Calculating Electricity Usage/Cost for a 3 phase motor

[JPWHI]

I want to determine the actual electricity usage and cost for a pump with a 3 phase motor. Do I use the sum of amp readings from each phase or the average of the amp reading from the three phases in my calculation?

 

[davidbeach]

You need a power meter. An ammeter might let you make a rough approximation of the kVA usage, but usage and cost is based on kW and you won't get there unless you know the phase angle between the volts and the amps, and the best way there is a power meter.

 

[CJCPE]

Look at

http://www.eng-tips.com/viewthread.cfm?qid=148109

 

[JPWHI]

Thank you

 

[CJCPE]

The best thing to do is to buy, borrow or rent a power meter.

Watts = Volts X Amps X 1.732 X Power Factor. For amps use the average of the amp readings from all three phases, but if they differ by very much there is a problem that needs fixing. You need to know the power factor, the phase angle between the volts and the amps. That might be 0.85 or higher if the motor is running near full load and 0.6 or lower if the motor is lightly loaded. It is difficult to estimate.

If you can get manufacturer's efficiency and power factor data for various loads, you can use the current measurement to estimate load and calculate power from that.

 

[itsmoked]

If it is a simple motor not requiring a bunch of other stuff to come on with it you can try this. Take a watt-hour reading from your plant meter. wait exactly 10 minutes and take another reading. Then change the state of the motor. Turn it on if it's been off, turn it off if its been on. Read the meter for exactly 10 minutes(or some other time period you select, the longer the more accurate). Then just do the math to see the difference between motor on/motor off. This can give you an exact value measured on the instrument that is actually billing you.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[rbulsara]

JPWHI:

To answer your basic question, you should use the average current of the three, if you are using a formula for 3 phase power.

3-ph kW= 1.732*V*A*power factor, where V is line to line voltage and A is ampere in one of the lines. This assumes a balance load , that is equal curent in all three lines. A motor would be such load for all practical purpose.

You can use a reasonable assumption of 0.8 pf for most applications. This will give you a fairly accurate results for your purpose but not precise.

If you really need to get more precise, you need a meter as suggested before. I believe more in accuracy than precision for cost comparisons.

 

[waross]

I was going to suggest using the utility meter, when I read itsmoked's post.
An easier and often more accurate reading may often be obtained using the kh factor of the meter.
It is quite easy on the older electro-mechanical meters. It is more difficult on an electronic meter.
Step one. Time about 10 or 20 turns of the meter disk.
Step two. Calculate the number of turns that the disk would make in an hour.
Step three. Look on the face of the meter for the kh factor. 7.2 is a common value, but it is by no means universal.
The kh is the number of watt-hours for one turn of the meter disk.
Step four. Multipy the kh factor by the number of turns per hour.
Step five. Divide by 1000 to convert watt-hours to kilowatt-hours.
respectfully

 

[DickDV]

Here's another useful estimation method. Rather than measure motor lead current, using a strobe light, measure the amount of slip in the shaft speed. This will be in direct proportion to the nameplate hp. That will give you a measure of how heavily the motor is loaded. Of course, this assumes a reasonably constant load.

Assuming that your electric power costs you 10 cents per kwhr and the motor is running in 24/7/50 duty, each motor hp costs you about $550 per year. If your power costs are different or the duty cycle is different, a simple proportion from the $550 figure will get you a good estimate.

For example: a 100hp motor nameplated 1760rpm running 24/7/50. The shaft speed is measured at 1770 rpm. This is 30 rpm slip on a motor rated at 40rpm slip at full load. Therefore, the motor load is 30/40=.75. That would be 75hp on this motor. 75 x $550 is $41250.

Let's say that your electricity costs are actually 12 cents per kwhr. Take the $41250 and multiply by 12/10 which yields $49500.

I'm sure the purists will have plenty of reason to pick this apart but, for estimating purposes, it is simple and reasonably accurate.

 

[rbulsara]

JPWHI:

I had one error in the formula, it is for Watts.
3-ph Watts, W= 1.732*V*A*power factor. For kW divide W by 1000.

I did not know, figuring cost of energy for a motor was that complicated. Any margin of error corrected by placing a meter etc. will be far undone by estimate of run hours.

 

[VxA]

1) take the average amps
2) multiply by square root of 3,
3) multiply by the voltage (460; or whatever -- from the motor's nameplate).
4) multiply by the power factor (from the motor's nameplate). Assume 0.85 if you don't have that data available (or better yet consult the manufacturer's data).
5) divide by one thousand
6) multiply by the number of anticipated operating hours per year.
7) multipy by the cost of electricity per kWh.

That should get you quite close assuming you have a constant speed motor with a more-or-less constant load. I see no reason to get much more complicated than that.

 

[waross]

Hello VxA;
Your formula works well for fully loaded motors which have the power factor listed on the nameplate.
As the power factor changes with load changes, the calculations for a motor running at less than full load get more complicated whether you see the reason or not.
respectfully

 

[VxA]

waross,

True enough.

I guess the real question is how accurate are you trying to get, and how much of a pain in the *** are you willing to put up with to get there.

With most facilities I would typically do any work in, it would be nearly impossible to use the utility meter method, as it would be nearly impossible to know the difference you are observing at the meter only corresponds to the motor in question (as so many of the other loads in the plant would be varying continuously). I can see where it would be a good idea if it were possible to assume all else equal.

I sort of think that if the motor is loaded somewhere between say 80% and 100% during the vast majority of its time in operation, the power factor can be assumed to be close to the nameplate value (assuming you are given a nameplate value). The increasing ratio of the magnetizing current (which is constant) to the total current would be counteracted by a decrease in the apparent power drawn as a result of the motor's leakage reactance. Granted, these two opposing quantities may not be equal, but again, it comes down to how accurate we are trying to be.

Thanks,

VxA

 

[waross]

Hi VxA
Good points.
The kwhr method using the kh factor has the advantage that you are reading watt-hours instead of kilowatt-hours. I get a lot of nuisance calls every summer from residential users that their power bill is too high. The meter test is a good two-way test. It shows what is being charged for the air conditioner, and if the reading is unrealistic, will show up a faulty meter. In one instance, a check did show up an erratic utility meter and the utility changed out the meter and issued a sizeable credit to my customer.
May I suggest a slight modification of your method?
Calculate the kilowatts with the nameplate current. Assume a power factor of 90%.
Measure the current and multiply the power by the ratio of actual current over nameplate current.
The accuracy of this method will degrade as the load drops, but it should be a usable estimate down to 60% or 50% load.
You may consider adjusting for voltage also. Multiply by actual voltage over nameplate voltage.
For the original poster, I have seen refurbished hwhr meters offered for as little as $8. If you can aquire any used kwhr meter, you need only monitor one phase. This will give you a much more accurate figure than any of the suggested methods of estimation. It will also compensate for actual run time and varying loads in the event of changes in the dynamic head and/or flow rate.
respectfully

 

[JPWHI]

Thanks to everyone