PowerSystems (1)

1Inductance of 3 Lines with Equilateral SpacingAssumptions:1. Transmission Line(so no neutral wire)2. Balanced 3 currents Ia + Ib + Ic = 0a= 2 X 10-7 [ IaLn 1/r+ IbLn 1/D + Ic Ln 1/D ] (wbt/m)a= 2 X 10-7 [ Ia Ln 1/r+ (Ib + Ic) Ln 1/D ] (wbt/m)(I I )= -IaDept. of Electrical Engineering Sandhya S.Sandhya S.(Ib + Ic)= -Iaa= 2 X 10-7 [ Ia Ln 1/r- Ia Ln 1/D ] (wbt/m)La= a / Ia = 2 X 10-7 Ln D/r] H/mb cDD D2Inductance of 3 Lines with Asymmetrical SpacingWhen in position 1a1= 2 X 10-7 [ Ia Ln 1/r+ Ib Ln 1/D12+ Ic Ln 1/D13] wbt/mWhen in position 2Dept. of Electrical Engineering Sandhya S.Sandhya S.When in position 2a2= 2 X 10-7 [ Ia Ln 1/r+ Ib Ln 1/D23+ Ic Ln 1/D21] wbt/mWhen in position 3a3= 2 X 10-7 [ Ia Ln 1/r+ Ib Ln 1/D31+ Ic Ln 1/D32] wbt/mEach of the conductors occupies the position of the other two conductors for 1/3rdof the total length of the conductorsAverage value of flux linkage of Conductor aa = (a1 + a2+ a3) / 3Contd3a= (2 X 10-7) / 3 [ 3 Ia Ln (1/r) + Ib Ln (1/D12 D23 D31) + Ic Ln (1/D12 D23 D31) ] wbt/mInductance of 3 Lines with Asymmetrical Spacing .. contdabcD 3113D 12D 23Dept. of Electrical Engineering Sandhya S.Sandhya S.(Ib + Ic) = -Ia a = (2 X 10-7) / 3 [ 3 Ia Ln (1/r) - Ia Ln (1/D12 D23 D31) ] wbt/m= (2 X 10-7) Ia Ln D12 D23 D31/ rLa= 2 X 10-7 Ln D12 D23 D31/ rH/m= 2 X 10-7 Ln ( Deq) / rwhereDeq =D12 D23 D313334Necessity of Transposition of ConductorsEven when the system is balanced, Ia+ Ib+ Ic= 0.MagVa = Vb = Vc & 120 Deg. apartWhen the three conductors of the 3 line are asymmetrically spaced Flux linkages and the inductances of different phases varies, which causes;Dept. of Electrical Engineering Sandhya S.Sandhya S.1. Unequal V drops in the three phases, even when Ia+ Ib+ Ic= 02. There is a transfer of power between the phases, represented by the imaginary terms of the expression for inductance.This shows that the magnetic field outside the three conductors is not zeroThis will cause induced voltages in adjacent electric circuits, particularly telephone lines, resulting in telephonic interference.5Unsymmetrically spaced throughout , but balanced IR = IR00, IY= IR -1200, IB= IR-24003 ConceptIawhen multiplied by k = 1 1200rotates the vector by1200Unsymmetrical SpacedRD121D31Dept. of Electrical Engineering Sandhya S.Sandhya S.IR= I 00Iy= I -1200= k2IIB= I -2400= k IIRIBIYYB23D23D316R= 2 X 10-7 [ IRLn1/r1+ IYLn 1/D12+ IBLn 1/D13]Y= 2 X 10-7 [ IYLn1/r2+ IB Ln1/D23+ IR Ln1/D21]B= 2 X 10-7 [ IB Ln1/r3+ IR Ln1/D31+ Iy Ln 1/D32 ]r1 = r2 = r3 R= 2 X 10-7 [ IR Ln1/r1+ k2IR Ln 1/D12+ k IR Ln1/D13]Put k2 = - 0.5 j/ 2(Cos + j Sin ) formk= - 0.5 + j/ 2 Unsymmetrical Spaced contd33RYBD12123D23D31Dept. of Electrical Engineering Sandhya S.Sandhya S.R= 2 X 10-7 IR [Ln 1/r1- (1/2) Ln (1/D12) - (1/2) Ln(1/D13) + (j / 2) [ - Ln(1/D12) + j Ln(1/D13)]= 2 X 10-7 IR[Ln (D12 D13 / r ) + (j/ 2 Ln(D12/ D13)Y= 2 X 10-7 IR[Ln (D23 D21 / r ) + (j/ 2 Ln(D23/ D21)B= 2 X 10-7 IR[Ln(D31 D32 / r ) + (j/ 2 Ln(D31/ D32)3333B 37Unsymmetrical Spaced contdLR= 2 X 10-7 [ Ln(D12 D13 / r ) + (j/ 2Ln(D12/ D13) ]LY= 2 X 10-7 [ Ln(D21 D23 / r ) + (j/ 2Ln(D23/ D21) ]LB= 2 X 10-7 [ Ln(D31 D32 / r ) + (j/ 2Ln(D31/ D32) ]333Dept. of Electrical Engineering Sandhya S.Sandhya S.3 Inductances are differentAlthough the voltages and current are balanced at one (sending) end, the voltages at the receiving end will be unbalanced due to unequal voltage drops in the 3 phases.8Is the mean of the distance from one point (P) to a group of other points,If the points 1, 2, 3 and 4 be situated on a circle and P be a point from which the distance is considered, the Geometric Distance is D= Geometric Mean Distance (GMD)D1D2D3D4 4Dept. of Electrical