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Faults - Scenario 4

This post demonstrates how to estimate the kiloAmpere Interrupting Capacity (kAIC) rating of the first circuit breaker at the secondary side of a transformer using the Infinite Bus Method. The method assumes infinite supply from the primary side with negligible line impedances.

Below is a modified version of Example D13 (Available Short Circuit Current) in Appendix D of the 2017 Philippine Electrical Code (PEC).

SITUATION

Two adjacent buildings are supplied by two different distribution transformers.

One transformer is single-phase rated 100kVA, 230V with 2.5% impedance. The other is three-phase rated 100kVA, 230V line-to-line and also has 2.5% impedance.

What minimum kAIC ratings are needed for the first circuit breakers down the line of each transformer to safely interrupt the maximum symmetrical fault currents?


ANALYSIS

1.) SINGLE-PHASE TRANSFORMER LOAD-SIDE SUB-CIRCUIT

Base Power, 1ph: S,base = 100kVA
Base Voltage, 1ph: V,base = 230V
Impedance, given: Z = 2.5%

1.1.) Base current, 1ph:

i,base = S,base / V,base
i,base = 100kVA /  230V
i,base = 434.78261 A

1.2.) Short-circuit current, per-unit:

V,pu = 1 pu
i,pu = (V,pu / %Z) * 100
i,pu = (1 / 2.5) * 100
i,pu = 40 pu

1.3.) Short-circuit current, 1ph:

i,sc = i,pu * i,base
i,sc = 40 * 434.78261 A
i,sc = 17,391.304 Amps
i,sc = 17.391 kiloAmps

1.4.) CB minimum interrupting capacity:

1-phase circuit breaker >= 18 kAIC

2.) THREE-PHASE TRANSFORMER LOAD-SIDE SUB-CIRCUIT

Base Power, 3ph: S,base = 100kVA
Base Voltage, LL: V,base = 230V
Impedance, given: Z = 2.5%

2.1.) Base current, 1L:

i,1ph = i,1L (assuming Y-config)
i,1L = i,base

i,base = S,base / (sqrt(3) * V,base)
i,base = 100kVA /  (sqrt(3) * 230V)
i,base = 251.02186 A

2.2.) Short-circuit current, per-unit:

V,pu = 1 pu
i,pu = (V,pu / %Z) * 100
i,pu = (1 / 2.5) * 100
i,pu = 40 pu

2.3.) Short-circuit current, 1L:

i,sc = i,pu * i,base
i,sc = 40 * 251.02186 A
i,sc = 10,040.874 Amps
i,sc = 10.041 kiloAmps

2.4.) CB minimum interrupting capacity:

3-phase circuit breaker >= 11 kAIC

CONCLUSIONS

The 1-phase circuit breaker needs a kAIC rating of at least 18 kiloAmps, while the 3-phase circuit breaker needs a kAIC rating of at least 11 kiloAmps.

For the same power and voltage ratings, three-phase breakers usually have lower interrupting capacities because the currents are distributed over three lines, while currents on single-phase breakers are concentrated on one line.

Comments

  1. Nice analysis but I heard lastime in other EE for SCC that the KAIC of Main CB that is 3phase that will be the same for its branches?'.is this correct?

    ReplyDelete
    Replies
    1. Hi. The answer for your question is in this next scenario: https://www.electricaldean.com/2020/02/faults-scenario-5.html

      To summarize:

      1. It is okay to use the same kAIC rating of the main CB for its branches IF AND ONLY IF there are no motor loads in the system.

      2. It is NOT okay to use the same kAIC rating of main CB for its branches IF there are motor loads in the system.

      3. Motors become generators during fault incidents and contribute to the fault currents in other branches. These additional fault currents increase the burden to be carried by all adjacent branch circuit breakers.

      4. When motors are present, the kAIC for the adjacent feeder and branch circuit breakers need to be higher than the main CB.

      Delete
  2. Hi sir can you have a topic for protection coordination and please explain the graph of the selective coordination.thanks and more power to your blog

    ReplyDelete
    Replies
    1. Hi. Yes, discussions on protection coordination are already being lined up. They will be tackled once the topic on Time-Current Curve is introduced.

      Keep on visiting the blog. Thanks!

      Delete
  3. Hi Sir, I have seen this blog of yours.

    I would like to ask; how can I calculate the required kAIC of a circuit breaker considering the load side?

    The load side has 10 units of .37 KW motors.

    ReplyDelete

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