# Power Factor - Practical Question

Discussion in 'General Electronics Chat' started by vg19, Sep 24, 2009.

1. ### vg19 Thread Starter New Member

Nov 3, 2004
6
0
Hey - I am trying to get a better idea regarding the notion of power factor, and power factor correction as it relates to power systems.

Here are my thoughts (I'm looking for confirmation that my intuition is correct):

In a power system with inductive loads, the power factor is lagging, as a result the reactive power drawn results in increased current through the transmission lines and as a result, more loses.

Here is where I get a bit confused. Inductive loads (induction motors/generations) NEED reative power to operate. As a result, wouldn't the notion of correcting the power factor to unity result in current and voltage in phase - meaning no reactive power?

Does the addition of parrallel capacitors in essence create o 'local' source of reative power that can alternate back and forth, as opposed to reactive power that must travel back to the 'distant' source, thus resulting in higher losses (through transmission lines)?

Hopefully, my little confusion here is clear. Any help is much appreciated!

2. ### JDT Well-Known Member

Feb 12, 2009
658
85
The current (wave) lags the voltage (wave). The power factor is less than 1.

Forget the idea of reactive power. Power is power. It's just that if the P.F of your equipment is less than 1, then it is drawing more current than it needs. This extra current causes losses in the supply cables. The suppliers don't like that.

You could say it creates a local supply of current. (Not sure it that's the best way of looking at it though.)

Hopefully that helps!

3. ### Ratch New Member

Mar 20, 2007
1,068
3
vg19,

I don't think you quite have a handle on reactive power. Reactive power is a consequence of induction and capacitance, not a "need". Reactive power takes energy from the source for half a cycle, and gives it back the next half cycle. Its net energy consumption for the whole cycle is zero. The source voltage and current used to supply and return the power can cause energy losses from the resistance present in the circuit. When the power factor is 1, the circuit is in resonance. This means that the inductance and capacitance take and return energy from each other at opposite times. This can greatly lower the current requirements to be supplied by the energy source, thereby reducing the resistance (IR) losses.

Ratch

4. ### S_lannan Active Member

Jun 20, 2007
247
2
On the reactive power to make a motor operate. Don't need to be too concerned with that.

The goal of power factor correction is to the power supply see V and I cophasor. What happens from then on really isn't a big deal, the circuit will operate as it was intended to.

When the circuit is at resonance no net reactive power will be drawn. The power factor will be 1.

When a circuit has a low power factor it doesn't actually USE any more power, it just adds to I^2R losses in the conductors, this was something that took me a long time to realize.

5. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
The power factor on a typical home is around 0.85 looking inductive. Power companies put huge capacitors on the poles along the power distribution lines to correct the phase shift, effectively creating a source to pump current back and forth to the house.

In cases where the home's PF is significantly worse than 0.85, the power company may install a VAR meter as well as a power meter on the home and charge for both kinds of power.

6. ### JDT Well-Known Member

Feb 12, 2009
658
85
They are charging you for the power that they lose on their lines, transformers, etc. Due to the extra current that they had to supply you because your PF was less than 1.

BTW. In most buildings and factories the load is inductive due to electric motors and fluorescent lamps with inductive ballasts. So a parallel capacitor is required to correct the power factor.

If you are a power supply utility sending lots of kilovolts hundreds of miles the capacity between the conductors in the lines causes a capacitive current to flow. This also is wasted current with a less than 1 power factor in the capacitive direction. Power engineers call this MegaVars - Mega Volt Ampere Reactive - MVAr. In this case a series inductor is required to correct the PF.

This is one of the reasons why overhead lines on pylons are preferred to underground cables (as well as the cost of installation). On overhead lines the wires are further apart and have lower capacity.

When long lengths of high voltage cables have to be used - especially under water. Often the power will be converted to DC and then re-converted to AC at the other end. DC - no capacitive losses.

This system is already used to connect France to Britain under the English Channel (about 70 miles) and many other places in the world.

Am I going off topic here?? Must stop.

7. ### subtech Senior Member

Nov 21, 2006
123
5
The answer to your first question is correct. The more inductive the load is, the more reactive current flow there is and the more losses there are. Also, don't forget that the more current flow there is, the larger all of the equipment has to be. All the way from the generator to the customers equipment.
That is an unnecessary expense. For the power company AND the customer.

Last edited: Sep 26, 2009
8. ### subtech Senior Member

Nov 21, 2006
123
5
The second question/answer is also correct to a degree. It takes a diagram to answer this effectively, but I'll try. If sufficient capacitance was placed half way between the offending customer and the generator, the from the perspective of the generator the power factor would indeed be 1 or unity.
However, if you were to measure the power factor downstream of the cap bank on the line, nothing will have changed. That is to say, if you had measured the power factor at the customers service entrance before the installation of the capacitors, and then again after, you would see no change. The correction in pf will be seen between the cap bank and the generator. It's hard for people to visualize this, it helps to sketch it out on a piece of paper and analyze it as a simple circuit with lumped quantities of L and R and C.

Last edited: Sep 26, 2009
9. ### subtech Senior Member

Nov 21, 2006
123
5
Your third and final question/conclusion is a VERY slippery slope. For some reason, most engineers today are taught that capacitors are VAr generators. Considering your choice of words, it appears that you have been taught this also. It is absolutely foolish and false to say that capacitors generate anything. They are static devices. It is much more simple and accurate to think and say that capacitors draw current from the source that is LEADING. This is of course the opposite of inductors, which draw current from the source which is lagging. So, up to the point where the capacitor is attached in the circuit, you will have leading current being drawn from the source and this leading current cancels the effect of any lagging current that is being drawn by any inductive loads.
Again, it is most helpful to draw this out on a sheet of paper showing the generator, the customers inductive load, and the capacitor connected halfway in between. Remove the cap and you've got a lagging load (inductive). Connect the caps and you've cancelled the effect of the lagging current but ONLY up to the point where the capacitors are connected in the circuit.
An important point to rmember here is the fact that for proper power factor correction, capacitors MUST be sized correctly. I'm sure most folks here know that already, but for us less experienced blokes, I thought it prudent to add the reminder.
(Do yourself a lifelong favor and remember this. Capacitors DO NOT generate anything. They draw leading current from the source.)

Hope all of this helps...

Last edited: Sep 25, 2009