Calculating Motor Operating Costs

[JUSTEINEE]

I was asked to give an estimated running cost of operating a 3 phase 75 HP motor for 80 hrs. I assumed at 480 Vac that the motor would have roughly 100 FLA.

My equation then became KW = 480 V * 100 A * 1.732 / 1000 (ignored power factor and demand) and came to roughly 83 KW.

I had a coworker check to verify my answer and he used a different method. He used 1 HP = .746 KW and came to roughly 56 KW. Does motor phase affect this equation?

Which method is correct when solving for KW in a three phase system and why?

Any insight would be apprecciated, Thanks.

 

[jumper]

430.252 says 96 amps for a 75 HP 3 phase 480 VAC MOTOR

 

[Jraef]

Your "cost of operating" has the word "cost" in there. Cost of electricity is for kW, not Amps. Amps is essentially irrelevant.

So yes, there are 746 watts/HP, that has nothing to do with amps, volts, phase etc. That is the power that is used. You are metered on kWH, (kilo Watt Hours), so to determine the cost of operating something electrical, simply take the watts (kW) x operating time (Hours).

Now, without measuring anything all you can REALLY estimate is the MAXIMUM POSSIBLE cost of operating that motor, because all you have is the RATING of that motor, not a measurement of how much power it is actually consuming. So 75HP x .746 = 55.95kW, x 80 hours = 4,476 kWH, x whatever your electricity rate is will give you the worst case operating cost. It will probably be less than that however, you will only be able to tell the exact amount by measuring.

 

[gar]

110107-2001 EST

You need to know what is the motor load, how constant, and duty cycle.

A useful method would be to measure the KWH for one typical 24 hour period.

If you want a wild guess at full mechanical load, then maybe use 1 KW/HP as an estimate, or 75 KW. Then you need to estimate the number of full load KW per day, possibly 16 hours. At 16 hours per day the consumption would be 16*1*75 = 1200 KWH/day, and if cost is $0.10/KWH the daily cost is $120.

You can change the cost all over the place based upon assumption you make.

.

 

[broadgage]

Note that the stated horsepower of a motor is normally the mechanical output of which the motor is capable.
In this case it is stated to be 75 HP which is indeed about 56KW.
The input must be more than this owing to the losses in the motor, in the absence of detailed information on the motor efficiency, I would allow approaching 10% for losses giving an input of about 61KW.
Multipying 61 by the cost of power per KWH would give the hourly operating cost.

This assumes that the motor is fully loaded, the actual loading may be much less and can only be determined by measurement.

 

[Besoeker]

110107-2001 EST

You need to know what is the motor load, how constant, and duty cycle.

A useful method would be to measure the KWH for one typical 24 hour period.

If you want a wild guess at full mechanical load, then maybe use 1 KW/HP as an estimate, or 75 KW.

I agree with you points on motor loading and duty cycle. But think that taking 1kW for 1 hp is an unduly pessimistic assumption about motor efficiency at the rating being discussed here.

 

[gar]

110108-0834 EST

Besoeker:

I agree that 1 KW is probably high for a motor of that size, but it is an easy number to use for a quick estimate.

There are too many other undefined parameters that a high estimate won't make much difference. Actual KWH measurements over a reasonable time are needed to get good information.

.

 

[Besoeker]

110108-0834 EST

Besoeker:

I agree that 1 KW is probably high for a motor of that size, but it is an easy number to use for a quick estimate.

Quite.
Of course, if the motors were rated in kW instead of the archaic HP, that would then eliminate the need for conversion factors.......
Runs for cover........

There are too many other undefined parameters that a high estimate won't make much difference. Actual KWH measurements over a reasonable time are needed to get good information.

I agree. With just the motor rating alone, all you can calculate is the maximum energy that could be consumed over the 80 hour period.

 

[JUSTEINEE]

Once again

Once again

So everyone missed my point. I guess i will spell out my question.

This is all theoretical. Only want Ballpark answers.

A 75 HP motor at 480 find KW.

If single phase 75HP * .746KW/HP = roughly 56 KW

If three phase 75 HP * .746KW/HP = roughly 56 KW

If three phase and assuming 100 FLA at 480V
KW = Volts *Amperes * Power Factor * 1.732

KW = 480 * 100 *Assume 1 (reality .8) *1.732 = 83 KW or (100 KW if power factor is used)


All i want to know is if you generically use the formual for 1 hp = 746 watts do you need to account for phree phase situation by multiplying by 1.732.

