Skip to main content

Capacitance of double circuit line, 6 mm radius

Short answer

Treat each phase as a two-conductor bundle. Equivalent mutual GMD =4.7082 m, self GMR =0.20717 m. Capacitance C =2*pi*epsilon0/ln(4.7082/0.20717)=17.81 nF/km. For 100 km, C_total=1.781 µF, Xc=1/(2π*50*1.781e-6)=1787 Ω. At 132 kV, total charging VAR = (132e3)^2/1787 =9.75 MVAR.

Find the capacitance per phase per km of the double circuit three phase line shown in figure below. The line is completely transposed and operates at a frequency of 50 Hz. Radius of each conductor is 6 mm. Also compute the capacitive reactance per phase and total VAR generated by the line of 100 km and operating at 132 kV. [Figure: vertical double-circuit arrangement; left column top to bottom a, b, c; right column top to bottom c', b', a'; vertical spacing 3 m between rows; horizontal distances a–c' = 5 m, b–b' = 6 m, c–a' = 5 m.]

For a transposed double-circuit line, the two conductors of a phase are treated as a two-conductor bundle. The capacitance per phase is C=2πε0/ln⁡(GMD/GMR)C = 2\pi\varepsilon_0/\ln(GMD/GMR), where GMD is the equivalent mutual distance between phase groups and GMR is the self-GMD of each phase group (using actual radius rr).

   a o------- 5 m -------o c'    --
                                  3 m
  b o-------- 6 m --------o b'   --
                                  3 m
   c o------- 5 m -------o a'    --

Data: r=6r = 6 mm =0.006= 0.006 m, vertical spacing 3 m, f=50f = 50 Hz, 132 kV, 100 km.

Distances (from geometry)

Horizontal offset between a and b is (6−5)/2=0.5(6-5)/2 = 0.5 m.

PairDistance (m)
ab, bc, a'b', b'c'32+0.52=3.0414\sqrt{3^2+0.5^2} = 3.0414
ab', a'b, bc', b'c32+5.52=6.2650\sqrt{3^2+5.5^2} = 6.2650
ac, a'c'6
ac', a'c5
aa', cc'62+52=7.8102\sqrt{6^2+5^2} = 7.8102
bb'6

Mutual GMD

DAB=(DabDab′Da′bDa′b′)1/4=(3.04142×6.26502)1/4=4.3651 mDBC=4.3651 m(by symmetry)DCA=(6×5×5×6)1/4=5.4772 mDm=4.3651×4.3651×5.47723=4.7082 m\begin{aligned} D_{AB} &= (D_{ab}D_{ab'}D_{a'b}D_{a'b'})^{1/4} = (3.0414^2 \times 6.2650^2)^{1/4} = 4.3651\ \text{m} \\ D_{BC} &= 4.3651\ \text{m} \quad (\text{by symmetry}) \\ D_{CA} &= (6 \times 5 \times 5 \times 6)^{1/4} = 5.4772\ \text{m} \\ D_m &= \sqrt[3]{4.3651 \times 4.3651 \times 5.4772} = 4.7082\ \text{m} \end{aligned}

Self GMD of each phase (for capacitance)

DsA=r⋅Daa′=0.006×7.8102=0.21648 mDsB=0.006×6=0.18974 mDsC=DsA=0.21648 mDs=0.21648×0.18974×0.216483=0.20717 m\begin{aligned} D_{sA} &= \sqrt{r \cdot D_{aa'}} = \sqrt{0.006 \times 7.8102} = 0.21648\ \text{m} \\ D_{sB} &= \sqrt{0.006 \times 6} = 0.18974\ \text{m} \\ D_{sC} &= D_{sA} = 0.21648\ \text{m} \\ D_s &= \sqrt[3]{0.21648 \times 0.18974 \times 0.21648} = 0.20717\ \text{m} \end{aligned}

Capacitance per phase per km

C=2πε0ln⁡(Dm/Ds)=55.63×10−12ln⁡(4.7082/0.20717)=55.63×10−123.1235=17.81×10−12 F/m=17.81 nF/km\begin{aligned} C &= \frac{2\pi\varepsilon_0}{\ln(D_m/D_s)} = \frac{55.63\times10^{-12}}{\ln(4.7082/0.20717)} = \frac{55.63\times10^{-12}}{3.1235} \\ &= 17.81\times10^{-12}\ \text{F/m} = 17.81\ \text{nF/km} \end{aligned}

Capacitive reactance for 100 km

C100=17.81×100=1781 nF=1.781 μFXC=12πfC=12π×50×1.781×10−6=1787.2 Ω\begin{aligned} C_{100} &= 17.81 \times 100 = 1781\ \text{nF} = 1.781\ \mu\text{F} \\ X_C &= \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 1.781\times10^{-6}} = 1787.2\ \Omega \end{aligned}

Total VAR generated

Phase voltage Vp=132/3=76.21V_p = 132/\sqrt{3} = 76.21 kV.

Ic=VpXC=76 2101787.2=42.64 AQ=3VpIc=VL2XC=(132×103)21787.2=9.749 MVAR\begin{aligned} I_c &= \frac{V_p}{X_C} = \frac{76\,210}{1787.2} = 42.64\ \text{A} \\ Q &= 3V_p I_c = \frac{V_L^2}{X_C} = \frac{(132\times10^3)^2}{1787.2} = 9.749\ \text{MVAR} \end{aligned}

Answer: C=17.81C = 17.81 nF/km per phase; XC=1787.2 ΩX_C = 1787.2\ \Omega per phase (100 km); total charging = 9.75 MVAR.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Power System Analysis I chapter 4 questions →