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# Calculating cost of running electric motor?

Posts: 2,809
I found an online converter but it said I was at \$2/month and that can't be right....

\$.0995440 per kwh

208v, 2-pole, 12 RLA

How much per day to run this motor? It's a 3/4 horse motor.

Thanks! Tim
Working on steam and hot-water systems isn't rocket science....it's actually much harder.

• Posts: 5,552
Hi Tim...

I think it depends upon the phases.

http://www.abrconsulting.com/Conversions/elec-con.htm#Watts

Single phase is easy. Amps times volts = watts

Multi phase has to account for other things and is not as simple.

ME
It's not so much a case of "You got what you paid for", as it is a matter of "You DIDN'T get what you DIDN'T pay for, and you're NOT going to get what you thought you were in the way of comfort". Borrowed from Heatboy.
• Posts: 2,809
Costs

This is a 3-phase building but motors are 2-pole. We used to call that 2-phase to denote 2-phases of a 3-phase system rather than calling it single phase 2-pole.

I wasn't converting to KWH. I came up with \$.30/ hr, \$7/ day, \$210/ month.

The motors currently (no pun intended) run 24/7 and they want to add thermostats to kill the blower on temp satisfy.
Working on steam and hot-water systems isn't rocket science....it's actually much harder.
• Posts: 4,158
edited February 2012
Watts

746 watts to 1 HP. Regardless of voltage.Ballpark \$1.20 a day.
• Posts: 5,552
edited February 2012
So....

746 * .75 (3/4 HP motor) times 24 times 30 divided by 1000 = 402.840 kilowatts

402.8 X \$0.09554 = \$38.48 per month.

No?

Doing V x A based on the URL (208V Single phase, 2 pole) comes up to around \$178.00 per month

ME
It's not so much a case of "You got what you paid for", as it is a matter of "You DIDN'T get what you DIDN'T pay for, and you're NOT going to get what you thought you were in the way of comfort". Borrowed from Heatboy.
• Posts: 604
Energy conversion

One horsepower is equivalent to 746 Watts, assuming 100% efficiency. 3/4 horsepower would convert to .75 x 746 = 560 Watts at 100% efficiency.

A real 3/4 HP motor might be 80% efficient, so then actual wattage consumption would be 560 W /.80 = 700 W.

Assuming \$.10 per KWh x .700 KW = \$.07 per hour.

So actual cost of operating the motor might be about 7 cents per hour or \$1.68 per day.
• Posts: 4,158
Thanks

Mark....Didn't have a calculator handy.
• Posts: 4,158
?

560/.80 or 560x.80?
• Posts: 2,809
Phases??

Does that take into account 2-phases? (208v)
Working on steam and hot-water systems isn't rocket science....it's actually much harder.
• Posts: 5,552
If we assume a 40% wire to water efficiency for a typical pump...

Then we divide by the inverse (.6) and come up with (\$178.00 /.6) = \$297.00 per month.

Looking more like Tim's numbers.

ME
It's not so much a case of "You got what you paid for", as it is a matter of "You DIDN'T get what you DIDN'T pay for, and you're NOT going to get what you thought you were in the way of comfort". Borrowed from Heatboy.
• Posts: 2,809
Motor duty

It runs an air handler.
Working on steam and hot-water systems isn't rocket science....it's actually much harder.
• Posts: 604
V x A

Mark, how did you come up with that number?

With an inductive load like a motor, you cant just multiply volts times amps without taking the power factor into account.

If you assume an efficiency of 40 % then take 560 watts divided by .4 = 1400 W

That gives 14 cents per hour or \$3.36 per day, or about \$100 per month
• Posts: 2,809
Phases?

So phases don't matter?
Working on steam and hot-water systems isn't rocket science....it's actually much harder.
• Posts: 604
Phases etc.

The type of motor or the number of phases doesn't matter in the wattage calculation. A multiphase motor just draws less current per phase for the same HP output as would as single phase motor.

If you are operating 208V across two phases of a 3 phase circuit, the motor is only acting as a single phase as that's what it sees

What really counts is the wattage consumed, which is what you pay for, which can be calculated from the 746 Watts per horsepower, divided by the efficiency.
• Posts: 604
Phases dont matter

No the phases don't matter in the wattage calculation when using the conversion factor from mechanical energy to electrical energy.

For example, a single phase 3/4 HP motor might draw 4 Amps at 208 across two phases, but an equivalent 3 phase motor only draws 2.4 Amps per phase. In the end they consume the same number of Watts if the load and efficiency is the same
• Posts: 4,158
calculated

the amperage and came up with \$1.78 a day at 100%.
• Posts: 2,809
Costs..

So we have \$1.78 to over \$3.00.

Who should I trust???