Engineering Sandhya S.If the number of points is increased, then the geometric mean distance is equal to the distance of the point P from the centre of the circle.Deductions:1. GMD from any point on a circle to all other points on the circle is equal to the radius of the circle.2. GMD between two circular areas is equal to the distance between their centres (circular area may be of different sizes)3. Self GMD of a circular area is equal to the radius line e(-1/4).4. If a circular area of radius r has equally spaced points around its periphery, the GMD among them is r(n-1)5. The self GMD of a rectangular area where sides are a & b is approximately equal to 0.2235 (a+b) for all ratios of a/b.Sandhya S.9Bundled Conductors Multiple ConductorsConsists of two or more individual conductors separated by spaces and supported by the same insulator. Each bundle conductor forms one equivalent phase conductor. Due to the cost involved, it is used in EHV ( > 230 kV) lines only. It increases the capacity. The spacing shall be such as to prevent excessive swinging.Dept. of Electrical Engineering Sandhya S.Sandhya S.Advantages:1. Reduced Reluctance2. Reduced Voltage Gradient3. Reduced Corona loss4. Reduced wireless interference5. Reduced surge impedance10AdvantagesReduced ReactanceSince the self GMD of the conductor is increased the reactance is reduced.L = 2 X 10 -7Ln(GMD/GMR)Theoretically there should be an optimum sub conductor spacing. Usually it is not greater than 5 to 6 times the diameter of the conductor.Voltage GradientSince the voltage gradient is reduced, it produces reduced radio interference and lesser Dept. of Electrical Engineering Sandhya S.Sandhya S.Since the voltage gradient is reduced, it produces reduced radio interference and lesser corona loss.Surge ImpedanceReduces as self GMD or GMR is increased. So L is reduced and C is increased. So the maximum power that can be transmitted (steady state stability limit) is increased.Bundled conductors are separated from each other by spacers at a distance of at least 30 cms. or more, whereas composite conductors almost touch each other.11To find GMR of a bundled conductordddTheGMR of bundled conductors can be found in the same manner as that of stranded conductors.For a two conductor dule! arran"ementD#b2 $% &r . d.'lDept. of Electrical Engineering Sandhya S.Sandhya S.dddd1( 1(3 conductor bundleD#b% &r . d. d'$ conductor bundle D#b& r. d. d.2d ')3 % &r . d. d'3% r d 32$l lll% 1.*)r d3l$%12Inductance of double circuit 3 lineDouble conductor 3 line has two parallel conductor per phase.Greater StabilityHigher Transmission CapabilityDouble-circuit towers are taller than single-circuit towers because the phases are arranged vertically and the lowest phase must maintain a minimum ground clearance, while the phases are arranged horizontally on single-circuit.Dept. of Electrical Engineering Sandhya S.Sandhya S.If two circuits are widely separated mutual inductance between circuits can be neglected . So net inductance is half the inductance of each circuit Leq= Leach. ckt.In actual practice the separation is not wide and the mutual inductance is not negligible.The method of GMD can be used to find the inductance per phase.13Inductance of double circuit 3 line .. ContdMethod:Take the conductors of one phase to be the strands of one composite conductor.It is desirable to have minimum inductance for maximum transmission capability. This is possible if GMD is small and GMR is large.Hence the individual conductors of a phase should be widely separated (for high GMR) and the distance between phases should be low or less.abcc a+bcac+b+b+a+c+c+b+a+ab#ection 1 #ection 2 #ection 3Dept. of