How do I calculate the time to fill a pipe with an air compressor?

[newbieciv]

Hello,

I would greatly appreciate some help on a problem proposed to me today. I need to figure out how long it will take to fill 1 mile (5280 ft) of 6" ID pipe to pressure test using a 425 CFM @ 350 psi compressor. The pipe needs to go from 0 psig to 100 psig at 80 deg F. Then using the same compressor, I need to fill 5.1 miles of the same pipe using the same compressor from 0 psig to 255 psig. I do not need to take into account any kind of pipe expansion. I am not sure what kind of compressor this is, but it would be some kind of mobil gas or diesel powered unit as this is literally out in the middle of nowhere.

 

[vpl]

School assignment?

Do you really think there will be any pipe expansion at those pressures?





Patricia Lougheed

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[newbieciv]

For surely not a school assignment. It is for a job we are bidding in southern california, I just don't know what I need to do. I used the ideal gas law assuming that everything but pressure and volume stay constant, and then manipulating the formula to say that volume (2) is equal to one, normalize the pressure to atmospheric by adding 14.7, and dividing the the flow of my compressor. I got a time in minutes out of that, but my boss says that is not good enough. The problem I have is that I don't know if I need to be looking at some kind of pump curve for my compressor and then integrating over pressure difference or something to try and account for the fact that as the volume fills up compressor is not able to push 425 cfm into the pipe anymore. The deeper I dig into this, the nastier it gets and I'm not sure where to start. Any help is appreciated.

 

[msquared48]

At first gance, from your title, i thought you were trying to stuff an air compressor into a pipe. I said "This looks intersting", and opened the thread. Darn...

Anyway, yes, you do have a variable flow problem as the fill rate of the conpresser will decrease as the pressure in the pipe increases. The pump manufacturer should have back pressure and flow rate curves to use. If not, call them...

The method for doing this should be similar to doing a backwater analysis in hydraulics.

Mike McCann
MMC Engineering

 

[toohotforme]

Hi newbieciv,
I don't visit here often and was browsing for discussions about a piping issue when I came across this. It may be too late for your needs but here is an approximation anyway.

Firstly, air is compressible so hydraulics (incompressible fluid) won't be valid. You need some sort of integration approach. I have assumed that the temperature in the pipe remains constant which is not strictly true, since it will fade to the temperature of the surroundings. The error will be small. This relies on thermodynamics knowledge.

The basis of the calculation is that work = integral Pressure x change in volume. If you drew a P-V graph (P on Y-axis and V on x-axis) the curve will slope from high pressure – low volume to low pressure – high volume. It will be a curve – not a straight line.

Integrating with pressure changing rather than volume changing gives a similar equation – you are taking a small horizontal slice through the work area rather than a thin vertical slice, with which to integrate.
Integrating the equation gives Work = P1 V1 ln (P2 / P1)

Do this for your pipe pressure test variables and then do the same for the compressor. The difference is that the compressor is delivering a volume per unit time. Time is what we want.
Divide the test pressure “Work” by the work per unit time of the compressor and you have seconds.

Unfortunately I work in SI units so you can’t just change my Excel psi type units and get the spreadsheet to update. I have however done both scenarios for you.
You could try PVn where n is typically 1.3 or 1.4 but the errors will be small. If you added a little time (say <10% ) to these answers for the cooling effect, you will be pretty close anyway.

If the pipe remained at the temperature of the compressor air discharge, the approach is valid. The whole pipe will not reach this temperature – not even close, so the heat transfer issue becomes trivial and the solution becomes “close enough.”