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### EE 369POWER SYSTEM ANALYSIS

Lecture 5

Development of Transmission Line Models

Tom Overbye and Ross Baldick

Reading

- For lectures 5 through 7 read Chapter 4
- we will not be covering sections 4.7, 4.11, and 4.12 in detail
- Read Section 1.5,
- HW 4 is problems 4.2, 4.7, 4.8, 4.10, 4.11, 4.14, 4.15, 4.17, 4.20, 4.23, 4.25 (assume Cardinal conductor; look up GMR in Table A.4; you might need to extrapolate other data); due Thursday 9/26.
- HW 5 is problems 4.26, 4.32, 4.33, 4.36, 4.38, 4.49, 5.1, 5.7, 5.8, 5.10, 5.16, 5.18; case study questions a, b, c, d, is due Thursday, October 3.

Inductance Example

- Calculate the inductance of an N turn coil wound tightly on a toroidal iron core that has a radius of R and a cross-sectional area of A. Assume

1) all flux is within the coil

2) all flux links each turn

3) Radius of each turn is negligible compared to R

Circular path Γ of radius R within the iron core encloses all N turns of the coil and hence links total enclosed current of Ie = NI.

Since the radius of each turn is negligible compared to R, all circular paths within the iron core have radius approximately equal to R.

Inductance of a Single Wire

- To develop models of transmission lines, we first need to determine the inductance of a single, infinitely long wire. To do this we need to determine the wire’s total flux linkage, including:

1. flux linkages outside of the wire

2. flux linkages within the wire

- We’ll assume that the current density within the wire is uniform and that the wire is solid with a radius of r.
- In fact, current density is non-uniform, and conductor is stranded, so our calculations will be approximate.

Two Conductor Line Inductance

- Key problem with the previous derivation is we assumed no return path for the current. Now consider the case of two wires, each carrying the same current I, but in opposite directions; assume the wires are separated by distance D.

To determine the

inductance of each

conductor we integrate

as before. However

now we get some

field cancellation.

Creates a

clockwise field

Creates counter-

clockwise field

Two Conductor Case, cont’d

D

D

R

Direction of integration

Key Point: Flux linkage due to currents in each conductor tend

to cancel out. Use superposition to get total flux linkage.

Left Current

Right Current

Many-Conductor Case

Now assume we now have k conductors, each with

current ik, arranged in some specified geometry.

We’d like to find flux linkages of each conductor.

Each conductor’s flux

linkage, lk, depends upon

its own current and the

current in all the other

conductors.

To derive the flux linkage for conductor 1, l1, we’ll be integrating from

conductor 1 (at origin) to the right along the x-axis.

Many-Conductor Case, cont’d

Rk is the

distance

from con-

ductor k

to point

c.

We’d like to integrate the flux crossing between b to c. But the flux crossing between a and c is easier to calculate and provides a very good approximation of l1k. Point a is at distance d1kfrom conductor k.

At point b the net

contribution to l1

from ik, l1k, is zero.

Line Inductance Example

Calculate the reactance for a balanced 3f, 60Hz

transmission line with a conductor geometry of an

equilateral triangle with D = 5m, r = 1.24cm (Rookconductor) and a length of 5 miles.

Conductor Bundling

To increase the capacity of high voltage transmission

lines it is very common to use a number of

conductors per phase. This is known as conductor

bundling. Typical values are two conductors for

345 kV lines, three for 500 kV and four for 765 kV.

Bundled Conductor Flux Linkages

- For the line shown on the left,
- define dijas the distance between
- conductors i and j.
- We can then determine lk for conductor k.
- Assuming ¼ of the phase current flows
- in each of the four conductors in
- a given phase bundle, then for conductor 1:

0.25 M

0.25 M

Bundle Inductance ExampleConsider the previous example of the three phases

symmetrically spaced 5 meters apart using wire

with a radius of r = 1.24 cm. Except now assume

each phase has 4 conductors in a square bundle,

spaced 0.25 meters apart. What is the new inductance

per meter?

Transmission Tower Configurations

- The problem with the line analysis we’ve done so far is we have assumed a symmetrical tower configuration.
- Such a tower configuration is seldom practical.

- Therefore in
- general Dab
- Dac Dbc

- Unless something
- was done this would
- result in unbalanced
- Phases.

Typical Transmission Tower

Configuration

Transposition

- To keep system balanced, over the length of a transmission line the conductors are “rotated” so each phase occupies each position on tower for an equal distance.
- This is known as transposition.

Aerial or side view of conductor positions over the length

of the transmission line.

Transposition Impact on Flux Linkages

“a” phase in

position “1”

“a” phase in

position “3”

“a” phase in

position “2”

Inductance Example

- Calculate the per phase inductance and reactance of a balanced 3, 60 Hz, line with:
- horizontal phase spacing of 10m
- using three conductor bundling with a spacing between conductors in the bundle of 0.3m.
- Assume the line is uniformly transposed and the conductors have a 1cm radius.

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