Calculating total system Impedance

[mbrooke]

In terms of the series impedance of a wire or cable run rather than a transmission line, it does in fact not make a difference, as the poster I replied to stated.

Right, but the quote said the ground wire would not carry current. Ground faults are by far the most important considering:

1) breaker must open fast enough to prevent electrocution. (0.4 seconds for 277/480, 0.8 seconds max for 120/208)
2) Ground wire must not overheat or melt
3) Its wise to reduce incident energy at the fault point.

These 3 are governed by the combined impedance of the phase and ground wires. With a low enough impedance all 3 are successfully covered.

Where as excessive voltage drop at a device is typically not a hazard.

 

[GoldDigger]

When you are doing the fault current calculations, you use the series impedance for the two runs of wire, using the table figures for each wire size, with no additional corrections for the use of two different sizes.
In your other thread you illustrate a cable with with three phase conductors, and reduced neutral and EGC. If you are talking about one 500 foot length of cable, then for a ground fault you will not need the series impedance of the phase conductor, the neutral and the EGC. Practically speaking the normal current will see the series impedance of phase and neutral (or just two phase runs, with a different driving voltage if it is a phase to phase short). And the fault current of a phase to ground fault will see the series impedance of phase and ground with the effective ground impedance being that of the wire EGC in parallel with all other effective ground paths.
Your initial calculation makes me wonder whether you would also assign some sort of significance to the sum of the three phase impedances.

 

[mbrooke]

When you are doing the fault current calculations, you use the series impedance for the two runs of wire, using the table figures for each wire size, with no additional corrections for the use of two different sizes.

Easy!