Equation 3.3
where: A = Metallic shield cross-sectional area, in circular mils
I = Short-circuit current in shield, in amperes
t = Time of short circuit, in seconds M = Constant; see Tables 3.5 and 3.6
A = I t M
Formula for Calculating A
Type of Shield (See Notes 1 and 2)
1. Wires applied either helically, as a braid or nds2
serving, or longitudinally with corrugations
2. Helically applied tape, not overlapped 1.27 nwb
3. Helically applied flat tape, overlapped (See Note 3)
4. Corrugated tape, longitudinally applied 1.27 [π (dis + 50) + B] b
Note 1. Meaning of Symbols
A = Effective cross-sectional area of shield B = L.C. tape overlap, in mils (usually 375) b = Thickness of tape, in mils
dis = Diameter over semiconducting insulation shield, in mils dm= Mean diameter of shield, in mils
ds = Diameter of wires, in mils
n = Number of serving or braid wires or tapes L = Overlap of tape, percentage
w = Width of tape, in mils
Note 2. The effective area of composite shields is the sum of the effective areas of the
components. For example: The effective area of a composite shield consisting of a helically applied tape and a wire serving is the sum of the areas calculated from formula 2 (or 3) and formula 1.
Note 3. The effective area of thin, helically applied overlapped tapes depends also on
the degree of electrical contact resistance of the overlaps. Formula 3 may be used to calculate the effective cross-sectional area of the shield for new cable. An increase in contact resistance may occur after cable installation during service exposed to moisture and heat. Under these conditions, the contact resistance may approach infinity where formula 2 would apply.
TABLE 3.2: Effective Cross-Sectional Area of Shield. Adapted from
Okonite Company, Engineering Data for Copper and Aluminum
Conductor Electrical Cables, 1998.
4bdm 100 2(100 – L)
Shield or Sheath Temperature °C at Conductor Temperature Rated Voltage (kV) 105 100 95 90 85 80 75 70 65 5 100 95 90 85 80 75 70 65 60 15 100 95 90 85 80 75 70 65 60 25 100 95 90 85 80 75 70 65 60 35 95 90 85 80 75 70 65 60 55 46 95 90 85 80 75 70 65 60 55 69 90 85 80 75 70 65 60 55 50
Note. The maximum conductor temperature should not exceed the normal temperature rating of the insulation used.
TABLE 3.3: Values of T1, Approximate Shield Operating Temperature, °C, at Various Conductor Temperatures. Source: Aluminum Electrical Conductor Handbook, 1989.
Shield Operating Temperature (T1), °C
Shield Material 100 95 90 85 80 75 70 68 60 55 50
Aluminum 0.039 0.040 0.041 0.042 0.043 0.044 0.045 0.046 0.047 0.048 0.049
Copper 0.059 0.061 0.062 0.063 0.065 0.066 0.068 0.070 0.071 0.073 0.074
TABLE 3.5: Values of M for the Limiting Condition Where T2= 200°C. (Thermoplastic Materials = HMWPE, LLDPE, PVC.) Source: Aluminum Electrical Conductor Handbook, 1989.
Shield Operating Temperature (T1), °C
Shield Material 100 95 90 85 80 75 70 68 60 55 50
Aluminum 0.057 0.057 0.058 0.059 0.060 0.060 0.061 0.062 0.063 0.063 0.064
Copper 0.087 0.087 0.088 0.089 0.091 0.091 0.092 0.093 0.094 0.096 0.097
TABLE 3.6: Values of M for the Limiting Condition Where T2= 350°C. (Thermosetting Materials = XLPE, EPR.) Source: Aluminum Electrical Conductor Handbook, 1989.
Cable Material in Contact With Shield T2, °C/°F
Cross-linked (thermoset) 350
Thermoplastic 200
Deformation-Resistant Thermoplastic 250
Note. The temperature of the shield is limited by the material in contact with
it. For example, a cable having a cross-linked semiconducting shield under the metallic shield and a cross-linked jacket over the metallic shield will have a maximum allowable shield temperature of 350°C. With a deformation-resistant thermoplastic jacket, it will be 250°C.
TABLE 3.4: Values of T2, Maximum Allowable Shield Transient Temperature, °C. Source: Aluminum Electrical
Determine the size copper wire shield required to carry a fault current of 10,000 amperes for 10 cycles for a 15-kV XLPE cable having an XLPE insulation shield and a deformation-resistant thermoplastic overall jacket.
EXAMPLE 3.3: Determine Minimum Shield Size for Known Through-Fault Current.
