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Pump/piping question

Pump/piping question

Pump/piping question

(OP)
I have two pumps operating in separate sumps and joining at tee and entering a tank through a single line. All lines are 2" O.D. HDPE SDR 11 pipe, shortly after the connection, the line enters a manhole and exits as a 4" SDR 11 HDPE Pipe. I have very little info on the pumps other than one was spec'ed at 55 GPM at 96' TDH and the other at 53 GPM at 85' TDH. I need to ensure this configuration is not a problem, especially in the 2" line before it expands to 4". Any insight on how to best check this? Thanks.

RE: Pump/piping question

Are you sure pipe is actually/exactly " 2" OD " (I only ask as this sounds a little strange to me, and actual OD and DR are important to determining ID, that is most important in flow calculations)?

RE: Pump/piping question

You have a parallel pump flow problem. You must solve it either graphically (plotting system curve on pump curves and adding flow at common head, or by a network solution, modeling the pumps with a curve fit equation.

When the combined flow enters a manhole, does the pressure go to atmospheric? That is not clear in your post. Is the manhole before or after the tank connection? A sketch would help.

RE: Pump/piping question

rconnor,
2-inch (nominal) HDPE has an OD of 2.375 inches. So SDR 11 would be 9/11ths of that or 1.94 inches. Someone may make a 2-inch OD HDPE, but I've never seen it in the catalogs.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.
The plural of anecdote is not "data"

RE: Pump/piping question

Thanks, that is a cute rule of thumb. For many years I know a quite large USA hdpe manufacturer actually advocated the "ID" of their "OD-controlled" pressure pipe be calculated by the formula:

ID = OD [1-2.12/DR]

While it is a fine point particularly in this case, applying this recommendation to the current example and assuming the OD is in fact 2.375" (not the 2" of the OP) and the DR is "11", the ID so-calculated however would be:

ID2 = 2.375 in [1 - 2.12/11] = 1.917 in

RE: Pump/piping question

DR=OD/t by definition. Solving for t = OD/DR

ID=OD-2t --> ID = OD - 2(OD/SDR) --> ID = OD*(1-2/SDR) --> ID = OD*(SDR-2)/SDR

Not a "rule of thumb" at all, just arithmetic.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.
The plural of anecdote is not "data"

RE: Pump/piping question

While I guess I shouldn't pre-suppose why a major manufacturer included the formula I quoted in their "Engineering Manual"(that gives a slightly smaller I'D); however, I see per standards the thickness you calculate is a "minimum" required thickness per the standard. Maybe they can't control this thickness perfectly, meaning to me pipe may very well have some smaller ID, to dependably meet the minimum thickness requirement?

RE: Pump/piping question

Listen to the Trashcanman...you have two pumps that are fighting each other. The higher head pump will pump a portion of its flow back through the lower head pump, unless a check valve shuts and deadheads the low pressure pump. Centrifugal pumps respond to the pressure they see at their discharge flanges. In terms of pipe size you probably want to keep your velocities around 10-15 ft/sec.

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