Location of Shunt Capacitors
The power factor improver capacitor can be installed in:
1. system bus
2. distribution point
3. load side
In such cases it is recommended to install capacitor bank with the feeder which provides
entire of this particular load. This scheme is known as branch capacitor bank scheme. As
capacitor bank is directly connected to feeder or branch, it does not help to reduce losses in
primary system from where the branch comes out.
The main reasons of complexity is in that case, different sizes and capacities of capacitor
bank are required to be installed depending upon the demand of individual load. To overcome
this difficulty, it is always preferable to install a bulk capacitor bank at high voltage bus
system instead of installing smaller capacitor bank at every load point. Although the control
over reactive power of the system is little bit compromised but still it is much practical
approach in the view of complexity and cost. So capacitor bank at load side and capacitor
bank at primary systems both have their own benefits. Depending upon the demand of
system, both schemes are used.
Distribution System Capacitor Bank
In distribution feeder capacitor bank are installed on pole to compensate reactive power of
that particular feeder. These banks are normally mounted on one of the poles on which the
distribution feeders run. The mounted capacitor banks are normally interconnected with over
head feeder conductors by means of insulated power cable. The size of the cable depends
upon the voltage rating of the system. The voltage range of the system for which pole
mounted capacitor bank can be install, may be from 440 V to 33 KV. The rating of capacitor
bank may be from 300 KVAR to MVAR. The pole mounted capacitor bank can be either
fixed unit or switched unit depending upon varying load condition.
EHV Shunt Capacitor
In extra high voltage system, the generated electrical power may have to be transmitted a
long distance via transmission line. During journey of power, sufficient voltage may be
dropped due to inductive effect of the line conductors. This voltage drop may be
compensated by providing EHV capacitor bank at EHV sub-station. This drop of voltage is
maximum at peak load condition; hence, the capacitor bank installed for in this case should
have switching control to make it off and on as when required.
Substation Capacitor Bank
When high inductive load has to be delivered from a high voltage or medium voltage
substation, one or more capacitor bank of suitable size should be installed at substation to
compensate inductive VAR of the entire load. These capacitors banks are controlled by
circuit breaker and provided with lightening arrestors. Typical protection scheme along with
protection relays are also provided.
Metal Encoder Capacitor Bank
For small and industrial subtraction indoor type capacitor banks may also be used. These
capacitor banks are installed in metal cabinet. This design is compact and bank requires less
maintenance. The uses of these banks are more compared to outdoor bank, as these are not
exposed to external environment.
Distribution Capacitor Bank
Distribution capacitor banks are normally pole mounted capacitor bank installed nearer to
load point or installed at distribution subtraction.
These banks do not help to improve power factor of primary system. These capacitors bank
are cheaper than other power capacitors bank. All types of protection schemes for capacitor
bank cannot be provided to a pole mounted capacitor bank. Although pole mounted cap bank
is outdoor type but sometimes it is kept in metal enclosure to protect from outdoor
environmental conditions.
Fixed Capacitor Bank
There are certain loads mainly certain industrial loads which need fixed reactive power to
meet power factor. In this type feeder fixed capacitor bank is used. These banks do not have
separate control system to switch ON or OFF. These banks run with feeders. The banks are
connected to the feeders as long as the feeders are live.
Switched Capacitor Banks
In high voltage power system, compensation of reactive power is mainly required during
peak load condition of system. There may be reverse effect if the bank is connected to the
system at mean load condition.
At low load condition, the capacitive effect of bank may increase the reactive power of the
system instead of decreasing it.
In this situation capacitors bank must be switched ON during peak load poor power factor
condition and must also be switched OFF during low load and high power factor condition.
Here switched capacitor banks are used. When a capacitor bank is switched ON it provides
more or less constant reactive power to the system. It helps to maintain desired power factor
of the system even at peak load condition. It prevents, over voltage of system during low load
condition as capacitor is disconnected from the system during low load condition. During
operation of bank, it reduces losses both of the feeders and transformer of the system as it is
directly installed at primary power system.
How to Calculate the Suitable Capacitor Size in Farads & kVAR for Power factor
Improvement
Once the power factor (cos θ1) of the installation and the power factor to be obtained (cos θ2)
are known, it is possible to calculate the reactive power of the capacitor bank necessary to
improve the power factor.



P – the installed active power
θ1 – the phase displacement angle before power factor correction
θ2 – the phase displacement angle to be obtained with the power factor correction the
power of the capacitor bank Qc is:
How to Calculate the Suitable Capacitor Size in Farads & kVAR for Power factor
Improvement
Example: 1
A 3 Phase, 5 kW Induction Motor has a P.F (Power factor) of 0.75 lagging. What size of
Capacitor in kVAR is required to improve the P.F (Power Factor) to 0.90?
