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System head curve

I'm currently trying to specify some centrifugal pumps in an application where the pumps will be pumping to a location where the elevation is below the pump discharge elevation. There are two horizontal pumps pumping into a common discharge header. The pumps will be pulling from a tank open to the atmosphere, with the water level about 3' above the centerline of the impeller.

There is a static rise of 50' after the pumps, then from that high point, the discharge pipe descends 250'. The total length of this 6" discharge line is approximately 2640'.

Here's my problem. I know how to calculate piping losses and come up with a system head curve. I am very familiar with doing this in which the static head is a positive value. How do I come with the static head in this situation given that the end of the discharge line is so far below the pumps?

Any help with this application is appreciated.

Thanks, Greg

Reply to
Gig
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Dear Gig:

Gig wrote: ...

Install a vacuum breaker at peak elevation. Break it into two pieces.

1) Pump from reservoir to peak elevation, 2) use Manning's equations on a partially full pipe, or choose a pipe diameter that will kill your available head at the required flow when full, without scouring the pipe lining.

Because you are describing a siphon, and about "~32 feet" down the other side (depending on system elevation), you will have destructive cavitation.

David A. Smith

Reply to
dlzc

Off the top (in other words I haven't looked for gotchas) the static head is 50 - 3 ft = 47 ft of the fluid density in question.

When the pumps top the syphon, the piping loss will vary with the flow rate. At low rates, the pipe will not fill downstream and atmospheric pressure will exist at the peak. At increasing rates, this pressure will occur further downstream. I imagine you could work the flow rate and loss numbers for atmospheric pressure at the pipe exit, for example. At this point, I think you would be on familiar territory with a head you can continue to work out for higher flow rates?? There is the usual question of laminar vs turbulent flow for your requirements.......

Any help?

Brian Whatcott Altus OK

Reply to
Brian Whatcott

Yes, Brian, thanks for the help. I failed to realize that at low flows the pipe on the downhill side would not be completely full. I had it in my head that the static head would be a negative value since the discharge is well below the level in the supply tank.

Thanks, Greg

Brian Whatcott wrote:

Reply to
Gig

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