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Pressure decay 1
- Thread starter fireguy
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1
- #2
I imagine the reason you have no responses is because the word "cue" should be "due." This "pressure drop" is going to be a function of the material characterisitics of the hose, the length of the hose, the diameter of the hose, and the flow rate.
Let's look at what happens as we pressurize the hose. First the hose will fill with water and during this process the pressure drop to the end of the hose will be the inlet pressure minus the pressure at the end of the hose assumed to be zero initially and remain there until the hose is full. Assuming the hose is shut-off at the outlet end, then there will be no signifcant drop from the inlet to the outlet once the hose is fully presurized with no flow even if the hose expands due to pressure.
My guess is we are talking about a fire hose. Now should you decide to open the outlet and allow flow through the hose the pressure drop will be calculated based on the diameter of the hose, the friction factor of the hose inner lining, the number of bends or kinks, and the flow rate. If we assume that there are no more than four ninety degree bends then, the hose has a three inch inside diameter and is lined with a smooth rubber, and is 100 feet long we can calculate a Cv for the hose using Crane Technical Paper 410. That Cv would be approximately 100. Since the pressure drop is equal to the square root the flow divided by the square root of the Cv the pressure drop would equal the square root of the flow divided by ten for the hose described here. For a 200 GPM flow rate the pressure drop would equal about 4 PSI. Since there are no signifcant changes in the diameter of the hose due to the pressure of the water in the hose once it is flowing there is no reason to be concerned with the pressure drop due to expansion during this time - there is none.
Let's look at what happens as we pressurize the hose. First the hose will fill with water and during this process the pressure drop to the end of the hose will be the inlet pressure minus the pressure at the end of the hose assumed to be zero initially and remain there until the hose is full. Assuming the hose is shut-off at the outlet end, then there will be no signifcant drop from the inlet to the outlet once the hose is fully presurized with no flow even if the hose expands due to pressure.
My guess is we are talking about a fire hose. Now should you decide to open the outlet and allow flow through the hose the pressure drop will be calculated based on the diameter of the hose, the friction factor of the hose inner lining, the number of bends or kinks, and the flow rate. If we assume that there are no more than four ninety degree bends then, the hose has a three inch inside diameter and is lined with a smooth rubber, and is 100 feet long we can calculate a Cv for the hose using Crane Technical Paper 410. That Cv would be approximately 100. Since the pressure drop is equal to the square root the flow divided by the square root of the Cv the pressure drop would equal the square root of the flow divided by ten for the hose described here. For a 200 GPM flow rate the pressure drop would equal about 4 PSI. Since there are no signifcant changes in the diameter of the hose due to the pressure of the water in the hose once it is flowing there is no reason to be concerned with the pressure drop due to expansion during this time - there is none.
- Thread starter
- #3
Actually I am dealing with PVC hose. I am looking for a way to calculate the pressure drop seen when the hose is flowing 0 CFM. We are looking at pressure decay tesing our product here at our plant for leaks but I'm not certain how much pressure drop we are going to be seeing when the hose on our reels expands. In essence what I am looking at is if I have a pressure vesel that is 10 cubic inches and pressureized to 100 PSI. What happens if I don't add any pressure but increase the volume to 50 cubic inches?
Hi,
1) This is not "pressure drop" as used in fluid mechanics, where it is (as Scorpio already explained), the pressure loss of fluid flowing through pipe due to friction.
See also:
2) In essence what I am looking at is if I have a pressure vessel that is 10 cubic inches and pressureized to 100 PSI. What happens if I don't add any pressure but increase the volume to 50 cubic inches?
P x V = constant
P1 x V1 = P2 x V2
P2 = 10 x 100 / 50 = 20 PSI
Of course this is only valid for a theoretical piston-cilinder system. Is sounds a bit dim, but it's prop. best to hook up a pressure gauge to the system. They are not that expensive.
Hope this helps,
MVD
1) This is not "pressure drop" as used in fluid mechanics, where it is (as Scorpio already explained), the pressure loss of fluid flowing through pipe due to friction.
See also:
2) In essence what I am looking at is if I have a pressure vessel that is 10 cubic inches and pressureized to 100 PSI. What happens if I don't add any pressure but increase the volume to 50 cubic inches?
P x V = constant
P1 x V1 = P2 x V2
P2 = 10 x 100 / 50 = 20 PSI
Of course this is only valid for a theoretical piston-cilinder system. Is sounds a bit dim, but it's prop. best to hook up a pressure gauge to the system. They are not that expensive.
Hope this helps,
MVD
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