Efficiency of a 3-phase motor. 3

[broc028]

Hi everyone,
This is my first time dabbling in the energy management field and I already find myself turning to a forum for some help!

I am trying to figure out the cost of running a three phase motor. I have ended up with:
Cost=Volts*Amps*PF*1.73/1000 * hours*cent/kWh
From what I have read this is good for a general indication of the cost. However would I be correct in saying that the efficiency of the motor is not taking into account and if so, could it be used in conjunction with this formula to give a more accurate running cost.

The motor is powering a conveyor and at the minute I am using the nameplate values.

 

[dpc]

If you are using the actual amps, volts and power factor, you will be computing the power INTO the motor. If you are using motor nameplate horsepower, that refers to power OUT of the motor (shaft power) and the motor losses must be added back in.

If you are using motor nameplate amps and power factor, that will give you the power into the motor assuming it is running at full load, which is generally not a good assumption.

Power factor, efficiency and amps will vary as the motor load changes.

 

[itsmoked]

The best way is a kWhr meter running for a couple of days. Takes all the guessing right out of it.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[jraef]

However would I be correct in saying that the efficiency of the motor is not taking into account and if so, could it be used in conjunction with this formula to give a more accurate running cost

So to be clear, since you say that you are using nameplate data, then yes, efficiency must be accounted for. If you are (or eventually will be) using measured data, then no, the efficiency will already accounted for.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
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[electricpete]

I dunno. Depends on what nameplate data is used.

If used FLA off nameplate and power factor from SWAG !?!), then no need to account for efficiency.

If used hp off of nameplate, then don't need to worry about voltage or amps (if voltage is assumed at nameplate), and efficiency is the only thing you need to account to determine full load input power = Pout / Efficiency with suitable unit conversions for HP <-> watts


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[peebee]

The motor's never going to pull nameplate amps unless it's loaded right on up to 100% of nameplate hp rating.

That's a bigger error than efficiency.

 

[LionelHutz]

Yes, if you're just going to use nameplate data then you might as well just use something like use hp x 0.746 / 0.95 (or actual motor efficiency if you have the number) to calculate kW. Either calculation done by using nameplate data will likely introduce more error than just assuming the efficiency is 95%.

If you actually measure the values in your formula you do not want to include efficiency because you want to calculate the power going into the motor since that what you pay for to run the motor. But then, a meter that will measure volts, amps and power factor should also measure power directly and would likely give a kWh number if you let it record for a while.

 

[sreid]

Remember this, however; Motors are pretty efficient to start with. Suppose you start with a 10 kW motor that's 95% efficient. You are losing 500 Watts. Now buy a motor that's 97.5% efficient. Now you're losing 250 Watts. You can get the same gain by turning off 2.5 100 Watt Light Bulbs.

 

[GregLocock]

Not many motors are 95% efficient. Very few indeed are 98% efficient.




Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[sreid]

Greg,

The math was an trending example. Larger motors are typically about 95% efficient at full load.

 

[electricpete]

True.

http://www.reliance.com/pdf/pdf/aced/W06326-A-A001.pdf

500hp 2-pole 96.7% efficient at 3/4 load

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Eng-tips forums: The best place on the web for engineering discussions.

 

[GregLocock]

Wow, I'm impressed. I suppose one point is that effectively cooling a less efficient motor is more difficult than designing an efficient motor in the first place.

Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[sreid]

One thing to note about suppliers of "Energy Efficient Motors."

1) Use larger diameter wire (less IR^2 losses) and increase the slot "Packing Factor." A slight increase in cost.

2) Use thinner and/or better Electrical Steel for the laminations. Hold the Lamiation Suppler's "Feet in the Fire" while negotiating cost.

3) Improve the Windage loss. Zero cost after re-design.

Charge Premium Prices for the Motors. See all Government Rule Changes as an opportunity to increase profit margins.

 

[Marke]

Hello broc028

The volts * amps * PF * Rt3/1000 gives you electrical input power which is your cost. If you now multiply this by the efficiency, you will have the mechanical output power.
As you are interested in the cost, you can use the electrical input power, measured or calculated using measured or rated values, or, you can use the motor rated power divided by the efficiency.
NB The rated values are at full load. As the shaft load reduces, the input power reduces and the cost goes down. The motor has a high efficiency down to less than half load.

Best regards,

Mark Empson
L M Photonics Ltd

 

[burnt2x]

To do away with inaccuracies in determining by how much your equipment eats (costs when running) try installing a power meter. You can read out from the PM the kWHr consumed (Bill's recommendation), Power factor, kVA, kVAr,per phase amps, per phase volts, per phase kW as well as line-neutral values, maximum values, etc. There are also PMs that offer THD(I), THD(V).
These power meters could easily pay back when used as a load management tool especially if your system is supplied by utilities which charge demand, penalize for low PF,and or employs "time-of-use" (TOU) schemes in their power rates.

 

[itsmoked]

(Bill's recommendation)

?

But you remind me that if they have a relatively steady conveyor loading they could use the plant PM to measure just the conveyor for, say, 30 minutes.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[waross]

You don't have to watch a North American kWHr meter for half an hour to get a reading. Usually less than a minute is required if you use the Kh.
This is the number of Watt-Hours per revolution of the disk.
The Kh value is printed on the meter face. I usually time 10 revolutions and take an average to get the time for one revolution.
3600 seconds per hour divided by the time in seconds for one revolution times the Kh equals the kilowatt demand on the meter.
Quality data at an affordable price.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[abudabit]

Don't forget inefficiencies in the controller. Take your measurements before the controller and not between the motor and controller.

 

[broc028]

Hi guys,
Thanks to everyone who replied (really sorry about the delay).
Some good info here, I should have plenty to work with.
Thanks again!