ECE 476 Power System Analysis Review 1 Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign [email protected].

ECE 476 Power System Analysis

Review 1

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

[email protected]

Announcements

• HW5 is 3.4, 3.10, 3.14, 3.19, 3.23, 3.60, 2.38, 6.9• It should be done before the first exam, but does not need

to be turned in

• First exam is Tuesday Oct 6 during class• Closed book, closed notes, but you may bring one 8.5 by

11 inch note sheet and standard calculators. • Last name starting with A to 0 in 3017; P to Z in 3013

2

Complex Power

*

cos( ) sin( )

P = Real Power (W, kW, MW)

Q = Reactive Power (var, kvar, Mvar)

S = Complex power (VA, kVA, MVA)

Power Factor (pf) = cos

If current leads voltage then pf is leading

If current

V I V I

V I

S V I j

P jQ

lags voltage then pf is lagging

(Note: S is a complex number but not a phasor)

3

Reactive Compensation

44.94 kV

16.8 MW

6.4 MVR

40.0 kV

16.0 MW

16.0 MVR

16.8 MW 16.0 MW 0.0 MVR 6.4 MVR

16.0 MVR

Key idea of reactive compensation is to supply reactive

power locally. In the previous example this can

be done by adding a 16 Mvar capacitor at the load

Compensated circuit is identical to first example with

just real power load4

Reactive Compensation, cont’d

• Reactive compensation decreased the line flow from 564 Amps to 400 Amps. This has advantages – Lines losses, which are equal to I2 R decrease– Lower current allows utility to use small wires, or

alternatively, supply more load over the same wires– Voltage drop on the line is less

• Reactive compensation is used extensively by utilities

• Capacitors can be used to “correct” a load’s power factor to an arbitrary value.

5

Power Factor Correction Example

1

1desired

new cap

cap

Assume we have 100 kVA load with pf=0.8 lagging,

and would like to correct the pf to 0.95 lagging

80 60 kVA cos 0.8 36.9

PF of 0.95 requires cos 0.95 18.2

S 80 (60 Q )

60 - Qta

80

S j

j

cap

cap

n18.2 60 Q 26.3 kvar

Q 33.7 kvar

6

Balanced 3 -- No Neutral Current

* * * *

(1 0 1 1

3

n a b c

n

an an bn bn cn cn an an

I I I I

VI

Z

S V I V I V I V I

7

Three-Phase - Wye Connection

• There are two ways to connect 3 systems– Wye (Y)– Delta ()

an

bn

cn

Wye Connection Voltages

V

V

V

V

V

V

8

Delta-Wye Transformation

Y

phase

To simplify analysis of balanced 3 systems:

1) Δ-connected loads can be replaced by 1

Y-connected loads with Z3

2) Δ-connected sources can be replaced by

Y-connected sources with V3 30Line

Z

V

9

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.

10

Line Resistance

-8

-8

Line resistance per unit length is given by

R = where is the resistivityA

Resistivity of Copper = 1.68 10 Ω-m

Resistivity of Aluminum = 2.65 10 Ω-m

Example: What is the resistance in Ω / mile of a

-8

2

1" diameter solid aluminum wire (at dc)?

2.65 10 Ω-m1609 0.084

0.0127m

mR

mile mile

11

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

12

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

13

Line Inductance Example

Calculate the reactance for a balanced 3, 60Hz

transmission line with a conductor geometry of an

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

0 1 1 1ln( ) ln( ) ln( )

2 'a a b ci i ir D D

a

Since system is assumed

balanced

i b ci i

14

Line Inductance Example, cont’d

a

0a

0

70

3

6

Substituting

i

Hence

1 1ln ln

2 '

ln2 '

4 10 5ln ln

2 ' 2 9.67 10

1.25 10 H/m

b c

a a

a

a

i i

i ir D

Di

r

DL

r

15

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.

16

Bundled Conductors, cont’d

14

12 13 14

1

12 1

1

14

15 16 17 18 2 3 4

1 19 1

geometric mean radius (GMR) of bundle

( ' ) for our example

( ' ) in general

geometric mean distance (GMD) of

conductor 1 to phase b.

