Pump efficiency e in hydraulic power equation (Q*P)/(600*e)

[J.Cho]

Hi, I am a graduate-level mechanical engineer.

I am working out a hydraulic power using the equation below:
(Q * P) / (600 * e) = kW
Where, Q: Flow rate (L/min), P: Pressure (bar), e: Pump efficiency

I wonder if the pump efficiency 'e' is the total pump efficiency.
I believe there are a couple of types of pump efficiency, such as mechanical efficiency, volumetric efficiency, total efficiency, etc.

I would appreciate it if I could get your advice on which efficiency shall be used for this equation.

Thank you.

 

[1503-44]

What type of pump?

Recips can have a volumetric efficiency.
Centrifugals do not.

Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Artisi]

As far as I'm concerned there is only one efficiency to be considered when calculating power requirements at the pump input drive, this will be generic information supplied by the manufacturer or from test results.
If you require power input to the motor assuming its electric motor driven you also need to account for the motor efficiency.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[LittleInch]

I can't calculate whether this is the same as I use but it seems about right.

This looks more like a centrifugal pump calculation to me, so you would simply use the efficiency given by the pump manufacturer which includes all efficicencies.

For a PD pump you're talking a different thing and it will vary for different flowrates based on the discharge pressure.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[pierreick]

Hi,
Consider this link to support your query:

https://www.linquip.com/blog/pump-efficiency/

Pierre

 

[1503-44]

Centrifugal have hydraulic efficiency, typ 0.50 to 0.85
Electric Motor efficiency typ 0.875 to 0.95
Vfd efficiency typ 0.95, if you have one
Gear box typ 0.95, if not direct drive.
Total eff is the sum of those

Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[saplanti]

1503-44,

I guess you meant “Total eff is the multiplication of those” in the last sentence.

 

[J.Cho]

I appreciate all your comments.

I was working out for a positive displacement pump, and the datasheet included two types of efficiency, which are mechanical and volumetric.
I thought these two types should be combined but not really sure if I was doing it correctly.
The multiplication of those efficiencies could work to determine the required hydraulic power, I believe.

Appreciate the clarification, guys.

Regards,
Joe

 

[LittleInch]

Correct.

But what you are calculating is the pump SHAFT power. Hydraulic power does not include the efficiency.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.