 Home Page News Search Contact Us      Cabling Riddle No.17 - Induced voltage calculation
Just wondering if you would be able to assist me in a calculation. Ive two separate circuits both rated at 150MVA/656Amps at 132kV at 50Hz. These are both laid in trefoil and are separated at a 1 meter center to center.
The cable conductor diameter is 47.2 mm and the overall sheath diameter is 96.9 diameter. The rated short circuits are 31500amps for three seconds.
Is there way to work out the induced voltage when one circuit is down and the other working at ratings.
Also if a fault were to occur on the working conductor, what would the voltage be present on the other cable.
Id really appreciate if you could be of help.
#1
Thu, March 19th, 2015 - 11:33
If that conductor is part of a circuit, the induced voltage will result in current flow.
This situation occurs during operation of metallic shielded conductors. Current flow in the phase conductors induces a voltage in the metallic shields of all cables within the magnetic field. If the shields have two or more points that are grounded or otherwise complete a circuit, current will flow in the metallic shield conductor.
So if your cabling system consist mentioned metallic two point grounded sheaths, you can ignore any mutual magnetic interferences between two parallel cable systems.

In multi-phase circuits, the voltage induced in any shield is the result of the vectoral addition and subtraction of all fluxes linking the shield. Since the net current in a balanced multi-phase circuit is equal to zero when the shield wires are equidistant from all three phases, the net voltage is zero. This is usually not the case, so in the practical world there is some “net” flux that will induce a shield voltage/current flow. In a multiphase of shielded, single-conductor cables, as the spacing between conductors increases, the cancellation of flux from the other phases is reduced. The shield on each cable approaches the total flux linkage created by the phase conductor of that cable.

In absence of any grounded sheaths, and in one phase to ground fault situation, occurring of some mutual magnetic interference and mutual induced voltage is possible.
The inductance of an electrical circuit consisting of parallel conductors, such as a single-phase concentric neutral cable may be calculated from the following equation:

XL=2πƒ(0.1404 log10S/r+0.153)×10−3

where XL=ohms per 1,000 feet
S=distance from the center of the cable conductor to the center of the neutral
S and r must be expressed in the same unit, such as inches.
The inductance of a multi-conductor cable mainly depends on the thickness of the insulation over the conductor.

You can use similar relation for an approximately calculation of  mutual reactance between two parallel circuits. Also you can calculate induced voltage as below:

U=ISC . XL    