How much compensation capacitor is required for the 250W high pressure sodium lamp?
The transformer efficiency is calculated as 0.8, 315KVA can load capacity of about 252KW, power factor is increased from 0.7 to 0.95 per KW capacity 0.691 kilovar, required capacity: 252 x 0.691 = 174 (kilovar) total current capacitance: I = C/1.732 * U = 174/ (1.732 x 0.45) = 223 (A).
Because of the high voltage sodium lamp ballast inductance element, the power factor is very low, generally about 0.5 COS. The general vicinity, energy consumption is about 20% ballast tube near the rated current of high voltage sodium lamp 150 Watt: I = P/U/COS (phi = 15030) /220/0.5 = 1.6 (A) if the capacitance of the power factor increased to 0.95 required: C = (3183 * I * sin) /U: sin = Reagan (0.95 - Phi COS Phi Square) C = (3183 x 1.6 x sqrt (0.95 - 0.5 * 0.5) /220 = 19 (F).
The compensation capacitor is not in line with the general 18----30uf compensation capacitor
