EE 369POWER SYSTEM ANALYSIS

Lecture 10Transformers, Load & Generator Models, YBus

Tom Overbye and Ross Baldick

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Announcements

• Homework 7 is 5.8, 5.15, 5.17, 5.24, 5.27, 5.28, 5.29, 5.34, 5.37, 5.38, 5.43, 5.45; due 10/22.

• Homework 8 is 3.1, 3.3, 3.4, 3.7, 3.8, 3.9, 3.10, 3.12, 3.13, 3.14, 3.16, 3.18; due 10/29.

• Homework 9 is 3.20, 3.23, 3.25, 3.27, 3.28, 3.29, 3.35, 3.38, 3.39, 3.41, 3.44, 3.47; due 11/5.

• Start reading Chapter 6 for lectures 11 and 12.

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Load Tap Changing TransformersLTC transformers have tap ratios that can be varied

to regulate bus voltages.The typical range of variation is 10% from the

nominal values, usually in 33 discrete steps (0.0625% per step).

Because tap changing is a mechanical process, LTC transformers usually have a 30 second deadband to avoid repeated changes to minimize wear and tear.

Unbalanced tap positions can cause “circulating VArs;” that is, reactive power flowing from one winding to the next in a three phase transformer.

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Phase Shifting Transformers

Phase shifting transformers are used to control the phase angle across the transformer.

Since power flow through the transformer depends upon phase angle, this allows the transformer to regulate the power flow through the transformer.

Phase shifters can be used to prevent inadvertent "loop flow" and to prevent line overloads by controlling power flow on lines.

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Phase Shifting Transformer Picture

230 kV 800 MVA Phase Shifting Transformer During factory

testingSource: Tom Ernst, Minnesota Power

Costs about $7 million,weighs about 1.2million pounds

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AutotransformersAutotransformers are transformers in which the

primary and secondary windings are coupled magnetically and electrically.

This results in lower cost, and smaller size and weight.

The key disadvantage is loss of electrical isolation between the voltage levels. This can be an important safety consideration when a is large. For example in stepping down 7160/240 V we do not ever want 7160 on the low side!

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Load ModelsUltimate 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 analysisTwo common models

– constant power: Si = Pi + jQi

– constant impedance: Si = |V|2 / Zi

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Generator ModelsEngineering models depend on the application.Generators are usually synchronous machines:

– important exception is case of wind generators,For generators we will use two different models:

– (in 369) a steady-state model, treating the generator as a constant power source operating at a fixed voltage; this model will be used for power flow and economic analysis.

– (in 368L) a short term model treating the generator as a constant voltage source behind a possibly time-varying reactance.

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Power Flow AnalysisWe now have the necessary models to start to

develop the power system analysis tools.The most common power system analysis tool is the

power flow (also known sometimes as the load flow):– power flow determines how the power flows in a network– also used to determine all bus voltages and all currents,– because of constant power models, power flow is a

nonlinear analysis technique,– power flow is a steady-state analysis tool.

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Linear versus Nonlinear Systems• A function H is linear if H(11 + 22) = 1H(1) + 2H(2)• That is:

1) the output is proportional to the input2) the principle of superposition holds

• Linear Example: y = H(x) = c x y = c(x1+x2) = cx1 + c x2

• Nonlinear Example: y = H(x) = c x2 y = c(x1+x2)2 ≠ c(x1)2 + c(x2)2

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Linear Power System ElementsResistors, inductors, capacitors, independentvoltage sources, and current sources are linear circuit elements:

1

Such systems may be analyzed by superposition.

V R I V j L I V Ij C

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Nonlinear Power System Elements•Constant power loads and generator injections are nonlinear and hence systems with these elements cannot be analyzed (exactly) by superposition.

Nonlinear problems can be very difficult to solve,and usually require an iterative approach. 12

Nonlinear Systems May Have Multiple Solutions or No Solution

•Example 1: x2 - 2 = 0 has solutions x = 1.414…•Example 2: x2 + 2 = 0 has no real solution

f(x) = x2 - 2 f(x) = x2 + 2

two solutions where f(x) = 0 no solution to f(x) = 013

Multiple Solution Example 3• The dc system shown below has two solutions for a value of

load resistance that results in 18 W dissipation in the load:

That is, the 18 wattload is an unknown resistive load RLoad

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Load LoadLoad

Load

Load

The equation we're solving is:

9 volts 18 watts1 +

One solution is 2Other solution is 0.5

I R RRRR

A different problem:What is theresistance to achieve maximumPLoad? 14

Bus Admittance Matrix or Ybus

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

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Ybus ExampleDetermine the bus admittance matrix for the networkshown below, assuming the current injection at eachbus 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.

