2. Power Factor Basics
• In all industrial electrical distribution systems, the major loads are resistive and inductive. Resistive loads
are incandescent lighting and resistance heating. In case of resistive loads the voltage (v) , current (I),
resistance relations are linearly related, i.e. V= I x R and Power (kW) = V x I
• Typical Inductive loads are A.C. Motors, Transformers and ballast –type lightings. Inductive loads require
two kinds of power : a) Active power (kW) to perform the work and b) Reactive power ( KVAr) to create
and maintain electro-magnetic field.
• The vector sum of the active power and reactive power make up the total (or apparent) power used
which is measured in KVA
3. • The ratio of KW to KVA is called the Power Factor which is always less than or equal to unity. Theoretically, if the loads have
unity power factor, maximum power can be transferred for the same distribution system capacity.
• However, as the loads are inductive in nature, with the power factor ranging from 0.2 to 0.9, the electrical distribution
network is stressed for the capacity at low power factor.
KVAR
Reactive Power
KW
Active Power
KVA
Total Power
KW, KVAr and KVA Vector Diagram
PF = KW / KVA
= Cos ф
Cos ф
• The active power (shaft power required) in KW and the reactive power required in (KVAr) are 900 apart vectorially in a pure
inductive circuit i.e. reactive power lagging active power. The vector sum of two is called the apparent power KVA.
4. A Starch Industry had installed a 1500 KVA transformer. The initial demand of the plant was 1160 KVA with power factor of 0.70 . The %
loading of transformer was about 78% ( 1160/1500 = 77.3 %) .
To improve the power factor and to avoid the penalty, the unit had added about 410 KVAr capacitor in motor load end. This improves
the PF to 0.89 and reduced the required KVA to 913 which is vector sum of KW and KVAr.
KVAR =828
KW=812
Cos ф =0.70
PF =812/1160
= 0.70
KVA=1160
KW=812
KVAR=
828-410=418
KVA=913
PF= 812/913
= 0.89
After improvement, the penalty will be avoided and the 1500 KVA transformer is now
loaded only to 60% capacity. The transformer can now also be loaded more in the future.
Cos ф =0.89
Example
5. Advantages of Power Factor Improvement
• Reactive component of the network reduced
and so also the total current in the system
from the source end.
• I2 R losses of the system are reduced because
of reduction in current.
• Voltage level at the load end is increased.
• KVA loading on the source Generators,
transformers and lines up to capacitor
reduces giving capacity relief.
• A high power factor can help in utilizing the
full capacity of the electrical system
6. Cost Benefits of PF improvement
• Reduction in KVA demand ( Maximum
demand) charges in utility bill.
• Distribution losses (KWH) reduced within the
plant network.
• Better voltage at motor terminals and
improved performance of motors.
• A high power factor eliminates penalty
charges imposed when operating at low
power factor.
• Reduction in investment on kVA rating of
transformers , Cables, switchgear etc as its
delivering load reduces.
7. KW / KVA = Cos ф1
KVAr / KVA = Sin ф1
KVAr/ KW = Sin ф1 / Cos ф1
KVAr/ KW = Tan ф1
KVAr = KW X Tan ф1
KVAr required to improve the power factor
from Cos ф1 to Cosф2 will be
KVAr = KW ( tan ф1 – tan ф2 )
Example: The utility bill shows an average power factor
of 0.72 with an average kw of 650 . How much KVAr is
required to improve the power factor to 0.95.
Cos ф1 = 0.72 , tan ф1 = 0.963
Cos ф2 = 0.95 , tan ф2 = 0.329
KVAr required = KW ( tanф1 – tanф2)
= 650 ( 0.963 -0.329 ) = 412 KVAr
• The primary purpose of capacitors is to reduce the maximum demand. Maximum benefits of capacitors is derived by
locating them as close as possible to the load.
• At this location , its KVAr are confined to the smallest possible segment, decreasing the load current .This in turn will
reduce power losses of the system substantially .
• Power losses are proportional to the square of the current . When power losses are reduced, the voltage at the motor
terminals increases ,thus motor performance also increases.
Selection and location of capacitors
8. Operational Performance of Capacitors
•This can be made by monitoring capacitor charging current via the rated charging current.
• The capacity of fused elements can be replenished as charging current.
•The portable analyzer can be used for measuring KVAr delivered as well to charging current.
•Capacitors consume 0.2 to 6.0 watt per KVAr, which is negligible in comparison to benefits
• Nameplates can be misleading with respect to rating . It is good to check
by charging current.
• Capacitor boxes may contain only insulated compound and insulated
terminals with no capacitor element inside.
• Capacitors for single phase motor starting and those used for lighting
circuit for voltage boost are not power factor capacitor units and these
cannot withstand power system conditions.
Some basic checks for the Capacitors :
9. • Induction motors are characterized by power factor less than unity i.e. lagging PF. It leads to lower overall
efficiency and higher overall operating cost.
• The capacitors are connected in parallel with motor to improve power factor.
• The impacts of PF correction include :
• reduced KVA demand (and hence reduced utility demand charges),
• reduced I2R losses in cables upstream of the capacitors (and hence reduced energy charges),
• reduced voltage drop in the cables (leading to improved voltage Regulation).
•This increases the overall efficiency of the plant.
• The size of capacitors required for a particular motor should not exceed 90% of the no load KVAr of the
motor.
Power factor correction of Motors
