Induction Motor Efficiency Vs Its Power Factor

"There is a motor which is 96.9 % efficient"

I have read first time of so much efficient Induction motor

"Why Losses in Low Power rating motors are greater than high power rating motors?"

Losses in Low Power rating are greater because of the reason that it have high rotor resistance. Due to high rotor resistance it has large slip and high I^2 R losses. Due to high losses they have relatively lower efficiency.

"Upon what factors Power factor of a motor depends?"

Induction motor is combination of inductive load and resistive Load. Mainly itis considered to be an inductive load. Its power factor depends on the following

Z = Stator winding impedance, Ω

XL = Stator winding reactance, Ω

L = inductance of winding, H

"If the Power coming to the motor have Low P.F does it effect Motor Efficiency? (There will be increse in (I*I* R) Losses?)"

Incoming Supply Low Power Factor Doesn't effect the motor efficiency.

From motor power equation P = 1.732 *V*I*P.F*Eff

For lower values of power factor current increases. When power factor is improved,

current decreases approximately with the same proportion

In case of Low P.F incoming supply Line Losses (Transmission) increases not the Motor Internal Losses (significantly)

"What Percentage are (I*I*R) losses to the total motor losses?"

I am not sure but it will be less than 35 % of the total motor losses. Also depend on the KW rating of the motor

"What Percentage are (I*I*R) losses to the total motor losses?"

I am not sure but it will be less than 35 % of the total motor losses. Also depend on the KW rating of the motor

I have read several papers identifying it at 25% of total losses on average.

Thank U very much for correcting me .

Motor Efficiency is Depicted as: η =0.7457 x hp x Load/Pi Where: η = Efficiency as operated in % Por = Nameplate rated horsepower Load = Output power as a % of rated power Pi = Three-phase power in kW The Efficiency in the above equation is related to the Load output. And if the input power factor is low it directly affects the output power and its efficiency also. The Current Decreases as Power factor increased mentioned in the earlier reply is seems very interesting, in general if power factor is improved the current is brought in align with the voltage to reach the real power of the system and not to reduce the current.

From Ur statement and equation what I understand is that efficiency depend upon the running Load on the Motor

The equation mentioned is

Efficiency ( %) = {0.7457 * HP *Load (Running Load)}/{Pi (Power Input)}

If I am not wrong then it will be ( In terms of KW)

Efficiency ( %) = {Po *Load }/{Pi }

Po = Power output in KW

Load= Running Load in Percentage

Pi = Input Power to the motor in KW

Consider the case of Motor-1

Motor-1 ( 7.5 KW, Efficiency 86.88 % ):

(LEROY SOMER: 7.5 KW, 400 V, 50 Hz, Y, 14 A, P.F 0.89, 1455 RPM)

If the motor is running at 100 % Load Then Po = 7.5 Kw (Name Plate Rating), Pi =

1.732*400*14*.89 = 8.6322 KW and efficiency at 100 % load will be according to equation 86.88 which is correct.

Now if we take that motor is running at 90 % of its rated load then efficiency will be

Efficiency ( % ) = (7500/8632)*(0.9) = 78.19 %

It means that efficiency is the function of load

Waiting for Ur reply

Yes, Its is true, the efficiency depends on the load of the motor. A no load motor will have a very bad efficiency because most of the power consumed will be the I2*R losses. Below is an extract of a book where the efficiency depending on the Power factor clearing the doubt which you raised in this thread: One often overlooked aspect of overall motor drive efficiency is the power factor. The power factor relates the shape of the current waveform drawn by a load to the sinusoidal voltage waveform supplied by the power company. If a load looks purely resistive, then the current drawn by the load is a sinusoid exactly in phase with the voltage waveform, and the power factor is unity (1). This is the most efficient condition. If the load appears to be inductive, as it does in many motors, the current will lag behind the voltage in phase, and the power factor will be less than one, according to the cosine of the phase angle. Capacitive loads, which cause the current to lead the voltage, also reduce the power factor below one. In either case, the energy supplied to the motor will not be used optimally. Since the peak (and shape) of the current sine wave does not line up with the peak of the voltage sine wave, the instantaneous product of voltage times current averaged over a full cycle is lower. This is called the true power and is measured in watts. Since the mains voltage is fixed, a higher current is required from the power company to compensate for the phase shift and deliver the same usable power to the motor, bringing the usable power (in watts) back up to the level required to do the desired mechanical work (in horsepower, for example).The product of this higher RMS current and the RMS voltage (measured in volt-amps) is called the apparent power. In many respects the power company has to build the infrastructure and pay for the higher apparent power, even though only the true power is doing useful work for the end user. This higher current means more losses as the power company generates and distributes the power. Power-line transformers can heat up and fail. Power losses go up as the square of the apparent power. A power factor of 0.7 means an apparent power of 1.4 times the true power, with nearly double the losses compared to a power factor of one.

The motor is designed to meet the requirements specified by the customer which include its torque, speed, type of enclosure, may be type of cooling and many more. Based on these factors designed makes his design. He attempts to make his motor as cost effective as possible by proper choice of materials and cooling methods.

Motor is made of magnetic materials (sheet steel) and copper. He can make motor more copper less steel combination or less copper more steel combination. Amount of copper decides copper losses and of amount steel decides iron losses.

Motor losses and be divided into two parts constant losses and load dependent losses. Lower the constant losses higher will be the efficiency of the motor. Constant losses include mechanical losses (bearing, brush, windage Fan losses) and the iron losses.

Why higher rating motors are more efficient? They are more efficient because they have lower percentage of mechanical losses (they have better cooling and therefore use less Kilogram of material per KW rated load).

Power factor is hardly a consideration for design of motor. As stated by many experts, power factor can be corrected out side the motor by suitable means. Motor has to have coils and the iron core and therefore it has to have inductive reactance, which cannot be eliminated under any circumstances. So designer is not bothered about its power factor. His interest is always to meet the customer specification at the lowest lost.

The comparisson of 2000kw motor with a 7.5Kw motor itself is not correct.The Higher the rating and RPM eff. is better.Higher speed motors like for example a2000kw 2or 4 pole compared to a 2000kw 6 or 8 pole will be more efficient due lower magnetizing current needed.Power factor will affect the eff as explained by other forum members.High efficiency and energy efficient motors are different.In energy eff. motors as per EURO EFF1 motors eff from 100 to 60% load is almost constant which is more important as motor maintains its eff close to full load eff even when there is a load fluctuation.Induction motors larger ratings like 5000kw and above have eff 96.5 to 97% to day.Power factor if needed can be corrected by adding capacitors of suitable rating.