# Motor Efficiency

### #1

Posted 13 June 2006 - 05:24 PM

Thanks

CS

### #2

Posted 13 June 2006 - 11:30 PM

Efficiency = Output Power/Input Power. Motors are rated on output power.

This is usually described as a percentage at full load. There fore if you had a 15 kW motor that is 90% efficient it would require 16.6 kW input power at full load. You should also be aware that a motors efficiency varies with load. A general rule of thumb is that larger motors are more efficient.

The level of efficiency can vary widely between manufactures so it is well worth investigating before purchasing equipment. The initial purchase price of a motor is a fraction of its total lifetime cost so money well spent on a more efficient motor will save plenty in the life time of the motor. An example may be of a motor that runs 18 hrs per day. One brand has an efficiency of 90% and another has an efficiency of 85%. If the motor is 15kW there is .98 kW difference in the input power so .98kW x 18 hrs per day = 17.6 kWh per day x 5 days a week x 52 weeks per year= 4586 kWh per year difference between the two motors. If the motor is on an industrial application then the hours run could be greater and so will the kWh difference. If we assume a cost of 10c/kWh then the cost difference per year is $458.

It does not take long to payback the initial extra expence of a high efficiency motor. If we assume that the motor will operate for 15 years then the savings are nearly $7000.

Ken

*An expert is one who knows more and more about less and less until he knows absolutely everything about nothing*

### #3

Posted 14 June 2006 - 12:13 AM

Regards,

GGOSS

### #4

Posted 14 June 2006 - 10:13 PM

15 kW 400V 50Hz 1460 rpm 29A p.f. .82. These are all rated at full load. We first need to calculate our kW input from this data so kW =√3 x V x I x cosΦ/ 1000 or 1.732 x 400 x 29 x .82 / 1000 = 16.47 kW. Efficiency = Input Power / Output Power so 16.47 / 15 = .91 or 91%. If you don’t have the efficiencies listed you can use the above to get a pretty good idea about different motors performance.

Ken

*An expert is one who knows more and more about less and less until he knows absolutely everything about nothing*

### #5

Posted 15 June 2006 - 02:34 AM

to kens, CSTEOH

Only one thing, you have calculated this efficiency for nominal mode, but really it is the very curved characteristic which starts from 0 when speed's 0 to max when speed's nominal.

### #6

Posted 15 June 2006 - 02:53 AM

Cheers

Ken

*An expert is one who knows more and more about less and less until he knows absolutely everything about nothing*

### #7

Posted 15 June 2006 - 05:44 AM

by mistake you wrote wrongly, right is:

Efficiency = Output Power / Input Power so 15 / 16.47 = .91 or 91%.

I would add some considerations on AC motor efficiency under non-sinusoidal main supply.

Where harmonics are present, there is to consider further losses due to presence of harmonic currents and related flux in the magnetic circuit. This flux generally make a negative action on output torque.

To reduce such power losses - due to undesired harmonics - it is suggested to add some harmonic filters, their cost should be less than the cost of energy saved during years (if motor is used for many hours every day).

Cheers

Mario

Mario Maggi - Italy - http://www.evlist.it - https://www.axu.it

### #8

Posted 15 June 2006 - 04:37 PM

Mr.Kens thanks for the details and the calculation, you are a life saver. There's one more thing that i can get hold of, can we actually calculate the power factor of an induction motor or do we need to used a power analyzer?

Regards

CS

### #9

Posted 15 June 2006 - 06:48 PM

You can calculate the power factor if you know the KVA input and the KW input.

The power factor is the KW/KVA. Note Input KW not rated KW which is shaft or output KW.

Somewhere, you will need to make some measurments.

Best regards,

Mark Empson | administrator

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | Soft Starters

### #10

Posted 15 June 2006 - 09:57 PM

Dear kens,

by mistake you wrote wrongly, right is:

Efficiency = Output Power / Input Power so 15 / 16.47 = .91 or 91%.

I would add some considerations on AC motor efficiency under non-sinusoidal main supply.

Where harmonics are present, there is to consider further losses due to presence of harmonic currents and related flux in the magnetic circuit. This flux generally make a negative action on output torque.

To reduce such power losses - due to undesired harmonics - it is suggested to add some harmonic filters, their cost should be less than the cost of energy saved during years (if motor is used for many hours every day).

Cheers

Mario

Thanks Mario, Not enough coffee yesterday

### #11

Posted 16 June 2006 - 12:57 PM

Thanks for your reply, i just wanna clarify one thing. I didn't really get it, to calculate pf=kw/kva right? You note that "input kw and not rated kw which is shaft or output kw" If i didnt get it wrongly, i should used input kw instead of rated kw. Im not really familiar with induction motors terms, if you don't mind can you pls explain to me the difference betweeen input kw, rated kw, shaft and output kw. Thanks.. To get pf why used input kw instead of rated kw??

Regards

CS

Hey everyone!!!

Actually this is my first time in LMPForum, just wanna let you guys know that im glad that i've join in to this forum. You guys are helpful and i can get alot of info that i need in here. If you guys don't mind i think from today onwards i'll be bugging you guys . I hope i can share some info with you guys to. By the way who's the creator of this website anyway?

Once again thanks alot.

Cheers to everyone

CS

### #12

Posted 06 March 2009 - 09:05 AM

**Just to check the formula kVA = kW / ( p.f * eff ) is correct?**

And kW = ( kVA * p.f ) / eff is also correct?

Thanks!

And kW = ( kVA * p.f ) / eff is also correct?

### #13

Posted 07 March 2009 - 08:58 AM

First, it is important to identify which KW you are referring to. There is mechanical output power and electrical input power.

