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power consumption and cost of operation for a 3phase motor

Teto (Mechanical) (OP) 
9 Apr 06 10:25 
Hi all, I'm here from the mechanical side. Need help on understanding power consumption on a 3 phase induction motor. I was trying to calculate electrical cost of operation on a 3 phase induction motor. I used the formula Kw=V*A*1.73*PF/1000 for my calculations out of a electrician's pocket reference book. I then multiplied my Watt hours (hours of operation) by our utiliy's Kw rate. In this case I have 208V and 24 amps. As I fiddled with the numbers in Excel and I asked around for a PF value I could use, It became apparent to me that I may not have a good understanding of how to calculate it, cost of operation. I say this because when I inserted a low PF value as opposed to a value closer to unity, I get a lower cost. If true, at 11.5 cents/Kwhr., ths motor will use 4.3KW at .5 PF and 6.9 KW at .8 PF. From what I remember reading, this cannot be right, as I thought the further away you get from unity (lower power factor), the more it should cost to operate the motor. Do I have this right or have I missed something? Also, if the power factor is unknown, is .8 good to use? One electrician said .15, that's not right is it? Thanks for looking at this, Teto 

The formula is correct, but meaningless for your application. The kW (not Kw) of the motor is determined by the load. At a given operating point, the motor will have a certain PF and efficiency; with the PF, efficiency, and voltage, you can calculate the current. A reduction in PF or efficiency will cause an increase in current.
Metered at the terminals of the motor, the PF of the motor won't affect the operating cost, but the efficiency will (lower efficiency means more power in for a given power out; lower PF only means more current in for a given power out). Metered further away from the motor, the higher current associated with lower PF or efficiency will cause greater I^{2}t losses, which result in higher operating costs. Lower PF can also result in PF penalties from the utility. 

Teto; If you know the kW value of the load and its daily hours of operation you should be able to turn those into kWhrs then multiply by your utility rate. Or are you reading an ammeter on a particular motor and trying to find its specific cost? With respect to your PF confusion. Think of it this way: You measure the current going to a motor with a clamp on ammeter. It measures ALL the current going into that motor. But depending on that motor's PF only a percentage of that current is going to production of useful mechanical energy output. Or as it is called 'kW'. So the worse(lower) the PF the less the percentage of the measured current is going in kW production. So the kWatts drop, sometimes dramatically with really poor PF. To continue the story, traditionally the power companies only charged you for the "work" energy they supplied(exactly the kWatts). In recent years they have decided that, "heck we have to supply bigger transformers, larger wires, and bigger generators" to supply those 'extra' nonkW amperes so why not charge for them. (As annoying as this is, it actually makes sense.) So even though you're thinking, "gee this motor draws less kW then I thought", with worse PF; your bill may be going up rapidly because your utility may charge a PF multiplier penalty. I believe most are and if not yet soon will be. Keith Cress Flamin Systems, Inc. < http://www.flaminsystems.com> 

Teto (Mechanical) (OP) 
9 Apr 06 16:07 
Keith, I was using the name plate rating of RLA . There is no FLA on the tag. It is an air conditioning unit. I was trying to furnish an idea on the cost of operation for the customer on a worse case scenario.
So then, should I calculate HP first with: Hp=(V*A*Eff*Pf*1.73)/746 then calculate KWhr with: KWhr=(Hp*(.766*(hours))/746 then the cost by: cost=KWhr*$/KWHr
If so, I still have a problem in that the specs do not show either the Pf or the EFF of the compressors. Any suggestions? Thanks Teto


