Lecture 3Three Phase, Power System Operation

Professor Tom OverbyeDepartment of Electrical and

Computer Engineering

ECE 476

POWER SYSTEM ANALYSIS

2

Reading and Homework

• For lecture 3 please be reading Chapters 1 and 2• For lectures 4 through 6 please be reading Chapter 4

– we will not be covering sections 4.7, 4.11, and 4.12 in detail though you should still at least skim those sections.

• HW 1 is 2.9, 22, 28, 32, 48; due Thursday 9/8• For Problem 2.32 you need to use the PowerWorld

Software. You can download the software and cases at the below link; get version 15.http://www.powerworld.com/gloversarma.asp

3

Three Phase Transmission Line

4

Per Phase Analysis

Per phase analysis allows analysis of balanced 3 systems with the same effort as for a single phase system

Balanced 3 Theorem: For a balanced 3 system with– All loads and sources Y connected– No mutual Inductance between phases

5

Per Phase Analysis, cont’d

Then– All neutrals are at the same potential– All phases are COMPLETELY decoupled– All system values are the same sequence as sources. The

sequence order we’ve been using (phase b lags phase a and phase c lags phase a) is known as “positive” sequence; later in the course we’ll discuss negative and zero sequence systems.

6

Per Phase Analysis Procedure

To do per phase analysis

1. Convert all load/sources to equivalent Y’s

2. Solve phase “a” independent of the other phases

3. Total system power S = 3 Va Ia*

4. If desired, phase “b” and “c” values can be determined by inspection (i.e., ±120° degree phase shifts)

5. If necessary, go back to original circuit to determine line-line values or internal values.

7

Per Phase Example

Assume a 3, Y-connected generator with Van = 10 volts supplies a -connected load with Z = -j through a transmission line with impedance of j0.1 per phase. The load is also connected to a -connected generator with Va”b” = 10 through a second transmission line which also has an impedance of j0.1 per phase.Find

1. The load voltage Va’b’

2. The total power supplied by each generator, SY and S

8

Per Phase Example, cont’d

First convert the delta load and source to equivalent

Y values and draw just the "a" phase circuit

9

Per Phase Example, cont’d

' ' 'a a a

To solve the circuit, write the KCL equation at a'

1(V 1 0)( 10 ) V (3 ) (V j

3j j

10

Per Phase Example, cont’d

' ' 'a a a

'a

' 'a b

' 'c ab

To solve the circuit, write the KCL equation at a'

1(V 1 0)( 10 ) V (3 ) (V j

310

(10 60 ) V (10 3 10 )3

V 0.9 volts V 0.9 volts

V 0.9 volts V 1.56

j j

j j j j

volts

11

Per Phase Example, cont’d

*'*

ygen

*" '"

S 3 3 5.1 3.5 VA0.1

3 5.1 4.7 VA0.1

a aa a a

a agen a

V VV I V j

j

V VS V j

j

12

Example 2.14

13

Example 2.21

14

Example 2.29

15

Example 2.44

16

Development of Line Models

Goals of this section are

1) develop a simple model for transmission lines

2) gain an intuitive feel for how the geometry of the transmission line affects the model parameters

17

Primary Methods for Power Transfer

The most common methods for transfer of electric power are

1) Overhead ac

2) Underground ac

3) Overhead dc

4) Underground dc

5) other

18

Magnetics Review

Ampere’s circuital law:

e

F = mmf = magnetomtive force (amp-turns)

= magnetic field intensity (amp-turns/meter)

d = Vector differential path length (meters)

= Line integral about closed path (d is tangent to path)

I =

eF d I

H l

H

l

l

Algebraic sum of current linked by

19

Line Integrals

Line integrals are a generalization of traditional integration

Integration along thex-axis

Integration along ageneral path, whichmay be closed

Ampere’s law is most useful in cases of symmetry, such as with an infinitely long line

20

Magnetic Flux Density

Magnetic fields are usually measured in terms of flux density

0-7

0

= flux density (Tesla [T] or Gauss [G])(1T = 10,000G)

For a linear a linear magnetic material

= where is the called the permeability

=

= permeability of freespace = 4 10

= relative permea

r

r

H m

B

B H

bility 1 for air

21

Magnetic Flux

Total flux passing through a surface A is

=

= vector with direction normal to the surface

If flux density B is uniform and perpendicular to an area A then

=

Ad

d

BA

B a

a

22

Magnetic Fields from Single Wire

Assume we have an infinitely long wire with current of 1000A. How much magnetic flux passes through a 1 meter square, located between 4 and 5 meters from the wire?