STEP 1. Determine the approximate shield operating temperature for 90°C
conductor temperature (which is the maximum temperature for normal operation of XLPE-insulated cables). From Table 3.3,
STEP 2. Determine the maximum allowable shield transient temperature for
the cable materials in contact with the shield, which in this case is deformation-resistant thermoplastic. From Table 3.4,
STEP 3. Determine the M value for a copper shield with T1equal to 85°C and T2equal to 200°C. From Table 3.5,
From Table 3.6,
Interpolation of these values for M yields M where T2= 250°C:
STEP 4. Calculate the required shield cross section for a fault duration of 10
cycles (0.167 seconds). Applying Equation 3.3,
Number of 14 AWG wires = 56,758 ÷ 4,110 = 13.8 (Use 14) T1= 85°C T2= 250°C M = 0.063 where T2= 200°C M = 0.089 where T2= 350°C M = × (0.089 – 0.063) + 0.063 M = (0.3333) × (0.026) + 0.063 M = 0.072 250 – 200 350 – 200 A =10,000 0.167= 56,758 circular mils 0.072
STEP 5. Determine the number and size of the wires necessary to equal or
exceed 56,758 circular mils. Table 3.2 shows that the effective cross- sectional area of a wire shield is equal to nds2, or the number of wires multiplied by the circular mil area of each wire. The number required for any specific wire size is simply the total cross section calculated in Step 4 divided by the individual wire circular mil area and rounded up to the nearest whole number:
Standard Practices
Most fuses begin to melt at approximately twice their continuous rating and series coil-operated oil circuit reclosers also tend to trip at approxi- mately twice their continuous rating. For these types of devices, it is typical to match the con- tinuous rating of the recloser or fuse to the con- tinuous rating of the cable. For electronically controlled reclosers or relayed circuit breakers, the equivalent continuous rating would be about one-half the trip rating. This general rule would not be used in the following situations:
• Where the maximum load expected on the cable is much less than the capacity of the cable, the protecting device can be reduced in size, improving protection of the cable as long as other coordination criteria can still be met.
• Where emergency overloads of the cable can be routinely expected, the fuse characteristics should be reviewed to make sure the over- load capability of the fuse is in line with the expected overload on the cable.
• In the areas where the cold-load pickup is substantially more than the maximum load current or where the duration of the cold-load pickup is long, it may be necessary to increase fuse sizes on the basis of operating experience.
Whatever the situation, the fuse or device curve should be kept below the thermal damage curve of the cable in question. This is rarely a problem except where a fuse might be
protecting several cables or several sections of decreasing- size cable. If the system engi- neer encounters such a problem, the obvious solution is to insert additional fuses wherever a conductor size change occurs.
PROTECTION AGAINST PAD-MOUNTED TRANSFORMER TANK RUPTURE Internal Faults as Cause of Rupture
Of the very small percentage of transformer tanks that fail by rupture, most rupture because of in- ternal faults. The magnitude of fault current is highest for a fault between the primary leads
inside a three-phase transformer or between the primary phase lead and ground inside a single- phase transformer. The next highest fault is when the primary windings short; the magnitude of this fault depends on the impedance of the windings between the fault location and the primary leads. The lowest magnitude of fault occurs because of a short in the secondary windings. The more wind- ings between the fault location and the primary side of the transformer, the lower the fault current.
The rupture can result from the energy re- leased within the tank and the resulting pressure. The energy, which is typically measured in joules, is proportional to the magnitude of the fault cur- rent squared multiplied by the time duration of the fault in seconds (I2t). Because tank rupture is usually caused by failure of the transformer winding, the transformer will need to be dis- carded or opened for repairs. Therefore, a com- mon solution to preventing tank rupture is to place a partial-range, current-limiting fuse under the oil. Although operating such a current-limit- ing fuse will require opening up the transformer tank to replace the fuse, this is not a problem because the tank will have to be opened anyway.
In addition, a dry-well canister or clip-mounted, partial-range, current-limiting fuse will provide the same result. Either of these can also be full range. The disadvantage of using a full-range, current-limiting fuse is that it will operate for all levels of fault current and is much more ex- pensive to replace than an expulsion fuse. The use of a bayonet fuse in se- ries with an under-oil current limiting fuse can overcome many of these disadvantages, since the replaceable element opens for low-level faults or overloads, and the current- limiting element opens for high-level faults.
Philosophy and Theory of Rupture Prevention
The basic philosophy of rupture prevention is to prevent ruptures for any and all fault conditions. The consequences of a rupture are as follows:
• Release of oil and the consequent environ- mental damage,