Hi, Ed, thanks for added me as Friend. For the sake of knowledge of all Z4E Members, I am attending to Ronnie suggestion to reply here. Thanks, Ronnie.
This topic is talking about U.S. NEC/NFPA Chapter 9 Tables - Table 9 Alternating-Current Resistance and Reactance for 600-Volt Cables, 3-Phase, 60 Hz, 75oC (167oF) — Three Single Conductors in Conduit.
As many of us know, U.S. and their dependencies, have their grid at 60Hz as near half of the world, while the other half of world have 50Hz grid - British standard. Always interesting to remind that Japan is the only country with both frequencies: Northern Japan (from Tokyo) at 50Hz and Southern Japan (from Tokyo) at 60Hz, where most of U.S. Military Bases in Japan are located - in Kwanto region around Tokyo there are the Grid Interconnection/Frequency Conversion stations.
Answering Ed's question: the NEC Table 9 is applicable for 50Hz SINCE some calc is applied. The Resistance (R, in ohms) of a conductor changes with temperature that can be represented as function R = fr(T), while the Capacitive Reactance (Xc, in ohms) and Inductive Reactance (Xl, in ohms) are both dependent of frequency:
- Capacitive Reactance, in ohms: Xc = fc(f) = 1/(2.pi.f.C) , but at the NEC Table 9 there is a note "Capacitive reactance is ignored, since it is negligible at these voltages";
- Inductive Reactance, in ohms: Xl = fl(f) = 2.pi.f.L , into consideration in the 2nd and 3rd columns of Table 9, running conductors inside of conduits;
- Total (Effective) Impedance, in ohms: Z = R.PF + Xl.sin(arccos(PF)) , where PF is the Power Factor of the circuit. Multiplying current by Total (Effective) Impedance gives a good approximation for line-to-neutral voltage drop;
- Where: pi = 3.14159, f is the frequency, C is the Capacitance, L is the Inductance;
As Inductive Reactance (Xl) was tested under 60Hz frequency for Table 9, we can get new (reduced) values for 50Hz frequency and consequently a new (reduced) Total (Effective) Impedance also for 50Hz, in this way:
- Inductive Reactance for 50Hz: Xl[50Hz] = Xl[60Hz] x 5/6 = Xl[60Hz] x 0.8333
- Total (Effective) Impedance for 0.85PF and 50Hz: Z[0.85PF,50Hz] = R x 0.85 + Xl[50Hz] x 0.527
Or from Table 9 values: Z[0.85PF,50Hz] = Z[0.85PF,60Hz] + 0.527 x ( Xl[50Hz] - Xl[60Hz] )
- Example for Table 9, second row: Size (AWG or kcmil) = 12, Xl (60Hz Reactance) for All Wires, in PVC, Aluminum Conduits = 0.177 Ohms (to Neutral per Kilometer @ 60Hz), so for
Xl (50Hz Reactance): Xl[50Hz] = 0.177 Ohms x 0.8333 = 0.1475 Ohms (to Neutral per Kilometer @ 50Hz)
- And in the same second row: Alternating-Current Resistance for Uncoated Copper Wires = 10.2 Ohms (to Neutral per Kilometer) the new
Effective Z (@50Hz) at 0.85 PF for Uncoated Copper Wires in PVC Aluminum Conduits:
Z[0.85PF,50Hz] = 10.2 Ohms x 0.85 + 0.1475 Ohms x 0.527 = 8.747 Ohms (to Neutral per Kilometer @ 50Hz);
I hope this helps. Regards.