400 V LV System - Short circuit calculations - Impedances Method

[Carlos Melim]

Good afternoon everybody.

I am involved in a project and I want to calculate the fault currents on several locations to validate the short-circuit power needed for each panel and the upstream trip switches nominal currents and its types A, B, C or D.

I use the impedances method and determine all the impedances ( resistances and reactances ), and add them up from the power transformer of the secondary substation, to the selected location.

To chose the protection devices short-circuit power, I consider a bolted three-phase short circuit;
For the the validation of the upstream trip switches nominal currents and types, I consider a single phase short-circuit, the fault current;

Please check an attached draft.

If the feeder cable of a panel is long I get a low fault current.
But if the feeder of a downstream panel is short the fault current is higher, even if the distance to the secondary substation is higher.

All the calculations that I´ve come across consider the three phase feeders of the panels as totally balanced. No neutral currents.
So the single phase short-circuit fault loop starts and ends on the upstream panel.

Is this calculation method valid?
What really happens to the neutral current during a single-phase short-circuit?
It should close on the neutral bar of the power transformer.

All help and enlightening will be much appreciated.

Best regards.

Carlos Melim

 

[che12345]

Dear Mr. Carlos Melim
Q1. "...I am involved in a project .... to calculate the fault currents on several locations ...."
A1. It appears that something is wrong ? L1=625m with If1 =207A; where L2=625+1=626m with If2=322A.
If If1 is "correct" at 207A, If2 is likely to be [the same or slightly lower] than If1; as the distance between them (is only 1m). The 1 m conductor size is likely to be smaller. It can/may be If2 approximately equal to If1, but impossible for If2 > If1.
Che Kuan Yau (Singapore)

 

[waross]

First, impedances are directed values and may not always be added directly.
For a rigorous solution:
Sum all of the resistances from the source (transformer) to the point at which you wish to calculate the current.
Sum all of the reactances from the source (transformer) to the point at which you wish to calculate the current.
From those sums, calculate the impedance.For each location of an assumed fault, repeat the calculations back to the source.
This is based on an assumed infinite supply to the transformer.
This method is conservative and will not become invalid in the event that the utility increases the available fault current sometime in the future due to switching or upgrades.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

The return conductor is different: up to panel 2 the return it is the neutral wire [If1= 207 A]. For the second [If2] the return, probably, is the grounding.