Many consumer grade pressure washers specify a minimum pressure for the supplied water, which is sometimes well above the low end of household water pressure. Our water pressure is often low and I'm looking for a way to verify without a pressure meter that it is adequate for this purpose.

Data from the utility company isn't useful. The pressure is very affected by the usage of other houses in the area, plus there are endless effects on the pressure between the water meter and the hose bib. However, I can use the water at off-peak times and test it at the hose at the time I want to use it.

The formulas I've found online tend to deal with what happens inside the pipe or between one point and another. They require all kinds of parameters that aren't really relevant to this problem. I'm only interested in what comes out the end of the hose.

I know that for pressure washer nozzles, the area of the hole relates the flow rate to the pressure, and one factor can be calculated from the other two. A garden hose is at a very different scale; characteristics that influence flow at high pressure on a tiny scale aren't likely to extrapolate well. Nevertheless, the same physical laws should apply.

The end of the hose is a 3/4" hole and I can measure the GPM coming out. Can the PSI of the system driving the water out of the hose be calculated from just those two factors?

I've looked at the standard model of a water tank with a hole in it, which would seem to represent this. However, the results were nowhere near the same ballpark as household water pressure.

  • $\begingroup$ what's wrong with a tyre pressure gauge? $\endgroup$
    – Solar Mike
    Commented Sep 5, 2017 at 12:31
  • $\begingroup$ could you point to a specific example (model) that specs a high input pressure? I believe some will actually draw from a tank (no pressure), but also on a quick search a bit surprised that most don't give input side requirements. In any case if you need pressure you need a certain pressure at a certain flow rate, so no you cant simply get that from the flow rate out of an open ended hose. $\endgroup$
    – agentp
    Commented Sep 5, 2017 at 14:16
  • $\begingroup$ @SolarMike, that's a great idea. I would need to make something to mate it to the hose, but it's is a promising avenue. $\endgroup$
    – fixer1234
    Commented Sep 5, 2017 at 19:01
  • $\begingroup$ @agentp, many models either spec it (usually buried in the manuals, not listed in the product info online), or the manufacturer will provide it in response to a query. Here's a popular one where the manufacture gave it in the online Q&A (50-60 PSI). I've only seen a few models that are capable of drawing from a tank (self-priming). Most tank setups use a small pump to feed the water to the pressure washer. (cont'd) $\endgroup$
    – fixer1234
    Commented Sep 5, 2017 at 19:26
  • $\begingroup$ The flow rate once connected will be what the device moves (different effective pipe size). The flow rate with the hose disconnected can be measured with a watch and a bucket and will be based on the diameter of the hose. The flow rate will be different, but the pressure driving the water out of the hose should be the same. Like the nozzle orifice, two of the three factors will be known. $\endgroup$
    – fixer1234
    Commented Sep 5, 2017 at 19:27

1 Answer 1


Pascal's law $\text{Pressure} = \rho gh$, $\rho=$ density of the fluid $g=$ gravitational force and $h=$ head pressure. To find head pressure you could use a clear tube and measure how high the water goes before it equalizes and stops flowing. Or just point it straight up, hold it against a wall and measure the highest point it makes the wall wet. releasing the fluid from the closed system of the hose isn't a very accurate way to do it but would probably get you close. Use $\rho = 1000\text{ kg/m}^3$, $g=9.8\text{ N/kg}$ (same as m/s2) and $h = m$ to get kg/m2 pressure. Generally macro level physics is 99% unit manipulation so getting good at understanding how they go together and cancel out will make memorization of the formulas a whole lot easier. the problem is usually with the gravitational constant you know 9.8 m/s2 but getting that into a figure you can conceptualize is difficult. That's why I always hated moment of inertia...

  • $\begingroup$ Thanks for responding. I had looked at the standard model of a tank with a hole in it to represent my view of the end of the hose, but couldn't get reasonable results from it. I suspect you're on the right track here, but I didn't follow how you got to kg/m^2. $\endgroup$
    – fixer1234
    Commented Feb 23, 2018 at 1:53
  • $\begingroup$ The unit for pressure would rather be [rho * g * h]=kg/m^3*N/kg*m=N/m^2=Pa $\endgroup$
    – Andrew
    Commented Feb 25, 2018 at 11:37
  • $\begingroup$ oops, forgot to go from kg to newtons but yea basically the formula boils down to N/m^3 * height (m) = N/m^2 which is a convertible pressure unit $\endgroup$
    – Dane P
    Commented Feb 26, 2018 at 17:28
  • $\begingroup$ The gravitational acceleration "g" is 9.81 m/s^2, 1 N = kgm/s^2, the multiplication of the units is (kg/m^2) x (m/s^2 x m) = (kgm/m^2s^2) = N/m^2.. $\endgroup$
    – r13
    Commented Apr 8, 2021 at 17:18

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.