the 3 phase high voltage transmission line · 19-oct-11 lecture 10 power engineering - egill...
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19-Oct-11
1Lecture 10 Power Engineering - Egill Benedikt Hreinsson
The 3 Phase High Voltage Transmission Line
19-Oct-11
2Lecture 10 Power Engineering - Egill Benedikt Hreinsson
Parameters of Transmission Lines
•Series Inductance (L)•Series Resistance (R)•Shunt Capacitance (C)•Shunt Conductance (G)
A 500 kV high voltage line
19-Oct-11
3Lecture 10 Power Engineering - Egill Benedikt Hreinsson
Elementary Inductance Calculations for Parallel
Conductors
19-Oct-11
4Lecture 10 Power Engineering - Egill Benedikt HreinssonElectric and Magnetic Field Around a Single Phase Line
19-Oct-11
5Lecture 10 Power Engineering - Egill Benedikt HreinssonMagnetic Field Inside and Outside a Single Conductor
2R
H dl I⋅ =∫
Amperes law applied to a circle with, radius x,
outside of the conductor:
22
IH x I Hx
ππ
⋅ = → =
x
Total current, I, is evenly distributed over the cross sectional area ( )
2IB x
xμ
π=
Magnetic field outside:
Amperes law applied to a circle, with radius y,inside of the conductor:
2
2yH dl IR
⋅ =∫y
2
2 222
y y IH y I HR R
ππ⋅
⋅ = → =
2( )2I yB y
Rμ
π⋅
=
Magnetic field inside:
19-Oct-11
6Lecture 10 Power Engineering - Egill Benedikt HreinssonMagnetic Field Density for Conductor of a Single Phase Line
The flux density as a function of distance across a cross section of 2 conductors is plotted below
19-Oct-11
7Lecture 10 Power Engineering - Egill Benedikt Hreinsson
Inductance and Magnetic Flux Outside 2 Parallel Conductors
2R 2R
D
Current direction into the picture
Current direction out of the picture
1 ln2 2
D RD R
Ro
R
B dx I D RL dxI I I x R
μμπ π
−
−⋅ ⋅Φ −= = = ⋅ =
∫∫
ln lnoD R DL
R Rμ μπ π
−= ≅
Therefore the inductance (per unit length) from both currents created by a magnetic flux
outside the conductors is:
Use integration to find the flux in a plane between conductors:The length alongside the
conductors is = 1
B,H
19-Oct-11
8Lecture 10 Power Engineering - Egill Benedikt HreinssonEnergy in the magnetic field inside the conductors
212
W L I= ⋅
The inductance L can be defined from the following relation:
...where W is the magnetic energy of the system and I is the current. The energy W in a magnetic field is given by the volume integral:
212
W H dvμ= ∫∫∫ ...where dv is a volume element.
22
1L H dvI
μ= ∫∫∫...and substituting 22
I xHRπ⋅
= ...we get for 1 conductor:
2
2 21 2
2I xL xdx
I Rμ π
π⋅⎛ ⎞= ⋅⎜ ⎟
⎝ ⎠∫∫∫
dvFrom these 2 equations we get:
19-Oct-11
9Lecture 10 Power Engineering - Egill Benedikt Hreinsson
Inductance and Magnetic Flux Inside 2 Parallel Conductors
2R 2R
D
Current direction into the picture
Current direction out of the picture
22
2 2 40
2 24
R
iIL x x dx
I Rμ π
π= ⋅ ⋅ ⋅ ⋅∫
4iL μπ
=
..the inductance (per unit length) from currents
inside both conductors is constant:
3 32 4 4
0 0
2 24
R R
iL x dx x dxR R
μ π μπ π⋅
= ⋅ =∫ ∫“2” for 2
conductors
For 2 conductors
3 4
0
14
R
x dx R=∫Since...
19-Oct-11
10Lecture 10 Power Engineering - Egill Benedikt HreinssonTotal Inductance for 2 Parallel Conductors (per Unit Length!)
2R 2R
D
0 01 ln ln4i o
D DL L LR g
μ μπ π
⎡ ⎤= + = + =⎢ ⎥⎣ ⎦
14g R e−= ⋅
g is called “the geometric mean radius” = GMR
Current direction into the picture
Current direction out of the picture
141 ln
4e=.. because
19-Oct-11
11Lecture 10 Power Engineering - Egill Benedikt HreinssonInductance with Different Conductor Radius
0
1 2
ln DLg g
μπ
=
14
1 1g R e−=1
42 2g R e−=
2R1 2R2
D
The geometric mean radius (GMR) for
conductor #1:
The geometric mean radius (GMR) for
conductor #2:
19-Oct-11
12Lecture 10 Power Engineering - Egill Benedikt HreinssonPartitioning of Inductances for Both Conductors
0
0
1 1 1ln ln4
1 1 1 1 1 1ln ln 2ln2 4 4
LR D
LR R D
μπ
μπ
⎡ ⎤= + −⎢ ⎥⎣ ⎦⎡ ⎤⎛ ⎞ ⎛ ⎞= + + + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦
1 2 122L L L M= + −
0 01 2
1 1 1ln ln2 4 2
L LR g
μ μπ π⎡ ⎤= = + =⎢ ⎥⎣ ⎦
012 21
1ln2
M MD
μπ
= =
We can decompose each inductance term into 2 factors corresponding to each of the 2 conductors. Each of the conductors cannot exist alone without the other. The current has to “come back”. Thereforeeach logarithm factor can only exist with a corresponding opposite logarithm factor!
19-Oct-11
13Lecture 10 Power Engineering - Egill Benedikt HreinssonVoltage Drop per Unit Length of Conductor for 2 Parallel Conductors
[ ]1 2 12
1 2 2 11 12 2 12
1 2
2dI dIV L L L Mdt dt
dI dI dI dIL M L Mdt dt dt dt
V V V
Δ = = + −
⎡ ⎤ ⎡ ⎤= + − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⇒ Δ = Δ −Δ
1 21 1 12
2 12 2 12
dI dIV L Mdt dtdI dIV L Mdt dt
Δ = +
Δ = +
M12
L1
ΔV
ΔV1
I1 = I
I2 = - I
ΔV2L2
1 2I I I I= = −If ..then:
19-Oct-11
14Lecture 10 Power Engineering - Egill Benedikt HreinssonVoltage Drop per Unit Length of Conductor for 2 Parallel Conductors
1 1 12 1
2 12 2 2
V L M IdV M L Idt
Δ⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥Δ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
1 1 12 1
2 12 2 2
V L M Ij
V M L Iω
Δ⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥Δ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
For sinusoidal voltages and currents:
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15Lecture 10 Power Engineering - Egill Benedikt Hreinsson
1 12 11 1
2 21 2 2 2
1 2
n
n
n nn n n
L M MV IV M L M I
j
V IM M L
ω
Δ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥Δ ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥
Δ⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦
Voltage Drop per Unit Length of Conductor for n Parallel Conductors
By induction, we can expand the previous relation to include any number of parallel conductors, assuming the sum of the currents is zero
1 2 3 ... 0nI I I I+ + + + =
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16Lecture 10 Power Engineering - Egill Benedikt Hreinsson
References and sources
• www.landsnet.is