Electric circuits are rated for electrical input power and mechanical systems for mechanical shaft power.

For the purposes of the description, let us identify these as KW

_{e}and KW

_{m}Where KW

_{e}is the electrical input power and KW

_{m}is the mechanical output power.

The power rating of an induction motor is usually the rated shaft power of that motor. (KW

_{m})

The efficiency of the induction motor is the ratio between the mechanical output power and the electrical input power.

eff = KW

_{m}/KW

_{e}

The power factor of an electrical circuit is the ratio between the electrical KW input to the KVA input, so pf = KW

_{e}/KVA

The mechanical shaft power is therefore KW

_{m}= KVA x pf x eff

The electrical input power is KW

_{e}= KVA x pf

So KVA = KW

_{m}/ pf / eff = KW

_{m}/ (pf x eff) = KW

_{e}/ pf

Be careful to ensure that you know whether you are referring to mechanical output power (usual with motors) or electrical input power.

Best regards,

Mark.

Mark Empson | administrator

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | Soft Starters

### #14

Posted 08 March 2009 - 03:50 AM

First, it is important to identify which KW you are referring to. There is mechanical output power and electrical input power.

Electric circuits are rated for electrical input power and mechanical systems for mechanical shaft power.

For the purposes of the description, let us identify these as KW

_{e}and KW

_{m}Where KW

_{e}is the electrical input power and KW

_{m}is the mechanical output power.

The power rating of an induction motor is usually the rated shaft power of that motor. (KW

_{m})

The efficiency of the induction motor is the ratio between the mechanical output power and the electrical input power.

eff = KW

_{m}/KW

_{e}

The power factor of an electrical circuit is the ratio between the electrical KW input to the KVA input, so pf = KW

_{e}/KVA

The mechanical shaft power is therefore KW

_{m}= KVA x pf x eff

The electrical input power is KW

_{e}= KVA x pf

So KVA = KW

_{m}/ pf / eff = KW

_{m}/ (pf x eff) = KW

_{e}/ pf

Be careful to ensure that you know whether you are referring to mechanical output power (usual with motors) or electrical input power.

Best regards,

Mark.

Hi Mark,

Thanks for your reply.

I am currently studying Electrical Engineering Level 5 (Advanced Trade), which is through Open Polytech in NZ, and sent away my paper which is on Standby Power Plants (and AC Motors). I don't know why I got tripped up on a question like this HAHA, being rather simple compared with the work I have done already, like the advanced AC/Power Factor Theory unit!!

I do understand your formulas and Im well aware with the (as i refer to them) OUTPUT and INPUT motor power formula for efficiency. Following is the question I was asked:

**Find the minimum kW rating of a standard induction motor needed to drive a synchronous generator with an output of 100kVA 0.8 pf, and an efficiency of 0.92 at full load.**

Well, I did work out that the answer would be the kWm (OUTPUT) power of the motor. I worked out the formula to be:

kW(m) = kVA * p.f * eff

kW(m) = 100,000 * 0.8 * 0.92 = 73,600W

The correct answer was in fact

**kW(m) = kVA * p.f / eff**

**kW(m) = 100,000 * 0.8 / 0.92 = 86,960W**

So obviously kVA(out) = kW(out) / p.f

**And therefore because the efficiency of the motor is less than 100%, the Output power of the motor has to compensate for that loss and must produce a higher power value, resulting in a higher rated motor to be used.**

Thanks again,

Shane

### #15

Posted 05 July 2009 - 10:40 AM

I think so too.

For motor efficiency, I think you need to discuss with the engineers from the manufacturer and talk with the experts. They will give us some good advice.

### #16

Posted 20 July 2009 - 07:05 PM

Our 15 kW motor name plate may have the following data

15 kW 400V 50Hz 1460 rpm 29A p.f. .82.

These are all rated at full load.

We first need to calculate our kW input from this data so kW =√3 x V x I x cosΦ/ 1000 or 1.732 x 400 x 29 x .82 / 1000 = 16.47 kW. Efficiency = Output Power / Input Power so 15 / 16.47 = .91 or 91%. If you don’t have the efficiencies listed you can use the above to get a pretty good idea about different motors performance.

According to this formula the lower p.f. has a motor, the more efficient it is.

As is known, p.f. rises with the load, so motor's efficiency falls ?

### #17

Posted 22 July 2009 - 08:29 AM

Actually the current stated in the name plate refers to full load current. The current drawn by the motor depends to the load attached. The output power will also change with respect to the input current. So at low p.f. (lightly loaded case) the motor might draw just enough current to establish flux in the core. This affects the output power as well. When computing the efficiency of the motor, one must always remember that the current (Iline) and p.f keeps changing until the motor is operating at full load.

efficiency = (1.732 x Vline x Iline x p.f) - power loss in motor /(1.732 x Vline x Iline x cos p.f)

Regards,

kana

### #18

Posted 22 July 2009 - 06:35 PM

1.7kW 220V 10A p.f. 0.8 (all at full load) should be considered more efficient than a motor

1.7kW 220V 10A p.f. 0.9 (all at full load).

### #19

Posted 22 July 2009 - 08:28 PM

efficiency = input gross power - power losses in iron, copper, etc/ input gross power

does not even consider output power on shaft

### #20

Posted 23 July 2009 - 01:28 AM

Stator & Rotor Copper losses depends to Iline.

Iron losses, rotational losses and stray losses are constant at all time.

power loss in motor = stator loss + rotor loss + iron loss + rotational loss + stray loss

input power = 1.732 x Vline x Iline x p.f

efficiency = output power (shaft power)/input power

Regards,

Kana

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