Question

Find the Recommend Size of Automatic Power Factor
Find the recommend size of automatic power factor, the capacitor bank to maintain Power Factor (PF) at 97%. Small Fabrication Facility has the following connected loads, - 10 units spot welder, 220V, 1PH, 60Hz, 20 Amps each. - 3 units bending machines, 5KW, 220V, 3PH, 60Hz. - 4 units hydraulic presses, 7.5KW, 220V, 3PH, 60Hz. - 2 units window type aircon, 2HP each, 220V, 1PH, 60Hz. - 20 pcs 2x40 watt fluorescent fixtures. - 2 units drying ovens, 100KW each, 220V, 1PH, 60Hz. The measured power factor as recorded from the electric bill is 86%. In the above table 0.86 - 0.97 use 0.343 (multiplying factor).

Answer 1

All the loads are connected in parallel with the supply 220V.

Voltage and Currents given are RMS.
Sum of RMS powers :
   spot welder = 10 units * 220 * 20 * 0.86 =  37.84 kW
   bending m/c = 3 units*5 kW * 3 phases = 45 kW
   Hydraulic press:  3 units*3 phases* 7.5 kW = 67.5 kW
   Window type Aircon: 2 units * 2HP* 1phase = 2 * 2*746.268= 2.985 kW
   Fluorescent fixtures:  20 * 2*40 W * 1 phase = 1.600 kW
   Drying Ovens:  2 units * 100 kW * 1 phase = 200 kW
   
Total power (real) drawn = 354.925 kW   at   power factor = 0.86
The power consumed in kVA  = 354.925 / 0.86 = 412.70 kVA
Billing is based on this kVA figure.

To change the power factor from 0.86 to 0.97 , from the table of multiplying factors we get: 0.343.  As we need to increase the power factor, we use a capacitor bank. 

 So the capacity of the Capacitor bank = 0.343 * 354.925 = 121.74 kVA

Now the billing will be for  354.925 /0.97 = 365.90 kVA
So the savings on the bill are on : 46.8 kVA