- #1
bhaarat316
- 3
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Here is the run down first, I'm trying to find a min mass flow rate for my project which needs to be 1 GPM, and max 4 GPM. Now water is going out of a 5 gallon jug, that has the top cut off, and the water is flowing into a PVC pipe. The area is already set at 1.5 inches, which I know is to big. I need to find the velocity of water at atm pressure, and then the diameter of the circle which the water will funnel through. I did some work I just need help making sure I did it right, basically a check over. think of it as a Deer Park 5 gallon bottle, like the ones for water dispenser, but the top cut off, and it flowing out the nozzle.
Man how do I use the Latex thing?
So, I know V=[itex]\sqrt{2*19.5 inches * 387.6 inches/s^{2}}[/itex]
V=122.95 inches/sec, 10.245 ft/s = 614.7 ft/min
Now the simple equation of Q=VA
1 GPM->.13[itex]ft^{3}[/itex]/min = 614.7ft/min * A
.03024 in^2 = A(min)
sqrt(.03024/∏)= r(min) = .09811 inches
We would have to reduce our inner diameter to .19622 inches.
Now is this right? I can't believe that? I was initial thinking it would be a DE since our mass flow would be varying, depending the height of the water, gravity would be pushing it through the main system.
Man how do I use the Latex thing?
So, I know V=[itex]\sqrt{2*19.5 inches * 387.6 inches/s^{2}}[/itex]
V=122.95 inches/sec, 10.245 ft/s = 614.7 ft/min
Now the simple equation of Q=VA
1 GPM->.13[itex]ft^{3}[/itex]/min = 614.7ft/min * A
.03024 in^2 = A(min)
sqrt(.03024/∏)= r(min) = .09811 inches
We would have to reduce our inner diameter to .19622 inches.
Now is this right? I can't believe that? I was initial thinking it would be a DE since our mass flow would be varying, depending the height of the water, gravity would be pushing it through the main system.