Electrical Engineering Sandhya S.between phases should be low or less.Type 1. Figure shows the three section of a double circuit 3 line with vertical spacing overatransposition cycle. The conductor a a are for the same phase.Since the conductors are not symmetrically placed, the conductors should be transposed.The three positions are as shown.14Inductance of double circuit 3 line .. ContdIn each case the conductors of 2phases areplaced diagonally opposite to each other and those of the 3rd phase is placed diametrically opposite to each other. This configuration gives high value for GMR. To calculate the inductance it is necessary to find the Deq or GMD and Dsor GMR for each phase.Dept. of Electrical Engineering Sandhya S.Deq=Dab. Dbc. DcaDab= Mutual GMD between phase a and phase b.= DabDabDabDab= ( d. g. d. g) = (d. g.) ab. ab.ab . abDbc= ( d. g. g. d) = (d. g.) bc. bc.bc . bcDca= ( 2d. h. h. 2d)ca. ca.ca . ca 4315Deq =Dab. Dbc. Dca= (d g) (d g) (2 d h)= 2d ghDsa = ( r f.f r ) = ( r f )Inductance of double circuit 3 line .. Contd33 1/61/21/31/61/41/21/41/Dept. of Electrical Engineering Sandhya S.Dsc= ( r f.f r ) = ( r f )1/41/2Dsb= ( r h.h r ) = ( r h )1/41/2Ds=Dsa. Dsb. Dsc3=( r f ) . ( r f ). ( r h ) 1/21/21/23=r Ifh 1/31/61/216Inductance of double circuit 3 line .. ContdWhat will be the equivalent radius of a bundled conductor having its part conductors of radius r on the periphery of a circle of diameter d if the number of conductors is 2,3,4 and 6?a) r =(r d ) =r d1/21/2 1/2 b)( r Id Id I)= r[ ( 3/2) d]1/31/31/3Dept. of Electrical Engineering Sandhya S.=rd (3 / 4) 1/3 2/3 1/3 c)r I =[ r . (d / 2) . (d / 2) . d )]=r d (1/2) 1/41/43/41/4d) r I=r(d/2) 6 = [ 6 r (d/2)] 1/6 5/6 1/6 51/6 17Inductance of double circuit 3 line .. ContdDetermine the inductance /km of a transposed double circuit 3 phase line shown in the figure. Each circuit of the line remains on its own side. The dia of the conductors is 2.532 cm?c+ab b+,.-m).* mSelf GMD of each conductor = 0.7788 X (2.532 / 2)= 0.00986 mbc= ab = 42+(0.75)2=4.067 mabI= 42+ 8.252= 9.1685 maaI = 82+7.52=10.965 mDab = 4.0697 X 9.168 X 9.168 X 4.0697= 6.108 ;Dab= Dbc4Dept. of Electrical Engineering Sandhya S.Self GMD of phase aDsa=0.00986X 10.965= 0.3288 =DscDSb=0.00986X9= 0.2989DS= DSaXDSbX DSc = 0.318 mInductance= 2 X 10 -4Ln ( 6.61 / 0.318)= 0.606 mH / kM / Phaseca+,.-m).* mabab bcDac= 8 X 7.5 X 7.5 X 8= 7.746Dm =Dab. Dbc.Dca=6.61 m4318Inductance of double circuit 3 line .. ContdDetermine the inductance of a double circuit line shown in figure. The self GMD of the conductor is 0.0069 m.ab=bc=32+ 52= 3.04 m ;ac = 6.0 mabI=32+62=6.708 maaI=62+ 5.52=8.14 mDab= 3.04 X 6.708 X 6.708 X 3.04=4.515= Dbc4c+a-.-mDept. of Electrical Engineering Sandhya S.Dab= 3.04 X 6.708 X 6.708 X 3.04=4.515= DbcDac= 6 X 5.5 X 5.5 X 6 = 5.745Dm =DabDbcDac =4.515X 4.515 X 5.745 = 10.821 Dsa= 0.0069X814 =0.2370= DScDSb = 0.0069X6.50=0.2117DS =DS1DS2DS3=0.228Inductance L =2 X 10 -7 Ln (10.821 / 0.228) =0.772 mH/kM4333bcb+a+-.-m(.- m19Inductance of double circuit 3 line .. ContdDetermine the inductance per phase per kilometer of a single circuit 460 kV line using two bundle conductors per phase as shown in the figure. The diameter of each conductor is 5 c.m.?Assuming the effect of transposition as too smallDS= 0.7788 X0.025 X0.4 = 0.08825DM=6.5X13.0X6.5=8.19 m-43Dept. of Electrical Engineering Sandhya S.Inductance /phase/km =2 X 10-4Ln (8.19 / 0.08825) = - 0.906 mH/ph/km20Inductance of double circuit 3 line .. ContdA 400 kV 3 phase bundled conductor line with 2 sub-conductors has a horizontal configuration as in figure. The radius of each conductor is 1.6 cm. a) Find the inductance/phase/km of the line.b) Compare