Power Factor
Correction and VAR
compensation
Week 8-9
What is power factor:
Percentage of electricity that is being useful for doing work and is defined as the ratio
between the ‘active or actual power’ in kW or W, to the ‘apparent power’ expressed in volt-
ampere or kilo volt-ampere
𝑃𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟 =
𝐴𝑐𝑡𝑖𝑣𝑒 𝑝𝑜𝑤𝑒𝑟
𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑒 𝑝𝑜𝑤𝑒𝑟
The apparent power also referred to as the total power supplied by the utility has two
components:
a- Productive component: powers the equipment and performs useful work and is
given in kW or W
b- Reactive power: generates magnetic field to produce flux necessary for the
operation of induction devices (AC motors, transformers, induction furnaces..etc.) and is
measured in kVar of Var. It produces no productive power
Industrial plants introduces inefficiencies into the electrical supply network by drawing
additional currents, called “inductive reactive currents”. Although these currents produce no
useful power, they increase the load on the supplier’s switch gear and distribution network
and on the consumer’s switchgear and cabling.
Typical un-improved power factor by industry
Industry
Power Factor
Industry
Power Factor
Arc welding
35-60
Paint manufacturing
65-70
clothing
35-60
Metal working
65-70
Chemical
65-75
Plastic
75-80
Cement
80-85
Office building
80-90
Calculation of the necessary reactive power:
This is done using:
1. Calculations and Power factor triangle method
2. Tables
3. Curves
Once the power factor 𝑐𝑜𝑠𝜑1 of the installation and the power factor to be obtained 𝑐𝑜𝑠𝜑2
are known, it is possible to calculate the reactive power of the capacitor bank necessary to
improve the power factor as:
𝑄𝑐 = 𝑃𝐿 × tan 𝜑1 − tan 𝜑2
Load PF= 𝑐𝑜𝑠𝜑1
Improved PF= 𝑐𝑜𝑠𝜑2
𝑃𝐿 = OB= kW-load capacity
OA=𝑆ȁ load capacity
OD=𝑆ȁ reduced capacity
Using tables
𝑘𝑊 = 𝑘𝑉𝐴1 × 𝑐𝑜𝑠𝜑1 , 𝑘𝑉𝐴𝑅1
= 𝑘𝑉𝐴1 × 𝑠𝑖𝑛𝜑1
∴ 𝑘𝑉𝐴𝑅1 = 𝑘𝑊 × 𝑡𝑎𝑛𝜑1
For correction to 𝑐𝑜𝑠𝜑2 :
∴ 𝑘𝑉𝐴𝑅2 = 𝑘𝑊 × 𝑡𝑎𝑛𝜑2
Therefore, the required correction capacitors
∴ 𝑄𝑐 = 𝑘𝑉𝐴𝑅1 − 𝑘𝑉𝐴𝑅2
∴ 𝑄𝑐 = 𝑘𝑊 × 𝑡𝑎𝑛𝜑1 − 𝑡𝑎𝑛𝜑2
Finally,
Multiplying factor
found in table
𝑄𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = 𝑄𝑐 × 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑓𝑎𝑐𝑡𝑜𝑟
Using curves:
In this case, choose the initial pf
value and select the current power
factor curve, the value of the
required capacitor is chosen from
the intersection of both values in
terms of the original kW load
Is it that simple?
Of course not!!!
Power factor correction in Distribution transformer:
 Transformers are often in constant service. It is advisable that power factor
correction is carried out by keeping into account the transformer reactive power
so that an average power factor equal to 0.9 on the MV side is guaranteed.
 Generally, the compensation power 𝑄𝑐 [kVar] in a transformer having a rated 𝑆𝑟
[kVA], shall not exceed the reactive power absorbed under minimum reference
load condition.
Apparent power kVA
Transformer
7.2-23 kV
24 kV
36 kV
No load
load
No load
load
No load
load
630
17
40.7
18.8
43.6
21.2
46
500
13.5
32.3
15.8
36.8
18
39
400
1.8
25.7
13.2
30
15.2
32
100
3.6
6.92
4.16
7.96
5.08
8.88
More than
double!
