Ground Shielding

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Ground-fault coordination should include MVcable shielding.

Feb 1, 1999 12:00 PM, DeDad, John A.

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Verification of shield performance under ground-fault conditions will prevent systemwide cable damage.

As we all know, the metallic shielding in a medium voltage (MV) cable provides the necessary uniform electricfield within the cable's insulation. What is not as well known is its other equally important function: that ofcarrying a portion of return ground-fault current. As such, your ground-fault protection coordination studyshould address two specific shielding questions.

* What is the magnitude of ground-fault current that will be carried by the metallic shields of the specific feeder,in respect to its raceway/cable configuration?

* Is the circular mil area of the metallic shield sufficient to carry the calculated ground-fault currents within theoperational time frame of the specific ground-fault relay and protective device?

Functions of metallic shield

In MV cable constructions, the metallic shield functions as an electrical stress reducer. That is, it keeps theelectrical field within the insulation uniform and evenly distributed, without any high stress points. The stresslines in an electrical field are always attracted to any grounded surface. Since the metallic shield surrounds theMV insulation and is grounded, all stress lines within the insulation radiate out uniformly from the copper oraluminum conductor, at the cable's center, through the MV insulation, to the metallic shield.

Shielding also protects the MV cable from induced voltages, such as from adjacent power conductors, andminimizes interference with communications circuits.

Of equal importance but not commonly known, the metallic shield also functions as a return path for ground-fault currents. Its physical construction affects its current-carrying capabilities much like a copper conductor:larger circular mil area decreases resistance and impedance, thereby increasing current-carrying capacity. TheInsulated Cable Engineers Association (ICEA) Publication No. P-45-482 details the specific shield performanceparameters required for various cable constructions.

The potential problem

Many industrial MV power distribution systems employ low-resistance grounding, which allows from 200A to2000A to flow during solid line-to-ground faults. In most cases, the metallic shielding on those MV cablesbeyond the fault carry part of the fault return current, which may return along the shields of other conductors orother equipment ground paths. When this happens, any MV cable shielding with inadequate current-carryingcapacity will overheat and cause cable damage.

Determining fault-current return paths and magnitudes

The typical MV feeder consists of a conduit or raceway, three shielded phase conductors, and a groundconductor. If one of the phases sustains a line-to-ground fault, the return fault current will divide itself amongthe metallic shields, ground conductor, and raceway (if metallic). The magnitude of current flow in each ofthese paths will vary inversely as to the impedance of the specific path. In other words, the higher the

impedance of the specific ground path, the less amount of fault current will flow in that ground path. Thecommon belief is that most of the ground-fault current returns on the faulted phase shield. This is true if theground fault is from a conductor to its own metallic shield. However, testing has shown that only 3 to 14% ofthe available ground-fault current will flow through each cable metallic shield. The percentages, shown in theTable on page 60, vary with the cable/conduit/ground wire configuration of the tested feeder. The majority offault return current will flow in the metallic conduit and ground wire since their impedances are much lower thanthat of typical shielding constructions. For ground-fault coordination study purposes, you can safely assumethat 15% of the available fault current will flow in each metallic shield.

Verifying shield short-circuit performance


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ICEA P-45-482 provides an equation you can use to determine the required cross sectional area of the metallicshield.

[M.sup.2] = [[I.sub.o].sup.2] t/[A.sup.2] (eq. 1) where

[I.sub.o] = fault current (amperes)

t = time of fault (seconds)

A = effective cross sectional area of shield (circular mils),

M = constant.

Depending on the MV cable voltage rating, configuration, and construction, the constant M will vary. For 90[degrees] C-rated cable with copper shielding and thermoplastic jacket (PVC or thermosplastic CPE), orimpregnated paper insulation, the constant M will be as follows.

* 0.063 for 5kV through 15kV and 25kV rated cables.

* 0.065 for 35kV to 46kV rated cables.

* 0.066 for 69kV rated cables.

