Power factor/efficiency measurement of motor 1

[mjpetrag]

I have a Goulds 1x2-10 1800 rpm 3 hp motor with a 9" impeller putting out 1200 GPH. It keeps tripping the overloads. The motor wasn't sized for curve runout, but still it pulls around 3.6 amps and overloads are 3.74. A small increase in flow trips it out.

So to live with it temporarily, I changed to a 7.75" impeller and it draws now about 2.5 amps. With the 9" impeller, the hydraulic hp is 2.1 at 1200 GPH. With the 7.75" impeller, the hydraulic hp is 1.5 hp.

So I tried to match this up with the motor hp.

W = v*i*1.73*pf*eff

With the 9" it becomes W=470*3.6*1.73*.95*.88=2450 W = 3.3 hp. I estimated pf and eff. But as you can see the power is much higher than the hydraulic hp.
Same issue with the 7.75" impeller
W=470*2.5*1.73*.95*.88=1700W = 2.28 hp, much higher than the 1.5 hp calculated

The fluid has a cp of 10 and a SG of .97

So either my pf and/or efficiency are wrong. I know they are fairly dependent on load, but how do you figure this out? Is 50% load on the motor the hydraulic hp divided by full load of motor ?

-Mike

 

[Artisi]

If you are using 88% for hydraulic efficiency - would suggest that is where the problem is - you need to check the pump curve for efficiency - you can do that by Google "Goulds 1x2-10" or similar.


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.)

 

[mjpetrag]

Actually the hydraulic power I posted above accounts for the pump efficiency of about 24.5% efficiency.

Basically I have the following for the 7.75" impeller:
Hydraulic HP = (20 GPM)*(69 ft)*(.97 SG)/(.245*3960) = 1.45 HP (Corrected for viscosity cP of 5)
Now of the motor side, it was drawing 2.5 Amps. I calculate that as
Motor HP = (470 V)*(2.5A)*SQRT(3)*(.88)*(.95)*.00134 = 2.28 HP

The Hydraulic HP and Motor HP should be equal after I account for losses. They are not matching in both cases. The only way I can make them match is by screwing with the PF and efficiency numbers for the motor.

Therefore, I pick a combination and say the motor efficiency is 82%, then my power factor needs to be 65%.

Motor HP = (470 V)*(2.5A)*SQRT(3)*(.82)*(.65)*.00134 = 1.45 HP

But I am unsure of how to get an accurate number for power factor and efficiency of the motor. I know these may be published with the motor, but I can't find them. They are typically written as factors of % FLA. Well if I'm 50% FLA, and the factors are causing that amp reading to be higher, am I really 50% FLA?

For example, I have this graph showing power factor on the Y axis, and % FLA on the X axis. If I'm 2.5 amps, and the FLA of the motor is 3.74, then I am at 67% FLA. The power factor would then be about .65. But as I open the control valve and draw more amps, the %FLA goes up, but the power factor goes up as well, so the power should not go up as quickly. Is that correct?

-Mike

 

[GPRnD]

Goulds has an online selector tool that you can use to verify pump selections and performance. It is free you just need to sign up.

http://www.gouldspumps.com/Tools-and-Resources/Pump-Selection-System/

My check says that the pump at 20 GPM and 69 ft should require just under 1.5 HP.

I don't normally deal so much with smaller motors but notes I have on them says the following are reasonable for a sub 5 HP motor:

50% load PF=72%
75% load PF=82%
100% load PF=84%

Motor Full load efficiency of 79% minimum per NEMA B

Given the above are average motor values and given the tolerance possible in the pump efficiency (particularly since you are operating quite far back on the HQ curve), your calculations seem reasonable.

The question I had is how are you measuring flow and head for the pump? What accuracy do you think you can measure them to ?

 

[mjpetrag]

Coriolis meter for flow . Pretty darned closed

-Mike

 

[Xalii]

I have the impression you are mixing your powers.
Pelec: W=1.73*V*I*PF= 2.5 kW (pf for this size, 2 poles, should be ~0.86) cf attached doc
Pmec = Pelec/effmotor = 2 kW (eff motor this size 2 pole should be 79%)
Then
Phydro= Pmec/effpump
Phydro 2.i*746=1.57 kW
Therefore I conclude that you have a effpump of 79% what seems quite logical
Pelec=Phydro/effmotor/effpump

The motor coeff were taken assuming a full load what should be the case with the 9" impeller.
For the 7.75" impeller the power should be reduced as suggested by GPRnD

 

[LittleInch]

mjp,

Your calculations seem correct, but for this size pump and motor you are almost certainly using generic pump and motor data and not specific data from your units.

