Engineering Question #3541
Richard, a 22 year old male from Halifax, N.S. asks on July 21, 2006,
How do you calculate water flow rate (volume/sec) based on water pressure and hole or pipe size? And is it the same based on water pressure created by gravity (water tower PSI) and pressure created by compression like a hole in the bottom of a boat which is 30ft underwater (underwater PSI)?
viewed 15270 times
answered on July 25, 2006
To answer the second question first, yes, there is only one kind of pressure. `Underwater PSI' and `Water tower PSI' are both created by gravity. We can calculate the flow rate using Newton's second law:
force = mass * acceleration,
Suppose we have a hole of area A, and the pressure differential across the hole is p. Then a force Ap acts on the water at the hole.
Suppose the water flows through the hole at a rate u m/s. Then in one second, a mass uAd flows through the hole, where d is the density of water in kg/m^3. This mass has been accelerated from rest to a velocity u in one second, so it has undergone an acceleration u m/s^2. So from Newton's second law,
Ap = u^2 A d
So u = square_root_of(p/d), And if you want the fluid flow rate as a volume/sec, it's
Q = A * square_root_of(p/d)
This analysis leaves some things out -- for example, it takes no account of the shape of the hole, or how thick the wall surrounding the hole is. These things are usually included in a numerical `K' factor, a number close to unity which can be looked up.
If instead of a short hole you've got a long pipe, you would need to calculate the pipe friction and subtract the frictional force from the force due to pressure. The friction produces an effective reduction in the pressure of
Delta p = dfLu^2/2D
where L is the pipe length, D its diameter, and f the friction factor, which depends on the flow rate and the roughness of the pipe's inner surface in a rather complicated way, usually summarised by a Moody diagram.
Add to or comment on this answer using the form below.
Note: All submissions are moderated prior to posting.
If you found this answer useful, please consider making a small donation to science.ca.