About Power Factor

What is Power Factor

To understand power factor, we’ll first start with the definition of some basic terms:

KW is Working Power (also called Actual Power or Active Power or Real Power).
It is the power that actually powers the equipment and performs useful work.

KVAR is Reactive Power.
It is the power that magnetic equipment (transformer, motor and relay) needs to produce the magnetizing flux.

KVA is Apparent Power.
It is the “vectorial summation” of KVAR and KW.

Let’s look at a simple analogy in order to better understand these terms….

Let’s say you are at the DELHI, and it is a really hot day. You order up among of your favorite beer. The thirst-quenching portion of your beer is represented by KW (Figure 1).

Unfortunately, life isn’t perfect. Along with your ale comes a little bit of foam. (And let’s face it…that foam just doesn’t quench your thirst.) This foam is represented by KVAR.

The total content of your mug, KVA, is this summation of KW (the beer) and KVAR (the foam).

So, now that we understand some basic terms, we are ready to learn about power factor:
Power Factor (P.F.) is the ratio of Working Power to Apparent Power.

P.F. = KW          
          KVA

Looking at our beer mug analogy above, power factor would be the ratio of beer (KW) to beer plus foam (KVA).

P.F. = KW         
      KW + KVAR

P.F. = Beer
       Beer + Foam

Thus, for a given KVA:-

• The more foam you have (the higher the percentage of KVAR), the lower your ratio of KW (beer) to KVA (beer plus foam). Thus, the lower your power factor.

• The less foam you have (the lower the percentage of KVAR), the higher your ratio of KW (beer) to KVA (beer plus foam). In fact, as your foam (or KVAR) approaches zero, your power factor approaches 1.0.

Our beer mug analogy is a bit simplistic. In reality, when we calculate KVA, we must determine the “vectorial summation” of KVAR and KW. Therefore, we must go one step further and look at the angle between these vectors.

The Power Triangle

P.F. = KW  =   COS θ
          KVA

KVAR   =   SIN θ KVA

KVA  = /KW2 + KVAR2         =        KV * I * _ / 3

Note that…in an ideal world…looking at the beer mug analogy:

• KVAR would be very small (foam would be approaching zeo)
• KW and KVA would be almost equal (more beer; less foam)

So….

In order to have an “efficient” system (whether it is the beer mug or Mac dragging a heavy load), we want power factor to be as close to 1.0 as possible.

Sometimes, however, our electrical distribution has a power factor much less than 1.0. Next, we’ll see what causes this.

What Causes Low Power Factor?

Since power factor is defined as the ratio of KW to KVA, we see that low power factor results when KW is small in relation to KVA. Remembering our beer mug analogy, this would occur when KVAR (foam, or Mac’s shoulder height) is large.

What causes a large KVAR in a system? The answer is…inductive loads.

Inductive loads (which are sources of Reactive Power) include:

• Transformers
• Induction motors
• Induction generators (wind mill generators)
• High intensity discharge (HID) lighting

These inductive loads constitute a major portion of the power consumed in industrial complexes.

Reactive power (KVAR) required by inductive loads increases the amount of apparent power (KVA) in your distribution system. This increase in reactive and apparent power results in a larger angle θ (measured between KW and KVA). Recall that, as θ increases, cosine θ (or power factor) decreases.

So, inductive loads (with large KVAR) result in low power factor.

Why Should we Improve My Power Factor ?

You want to improve your power factor for several different reasons. Some of the benefits of improving your power factor include:

1. Lower utility fees by:

A. Reducing peak KW billing demand

Recall that inductive loads, which require reactive power, caused your low power factor. This increase in required reactive power (KVAR) causes an increase in required apparent power (KVA), which is what the utility is supplying.

So, a facility’s low power factor causes the utility to have to increase its generation and transmission capacity in order to handle this extra demand.

By raising your power factor, you use less KVAR. This results in less KW, which equates to a dollar savings from the utility.

B. Eliminating the power factor penalty

Utilities usually charge customers an additional fee when their power factor is less than 0.95. (In fact, some utilities are not obligated to deliver electricity to their customer at any time the customer’s power factor falls below 0.85.) Thus, you can avoid this additional fee by increasing your power factor.

2. Increased system capacity and reduced system losses in your electrical system

By adding capacitors (KVAR generators) to the system, the power factor is improved and the KW capacity of the system is increased.

For example, a 1,000 KVA transformer with an 80% power factor provides 800 KW (600 KVAR) of power to the main bus.

1000 KVA = / (800 KW)2  +  ( ? KVAR)2

KVAR = 600

By increasing the power factor to 90%, more KW can be supplied for the same amount of KVA.

1000 KVA = / (900 KW)2  +  ( ? KVAR)2

KVAR = 436

The KW capacity of the system increases to 900 KW and the utility supplies only 436 KVAR.

Uncorrected power factor causes power system losses in your distribution system. By improving your power factor, these losses can be reduced. With the current rise in the cost of energy, increased facility efficiency is very desirable. And with lower system losses, you are also able to add additional load to your system.

3) Increased voltage level in your electrical system and cooler, more efficient motors

As mentioned above, uncorrected power factor causes power system losses in your distribution system. As power losses increase, you may experience voltage drops. Excessive voltage drops can cause overheating and premature failure of motors and other inductive equipment.

So, by raising your power factor, you will minimize these voltage drops along feeder cables and avoid related problems. Your motors will run cooler and be more efficient, with a slight increase in capacity and starting torque.

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