Pump power calculation for non water application 1

[jddj]

Hello,

I'm trying to determine the power required to operate the pump at the indicated point on the pump curve. This pump is operated by a VFD. The curve shown is based off of water, the fluid I am working with has a SG of 1.2. I am using the equation

Brake HP = (GPM*Head*SG)/(3960*Efficiency)

I Know the GPM value of 6800 GPM, the head value of 138 feet and the SG of 1.2.

I'm having issues trying to determine what to use for the efficiency, I have been told two different things, one was that I can use the efficiency from the curve (about 78%) based off of water because the SG of 1.2 value accounts for the reduced efficiency, and another person told me I need to use a different efficiency for the fluid I am using.

So my question is which is correct? can I use the the efficiency value directly off the pump curve for water, or do I need to determine a new efficiency and if so How so I go about that?

Thanks for the help.

 

[TugboatEng]

Just multiply the y-axis (head) by 1.2. The chart should otherwise be the same.

 

[LittleInch]

I agree. Only if your viscosity goes a lot higher than water ( say 50cP +) then you might need to adjust the curve and the efficiency.

If you're using a VFD though beware that this might alter the efficiency and that curve is only good for the exact speed it was calculated for. Head varies approximately the square of the pump speed.

"because the SG of 1.2 value accounts for the reduced efficiency, " errr, no. The SG means that your pressure and your power will be higher than water, but doesn't affect the efficiency

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

 

[jddj]

TugboatEng & Littleinch, thanks for the reply's. maybe I miss quoted the "because the SG of 1.2 value accounts for the reduced efficiency, " but basically they were saying I do not need to use a different efficiency. So when you say " Just multiply the y-axis (head) by 1.2. The chart should otherwise be the same" are you saying I would go 138 * 1.2 = 165.6, then read the efficiency at the crossing of 165.6 with 6800 GPM which is off the graph but looks like it would be about half way between 75% and 80%?

Thanks,

 

[1503-44]

SG different than water do not change the pump curve Head vs flow. Your pump curve remains the same.

Different SG changes the power needed. Power, if shown on a pump curve, is (almost) always based on water. With a SG of 1.2, multiply the power based on water by 1.2 to find the power you will need to pump a 1.2 SG fluid. The equation is correct. Brake HP = (GPM*Head*SG)/(3960*Efficiency)

Efficiency does not change with SG, but as LittleInch says, efficiency can vary with viscosity. Use the efficiency as shown on your pump curve. Modify it for viscous effects only if you need to.

Clear now?

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

 

[jddj]

1503-44, yes that is clear thank you for the reply!

 

[FacEngrPE]

I think saying the head should be multiplied by SG is a bit confusing.

Head on a pump curve (Head VS Flow) is adjusted when determining the corresponding gauge pressure.
I agree the OP proposed calculation matches the pump laws.

The NPSH values on the pump curve should be adjusted to gauge pressure in the same manner.

Had the power lines been drawn on the pump curve chart they would need to have their values increased by the OP's power equation.

 

[1503-44]

I think saying "head on a pump curve is adjusted" is confusing. Pressures corresponding to head can be calculated. Head, no. It is never "adjusted". It is what it is.

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

 

[pierreick]

Hi,
Consider this document to support your work , present and future .
Good luck
Pierre