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 Posted by: GJensen ® 08/22/2006, 08:51:34 Author Profile eMail author Edit Good Morning, I am attempting to calculate what volume and height of water is required to provide a minimum .4 gallons per minute pressure for at least 15 minutes to two nozzles. There is no auxillary pump. The only available pressure is that which is created by the volume of water above the nozzles. The total area of water release in the nozzles is approximately 0.044 sq. in.(very small). Pressure would obviously start higher and gradually decline as water is depleted from storage vessel.What is the approximate volume and height required to maintain the .4gpm for the full 15 minutes? (6 gallons is not the correct answer.)Thanks for any and all insight. GJ

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 Posted by: zekeman ® 08/28/2006, 11:20:49 Author Profile eMail author Edit Homework problem? I'll assume not You have to start with a vessel area, A as large as practical to get the minimum vol. Next you assume the height x which is to be solved. First, you get the minimum height,xmin you need to deliver 0.4gpm or about .05ft3/minute or .000833ft3/sec. if the nozzle area is "a" vmin=.00083/a and from Bernouli xmin=vmin^2/2g=(.00083/.044/144)^2/2g=2.716^2/2g=.115 ft Now invoking d/dt(Ax)=-av=-a*sqrt(2gx) which says tha the rate of change of volume is the rate at which it is spewed out of the nozzles, and so (A/a)/sqrt(2g)*dx/sqrt(x)=-dt Integrating both sides with the x limits of x to xmin, I get A/a/sqrt(2g){sqrt(x)-sqrt(xmin}=t=15*60sec I'll let you solve this for x after assuming a value for A. The container volume is xA

 Posted by: Scooter ® 08/22/2006, 11:26:01 Author Profile eMail author Edit I'm not sure volume & height are going to "keep" your flow at a true .4GPM's. Height will give you Head Pressure, but as the Height decresses, so does Head Pressure. Is there some reason you can't just feed this nozzle from,,,like a 55 gal drum or some other container ?