- **Thermodynamics and Fluid Mechanics**
(*http://physicshelpforum.com/thermodynamics-fluid-mechanics/*)

- - **Need help with water flow rates**
(*http://physicshelpforum.com/thermodynamics-fluid-mechanics/6829-need-help-water-flow-rates.html*)

Need help with water flow ratesFrom a reservoir we have an irrigation line that is about 25 feet below the surface water. The pipe is 12" in diameter and runs 3000 feet to a 10" diameter pipe that runs another 1370 feet to the outlet. The elevation drop from the origin of the pipe to the 10" pipe is about 35 feet. From the start of the 10" pipe to the final outlet it drops about another 40 feet. I was told if I multiplied the total drop X .433 I would be able to determine the PSI. 1. is that correct? 2. From this information can I figure out the gallons per minute? Can someone give me a formula? I am trying to determine what gpm I would get using smaller pipe and wanting to know if PSI formula is correct? I hope this is enough information. I am not sure if the water in the pipes is totally full (12" or 10" of water) but am guessing it is at least half to 3/4 full all the time. Any ideas??? |

Not sure where you got the 0.433 factor from, but I assume you must have an engineering handbook that provides estimates of the loss of head pressure due to losses attibutable to pipe diameter and length. Assuming that the 0.433 factor is good, then the velocity of water flow at the outlet would be v = sqrt(2gh) where 'h' is the adjusted total head height of 25 feet + 35 feet + 40 feet. This gives 'v' in feet per second. From this you can calculate flow in GPM given the area of the outlet pipe in ft^2: GPM = v (ft/s) x 60 (s/min) x A (ft^2) x 7.48 (gal/ft^3) |

All times are GMT -7. The time now is 09:15 AM. |

Copyright © 2016 Physics Help Forum. All rights reserved.

Copyright © 2008-2012 Physics Help Forum. All rights reserved.