Overall Efficiency of pumps in parallel

[arm83]

Hey everyone,

I came across an equation to calculate the efficiency of pumps in parallel. It is as follows:

H*(specific gravity) x Sum of pump flow rates (gpm)
____________________ ____________________________
k Sum of total power to pump(hp)

where k= 3960

What I would like to know is what would be an acceptable value for the efficiency of a parallel system?

 

[Artisi]

Acceptable efficiency depends largely on the pump style and size of the pump unit together with the drive unit efficiency.

Pumps can range from a lot less than 50% for small units , solids handling unit etc up to 95% for large custom designed units, likewise electric motor drive units can range mid 80 to the high 90%.

What flows are you talking about - this can dictate an idea for pump / motor efficiency.

 

[BigInch]

I think you really have to answer that question by asking yourself if you're happy with your present energy consumption and if there exist an alternative which could reduce that energy consumption and overall system operating cost, keeping in mind that sometimes high efficiency units also mean high maintenance costs.

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[arm83]

There are three pumps in parallel. Two of these pumps have a rated flow of 100 usgpm with a corresponding head of 150 psi while one pump has a rated flow of 900 usgpm with a corresponding head of 170 psi. When applying the efficiency equation, using the above values, I got 60% efficiency.Is that acceptable ?

 

[BigInch]

What's the efficiency at BEP for each pump?

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[Artisi]

What you are telling us in meaningless as we have no idea where the pump are operating on their individual performance curve and what the pump characteristics are.

With any hydraulic problem, the first step for me is to develop a flow/head curve for the static head /pipework system, this doesn't have to be too precise but sufficient to give a feel for the system that the pump/s need to operate against.
Once you have establish this you can then super-impose the pump/s performance over the "head loss curve" to asertain the efficiency of the pump units.
Without this it is like asking - "how long is a piece of string"

The question you have proposed is meaning less and there is no answer.

 

[picardd]

Along with this question i have a quick ammendment to it. When calculating the power out of a parallel pump system what do u take as the value for the pressure head? I have a system with two pumps in parallel, identical pumps, identical power to the system, however i cannot seem to find an equation for the power out. i suppose i could use the efficiency equation posted above but that still leaves the question as to the value of H. How is H for the system calculated?

 

[Artisi]

H for the system can be measured by pressure gauge across the pumps while the pumps are in operation - both pumps should give exactly the same H - any discrepancy will be gauge error not pump difference.

 

[BigInch]

What Artisi says is true for identical parallel pumps, but you have said your pumps are different. Also be careful about calculating heads, as it should be differential heads.

It may also be that the suction of the two different types of pumps vary. Are the pumps feed from the same suction line? If not any difference in suction pressure must be included. BTW, as Artisi said, your discharge pressures should be roughly the same, except for any discharge header flow losses between pump inlet points. I note you have both 150 and 170 psi.

The pumps use energy to produce a differential head between the suction and discharge. You have only told us what I believe to be the discharge pressure, not the differential head. Differential head must be calculated from the difference between discharge pressure and suction pressure. If suction pressures are different, you have to do the efficiency calculations for each pump independently.

To calculate the efficiency of different parallel pumps, or identical parallel pumps with different differential head outputs, I would use a weighted average of the sum of the individual efficiencies.

For a pump, the efficiency equation is,

A pump's differential head (in feet) is calculated from suction and discharge pressures as,

dH_ft = (disch_press_psig - suction_press_psig) * 144 / 62.4 / SG

where,
SG is the fluid's specific gravity (=1.0 for water).

Hydraulic power (power to be delivered to the fluid) for each pump is,

Power_Hyd = Q_gpm * dH_ft * SG * 62.4 / 60 / 7.4805

where,
Q_gpm is the pump's flowrate in usgpm.

Efficiency is Power_Hyd / Input_Power

I would then multiply each pump's efficiency by its flowrate, then sum that for all pumps and divide by total flowrate.

I put a small Excel spreadsheet here for you that does the above calcs at,

http://rapidshare.com/files/43356841/pump_group_efficiency.xls.html

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