This certainly applies to the pump where as you seem to be operating at the far left of the curve, the stated efficiency (already not great at 25%) could be even lower in reality, especially if you've got a smaller unit. If you have a pump efficiency of 14% it all works. My guess is that neither the pump nor the motor are as efficient as you appear to think and hence the gap between your pump shaft power requirement and your motor shaft output is shared between them.

Little things like the bearing used or shaft seals (are they packed glands?) can make a lot of difference to a small pump / motor combination.

The only way you find out which is which is to bench test the pump and the motor separately using accurate testing spec equipment - is it really worth it?? Next time allow a suitable margin when running this size of unit, especially when you're operating a long way from BEP of the pump. Unless you pay a fortune, the pump vendor will not individually test small pumps as it doubles the cost, so what you have is an average or computed efficiency based on a few tests and a few points.

Out of interest where does 470V 3 phase come from?? I've seen all sorts of voltages all over the world, but never 470V. This might mean you're actually using a motor designed for some other more common voltage (400/415 or 480 or 600) and just de-rated or qualified for more than one voltage.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[bimr]

mjp,

It is probably less expensive to install an orifice plate or cock a valve closed on the discharge piping than to change the pump impeller. Both of these options will end pump runout.

 

[77JQX]

I agree with Littleinch. I suspect your pump efficiency value of 24.5% is wrong. How did you determine that value? I don't think I've ever seen so low an efficiency published at pump run-out, although I admit I don't look at small pumps.

 

[Artisi]

So far lots of academic discussion and even more wheel spin all of which is meaningless without knowing where the pump is running on its curve. Without a pump curve, flow, head, NPSHa etc. nothing can be resolved with any certain.
A curve is available from Goulds website, so OP how about something meaningful.

.

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.)

 

[bimr]

Most electric motors are designed to run at 50% to 100% of rated load. Maximum efficiency is usually near 75% of rated load. Thus, a 10-horsepower (hp) motor has an acceptable load range of 5 to 10 hp; peak efficiency is at 7.5 hp. A motor’s efficiency tends to decrease dramatically below about 50% load.

Overloaded motors can overheat and lose efficiency. Many motors are designed with a service factor that allows occasional overloading. Service factor is a multiplier that indicates how much a motor can be overloaded under ideal ambient conditions. For example, a 10-hp motor with a 1.15 service factor can handle an 11.5-hp load for short periods of time without incurring significant damage. Although many motors have service factors of 1.15, running the motor continuously above rated load reduces efficiency and motor life.

https://www1.eere.energy.gov/manufacturing/tech_assistance/pdfs/10097517.pdf

A power factor of 95% is too high. The power factor should be less than 70% at FLA. You are probably operating at 60% PF.

The motor efficiency is approximately 88% at 50 to 100% load.

Not sure what pump you are operating, but for the Goulds Size 1 x 2-10 in the attachment, you should be operating at approximately 36% pump efficiency.

 

[ScottyUK]

AAgree with bimr apart from the comments about power factor. Your full-load PF should be much better (higher) than 0.7, and a 0.6 PF on a laden motor is abnormally low even for a high pole-count motor designed for low speed operation.

 

[LittleInch]

mjpetrag,

Of course we've forgotten to ask the obvious question - Are you really really sure (i.e. you yourself have witnessed it) that the pump is going round the correct way? Very easy to get it wrong on a 3 phase motor and would explain a lot of the issues raised....

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[Artisi]

Think there are more unknowns than direction of rotation, initial post said 1200GPH for a 2x1x10 would probably be left on the curve, however the OP also said in the initial post that the motor wasn't sized for curve runout, so where is the pump operating?

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.)

 

[GPRnD]

On a 3196 the impeller is threaded onto the shaft, so operating with reverse rotation tends to be short and painful.

 

[Artisi]

Luckily, at 3hp usually not all that catastrophic

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]

GPRnD - Fair enough. Can you download the LF version - I can't get the PSS to work - just keeps dumping me into a Java page

all depends if it is a standard 3196 (the pump curve noted above) where 20 gpm is at the left of the curve or the LF (low flow) version where 20GPM is on the right of the curve...

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[mjpetrag]

http://i.imgur.com/xcjiOpR.png

Both pump curves are attached.

-Mike

 

[mjpetrag]

Actually, the pump was running backward a while ago, but the impeller didn't spin off (surprisingly). However it was with the 9" impeller and it was drawing about 3.5 Amps. I guess the curves of the impeller going "against the grain" cause a high amp draw. I was surprised that it still pumped for a while before tripping out.

-Mike

 

[GPRnD]

LittleInch, sorry you had issues with the Java.

PSS uses it to run but it does seem sensitive to the version of Java installed. All I can suggest is trying a different version.