Voltage drop with impedance on parallel conductors

[Anderman]

Guys,

I have been looking at the voltage drop issue and have a question about the following situations:

To get voltage drop using the IEEE 141 formula and R and I values from Table 9 in the NEC:

Vd=V+IRcos@+IXsin@-sqrt(V^2-(IXcos@-IRsin@)^2)


Looking at large installations (300HP and above), we usually use 2 conductors per phase because it is an easier installation. These parallel conductors will have half the R value, but won't the X (impedance) be double? Can we just use this formula and double the impedance and half the resistance?

Thanks,
Anderman

 

[winnie]

You can't use an approach that simple, because the inductive reactance will not simply double. The reactive impedance will depend strongly on the geometry of the conductors in the conduit.

Inductive reactance is caused by the magnetic field created as current flows through the conductors. Since current is flowing in balanced circuits, the current from one phase is balanced by the currents in the other phases, canceling the bulk of this magnetic field. The inductive reactance is caused by what is left of the magnetic field, mainly _between_ the conductors. Depending upon how the various parallel conductors are sitting relative to the other phases, they may not see each other magnetically, or they may be closely coupled increasing the inductance.

Think about the two extreme cases: two separate conduits, each with a complete three phase set, versus a single conduit, where both conductors for each phase are twisted together and remain side by side for the entire run.

-Jon

P.S. As a terminology note: impedance is the combination of resistance and reactance.

 

[bob]

Guys,

I have been looking at the voltage drop issue and have a question about the following situations:

To get voltage drop using the IEEE 141 formula and R and I values from Table 9 in the NEC:

Vd=V+IRcos@+IXsin@-sqrt(V^2-(IXcos@-IRsin@)^2)


Looking at large installations (300HP and above), we usually use 2 conductors per phase because it is an easier installation. These parallel conductors will have half the R value, but won't the X (impedance) be double? Can we just use this formula and double the impedance and half the resistance?

Thanks,
Anderman

The parallel reactance will be about 1/2 the value for 1 conductor. For most calculations the formula can be reduced to V= IRcos@+IXsin@.
V = VD from line to neutral