Pressure decay
Pressure decay
(OP)
Anyone know how to calculate the pressure drop you'll see when pressurizing a hose cue to hose expansion?
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS Come Join Us!Are you an
Engineering professional? Join Eng-Tips Forums!
*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail. Posting GuidelinesJobs |
|
RE: Pressure decay
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.
RE: Pressure decay
RE: Pressure decay
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: www.carf-engineering.com
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
RE: Pressure decay
is only valid for gasses, not for fluids.