Pump efficiency estimate based on temperature 2

[Fortmac2015]

Hi,

I'm working on a project to estimate pump efficiency where pump temperature, pressure, and flow are the only readily available data. I came across a formula in "Forsthoffer's Best Practice Handbook for Rotating Machinery" that relates pump efficiency to temperature difference. In this book (link below) the author says pump head is calculated from data in (m*kgf)/(kgM) OR (ft*lb_f)/(lb_M). This doesn't make sense to me as I've only ever seen head calculated in units of length and I can't find information online.

Has anyone seen head measured in the units described in the handbook or understand why the author would have used this method?

https://books.google.ca/books?id=9l3F3qovu9EC

Page 77
(I can post a screenshot of the link if it cannot be accessed)

Any comments would be appreciated!

Thanks,

 

[georgeverghese]

You've got it all out of context - this formula on page 77 gives you the temp rise in the fluid when the pump is running at dead head (or some low flow), given the pump eff at this point. Read the context on page 76 - this page doesnt show up online.

 

[LittleInch]

Please add screen shot as it now won't preview the page in question.

However I did see it earlier and it's far from clear what M stood for or where the kgf came from. Missing the page 76 is fairly crucial.

Anyway you shouldn't need it.

If you take the assumption that the inefficiency is translated to temperature difference, just follow the energy.

You can calculate shaft power at 100% eff using flow, head and density.
You can presumably obtain actual shaft power.
You can calculate the energy rise due to temperature increase.

Then you have a difference

Or just play around with the efficiency until the shaft power required matches the actual shaft power, but then you do need to account for other losses in the pump like bearing friction etc.

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

 

[TenPenny]

That looks to me to be kgforce and kgmass and pounds force/pounds mass, the relationship being a reference to whether or not you are in a place where the force of gravity is not equal to 9.8...m/s^2.

I'm assuming (without having read the book), that he's showing the derivation and using this to keep units in some sort of rational progression as he explains what he's doing.

 

[Fortmac2015]

Hi,

Please see the attached picture with the pages being referenced.
Image is from books.google.ca

http://files.engineering.com/getfil...ae0-4932-8688-c1bf1bc678e5&file=Untitled2.jpg

 

[LittleInch]

Pump head is not the correct term. I think what he is trying to do is incorporate the density into the equation somehow but it's not written very well and is confusing.

Not sure if it's actually correct....

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

 

[Fortmac2015]

Thank you all for your comments!


@georgeverghese
On page 76 it states that "pump efficiency is low at low flows" which makes sense based on pump performance curves. But to me it doesn't seem like the author is saying the equation can only be used in cases when the pump is running at low flow. Do units given for calculating pump head make sense to you?

@LittleInch
Thank you for the suggestion!
Unfortunately due to a fluid coupling between the motor and pump, calculating the shaft power is very difficult based on my discussions with the coupling vendor. Hence obtaining actual shaft power as you suggested doesn't seem feasible at this time.

@TenPenny
The author doesn't seem to show derivation of his equations, units, or constants which is making things a little more difficult.

@LittleInch
I agree. I don't think pump head is the correct term. Without the derivation is hard to understand the rational and meaning.

 

[LittleInch]

One more thought. Why do you not have access to the pump curve?

If you can't accurately get shaft power in its not really possible to guage efficiency.surely?

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

 

[TenPenny]

I'm sure that's a conversion factor to get from kg mass (used for the specific heat portion to convert power into the temp rise in the pump) to the kgforce used in the pump power calculation. But to be honest, I'd be tempted to throw that book in the garbage, because the formula uses 337,100 as a factor, and then below explains that the 367,100 is a conversion factor.

With such poor editing, it would be impossible to understand what it is supposed to say.

In the Goulds literature, temp rise in F is calculated by 5.09 * (shut off HP) / (gpm * SG * SH).

Which uses hp input instead.

 

[Fortmac2015]

@LittleInch
I do have access to the pump curve. I want to compare the calculated efficiency to the ones from the tested pump curve provided by the manufacturer.
The actual shaft power is difficult with the torque converter

@TenPenny
Do you have a copy of that literature you could post here?

