POWER FACTOR & NEED FOR
IMPROVEMENT
by
Engineer Haji Mumtaz
Power Factor Basics
1. Power factor is the percentage of electricity that is being used to
do useful work
2. power factor= active power/apparent power
3. Apparent power has two components:
Productive power
Reactive power
Any industrial process using electric motors (to drive
pumps, fans, conveyors, refrigeration plant etc.)
introduces inefficiencies into the electricity supply
network by drawing additional currents, called "inductive
reactive currents".
Resistive Loads====> Real
Power ( kW )
Voltage & Current are in phase
P = VI
Inductive / Capactive Loads
=====> Reactive Power ( kVAR )
Current Lags Voltage and Vice Versa P = VI
Reactive
Power
Real Power, Reactive
Power , Apparent Power
Power Factor
Power Factor is the ratio of Real Power to Total Power
Lassi Analogy
Power Factor = kW
kVA kVA
R
Pf= kW
kW+kVAR
kVA
= Lassi
Lassi + Foam kW
DISADVANTAGES OF LOW
POWER FACTOR
Indirect costs of low power factor
Direct costs of low power factor
Loss in distribution capacity
i. Larger Investment
ii. Transformers
iii. Large size conductors
BENEFITS OF POWER
FACTOR CORRECTION
Benefit 1 - Reduce Utility Power Bills
Benefit 2 - Increase System Capacity
Benefit 3 - Improve System Operating Characteristics (Gain
Voltage)
Benefit 4 - Improve System Operating Characteristics (Reduce Line
Losses)
EQUIPMENTS CREATING
POOR POWER FACTOR
Lighting Incandescent Lamps
Fluorescent Lamps
Mercury Vapor Lamps
Distribution Transformer
Induction Motors
Synchronous Motors
Power Factor Correction
• Power factor correction is achieved by the
addition of capacitors in parallel with the
connected motor circuits and can be
applied at the starter, or applied at the
switchboard or distribution panel.
Our Goal is to achieve as near and similar
as possible the power factor resembling a
Resistive Load, near to 1 , or unity power
factor
Parallel Capacitor Banks
Supplied Voltage – 240 Volts Single phase.
Motor input – 10 KW
Power Factor – 0.65
Current (I1) = Power (kW)__ = 10000__ = 64.1 Amp.
Volts (V) * PF 240*0.65
If the power factor of the motor is increased to 0.9 the current
drawn by the motor shall be –
Current (I2) = Power (kW)__ = 10000__ = 46.3 Amp.
Volts (V) * PF 240*0.9
Thus, as the power factor decreases the current required for the same
value of active, or useful, power increases. The result is that the sizes of the
equipment, like the switchgear, cables, transformers, etc., will have to be
increased to cater the higher current in the circuit. All this adds to cost
AUTOMATIC SWITCHING OF
CAPACITORS
Questions & Answers