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Pump Curve - Outlet Pressure or Pressure differential 1

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colar

Mechanical
Jun 28, 2001
26
I am having a brain cramp today. Haven't worked with pumps much for a few years.
If I want to use a pump curve to estimate the flow in an existing pump system, and I can measure the pressures at the pump, do I use the difference in pressure (Pout-Pin) to correlate to the pump curve, or do I just use the outlet pressure?

My brain is telling me it should be the differential, but can't find any literature off hand that confirms this for me.
 
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Use the difference of pressures with the specific gravity of the fluid to convert differential pressure to head of water and you can correlate that to a standard pump curve.

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"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
BigInch has given you the goods, but don't forget to adjust gauge pressures back to the pump centreline - can be significant.
 
Have to agree with BigInch as always, to continue the thought, why differential pressure. A centrifugal pump adds energy to the fluid, in simple terms by flinging it out into the pump casing volute. This additional of energy is added to the energy the fluid already posses at the inlet. Hence the differential head will match the pump curve.

Mark Hutton


 
Thanks for the help. A related question. This pump has a flooded inlet via a raised tank open to atmosphere and I want to find out if the shut off pressure matches the pump curve supplied. We run the pump with the outlet valve closed. Since there is no flow through the pump there are no inlet losses. Do I still need to measure the inlet pressure and subtract this from the outlet pressure to obtain the shut off head? What confuses me slightly is that when the pump is running (pumping from one tank up to the next) both ends are open to atmosphere and so the atmospheric pressures should cancel out. But when the valve is shut off, the inlet side is exposed to atmospheric pressure but not the outlet. Does this mean that my shut off head measured at the outlet is larger due to NPSHa?
 
No problem. Both pressure gages measure the difference between pipe internal pressure and external atmospheric pressure, so its automatically cancelled out.

Use absolute pressures in your calculations to see how they do it.



**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
Just to fit the pieces together, would the outlet gage alone be showing pump pressure+fluid height+atmospheric pressure working above the tank? Or is the Patm cancelled out? (for the shut off scenario).
 
Pump:
Pump Disch Gage Press psig = Pump Suction Gage Press psig + Pump Diff H_ft * SpGrav * 62.4_pcf / 144

If there is a Suction Tank:
Pump Suction Gage Press psig = Suction_TankPress_psig + LiquidHt_overPumpCL_ft * SpGrav * 62.4_pcf - suction pipe pressure loss due to flow psig.
If tank is open to atmosphere, then tank pressure = 0 psig
If tank is closed, then tank pressure psig = whatever the gage says. That could be negative, if there is a vacuum in the tank.

If there is a Disch Tank:
Pump Discharge Gage Press psig = Disch_TankPress_psig + LiquidHt_overPumpCL_ft * SpGrav * 62.4_pcf - discharge pipe pressure loss due to flow psig.
If tank is open to atmosphere, then tank pressure = 0 psig, if closed, then = whatever the gage says.

If there is not a discharge tank:
Pump Discharge Gage Press psig = Discharge Pipe Gage Press psig


**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
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