Design and costing of a Harmonic Filter.

Mr. Sachin V. Shelar. ( R & D Engineer )

Madhav Capacitors Pvt. Ltd., Pune.

A harmonic filter essentially consists of a power capacitor, a tuning reactor and its control gear. It will act in parallel with an untuned basic power factor improvement capacitor bank.

For the designing purpose we require some data such as load details, existing power factor, required new power factor, total harmonic distortion, individual harmonic details etc. This data can be obtained from a harmonic analyser.

First calculate the total KVAR required to improve the power factor to the desired value. Then distribute the total KVAR into two parts, a normal capacitor bank (untuned) for power factor correction and a filter bank (tuned) for filtration purpose. This is explained in detail in the example given below.

We consider a practical example of a rolling mill, circuit diagram is as shown in Fig. 1.

Fig.1 Line diagram of the load.

Data obtained from harmonic analyser at 30% load :

Load in KW = 606.20.

Power factor = 0.42 lag.

ITHD = 30.47.

5th harmonic current = 39.25 Amp.

Assuming required power factor to be 0.97, we calculate the total KVAR required to raise the power factor from 0.42 to 0.97.

KVAR required = KW (tanf 1- tanf 2)

KW = 606.20

Cosf 1 = 0.42 tanf 1 = 2.16

Cosf 2 = 0.97 tanf 2 = 0.25

KVAR required = 606.20 (2.16 - 0.25)

= 1157.80 KVAR.

1158 KVAR.

Thus we require 1158 KVAR to raise the power factor to 0.97.

From the readings it can be seen that, ITHD is 30.47 but maximum harmonic current distortion as recommended by IEEE specifications C-519-1992 is 4.0.

Out of 1158 KVAR to be installed we employ 30% KVAR towards filter duty and remaining KVAR for power factor correction.

Thus,

Filter KVAR = 30% of 1158.

= 347.4 KVAR.

Filter KVAR 350 KVAR.

Design of filter :

The basic diagram is as shown in Fig. 2.

Fig. 2 - Basic circuit diagram.

We consider a star connected capacitor bank.
  1. Capacitance per phase :
    2.

    C = 9.21 mfd/Ph.
  2. Capacitive reactance at 50Hz :
    3.

    XC50 = 345.80 W /Ph.
  3. Capacitive reactance at 250Hz :
    4.

    XC250 = 69.16 W /Ph.

    For resonance at 5th harmonic we should have,

    XL250 = XC250.

    XL250 = 69.16 W /Ph.

     

    From above we can write,

    XL50 = 13.82 W /Ph.

    So,

    L = 44 mH.

  4. Capacitor current at 50 Hz.

    5.

    IC50 = 18.37 Amp.

    5th harmonic current from the data is ,

    I250 = 39.25 Amp.

    40 Amp

  5. RMS current is given by,

    6.

    IRMS = 44 Amp.

    Fig. 3 Parameters.

  6. KVAR rating of Reactor :

    7.

    Voltage across reactor,

    VL = VL50 + VL250

    VL50 = Voltage drop across reactor due to fundamental

    current.

    = I50 x XL50

    = 18.37 x 13.82

    VL50 = 253.87 V.

    VL250 = Voltage drop across reactor due to 5th harmonic

    current.

    = I250 x XL250

    = 40 x 69.16

    VL250 = 2766.4 V.

     

    Total voltage across reactor (VL) :

    VL = VL50 + VL250

    = 253.87 + 2766.40

    = 3020.27

    VL 3020 V.

    KVAR rating of single phase reactor :

    = VL x IRMS

    = 3020 x 44.

    = 132.88 KVAR/Ph.

    133 KVAR.

    We use 3 such reactors.

  7. KVAR rating of capacitor.

    8.

Voltage across capacitor,

VC = VC50 + VC250

VC50 = Phase voltage = = 6350 V.

VC250 = I250 x XC250 = 40 X 69.16.

= 2766.4 V.

VC250 2766 V.

Total voltage across capacitor (VC) :

VC = VC50 + VC250

= 6350 + 2766

VC = 9116 V.

KVAR rating of capacitor :

= VC x IRMS

= 9116 x 44.

= 401.10KVAR/Ph.

401 KVAR.

 

 

 

 

Costing :

  1. Capacitor bank without filter :

    2.

 

Without filter bank circuit diagram.

 

Sr. No.

Item

Quantity

Cost in Rs.

1

11 kV Outdoor VCB.

1 No.

2,10,000

2

Breaker panel.

1 No.

1,80,000

3

3 Phase isolator

1 No.

26,000

4

2.3 KVAR, Gap cored series reactor.

1 No.

1,00,000

5

1158 KVAR capacitor bank including structure and fuses

1 No.

1,70,000

Total =

6,86,000

 

2. Capacitor bank with filter :

 

With filter bank circuit diagram.

 

Sr. No.

Item

Quantity

Cost in Rs.

1

11 kV Outdoor VCB.

1 No.

2,10,000

2

Breaker panel.

1 No.

1,80,000

3

3 Phase isolator.

2 No.

52,000

4

1.6 KVAR, Gap cored series reactor.

1 No.

85,000

5

808 KVAR main capacitor bank.

1 No.

1,16,000

6

399 KVAR tuning reactor.

1 No.

5,05,000

7

1203 KVAR capacitor.

(Harmonic filter)

1 No.

2,18,000

Total =

13,66,000

 

Conclusion :

When normal power factor correction bank is converted into the filter bank the cost almost becomes twice, but with a load of 2000 KW and 40 amp of 5th harmonic current at 11 kV it can be justified.

This particular consumer has been trying to maintain his power factor near 0.85 by -

He burns and replaces his LT capacitors within 2-3 months and then pays a heavy penalty until a new set of capacitor is installed.

  1. Cost of LT capacitors burnout:

    2. Rs. 4,20,000/-, which includes cost of switchgear, cables, installation etc.

  2. Penalty over an average monthly consumption of 600 kW x 18 hrs x 25 days per month at an energy rate of Rs. 31/KVAR. Power factor maintained by other means 0.75. penalty threshold 0.85.

    3. His monthly electricity bill is Rs. 10,60,000/-. His penalty will come to Rs. 1,06,000/- per month.

  3. His losses due to mal operation of microprocessor based equipment's within his premises, those due to burning out of cables , due to burning out of motor winding etc. are not included in above.

Reference :