European standard
Power factor correction in Distribution transformer:
3 phase transformer with transformation ratio of 6.6kV/ 433V or 11 kV/433 V
Apparent power kVA
Transformer
Core losses
(W)
Copper
losses (W)
% magnetizing
current
1000
1770
11820
1.2
500
1030
6860
1.53
 Deriving from the nameplate characteristics of the transformer the percentage no-load
current 𝐼𝑜 %, the percentage short-circuit voltage 𝑢𝑘 %, the iron losses 𝑃𝑓𝑒 and the copper
losses 𝑃𝑐𝑢 [kW], the required compensation power results to be about:
𝑄𝑐 =
2
𝐼𝑜 %
× 𝑆𝑟
100
≈
2
− 𝑃𝑓𝑒
+
𝐾𝑙2
×
2
𝑢𝑘 %
× 𝑆𝑟
100
2
− 𝑃𝑐𝑢
𝐼𝑜 %
𝑢𝑘 %
× 𝑆𝑟 + 𝐾𝑙2 ×
× 𝑆𝑟
100
100
Where 𝑘𝐿 is the load factor, defined as the ratio between the minimum reference load and
the rated power of the transformer
Power factor correction in distribution transformer:
𝑇𝑎𝑏𝑙𝑒 1: 𝑄𝑐 [kVar] to be connected to the secondary winding of an ABB transformer
Power factor correction in distribution transformer:
USA recommendation practice: in case
of use of fixed capacitor banks placed
on the secondary transformer windings,
capacitors should be sized to 40-67%
from transformer apparent power rating
Tables are available for both
power transformers and
distribution transformers
Egyptian code
Power factor correction in Three phase induction motor:
Power factor versus speed performance. Better
PF of 3-phase IM versus rated power output
power factor is achieved at higher speed
in kW (or HP). Power factor improves with
higher motor power rating
Power factor correction in Three phase induction motor:
As the rotor is loaded an increasing resistive
component is reflected from rotor to stator,
increasing the power factor.
Power factor improves with higher
loading for the same power rating.
Power factor correction in Three phase induction motor:
Power factor correction capacitors can be
calculated using the following:
=
𝐾𝑤 × 𝑎𝑐𝑡𝑢𝑎𝑙 % 𝑙𝑜𝑎𝑑𝑖𝑛𝑔 × 𝑐𝑜𝑟𝑟. 𝑓𝑎𝑐𝑡𝑜𝑟
𝑒𝑓𝑓𝑒𝑐𝑖𝑒𝑛𝑐𝑦 𝑎𝑡 𝑎𝑐𝑡𝑢𝑎𝑙 𝑙𝑜𝑎𝑑
Correction factor: from tables
% loading: actual load / full load (usually 75%
from full load)
Efficiency at actual load: from chart
Tables for individual power factor
correction for three phase induction
motors
How much can I save by installing power capacitors?
 Power capacitors used for power factor correction provide many benefits:
 Increase system capacity
 Reduce utility power bills
 Improve system operating characteristics (voltage gain)
 Improve system operating characteristics (reduce line losses)
 Increase system capacity:
PF improvement releases system capacity and
permits additional loads (motors, lighting
..etc.) to be added without overloading the
system. Capacitors reduce the current drawn
from the power supply, less current means
less load on transformers and feeder circuits,
leading to more investment in other devices
such as transformers.
Switched capacitor panel
How much can I save by installing power capacitors?
 Reduce utility power bills:
Electricity bill contains kW and kVAr of power plant. While reactive power
doesn’t register on kW demand or kW hour meters, the utility’s transmission
and distribution system must be large enough to provide the total power.
Utilities have various ways of passing along the expense of larger generators,
transformers, cables, switches..etc. Capacitors will save money no matter how
the utility bills on power. Utility charges according to the kW demand and add
a surcharge or adjustment for power factor. The adjustment may be a
multiplier applied to kW demand.
If the power factor was 0.84, the utility
𝑘𝑊 − 𝑑𝑒𝑚𝑎𝑛𝑑 ∗ 0.9
𝑛𝑒𝑤 𝑢𝑡𝑖𝑙𝑖𝑡𝑦 𝑏𝑖𝑙𝑙 =
𝑎𝑐𝑡𝑢𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟
would require
𝟎.𝟗
𝟎.𝟖𝟒
= 𝟏. 𝟎𝟕 kW –normal
demand or 7% increase in utility billing
How much can I save by installing power capacitors?
 Improve system operating characteristics (voltage gain):
Good PF provides “stiffer” voltage, typically a 1-2% voltage rise can be
expected when PF is brought to ±0.95. Excessive voltage drop make your motors
sluggish, and cause them to overheat.
Low voltage also interferes with lighting, the proper application of motor controls and
electrical and electronic instruments.
An estimation of the voltage rise from the improved power factor with the installation of
power capacitors can be made using the following:
% 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑖𝑠𝑒 =
𝑘𝑉𝐴𝑟 𝑜𝑓 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟𝑠 × % 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒
𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 − 𝑘𝑉𝐴
How much can I save by installing power capacitors?
