EEL3 EE L303 03:: Powe Powerr Eng Engin inee eeri ring ng - 1 Transmis Tran smission sion Line Line Inducta Inductance nce Calculat Calculation ion – Part 2
Course Coordinator: A. R. Abhyankar
© A. R. Abhyankar, IIT Delhi (2011)
1
Generic Formula To Calculate Inductance
© A. R. Abhyankar, IIT Delhi (2011)
2
Inductance of Bundled / Composite Conductors Single Phase Line
Each conductor in bundle ‘X’ carries i/n current Each conductor in bundle ‘Y’ carries -i/m current
© A. R. Abhyankar, IIT Delhi (2011)
3
7
Applying above equation for conductor ‘a’ of bundle X, flux linkages of conductor ‘a’:
λ a
= 2 ×10
−7
− 2 × 10 −7
λ a
i ⎛
⎞ ⎜⎜ ln ⎟⎟ + ln + ln + L + ln n ⎝ r 'a Dab Dac Dan ⎠ i ⎛ 1 1 1 1 ⎞ ⎜⎜ ln ⎟⎟ + ln + ln + L + ln m ⎝ Daa ' Dab ' Dac ' Dam ⎠ 1
−7
= 2 ×10 i ln
1
m
1
1
Daa D ' ' ab D ac ' L Dam
n
r 'a Dab Dac L Dan
© A. R. Abhyankar, IIT Delhi (2011)
4
La =
λ a
i
= 2n ×10
−7
m
ln
Daa D ab D ac ' L Dam ' '
n
n
r 'a Dab Dac L Dan
Similarly, inductance of conductor ‘b’:
Lb =
λ b
i
= 2n × 10
n
−7
m
ln
Dba D bb D bc ' L Dbm ' '
n
Dab r 'b Dbc L Dbn
Average inductance of each of the conductor of bundle X:
Lav =
La + Lb + Lc + L + Ln n
© A. R. Abhyankar, IIT Delhi (2011)
5
Bundle ‘X’ is composed of ‘n’ conductors in parallel. The inductance of bundle ‘X’ is:
L X =
L X = 2 × 10
−7
mn
ln
Lav
=
La + Lb + Lc + L + Ln
n
n
2
( Daa Dab Dac L Dam )( Dba Dbb Dbc L Dbm )L( Dna Dnb Dnc L Dnm ) n (r ' D D L D )( D r ' D L D )L ( D D D L r ' ) a ab ac an ba b bc bn na nb nc n '
'
'
'
'
'
'
'
'
2
10
L X = 2 ×10
−7
ln
GMD
(Geometric Mean Distance)
GMR
(Geometric Mean Radius)
© A. R. Abhyankar, IIT Delhi (2011)
6
GMRX
GMRY GMD
GMD = mn ( Daa D ' ' ' ' ' ' ab D ac ' L Dam )( Dba D bb D bc ' L Dbm )L ( Dna D nb D nc ' L Dnm ) GMR X = n (r 'a Dab Dac L Dan )( Dba r 'b Dbc L Dbn )L ( Dna Dnb Dnc L r 'n ) 2
© A. R. Abhyankar, IIT Delhi (2011)
7
Problem: Find GMR of three symmetrically placed conductors of ‘go’ path of a single phase line as shown in the figure. r = 2 cm
50 cm
50 cm
50 cm
r ' = 2e
−1 / 4
= 0.7788r = 1.56cm
GMR = (1.56 × 50 × 50)
1 / 3
= 15.7cm
© A. R. Abhyankar, IIT Delhi (2011)
8
Problem: One circuit of single phase transmission line is composed of three solid 0.25 cm radius wires shown as side ‘X’ in the figure. The return circuit depicted as side ‘Y’ is composed of two 0.5 cm radius wires. Find inductance of each side of line in H/m. GMD between X and Y
9m
a
d
b
e
GMD = 6 ( Dad Dae )( Dbd Dbe )( Dcd Dce )
6m
GMR for Side X
GMR X = 9 ( Daa Dab Dac )( Dba Dbb Dbc )( Dca Dcb Dcc )
6m
c X
Where, Y 2 Daa = Dbb = Dcc = r ' = 0.7788r = 0.7788 × 0.25 ×10 −
© A. R. Abhyankar, IIT Delhi (2011)
9
Problem: One circuit of single phase transmission line is composed of three solid 0.25 cm radius wires shown as side ‘X’ in the figure. The return circuit depicted as side ‘Y’ is composed of two 0.5 cm radius wires. Find inductance of each side of line in H/m. 9m
GMR for Side Y
a
d
b
e
GMRY = 4 ( Ddd Dde )( Dee Ded )
6m
Where, −2 Ddd = Dee = r ' = 0.7788r = 0.7788 × 0.5 ×10
6m
c X
Y
© A. R. Abhyankar, IIT Delhi (2011)
10
Problem: One circuit of single phase transmission line is composed of three solid 0.25 cm radius wires shown as side ‘X’ in the figure. The return circuit depicted as side ‘Y’ is composed of two 0.5 cm radius wires. Find inductance of each side of line in H/m. 9m
a
− L X = 2 ×10 7 ln
d
6m
b
−7 LY = 2 ×10 ln
e
6m
GMR X GMD GMRY
L = L X + LY
c X
GMD
Y
© A. R. Abhyankar, IIT Delhi (2011)
11
Three Phase Symmetrical Bundled Line La = 2 ×10
−7
ln
GMD GMR
GMD ≈ D GMR = (r '1 d 12 d 13 d 14 )
1 / 4
© A. R. Abhyankar, IIT Delhi (2011)
12
Inductance of Three Phase Transposed Line
© A. R. Abhyankar, IIT Delhi (2011)
13
a: Position 1, b: Position 2, c: Position 3
