Cable shielding metal mesh shielding effectiveness engineering calculation

It is well known that the cable shielding layer includes both metal shielding and non-metal shielding, and the type of shielding used depends on the type of cable. In order to shield and homogenize the electric field, the power cable carries the fault current, usually in the form of a metal shield. In the national standard GB/T12706 "rated voltage 1kV (Um = 1.2kV) to 35kV (Um = 40.5kV) extruded insulated power cable and accessories" only stipulates: "Metal shielding should be made of one or more metal belts, Metal braid, concentric layer of wire or a combination of wire and metal strip." "The copper strip shield consists of a layer of soft copper strip wrapped around the package, or a double-layer soft copper strip can be wrapped around the gap." The nominal thickness of the single-core cable copper strip is ≥0.12mm, and the nominal thickness of the three-core cable copper strip is ≥0.10mm.” “The copper wire shield consists of a thinned soft copper wire with a reverse-wrapped copper wire or copper on the surface. The belt is tight, and the average gap of adjacent copper wires should be no more than 4mm."

Metal strip or wire shield is mainly responsible for a part of the fault current in a certain period of time in the event of a short circuit, to avoid thermal breakdown caused by the insulation under the influence of excessive current. The premise is that the metal shield must have a solid grounding measure, and the geometric cross-sectional area of ​​the metal shield should meet the corresponding electrical requirements.

When the voltage level is lower than 35kV or the nominal cross-sectional area of ​​the conductor is less than 500mm2, the national standard GB/T 12706 does not specify the scope of use of the metal strip or wire shield. The copper strip shield structure is used in China without special requirements; DIN VED 0276 and AS/NZS 1429.1 require that the metal shield of the cable be constructed with a copper wire shield and specify the geometric cross-sectional area or electrical requirements of the copper wire shield. The main reason is that domestic cables mostly adopt small resistance grounding method, and copper belt shielding can meet the requirements of carrying fault current; foreign cables mostly adopt direct grounding method, and copper wire shielding is required to meet the requirements of carrying fault current. So, how do you calculate the fault current that can be carried by the copper strip and copper wire shield structure? What issues should be paid attention to during the calculation process?

Allowable calculation of fault current

Before the calculation is performed, the meanings of the following symbols need to be explained:

A—considering the thermal properties of the surrounding or adjacent materials, (mm2/s) 1/2; B—considering the thermal properties of the surrounding or adjacent materials, mm2/s; F—imperfect thermal contact factor; I— The effective value of the fault current allowed during the short circuit, A; IAD - fault current calculated on the adiabatic basis during short circuit, A; K - carrier fluid material constant; M - thermal contact factor, S - 1/2; S - carrier fluid Geometric section, mm2; n—the number of layers or the number of single wires; d—the diameter of the monofilament, mm; t—the duration of the short circuit, s; w—bandwidth, mm; the reciprocal of the temperature coefficient of resistance at β—0°C, K Δ—metal sheath, shield or armor layer thickness, mm; ε—factor of heat loss in the adjacent layer; θf—terminating temperature, °C; θi—initial temperature, °C; ρ2, ρ3—metal sheath , medium or thermal resistance around the shielding layer or armor layer, Km/W; ζ1—shield layer, metal sheath or armor layer specific heat, J/K.m3; ζ2, ζ3—shield layer, metal sheath or armor The medium around the layer is hotter than J/K.m3. The specific values ​​of the above symbols can be found in IEC60949, and I will not explain them here.

According to IEC60949, the allowable fault current is known:

When calculated by adiabatic method, ε=1.

Note: When the continuous fault current time and conductor cross-sectional area <0.1s/mm2, the increase of fault current can be neglected, calculated by adiabatic method, otherwise it should be calculated by non-adiabatic method.

Allowable fault current in case of adiabatic:

It can be seen from the above formula that the ability to carry the fault current is mainly related to the carrier fluid material, the geometric cross-sectional area of ​​the carrier fluid, the duration of the short circuit, the starting temperature of the carrier fluid, and the termination temperature. Therefore, how to determine the geometric cross-sectional area of ​​the carrier fluid and the carrier fluid start temperature and termination temperature are the key to calculate the metal shield fault current. a) Geometric cross-sectional area of ​​the carrier fluid: Wire shielding:

According to IEC60949, the geometric cross-sectional area of ​​the carrier fluid of the wire shield is the geometric cross-sectional area of ​​a single wire multiplied by the number of wires.

Metal tape shielding:

It can be seen that the geometric cross-sectional area of ​​the metal-loaded fluid is only related to the width, thickness and number of layers of the metal strip, regardless of the overlap coverage of the metal strip shield. That is to say, the overlapping metal strips have the same ability to carry fault currents as the gap-wrapped metal strips. b) Starting and ending temperatures of the carrier fluid:

The termination temperature of the carrier fluid in the event of a short circuit can be selected by referring to IEC 60986 "Short-circuit temperature limits of electric cables with rated voltages from 6kV (Um = 7.2kV) up to 30kV (Um = 36kV)"; The standard stipulates that, for example, the copper wire shielding structure is usually calculated according to the experience of 55 ° C as the starting temperature. It is worth noting that for overseas products, customers usually specify the starting temperature at which the metal shield structure is short-circuited, such as Australian Standard AS/NZS 1429.1 “Electric cables—Polymeric insulated Part 1: For working voltages 1.9/3.3 ( 3.6) kV up to and including 19/33(36)kV” clearly states that the fault current of the metal shield shall be calculated by the adiabatic method specified in IEC60949, and the shielding starting temperature shall be 80 °C. At this time, if the geometrical cross-sectional area of ​​the wire shield is calculated according to 55 ° C, serious errors will occur, which may easily cause customer complaints or cable failure.

Calculation of the factor ε value

When the non-adiabatic method is used to calculate the allowable fault current of the metal shield, the ε value needs to be calculated. To highlight the method for determining the metal shield structure of the cable, only the copper strip shield and the copper wire shield are illustrated: a)

Copper tape shielding ε value calculation:

When I first contacted the metal strip to shield the fault current calculation, I mistakenly interpreted (s-1/2) in Equation 2 above as the square root calculation for the short-circuit time of the carrier fluid, but it is not.

It can be seen from Equation 1 that the factor ε of heat loss in the adjacent layer should be a constant without units. If the short-circuit time of the carrier fluid of (s-1/2) in Equation 2 is calculated by square root, the M value without unit will be obtained, and the M value without unit is multiplied by t in Equation 1, which will result in An ε value with s1/2 units causes a calculation error. b) Calculation of copper wire shielding ε value:

For copper wire shielding:

When considering the thermal imperfect contact factor, the F recommendation value is 0.7. 0.7 is also the calculated value taken by the author when initially contacting the wire shield fault current calculation. However, it is stipulated in IEC 60949 that the F value is 0.7 when the gap between the shielding wires is not less than one wire diameter and the wires are all embedded in the non-metal material. When there is an extruded tube outside the shield wire and there is an air gap between the single wires, the F value is taken as 0.5. Australian medium voltage cables are all made of copper wire shielding structure. I have discussed with Australian technicians and agreed that the imperfect contact factor of 0.5 is more accurate and reasonable.

to sum up

Conductor, metal shield, metal sheath layer, armor layer allow fault current calculation process according to IEC60949, the starting temperature and termination temperature of the carrier fluid in the case of short circuit are selected according to IEC60986, and the calculation can be accurately performed according to the corresponding conditions. The allowable fault current value. Through my own experience, the author will share the mistakes with you to avoid the recurrence of similar problems.