossible for him to give it. Here is "why": I'm French, so I've many fire hydrant data but from France. And when applying them to the formulas, the result was wrong...

In fact, the problem is not the formula but the way we measure the pressure and from vocabulary confusion.

In France, firefighter consider that a fire hydrant must provide a flow rate of 60m3 per hour, so 1000 liters per minute, at a pressure of 1 bar.

In order to check that, here is how we do: - Put a manometer and a valve at hydrant output - Open water - Close slowly the valve in order to increase pressure - When pressure is at 1 bar, measure flow rate which must be higher than 1000 liters per minute

This mean we change the diameter, in order to get 1 bar. So the formular cant' applied as i fact, we don't know the diameter we have.

This explain also why , in the USA, some fire hydrants flow 7000 liters per minutes when in France they flow only 2000. But in the USA, they flow at a low dynamic pressure when in France the flow is measured at 1 bar dynamic pressure.

Best regards to you all Peter


The discrepancy is that the pressure as measured by the Pitot tube is not just the kinetic energy term of the pressure, but instead is a combinaiton of static pressure and the kinetic energy term.

See if pages 16-34 of the following reference are helpful, though not metric:


  • $\begingroup$ Thanks. I'll continue searching because this doc gives the first formula I used and results are the same. I'm waiting for more data from French hydrant to check the result. $\endgroup$
    – Peter
    Commented Sep 26, 2014 at 18:34

I am here to answer your question more than 2 years later. In addition to what you said, I guess the biggest issue is this: The equation you posted is Bernoulli's equation. For that to work and be valid, you flow should be laminar (have determined streamlines), steady, inviscid and you have to apply the formula along one of the streamlines.

Of course, you know that water going out the fire hydrant is nothing close to the necessary conditions. It is turbulent and has to overcome a lot of viscous stresses too. The biggest pressure drop will also be at the exit of the fire hydrant. All in all, the formula you attempted using is not valid in your case.


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