Thanks for all the input!
Working on steam and hot-water systems isn't rocket science....it's actually much harder.
• Posts: 604
Best Guess

I would say you would be pretty close at between \$1.50 and \$2.00 per day. Just because you have a 3/4 HP motor installed does not mean that its running at full load. If it is not, it will draw less current and use fewer watts.

Without knowing the actual efficiency of the motor, all these calculations are approximate anyway, and without knowing the power factor of the motor, V x A will not give a really accurate answer.

So to be safe, just call it less than \$2.00 a day?
• Posts: 4,158
lol

Don't trust the converter that said \$2.00 a month.
• Posts: 5,552
Sorry Mike...

That's what I get for taking a gross swipe at the math. I took my base wattage conversion to dollars and divided by the inverse, thinking net profit margin :-)

Tim, blower motors, as Mike pointed out down below, are a different beast. You must test it for amperage under load. Don't trust the nomenclature plate.

ME
It's not so much a case of "You got what you paid for", as it is a matter of "You DIDN'T get what you DIDN'T pay for, and you're NOT going to get what you thought you were in the way of comfort". Borrowed from Heatboy.
• Posts: 604
Motors and Watts

Mark, I think some of the varied answers may be due to Tim's given 12 RLA spec. This might be the locked rotor current rating, not the actual operating current.

I work with a lot of large B&G motors. A single phase 3/4 HP motor at 208V draws about 3.5 to 4 Amps at full load. This current would give answers consistent with the watts per HP method.

Here is a good article which goes into detail about calculating motor wattage loads:

http://www.motorsanddrives.com/cowern/motorterms16.html
• Posts: 3,339
confused

I am now officially confused. I understand that it is complicated figuring 3 phase motors off the name plate.

If I put an amp meter on the running motor.

Will volts x amps x square root of 3 get me there? Haven't the other factors already played out because I am measuring actual draw?

A typical light commercial circ might be 208 volts (120 x 1.73) x 10 amps = 2080 watt (2.08 kw) ?
"If you can't explain it simply, you don't understand it well enough"
Albert Einstein
• Posts: 331

Zman,

Measuring the current draw of the motor will give you the current of the motor.  I know, profound statement.  For a balanced three phase load, phase current times the phase to phase voltage times the square root of three will give you the apparent power.

Apparent power IS related to actual power, but they are usually not the same.

Apparent power is the vector sum of the actual power, which does real work, and the reactive power, which is the power that the load stores for a moment and then gives it back to the power company.  The ratio between real power and apparent power is the Power Factor.

For what you are doing, I think the apparent power is probably close enough for estimating cost savings.  If this was a large motor, let's say greater than 20 HP, then it would make sense to calculate real power.

current X phase to phase voltage X sqr root 3 divided by 1000 = KVA  (KW).

KW X 24 Hrs x cost per KW Hr = cost per day.

Aside from energy cost savings, will starting and stopping the motor lead to increased maintenance costs and shorter motor and connected mechanical load life?
• Posts: 3,339
Thank You

Thanks Larry. I am just trying to calculate savings. That makes sense.
"If you can't explain it simply, you don't understand it well enough"
Albert Einstein
• Posts: 604
Confusion

Zman, as Larry says the line to line voltage times the measured current times the square root of 3 will give the total apparent power.

But in your example, you're using the line to neutral voltage instead. So in your example 120V x 1.73 = 208v line to line.  Then 208V x 1.73 x 10 = 3600 Watts.

If you use the line to neutral voltage 120V,  the apparent power is just 3 x 120V x 10A = 3600 Watts. The total power is just the sum of the individual phase power's if the phase loads are identical and balanced as it would be with a motor.
• Posts: 3,339
confused

So 1.73 is used to convert from line to neutral and then again in the equation?
"If you can't explain it simply, you don't understand it well enough"
Albert Einstein
• Posts: 604
Yes

Yes, 1.73 (square root of 3) is first used to convert line to neutral to line to line and then again in the power calculation.  120V x 1.73 = 208 (line to line)

When using line to neutral directly and simplifying the two equations, 1.73 x 1.73 = 3 so the power equation simply becomes 3 times (line to neutral) voltage times the current.

As an example, a B&G PD-37  3/4 HP 3 phase motor draws 2.4 Amps per phase.

120V x 2.4A = 288 Watts per phase. Each phase draws the same current, so the total power is 288 W x 3 phases = 864 W.  (120V x 2.4A x 3)

You get the same result by using 208V x 2.4A x 1.73 = 864 W.
• Posts: 1
Infographic

I realize this thread is a bit old, but for anyone still reading this, this infographic about calculating motor operation costs might be helpful...

http://www.groschopp.com/energy-efficient-motor-operation-costs/
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