Motor Loading Question 6

[nightfox1925]

What are the disadvantages if a motor is running with a significantly less Mechanical Load? For example, a 50HP motor was out of service and what is available at stock is a 100HP motor and this will be used to run the load driven originally by a 50HP motor assuming all other parameters are the same in terms of speed, frequency and voltage.

Is there any baseline limit of oversizing a motor with respect to its driven load?

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[davidbeach]

You'll get a really low power factor and the system losses associated with that power factor, but it won't harm the motor.

 

[itsmoked]

Nope. Won't hurt a thing. Except your pocket book!!!

It will require a substantially more expensive motor.
A larger more expensive starter.
A more expensive breaker.
Larger contactor.
More space for the motor.
More load carrying ability for mounts.
Larger wire.
Bigger conduit, possibly.
It will also hit your power system with a bigger inrush. Dim the lights more because it will hit your power network with a bigger electrical disturbance.

Your solution is really um..ah. Lame. I could see if you needed a 34.7HP motor and the closest was a 40HP so be it. But 50HP is a very standard common motor size. Could this not be purchased at about 50 different places?

Find a useful vendor?

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[jraef]

I wouldn't say "won't hurt a thing" without qualifying that when you go to start it, it will START like a 100HP motor, not a 50HP one. Breakers may trip, fuses blow etc. But perhaps more importantly, your wire size will not likely be legal any longer. The Code makes no allowances for loading of a motor.

 

[nightfox1925]

In one of the projects, there several product transfer pumps pumping liquids from different storage tanks. These pumped liquids flow to associated day tanks located near the loading platforms wherein these liquids from different day tanks are discharged to smaller pumps for loading purposes. The client wishes to demolish these day tanks and associated discharge pumps and connect the loading platforms directly to the storage pumps. Our process people are in the event of conducting hydraulic calculations for a new pump station between the storage tank farms and the loading platforms. They are intending to re-use the same transfer pumps if possible. If in the event that what come out as a HP requirement is lower than the transfer pumps per calculation, I am expecting them to suggest to re-use the same transfer pumps which would be the same case scenario as the one I posted earlier.

Based from your feed backs, the client would be paying like running a 100HP for a 50% loading due to system losses as a result of low power factor. Higher maintenance cost as well, since these motors are pretty old.

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[itsmoked]

Thanks for further info.

Your conclusion is NOT precisely correct though.

You would be paying only for the power used, which may still be 50HP.

It would be more only if the power factor increases enough to be a problem to your utility, if they actually care about that.(They might - they might not.)

More correctly you would be paying to install larger pumps. Not actually much more to run them.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[nightfox1925]

It would be more only if the power factor increases enough to be a problem to your utility, if they actually care about that.(They might - they might not.)

Operating a motor at low power factor will have an effect to the overall system power factor...we will check on that with respect to the other loads in the system.

These motors are already existing hence, existing starters are already available.

I have another question, if we operate our 100HP at 50% loading (50% load factor), we get the demand kW rating of 0.5 x 100 x 0.746 = 37.3kW, is the equivalent kVAR (power used to magnetize the motor circuit) of the motor constant? If this true, then at 50% loading, I will be getting a Power Factor of approximately = cos(arctan(38.63/37.3))= 0.69. Any comments?

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[edison123]

Keith

A LPS for your posts which covered just about everything about the op's query.

*Even inside a hard-boiled egg, there is a golden heart!!!*

 

[itsmoked]

Thanks Edison!

Fox; I understand the situation now. If the motors are old and you're going to do a bunch of new work/remodeling to re-arrange the whole place I'd probably just buy new motors. Otherwise, yeah, if you have the motor starters 'in hand' and all this would amount to just a bunch of conduit and wire thrashing then I'd go ahead and use the bigger motors. The payback even considering maybe a poor PF charge would be a long time replacing a bunch of $3500/50HP motors I suspect. If just the PF is a hassle, or becomes, you can add capacitors to correct it.

Remember the motor takes power to 'do its thing' rotate at the designed speed. Bearings, fan, windage, I²R losses, I²R losses due to increased current from reduced PF. So even if you aren't pulling the full load from the motor you're still paying for theses losses. Hence the efficiency hit.

If these pumps run all day you may want to do the numbers on reduced eff as that could add up in a year or two.

Someone want to address the kvar question please?

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[electricpete]

If you neglect the series leakage reactance of the equivalent circuit, then the KVAR drawn does not change as load varies and the calculation suggested above is approximatley correct. In fact the KVAR drawn from the power supply increases slightly as load increases due to the series leakage reactance. So if you took full load power factor and tried to estimate 50% power factor assuming constant kvar, you would overestimate the kvars and underestimate the powerfactor.

=====================================
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[nightfox1925]

Is there any empirical formula to estimate the equivalent power factor at reduced loading say 50% loading?

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[nightfox1925]

I agree with you electricpete. The % Power Factor at different loadings are tested and published in the motor data sheets as 50%, 75% amd 100%. By the way, one more important point for me to see in my application as a result of re-using a higher rated motor is the fluid hammering effect in addition to the losses due to lower efficiency (to my client's electric bill).

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[nightfox1925]

Thank you guys for the vital information and commentary. About the KVAR, P.F. if there is a way to calculate them at reduced loading for estimation purposes would be a great info for me.

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[Marke]

Hello nightfox1925

I would just like to confirm some of the statement made above.

1) Irrespective of the connected load, a motor provides a "load" on the supply during start that is dependent on the motor size. If you used smaller motors, the starting draw would reduce for the same driven load.

2) The magnetizing current of the motor is essentially independent of load, so the amount of power factor correction required is a function of the motor characteristics, not the load. - A larger motor will require more correction.

3) The efficiency of a larger motor at less than full load can be higher than the correctly sized motor at full load. - An older motor is probably going to be a lower efficiency than a modern motor. For two equal motors of different sizes, there is not usually an efficiency penalty by using an oversized motor unless it is significantly oversized.

The biggest issue that I would consider, is that as the motors are older, it is possible that the lower running efficiency (due to older design rather than size), may increase the running costs appreciably. Using newer designed motors (higher efficiency) may actually give a reasonable payback in energy savings.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[nightfox1925]

2) The magnetizing current of the motor is essentially independent of load, so the amount of power factor correction required is a function of the motor characteristics, not the load. - A larger motor will require more correction.

Does this confirm the validity of the equivalent power factor at 50% loading calculation I presented above as an estimate?

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[waross]

Use a power triangle to estimate power factor.
The KW will be the base of a right triangle. The KVAR (kilovolt amps reactive) will be the altitude and the KVA will be the hypotenuse. The KVAR or altitude of the triangle changes very little and can be assumed to be constant for estimations. As the KW or base of the triangle is changed it is easy to see the effect on the KVA or hypotenuse.
Remember that power factor is KW/KVA.
As an example, a motor running at 90% PF will have the ratios of KW-100, KVA-111.1 (100/0.9), and KVAR 48.
At 50% load the KW will be 50, the KVAR will still be 48 and the KVA will be 69.6 The power factor will now be 50/69.6=71.8%
If the power factor is corrected to 100% it will remain at about 100% regardless of loading. If it is corrected to less than 100% the percentage power factor will vary and must be calculated for each loading.
respectfully

 

[nightfox1925]

Thanks waross, that was very helpful as well.

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