 

[LittleInch]

Then just use what the pump should be doing in terms of efficiency or shaft power with what you have now. Assume all heat rise is due to the inefficiency and you're away.

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

 

[TenPenny]

Here is a copy of the page. This page is specifically referring to a magnetic drive pump, where you have to concern yourself with the temperature rise in the pump end, but also in the liquid that cools the mag drive assembly, and that why they have this guideline. So with a mag drive pump, if you're dealing with hot liquids, you have to ensure that the temp rise in the casing, PLUS the temp rise due to the recirculation through the magnets, won't boil the liquid. The same formula applies to a non-mag drive pump (obviously without the recirc portion through the mag drive), but it's extremely rare that the increase in temperature of the pumped liquid ever is a concern, so it's hardly ever addressed.

 

[georgeverghese]

I've yet to make sense of the 337100 (or 367 100) in the SI units version of this expression. Will get back in a few hours if I crack this one.

In my experience in doing these calcs on past projects ( which were to compute the temp rise at low flow, unlike your case where you want the back calculate eff from temp rise), you get a perceptible, measurable temp rise only at low flows coincident with low eff for centrifugal pumps( which is what this book says also). At higher flows, the dT is perhaps measurable but we are talking about very small dT here (perhaps less than 0.5degC). Which makes getting accurate dT values rather challenging at higher flows. Hence, this approach for extracting pump eff is possibly approximate only at low flows , at or near min flow up to near dead head.

 

[georgeverghese]

In my deriviation of this expression in SI units from first principles, the constant 337100 is replaced with 1000, and the numerator is multiplied by g, the gravitational constant.

Here, pump developed head is in metres, eff is pump hydraulic eff expressed as a fraction, and Cp is in kJ/kg/degC.

Temp rise across the pump dT = (g.h/(1000.Cp)).((1/eff)-1), where g is usually 9.807m2/sec.

If pump head is expressed as Nm/kg, which is the same as J/kg or kgf.metres / kg, then use this value directly in the numerator and delete multiplication by g. Most pump vendors do not use Nm/kg for expressing head; Nm/kg is usual practice for expressing centrifugal compressor head only.

So for example, when we have a developed head of 100metres at eff=0.6 for water with a Cp of 4.18kJ/kg/degC, then dT=0.16degC, which is hardly measurable with run of the mill devices.

 

[LittleInch]

A star to that man, great answer. You tend to look at this only if the efficiency has fallen by a considerable amount.

I'm intrigued as to why you have a fluid coupling though??

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

 

[TenPenny]

And with georgeverghese's explanation, the mention of 'g' explains the original question with the conversion of kgf and kg.

I wish people would stick to SI units, it makes things much more clear.

 

[Fortmac2015]

Thank you all for your comments!

@LittleInch
I'm not sure if I misunderstood what your asking but we use the fluid coupling to control speed of the pump. Its considered a VSD that transfers the motor power to the pump, hydro-dynamically.

@TenPenny
Thanks for attaching the page! Apparently there is a large range for the temperature profile on the pump casing I'm working on which I didn't realize.

@georgeverghese
You bring up a good point about the equation only being usable in low flow conditions. I'll do some testing with what data I have but I don't know if our devices can measure to an accuracy better than 0.16degC.
Also, thanks for taking the time to derive the equation! I rearranged it and it aligns with the equation I found on this website:

http://www.pumptraining.com/temperature-rise

 

[LittleInch]

No that makes sense.

Still don't understand though why you want to try and find efficiency? with a fluid coupling in the way it's not like you can complain that you're spending more on power than you thought so what's the point?

Doing a pump test for flow rates at 0, 20%, 40% etc to create your pump curve to compare to the as tested version may show some difference which implies wear or damage, but if you can't get an accurate shaft power then you can't get efficiency. Temperature rise is too low to accurately measure and who runs their pump at low flow?

So maybe you need to break the news to someone that unless they fit a torque measurement device with a speed counter on the pump inlet shaft then it's not really feasible to get efficiency with only pressure, flow and temperature measurements.

But WHY do they want to know????

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