I understand you are only charged for real power usage (KW) unless you are on a commerial site where your rate structure can ding you for you KVA usage (Low power factor = high ding). Amps doesn't matter you just want KW (I was using a generic voltage and Current to find KW).

 

[broadgage]

1 HP=746 Watts regardless of single phase or three phase.
In many cases though the HP is the mechanical output, not the electrical input which must be greater since the motor must be less than 100% efficient.

As a rough estimate I would add about 10% to the output in order to estimate the input.

If the motor is not fully loaded then measurement by use of a KWH meter is the best option.

 

[gar]

110114-1708 EST

JUSTEINEE:

Read broadgage's comments very carefully and understand what is being said.

You have a hang-up on some point that is not clear to most of us.

Power output = power input * efficiency

This has nothing to do with voltage, power factor, three phase, room temperature, speed, the phase of the moon, sunspot activity, or anything else other than as these factors influence the efficiency.

KW and HP are both units of power, the rate of doing work, and there is a linear equation that relates one to the other. KWH and horsepower-hours are units of energy, or work.

.

 

[JUSTEINEE]

"1 HP=746 Watts regardless of single phase or three phase.
In many cases though the HP is the mechanical output, not the electrical input which must be greater since the motor must be less than 100% efficient.

As a rough estimate I would add about 10% to the output in order to estimate the input.

If the motor is not fully loaded then measurement by use of a KWH meter is the best option."

Thanks. Its kinda clearer now.

 

[JUSTEINEE]

My hang up was if 1 hp = .746 kw and 75 HP=56 KW and 56 KW = V*I*PF*1.732 then why does V*I*PF*1.732 not equal 56KW. Equations should be interchangable.

Single Phase

KW = I x E x PF/1000

HP = I x E x Eff x PF/746

HP = KW x Eff/746

Three Phase

KW = I x E x 1.73 PF/1000

HP = I x E x Eff x 1.73 x PF/746

HP = KW x Eff/746


Whether Single phase or three phase HP = KW x EFF/746

I suppose the differences in our assumed numbers were due to Efficiency and Power factor. If i used a power factor of .67 i would arrive at the same answer of 56 KW.

I suppose from now on i will just use the 1 hp = .746 KW for my calcs.

 

[Finite10]

Without informatics from a recording data logger I'd probably go at it with FLC like this, if it were me;

Table 430.250 Squirrel cage(?) 460V 75hp = 96A FLC
96A x eff (* footnote says PF 80% = 125%, PF 90% = 110%) = adjusted FLC

adjusted FLC x 480V = W (real, not apparent)

W/1,000 = KW

KW is a rate of power, KWH is a rate of power use over time (t = 1hr), Kw/hr. 1 KW for 1 hr = 1 KWH.

KWH x Cost of power from utility per KWH

 

[gar]

110114-1941 EST

JUSTEINEE:

I think I now see your hang-up.

From your post #13:

HP = KW x Eff/746

You continue to relate HP to the motor output power and WATTS to motor input power. Erase this concept. Horsepower and watts are the same kind of animal, they are are both power and related to each other by the unit conversion equation
W = 746 * HP .

Your equation is wrong unless you write it as
Motor Output HP = Input Electrical power * Efficiency / 746
If you do not explicitly define your variables, then the equation can mean all sorts of things.

When you use a variable, such as HP, that has the implication of power, and relate it to some other variable, such as W, that also implies power, then this implies a conversion factor formula relating the two units.

At either the output or input of the motor you can express the power at that point in WATTS or HP or ERGS/second or many other units.

Your confusion seems to arise from standard usage in the US where we typically describe the power output of a motor in HP and the motor input in WATTS ( more likely VA * power factor ) and thus you are confused in the meaning of these units.

.

 

[Jraef]

HP = V x A x PF x Eff x 1.732

75HP = 480 x 100 x .8 x .9 x 1.732 = 59.8kW

or

75HP x .746 = 55.95kW

Close enough to account for margins of error.

As everyone has said, kW is kW, makes no difference if it is 1 phase, 2 phase, 3 phase or 17 phase. You were coming up with a grossly higher number because you were leaving out part of the equation.