Solution :
Motor input = P = 5 kW
Original P.F = Cosθ1 = 0.75
Final P.F = Cosθ2 = 0.90
θ1 = Cos-1 = (0.75) = 41°.41; Tan θ1 = Tan (41°.41) = 0.8819
θ2 = Cos-1 = (0.90) = 25°.84; Tan θ2 = Tan (25°.50) = 0.4843
Required Capacitor kVAR to improve P.F from 0.75 to 0.90
Required Capacitor kVAR = P (Tan θ1 – Tan θ2)
= 5kW (0.8819 – 0.4843) = 1.99 kVAR
And Rating of Capacitors connected in each Phase
1.99/3 = 0.663 kVAR
Example 2:
An Alternator is supplying a load of 650 kW at a P.F (Power factor) of 0.65. What size of
Capacitor in kVAR is required to raise the P.F (Power Factor) to unity (1)? And how many
more kW can the alternator supply for the same kVA loading when P.F improved.
Solution :
Supplying kW = 650 kW
Original P.F = Cosθ1 = 0.65
Final P.F = Cosθ2 = 1
θ1 = Cos-1 = (0.65) = 49°.45; Tan θ1 = Tan (49°.45) = 1.169
θ2 = Cos-1 = (1) = 0°; Tan θ2 = Tan (0°) = 0
Required Capacitor kVAR to improve P.F from 0.65 to 1
Required Capacitor kVAR = P (Tan θ1 – Tan θ2)
= 650kW (1.169– 0) = 759.85 kVAR
P.F = Cosθ = kW/kVA
kVA = kW / Cosθ
= 650/0.65 = 1000 kVA
When Power Factor is raised to unity (1)
No of kW = kVA x Cosθ
= 1000 x 1 = 1000kW
Hence increased Power supplied by Alternator
1000kW – 650kW = 350kW
How to Calculate the Required Capacitor bank value in both kVAR and Farads?
(How to Convert Farads into kVAR and Vice Versa)
Example: 3
A Single phase 400V, 50Hz, motor takes a supply current of 50A at a P.F (Power factor) of
0.6. The motor power factor has to be improved to 0.9 by connecting a capacitor in parallel
with it. Calculate the required capacity of Capacitor in both kVAR and Farads.
Solution:
Motor Input = P = V x I x Cosθ
= 400V x 50A x 0.6
= 12kW
Actual P.F = Cosθ1 = 0.6
Required P.F = Cosθ2 = 0.90
θ1 = Cos-1 = (0.60) = 53°.13; Tan θ1 = Tan (53°.13) = 1.3333
θ2 = Cos-1 = (0.90) = 25°.84; Tan θ2 = Tan (25°.50) = 0.4843
Required Capacitor kVAR to improve P.F from 0.60 to 0.90
Required Capacitor kVAR = P (Tan θ1 – Tan θ2)
= 5kW (1.3333– 0.4843) = 10.188 kVAR
Solving for capacitance:
kVAR = 10.188
We know that;
IC = V/ XC
Whereas XC = 1 / 2 π F C
IC = V / (1 / 2 π F C)
IC = V * 2 pi * F *C
= (400) x 2π x (50) x C
IC = 125663.7 x C
And,
kVAR = (V x IC) / 1000 … [kVAR =( V x I)/ 1000 ]
= 400 x 125663.7 x C
IC = 50265.48 x C … (ii)
Equating Equation (i) & (ii), we get,
50265.48 x C = 10.188
C = 10.188 / 50265.48
C = 2.0268 x 10-4
C = 202.7 x 10-6
C = 202.7μF
Example 4
What value of Capacitance must be connected in parallel with a load drawing 1kW at 70%
lagging power factor from a 208V, 60Hz Source in order to raise the overall power factor to
91%.
Solution:
P = 1000W
Actual Power factor = Cosθ1 = 0.71
Desired Power factor = Cosθ2 = 0.97
Required Capacitor kVAR = P (Tan θ1 – Tan θ2)
=783 VAR (required Capacitance Value in kVAR)
Current in the Capacitor :
IC = QC / V = 783 / 208 = 3.76A
And
XC = V / IC = 208 / 3.76 = 55.25Ω
C = 1/ (2 π f XC)
C = 1 (2 π x 60 x 55.25)
C = 48 μF (required Capacitance Value in Farads)
Note:
Good to Know:
Important formulas which is used for Power factor improvement calculation as well as used
in the above calculation
Power in Watts
kW = kVA x Cosθ
kW = HP x 0.746 or (HP x 0.746) / Efficiency … (HP = Motor Power)
kW = √ ( kVA2– kVAR2)
kW = P = VI Cosθ … (Single Phase)
kW = P =√3x V x I Cosθ … (Three Phase)
Apparent Power in VA
kVA= √(kW2+ kVAR2)
kVA = kW/ Cosθ
Reactive Power in VA
kVAR= √(kVA2– kW2)
kVAR = C x (2 π f V2)
Power factor (from 0.1 to 1)
Power Factor = Cosθ = P / V I … (Single Phase)
Power Factor = Cosθ = P / (√3x V x I) … (Three Phase)
Power Factor = Cosθ = kW / kVA … (Both Single Phase & Three Phase)
Power Factor = Cosθ = R/Z … (Resistance / Impedance)
XC = 1/ (2 π f C) … (XC = Capacitive reactance)
IC = V/ XC … (I = V / R)
Required Capacity of Capacitor in Farads/Microfarads
C = kVAR / (2 π f V2) in microfarad
Required Capacity of Capacitor in kVAR
kVAR = C x (2 π f V2)