waross (Electrical) 
9 Apr 06 16:17 
Hi Teto I'll try to restate the advice and information that davidbeach and itsmoked have provided. "Kw=V*A*1.73*PF/1000" This formula ignores efficiency. Probably 90% or less. Give the book to the electrician that gave you .15 as a typical power factor. They deserve each other. You pay for the energy that you use. The energy is a combination of work done,(Actual horspower used, not the motor rating.), and the losses in the motor (Heating). The power factor is based on the magnetizing current demanded by the motor. The magnetizing current is at a phase angle of 90 degrees to the voltage and cannot be added directly to the current that is doing work. These two currents form the base and altitude of a right triangle and may be added vectorily with Pythagoris' theorem. This combined current (hypotenuse)is what you measure with your ammeter. The base divide by the hypotenuse is the power factor.
The magnetizing current is relatively constant. The in phase current varies as both the load and the losses vary. As a result, the power factor varies as the load varies but not proportionately. At no load a motor may have a power factor of .15 or .2 . At full load the power factor may be .9 As to your spread sheet confusion. By changing the power factor in a spread sheet, you are inadvertantly changing other factors. When you change the power factor, you are implying that the work done by the motor is changing. By using a low power factor you have implyed that most of the current is magnetizing current and comparatively little work is being done. With very little work being done, the energy consummed and the cost are low. Between davidbeach, itsmoked and myself, I hope we have helped explain why your calculations were confusing.
You are probably still wondering how to estimate the cost of running the motor. Method #1. Estimate the actual load in horsepower on the motor. If the current draw is close to rated current you may use a simple ratio of horsepower times actual current divided by rated current to estimate the actual load horsepower. The less the load, the more inaccurate this method is. Divide the estimated load by 3/4(HP to KW) and by .9 or .8 (Efficiency) and the result is kilowatts.
Method #2. It is some times possible to turn off everything except the motor under test. If so, go out to the watthour meter. Time a revolution of the meter disk with only the motor running. (I usually time 10 revolutions and divide by ten.) Your answer will probably be in seconds. Divide this figure into 3600 to determine revolutions per hour. Now look for a factor labelled Kh on the meter dial. This is the watthours per revolution. Dvide by 1000 to get watthours. This is the number that the utility will base their charges on. respectfully 

jraef (Electrical) 
9 Apr 06 17:18 
Assuming that your AC unit does not have a HP or kW rating on it and you have only the RLA to go by, then your formula is somewhat correct other than the fact that PF doesn't factor into your pure kWh number. HP and kW for motor power is either a mechanical rating, so the V*A*PF*Eff is a way to determine the mechanical power of the motor when all you know are electrical facts, or conversely, a way of determining electrical power consumption if all you know are the mecahnical facts (torque and speed). For the purposes of billing however, all you really care about are the electrical facts, measured voltage and amperage at expected load. There are a couple of issues however. 1) In case you didn't notice it when pointed out in the other posts, you are calculating kW, converting to HP, then reconverting back to kW again to get to kWh!. Just leave your calcs in kW, forget the "746" steps. 2) Technically there is no difference between RLA and FLA except as follows. Motor nameplates are supposed to show the load rating of the motor in HP or Watts (kilowatts), and in that case RLA or FLA is a rating of the Amps when at full load. Some OEM equipment mfrs however don't like to do that because customers will mistakenly use that HP information to judge them against competitors, who may be using a larger HP motor that is not really doing any more work than their smaller one because it isn't fully loaded. So if their equipment essentially always runs at the same load, such as AC units, they compromise by showing only the RLA as an actual measured value done by the equipment mfr. If that is the case in your AC unit, then for your purposes it isn't necessary to factor in the Efficiency to get HP, all you need is voltage and RLA to get "billable" power, or kW, then operating time to get kWh. The losses of efficiency are already accounted for in the measurement of actual RLA, and power factor may or may not cost YOU anything. Power factor is certainly an issue when it comes to your utility bill, and therefore technically should be included in a cost of operation calculation. That, however, is also dependent on whether or not your site's overall PF is poor. In some cases it's corrected in bulk so that you are not in fact penalized. In other areas the utilities don't care and don't charge you for PF, such as it is here with PG&E in California. EngTips: Help for your job, not for your homework Read FAQ731376 

I agree with jraef no eff number comes into this at all. However jraef this is what I am gathering from a read of your response, Tito should use: Cost = [V*A*1.73/1000]* h * $/kWh In his case: Cost = [V * RLA * 1.73/1000] * h * $/kWh Specifically: Cost = [208 * 24 * 1.73/1000] * h * 0.115/kWh Why not assume PF = .85? And do this: Cost = [208 * 24 * 1.73 * .85/1000] * h * $/kWh ???? Keith Cress Flamin Systems, Inc. < http://www.flaminsystems.com> 

waross (Electrical) 
9 Apr 06 22:40 
Hi Guys; Sorry about the efficiency, I stand corrected.
"Why not assume PF = .85?" We don't know if the 24 amps is measured or rated. We don't know how heavily loaded the motor is. 85% should be close enough, if the 24 amps is nameplate amps and the motor is fully loaded. If it is measured amps we have to know rated amps or rated horsepower to estimate the percentage load and the resulting power factor. Teto, What is the motor driving? Is the 24 amps measured or rated? respectfully. 

