Download presentation

1
**EE 369 POWER SYSTEM ANALYSIS**

Lecture 5 Development of Transmission Line Models Tom Overbye and Ross Baldick

2
**Reading For lectures 5 through 7 read Chapter 4 Read Section 1.5,**

we will not be covering sections 4.7, 4.11, and 4.12 in detail Read Section 1.5, HW 4 is chapter 4 case study questions A through D, and Problems 2.31, 2.41, 2.43, 2.48, 4.1, 4.3, 4.6, due Thursday 9/26. HW 5 is Problems 4.9, 4.11, 4.13, 4.18, 4.21, 4.22, 4.24, 4.25 (assume Cardinal conductor and look up GMR in Table A.4); due Thursday 10/2.

3
Substation Bus

4
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.

5
**Inductance Example, cont’d**

6
**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.

7
**Flux Linkages outside of the wire**

8
**Flux Linkages outside, cont’d**

9
**Flux linkages inside of wire**

10
**Flux linkages inside, cont’d**

Wire cross section x r

11
**Line Total Flux & Inductance**

12
**Inductance Simplification**

13
**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. D To determine the inductance of each conductor we integrate as before. However now we get some field cancellation. Creates counter- clockwise field Creates a clockwise field

14
**Two Conductor Case, cont’d**

Direction of integration D R 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

15
**Two Conductor Inductance**

16
**Many-Conductor Case Now assume we now have n conductors,**

with the k-th conductor having current ik, and 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. For example, 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.

17
**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 d1k from conductor k. At point b the net contribution to l1 from ik , l1k, is zero.

18
**Many-Conductor Case, cont’d**

19
**Many-Conductor Case, cont’d**

20
**Symmetric Line Spacing – 69 kV**

21
**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 (Rook conductor) and a length of 5 miles.

22
**Line Inductance Example, cont’d**

23
**Line Inductance Example, cont’d**

24
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.

25
**Bundled Conductor Flux Linkages**

For the line shown on the left, define dij as 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:

26
**Bundled Conductors, cont’d**

27
**Bundled Conductors, cont’d**

28
Inductance of Bundle

29
**Inductance of Bundle, cont’d**

30
**Bundle Inductance Example**

Consider 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? 0.25 M

31
**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

32
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.

33
**Line Transposition Example**

34
**Line Transposition Example**

35
**Transposition Impact on Flux Linkages**

“a” phase in position “1” “a” phase in position “3” “a” phase in position “2”

36
**Transposition Impact, cont’d**

37
**Inductance of Transposed Line**

38
**Inductance with Bundling**

39
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.

40
Inductance Example

Similar presentations

OK

1 hi at no doifpi me be go we of at be do go hi if me no of pi we Inorder Traversal Inorder traversal. n Visit the left subtree. n Visit the node. n Visit.

1 hi at no doifpi me be go we of at be do go hi if me no of pi we Inorder Traversal Inorder traversal. n Visit the left subtree. n Visit the node. n Visit.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on introduction to object-oriented programming in c++ Ppt on world teachers day Ppt on functions of ms excel Ppt on beer lambert law spectrophotometry Ppt on use of body language in communication Ppt online examination project in java Bcd to 7 segment display ppt online Ppt on life in prehistoric times Ppt on motivation in hindi Knowledge based view ppt on mac