1. Calculation for capacitive power requirement for achieving desired
power factor.
Author: Prasad Gore
E: prasad@electricalsciencesolutions.com
Electrical Science Solutions, April 2015.
Abstract: There has been always search on
capacitor ratings requirement for achieving
desired power factor, and lot of online tools
are available. Also lot of technical papers
available on the issue. This paper explains
how the calculations can be done to arrive
on kvar requirement.
I Introduction
Power (va) is consist of two parts, the real
part i.e. active power(w), and another
imaginary part i.e. reactive power (var).
Almost in every industry the nature of load is
inductive and hence has lagging power factor
thereby require higher VA demand from
utilities. Installation of correct size power
capacitors can improve power factor, reduce
demand, improve voltage regulation, reduce
line losses , save PF penalties and some of
other indirect benefits.
II Complex plane
Fig. 1 represent a complex plane, where,
a‐ Real part
b‐ Imaginary part
z‐ Complex number.
Ф‐angle between a & b
z = a+jb,
In polar form, z L ф, Using trigonometry,
We get,
a = zcos ф and b=zsin ф.
z = zcos ф + j zsin ф ……………. (II.1)
III Voltage, Current & Power in complex
plane.
In Fig 2,
I ‐ Line current in amps
V‐ Line Voltage/ Network voltage in Volts
I = Icos ф –j I sin ф ………… ( From II.1)
I*= Icos ф + j I sin ф ( Conjugate)
V = V
We know that,
S = V x I*, ( S = Apparent Power)
S = V x (Icos ф + j I sin ф)
S = VI cos ф + j VI sin ф.
a=zcos ф
b=zsin фb
z
Fig.1
ф
I
Va=Icosф
b=Isinф
+j
-j
+j
-j
Fig.2
P=w=VI cosф
S=VI
Q=var=VIsinф
ф
-j
+j
Fig.3

2. Fig 3 is called power triangle. It shows, real
part i.e. active power P = VI cos ф
And imaginary part i.e.
reactive power Q = j VI sin ф.
Magnitude IQI = VI sin ф
So here this way the reactive power can be
measured using complex plane, when we
have the data of current I, Voltage V and
Power Factor.
Also, As we know we are compensating for
var by adding capacitors and we are not
compensating the active power required,
hence active power (w) remains constant and
va will vary as per var compensation. ‐‐‐‐‐(III.1)
using trigonometry , We get,
var = w tan ф ‐‐‐‐‐‐‐‐‐‐‐‐‐‐(III.2)
So here data required is active power P and
Power factor to calculate var.
IV Reactive power var required to achieve
desire power factor
Above mathematical expression gives var
required for complete compensation and
achieving unity Power factor. From above
expression, we can deduce var required for
certain desired power factor.
Fig 4 shows,
Q1= Initial total reactive power var required,
Q2= Final reactive power var required,
Q3= Reactive power var added to reach final
var,
ф1 = Initial angle between w & va
ф2 = Final angle between w & va
From fig we can understand,
Q3 = Q1 – Q2
From III.1 and III.2 we can express above ,
Q3 = w tan ф1 – w tan ф2
= w (tan ф1 – tan ф2) ……….. (IV.1)
So with above if we know active power, initial
power factor and desired power factor, we
can calculate reactive power var required for
achieving final power factor.
V Selection of Capacitor voltage for required
reactive power var
Capacitors are rated in (Qn) ‐ rated kvar
output at (Vn) rated capacitor voltage.
We know that,
Qn = Vn Isin ф …………Capacitor rated power
Therefore,
Qn ἀ Vn2
So here if we have (V1) a network voltage,
then what amount of (Q1) reactive power can
be delivered by (Qn) rated capacitor which
has (Vn) rated voltage.
As per proportionality,
Qn ἀ Vn2
Q1 ἀ V12
Therefore,
Q1= Qn x V12
/Vn2
………………..(V.1)
Vice versa,
We know (Q1) which is required reactive
power to be delivered; we can easily select
Qn from eq. (V.1)
VI Example
Consider a load of 1000 kW at 0.85 power
factor, what is the kvar requirement of load to
reach Power factor of 0.95, how much kvar
w
Q1
Q3
Q2
S1
S2
ф2
ф1

3. capacitor required of rating at 440V when
network voltage is 415V
=>
Kw= 1000
cos ф1 = Initial PF= 0.85
cos ф2 = Desired PF= 0.95
Q= ?
V1= Network Voltage=415V
Vn= Capacitor rated voltage= 440V
Qn= Capacitor kvar rating
ф1= cos –
(ф1) ф2= cos –
(ф2)
ф1= 31.78 ° ф1= 18.19 °
Q= Kvar= kw (tan ф1 – tan ф2) ….. (from IV.1)
= 1000 * ( 0.6195‐ 0.3286)
Q = 290.91 Kvar
Now,
From eq. V.1
Qn = Q1 x Vn2
/ V12
= 290.91 x 440 2
/ 415 2
Qn = 327.02 Kvar