Download How to Calculate KVAR PDF

TitleHow to Calculate KVAR
TagsPhysics Electrical Engineering Physics & Mathematics Capacitor
File Size422.7 KB
Total Pages4
Document Text Contents
Page 1

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

Similer Documents