motor efficency vs. power factor? 1

[Inky29]

I'm trying to make sure that I have a concept understood and I'm kind of new to the electrical engineering field.

If you have a 20hp motor running at full load producing 20hp of torque and the motor is rated as 90% efficient am I correct in saying that the motor is actually having to pull 16.4 kw in order to produce 14.9kw worth of work?

If that is correct then how does a PF of .60 vs a PF of .90 effect the motor?

 

[Marke]

Hello Inky29

Yes that is correct, the input power to the motor increases as the motor efficiency falls.

The power factor will not affect the input power, but it will affect the input current. If the motor input power is 16.4KW at a power factor of 1, it will still be the same with a power factor of 0.6 The current however will increase as the power factor drops. This is due to the addition of either reactive currents, or harmonic currents or both. In the case of an induction motor fed from a standard supply, the additional current is inductive and does not add to KW usage.
Best regards,

Mark Empson

http://www.lmphotonics.com

 

[Inky29]

Thanks Mark.

So if all my understanding is correct then what someone is paying for is the true power and not the KVA? And the electric companies just have an additional charge for the increased current being used to deliver the power as a result of a low power factor?

 

[Marke]

Hello Inky29

Yes, that is correct. The standard electrical meter measures true KWHr and that is what is charged for. There is often an additional tarrif that has many derivatives depending on where you are located, but it ranges from a surcharge for a poor power factor, to a maximum demand tarrif. It is essentially designed to minimise the line losses due to a poor power factor and also from a high maximum demand.
Some of our local (New Zealand) tarrifs are based on maximum demand (Amps) and are metered using a thermal meter. You pay a tarrif based on your highest current draw in a three month period, and you pay that rate for the full period. i.e. a single high current draw for say half an hour will load your power bill for three months. Another scheme is coincident peak charging where you pay an additional surcharge if your maximimum demand coincides with the maximum load on the power distributor. A major incentive to draw a constant load 24/7

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[jbartos]

Suggestion to Inky29 (Electrical) Dec 11, 2003 marked ///\\If you have a 20hp motor running at full load producing 20hp of torque and the motor is rated as 90% efficient am I correct in saying that the motor is actually having to pull 16.4 kw in order to produce 14.9kw worth of work?
///Yes, this is correct for the motor rated values only since the efficiency is dependent on the motor loading. If the motor loading is 50% for example, the efficiency will drop according to the manufacturer provided efficiency versus loading curve. At low loads, the motor become less efficient. If the motor shaft load is equal to zero, the motor efficiency is equal to zero.
Visit

http://powerelec.ece.utk.edu/pubs/tiajan98-mag.pdf

for: Fig. 1 Efficiency versus load curves.
\\
If that is correct then how does a PF of .60 vs a PF of .90 effect the motor?
///It depends what is meant. The power factor is also dependent on the motor load. It decreases rapidly with the small load. The magnetizing branch of the motor significantly influence the power factor decrease for small motor loading. Sometimes, power factor versus motor loading curves are available from some manufacturers.\\

 

[brucecranene]

In answer to Marke. The extra current may be inductive, but it does contribute to kw usage in the I2R losses area. These losses are found from the motor to the meter in the way of heat. Put a capacitor at the load and the power factor will be increased and the kw will be reduced due to I2R losses, not the reduction of reactance.

 

[Marke]

Hello brucecranene

Yes that is correct, but if the KW reduction is significant, you are using the wrong sized cable as the losses due to the work component of the current will be very high!!
The major benefit of power factor correction is to reduce the i2R losses in the supply and that does not accrue on your kwhr meter.
Best regards,

Mark Empson

http://www.lmphotonics.com

 

[brucecranene]

Marke. What do you mean if the savings in kw is too significent the wire is too large. What is the definition of supply. The work component remains the same in the pf curve when the cap is introduced. The reduction is in the magnizing current, the current is reduced from the capacitor thru the meter to the utility. I have placed caps that have changed the kw from 1 to 36%. The motor current and voltage remains the same with no need to reduce the wire size?

 

[Marke]

Hello brucecranene

Perhaps the easiest answer is by example, so let us look at a 75KW motor operating on 400V
At full load, the current of the motor is 135 A and the power factor is 0.89

If the cables are sized for a 5% voiltage drop, there will be a power loss in the cables of 4.67KW. If we correct the motor (at the motor) to a power factor of 0.95, the power loss in the cables will reduce. The current drops to 126A and the power loss to 4.07KW. This is a saving of 603 Watts. - a very small percentage!!
If we increase the cable size, the voltage drop will reduce and so the savings will also reduce. i.e. select the cable for a 3% voltage drop, the cable losses will now be 2.8KW and the savings from the addition of correction at the motor would be 362 watts.

In this case, if you are able to make a significant saving from the addition of power factor correction, the cable must be undersized to give a voltage drop of over 5%

If you consider a situation where a motor is operating under very light load, then the saving as a percentage will be higher, but the watts loss will still be small if the cable is appropriately sized.

In the case above, under open shaft conditions, the current would be in the order of 34 amps and the cable loss would be in the order of 0.292KW Power factor correction would reduce this significantly because the open shaft current is primarily reactive, but the starting value is already low relative to the motor rating.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[welshwizard]

The power factor issue only relates to ac machines
The windings represent an inducitve load and this produces a lagging power factor
a pf = 1 is perfect! and 0.9 is better than 0.6 etc
The circuit must be capable of supplying this unwanted current so cables and switchgear must be sized up to carry it

You can improve the pf by installing pf correction capacitors. They produce a leading current which off sets the lagging current produced by the motor windings

Quite ofen your electricity tariff may have a pf penalty clause in the billing structure

 

[jbartos]

Suggestion: Visit

http://www.eng.uts.edu.au/~joe/subjects/ems\ems_ch13_ppt.pdf

for: Power factor compensation at synchronous machines

http://www.ee.und.ac.za/coursemain/DNE3PS2/notes/3PS2_2K2_4.PDF

for: Power factor correction

http://triasinnovations.com/techinfo.asp

for: Power factor versus motor load curves and power factor compensation
etc. for more info