( )

(

b

bb

b

b b b ab

c

R

r d d d

r d d

D

d d d d D D D D

D d d

14

,10 1,11 1,12 2 3 4) c c c acd d D D D D

17

Inductance of Bundle, cont’d

0a 1

But remember each bundle has b conductors

in parallel (4 in this example). So

L / ln2 b

DL b

R

18

Bundle Inductance Example

0.25 M0.25 M

0.25 M

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?

2 3

13 4

b

70a

1.24 10 m ' 9.67 10 m

R 9.67 10 0.25 0.25 2 0.25

0.12 m (ten times bigger!)

5L ln 7.46 10 H/m

2 0.12

r r

19

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.

20

Inductance of Transposed Line

13

m 12 13 23

0 0a

70

Define the geometric mean distance (GMD)

D

Then for a balanced 3 system ( - - )

1 1ln ln ln

2 ' 2 '

Hence

ln 2 10 ln H/m2 ' '

a b c

ma a a

m

m ma

d d d

I I I

DI I I

r D r

D DL

r r

21

Inductance with Bundling

b

0a

70

If the line is bundled with a geometric mean

radius, R , then

ln2

ln 2 10 ln H/m2

ma

b

m ma

b b

DI

R

D DL

R R

Line Conductors, cont’d

• Total conductor area is given in circular mils. One circular mil is the area of a circle with a diameter of 0.001 = 0.00052 square inches

• Example: what is the area of a solid, 1” diameter circular wire? Answer: 1000 kcmil (kilo circular mils)

• Because conductors are stranded, the equivalent radius must be provided by the manufacturer. In tables this value is known as the GMR and is usually expressed in feet.

23

Review of Electric Fields

eA

2

To develop a model for line capacitance we

first need to review some electric field concepts.

Gauss's law:

d = q (integrate over closed surface)

where

= electric flux density, coulombs/m

d = differential

D a

D

a

2

e

area da, with normal to surface

A = total closed surface area, m

q = total charge in coulombs enclosed

24

Gauss’s Law Example

•Similar to Ampere’s Circuital law, Gauss’s Law is most useful for cases with symmetry.•Example: Calculate D about an infinitely long wire that has a charge density of q coulombs/meter.

Since D comes

radially out inte-

grate over the

cylinder bounding

the wireeA

d 2 q

where radially directed unit vector2

D Rh qh

qR

r r

D a

D a a

25

Line Capacitance, cont’d

aa a

To eliminate mutual capacitance we'll again

assume we have a uniformly transposed line.

For the previous three conductor example:

q 2ince q = C

ln

a

a

V V

S V CDVr

26

Bundled Conductor Capacitance

1

1cb 12

Similar to what we did for determining line

inductance when there are n bundled conductors,

we use the original capacitance equation just

substituting an equivalent r

Note fo

adius

r t

( )

he

Rn

nrd d

b

capacitance equation we use r rather

than r' which was used for R in the inductance

equation

27

Line Capacitance, cont’d

1

m

13

m

1cb 12

-12o

For the case of uniformly transposed lines we

use the same GMR, D , as before.

2

ln

where

D

R ( ) (note r NOT r')

ε in air 8.854 10 F/m

n

mcb

ab ac bc

n

CD

R

d d d

rd d

28

Line Capacitance Example

•Calculate the per phase capacitance and susceptance of a balanced 3, 60 Hz, transmission 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.

29

Line Capacitance Example, cont’d

13

13

m

1211

c 11

8

(0.01 0.3 0.3) 0.0963 m

D (10 10 20) 12.6 m

2 8.854 101.141 10 F/m

12.6ln

0.09631 1

X2 60 1.141 10 F/m

2.33 10 -m (not / m)

cbR

C

C

30

Transmission Line Equivalent Circuit

•Our current model of a transmission line is shown below

For operation at frequency , let z = r + j L

and y = g +j C (with g usually equal 0)

Units on

z and y are

per unit

length!