the inductance of the line having the same cross section area of the conductor of each phase.rI= 0.7788X 1.6=1.246 cmGMR=rId . r I. d4Dept. of Electrical Engineering Sandhya S.GMR=rId . r I. d= 1.246X45 =7.49 cmDab= Dbc= 12m (12 + 0.45) X 12 X (12.0 - 0.45)= 11.996 mDca=24 (24 + 0.45) X 24 X (24.0 - 0.45)=23.998 mDeq = 11.996X23.998X11.996 =15.115 mInductance =0.4605Log (Deq/ Ds)= 0.4605 Log (15.115 / 7.49)= 1.06 mH/phase/km4424321Inductance of double circuit 3 line .. ContdTaking the centre to centre distance as DabDbcDca .Deq= 12 x 12 x 24=15.119Conductor having same cross section area.r2=(2 X . r12) / . =2 r1 =2 X 1.6= 2.2624 cmr2= 0.7788X2.2624=1.765 cm3Dept. of Electrical Engineering Sandhya S.Deq=15.119L =2 X 10 -7Ln (15.119X 100) / (1.762)= 1.351 mH/kmThis the inductance of the bundled conductor line is less than that of the line with single conductor/per phase.22Inductance of double circuit 3 line .. ContdA bundled conductor line has 4 conductors per bundle. The 4 sub-co0nductorsare placed at the corners of a sqaure of sides 25 cm. The radius of each sub-conductor is 1.573 cm. Find the GMR of the configuration.DS for each sub-conductor= 0.7788X 1.573= 1.225 cmDSb=(1.225x 25x 252 ) 4=12.826 cm 16Dept. of Electrical Engineering Sandhya S.A 3 phase transmission line has a horizontal configuration of 6m between adjacent conductors and 12 m between outer conductors. The radius of each conductor is 1.81 cm. Find the inductance per phase per km of the line.rI =0.7788X 1.81=1.4096Deq=6X6X12= 7.5595 ML = 2x 10 -7Ln(Deq /r )=2X 10 -7Ln (7.5595/1.4096 X 10-2 )= 1.2569mH/phase/km323Capacitance of Transmission LinesSince conductors are maintained at different potentials, there will be electric charges over them.Consider one straight, bound, infinitely long conductor- having instantaneous charge of q coulomb/m length.- There will be no charge or electric field inside the conductor (unlike electromagnetically set up flux lines)Dept. of Electrical Engineering Sandhya S.(unlike electromagnetically set up flux lines)- i.e. all of the charge will be uniformly distributed over the outside surface.Effect: This will give rise to a field outside the core as if the charge were concentrated on the surface of the conductor.Proximity effect neglected----Sincethe spacingbetween conductors is large compared to the conductor radius.Equi-potential surfaces will be concentric with the conductor.24Let the Electric flux lines emanating from 1mlength of the conductor q coulombs.Electric flux density D is constant along any equipotenial surface having radius x metre.DX = qcoulomb/sq.m ( Area A = 2 .x1 )2 . x 1Electric field intensity = Ex=(Dx/ /) = (q / 2 .x. /) V/mWhere / = / / =(1 / 36 . x 109) /Capacitance of Transmission LinescontdDept. of Electrical Engineering Sandhya S.Where / = /0/r =(1 / 36 . x 10 ) /r/0= (1 / 4 . x 9x109) /r=8.854 x 10-12F/mEx= q =(1.8 x 1010) . qV/m2 . x1/rx /r4 . x 9 x 109=(1.8 x 1010) qV/m/rxElectric field intensity E at any point is equal to negative of the potential gradient at that point.2 1 m . 25Ex =- (dV/dX)So potential at any point R above X.DDCapacitance of Transmission Linescontd0 D!r*1!Dept. of Electrical Engineering Sandhya S.VRX=VR- VX= DR Exdx = DR1.8 x 1010[q / x] dx= 1.8x 1010q [ Ln x ] rDx= 1.8x 1010 . q. Ln [Dx/r]DxDx26Capacitance of 2 infinitely long parallel straight round conductors (Single Phase line)Let the conductors be ofradii raand rbrespectively. D-- spacing between conductors & surface charges of +qaand +qbcoulomb/mSince dielectric medium is an isotropic medium , the principle of superposition theorem can be applied to find the potential difference between them.Vab due to qa =[qa/ (2 ./)] . Ln (D/ra)VoltsDept. of Electrical Engineering Sandhya S.Vab due toqb=[qb/(2 ./)] . Ln (rb/D)VoltsVabTotal= (1/2./) [ qa Ln (D/ra) + qbLn (rb/D)= (1/2./) [ qaLn (D/ra) - qa Ln (rb/D) ]= (qa/2./) [Ln (D/ra) +Ln (D/rb) ]= (qa/2./) . Ln (D2/ rarb)= (qa/./) . Ln (D / rarb)rarb D2a2bA Bc27Capacitance of 2 infinitely long parallel straight round conductors (Single Phase lineqa./qaLn D= LnD F/m. / rarb rarbCab= Q/V =Where / = /0/r= 1 for airIf ra= rb=r/0= ( 1/ 36 . X 109 )b a34e 5 4e6 6Dept. of Electrical Engineering Sandhya S.Cab=(. /0 /r) =1 7F / kmLn D/r 36Ln D/r=10-9 F/m36Ln D/rAs capacitance to neutral is twice line to line capacitance Cn= Can= Cbn = 2 CabCn= 2 . / / Ln (D/r)=0.555 / Ln (D/r) 7F/km 1+6e2% 1+63 1+6% 2+6 6e2%6+286 is 6n = 2 6ab613 623 636n% 2 6ab28Capacitance of a 3 Phase line with equilateral spacingAssumptions:1. There is no other charged surface in the vicinity. So qa+ qb +qc= 0 . Neglect effect of earth2. Since the distance between the conductors are large compared to the radii, the charges are uniformly distributed over the surface.V drop between any two conductorsDept. of Electrical Engineering Sandhya S.Vgedrop between any two conductorsVab=1 qa Ln (D/r)+ qbLn (r/D)+ qc Ln (D/D)2 . /Vac=1 qa Ln (D/r)+ qbLn (D/D)+ qc Ln (r/D)2 . /29Capacitance of a 3 Phase line with equilateral spacingAdding Vaband Vacand substituting qb+ qc= -qaVaband Vac =1 2 qa Ln (D/r)- qaLn (r/D)2 . /=1 2 qa Ln (D/r)+ qaLn (D/r)2 . /=1 3 q Ln (D/r)=3 VV + V = 3 VDept. of Electrical Engineering Sandhya S.=1 3 qa Ln (D/r)=3 Van2 . /Vac+ Vab= 3 VanVan=qaLn (D/r)2 . /0/r118 Ln (D/r)Cn= qa=2 Vanr Ln (D/r)F/m 7F /km =30Unsymmetrical Spacing But Transposed132abcFor a untransposed line, the capacitance from each phase to neutral are unequal. Transposition results in the same average capacitance to neutral to various phases over the entire length of the line.Dept. of Electrical Engineering Sandhya S.To determine the capacitance it is sufficient to find the capacitance of the transposed line as the average value of one phase of the same line correctly transposed.31Potential Difference Between Two Conductors ofA Group of Charged ConductorsbcDabDbcIf m number of conductors are arranged so that they are parallel to each other, the Voltage between any two of them can be found asV12 = [q / 2 . / ] X Ln (D2/D1)repeatedlyi.e. Due to the charge on each of the Dept. of Electrical Engineering Sandhya S.amDDcmDbmDacconductors on the 2 conductor positions are considered.Assumptions:1. Distance between the conductors is large compared to their radii2. Earth is far away to have negligible effect3. There is no other charge in the vicinityContd32By assumption 3, the sum of the charges on all conductors is equal to zero.If the spacing between the conductors is large, compared to the radii, the charge distribution over the surface of the conductors will be uniform.Potential Difference Between Two Conductors ofA Group of Charged Conductors - contd.Dept. of Electrical Engineering Sandhya S.33Effect of Earth on the Capacitance of Transmission LinePresence of earth alters the electric field of the line and hence the capacitance of the line is affected.Assuming earth to be a perfect conductor, the ground surface is an infinite horizontal plane, abundant in both +ve and ve charges. So the electric field is affected by the presence of this surface.Dept. of Electrical Engineering Sandhya S.GNDsurface acts as an equi-potentialsurfacethe lines of charges or electric flux line will enter this surface at right anglesA potential difference exists between the charged conductor and the earth.34Effect of Earth on the Capacitance of Transmission Line ContdIf another