(04-23-2024, 12:26 AM)PhilOfPerth Wrote: Maybe I'm missing something, but I believe the reactance of a circuit, however complex the circuit,, is dependent ONLY on frequency and the inductance ( or capacitance) of its components (2*PI*F*L or 1/(2*PI*F*C). Without seeing your actual setup, I would have thought it should be just a matter of replacing the frequency component in the fomulas. Maybe it's differrent in large-scale systems?You are mostly correct about the reactances; the equations also take into account the inductive effect of steel conduits on the reactances. But you must remember that the impedances have been calculated in this method for 60 Hz systems, so the equations are already normalized for 60 Hz AC. AC circuits can be a bear to calculate in situ (I'm also an amateur radio operator, so I am accustomed to working in the frequency domain), but when one is operating at one singular frequency it is a simple exercise to reduce the AC equations to simple algebraic equations. So, the C values I have for the conductors are all calculated for 60 Hz systems, which is in use in the entirety of North America. Canada utilizes some different voltages than the US (the highest Canadian low-voltage industrial system is 575Y/347V, while in the US the highest low-voltage system is 480Y/277V.) Working with a singular line frequency is the only thing that makes electrical power engineering tolerable, because all the pertinent equations can be reduced to simple equations.
The high-dollar five-figure analysis programs can work in any system, but spending $30k on software isn't always the most efficient way to get a quick answer.
It's not the having, it's the doing.