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1. Our energy meter for 33 KV line was replaced last year. The new meter was running slowly and the kwh readings were low along with the low power factor of 0.70. We informed officials about low readings and requested them to check the meter. They came and confirmed that the everything was ok. Since then they were billing according to the readings and we were paying.They sent a supplementary bill this year stating that the meter readings were not corresponding with their check meter so that we should pay the difference. We asked them to make sure that the meter readings are matching with the check meter. They have replaced the meter again . Now the meter readings are ok but the P.F is still .65Now they are saying that our capacitor bank is not correct. In fact there is no change in our capacitor bank. It was giving us .96 PF before they changed the meter first time.Our Capacitor bank works on 33kv line for Electrical Arc Furnace only. There are 4 capacitors ( 416.67 KVAR each) on each phace connected parallel. The furnace transformer is 12 MVA (33KV/450V). I request you to kindly assist us to solve this problem.Thanks,NVRSrinivas
Have you installed your own check meter in order to verify municipal readings? Picked similar problem when we installed check meter for 11Kv supply.
Assamo Energy Solutions.
make sure to check this with an analysor.
Mayby you have this kind of instrument yourself or you can rent one.
What could be the course is that the new energy meter is wrong connected.
It is also possible that there is a high THD (distorsion / harmonics) what can be the course of a low PF.
The old meter registered mayby only cos phi and thats something different.
good luck with this case.
Jan.
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.
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