Direction of H is givenby the “Right-hand” Rule

Easiest way to solve the problem is to take advantage of symmetry. For an integration path we’ll choose acircle with a radius of x.

23

Single Line Example, cont’d

0

5 00 4

70

5

4

22

25 5

ln 2 10 ln2 4 4

4.46 10 Wb

2 10 2B T Gauss

x

A

IxH I H

xB H

IH dA dx

xI

I

x

For reference, the earth’smagnetic field is about0.6 Gauss (Central US)

24

Flux linkages and Faraday’s law

N

i=1

Flux linkages are defined from Faraday's law

dV = where V = voltage, = flux linkages

The flux linkages tell how much flux is linking an

N turn coil:

=

If all flux links every coil then

i

dt

N

25

Inductance

For a linear magnetic system, that is one where

B = H

we can define the inductance, L, to be

the constant relating the current and the flux

linkage

= L i

where L has units of Henrys (H)

26

Inductance Example

Calculate the inductance of an N turn coil wound tightly on a torodial 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

27

Inductance Example, cont’d

0

0

20

2 (path length is 2 R)

H2

2

H2

e

r

r

r

I d

NI H R

NIB H H

RAB N LI

NINAB NA

R

N AL

R

H l

28

Inductance of a Single Wire

To development 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 has a radius of r.

29

Flux Linkages outside of the wire

R0A r

We'll think of the wire as a single loop closed at

infinity. Therefore = since N = 1. The flux linking

the wire out to a distance of R from the wire center is

d length 2Idxx

B a

30

Flux Linkages outside, cont’d

R0A r

R 00r

d length 2

Since length = we'll deal with per unit length values,

assumed to be per meter.

ln2 2

Note, this quantity still goes to infinity as R

Idxx

I Rdx Imeter x r

B a

31

Flux linkages inside of wire

Current inside conductor tends to travel on the outside

of the conductor due to the skin effect. The pentration

of the current into the conductor is approximated using

1the skin depth = where f is

f the frequency in Hz

and is the conductivity in mhos/meter.

0.066 mFor copper skin depth 0.33 inch at 60HZ.

fFor derivation we'll assume a uniform current density.

32

Flux linkages inside, cont’d

Wire cross section

x

r

2

2

2

Current enclosed within distance

x of center I

2 2

e

ex

xI

rI Ix

Hx r

2 30

inside 2 2 40 0

Flux only links part of current

2 82

r r rIx x Ixdx dx I

r r r

33

Line Total Flux & Inductance

0 0

0

0

(per meter) ln2 8

(per meter) ln2 4

L(per meter) ln2 4

Note, this value still goes to infinity as we integrate

R out to infinity

rTotal

rTotal

r

RI I

rR

Ir

Rr

34

Inductance Simplification

0 0 4

0 4

Inductance expression can be simplified using

two exponential identities:

aln(ab)=ln a + ln b ln ln ln ln( )

b

ln ln ln ln2 4 2

ln ln2

r

r

a

r

a b a e

RL R r e

r

L R re

0

4r

ln2 '

Where r' 0.78 for 1r

Rr

r e r

35

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 R.

R

Creates counter-

clockwise field

Creates a

clockwise field

To determine the

inductance of each

conductor we integrate

as before. However

now we get some

field cancellation

36

Two Conductor Case, cont’d

R R

Direction of integration

Rp

Key Point: As we integrate for the left line, at distance 2R from the left line the net flux linked due to the Right line is zero!Use superposition to get total flux linkage.

0 0left

For distance Rp, greater than 2R, from left line

ln ln2 ' 2

Rp Rp RI I

r R

Left Current Right Current

37

Two Conductor Inductance

0left

0

0

0

0

Simplifying (with equal and opposite currents)

ln ln2 '

ln ln ' ln( ) ln2

ln ln2 '

ln as Rp2 '

ln H/m 2 'left

Rp Rp RI

r R

I Rp r Rp R R

R RpI

r Rp R

RI

r

RL

r