|
SES is always exploring new measurement and analysis techniques. In recent months, SES has completed the following projects, which have generated new ideas:
· St. Leon Grounding Analysis (Manitoba, Canada) · NRI 345 kV Line / M&N Pipeline Maine (New Brunswick, Canada) · Eastern Region Water Supply Facility Raw Water Main (Orange County, Florida) · Special Operating Procedure: Work at Mine Substations During Frozen Soil Conditions (Alberta, Canada) |
|
CDEGS has never been wrong before! Is this a data entry error? Better just send the files to SES and let the Ph.D.’s over there deal with the enigma.
Faster than you can say “I knew there was something fishy going on,” you receive the curt reply: the computer model is fine and the results are normal. Hmmm.
Now you wonder what to do next. You are sure that there is something wrong with the results, yet… CDEGS… SES support…
Frankly, these results have troubled some SES researchers as well. Unfortunately, a rigorous analysis of a real cable sheath requires a fierce battle with Bessel functions, which still leaves one wondering if something hasn’t been overlooked or inaccurately calculated. On the other hand, a simple example can be created which allows one to obtain similar results and, if not gain an intuitive understanding as to why the system behaves the way it does, at least convince oneself that the observed behavior is physical.
Consider the following equivalent circuit for the system illustrated above, which has been simplified: the sheath has been hollowed out and replaced by two very thin concentric cylindrical shells, one representing the inner surface of the sheath and the other representing the outer surface of the sheath. Furthermore, the earth has been replaced by a third very thin concentric cylindrical shell, which represents the external current return path, so that we do not have to be concerned with the complexities of Carson’s equations, which apply when the earth is involved. |
|
The inner radius of the sheath is r1, its outer radius is r2, and the radius of the external return shell is r3. The region between the two surfaces of the sheath is assigned a relative permeability (with respect to free space) of µr, whereas that between the sheath and the external return path is 1. The resistance per unit length of the sheath inner surface is R1, that of the sheath outer surface is R2 and that of the external return path is R3. The current flowing in the cable core is I0, that in the sheath inner surface is I1, that in the sheath outer surface is I2 and that in the external return path is I3.
The question we would like to answer is the following: is it possible, by adjusting the values of r1, r2, r3, µr, R1, R2, and R3, to obtain a system in which the magnitude of the current flowing in the sheath is actually greater than the current flowing in the core? This would certainly be true if the phase angle between I3 and I0 turned out to be less than 90 degrees: this is what we will seek to determine.
A full derivation is beyond the scope of this article. What we will do is set up the basic equations, present the resulting relationship between I3 and I0, and then provide results from one example. The magnetic field density, B, at any location in our cylindrical system, is given by the following expression:
B = µ ∑Ii / 2πr
where µ is the magnetic permeability at the location where B is computed, ∑Ii is the sum of all currents flowing in conductors that are closer to the core than the location at which B is being computed, and r is the distance from the center of the core to the location at which B is being calculated.
The magnetic flux, Ф, between any two concentric conductors is the surface integral of the flux density B between the two conductors. Note that the flux lines are concentric with the cable elements. If we calculate the magnetic flux between the inner and outer surfaces of the sheath, Ф1, per unit length of cable, then we obtain the following:
Similarly, for the magnetic flux per unit length between the outer surface of the sheath and the external return path, Ф2:
These magnetic fluxes are time-varying and therefore induce an electromotive force in any path surrounding them, equal to the time derivative of the magnetic flux. If we look at the path formed by the inner and outer surfaces of the sheath (and the interconnections between them at both ends of the cable), then the electromotive force per unit length in this path, EMF1, will be as follows:
where f is the operating frequency of the system. Similarly, for the path formed by the outer surface of the sheath and the external return path (and the interconnections between them at both ends of the cable), the electromotive force per unit length, EMF2, will be as follows:
Now apply Kirchoff’s laws. The sum of voltage drops, including the induced emf, around the path made up of the inner and outer surfaces of the sheath must be zero. Thus:
If we define
Similarly, for the path made up of the outer surface of the sheath and the external return path:
Conservation of current imposes the following condition: I1 + I2 = - (I0 + I3) [3] Combining Equations [1], [2], and [3], the following relation can be obtained:
If the phase angle of this quantity can be made less than 90 degrees, then I3 will be less than 90 degrees out of phase with I0 and the sum of I3 and I0 will therefore be greater than I0. From [3], the magnitude of the sum of I3 and I0 is equal to the magnitude of the sum of I1 and I2, the total sheath current. Thus, if the phase angle of the expression above is less than 90 degrees, then the total sheath current will be greater than the core current.
Examining the expression above, we see that when R2 is small compared with (R3+Z2) and (R1+Z1), then 1/R2 becomes large compared with 1/(R3+Z2) and 1/(R1+Z1) and dominates the expression (1/R2 + 1/(R3+Z2) + 1/(R1+Z1)) in the denominator. Because 1/R2 has a phase angle of zero, (1/R2 + 1/(R3+Z2) + 1/(R1+Z1)) will have a phase angle close to zero. If Z2, which is purely inductive, is significantly larger than R3, which is purely resistive, and if Z1, which is purely inductive, is significantly larger than R1, which is purely resistive, then the denominator will have a phase angle between 90 and 180 degrees. The numerator has an angle of -180 degrees. The entire fraction will therefore have a phase angle of between -270 and -360 degrees (i.e., between 0 and 90 degrees). We have demonstrated, then, that the sheath current can be greater than the core current.
Let us now determine what values for the seven system parameters maximize the sheath current (i.e., | I1 + I2 | or | I0 + I3 |). Each of the parameters is varied independently through the following ranges of values and the combination resulting in the maximum sheath current is retained:
The values of r1, r2, and r3 are increased linearly, in 21 equal steps. The values of the other four parameters are increased exponentially, in 21 steps.
It is found that the worst case combination, for the equivalent circuit we have created, corresponds to a small diameter sheath, with minimal spacing between the inner and outer surfaces, a maximal distance to the external return path, a maximal magnetic permeability of the material between the sheath’s inner and outer surfaces, a high resistance inner sheath surface and minimal resistances for the outer sheath surface and external return path.
The maximum sheath current obtained with the combination of parameters listed in the rightmost column of table above is 1.02 Đ5.6 x I0.
This demonstrates that concentric ground return paths tend not only to carry a large fraction of the fault current back to the source: they can carry back even more! |
||||||||||||||||||||||||||||||||||
Physically ImpossibleRobert D. Southey, Y. Yang and J. Liu |
|
That’s physically impossible! More Current Flowing on the Cable Sheath than in the Cable Core. You’ve just run TRALIN and SPLITS. Contrary to your engineering intuition, SPLITS claims that more current returns to the source on the grounded sheath of your faulted cable than is flowing in the cable core. You sketch the system and scratch your head: |
