Hydraulic Grade Line Conversion

[canman]

Help!!

I have a flow test from a local water district in Southern California that provides the following information:

HGL - 830
Residual Pressure - 90
Outlet Size (inches) - 4
Outlet Coefficient - 0.9
Pitot Pressure - 40
Observed Flow - GPM - 2717

My question is how do I convert the HGL of 830 to a static pressure? I have not run into this situation before and I'm curious. Every other flow test I have recieved has given static in lieu of HGL.

Thanks!!

Canman

 

[cvg]

you need to know the elevation at the hydrant. subtract from the hgl to get feet of water and convert that to psi

 

[hydrae]

Canman
You need the elevation where the test was perfomed.
HGL = Pressure*2.31 + elevation
Given the residual pressure was 90, the elevation is not more than 620.
The district should have given both static and residual pressures in the same units.
Hydrae

 

[cvg]

i would guess that both residual and pitot pressure were also reported in feet. 90 psi is awful high for a residual pressure, but converted from feet to psi is only 39 psi which is just about right.

 

[hydrae]

cvg
With all respect, I think the numbers are in psi,
to get 2717 gpm out of single 4 inch nozzle, that will be 69 ft/sec, and a residual of 90 psi is about right to get that kind of flow.
Hydrae

 

[cvg]

As a general rule available (fire) flows are calculated based on what the hydrant can provide when pulled down to a residual pressure of 20 psi. 90 psi residual at 2,700 gpm could only be the residual if you are pulling off a very high pressure fire line, not from an ordinary watermain - which would likely not have a pressure higher than 120 psi.

The following are definitions pulled right off the internet. In order for these formulas to make sense, one needs to understand specific meaning of the following terms. (Most are pretty basic but a couple are generally misunderstood.)

STATIC PRESSURE
Pressure reading before water flows.

RESIDUAL PRESSURE
Pressure reading while water is flowing (from an outlet other than the flow outlet.)


PITOT PRESSURE
Reading taken by a pitot gauge inserted into the center of the flowing outlet, at a distance away from the lip of the outlet of about half the nozzle's diameter.

 

[canman]

Thanks for the replies. The existing elevation at the fire hydrant nozzle is about 618'(feet). Thus the static pressure should be about 92psi using your equations. The residual and pitot pressure I gave earlier were in psi. The existing main in the street is a 12" pipe. Using the above information the available gpm would be over 18,000 at a residual pressure of 20 psi (theoretically). Sounds too high to me, am I missing something?

Canman

 

[hydrae]

Canman
If there is truely just a 2 psi drop at 2717 gpm, yes the 18k would be about right. Three ways I can think of that could happen. The following objects should be within 300 ft as a single connection or 1000 ft as a looped connection

Is the 12" near a larger pipe say 30" or greater,
Is it next to a reservoir?
Is it near a prv, the prv will open up to maintain a constant downstream pressure at any flow up to 80% of the wide open pressure drop of the valve, thus giving a false calculation of flow at 20 psi. (this one I most suspect)

Hydrae

 

[cvg]

It sounds way too high! Not likely you could get that kind of flow through a 12 inch main. I would double check the reported residual pressure as it should be considerably lower than the static.