1

We suspect that there is too much pressure in the water system of a friend's house.

I checked the pressure regulator, the pressure gauge shows more than 4 bar (which AFAIK is way too much). However, there is water inside so it is probably no longer reliable.

enter image description here

I was wondering if there is a simple way to estimate water pressure (e.g. in bar) by filling a bucket of water (e.g. 10L) and measuring the time it takes to fill it?

EDIT : it's not from heater. It's a domestic water system.

EDIT2 : I did the following test: the main valve before the pressure regulator get closed and I opened the kitchen tap: the pressure dropped to 1.9 bar, which I think is excessive (I was expecting something like 0.5 bar). This indicates that the pressure gauge is damaged, probably due to the excessive pressure.

EDIT3 : I replaced the pressure gauge (the new one can go up to 10 bar). After replacement, it showed 4 bar. After some research, a leak was found in the basement (it seems fairly new). The leak was repaired and the pressure increased to 6 bar. I played with the pressure regulator (to try to lower pressure): no matter how it was set up, the pressure went up to 8 bar (which indicates that it was faulty and needs to be replaced too).

7
  • Is this a heating system or a domestic water system?
    – d.george
    Jan 23, 2022 at 16:24
  • 1
    No. You cannot measure pressure using flow, even with a calibrated hole, because once water starts to flow, the pressure will drop. You need a pressure gauge.
    – mkeith
    Jan 24, 2022 at 2:38
  • 1
    Also, I don't know what typical pressure is where you live, but here in California, 4 bar is not considered excessive. It is on the high side of normal. We measure in PSI. 4 bar is about 60 psi. The water pressure in my house is about 70 psi (more than 4 bar).
    – mkeith
    Jan 24, 2022 at 2:42
  • 1
    In the UK, legal min/max is 1 to 5.5 bar, 'normal' is between 2.5 & 4.
    – Tetsujin
    Jan 24, 2022 at 9:10
  • 1
    A pressure gauge that screws onto a garden hose faucet is pretty cheap. Screw on, open the valve, have all the taps in the house closed - done. At some point, replace the gauge filled with water - it is no longer reliable since at least one seal is gone...
    – Jon Custer
    Jan 24, 2022 at 19:50

2 Answers 2

2

Flow is not pressure.*

If that's a heating system, bleed the radiators to drop the pressure & make sure the loop is switched off [at both ends]. A modern [non-tanked] heating system should be a closed loop, not permanently connected to the mains supply.

*Pressure is what causes the water to flow, but compare how long it takes the pressure behind a tap filling a bucket to how fast you can empty it.

1

It's been a long time since college, but here's my attempt:

Assume you have a round, horizontal pipe you're getting the water out of, and that the flow is great enough that the water exiting the pipe fills it completely. And that the pressure remains constant.

First, measure the time it takes for a known volume of water to exit the pipe (i.e. fill a bucket). Volume divided by time gives you volumetric flow rate, and volumetric flow rate divided by area gives you velocity. The pipe being round, the area is π*r² (where r is the inside radius), so now we know the velocity just inside the exit of the pipe. I'll call it vₚ.

Next, letting the water jet out of the pipe, you need to measure the distance the jet travels and the height between the exit of the pipe and where you measured that distance. The water is like a projectile here, so travel time is √(2∙height/g) and horizontal distance is (velocity)∙(travel time), so putting the two together velocity = (distance)∙√(g/(2∙height)), where g is the gravity of Earth = 9.81 m/s². I'll call this, the velocity just outside the pipe, vₐ.

Now, we have Bernoulli's equation saying that v²/2+g∙z+p/ρ must be the same for the point just inside the pipe and the point just outsize the pipe. I've chosen the points so there is no change in height, so the z terms will cancel out. Since the point outside the pipe is at atmospheric pressure, p there is zero. What's left gives p = ρ/2 ∙ ((vₚ)² - (vₐ)²), where p is the pressure in the pipe, ρ is the density of water (~1 gram/cm³), and the velocities are as found above. (And be careful with the units, they will be a little tricky.)

This neglects air resistance when calculating vₐ and losses at the pipe opening (discharge coefficient) using Bernoulli's equation. But it should be in the ballpark....

1
  • 2
    This is approximation would only work if you have the "spherical cow" variety of plumping systems. You are computing the pressure at the opening, but that's not going to be the same as the static pressure of the system, as you'll have pressure losses along the pipe, dependent (often non-linearly) on the flow rate. Your results won't be in the same ballpark unless you have huge pipes with minimal losses. Feb 1, 2022 at 5:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.