KWH calculation

[mull982]

Hi all!

I was asked to calculate KWh values for all the motors within our plant. I figured I'd go about calculating the values by gathering the data on each motor then multiplying the Volts x Amps to get the KVA of each motor. Once I had the KVA for each motor, I was going to then multiply it by the power factor of the motor to get the total KW consumption of the motor. Once I had the KW for the motors I would then multiply this by the hours (was asked for consumption for a year with a runtime of 92%) to get my total KWH for each motor over the course of a year. Is this the correct way for going about calculating the KWh for each motor over the course of a year.

Does the 92% runtime mean to take 92% of the hours in a year to use as my number of hours for the motor running?

One las question, I cant find a Powr Factor for two of the motors which are .75 h.p. Is there an average or ideal power factor I can use for these motors in my calculations, since I cant find the actual power factors.

Thank you

mull982

 

[kingpb]

The 92% would be for whatever the maximum runtime would be. So, if the plant only operates on a 8-5 shift, M-F then it would be 92% of that, minus holidays, etc. I think 92% is used becasue you will have, on avg, some downtime for maintenance. Obviously, if you operate 24/7/365 the total hrs are different.

Typically, a good averge pf is 0.85, for those you don't know. You can also look in a manufacturers catalog for a replacement motor and use that value as well.

If your gathering all this data, you may want to update your one-lines as much as possible at the same time.

 

[suemarkp]

Another variable is the motor amp load. In some cases, it will draw what the nameplate says. But the amp draw is related to the load you're putting on the motor. Pump motors will vary because of the amount of pressure they have to work against (site install specific). Saws only draw their FLA when cutting wood (or perhaps a little over if you force it) and much less when just spinning the blade. So consider the task the motor is performing and the duty cycle of that task (e.g. a sawyer spends 40% of his time cutting and 60% moving wood around before and after cutting, which means the amp draw is the idle value 60% of the time and the loaded value 40% of the time).

So if you can, I would try to measure the amp draw of the motors instead of using the nameplate value. Or at least do this for each category of motor you have and see how actual compares with nameplate. You didn't say what kind of plant this is, so I don't know if this will be a major issue.

 

[motorcontrol]

Mull982,
I agree with suemarkp. Motor does not always run @ FLA. I do not know what type of plant or how many motors you are trying to calculate but it may be fessible to look into kwh meter. I do not know what do they cost...Some meters will read pf too...

 

[iwire]

Ditto on the real vs motor tag power consumption discrepancy.

Motors are not usally run at max capacity and you will only see the FLA current if the motor is producing 100% of it's rated HP.

Sort of a power meter being installed you could take voltage and current measurements at the motors using a True RMS meter. Of course the redding will need to be taken while the motor is in it's typical operating condition. Warmed up with the normal amount of mechanical load.

 

[motorcontrol]

If the load is repetitive everyday then he can use each days data & use it as average per day.....

 

[jmonty696]

During my last 25 years of collecting kwh readings for extruder, motors and heater, etc, the only good way that had any accuracy to it was to use a kwh meter that you could connect up and leave it there for at least 15 minutes or until the motor cycled as many times as practical. Air compressors and Chiller compressors vary in load so much and depending on how many pieces of equipment are running you need some good cycle times. Grinders such as in plastic bottles manufacturing generally surprise many people when I say that that 30 or 50 hp grinder only uses on the average of 5 to 10 kw per hour. The HP is for torque when needed to eliminate lock ups.
If you calculate the motor kw without actually measuring the average kw it may be tough later to go back to the person asking for the kw load of the motors and if he calculates the total plant kwh usage the motors will generally overextend the actual billing.
I purchased a pretty good kwh meter that does kw, kwh, pf, kva, kq, volts, amps, for single phase and for three phase and it cost $2000. I use it constantly in my travels around to my different plants for energy conservation and calculating ROI to help open managements eyes to how much they are wasting and the bottom line.
Good luck and have fun while do it with confidence of real power measuring.

 

[derek22r]

Why is it not accurate to calculate the KW by simply multiplying the hp by .746.

 

[Bob NH]

Why is it not accurate to calculate the KW by simply multiplying the hp by .746.

1. The load is hardly ever loading the motor to the nameplate horsepower, so the motor is not using rated power.
2. The motor never runs at 100% efficiency.

Motors are sometimes (usually in metric countries) rated in kW shaft power where 1 HP = 0.746 kW shaft power, but that is not the same as kW electrical power used by the motor.

 

[rattus]

Three Reasons:

Three Reasons:

Why is it not accurate to calculate the KW by simply multiplying the hp by .746.

1. Motors are not 100% efficient.

2. Motors typically do not run 100% of the time.

3. Motors are seldom loaded to 100% of their ratings.

 

[jim dungar]

Why is it not accurate to calculate the KW by simply multiplying the hp by .746.

Horsepower is a measurement of output power. If you multiply HP * .746 you will end up with output watts.

Calculating input watts require you to know the efficiency and the loading of the motor.

As Rattus has alluded to, the efficiency of a motor is heavily dependent on it's loading.

 

[LarryFine]

Yes, but if a pair of otherwise identical motors, one single-phase and one three-phase, can create identical horse-power while consuming identical electrical power, there is no clear advantage of three-phase over single-phase; they are of equal efficiency.

That's my take on it, anyway.