About Power Factor

_{......If Power factor is low ours power consumption is Low.....}

Your power consumption depends on your load. The motor will have to deliver that much power anyway. If you have correctly selected the motor, it will deliver that power at about 0.8 pf. Your power consumption will be low if you disconnect the load.

The motor itself will deliver its rated power at about 0.8 pf. You connect a capacitor across it to reduce the current drawn overall, so that your system pf is closer to unity.

Simple terms, if you have a 5000 W motor running with a power factor pf 0.5, the motor will be taking 10,000 W from the mains supply.

By having the motor at such a bad power factor, all the switch gear and cables will need to be rated to carry 10,000 W, this will increase the size and cost of the installation and you will be fined by the supply authority.

If the 5000 W motor was running at a power factor of 0.90, the motor would only require 5555 W from the supply, with this you can reduce the cable size and switch gear from above.

10,000 W 10,000 VA

Hope you are clear now.

Another way to help through the formula:

Rewriting the formula :

W = 3^1/2*V*I*PF

or

Real Power (W) = Apparent power (VA) * P.F.

Apparent Power (VA) = Real Power / P.F.

Thus at a lower power factor, your power consumption (VA) will be higher (to get the required real power).

Pardon the dissertation but:

Your equation is not quite correct for a three phase motor:

Apparent Power (VA) = 3 x VA per phase

But, VA per phase = Iph x Vph/sqrt of 3

Therefore, Total VA = 3 x Iph x Vph/sqrt of 3

Mult top and bottom by sqrt 3 and we have familiar form of:

Total VA = Iph x Vph x sqrt 3 which is a little different.

And finally Real Power in Watts = Total VA from above x power factor

FYI: Explaining real power and reactive power and power factor and their effects:

AC Power consists of two components including real power ( watts)and reactive power (Vars). Real power is used to do actual work; reactive power only provides energy for inductive components of the load mainly for magnetizing transformer cores or motor winding cores. The reactive power component just keeps pumping back and forth between the load and the source without doing any work.

Both watts and vars together heat up the conductors in the feeders and puts load on the power supply equipment like generators and transformers. But the reactive power (vars) is a kind of overhead power.

Power factor (PF)= watts/total VA,

...where total VA is apparent power (volt-amperes) equal to vector sum of both watts and vars. Therefore, keeping the vars low in the denominator of the equation helps to keep the resulting PF high and at the same time reduces conductor and transformer sizes and reduces wear and tear on generators. Unity PF = 1 and apparent power = watts and there are no vars and therefore no wasted power.

The utility normally bills on watts, but there is extra wear on their equipment if there are a lot of vars. That is why the utility wants you to have high power factor, and if you do not they want paid for their equipment and conductors and will charge you a penalty.

Summary- Higher PF (closer to 1) = less vars, smaller transformers, smaller feeders, less heat release to the space, less wear on generators, lower voltage drop, and possibly lower billing from the utility.

_{Pardon the dissertation but:}

_{Your equation is not quite correct for a three phase motor:}

_{Apparent Power (VA) = 3 x VA per phase}

_{But, VA per phase = Iph x Vph/sqrt of 3}

_{Therefore, Total VA = 3 x Iph x Vph/sqrt of 3}

_{Mult top and bottom by sqrt 3 and we have familiar form of:}

_{Total VA = Iph x Vph x sqrt 3 which is a little different.}

Er.... pardon, you are wrong ! Power in 3Φ = √3*I_line*V_line

or = 3*I_ph *V_ph

Please check and come back, thanks.

I apologize.

I was incorrect in my last response.

I was thinking line to line voltage in the formula instead of line to neutral. Your formula is correct if you use line to neutral voltage and individual phase current. In that case the total power is simply three times the power in each of the individual phases which is E x I x PF as you stated

I will try not to be too quick to criticize in the future.

Have a good day.

The objectives of power factor improvement is as follows

1. Better voltage regulation

2. minimize circulating Vars and thus reduce losses in the distribution system.

For the same load current consumption increases if pf is low. In turn it increases copper loss in utility's as well as consumer's cables,require larger cables,larger equipment,voltage drop etc. Also different modes of pf like displacement pf,true pf,harmonic pf should be studied.

Electrical energy is measured in KVA.Where as load is measured in KW.Realation between the two is

KW = KVA*PF

Consider load of 100KW at p.f of 0.6 lag.Euivalent KVA will be 100/0.6 =166.66KVA.

In this case 166.66KVA will be drawn from the mains.Now we connect a Shunt Capacitor and improve power factor to 0.95lag.

Now KVA requirement will be =100/0.95=105.26 KVA.

Saving in energy is 166.66-105.26=61.39KVA.This saved energy can be utilised for connecting additional load.

By improveing P.F ,for same load current drawn from mains is reduced ,as such reduction in losses.

