Motor Energy calculation... 4

[cfordyce]

If I have a 7.5 hp 575VAC motor runing all day and I know what the cost of electricity is (12 cents US per KWH) can I roughly estimate my cost by simply multipying the motor power rating (7.5 HP) x (746 W/HP) x (24 h/day) x ($0.12 /KWH) = $16.12/ day operating cost???

U have to know the motor load though, right? cause it may not draw its rated current under normal operations.. is there a % used to guestimate???

 

[dpc]

That will get you pretty close. You also need to account for the motor efficiency since the 7.5 hp is based on the mechanical output power. So your electrical input power is more like (7.5 * .746)/.9 or about 6.2 kW. The power factor of the motor will create some additional resistance losses in the system due to the higher current, but this will not be of the same order of magnitude.

The load factor can vary widely - from 20% to 110%. I don't think we can be of much help in that area unless we know more about the load.

Hope that helps.

 

[electricpete]

I think dpc covered it pretty well.

If you want to determine load, you can attempt to measure speed OR current OR [flow rate AND dp...assuming this is driving a pump].

If you obtain a current measurement, it would also help to know the nameplate full load amps and the nameplate rpm.

The ideal situation is obviously to measure electrical power with a coordinated current, voltage, phase-angle measuring device...I'm guessing that's not practical.

 

[jbartos]

Suggestion: It is also good practice to check the motor terminal voltage, which should be close to the motor nameplate voltage. If the motor terminal voltage is lower, e.g. 10%, the motor efficiency will be lower. This means the higher cost of operation as indicated in above postings. Voltage Total Harmonic Distortion (THD) greater than 5% can also cause the motor to run less efficiently due to higher heat losses RI**2 (Watt losses), and potentially reduce the motor life cycle.

 

[cfordyce]

Thanks alot guys ( or girls ). I believe that the motor is driving a pump for a new steam boiler they are looking at purchasing. Would it be safe to assume that the designers of the boiler would have sized the motor apppropriatly for the load and that it will be running fairly close to full load? So my calculations would be a decent guess...

 

[electricpete]

According to EPRI, typical design practice for "large power plant motors" for about 133% of their "max" (full plant power) load which also happens to be their normal steady state load (baseload plant). For a motor sized that way it's safe to assume the average motor load would be about 1/133%=75% of nameplate. If you have a less predictable load which varies more above and below it's average, I would think the motor would be oversized further above it's average load (perhaps 200% of average load?) and the average motor load might be on the order of 50% of motor nameplate. There also might be several auxiliary power plant pumps which would fall in this category.

It's a wild guess... as dpc mentioned the wide range 20-110% surely might be encountered in the real world.

 

[REDDOG]

Do not forget the monthly DEMAND CHARGE.
Assume 1kw input per HP, Demand charge (7.5 KW)($10.00/KW)=
$ 75.00 to start the motor for the monthl. Total electric bill would be the sum of KWH plus the KW Demand Charge.

 

[peebee]

Your $0.12 energy cost sounds high - you might want to check that. I believe .05 to .08 is more typical.

 

[cfordyce]

Your right about the electricity cost being too hight... should be .06 right now.. thanks.

 

[jbartos]

Suggestion: Regarding pump-motor set and motor selection visit

http://www.tifac.org.in/offer/tlbo/rep/st152.htm