N11830 - Line Parameters Calculation

CONTRACT : N-11830 132 KV OHL AL HAYL SUBSTATION TO KALBA SUBSTATION

CALCULATION OF LINE PARAMETERS

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General Details :

Line type : Three phases, Double circuit, Twin conductor per phase, one ground wire (OPGW).

Operating Voltage: 132 KV, 50 Hz

Line Lengths : 5.380 Kms

Conductor details : Type : ACSR “DOVE”

Diameter : 23.55 mm

Area of cross section : 328.52 mm2

DC resistance @ 200C: 0.09745 Ω/km Ground Wire details : Type : ‘OPGW – 32 fibers’ - Prysmian

Diameter : 18.86 mm

Area of cross section : 187 mm2

DC resistance @ 200C: 0.250 Ω/km

Distance (mm) Between Conductors/OPGW: (Refer sketch – annexure 1)

d12 = 6900 d1 I = 20739

d23 = 7100 d2 II = 14600

d31 = 14000 d3 III = 20739

d1 II = 16058 d1S = 14408

d2 III = 16058 d2S = 20710

d3 I = 20739 d3S = 27692

Note : Subscripts 1,2,3 indicates conductors of ckt 1 Subscripts I, II, III indicates conductors of ckt 2

Subscript ‘S’ indicates earthwire



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Calculation of Geometric Mean Conductor Spacings (mm) reqd for parameter calculations :

dm = 3√(d12.d23.d31) = 8819 mm ………………(1)

Dm = 3√((d12.d23.d31.d1 II.d2 III.d3 I) / (d1 I.d2 II.d3 III)) = 8359 mm ………………(2)

ds = 3√ (d1s.d2s.d3s) = 20217 mm ………………(3)

DAB = 6√(d1 I.d2 II.d3 III.d1 II.d2 III.d3 I) = 17962 mm ………………(4) Geometrical mean radius of bundled conductor :

Equivalent radius of bundled conductor, r = n√ (n.r’.rTn-1) = 68.64 mm

Where, n = Number of sub conductor = 2 r’ = Radius of sub conductor = 11.78 mm aT = Distance between bundled conductor = 400 mm rT = Radius of circle passing through = 200 mm midpoints of strands of a bundle A.C Resistance for Twin Bundle :

R = (0.0974 + (0.0974 x 4%))/2 = 0.099 Ω/ (km.phase) ………………….(A)

(4% considered for sag and skin effect.)

Inductive Reactance, XL :

Positive and Negative sequence reactance:

XL = ω (µ0/2π) ln (Dm/0.779 r) Ω/ (km.phase)

Where,

ω = Angular Frequency (S-1) = 2πf = 314 S-1 Hz f = Operating frequency (Hz) = 50 Hz Dm = Mean Spacing, as per (2) , mm = 8359 mm µ0 = Magnetic constant (H/km) = 4π x 10-4 = 126 x 10-5 H/km

r = Equivalent conductor radius (mm) = 68.64 mm (refer above) 0.779 = ∈-1/4 = Constant



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XL = 314 (126 x 10-5/2π) x ln ((8359)/(0.779 x 68.64))

XL = 314 x 2 x 10-4 x ln 156.33 = 0.0628 x 5.052

XL = 0.317 Ω/ (km. phase) ………………….. (B)

Positive and Negative sequence impedance (Z1) :

Z1 = R + j XL = 0.099 + j0.317 Ω/ (km.phase) ( from (A) & (B) )

Zero Sequence Resistance R0 : R0 = R + 3 (µ0/8)ω Ω/ (km.phase) Where,

R = Resistance, as per (A), Ω/km = 0.099 Ω/km µ0 = Magnetic constant (H/km) = 4π x 10-4 = 126 x 10-5 H/km

ω = Angular Frequency (S-1) = 2πf = 314 S-1 Hz f = Operating frequency (Hz) = 50 Hz R0 = 0.099 + 3 ((126 x 10-5) /8)) x 314 R0 = 0.099 + 0.1484 = 0.247 Ω/ (km.phase) …………………….(C) Zero Sequence Reactance X0 : X0 = 3 µ0 f ln δ Ω/ (km.phase) 3√(0.779 x r x dm

2) Where,

µ0 = Magnetic constant (H/km) = 4π x 10-4 = 126 x 10-5 H/km

f = Operating frequency (Hz) = 50 Hz r = Equivalent conductor radius (mm) = 68.64 mm dm = Mean Spacing, as per (1) , mm = 8819 mm δ = Effective depth of penetration = 2070 x 103, (given) of AC current into earth (mm)

(1) Frequency = 50 Hz (2) Dry sand



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X0 = 3 x 126 x 10-5 x 50 x ln 2070 x 103 Ω/ (km.phase) 3√ (0.779 x 68.64 x 88192) X0 = 18900 x 10-5 ln 2070 x 103 Ω/ (km.phase) 1608 X0 = 0.189 x 7.16 Ω/ (km.phase) X0 = 1.35 Ω/ (km.phase) …………………….(D)

Zero sequence impedence (Z0) :

Z0 = R0 + j X0 = 0.247 + j1.35 Ω/ (km.phase) (from (C) & (D) )

Mutual Impedance of line conductor – Earth and Earthwire/OPGW - Earthloop (Z’sm) :

