1. Real power (P) by which its unit is expressed in watts (W).

2. Reactive Power by which its unit is expressed in VAR

3. Apparent power (S) by which its unit is expressed in volt-amperes (VA).

The relationship of these three are represented by a power triangle.

In relation to this diagram we define power factor as the ratio of the real power to the apparent power, that is

The power drawn by the AC system depends on the connected loads whether those are purely resistive, inductive or capacitive. Purely resistive loads are drawn mostly by loads used in heating, passive resistor used in motor control starter and others.

But mostly if we consider the overall system particularly in industrial applications purely resistive system is very rare to exist. But before going further I would like to show first the graphical representation of power drawn by typical electrical system for more understanding.

**power factor = real power/ apparent power**

or we can expressed it as the cosine of the angle ( Ø ) of inclination from a geometrical perpective, that is

**power factor = cosine Ø**

**REACTIVE POWER WITH INDUCTIVE AND CAPACITIVE LOAD**The power drawn by the AC system depends on the connected loads whether those are purely resistive, inductive or capacitive. Purely resistive loads are drawn mostly by loads used in heating, passive resistor used in motor control starter and others.

But mostly if we consider the overall system particularly in industrial applications purely resistive system is very rare to exist. But before going further I would like to show first the graphical representation of power drawn by typical electrical system for more understanding.

**Inductive load**such as electric motors, lighting ballasts, welding machines and other equipments that has winding will draw inductive load.**Capacitive load**drawn mostly by a capacitor. There are also equipment like synchronous motor that draws leading power factor.

This diagram describes the power drawn by a typical electrical system. The positive angle represents capacitive system (leading power factor) while the negative angle represents inductive system (lagging power factor). In AC system the current and voltage are propagating as wave signals and the two quantities may or may not be displaced with each other. If they are displaced with each other, then either the voltage or the current will come first.

- If the current waveform comes first before the voltage, the current in this case leads the voltage by some degree of angle and this system will draw leading power.

However if we reverse the situation when the voltage comes first, the voltage in this case lags behind the current by some electrical degrees then that system is an inductive system and has lagging power factor.

Ideally the capacitor will draw purely leading current thus drawing purely leading power. This can be represented using the following vector diagram. Theoretically the current in this system leads the voltage by exactly 90 degrees

**THE POWER FACTOR CORRECTION USING A CAPACITOR**There are two methods where a system with low power factor can be corrected but I will discuss here the most commonly used method of power factor correction that is by way of installing a capacitor.

__Consider this example:__

A 3-phase, 440 volts electrical system is taking 10 MVA of power has an overall power factor of 0.6 lagging. Determine the rating of the capacitor in order to raise the power factor into .90?

- In this example we know can determine the power factor angle by simply taking the inverse cosine of the given power factor: Ø = (inv) cos (0.6) = -53.13 degrees. Therefore the vector diagram is represented as follows:

- The values were obtained using simple trigonometric calculation: P= 10 MVA x cos (-53.13 deg.) = 6 MW Q= 10 MVA x sin (-53.13 deg) = -8 MW

*Note: The negative quantity does not mean that it draws negative power, that is just a representation as a vector quantity. The negative value means the vector is directed downwards or below the X-axis.*

Refer to the above diagram we can conclude that we can have the same real power (P) but with different apparent power (S) and that values are dependent on the inductive power in the system.

Taking trigonometric calculation, we can get:

- By adding the capacitor that draws pure reactive power (+90 degrees) as shown in the highlighted vertical line, and take the vector addition using head-to-tail method, the new resultant apparent power is S (new) with Ø2 be the new power factor.
- The new power factor should have a value of 0.9 base on the requirement of the problem. And the new power factor angle must have a value of (inv) cosine (.9) = 25.84 electrical degrees.
- The degree of separation between the real power and the apparent power has obviously decreased from -53.13 electrical degrees to -25.84 electrical degrees.

Taking trigonometric calculation, we can get:

- Q (new) = 6 MW x tangent (25.84deg) = 2.9 MVAR S (new) = 6 MW/ cosine (25.84deg) = 6.67 MV.
- MVAR (added) = MVAR (old) - MVAR (new) = 8 MVAR - 2.9 MVAR = 5.1 MVAR
- Therefore we need a capacitor with a rating of 5.1 MVAR @ 440 VOLTS

- In real application these capacitors are divided into several steps with the use of power factor controller in order to prevent overshooting that will lead into a system with leading power factor.
- For example a design engineer decided to divide the 100 MVAR into 10 steps, that means he has to used a power factor controller with 10 discrete outputs and can control 10 magnetic contactors.
- And these contactors would serves as a magnetic switch directed to 10 capacitors.

Ideally the factor should be close to unity or 1.0 [cos 0 = 1.0] , that is the power factor angle is held to be zero. In this case the real power is equal to the apparent power thus making the system efficient. Sometimes the occurrence of the low power factor situation happened only during full blast operation of the plant.

**HOW POWER FACTOR AFFECTS THE SYSTEM.**

- Power factor correction can make our electrical system more efficient. Our electric meters operate according to the quantity of current that passes thru it.
- In the same case the cable stress that serves as a conductor of electric current from load to source is reduced.

P = V x I x pf -----> it follows that

I = P / [ V x pf ] ----> in order to minimize I then we should let pf closer to 1.0.

**IMPORTANT THINGS TO CONSIDER**

- Before we correct our system power factor we have to know first the level of
__harmonics__of our system because as we know capacitors are sensitive to heating caused by harmonics. - We may need to undergo power quality audit first if we are suspecting a significant level of harmonics in our system.
- After determining the level of harmonics choose the type of capacitor that suits your need.
- The manufacturer or the supplier may provide you a whole unit where there are already capacitors, controllers and harmonic filters that is compatible to your system.

## 2 comments

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