Most loads in modern electrical distribution systems are inductive. Examples include Motors, Transformers, and Induction furnaces. Inductive loads need a magnetic field to operate. Inductive loads require two kinds of current: Power-producing current (or working current) is that current which is converted by the equipment into useful work such as creating heat, light, motion ,machine output . The Unit of measurement of the power produced is the kilowatt (kw).

1. Plant Engineers shall give importance to Power
factor
PresentedBy-
Vineet Shekhar

2. Introduction
 Most loads in modern electrical distribution systems are inductive.
Examples include Motors, Transformers, and Induction furnaces. Inductive loads need a
magnetic field to operate.
 Inductive loads require two kinds of current:
Power-producing current (or working current) is that current which is converted by the
equipment into useful work such as creating heat, light, motion ,machine output .
The Unit of measurement of the power produced is the kilowatt (kw).

3. Magnetizing current (also known as wattless, reactive, or non-working current) is that
current which is required to produce the flux or to sustain the magnetic field necessary
to the operation of induction devices.
Without magnetizing current, energy could not flow through the core of a
transformer or across the air gap of an induction motor.
The Unit of measurement of magnetizing volt-amperes is the kilovar (kVAR).
 Working power consumes watts and can be read on a wattmeter. It is measured in
kilowatts (kW).
 Reactive power doesn’t perform useful “work,” but circulates between the generator
and the load. It places a heavier drain on the power source, as well as on the power
source’s distribution system. Reactive power is measured in kilovolt-amperes-
reactive (kVAR).
 Working power and reactive power together make up apparent power. Apparent
power is measured in kilovolt-amperes (kVA).

4. Fig.1 KW Power
Fig.2 kVAR Power
Fig.3 kVA Power
Fig.4 Power Triangle

5. We can also elaborate it as,
Power factor may be expressed as the ratio of power-producing current in a circuit to
the total current in that circuit. Another definition of power factor, which is generally
more useful, is the ratio of kW or working power to the total kVA or Apparent power.
Thus,
 Power factor = kW / kVA
 kW = kVA × pf
 kVA = kW / pf
Stated another may, the power factor is that factor by which the apparent power must be
multiplied in order to obtain the working power.

6. Should we concerned about low power
factor?
 Low power factor means you’re not fully
utilizing the electrical power you’re paying
for. As the triangle relationships in Figure
demonstrate, kVA decreases as power factor
increases. At 70% power factor, it requires
142 kVA to produce 100 kW. At 95% power
factor, it requires only 105 kVA to produce
100 kW. Another way to look at it is that at
70% power factor, it takes 35% more current
to do the same work

7. What are causes of low power factor?
 Generally, low power factor could be due to partially loaded induction motors.
Frequently drives are “overmotored,” i.e., the motor is selected to handle the
largest load but usually operated at less than full load.
 Low leading power factor conditions are most commonly caused by fixed
capacitors connected to the system during low load periods (off shift periods
when inductive loads are turned off).
Power capacitors provide many benefits:
• Reduced electric utility bills
• Increased system capacity
• Improved voltage
• Reduced losses
How much can we save by installing power capacitors?

8. Correcting your power factor
The first step in the process of correcting your power factor is identifying what is
causing the low power factor. This information will be extremely important in
determining the right approach for bringing your power factor closer to unity.
There are many different strategies that can be used individually or in
combination to correct low power factor. A few of these strategies are:
 Install capacitors in the distribution system
 Minimize operation of idling or lightly loaded motors
 Install variable frequency drive (VFD) systems to lightly loaded induction
motors
 Install new motors that will be operated near their rated capacity
 Replace lightly loaded motors with motors sized to be operated near their rated
capacity
 Avoid operation of equipment above its rated voltage.

9. You can improve power factor by adding power
factor correction capacitors to your plant
distribution system.
When apparent power (kVA) is greater than
working power (kW), the utility must supply the
excess reactive current plus the working current.
Power capacitors act as reactive current
generators. By providing the reactive current,
they reduce the total amount of current your
system must draw from the utility
Fig. Capacitors as kVAR generators

10. 95% power factor provides maximum benefit
Theoretically, capacitors could provide 100%
of needed reactive power. In practical usage,
however, power factor correction to
approximately 95% provides maximum
benefit. The power triangle in Figure shows
apparent power demands on a system
before and after adding capacitors. By
installing power capacitors and increasing
power factor to 95%, apparent power is
reduced from 142 kVA to 105 kVA—a
reduction of 35%.
Fig. Required Apparent power before and
after adding capacitors