Determining Currents of Cable Sheaths by means of Current ...

1 Determining Currents of Cable Sheaths by means ofCurrent Load Factor and Current reduction FactorI. Sarajcev, M. Majstrovic, Member, IEEE, and R. GoicAbstract- A new method for Determining sheath Currents asthe consequence of electromagnetic coupling during line-to-ground short circuit is described in this article. This method isbased on using both the Current load factor and the currentreduction factor. The model cables are selected to representmajor constructions encountered in practice.

2 The Cable lineconsists of three single-core cables laid in trefoil formation andtouching each other. Cable Sheaths are grounded at both load factor and Current reduction factor arecharacteristic data of the analyzed Cable line. The methodpresented in this paper is easy to words- Cable , Current , Current factor, load factor, sheath ,sequence componentI. INTRODUCTIONLine-to-ground short circuit is an unsymmetrical and voltages can be shown by sequence componentphasors.

3 Sequence components of Currents flow through phaseconductors and through the other active parts of thetransmission network. The Cable line is an element of thedirect-grounded transmission network and consists of threesingle-core power cables. Their conductive Sheaths aregrounded at both ends. Currents through Cable Sheaths flowduring line-to-ground short circuit. These Currents consist oftwo components. The first component occurs as theconsequence of increased potential of grounding grids.

4 Thesecond component occurs as the consequence ofelectromagnetic couplings and is analyzed in this paper. So farmany methods have been developed to calculate thesecurrents. We propose a new method based on both the currentload factor of the Cable sheath ( ) and the Current reductionfactor of the Cable line (k). The proposed method is presentedin this of the factor is presented in [1]. It iscalculated for symmetrical three phase Currents (positiveand/or negative sequence components) of the analyzed cableline.

5 Three single-core cables in trefoil formation have equalfactor . It is calculated as follows: iipddpIIII== (1)whereidI,I- positive and negative sequencecomponents of the phase conductorcurrent of the single-core Cable L1,respectively, ipdpI,I- positive and negative sequencecomponents of the sheath Current of thesingle-core Cable L1, respectively,The factor k depends on the Current that flows through theearth. It is calculated as follows, [2]:oeI3Ik=(2)whereeI- Current that flows through the earth,oI- zero sequence component of the Current that flowsthrough the phase conductor during line-to-ground short circuit in the transmission is calculated as follows.

6 III(31I3L2L1Lo++= (3)where, 1LI, 2LI and 3LI are phase Currents of single-corecables L1, L2 and L3, THEORETICAL BASIST hree-phase Cable line consisting of three single-corecables laid in trefoil formation is shown in Theirconductive Sheaths are grounded at both 1. Single-core cables laid in trefoil formationAccording to [1] and [2] factors and k are calculated asfollows:pco1ppcorDn2jRrDn2jll + = (4))fDr658(n2j83R3Rk32cpoo1p1p + +=l(5)whereRp1- single-core Cable sheath resistance per length unit, /m,rp -mean radius of single-core Cable sheath , m,Dc - outer diameter of single-core Cable , m, - earth electrical resistivity, m, o- air permeability, o= 4 10-7 Vs/Am, - angular frequency of the Current in phase conductorsand Cable Sheaths .)

7 Angular frequency is calculated asfollows: = 2 f(6)wheref - Current frequency, Cable line connected with active networks A and B isshown in Fig. 2. Networks A and B are direct-grounded. Line-to-ground short circuit is in the network B at the phaseconductor of L1. Currents flow through phase conductors(1LI, 2LI and 3LI), through Cable Sheaths (1 LpI, 2 LpI and3 LpI) and through the earth (eI ) during short circuit. Theyare shown Fig. 1LI, 2LI and 3 LIare known from the shortcircuit analysis. They can be shown by sequence componentsas follows [3]: = 3L2L1L22idoIIIaa1aa111131 III (7)where o120jea=.

8 Fig. 2. Currents of the Cable line during line-to-ground short circuitThe Current 3LI usually equals (8)Substituting from (8) into (7) yields:)I2I(31I2L1Lo+= (9))II(31II2L1 Lid == (10)According to (1) and (10) positive and negativecomponents of the sheath Current are as follows:)II(3I2L1 Ldp = (11))II(3I2L1 Lip = (12)Besides Currents dpI and ipI the zero sequencecomponent of the sheath Current opI flows through the cablesheath.. According to Fig. 2 it follows:opeoI3II3+= (13)Substituting from (2) into (13) it becomes:oopI)k1(I = (14)According to (9) and (14) it follows:)I2I(3k1I2L1 Lop+ = (15) Currents 1 LpI, 2 LpI and 3 LpI can be calculated by the nextmatrix equation, [3]: = ipdpop223Lp2Lp1 LpIIIaa1aa1111 III(16)Substituting from (11), (12) and (15) into (16) yields.

9 2L1L1 LpI3)k1(2I32k1I + + = (17)2L1L3Lp2 LpI3)k1(2I3k1II + + == (18)In case the network B is a passive network, phase currents2 LIand 3LI equal (19)Equations (17) and (18) become:1L1 LpI32k1I + = (20)1L3Lp2 LpI3k1II == (21) Currents 1 LpI, 2 LpI and 3 LpI, calculated by equations (17),(18), (20) and (21), are an outcome of electromagneticcoupling. They are calculated by both the Current load factor( ) and the Current reduction factor (k). The Current loadfactor includes the electromagnetic coupling of positive andnegative sequence components of the Currents .

10 The currentreduction factor includes the electromagnetic coupling of zerosequence components of the A NUMERIC EXAMPLEThe method described in this paper can be applied intransmission and distribution Cable networks. The Cable line of110 kV is chosen for the numerical example. It consists ofthree single-core cables of AXLJ 1x1000/95 mm2 , [4]. Cabledata are: DC = 85 mm, rP = 38 mm and RP1 = m cables are laid in trefoil formation and touch eachother. Their conductive Sheaths are grounded at both earth resistivity = 500 m.