Hope matter is clear.

All well and good, but here is what you wanted to know.

"Is LPF penalty only the reason to increase PF?"

Yes, for the most part, although "increase" is not really as important as "increase to correct", because if they penalize you for anything below .95PF, and you "increase" from .80 to .90PF, you are STILL going to be penalized.

Here is how it really works. Your kWh BILLING from the power supplier will not be significantly affected by improving the PF, so you having poor PF is not going to result in offsetting revenue for the power company. But because their COST TO DELIVER IT TO YOU is affected, they penalize you for having poor PF as a way to make up for that. Money is money, so it does benefit you to correct your PF. But correcting PF is not reducing energy consumption for the average user. For the most part, the reduction in copper losses inside of your own facility is severely overblown by people who promote PF correction as a method of energy savings. It only becomes significant for very large consumers who distribute power inside of a large facility.

The question asked on power consumption is related to power factor and billing be based on the power consumption. Then why do we need to improve the power factor.

All the discussions on the forum were trying to present mathematical modeling of induction motor power consumption. But the focal point of the question is why to improve the power factor is left unanswered.

Whether you improve the power factor or not but billing will be remain same to you provided meter connected is kwh meter. Then question is why we need to improve power factor.

There are two parties involved in the electrical power system one is the power supplier who is called local electrical Utility co. and second party is consumer that is you and your Induction motor.

If the electrical infrastructure (Generators, transformers, transmission lines) meets or balance the power consumption of the area load, Utility will be in profitable to provide the power to consumer if not they are the looser. No Electrical Utility co. accepts the loss in the business.

The power factor improvement will lower the burden on the electrical infrastructure and hence Electrical Utility co. does not have objection for very low power factor as long as power factor is improved at load end.

Let us take a small example (Utility and only one consumer) to understand the whole thing.

Case-1: Consumer load details are 1000Kwatt at unity power factor: then Utility requires electrical infrastructure as follows;

Generator capacity in KVA= 1000KW/1.0 = 1000KVA and adding 10% transmission loss. The Actual Generator capacity = 1100KVA.

Similarly 1100MVA Transformer capacity and Transmission shall meet 1100MVA capacity.

Case-2: Now the consumer has same 1000KW load but at 0.5 power factor: Then Utility requires electrical infrastructure as follows;

Generator capacity in KVA= 1000KW/0.5 = 2000KVA and adding 10% transmission loss. The Actual Generator capacity = 2200KVA.

Similarly 2200MVA Transformer capacity and Transmission shall meet 2200KVA capacity

From this it is clear that Electrical Utility Co. has to invest nearly double the cost in second case for same billing to consumer. This is a big loss and Electrical Utility co. cannot accept but applies heavy penalty or disconnect the electrical supply or connects KVA Meter instead of KW meter.

To avoid the situation arises from Case-2, Electrical Utility co. asks the consumer to improve the power factor from 0.5 to 1.0.

This is done as follows: Owing to 0.5 Factor of 1000KW load, it draws from the utility real power of 1000KW and reactive power of 1732KVAR (inductive). Then load consumes 1000KW from 2200KVA and return back +1732KVAR to Electrical Utility. Electrical Utility does not bear the returning of big +1732KVAR to Electrical Network as non-real power. This creates the followings:

· It needs Generator capacity double

· Transmission losses will be 4 times

· Electrical Infrastructure cost will be double

Electrical Utility asks to impose the consumer to connect a big capacitor load (for the sake of example) whose KVAR value equal to-1732KVAR. There by, Inductive +1732KVAR of load is neutralizes by local connected capacitor -1732KVAR. The net Electrical Utility co. network supplies to load will be only 1000KW as in the case-1. OR

At load Terminal:

There are two loads as:

· 1000KW+1732KVAR( Of the load) that is 1000KW at 0.5PF

· -1732KVAR( Of the Capacitor load)

· Sum of the two loads at load terminal be 1000KW only as Inductive and capacitive Reactive KVARs get cancelled. As load inductive reactive KVAR will be supplied locally by local capacitive reactive KVAR but not from Electrical Utility network.

· No return of non-real power of +1732KVAR to network or Utility

· Electrical Utility Network has to supply only 1000KVA at 1PF or 1000KW.

The net load to be supplied by electrical Utility will be 1000KW only. Now, Electrical Utility network will be as in the case of CASE-1. after power factor improvement in spite of load of 1000KW at 0.5PF.

==>>"The question asked on power consumption is related to power factor and billing be based on the power consumption. Then why do we need to improve the power factor."

Better voltage Regulation and Minimization of losses and hence minimization of generation cost.

Also we can say"energy saving". Wasting energy is like throwing food by rich people while millions starve.

See this thread. Example has been given to illustrate the concept of P.F. [Power factor]