Hey waross the OP stated air conditioning. The actual load on an A/C motor moves all over the place from 110% down to 25% depending on the head pressure of the moment. Since the actual HP demand is far from constant, the best estimate would probably be the RLA and a guess at the PF. Keith Cress Flamin Systems, Inc. < http://www.flaminsystems.com> 

Maybe you could look at it from the perspective of the unit's size. A one ton air conditioner is 12000 btu/hr. 12000 btu/hr is 3.51 kW.
I agree with Keiths last formula. His guess on power factor is probably close enough too. If you want to assume worst case then use a power factor of around 0.88 to 0.9 and you'll get an even higher kW number. Not including efficiency in that calculation is correct. Efficiency measures the power lost in the system but you still have to supply this power.


Teto (Mechanical) (OP) 
10 Apr 06 10:09 
I LOVE THIS FORUM!!! LionelHutz's approach makes it so much easier. So would it be applicable between single and three phase? I worked a few scenarios and the KW was not that much different. Thanks Teto 

Yes, power is power. But, I believe the efficiency of a 1phase motor is a fair bit lower than a 3phase motor and the BTU calculation doesn't include motor efficiency. A 3phase motor typically reaches at least 95% efficiency so it's not much for extra loss but a 1phase motors losses may be more significant. Me not thinking it through didn't tell you to add another 5% or so for motor losses. In the BTU case you have output power which is input power  losses.
At any rate, it sounds like your calculations are jiving so you're likely producing about as good an estimate as possible without actually knowing more info or just measuring the power.


CJCPE (Electrical) 
10 Apr 06 12:03 
If you look in the National Electrical Code, you will find tables giving typical full load currents for various types of motors. For a 208 volt, threephase motor, 24.2 amps is listed for a 7.5 Hp motor. You might estimate that your air conditioner represents a 7.5 Hp load. From the condensed version of NEMA MG1, you can find that the efficiency of an energy efficient 7.5 Hp motor should be about 88%. The kW load could be estimated to be 7.5 Hp * .746 /.88 = 6.3 kW.
I don't think you can estimate energy use by converting the BTU rating of the air conditioner to kW. The BTU rating tells how much heat the unit can move. That requires less energy than it takes to actually produce that much heat. 

CJCPE; I agree! Going with the TonsbtukW scheme will have you missingbyamile. If you are going to mess with this any more than my last posted equation, STOP! Go get a clampon watthour meter and let it run for a month. There will be no question about anything then. Keith Cress Flamin Systems, Inc. < http://www.flaminsystems.com> 

Teto (Mechanical) (OP) 
10 Apr 06 14:28 
Ok, I'll use itsmoked's last equation for KW per hour. I understand now that when I fiddled with Excel, I ignored the change in other parameters therefore the confusion on PF. Thanks to all , I always appreciate all the postings on this forum. Teto 

waross (Electrical) 
10 Apr 06 14:50 
I went through the posts several times and somehow missed Teto's second post every time. I apologise for any confusion I may have caused. (P.S. Keith, where can I get a copy of the official apology forms??) I'll support Keith's formula. respectfully 

jghrist (Electrical) 
11 Apr 06 13:59 
I don't think the single phase question got answered. For single phasetoneutral load, you would use the phasetoneutral voltage and not use the 1.73 multiplier:
Cost = [V_{ØN}*A*PF/1000]* h * $/kWh
Note that the phasetophase voltage is phasetoneutral voltage times 1.73, so the three phase formula in terms of phasetoneutral voltage would be
Cost = [V_{ØN}*1.73*A*PF*1.73/1000]* h * $/kWh or Cost = 3*[V_{ØN}*A*PF/1000]* h * $/kWh


Yes jghrist; but the OP specifically stated 3phase. Though I like your SP equation! Keith Cress Flamin Systems, Inc. http://www.flaminsystems.com 

To respond to LionelHutz, be careful with conversion factors. 3.51 kW is "kW of cooling", which is NOT the same as power consumption kW. 12000 Btu/hr = 1 ton = 3.51 kW OF COOLING. Generally, chillers and other cooling equipment range from about 0.7 (for new highefficiency) upto 1.3 (for old inefficient) kW of power consumption per ton of cooling. 