31

Derivation of V, I Relationships

We can then derive the following relationships:

( )

( ) ( )

dV I z dx

dI V dV y dx V y dx

dV x dI xz I yV

dx dx

32

Setting up a Second Order Equation

2

2

2

2

( ) ( )

We can rewrite these two, first order differential

equations as a single second order equation

( ) ( )

( )0

dV x dI xz I yV

dx dx

d V x dI xz zyVdxdx

d V xzyV

dx

33

V, I Relationships, cont’d

2 2

Define the propagation constant as

where

the attenuation constant

the phase constant

Use the Laplace Transform to solve. System

has a characteristic equation

( ) ( )( ) 0

yz j

s s s

34

Determining Line Voltage, cont’d

1 2

1 2

1

1 2

2

c

( ) cosh( ) sinh( )

(0) cosh(0) sinh(0)

Since cosh(0) 1 & sinh(0) 0

( )sinh( ) cosh( )

( ) cosh( ) sinh( )

where Z characteristic

R

R

R RR

R R c

V x K x K x

V V K K

K V

dV xzI K x K x

dx

zI I z zK I

yyz

V x V x I Z x

zy

impedance

35

Lossless Transmission Lines

c

c

c

For a lossless line the characteristic impedance, Z ,

is known as the surge impedance.

Z (a real value)

If a lossless line is terminated in impedance

Z

Then so we get...

R

R

R c R

jwl ljwc c

VI

I Z V

36

Lossless Transmission Lines

2

( ) cosh sinh

( ) cosh sinh

( )( )

V(x)Define as the surge impedance load (SIL).

Since the line is lossless this implies

( )

( )

R R

R R

c

c

R

R

V x V x V x

I x I x I x

V xZ

I x

Z

V x V

I x I

If P > SIL then line consumes

vars; otherwise line generates vars.

37

Transmission Matrix Model, cont’d

S

S

VWith

I

Use voltage/current relationships to solve for A,B,C,D

cosh sinh

cosh sinh

cosh sinh

1sinh cosh

R

R

S R c R

RS R

c

c

c

VA B

IC D

V V l Z I l

VI I l l

Z

l Z lA B

l lC DZ

T

38

Equivalent Circuit Model

The common representation is the equivalent circuit

Next we’ll use the T matrix values to derive the

parameters Z' and Y'.

39

Equivalent Circuit Parameters

'' 2

' '1 '

2

' '2 2

' ' ' '' 1 1

4 2

' '1 '

2' ' ' '

' 1 14 2

S RR R

S R R

S S R R

S R R

S R

S R

V V YV I

ZZ Y

V V Z I

Y YI V V I

Z Y Z YI Y V I

Z YZ

V V

Z Y Z YI IY

40

Three Line Models

(longer than 200 miles)

tanhsinh ' 2use ' ,2 2

2 (between 50 and 200 miles)

use and 2

(less than 50 miles)

use (i.e., assume Y is zero)

ll Y Y

Z Zll

YZ

Z

Long Line Model

Medium Line Model

Short Line Model

41

Power Transfer in Lossless Lines

21 1 2

12 12 12

12 12

1 212 12

If we assume a line is lossless with impedance jX and

are just interested in real power transfer then:

90 90

Since - cos(90 ) sin , we get

sin

Hence the maximu

V V VP jQ

Z Z

V VP

X

1 212

m power transfer is

Max V VP

X

42

Ideal Transformer

• First we review the voltage/current relationships for an ideal transformer– no real power losses– magnetic core has infinite permeability– no leakage flux

• We’ll define the “primary” side of the transformer as the side that usually takes power, and the secondary as the side that usually delivers power.– primary is usually the side with the higher voltage, but

may be the low voltage side on a generator step-up transformer.

43

Ideal Transformer Relationships

1 1 2 2

1 21 1 2 2

1 2 1 1

1 2 2 2

Assume we have flux in magnetic material. Then

= turns ratio

m

m m

m m

m

N N

d d d dv N v N

dt dt dt dtd v v v N

adt N N v N

44

Current/Voltage Relationships

'1 1 2 2

1 2 1 2'

1 2 12

1 2

1 2

If is infinite then 0 . Hence

1or

Then

0

10

N i N i

i N i NN i N ai

av v

i ia

45

Transformer Equivalent Circuit

Using the previous relationships, we can derive an

equivalent circuit model for the real transformer

' 2 '2 2 1 2

' 2 '2 2 1 2

This model is further simplified by referring all

impedances to the primary side

r e

e

a r r r r

x a x x x x

46

Simplified Equivalent Circuit

47

Calculation of Model Parameters

• The parameters of the model are determined based upon – nameplate data: gives the rated voltages and power– open circuit test: rated voltage is applied to primary with

secondary open; measure the primary current and losses (the test may also be done applying the voltage to the secondary, calculating the values, then referring the values back to the primary side).

