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I have a problem with calculations with multiple branches. Hope someone can help.

The system is a switchboard with four generators divided by a bus tie. Two generators on each main bus.
The bus tie is set to trip at 0.3 seconds and the generator breakers at 0.5 seconds.

With a short circuit on the main bus bars, the 3-phase current is calculated to 49kA. Incident energy is about 39 cal/cm2 and arc boundary 658 cm. This is at 0.3 seconds arc duration.

The remaining fault after bus tie tripping is 24,5 kA for 0.2 seconds. Incident energy about 13 cal/cm2 and boundary 300 cm.

How do I create a warning label for the main switchboard? Do I add up the incident energies and calculate new boundary zone based on that incident energy (39 + 13 cal/cm2)? Or is it as simple as saying that the worst case is 658 cm with 39 cal/cm2 and that the incident energy after bus tie tripping is 13 with a 300 cm boundary?

Author:	Beo [ Thu Jan 24, 2013 9:08 am ]
Post subject:	Calculation with multiple branches
I have a problem with calculations with multiple branches. Hope someone can help. The system is a switchboard with four generators divided by a bus tie. Two generators on each main bus. The bus tie is set to trip at 0.3 seconds and the generator breakers at 0.5 seconds. With a short circuit on the main bus bars, the 3-phase current is calculated to 49kA. Incident energy is about 39 cal/cm2 and arc boundary 658 cm. This is at 0.3 seconds arc duration. The remaining fault after bus tie tripping is 24,5 kA for 0.2 seconds. Incident energy about 13 cal/cm2 and boundary 300 cm. How do I create a warning label for the main switchboard? Do I add up the incident energies and calculate new boundary zone based on that incident energy (39 + 13 cal/cm2)? Or is it as simple as saying that the worst case is 658 cm with 39 cal/cm2 and that the incident energy after bus tie tripping is 13 with a 300 cm boundary?

Author:	Beo [ Thu Jan 24, 2013 11:25 pm ]
Post subject:
Some more thoughts.. For simplicity lets say the short circuit on the main switchboard is 50kA. 12.5 kA from each of the four generators. Bus gap 20mm, ungrounded. 690V. The bus tie reacts to 25kA. Trips at 0.3 seconds. The generator breakers trip at 0.5 seconds. If I now calculate with 50kA for 0.3s and the remaining 25kA for 0.2s I get a boundary of 6.6m and 3m. Totalling 9.6m. If I ignore the bus tie and let 50kA for 0.5s I get 9.4m. I then considered adding the normalized incident energy. 18 J and 8.5 J. That gives me 26.5J for a totalt of 0.5s and 12.3meters boundary. Then, I considered adding the incident energy from each breaker. 25kA through the bus tie for 0.3s and 12.5kA through two generator breakers for 0.5s. This gives me about 11meters. Finally, I add the incident enerrgy normalized for each breaker. 8.5 J for the bus tie and 4.1 for each generator breaker. Total of 16.7 J for 0.5s gives 9meters boundary. What is correct?

Author:	Jim Phillips (brainfiller) [ Sun Jan 27, 2013 8:11 am ]
Post subject:
Beo wrote: I have a problem with calculations with multiple branches. Hope someone can help. The system is a switchboard with four generators divided by a bus tie. Two generators on each main bus. The bus tie is set to trip at 0.3 seconds and the generator breakers at 0.5 seconds. With a short circuit on the main bus bars, the 3-phase current is calculated to 49kA. Incident energy is about 39 cal/cm2 and arc boundary 658 cm. This is at 0.3 seconds arc duration. The remaining fault after bus tie tripping is 24,5 kA for 0.2 seconds. Incident energy about 13 cal/cm2 and boundary 300 cm. How do I create a warning label for the main switchboard? Do I add up the incident energies and calculate new boundary zone based on that incident energy (39 + 13 cal/cm2)? Or is it as simple as saying that the worst case is 658 cm with 39 cal/cm2 and that the incident energy after bus tie tripping is 13 with a 300 cm boundary? This is the most straight forward solution method. Look at it in steps is (for now) about the only way to go. We had some language in the next draft of IEEE 1584 about this stepped approach. As you show in your post, the sum of the two steps is what the total calculated incident energy could be. There is a bit of a wrinkle, generator short circuit current tends to decrease rather quickly. This requires a very complex solution and most chose to ignore it. It would necessitate using generator decrement curves based on sub transient reactance and other data. Hope it helps!

Author:	Beo [ Mon Jan 28, 2013 5:01 am ]
Post subject:
Thanks. What about the boundary? If I add them I get a higher value than leaving the entire fault for 0.5 s.

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