conductor of equal size with equal and opposite charge is assumed to be present below the GND surface at a distance equal to the height of the conductor above GND surface, the flux distribution will be as in the figure.The flux lines intersect the GND surface at right angles.Dept. of Electrical Engineering Sandhya S.GND surface can now be removed by replacing the GND with the presence of an image conductor (fictitious conductor) which has a charge equal and opposite to that of the conductor.This arrangement can be extended to any number of conductors.35Capacitance of a single conductor lineFrom fundamentals of capacitanceC=. /0= . /0 F/mLn (D/r)Ln (2h/r)Dept. of Electrical Engineering Sandhya S.Capacitance w.r.t GND=2 . /0F/mLn (2h/r)36Capacitance of a Single Phase Line (Considering the Influence of Earth)C12= 1 q1LnD12+ q2LnD22+ q3LnH32+ q4LnH422 ./ D11D21H31 H41q1= q3=+q;q2= q4=-qC12= 1 q1LnD12- q2Lnr+ qLnH32- qLnH422 ./ r D21 H31 H41C12= 1 qLnD12H32- q Lnr H42Dept. of Electrical Engineering Sandhya S.C12= 1 qLnD12H32- q Lnr H422 ./ rH31 D21 H41= 1 qLnD12 H32H41./ rH31H42xC12=q/V=./LnD12 H32H41rH31H42x37Capacitance of a Single Phase Line (Considering the Influence of Earth)contd..IfH32andH41approach unity gives capacitance without presence of GND.H31H42H32 = 2H=H41;H31=D2+ 4H2; D12= DDept. of Electrical Engineering Sandhya S.Cn= 2C1238Capacitance of a Single Phase Line (Considering the Influence of Earth)contd..Consider the case of a 2 conductor line with the following data. H32= H14= 20m; D12= D34= 3 m; r = 5.25mmH42= H31= 202 + 32 =20.2 mDept. of Electrical Engineering Sandhya S.The capacitance increased only by 7% on account of the presence of GND. Hence, here the effect of GND is negligible.39Effect of Earth on the Capacitance of a 3 Phase Transmission LineAssuming transpositionDept. of Electrical Engineering Sandhya S.9RYaverage:n &D12 D23 D31)r3 qR:n;62 ;53 ;41 ;61 ;52 ;43+qY+qB40Effect of Earth on the Capacitance of a 3 Phase Transmission LineH42= H53,H63=H41,H51=H62+VRYAssuming balanced Dept. of Electrical Engineering Sandhya S.system as before41Effect of Earth on the Capacitance of a 3 Phase Transmission LineInferences:Equation shows that the cap/phase is increased on account of the presence of GND.If the line is High above GND the diagonal distances will be nearly equal to the vertical distances.Dept. of Electrical Engineering Sandhya S.H52= H51, H53=H52,H41=H43so that the 2ndterm of the denominator vanishes42Capacitance of Bundled Conductor LinesBundled conductor lines with two sub conductor per phaseDept. of Electrical Engineering Sandhya S.The two sub-conductors of each phase are parallelThe charge in each bundle is considered to be equally divided between two sub-conductors of the bundle.Since (D12 >> d) the distance D12can be used in place of (D12-d) and (D12+d)43Capacitance of Bundled Conductor LinesDept. of Electrical Engineering Sandhya S.This equation is similar to the equation for inductance. SO assuming transposition and proceeding in the same way before,is the Dsbfor 2 conductor bundle, except that the outside conductor radius has replaced effectie radius r. For 3 conductor and 4 conductor bundle the same method can be used.44Double Circuit Three Phase LinesThroughout the analysis a similarity can be observed between expressions for inductance L and Capacitance C.So the capacitance of transposed double circuit line can also be found by modified GMD method.Double Circuit Three Phase Lines with Equilateral SpacingDept. of Electrical Engineering Sandhya S.SpacingThe conductors will come arranged at the vertices of a regular hexagon.45Double Circuit Three Phase Lines with Equilateral Spacinga c