λ a1
= 2 ×10
−7
⎛ 1 1 1 ⎞ ⎜⎜ ia ln + ib ln ⎟⎟ + ic ln r ' D12 D31 ⎠ ⎝
a: Position 2, b: Position 3, c: Position 1
λ a 2
= 2 ×10
−7
⎛ 1 1 1 ⎞ ⎜⎜ ia ln + ib ln ⎟⎟ + ic ln r ' D23 D12 ⎠ ⎝
a: Position 3, b: Position 1, c: Position 2
λ a 3
= 2 ×10
−7
⎛ 1 1 1 ⎞ ⎜⎜ ia ln + ib ln ⎟⎟ + ic ln r ' D31 D23 ⎠ ⎝
© A. R. Abhyankar, IIT Delhi (2011)
14
Average flux linkage of ‘a’
λ a
λ a
=
2 × 10 3
−7
=
λ a1
+ λ a 2 + λ a 3 3
⎛ ⎞ 1 1 1 ⎜⎜ 3ia ln + ib ln ⎟⎟ + ic ln r ' D12 D23 D31 D12 D23 D31 ⎠ ⎝
λ a
−7
= 2 ×10 ia ln
3
D12 D23 D31 r '
© A. R. Abhyankar, IIT Delhi (2011)
15
Average inductance per phase:
La = 2 ×10
−7
ln
Deq r '
H / m
Deq = 3 D12 D23 D31
© A. R. Abhyankar, IIT Delhi (2011)
16
Problem: A single circuit three phase transposed line operated at 60 Hz has configuration as shown in the figure. Radius of each of the conductor is 0.0373 meter. Calculate per phase inductance.
20
20
38
GMR = r ' = 0.7788 × 0.0373
L = 2 ×10
−7
GMD = D ln
GMD GMR
eq
=3
20 × 20 × 38
= 24.8
H / m
© A. R. Abhyankar, IIT Delhi (2011)
17
Calculation for Unsymmetrically Bundled Transposed Three Phase Line
© A. R. Abhyankar, IIT Delhi (2011)
18
GMR R = i (r 'a Dab Dac L Dai )( Dba r 'b Dbc L Dbi )L ( Dia Dib Dic L r 'i ) 2
Product of distance between each conductor of R phase to every other conductor of phase R
GMRY = j (r 'a Dab Dac L Daj )( Dba r 'b Dbc L Dbj )L ( D ja D jb D jc L r ' j ) 2
Product of distance between each conductor of Y phase to every other conductor of phase Y
GMR B = k (r 'a Dab Dac L Dak )( Dba r 'b Dbc L Dbk )L ( Dka Dkb Dkc L r 'k ) 2
Product of distance between each conductor of B phase to every other conductor of phase B
GMR phase = 3 GMR R ⋅ GMRY ⋅ GMR B © A. R. Abhyankar, IIT Delhi (2011)
19
GMD RY = ij ( Daa D ab D ac ' L Daj )( Dba D bb D bc ' L Dbj )L ( Dna D nb D nc ' L Dij ) ' ' ' ' ' ' i: total conductors in phase ‘R’ line j: total conductors in phase ‘Y’ line Product of distance between each conductor of R phase and every conductor of Y phase
( ' nb D ) GMDYB = jk ( Daa D ' ' ' ' ' ab D ac ' L Dak )( Dba D bb D bc ' L Dbk )L Dna D nc ' L D jk j: total conductors in phase ‘Y’ line k: total conductors in phase ‘B’ line Product of distance between each conductor of Y phase and every conductor of B phase
GMD BR = ki ( Daa D ab D ac ' L Dai )( Dba D bb D bc ' L Dbi )L ( Dna D nb D nc ' L Dki ) ' ' ' ' ' ' k: total conductors in phase ‘B’ line i: total conductors in phase ‘R’ line Product of distance between each conductor of B phase and every conductor of R phase
GMD phase = 3 GMD RY ⋅ GMDYB ⋅ GMD BR © A. R. Abhyankar, IIT Delhi (2011)
20
GMR phase = 3 GMR R ⋅ GMRY ⋅ GMR B GMD phase = 3 GMD RY ⋅ GMDYB ⋅ GMD BR
L phase = 2 ×10
−7
ln
GMD phase GMR phase
© A. R. Abhyankar, IIT Delhi (2011)
H / m
21
Use of Tables To Calculate Inductive Reactance X L = 2π fL = 2π f × 2 × 10
−7
X L = 2.022 × 10 −3 f ln
ln
GMD GMR
GMD GMR
Ω / m
Ω / mi
This value for a stranded conductor is given in inches in standard table
Use GMR as given in table instead of r’ © A. R. Abhyankar, IIT Delhi (2011)
22
−3
X L = 2.022 × 10 f ln
−3
X L = 2.022 × 10 f ln
1
GMR
GMD GMR
Ω / mi
+ 2.022 ×10 −3 f ln GMDΩ / mi Xd
Xa inductive reactance at 1 ft spacing
inductive reactance spacing factor
These values are directly given in the standard data table © A. R. Abhyankar, IIT Delhi (2011)
23
Important In standard data table, GMR of stranded conductor (as shown) is given (in inches). In this case, use this GMR value in all formulae instead of r’ (or 0.7788 x r) to calculate GMR of bundle r
If conductor is a solid conductor, with radius given as r, r’ = 0.7788 x r should be used in all formulae to calculate GMR of bundle
© A. R. Abhyankar, IIT Delhi (2011)
24
