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Question:

How to calculate (or build) the capacitor bank for 900 KW generator having 1565 Amps capacity, voltage 380 V?

How to calculate (or build) the capacitor bank (power factor/correction improvement) for 900 KW generator having 1565 Amps capacity, voltage 380 V to 400 Vac ?How many ways / methods to do that ? which one would be best cheapest ?thanks

Answer:

It's viable that the capacitor financial institution would help in supplying commencing inrush present to the motor, but it might frequently be pleasant if the capacitor is remoted from the circuit as quickly as the motor started. Considering that the motor at nominal mechanical load would have really high vigour element, the capacitor might influence in extra current output, overloading the generator and may intent a voltage regulating main issue.
Rated apparent power A 1565*660 1.033 kVA rated power factor p.f. nom. P/A 900/1.033 0.87 φ arccos 0,87 0,513 rad tanφ 0.563 rated reactive power Q P tanφ 900*0,563 507 kVAR Let's assume minimum p.f. being 0.7!!!!.then tan φ1 1 Without any power factor correction (improvement) the relevant power P1 will be : P1 A*0.7 723 kw and the relevant Q1 : Q1 P1*tan φ1 725 kVAR. If you want to get improved power factor 1.00, then u need a capacitor bank Q1 worth 725 kVA @ 380 V, then : 725*10^3 380^2/Xc phase impedance Xc 380^2/725*10^3 0.20 ohm C 10*6/(ω*Xc) 10^6/62.8 15.9*10^3 μF

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