 Improve system operating characteristics (reduce line losses):
Improving PF at the load point shall relieve the system of transmitting reactive
current. Less current shall mean lower losses in the distribution system of the
facility since losses are proportional to the square of the current. Therefore, the
fewer kW-hr needed to be purchased from the utility
An estimation of the power losses can be made using the following:
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑃𝐹
% 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑝𝑜𝑤𝑒𝑟 𝑙𝑜𝑠𝑠𝑒𝑠 = 1 −
𝑖𝑚𝑝𝑟𝑜𝑣𝑒𝑑 𝑃𝐹
2
× 100
Types of power factor compensation:
1.
Individual correction
2.
Group compensation
3.
Centralized compensation
4.
Combined compensation
5.
Automatic compensation
 Individual correction
 Applied directly at the terminals of the load which demand reactive power.
 Simple, capacitor and load can use the same protective devices against over current and
are connected and disconnected simultaneously.
 Advisable in the case of large equipment with constant load and continuous operation
 Used with motors and lamps
 Individual correction
 In case of direct connection (diagram 1 and 2), after the motor disconnection from the
supply, the motor will continue to rotate (residual KE) and self excite with the reactive
energy drawn from the capacitor and may turn into an asynchronous generator and
might damage with over voltage
 With this type of correction the network on the supply side of the load works with a high
power factor, on the other hand, this solution is costly
 Delta connected capacitors for 3rd harmonic cancellation
 Group correction
 Improving locally the power factor of groups of loads having similar functioning
characteristics by installing a dedicated capacitor bank.
 Power factor is improved only by upstream the point where the capacitor bank is
located.
 The capacitor requires its own switching device.
 Centralized correction
 System with permanently changing loads, several compensation units are usually
installed centrally in the main distribution switchboard.
 Switched ON and OFF by a controller depending on the respective reactive power
demand.
 Rated for the highest load, thus the total compensation power installed is lower than for
the other types of compensation. However, the load on the distribution system is not
reduced, and there must be enough room for all capacitor units in the main distribution
panel.
 Combined PF Compensation
 Derives from a compromise between the two solutions of individual and
centralized power factor correction and it exploits the advantages they offer.
 In this case, distributed compensation is used for high power electrical
equipment and the centralized modality for the remaining part.
 Combined solution is used in installations where large equipment only are
frequently used, in such circumstances their power factor is corrected
individually, whereas the PF of small equipment is corrected by the
centralized modality.
 Automatic PF Compensation
 Automatic correction is used in installations with variable absorption of reactive power.
This system monitors the installation and automatically switches different capacitor
banks.
 This system required sensors, intelligent system comparing measured and desired
operation to connect/disconnect capacitors, and electric power boards with switches and
protective devices.
Power factor correction: Common Problems
1- Harmonic and Resonance:
• Capacitors connected to induction motors increase the chance of resonance
between the power factor capacitors and the motor’s inductive reactance.
• This issue makes the selection and filters design even more difficult.
• For these reason, NEMA standards (NEMA MG 1-1993 section 14.43.4)
have recommended not to use individual correction technique in presence of
large number of induction motors.
𝑓𝑟𝑒𝑠 =
1
2𝜋 𝐿𝐶
Increasing C means that the square root increases so that the resonant frequency
reduces and could approach the 50 Hz. Non linear loads introduce currents of
harmonic ferq. 150-300 Hz so it even becomes nearer to the boundaries of
resonance
Power factor correction: Common Problems
2- Self Excitation:
• Capacitors are connected in parallel to motors
where both are fed/disconnected from the same
source together.
• Motors store energy in their rotating mass while
capacitors stores energy in their electric field.
• When motors are disconnected (with capacitors) Capacitor and motor magnetization
from the power source, the motor will continue to
rotate due to the energy stored in its inertia. At
this point energy is being exchanged between the
capacitor and the motor.
• Current will pass in the circuit between the motor
and capacitors and this current depends on the
voltage difference between both.
curve for self excitation
Power factor correction: Common Problems
 assume motor terminal voltage is 460 v and that the
capacitors used are 6 kVAr represented by the
straight line shown. The intersection point between
the capacitor load line and the magnetizing curve
shows that when the motor is disconnected both the
motor and capacitor will have the same voltage.
 Assume another capacitor is used of 14.4 kVAr. At
460 volts, motor draws reactive current of 8 A while
capacitor draws 18 A.
 When both are disconnected from source, the motor
terminal voltage will rise suddenly to 680 V. Leading
to that the capacitor tries to discharge this
“unbalanced” energy between capacitor and motor
leading to rise in motor terminal voltage.