Motor Loading Question
Motor Loading Question
(OP)
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?
Is there any baseline limit of oversizing a motor with respect to its driven load?
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RE: Motor Loading Question
RE: Motor Loading Question
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
RE: Motor Loading Question
RE: Motor Loading Question
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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RE: Motor Loading Question
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
RE: Motor Loading Question
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?
GO PLACIDLY, AMIDST THE NOISE AND HASTE-Desiderata
RE: Motor Loading Question
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!!!*
RE: Motor Loading Question
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
RE: Motor Loading Question
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RE: Motor Loading Question
GO PLACIDLY, AMIDST THE NOISE AND HASTE-Desiderata
RE: Motor Loading Question
GO PLACIDLY, AMIDST THE NOISE AND HASTE-Desiderata
RE: Motor Loading Question
GO PLACIDLY, AMIDST THE NOISE AND HASTE-Desiderata
RE: Motor Loading Question
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
RE: Motor Loading Question
Does this confirm the validity of the equivalent power factor at 50% loading calculation I presented above as an estimate?
GO PLACIDLY, AMIDST THE NOISE AND HASTE-Desiderata
RE: Motor Loading Question
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
RE: Motor Loading Question
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