Rate of Flow Through a Pipe

Date: 06/30/99 at 18:05:17
From: Joe Pitchford
Subject: Water volume

Dr. Math,

How much water or fluid would be dispensed from a line 3/8 of an inch 
in size at a pressure of 150 psi after 1 hour? I would like to see the 
formula.

Thanks
Joe

Date: 07/01/99 at 11:30:15
From: Doctor Fwg
Subject: Re: Water volume

Dear Joe,

If the water is flowing through a "line," you must specify the line 
length to find a good estimate for the answer. The temperature of the 
water is also important, but is not too critical. Write back with 
additional information, and I will send you a more complete answer.

With best wishes,

- Doctor Fwg, The Math Forum
http://mathforum.org/dr.math/   

Date: 07/01/99 at 16:08:33
From: Bruce Pitchford
Subject: Re: Water volume

Dr. FWG,

Thanks for your reply. The length would be about 3 ft., and the water 
temperature would be around 70 degrees. Please include the formula so 
I can show it to my dad. Thanks again.

Joe

Date: 07/02/99 at 16:36:11
From: Doctor Fwg
Subject: Re: Water volume

Dear Joe,

Here is an answer - I hope it is what you were looking for.

Flow (Laminar) of Liquid through Pipe:

The appropriate equation for laminar flow (i.e., not turbulent) of a 
liquid through a straight length (L) of pipe or tubing is:

     FR = (Pi (R^4) (P - Po))/(8 N L)

where FR is the volumetric flow rate of the liquid (e.g., gal/sec),
Pi = 3.14159..., R is the radius of the pipe or tube, Po is the fluid 
pressure at one end of L, P is the fluid pressure at the other end of 
L, N is the fluid's viscosity, and L is the length of the pipe or 
tube. The temperature dependency here is in N because the fluid's 
viscosity depends on its temperature. A significantly different 
equation must be used for gases.

In any case, this equation can be used to predict laminar flow rates 
for water passing through a pipe. However, in the case you mentioned 
the flow is probably turbulent and the water is probably also coming 
out of a valve so this equation may not be extremely accurate for 
those conditions. But I think it will give a good approximate answer, 
especially if you can accurately measure the pressure difference along 
any length L of the pipe in question while the water is flowing. 
Remember too that the pressures at each end of L are the pressures 
that exist when the fluid is flowing, not the static line pressure 
when there is no flow. Another problem with this approach is that the 
pressures at each end of L may be difficult to distinguish from one 
another unless L is very long. In other words: P - Po will appear to 
be equal to zero in many cases unless you have two very precise 
pressure gauges (mounted at each end of L) or unless L is very long. 
It would actually be better to try to measure the pressure difference 
along L (using a differential pressure gauge) rather than the 
individual pressures, because a differential pressure gauge would tend 
to give you a much more accurate result, and you only need to know the 
pressure difference here anyway.

If you wanted to actually measure the water flow for your situation, 
you could use a timer (i.e., stopwatch) to record how long the system 
in question must be "open" in order to fill a five-gallon pail. Then 
you could calculate the flow pretty accurately by dividing the 5.0 gal 
volume of fluid collected by the measured time. This would be the best 
way to get a reasonably accurate answer because you don't have to make 
any assumptions about the flow conditions, you don't need to know the 
water's viscosity, and you don't need to make any other measurements 
(for instance, pressures at each end of the line of length L).

Some additional data for possible use in calculations:

     150 psig = 164.7 psia = 11.2 atm = 1.13 x 10^7 dyne/cm^2

     0.0 psig = 1.0 atm = 0.10 x 10^7 dyne/cm^2

(If the upstream pressure is 150 psig and the other end of the pipe is 
open to the atmosphere, the downstream pressure is 1.0 atm.) 

     L = 3.0 ft = 91.44 cm

     N (for H2O) = 0.01 dyne sec/cm^2 

     R = (3/8)/2 inches = 0.476 cm  

     FR = (Pi (R^4) (P - Po))/(8 N L)
        = ((3.14159)(0.476)^4(1.13 - 0.01) x 10^7)/(8 (0.01)(91.4))

If the equation above is used, FR will come out in cubic 
centimeters/sec. Remember that 1000 cm^3 = 1.0 liter and 28.3 liters = 
1.0 cubic feet, and 1.0 cubic feet = 7.48 gal. Remember also that the 
equation above is only really good for laminar flow, which probably 
doesn't apply here. 

I hope this answer is of interest to you.

With best wishes,

- Doctor Fwg, The Math Forum
http://mathforum.org/dr.math/