Course: ELEC 423 Electric Power distribution

Example on Design of Passive Harmonics Filter

It is desired to convert an existing 600 kVAR shunt capacitor bank to a passive harmonic filter that

traps the 5th harmonics component. The capacitor is connected at 11 kV bus for power factor

improvement. Find the parameters of the passive harmonic filter.

Solution: Here we have V=11 kV and the harmonic order we want to remove is 5th harmonic order

(h= 5). Given that QC is 600 VAR and the total load is 3000KVA. The impedance of the capacitor at

the fundamental frequency is:

Ω=== 667.201600

)11000( 22

capC Q

VX

The impedance of the inductor at the fundamental frequency is

Ω=== 0667.8)5(667.20122h

XX C

L

The shunt impedance of the harmonic filter at 50 Hz is

6003.193=−= LCf XXX

Applying IEEE/ANSI standard for the capacitor element of the designed harmonics

After installing the filter, the 5th harmonic order will be reduced. In this case, we can find the

amount of current that is going to pass through the filter. The current through the harmonic filter is

AkVXV

If

LNf 8039.32

6003.1933/11

1 ===

The 5th harmonic current to which the harmonic filter will be subjected is

==3*11

3000)05.0(5 kVI f 7.87 A

We can also calculate the effective current flowing through the harmonic filter by:

AIII ffeff 7365.33)( 25

21 =+=

The fundamental frequency voltage across the capacitor is

VXIV cfc 6895.661511 ==

The fundamental frequency voltage across the reactor is

VXIVV xfrqactor 6895.66153 11 =−=

VX

IV cfc 42.317

5667.20187.7

555 ===

VV totalc 3.6623)42.317689.6615( 22)( =+=

The ANSI standards states that the value of Vc(total) should be less than 110 % of the nominal

voltage which is 12000 V, so it is within the capacitor rating.

kVARKVARKVAR

kVARIVKVAR efftotal

3.670*3

43.223*

13

1

==

==

φφ

φ

It should also be verified that the total KVAR production of the capacitor is less than 135 %

of the nominal value, which is an ANSI defined limit. This value of exceeds the nominal KVA rating

by 11.7 %, so it is within the acceptable range.