Normally the power factor of any industrial unit is less than unity e.g 0.8 or 0.99 lagging. What will be the effect of the leading power factor on motor effeciency?

Can some body give me the detail calculation on this question.

 

When you connect a motor on a leading power factor you will blow up the motor if it is use on a high inertia load.
When you stop the motor, the load will drive the motor and the motor will self energize itself. If the power factor is leading enough, it will produce high voltage that will blow up the motor insulation.

 

The supply power factor has nothing to do with a motor. The motor will work with the usual low power factor. This will only try to bring down (your case bring up) the overall system power factor. Only when a leading power factor equipment and lagging power factor equipment happen to be connected in SERIES, the voltage across the individual equipment will increase and in case of resonance blow both the equipment due to very high voltage developed.
When a capacitor bank above 200 Kvar is connected to the line, one must ensure that there is sufficient inductive load (atleast 30 %) to avoid ringing effect on the line and there by avoiding high voltage build up on the line and destroying the capacitors.
Best regards.
Sekar

 

The formula you are looking for is really simply a guideline: The amount of "floating" capacitors added to any distribution system should not exceed 25% of the power transformer KVA capacity or the voltage could rise above safe limits and cause an over voltage condition. Motor effeciency is a factor of load and it increases as the load increases. A motor at no-load has voltage and current applied but no useful work is being done. You increase motor effencieny by adding a load to the motor.

 

Power factor does not "effect" operation of an induction motor! Instead, the motor "causes" it!

The electrical input to an ordinary induction motor is comprised of two components. One is called the "real" power (W) component. It provides the power for producing shaft output as well as any electrical or mechanical power losses. The second is called the inductive or "reactive" power (VAR) component. It produces the magnetic-flux that transfers power from the stator circuit via the motor's air-gap into the
rotor circuit. Because the "reactive" component can only be inductive, it lags the "real" component by 90 electrical degrees.

Vector addition of the real and reactive components is called the apparent power (VA) input. It is equal to the square-root of the sum of the real and reactive components squared: VA = Sqrt [W^2+VAR^2]. Input power factor, then, is defined as the ratio of real power to apparent
power, or W/VA.

In conclusion, the induction motor isn't "effected" by power factor, but instead, is responsible for it.

If additional information is required please contact me at [email protected].

Regards,
Phil Corso, PE
(Boca Raton, FL)

 

Any good books on electrical principles will provide you will additional details, but here is a summary:

The Power Factor is the ratio (k) [referred also as cos phi] always smaller than 1 by which you multiply the (V.A) to obtain the real power (Pr) consumed by any AC device or what is really converted to mechanical or usable energy.

For a single phase device

Pr =3D VA x k (cos phi)=20

So what is k (cos phi)?
Any AC device reacts to the variations of the AC power to which it is submitted. A resistive circuit will have the effect of reducing the
intensity (A) but does not affect the phasing. An inductive circuit will introduce (V) phase lag. Therefore the intensity (A) is not in phase
with voltage (V). In other words, cos phi is the phase variance between the voltages present to the terminals of an AC device and the current that pass thru it.

If there is no phase variance (0 degrees) than cos 0 =3D 1.

What does a poor power factor means for the Supplier?

- Ineffective use of transmission lines.
- Ineffective use of generators since the maximum intensity does not
match the maximum power (watts) used.
- Loss of productivity since more resources (coal, water, ---) is
required to produce the same amount of real power used.

What does a poor power means for the User?

- Increase of thermal loses in the installed devices.
- Larger capacity supply line, transformer, power usage therefore added
costs.
- Increase of the cost of use for electricity.

Hope this helps.
Jeff Gazidis
Senergis Technologies Inc.