Z’sm = (π µ0)/4 x f + j µ0 f x in (δ/ds) Ω/ (km.phase)

Where,


f = Operating frequency (Hz) = 50 Hz ds = Mean Spacing, as per (3) , mm = 20217 mm δ = Effective depth of penetration = 2070 x 103, (given) of AC current into earth (mm)


Z’sm = (3.14 x 126 x 10-5)/4 x 50 + j 126 x 10-5 x 50 x ln (2070 x 103/20217) Z’sm = 0.049 + j 0.291 Ω/ (km.phase) …………………….(E)



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Mutual Impedance of Earthloop systems (Z’m) : Z’m = (π µ0)/4 x f + j µ0 f x ln (δ/dAB) Ω/ (km.phase)

Where,


f = Operating frequency (Hz) = 50 Hz dAB = Mean Spacing, as per (4) , mm = 17962 mm δ = Effective depth of penetration = 2070 x 103, (given) of AC current into earth (mm)


Z’m = (3.14 x 126 x 10-5)/4 x 50 + j 126 x 10-5 x 50 x ln (2070 x 103/17962) Z’m = 0.05 + j 0.299 Ω/ (km.phase) …………………….(F) Impedance of Earthwire/OPGW Earthloop (Z’s) : Z’s = Rs + µ0 ω + jω µ0 (ln δ/rs + µr/4a) 1 8 2π Where,

Rs = Resistance of earthwire (Ω/km) = 0.250 Ω/km


ω = Angular Frequency (S-1) = 2πf = 314 S-1 Hz f = Operating frequency (Hz) = 50 Hz rs = Radius of earthwire (mm) = 9.43 mm

µr = Relative permeability = 1

a = Factor (No. of earthwire) = 1 δ = Effective depth of penetration = 2070 x 103, (given) of AC current into earth (mm)




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Z’s = 0.250 + 126 x 10-5 x 314 + j 314 126 x 10-5 ( ln 2070 x 103

+ 1 ) 1 8 2 x 3.14 9.43 4x1 Z’s = 0.250 + 0.049 + j0.063 (12.299 + 0.25)

Z’s = 0.299 + j 0.790 Ω/ (km.phase) …………………….(G)

Zero Sequence Impedence (ZO) :

1. Single Circuit Line without Earthwire/OPGW (Z’OE) : Z’OE = R0 + j X0 Ω/ (km.phase)

Z’OE = 0.247 + j1.35 Ω/ (km.phase) …………………….(H)

2. Double Circuit Line without Earthwire/OPGW (Z’OD) : Z’OD = Z’OE + 3 Z’m Ω/ (km.phase) Z’OD = 0.247 + j1.35 + 3 (0.05 + j 0.299) Z’OD = 0.247 + 0.15 + j1.37 + j0.897 Z’OD = 0.397 + j2.267

Z’OD = 0.397 + j2.267 Ω/ (km.phase) …………………….(I)



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3. Double Circuit Line with Earthwire/OPGW (Z’ODS) :

Z’ODS = Z’OD - 6 x ((Z’sm)2 / Z’s) Ω/ (km.phase) Z’ODS = 0.397 + j2.267 - 6 x ((0.049 + j 0.291)2 / (0.299 + j 0.790)) Z’ODS = (0.397 + j2.267) + (0.017 - j0.618)

Z’ODS = 0.414 + j1.649 Ω/ (km.phase) …………………….(J)

Capacitive Reactance Xc : Xc = 1/(ω x 2π x εo) x ln (Dm / r) Ω/ (km.phase)

Where,

ω = Angular Frequency (S-1) = 2πf = 314 S-1 Hz f = Operating frequency (Hz) = 50 Hz εo = Permittivity of free space (F/km) = 0.886 x 10-8 Dm = Mean Spacing, as per (2) , mm = 8359 mm r = Equivalent conductor radius (mm) = 68.64 mm (refer above) Xc = 1/(314 x 2 x 3.14 x 0.886 x 10-8) x ln (8359 / 68.64) Xc = 1/(1747 x 10-8) x 4.80 Xc = 274756.7 Ω/ (km.phase)

Xc = 274 kΩ/ (km.phase) …………………….(K)



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Equivalent Capacitance of a Conductor C’b : C’b = 1 / (ωXc) Where,

ω = Angular Frequency (S-1) = 2πf = 314 S-1 Hz f = Operating frequency (Hz) = 50 Hz Xc = 274 kΩ/km/phase C’b = 1 / (314 x 274)

C’b = 18.30 Nano farad/km …………………….(L)

SUMMARY:

Sl.No PARAMETERS VALUE

1. Resistance 0.247 Ω/ (km.phase)

2. Positive and Negative sequence Impedance (Z1) 0.099 + j0.317 Ω/ (km.phase)

3. Zero Sequence Impedance (ZO) :

Single Circuit Line without Earthwire/OPGW (Z’OE) 0.247 + j1.35 Ω/ (km.phase)

Double Circuit Line without Earthwire/OPGW (Z’OD) 0.397 + j2.267 Ω/ (km.phase)

Double Circuit Line with Earthwire/OPGW (Z’ODS) 0.414 + j1.649 Ω/ (km.phase)

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N11830 - Line Parameters Calculation