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At power transformer Meduim to Meduim voltage, What is the suitable rated circuit breaker at both sides for transformers below:
1- 33/11 KV, 12000 KVA
2- 33/3.3 KV, 12000 KVA
3- 33/3.3 KV, 8000 KVA
4- 11/3.3 KV, 3000 KVA
How its calculated?

2007-12-26 02:03:30 · 1 answers · asked by AtheelKK 1 in Science & Mathematics Engineering

1 answers

Fault Duty:
The "low side" fault current can be calculated using the tool at the link below. They provide a great explanation and work through an example. It assumes an infinte bus and the only missing piece of information you have not provided is the transformer's impedance. For the transformer sizes you have listed, a reasonable assumption is Z = 7%.

Assuming Z= 7%
12,000 kVA at 11 kV = 9008 amps
12,000 kVA at 3.3 kV = 30,026 amps
8,000 kVA at 3.3 kV = 20,017 amps
3,000 kVA at 3.3 kV = 7506 amps.

Some common breaker and recloser fault duty ratings are 12.5 kA and 16 kA. There are other ratings available from 25 kA up through 40 kA.

The high side breaker is dependent on the available fault current of the 33 kV system.

Settings:
The actual relay setting are going to be based on the allowable loadings used by the customer. The "high side" breaker needs to handle the magnitizing inrush currents without tripping. These inrush currents are estimated at 25x nameplate for 0.01 seconds and 12x nameplate for 0.10 seconds.

2007-12-26 02:28:06 · answer #1 · answered by Thomas C 6 · 0 0

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