 

[kwired]

Without informatics from a recording data logger I'd probably go at it with FLC like this, if it were me;

Table 430.250 Squirrel cage(?) 460V 75hp = 96A FLC
96A x eff (* footnote says PF 80% = 125%, PF 90% = 110%) = adjusted FLC

adjusted FLC x 480V = W (real, not apparent)

W/1,000 = KW

KW is a rate of power, KWH is a rate of power use over time (t = 1hr), Kw/hr. 1 KW for 1 hr = 1 KWH.

KWH x Cost of power from utility per KWH

That footnote applies to the columns for the synchronus motors not a squirrel cage induction motor.

 

[Finite10]

That footnote applies to the columns for the synchronus motors not a squirrel cage induction motor.

Thanks kwired, I whiffed that eh?

It's also easy for me to miss what gar posted when I don't slow down;
"When you use a variable, such as HP, that has the implication of power, and relate it to some other variable, such as W, that also implies power, then this implies a conversion factor formula relating the two units."

What's a good field book for motors? I hear Dewalt does a good job but never used one. Ugly's style is a little hard to find things in, but it is compact for the tool bag.

 

[kwired]

Thanks kwired, I whiffed that eh?

It's also easy for me to miss what gar posted when I don't slow down;
"When you use a variable, such as HP, that has the implication of power, and relate it to some other variable, such as W, that also implies power, then this implies a conversion factor formula relating the two units."

What's a good field book for motors? I hear Dewalt does a good job but never used one. Ugly's style is a little hard to find things in, but it is compact for the tool bag.

I use mostly Square switchgear, breakers, controls. Ask a distributor for a motor slide chart calculator. I don't have any new ones but would guess they still have them.

You slide the chart to match the motor horsepower and then follow down to the correct voltage. Here it tells you full load current (based on NEC tables), minimum copper wire size (75 deg), recommended thermal magnetic breaker trip setting, Catalog number of of suggested fusible switch, size of dual element time delay fuses needed, catalog number of NEMA magnetic starter, and catalog of thermal overload elements. (I don't like to use the overload selection because using the actual load marked on the motor is the best way to protect it) It has all standard motor sizes from 1/2 - 200 hp three phase and turn it over and the other side has 1/6 to 10 hp single phase.

I have seen similar slide charts from other manufactureres in the past. Is faster than using any book I have seen. Especially if designing a single circuit install.

 

[gar]

110115-1618 EST

Following is some data from an "Induction Motor Performance Sheet" for a 35 HP 1800 RPM 3 Phase 440 V motor from Bailey and Gault.

[FONT="Courier New"]
Approx load           0.25       0.5        0.75        1.00        1.25

Stator current          17        25          34        43.5          54
Fixed loss            1960      1960        1960        1960        1960
Total stator loss     2062      2183        2372        2636        3000
Input to motor        9850    17,000      23,900      31,200      38,500
Output                7710    14,561      20,921      27,484      33,814
Horsepower            10.3      19.5        28.0        36.8        45.3
Efficiency           0.782     0.856       0.875       0.880       0.870
Power factor         0.758     0.891       0.924       0.924       0.938 [/FONT]

Note: All power is in watts except for the one line in horsepower, and that converts exactly using 746 watts per horsepower. Also the fixed loss is 1960 W. Stator loss at no load is about 10.5*10.5*0.35 = 39 W. So no load power consumption is about 2000 W.

The answer to the original question posted, Calculating Motor Operating Costs,

I was asked to give an estimated running cost of operating a 3 phase 75 HP motor for 80 hrs. I assumed at 480 Vac that the motor would have roughly 100 FLA.

Ignore this type of assumption.

Work from the 75 HP. Determine what can be an estimated load profile.

The simplest is full load. Using this as the assumption, then determine the efficiency at full load. If not available, then estimate from other comparable motors.

As an example use the efficiency from my above table of 0.88 . And calculate output power as 75*746 = 55,950 W. Using efficiency the input power is 55,950/0.88 = 63,580 W.

For 80 hours at full load and $0.10/KWH the result is 63.58 * 80 * 0.1 = $508 .

Also an estimated cost to idle the motor is about 5.2 % of full load, based on my above chart, or about 63.58 * 0.052 * 0.10 = $3.31/hour.

If you have a complex load profile, then it is necessary to break it into pieces, and calculate these separately and add together. Quite likely a watt-hour meter would be the better approach.

.