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:

19-Oct-11

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+ + + + =

19-Oct-11

16Lecture 10 Power Engineering - Egill Benedikt Hreinsson

References and sources

• www.landsnet.is