p.f calculation for transformer
Posted 14 October 2002 - 11:23 AM
T1: 1 phase 1302KVA, 50HZ, primary 11kv, 118A, secondary 330-1210v, 1207A with Z%= 11
T2: 1 phase 1500KVA, 50HZ, primary 1119kv, 1250A, secondary 2*93.3 v, 2*7500A with Z%= 6
B/w two connection i.e. secondary of T1 and Primary of T2 we have 3 individual cap banks, each rating is
Qn = 300KVAr, Un= 1200V, Cn=663microF, Ic = 250A, Hz = 50
The load at secondary of T2 is an arc furnace with max load of 850KW.
As I know through the seimens hand book that the cap rating for transform is near to the 80-90% of the reactive full load power of the transform. With formula
Qtr = Q0 + Uz/100% (S/Sn) ^2 * S
Where Qtr = transformer reactive power, Q0 is the transformer no load power, S = apparent power that a transformer is supplied, Sn = rated power for transformer, Uz is the % impedance of transformer.
Here in our case
Q0 from seimens graph = 27 Kvar
S= 1302 for full load
Thus Qtr = 27 + 11/100 (1302/1302)^2 * 1302 = 170KVAR
Which is not fulfill in our case, can you suggest me what is the exact relation to calculate the cap bank KVAr and step required to control the p.f in case of transformer, is the bank rating equal to kva rating of transformer?
Is there any effect on the secondary voltage of the transform if the cap bank is over rated i.e. if load on transform is reduced?
Posted 16 October 2002 - 09:24 AM
A transformer has two components of current flowing, one is the resistive current and the other is the inductive or reactive current. On a single transformer, the reactive, or magnetising current is constant and independant of load. There is an additional reactive component due to the leakage reactance of the transformer, and this does vary with the load. Unlike an induction motor, the dangers associated with over correction are essentially eliminated due to the fact that the transformer does not act as a generator. You can however, get an elevated voltage due to correction but this is not major.
With a transformer, I would suggest that you measure the current with the secondary open circuit. This current is essentially magnetisng current and can be fully corrected. As the load is increased, the reactive component will also increase and the power factor will begin to lag. You could add more correction when the load is high, but it depends on how accurate you actually need to be. Automatic correction is probably the best solution for your application, because you are varying the reactive current in the second transformer by varying the voltage.
Your load of an arc furnace will probably have a very poor power factor, not just due to the reactive current, but also due to the distortion of the current waveform. This distortion can not be reduced by the use of power factor correction, so you may still have a poor power factor and high KVA demand.
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Posted 24 October 2002 - 01:17 PM
thanks for nice explaination regarding the subject matter .. still plz tell me , is there any calulation for cap bank for transformer , how can we select the cap bank for transformer
Posted 21 May 2004 - 06:28 AM
I am your new friend can you help me about transformer
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