Transformer Power Factor Calculation

Many questions are being asked about the power factor and the definition of power factor. I hope everyone’s question will be clear today. I am presenting this article on power factor in my own language. Let me tell you a complicated thing at the beginning, if there is a coil connection to a load, then that load is lagging. And if there is a capacitor connection in a load, then that load is running in Leading. If the loads go into excess lagging or leading then neither will be good. This will increase the electricity bill, the machine will be damaged, efficiency will be reduced. If the load is in the unit then the electricity bill will not come more and the machine will not be damaged. Now we will understand how the load is moving. Lagging, Leading or Unity. I’ll do a math. I hope you can understand this total system if you understand the mathematics carefully. This is usually the case when you are asked about the power factor in any job interview.

The question is – the total load in your factory is 20,000 Watt. Show how many quality capacitors you can use to make Unity Power Factor or PFI by calculation.

Total load at the factory: 20,000 / 1,000 = 20 kWatt. Now we get KVA from kilowatts, 20 / 0.8 = 25 kVA. My total load is 25 kVA. My input transformer will be 50% more than that. Because I can increase the load more in the future. So we need- (50% of 25 + 25) = 37.5kVA. We are taking it to 37kVA. Then I have to put 37kVA transformer in the input. First we have to find out the power factor.

Since P = VIcos$

Or, Load kW = Input Transformer kVA * cos$ .

So power factor, cos $ = Load kW / kVA = 20/37 = 0.54 Lagging

You may ask, why lagging here? Because the load of coil connection is high in the factory. Such as 3-phase motors. Now I need to bring this 0.54 lagging value to mean unity or close to 1. For this you need to connect the static capacitor. Whose value do I have to find out? Here, the input voltage is 400V. Because 3-phase 400V will come from the transformer. Frequency 50 Hz since Cos $ = 0.54 Lagging

Then current, I = kW * 1000 / V * PF * 1.73 = 20 * 1000/400 * 0.54 * 1.73 = 53.45A

Now, at 0.54 lagging Power Factor P = 1.73 * VICos $ = 1.73 * 400 * 53.45 * 0.54 = 19973Watt = 20 kW

Now, Cos $ 1 = 0.54 or, $1 = Cos-1 (0.54) or, $1 = 57.31 degrees.

Since the ideal value of the power factor is 0.95, I have to bring this to mean. So, Cos $ 2 = 0.95 or, $2 = Cos-1 (0.95) = 18.19 degrees now kVAR = kW (tan $ 1-tan $ 2) = 20 (tan57.31-tan 18.19) = 20 * 1.22 = 24.4Kvar

Now, Ic = kVAR / V = ​​24.4 * 1000/400 = 61

Again, Ic = V * 2πfc or, V * 2πfc = Ic or, capacitance, C = 61 / V * 2πf = 61/400 * 2 * 3.1416 * 50 = 0.00048542 Farads = 485.42 micro Farads.

So we need to connect the capacitor parallel to the value of 485.42 micro Farads. In PFI the capacitor has to be connected at parallel all the time. PFI cannot be made with 1 of these 485 quality capacitors at once. In that case PFI has to be made by connecting these capacitors in parallel at 3 o’clock 100 and 1 o’clock 85. In this article I have used the “$” symbol as theta.

Edited By
Jeion Ahmed
B.Sc. in Electrical & Electronic Engineering (EEE)
Chittagong University of Engineering & Technology (CUET)

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