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To: supercat

Yes, you are right that the power increase is to compensate for head loss. When you are pumping water into Lake Pontchartrain, you have to operate the pump at a head to compensate for the difference in height being pumped and for the head loss in the pipes. The pressure of the water exiting will be same as the medium that it is entering (not at a much higher pressure). If at any point in time it is higher, the flow would increase until it was equal at exit. When flow increases, head loss goes up. Therefore, there is a balance between flow rate, head loss, and pump head. Head loss (friction in the piping and the energy required to change the direction of water in horizontal turns of the pipe) is still the controlling factor. The power needed is still proportional to the cube of the volumetric flow rate.


2,018 posted on 08/29/2005 11:06:35 PM PDT by burzum
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To: burzum
Head loss (friction in the piping and the energy required to change the direction of water in horizontal turns of the pipe) is still the controlling factor. The power needed is still proportional to the cube of the volumetric flow rate.

There are two forces one has to work against: head loss and gravity. In any given system, I would expect the power requirement to be approximately k1*V^3 + k2*V where V is the volumetric flow rate and k1 and k2 are proportionality constants. If V is much less than sqrt(k2/k1), the linear term will dominate. If V is much greater than sqrt(k2/k1), the cubic term will dominate. If V equal or nearly equal to sqrt(k2/k1), both terms will contribute significantly.

2,164 posted on 08/29/2005 11:44:06 PM PDT by supercat (Sorry--this tag line is out of order.)
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