Rate of Flow Through a PipeDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/