- #1
charlieroper
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Can anyone tell me how to calculate the kenetic energy loss in the pressure of a fluid by forcing it to change direction 180 degrees? I need to know the pressure loss in a single reversal so I can start to figure how many I need. I’ve asked this question of your fluid power engineers, but they tell me it basically boils down to trial and error. But I need a reasonable point at which to start a trial.
I need to lower the pressure of a 9.5 Liter/minute stream of .453 Kg/liter water 280 Kg/cm through a minimum passage area of 3.17 mm. I can’t do this with a single orifice because of anticipated solids in the fluid which would quickly clog such a tiny opening. I wish to harass and hector the fluid as much as possible by passing it through a course of abrupt right angle turns using the fluid’s inertia and other properties to accomplish a series of pressure drops. Is it possible working with this small volume? Does anyone have a reasonably non graduate student level formula which takes into account those factors needed to calculate the pressure losses? Thanks to anyone with the education to size this thing up.
I need to lower the pressure of a 9.5 Liter/minute stream of .453 Kg/liter water 280 Kg/cm through a minimum passage area of 3.17 mm. I can’t do this with a single orifice because of anticipated solids in the fluid which would quickly clog such a tiny opening. I wish to harass and hector the fluid as much as possible by passing it through a course of abrupt right angle turns using the fluid’s inertia and other properties to accomplish a series of pressure drops. Is it possible working with this small volume? Does anyone have a reasonably non graduate student level formula which takes into account those factors needed to calculate the pressure losses? Thanks to anyone with the education to size this thing up.
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