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How to Calculate the Suitable Capacitor Size in Farads & kVAR

for Power factor Improvement (Easiest way ever)

How to Calculate the Suitable Capacitor Size in Farads & kVAR for Power factor

Improvement (Easiest way ever)

Hi there! With a very important tutorial.. I hope you will find it very useful because I have already spent

two days to prepare this article. I think all of those who have sent messages and mails about the topic

will never ask again if they follow these simple methods to calculate the proper Size of Capacitor bank in

kVAR and micro-farads for power factor correction and improvement in both single phase and three

phase circuits. I think it’s too much..

Now let’s begin...

Consider the following Examples.

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 #1 (By Simple Table Method)

Motor Input = 5kW

From Table, Multiplier to improve PF from 0.75 to 0.90 is .398

Required Capacitor kVAR to improve P.F from 0.75 to 0.90

Required Capacitor kVAR = kW x Table 1 Multiplier of 0.75 and 0.90

= 5kW x .398

= 1.99 kVAR

And Rating of Capacitors connected in each Phase

1.99/3 = 0.663 kVAR

Solution # 2 (Classical Calculation Method)

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 #1 (By Simple Table Method)

Supplying kW = 650 kW

From Table 1, Multiplier to improve PF from 0.65 to unity (1) is 1.169

Required Capacitor kVAR to improve P.F from 0.65 to unity (1)

http://www.electricaltechnology.org/2013/11/How-to-Calculate-Suitable-Capacitor-Size-for-Power-factor-Improvement.html

http://www.electricaltechnology.org/2013/11/How-to-Calculate-Suitable-Capacitor-Size-for-Power-factor-Improvement.html

How to Calculate the Suitable Capacitor Size in Farads & kVAR

for Power factor Improvement (Easiest way ever)

How to Calculate the Suitable Capacitor Size in Farads & kVAR for Power factor

Improvement (Easiest way ever)

Hi there! With a very important tutorial.. I hope you will find it very useful because I have already spent

two days to prepare this article. I think all of those who have sent messages and mails about the topic

will never ask again if they follow these simple methods to calculate the proper Size of Capacitor bank in

kVAR and micro-farads for power factor correction and improvement in both single phase and three

phase circuits. I think it’s too much..

Now let’s begin...

Consider the following Examples.

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 #1 (By Simple Table Method)

Motor Input = 5kW

From Table, Multiplier to improve PF from 0.75 to 0.90 is .398

Required Capacitor kVAR to improve P.F from 0.75 to 0.90

Required Capacitor kVAR = kW x Table 1 Multiplier of 0.75 and 0.90

= 5kW x .398

= 1.99 kVAR

And Rating of Capacitors connected in each Phase

1.99/3 = 0.663 kVAR

Solution # 2 (Classical Calculation Method)

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 #1 (By Simple Table Method)

Supplying kW = 650 kW

From Table 1, Multiplier to improve PF from 0.65 to unity (1) is 1.169

Required Capacitor kVAR to improve P.F from 0.65 to unity (1)

http://www.electricaltechnology.org/2013/11/How-to-Calculate-Suitable-Capacitor-Size-for-Power-factor-Improvement.html

http://www.electricaltechnology.org/2013/11/How-to-Calculate-Suitable-Capacitor-Size-for-Power-factor-Improvement.html