– short circuit test: with secondary shorted, apply voltage to primary to get rated current to flow; measure voltage and losses.

48

Per Unit Calculations

• A key problem in analyzing power systems is the large number of transformers. – It would be very difficult to continually have to refer

impedances to the different sides of the transformers

• This problem is avoided by a normalization of all variables.

• This normalization is known as per unit analysis.

actual quantityquantity in per unit

base value of quantity

49

Per Unit Conversion Procedure, 1f

1. Pick a 1 VA base for the entire system, SB

2. Pick a voltage base for each different voltage level, VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral.

3. Calculate the impedance base, ZB= (VB)2/SB

4. Calculate the current base, IB = VB/ZB

5. Convert actual values to per unitNote, per unit conversion on affects magnitudes, not the angles. Also, per unit quantities no longer have units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)

50

Per Unit Example

Solve for the current, load voltage and load power

in the circuit shown below using per unit analysis

with an SB of 100 MVA, and voltage bases of

8 kV, 80 kV and 16 kV.

Original Circuit51

Three Phase Per Unit

1. Pick a 3 VA base for the entire system,

2. Pick a voltage base for each different voltage level, VB. Voltages are line to line.

3. Calculate the impedance base

Procedure is very similar to 1 except we use a 3 VA base, and use line to line voltage bases

3BS

2 2 2, , ,3 1 1

( 3 )

3B LL B LN B LN

BB B B

V V VZ

S S S

Exactly the same impedance bases as with single phase!

52

Three Phase Per Unit, cont'd

4. Calculate the current base, IB

5. Convert actual values to per unit

3 1 13 1B B

, , ,

3I I

3 3 3B B B

B LL B LN B LN

S S S

V V V

Exactly the same current bases as with single phase!

53

Three Phase Transformers

• There are 4 different ways to connect 3 transformers

Y-Y D-D

Usually 3 transformers are constructed so all windings

share a common core54

3f Transformer Interconnections

D-Y Y-D

55

Load Models

• Ultimate goal is to supply loads with electricity at constant frequency and voltage

• Electrical characteristics of individual loads matter, but usually they can only be estimated– actual loads are constantly changing, consisting of a large

number of individual devices– only limited network observability of load characteristics

• Aggregate models are typically used for analysis• Two common models

– constant power: Si = Pi + jQi

– constant impedance: Si = |V|2 / Zi

56

Bus Admittance Matrix or Ybus

• First step in solving the power flow is to create what is known as the bus admittance matrix, often call the Ybus.

• The Ybus gives the relationships between all the bus current injections, I, and all the bus voltages, V,

I = Ybus V

• The Ybus is developed by applying KCL at each bus in the system to relate the bus current injections, the bus voltages, and the branch impedances and admittances

57

Ybus Example

Determine the bus admittance matrix for the network

shown below, assuming the current injection at each

bus i is Ii = IGi - IDi where IGi is the current injection into the

bus from the generator and IDi is the current flowing into the load

58

Ybus Example, cont’d

1 1

2 2

3 3

4 4

We can get similar relationships for buses 3 and 4

The results can then be expressed in matrix form

0

0

0 0

bus

A B A B

A A C D C D

B C B C

D D

I Y Y Y Y V

I Y Y Y Y Y Y V

I Y Y Y Y V

I Y Y V

I Y V

For a system with n buses, Ybus is an n by n

symmetric matrix (i.e., one where Aij = Aji)

59

Ybus General Form

• The diagonal terms, Yii, are the self admittance terms, equal to the sum of the admittances of all devices incident to bus i.

• The off-diagonal terms, Yij, are equal to the negative of the sum of the admittances joining the two buses.

• With large systems Ybus is a sparse matrix (that is, most entries are zero)

• Shunt terms, such as with the line model, only affect the diagonal terms.

60

Modeling Shunts in the Ybus

from other lines

2 2

Since ( )2

21 1

Note

kcij i j k i

kcii ii k

k k k kk

k k k k k k k

YI V V Y V

YY Y Y

R jX R jXY

Z R jX R jX R X

61

Two Bus System Example

1 21 1

1 1

2 2

( ) 112 16

2 0.03 0.04

12 15.9 12 16

12 16 12 15.9

cYV VI V j

Z j

I Vj j

I Vj j

62

Power Flow Analysis

• When analyzing power systems we know neither the complex bus voltages nor the complex current injections

• Rather, we know the complex power being consumed by the load, and the power being injected by the generators plus their voltage magnitudes

• Therefore we can not directly use the Ybus

equations, but rather must use the power balance equations

63

Power Balance Equations, cont’d

** * *

i1 1

S

This is an equation with complex numbers.

