# Economics of harmonic filters for large power systems #### WHAT ARE HARMONICS ?

Electrical loads can be divided into linear and non-linear loads. A linear load is one which draws a sinusoidal current when subjected to a sinusoidal voltage as shown in Fig 1 (a). A pure resistance, capacitance, inductance or a combination of these elements form a linear load.

A non-linear load draws a non-sinusoidal current, when subjected to a sinusoidal voltage as shown in Fig 1(b).

Linear loads draw currents that are proportional to applied voltages. For example incandescent lighting, heating and motor loads. Non-linear loads draw currents for only a part of voltage cycle and introduce harmonics.

Any non-sinusoidal current can be mathematically resolved into a series of sinusoidal components (fourier series). The first component is called fundamental component and the remaining components whose frequencies are integral multiple of the fundamental frequency are known as harmonics. In India, fundamental frequency is 50 Hz. Hence, 2nd harmonic will have a frequency of 100 Hz, 3rd harmonic will have 150 Hz and so on. A diagrammatic representation of the various harmonics of a non-sinusoidal current is shown in Fig 1(c). Here, f1 = fundamental frequency, f5 = Fifth harmonic, f7 = Seventh harmonic, f11 = Eleventh harmonic. The figure also shows the resultant non-linear current.

#### SOURCES OF HARMONICS

Some of the non-linear loads which generate Harmonics are given as below.

• Semiconductor devices like power rectifiers, power inverters, UPS, telecommunication equipments, computers etc.
• Variable frequency drives.
• Saturated transformers.
• Induction and Electric arc furnaces.
• Gas discharge lamps.
• Fluorescent lamps.

#### ILL EFFECTS OF HARMONICS

Table -1 indicates the ill effects of harmonics on power system equipments.

#### FUNCTIONS OF HARMONIC FILTER – ‍

The objective of the shunt type harmonic filter is to shunt harmonic current from the load into the filter, by providing low impedance path compared to total impedance of source for a particular order of harmonic frequency for which it is tuned,  thereby reducing the amount of harmonic current that flows into the power system.

To achieve above requirements, first order shunt type tuned passive filter is suitable and economical for large power system applications. This type of filter is suitable for power systems of more than 500 kVA, where harmonic frequencies of the order of 5th and above are dominant.

The first order shunt type tuned passive filters is a series combination of resistance, inductance and capacitance (RLC) which is connected in parallel to the connected loads of power system, as shown in Fig-4.

It can be understood that design and installation of the harmonic filters shall result in improvement in power quality for good health and long life of power capacitors & eliminate ill effects of harmonic current as indicated in table-3, improvement in power factor (pf) and reduction in KVA demand from supply side. This will in turn result in reduction in fixed charges in electricity bill and incentive from DISCOM if pf > 0.95.

In our analysis we shall consider a simple harmonic filter of 400 kVAr rating and capable to eliminate 5th order harmonic currents. The separate harmonic filters shall be required to be connected in parallel to filter out harmonic frequencies of higher orders.

#### ECONOMICS OF INSTALLING HARMONIC FILTER‍

For economical analysis and calculation of savings by installing harmonic filter for 5th order harmonic current, let us assume certain parameters as given below. Phasor diagram for the following parameters is given in fig-5.

As shown in fig-4, the harmonic filter is connected in star configuration.

Line to line source voltage (VLL) = 415 V

Line to neutral voltage at terminals of harmonic filter (VPH) =                  415 V/1.732 = 239.6 V

Reactive power fed by of the harmonic filter (Qhf) = 400 kVAr

Order of harmonic frequency (h) = 5

Fundamental frequency (f) = 50 Hz

Initial pf (pf1) = 0.8, initial power factor angle = Φ1

final pf (pf2) = 0.98, final power factor angle = Φ2

Active power demand of power system  = P kW

Reactive power requirement of connected loads before use of harmonic filter = Q1 kVAr

Reactive power requirement of connected loads after installation of harmonic filter = Q2 kVAr

Apparent power requirement of connected loads before use of harmonic filter = A1 kVA

Apparent power requirement of connected loads after use of harmonic filter = A2 kVA

##### Calculation for economical analysis is given as below.

Initial pf angle (Φ1) = Cos-1pf1 = Cos-1 0.8 = 36.8 degree

Final pf angle (Φ2) = Cos-1pf2 = Cos-1 0.98 = 18.2 degree

We have specified Qhf = Q1 - Q2 = 400 kVAr

Q1 = P tan Φ1 = P tan (Cos-1 0.8) kVAr

Q2 = P tan Φ2 = P tan (Cos-1 0.98) kVAr

Q1 – Q2 = P {tan (Cos-1 0.8) – tan (Cos-1 0.98)} kVAr

• Q1 – Q2 = P X 0.547 kVAr
• Qhf = P X 0.547 kVAr
• P = Qhf/0.547 = 400/0.547 kW = 731.26 kW

A1 =  P/0.8 = 731.26/0.8 = 914 kVA

A2 =  P/0.98 = 731.26/0.98 = 746.2 kVA

Reduction in kVA demand (A) = A1 – A2 =167.8 kVA

Monthly tariff for fixed charge based on kVA demand = Rs  240/kVA

Savings per month due to Reduction in kVA demand = 167.8 kVA * Rs 240/kVA = Rs. 40,272

Savings per annum due to Reduction in kVA demand = Rs 40272 * 12

• Savings per annum = Rs 4,83,264.

We get incentive from DISCOM for improvement in power factor above 0.95 @ of 10 % of monthly bill amount for every 0.1 increment in power factor above 0.95. This saving is calculated as below.

Tariff for energy charges for industrial consumers varies with time of day (TOD) depending upon peak hours or off peak hours of load demand.

In this way, present average tariff for energy charges for industrial consumers = Rs 6.8/unit.

Active (kW) load demand for this typical power system has already been calculated as 731.26 kW.

Hence, consumption of units per day = 731.26 kW * 24 hours = 17,550 units.

Consumption of units per month = 17550 * 30 = 5,26,500 units

Consumption of units per year = 17550 * 365 = 64,05,838 units.

Energy charges per month (CE) = Rs 6.8 * 5,26,500 = Rs 35,80,200

Energy charges per year = Rs 6.8 * 64,05,838 = Rs 4,35,59,695

= Rs 43.5 million.

Fixed charges per month (CF) = A2 kVA* Rs 240/kVA

• CF = Rs 746.2 * 240 = Rs 1,79,088
• Total monthly bill CM=CE+CF=Rs 3580200+Rs179088
• = Rs 37,59,288

Power factor incentive per month = Rs (pf2 – pf1) * CM

• = Rs (0.98 – 0.8) * 3759288 = Rs 6,76,672
• Total savings = savings due to kVA demand reduction + pf incentive
• = Rs. 40,272 + Rs 6,76,672
• Total savings per month = Rs 7,16,944
• Total savings per year = Rs 7,16,944 * 12 = Rs 86,03,328

##### Calculation of payback period

Approximate cost of the Harmonic Filter of 400 kVAr = Rs 15,00,000

Cost of filter includes interest on investment.

Payback period in months = Cost of filter*12/ annual savings

• = Rs 15,00,000*12/ Rs 86,03,328 months
• Payback period in months = 2.0 months

Now, it can be understood that installation of harmonic filter shall provide two fold advantage. First advantage is annual savings in electricity bill @ Rs 86,03,328 due to improvement in power factor and second advantage is improvement in power quality of power system parameters thereby enhancement in life & performance of power system equipments.

This analysis shall create awareness among consumers about importance of power quality and power factor improvement by installation of harmonic filters.