Harmonic filters are used to eliminate harmonic distortion caused by excess currents in and out of appliances. It can prevent large quantities of harmonics from causing damage to equipment, downtime of operation, and preventing an increase in operating costs. In this article, we take a deep dive into understanding the economics behind a Harmonic filter
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.
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 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.
Some of the non-linear loads which generate Harmonics are given as below.
Table -3 indicates the ill effects of harmonics on power system equipment.
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.
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.
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
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
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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