Equivalent inductance of parallel substations

[rail1996]

When considering DC short circuit faults of an overhead line system occuring immediately outside one of the substations, contributions from the adjacent substations should be taken into account.

If I take these as three RL circuits, i.e the local, and the two adjacent circuits with

1. R1 & L1
2. R2 & L2
3. R3 & L3

Can you advise how to calculate the equivalent inductance of these three circuits connected in parallel?

Thanks in advance

 

[rsherry]

Depends what you are trying to do with the result to some extent. If you are lookng to know the DC components of the fault currents then the only accurate way is to calculate the individual infeeds and sum them. If you do a standard 'parallel' calculation of the impedances you will find the L/R ratio of the combination (equivalent impedance) is much lower than that of the three branches separately and the DC component from the equivalent will not be correct. Is this what you are asking ?

 

[rail1996]

Sorry, I should have made this clear.

This is actually for a DC distribution system. I suppose paralleling these impedances should lower the overall inductances.

Would the below formula be correct to use?

Leq = (Rp/R1)^2 * L1 + (Rp/R2)^2 * L2 + (Rp/R3)^2 * L3

where Rp = 1/(1/R1 + 1/R2 + 1/R3)

 

[DanDel]

rail1996. I believe this question was posted before recently, but I can't find it. Perhaps it was removed.
I have to assume that you are using a Thevenin or Norton Equivalent to calculate DC fault current for a railway supply system.
Parallel R and L values are calculated the same:
1/Rt = 1/R1 + 1/R2
1/Lt = 1/L1 + 1/L2

I suspect that you are concerned with the calculation of not only the DC fault current, but the rise time based on the circuit time constant, which is not quite as simple as using the above equation. The problem is that each source circuit has its own different rise time which usually puts the peak fault current at different times for each source(or multiple peaks/time constants).
The conservative method for fault current analysis, however, is to assume that all sources peak at the same time, and use the equivalent R and L values. This may not be totally accurate for rate-of-rise relay calculations, but the only other option is to calculate individual source time constants and plot using time-domain analysis.

 

[rail1996]

DanDel,

Yes, you are correct that I did post this question recently but for some reason it was removed.
You are also spot on that I'm working on the fault currents for a railway line. As you have suggested, I think I can assume all sources peak at the same time as the R-T units all have the same rating and characteristics.

 

[jbartos]

Comment: Since information given in the original posting is missing data, e.g. configuration distances of conductors, then it would be better to follow a suitable reference for calculations, e.g.
C. L. Wadhwa "Electrical Power Systems," 2nd Edition, John Wiley & Sons, 1991,
Chapter 2 Line Constant Calculations
Notice that for AC, R resistance and G conductance are least important since they do not affect much the total equivalent impedance of the line and hence the transmission capacity, however for DC, R and G do count except transients where they again may be omitted.