Sometimes we would like an equivalent set of real

power equations. These can be derived by defining

n n

i i i ik k i ik kk k

ik ik ik

i

V I V Y V V Y V

Y G jB

V

jRecall e cos sin

iji i i

ik i k

V e V

j

64

Real Power Balance Equations

* *i

1 1

1

i1

i1

S ( )

(cos sin )( )

Resolving into the real and imaginary parts

P ( cos sin )

Q ( sin cos

ikn n

ji i i ik k i k ik ik

k k

n

i k ik ik ik ikk

n

i k ik ik ik ik Gi Dik

n

i k ik ik ik ik

P jQ V Y V V V e G jB

V V j G jB

V V G B P P

V V G B

)k Gi DiQ Q

65

Gauss Iteration

There are a number of different iterative methods

we can use. We'll consider two: Gauss and Newton.

With the Gauss method we need to rewrite our

equation in an implicit form: x = h(x)

To iterate we fir (0)

( +1) ( )

st make an initial guess of x, x ,

and then iteratively solve x ( ) until we

find a "fixed point", x, such that x (x).ˆ ˆ ˆ

v vh x

h

66

Gauss Iteration Example

( 1) ( )

(0)

( ) ( )

Example: Solve - 1 0

1

Let = 0 and arbitrarily guess x 1 and solve

0 1 5 2.61185

1 2 6 2.61612

2 2.41421 7 2.61744

3 2.55538 8 2.61785

4 2.59805 9 2.61798

v v

v v

x x

x x

v

v x v x

67

Gauss Power Flow

** * *

i1 1

* * * *

1 1

*

*1 1,

*

*1,

We first need to put the equation in the correct form

S

S

S

S1

i i

i

i

n n

i i i ik k i ik kk k

n n

i i i ik k ik kk k

n ni

ik k ii i ik kk k k i

ni

i ik kii k k i

V I V Y V V Y V

V I V Y V V Y V

Y V Y V Y VV

V Y VY V

68

Gauss Two Bus Power Flow Example

•A 100 MW, 50 Mvar load is connected to a generator •through a line with z = 0.02 + j0.06 p.u. and line charging of 5 Mvar on each end (100 MVA base). Also, there is a 25 Mvar capacitor at bus 2. If the generator voltage is 1.0 p.u., what is V2?

SLoad = 1.0 + j0.5 p.u.

69

Gauss Two Bus Example, cont’d

2

2 bus

bus

22

The unknown is the complex load voltage, V .

To determine V we need to know the .

15 15

0.02 0.06

5 14.95 5 15Hence

5 15 5 14.70

( Note - 15 0.05 0.25)

jj

j j

j j

B j j j

Y

Y

70

Gauss Two Bus Example, cont’d

*2

2 *22 1,2

2 *2

(0)2

( ) ( )2 2

1 S

1 -1 0.5( 5 15)(1.0 0)

5 14.70

Guess 1.0 0 (this is known as a flat start)

0 1.000 0.000 3 0.9622 0.0556

1 0.9671 0.0568 4 0.9622 0.0556

2 0

n

ik kk k i

v v

V Y VY V

jV j

j V

V

v V v V

j j

j j

.9624 0.0553j71

Gauss Two Bus Example, cont’d

2

* *1 1 11 1 12 2

1

0.9622 0.0556 0.9638 3.3

Once the voltages are known all other values can

be determined, such as the generator powers and the

line flows

S ( ) 1.023 0.239

In actual units P 102.3 MW

V j

V Y V Y V j

1

22

, Q 23.9 Mvar

The capacitor is supplying V 25 23.2 Mvar

72

Slack Bus

• In previous example we specified S2 and V1 and then solved for S1 and V2.

• We can not arbitrarily specify S at all buses because total generation must equal total load + total losses

• We also need an angle reference bus.• To solve these problems we define one bus as the

"slack" bus. This bus has a fixed voltage magnitude and angle, and a varying real/reactive power injection.

73