Power System Basics

Economics of Harmonic filters for Large Power Systems

In this article, we take a deep dive into the functions of Harmonic Filters in Large Power Systems and its day to day applications.

What are Harmonics?

Electrical loads can be divided into linear and non-linear loads. A linear load 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.  

Waveform of a Linear Load

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

Waveform for a Non Linear Load

Linear loads draw currents that are proportional to applied voltages for example incandescent lighting, heating, and motor loads. Non-linear loads draw currents only a part of the 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 the fundamental component and the remaining components whose frequencies are integral multiples of the fundamental frequency are known as Harmonics.  In India, the 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. 


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 equipment, computers, etc. 
  • Variable frequency drives. 
  • Saturated transformers. 
  • Induction and Electric arc furnaces. 
  • Gas discharge lamps. 
  • Fluorescent lamps.

Ill effects of Harmonics  

Table -3 indicates the ill effects of harmonics on power system equipment.

Ill effects of harmonics on power system equipment.

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 a low impedance path compared to the 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 the above requirements, the 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 the 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 the supply side. This will in turn result in a reduction in fixed charges in electricity bills 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. 

Series connection of RLC to a parallel load

Economics of installing Harmonic Filter

For economical analysis and calculation of savings by installing a 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. 

Phasor diagram for different parameters

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

The 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 

Rective 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 an incentive from DISCOM for improvement in power factor above 0.95 @ of 10 % of the monthly bill amount for every 0.1 increments 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 the installation of a harmonic filter shall provide a two-fold advantage.

The first advantage is annual savings in electricity bill @ Rs 86,03,328 due to improvement in power factor and the second advantage is its improvement in power quality of power system parameters thereby enhancement in life & performance of power system equipment.

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


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