practical hydraulics question

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Feil
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Location: Illinois, USA

practical hydraulics question

Post by Feil »

My pump draws from a source pressure that reads 60 PSI with an unknown flow rate.
I open a line at 250 gallons per minute and observe a 5 PSI pressure drop: 60 to 55 PSI.
I open two new lines and so double my flow to 500, and then triple my flow to 750.
Does my pressure go as:

60, 55, 40, 15 PSI? (60 - 5, then 60 - 5*22, then 60 - 5*32), supplying all three lines with a good reserve pressure?

Or as:
60, 55, 35, pressure failure (60 - 5, then 60 - 5 - 5*22, then 60 - 5 - 5*22 - 5*32), ruining everybody's day?
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madd0ct0r
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Re: practical hydraulics question

Post by madd0ct0r »

you want this bit from: Image

taken from:
https://www.pipeflowcalculations.com/pi ... -pipes.php


For your case, pipe length and friction coeffecient is constant. so you can probably calculate them from the pressure drop from a single line opening.

Internal Diameter will probably be smaller then the pipe itself, it's the smallest diameter in the chain that matters, so probably a valve on the pump.

You can guesstimate those constant unknowns and run a sensitivity check on various combinations. The cube means that the internal pipe diameter will be a much more sensitive factor to mistakes, but if you have a few hundred meters of pipe it may well be that the fL combination has a bigger magnitude effect.
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Feil
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Joined: 2006-05-17 05:05pm
Location: Illinois, USA

Re: practical hydraulics question

Post by Feil »

Should be Option 1 exactly, then. Since my guage measures before the pump, everything's constant but Q and we can stuff everything other than (Q/Q1)^2 into a proportionality constant. Where Q1 is whatever flow we measured the first delta-P with.
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