Where to provide Break Pressure Tank in water pipeline
Where to provide Break Pressure Tank in water pipeline
(OP)
I am designing a gravity flow system for a rural area. The elevation difference b/w the source and point of use is 1004 feet (306 m). I will provide a 1.5" Schedule 40 PVC pipe. I would like to know what is the ideal static pressure for pipe and fittings where I should locate a break pressure tank in the system.
Also will pressure reducing valves serve the same purpose as break pressure tanks?
Thanks
Also will pressure reducing valves serve the same purpose as break pressure tanks?
Thanks





RE: Where to provide Break Pressure Tank in water pipeline
Assuming that you are going up, pressure control valves won't help. They cannot reduce the pressure below the "no flow" static pressure imposed by the increase in elevation.
From "BigInch's Extremely simple theory of everything."
RE: Where to provide Break Pressure Tank in water pipeline
You may be able to use pressure reducing valves in combination with a break tank.
If you locate a reservoir at the midpoint, you will have 500 ft of water pressure at the bottom. You can then use pressure reducing valve(s) to drop the pressure at the point of use to drop the pressure to 90 ft to 160 ft.
Break tanks also serve as reservoirs.
Use of 1.5-Inch Schedule 40 PVC pipe will allow a flow of around 30-35 gpm when using a reasonable pipe velocity.
RE: Where to provide Break Pressure Tank in water pipeline
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If you use sch 40 PVC, the pressure limitations will require at least 2 break tanks as you drop the 1000 ft elevation.
RE: Where to provide Break Pressure Tank in water pipeline
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I was thinking of providing 1 break pressure tank at 150 - 200 psi below the static line and a storage tank at similar difference below the new static line (starting at break pressure tank).
Is 200 psi reasonable or should I be more conservative?
Also can I replace the 1st break pressure tank with a pressure reducing valve that will essentially reduce the maximum (static) pressure and readjust the static line?
Thanks
RE: Where to provide Break Pressure Tank in water pipeline
Have you considered using HDPE? HDPE is rated for higher pressures, is available in coils, and can be trenched in.
You can use pressure reducing stations rather than break tanks.
You will have a gravity line down 400 ft to the first pressure reducing station. At that station, you can drop the pressure down from 200 psi to maybe 30 psi. Now the pipeline will be pressurized down the slope.
You can drop down another 400 ft to a second pressure reducing station. If you drop the pressure here to 30 psi, then you will have 116 psi at the bottom.
RE: Where to provide Break Pressure Tank in water pipeline
RE: Where to provide Break Pressure Tank in water pipeline
@bimr
Not sure if I understood your last post about locating the the pressure reducing stations. Please let me know if my understanding is correct~
So basically I need to locate minimum 2 pressure reducing valves along the pipeline. The 1st would be 400 ft (173 psi) below the original static line. This will be the max (static) inlet pressure into the valve. I can set the outlet pressure @ about 30 psi. The outlet pressure is the residual head at the valve and the new static line and HGL will begin at 30 psi (70 ft) above the valve elevation. Then I drop another 400 ft (173 psi) below the new static line and set the outlet (residual) pressure at 30 psi. The 2nd static line would begin about 70 ft above the 2nd valve elevation. Therefore, the final static head at the end of the line will be [1000-(400-70)-(400-70)] = 340 ft or 147 psi.
I am not sure how you got 116 psi number in the end? Can you please explain?
Thanks
RE: Where to provide Break Pressure Tank in water pipeline
You can adjust the distance between pressure reducing stations based on your application. I was just trying to keep the pressure between 30 and 200 psi.
The pressure at the top of the 2nd reducing station will be 30 psi. When you drop the 200 ft (86.6 psi), you now have 116 psi. I think you added the residual 30 psi twice. [1000-800)+(70)] = 117 psi.