water pressure into house

[MasterPlumbersKeymaster]

The water pressure coming into my house is at 60 psi and I would like to know how to increase that to about 75 psi.
Thanks for any help you can give me.
Mike

[RickParticipant]

BigMike,

First, need a little more info.

Do you have a well or public water? How do you know the pressure and where did you measure it? Do you have a pressure regulator on your main supply already?

Rick

[GuestParticipant]

I have public water,I measured the pressure with a pressure gauge at the furtherest point from the water meter.There is no pressure regulator.

[fourth yearParticipant]

In that case the only way to increase the pressure is to install a booster pump system on the incoming line.

[RickParticipant]

ONE WAY to increase the pressure is to install a booster pump.

ANOTHER WAY is to increase the size of the main feed pipe to your house. If you have a 3/4″ main feed to your house, you can get higher pressure by increasing it to a 1″ line.

Perhaps we could help more if we knew what you are trying to accomplish by increasing the pressure from 60 to 75 psi. It may be possible to address that target from a different route.

Rick

[fourth yearParticipant]

If the city pressure is 60#, a larger main may give more volume but not more pressure, unless they are employing some new law of physics that has not previously been discovered. And even that would only happen if the existing pipes are unable to provide full flow to the fixture(s).

[RickParticipant]

OK, let’s review our physics, P1V1=P2V2. Are we in agreement so far? It is not new physics, Sir Charles Boyle (Boyle’s Law) lived over 300 years ago (1627 – 1691)

Now lets assume, as you stated, public water pressure is 60psi = P1, and assume 1” diameter pipe id. So volume of a 1” linear piece of this pipe is Pi (3.1416) (.5)(.5)(radius squared) 1(length).1416 0.25 1 = 0.7854 = V1

Now lets’s calculate V2 for ¾” diameter pipe (radius is .375”), also 1” long, so V2 is Pi (3.1416) (.375)(.375)(radius squared) 1(length).1416 0.140625 1 = 0.4418 = V2

So now, we have P1=60, V1=0.7854, V2=0.4418, and solving for P2,
P2 = (P1V1)/V2 or (60 0.7854) / 0.4418, which yields calculated P2 of 106.67 psi.

[Author]

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