For 90 [degrees] C-rated cable with copper shielding and thermoset jacket [neoprene, hypalon, or thermoset

polyethylene (CPE), the constant M will be as follows.

* 0.089 for 5 kV through 15kV and 25kV rated cables.

* 0.090 for 35kV to 46kV rated cables.

* 0.091 for 69kV rated cables.

To find out various performance characteristics of your proposed shield construction, you insert the propervalue for the constant M into Equation 1 and then solve this equation for the characteristic in question. Forexample, if you want to know the maximum amount of time (t) that a given fault current can flow in a givenshield, you can use the following equation:

t = [(MA/[I.sub.o]).sup.2] (eq. 2)

Or, you can determine the maximum fault current ([I.sub.o]) that can flow in a given shield for a given amount

of time:

[I.sub.o] = MA/[square root of t] (eq. 3)

Or, you can determine the shield effective cross-sectional area (A) required to withstand a given fault currentfor a given time:

A = [I.sub.o] [square root of t] / M (eq. 4 )

As you can see, by inserting any known two of the three variables, you can verify that the copper shield in eachof your MV feeders is capable of carrying its portion of fault current within the time requirements dictated by therespective ground-fault relay and overcurrent protection device.

Adapting shield construction to required performance

You may find that standard shield constructions may not have sufficient cross-sectional area to handle the fault

current for the cable in question. In this situation, you can then consider the following construction options thatmay satisfy the more demanding requirement.

* No. 14 AWG concentric wire strands in lieu of the standard No. 18 AWG sizing.

* Helically-applied 5-mil tape with 17% overlap.

* One layer of overlapped, helically-applied, 5-mil tape plus a second layer of reverse lay overlapped, helically-applied, 5-mil tape.

* Corrugated longitudinal 8-mil tape.


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You can get effective cross-sectional area values for each of the above constructions, per cable size, from thecable manufacturer. By plugging this information into Equations 2, 3, or 4, you will obtain the best option for thespecific feeder in question.

Sample problem

Fig. 1 shows a single-line diagram of a typical MV distribution system, where a distribution substation is fed bycable from a main switchgear lineup. Ground-fault protective relays are also shown. Let's analyze this system

closer.

What would happen if Relay 50G (for Feeder No. 1) fails? Well, Relay 251G would have to clear the faultbefore the shield in Feeder No. 1 is damaged.

Looking upstream of Relay 50G, we see that the operating time of Relay 251G at maximum ground fault is 0.4sec. With Relay 50G's operating time of 0.1 sec, Feeder No 1's shield must be capable of withstanding itsportion of the total ground-fault current of 1000A for 0.5 sec without sustaining damage. This portion is 15% of1000A, or 150A.

Suppose Feeder No. 1 consists of three No. 4/0 AWG, 15kV, shielded conductors in a rigid steel conduitwithout a ground wire. Its shield must be able to withstand 150A of ground-fault current for 0.5 sec. Similarly,Feeder No. 2 shield must withstand 150A of ground-fault current for 1.3 sec (0.5 sec for Feeder No. 1 plus 0.8sec for operation of Relay 351G).

By inserting these ground-fault current and time duration values into equation 4, you can determine therequired shield circular mil area. By comparing this value with the actual shield circular mil area of the cable inquestion, you can confirm the adequacy of the shield construction.

Shield withstand-limit curves

In situations where numerous calculations are required, especially in coordination studies involving many MVfeeders, you can use shield withstand-limit curves that you prepare yourself. What's involved?

First, you obtain actual circular mil area values of metallic shields per cable voltage and shield and cableconstruction.

Second, you plug in these values, along with either incremental time values or ground-fault current values, intoequations 2 or 3. The solutions are then plotted on log-log graph paper and joined together to form a specificshield withstand-limit curve, as shown in Fig. 2.

By verifying that the required shield withstand falls below and to the left of the limit curve, you confirm theadequacy of the proposed shield construction.

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