Low power factor in arc furnace operation
1. the power meter installed by authority is to read a cumulative kWh, kVARh and KVAh from various appliances.
2. take a sample of reading of power meter as mentioned below:
Date: Time: kWh kVARh kVAh
25.09.2013 12:00:00 X Y Z
26.09.2013 12:00:00 A B C
Difference: 1 day 24 hour (X-A) (Y-B) (Z-C)
tan(Φi) = (Y-B)/(X-A) = R i.e. assume R = 0.7
Φi = arctan[(R)] = 34.99
cos(Φi) = cos(34.99) = 0.82
value of capacitor bank, kVAR = kW [tan(Φi) – tan(Φf)]
assume final cos(Φf) = 0.96,
Φf = arcos(0.96) = 16.26
therefore tan(Φf) = tan(16.26) = 0.292
The value of capacitor bank, KVArRh = kWh [tan(Φi) – tan(Φf)]
= (X-A)*[0.7 – 0.292)
= (X-A)*0.408
=0.408(X-A)
the value of capacitor bank, kVAR = kVARh/(24 houurs)
=0.408*(X-A)/24 = 4*416.67 = 1666.68
therefore (X-A) = 4*416.67*24/0.408
= 98040 kW
therefore, the difference in (X-A) in terms of kWh = 98040*24 = 2352960 kWh
Therefore, I suggested that reading of kWh, kVARh shall be take every hours and stated date and time taken of reading.