OK, area $A$ is just $\pi D^2/4$. The real question is: What is $C$? It depends on the shape of the orifice (and Reynolds number).
There are some quick-and-dirty approximations here.
It depends on the orifice geometry, like whether its edges are rounded or sharp.
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
Best Answer
This sounds like a centrifugal pump. If so, this type of pump has a published pump curve, which tells what flow rate can be achieved relative to discharge pressure. For an intro to pump curves, see https://blog.craneengineering.net/how-to-read-a-centrifugal-pump-curve.
Regarding the concepts involved (no equations), it is a known fact that the pressure drop in a pipe is proportional to the flow rate squared (assume turbulent flow). This means that if you double the flow rate in a pipe of given length, diameter, etc., you will find that the pressure drop through the pipe increases by a factor of 4. The same will also apply for the pump, which will have its own pressure drop.
For low flow rates, the nozzle on the end of the water line will be constricted, so most of the pressure drop will be taken at the nozzle, and all pressures before the nozzle will be relatively high. For higher flow rates, the piping and pump will experience more pressure drop, and the pressure at the nozzle will be lower as a result.
The highest achievable pressure to be obtained from a centrifugal pump occurs when the flow rate is zero, which is known as "dead head" pressure. This maximum pressure depends on the pump's impeller diameter and rpm. For an electrically driven centrifugal pump that does not have a variable speed driver connected to it, the rpm is fixed, so increases or decreases in dead-head pressure are achieved by changing the impeller diameter.
Obviously, nobody would run the pump at its dead-head pressure because no flow rate occurs under these conditions. The pump is expected to deliver a desired flow rate at a specified nozzle pressure. If you find that the pressure is too low at the desired flow rate, this can be fixed by buying and installing a larger diameter pump impeller, if you are not already at the maximum impeller size that the pump case can handle.