Motor Running Load Calculation

Well, if nothing else, that was completely humorous.

Not very much, I wouldn't think. The efficiency of the old motor was lower at the low load, but not vastly. If you're just paying per kWh it would take a long time to get your money back, if the motor didn't need replacing as worn out.

But if part of your tariff is based on kVah, you'll have bigger savings. A meter would give you all the data you need to work it out, can you hire one? It would have been nice to check the old motor too, but you can back-calculate for that as (presumably) the output power to the load is unchanged.

Give nameplate data(speed/Amp etc) of both motors. What is the difference between de-coupled and no-load conditions?.

Just in passing, notice that your incremental load was 10A for both motors. I have to go now but, if no-one else has done the maths before I get back, i'll look at it after morning service (which is actually in the forenoon)!

JRaef gave an excellent answer to the question you asked, that is how much power you saved. If you replaced the motor to reduce demand on your power lines, then you're done. If you really are concerned about money saved, then the question would have to include energy saved. To answer that, we need to know the times running loaded, unloaded, and off.

I do agree with the JReaf and very thankful to him. He explains the things very nicely. All his assumptions are matching with my plant condition.

I need more ....

The kw load drawn by 37 kw motor was calculated on the basis of the loading (amps) of 22kw motor.

Is the only way to know the kw of under-loaded motor, like my 37 kw motor, to install energy meter?

Energy meter installed in that section (MCC incomer) shows 90 kwhr reduction per day. Means 90kwhr/24 hrs = about 4 kw reduction in load. I need to run this motor continuously.

Why don't you give the name plate details of both motors?. This is not a quiz competition.

746 Watts = 1 horsepower

Old motor = 37KW = 49.6 HP

And: per your loading numbers of 40A x 415V x 1.76/1000 = 29KW/746W => 39 HP load demand due to motor loss and load dynamics.

New motor = 15KW = 20.1 HP

And: Per your new numbers of 22A x 415V x 1.76/1000 = 16KW/746W => 21.5 HP Load

Unless my calculations are wrong the new motor is at least 1.4 HP too small, will not meet the load requirement, is operating in an overloaded condition where the PF will not matter, and your efficiency is lost.

No. Your formula is leaving out power factor.

you cannot make any calculations of power in an AC motor based only on amps and voltas without factoring in Power Factor somewhere. kW = V x A x PF

And PF will have been low for the bigger motor running light. Indeed, with the same load at the same speed, the HP and hence the actual kW (not the kVA) will have been the same!

Yes, thanks for pointing it out.

I did not factor the PF into the equation as it is actually unknown in either motor's case.

I just wanted to point out that his actual measured running current and voltage values indicate that the application apparently requires more horsepower (KW) than the new motor is capable of producing.

In the name plate of motors they should give current at no-load,25%,50%,75% load and pf at 25%,50%,75% and 100% of load so that consumer don't need to do guesswork or wrong/approximate calculations.

Great Idea, but don't hold your breath 'till they do it...

Yeah, there's only full-load figures on the nameplate, maybe at 2 or 3 voltages and for star and delta connection. He should be able to get the rest of it from motor data sheets though.

In some countries you may not get motor data sheets if you buy from a dealer. When you order from manufacturer or an appointed dealer only you will get technical data in detail.

Used to be able to get detailed data sheets from the manufacturer, for the full range for sales purposes, not for a particular one you'd bought. I haven't tried asking recently but I doubt if they're as forthcoming these days. But you might get it from the manufacturer's website.