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Ybus Example, cont’d

1 1 1

1 31 21 12 13

1 1 2 1 3

1 2 3

2 21 23 24

1 2 3 4

By KCL at bus 1 we have

1( ) ( ) (with )

( )Similarly

( )

G D

A B

A B jj

A B A B

A A C D C D

I I IV VV VI I I

Z Z

I V V Y V V Y YZ

Y Y V Y V Y V

I I I IY V Y Y Y V Y V Y V

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Ybus Example, cont’d

bus

1 1

2 2

3 3

4 4

We can get similar relationships for buses 3 and 4The results can then be expressed in matrix form

0

00 0

A B A B

A A C D C D

B C B C

D D

I Y Y Y Y VI Y Y Y Y Y Y VI Y Y Y Y VI Y Y V

I Y V

For a system with n buses, Ybus is an n by n symmetric matrix (i.e., one where Ybuskl = Ybuslk).

From now on, we will mostly write Y for Ybus, but be careful to distinguish Ykl from line admittances.18

Ybus General Form •The diagonal terms, Ykk, are the “self admittance” terms, equal to the sum of the admittances of all devices incident to bus k. •The off-diagonal terms, Ykl, are equal to the negative of the admittance joining the two buses.•With large systems Ybus is a sparse matrix (that is, most entries are zero):–sparsity is key to efficient numerical calculation.

•Shunt terms, such as in the equivalent line model, only affect the diagonal terms. 19

Modeling Shunts in the Ybus

from other lines

2 2

Since ( )2

21 1Note

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 jXYZ R jX R jX R X

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Two Bus System Example

1 21 1

1 1

2 2

( ) 1 1, where 12 16.2 0.03 0.04

12 15.9 12 1612 16 12 15.9

cYV VI V jZ Z j

I Vj jI Vj j

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Using the Ybus

bus

1bus bus

1bus bus

If the voltages are known then we can solve for the current injections:

If the current injections are known then we can solve for the voltages:

where = is the bus impedan

Y V I

Y I V Z I

Z Y ce matrix.

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Solving for Bus Currents

*1 1 1

For example, in previous case assume: 1.0

.0.8 0.2

Then12 15.9 12 16 1.0 5.60 0.70

12 16 12 15.9 0.8 0.2 5.58 0.88Therefore the power injected at bus 1 is:

1.0 (5

j

j j jj j j j

S V I

V

*2 2 2

.60 0.70) 5.60 0.70

(0.8 0.2) ( 5.58 0.88) 4.64 0.41

j j

S V I j j j

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Solving for Bus Voltages

1

*1 1 1

As another example, in previous case assume 5.0

.4.8

Then

12 15.9 12 16 5.0 0.0738 0.90212 16 12 15.9 4.8 0.0738 1.098

Therefore the power injected is

(0.0738 0.

j j jj j j

S V I j

I

*2 2 2

902) 5 0.37 4.51

( 0.0738 1.098) ( 4.8) 0.35 5.27

j

S V I j j

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Power Flow AnalysisWhen 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 and their voltage magnitudes.

Therefore we can not directly use the Ybus equations, but rather must use the power balance equations.

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Power Balance Equations

1

bus

1

From KCL we know at each bus in an bus systemthe current injection, , must be equal to the currentthat flows into the network

Since = we also know

i

n

i Gi Di ikk

n

i Gi Di ik kk

i nI

I I I I

I I I Y V

I Y V

*The network power injection is then i i iS V I26

Power Balance Equations, cont’d*

* * *

1 1

This is an equation with complex numbers. Sometimes we would like an equivalent set of realpower equations. These can be derived by defining

n n

i i i i ik k i ik kk k

ik ik ik

i

S 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

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Real Power Balance Equations* *

1 1

1

1

1

( )

(cos sin )( )

Resolving into the real and imaginary parts

( cos sin )

( sin cos

ikn n

ji i i i ik k i k ik ik

k kn

i k ik ik ik ikk

n

i i k ik ik ik ik Gi Dikn

i i k ik ik ik ik

S P jQ V Y V V V e G jB

V V j G jB

P V V G B P P

Q V V G B

)k Gi DiQ Q

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Power Flow Requires Iterative Solution

bus

In the power flow we assume we know and the. We would like to solve for the values .

The difficulty is that the following nonlinear equation (solve for the values given ) has no closed

i

i

i i

SV

V S

Y

** * *

1 1

form solution:

Rather, we must pursue an iterative approach.

n n

i i i i ik k i ik kk k

S V I V Y V V Y V

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Gauss (or Jacobi) IterationThere are a number of different iterative methodswe can use. We'll consider two: Gauss and Newton.

With the Gauss method we need to rewrite our equation in an implicit form: ( ).Our goal is to f

x h x

(0)

( 1)

ind that satisfies this equation.To seek a solution we first make an initial guess of ,

which we call ,and then iteratively plug into the right-

hand side to evaluate an updated guess (v

xx

x

x h x ( ) ), until we are close to a "fixed point," , such that ( ).ˆ ˆ ˆ

v

x x h x30

( 1) ( )

( ) ( )

( 1)

Gauss Example: To solve 1 0, rearrange in the form

( ), where ( ) 1 . Iteration is: 1 .

That is, plug current iterate into: 1 ; the answer is

the next iterate ; repea

v v

v v

v

x x

x h x h x x x x

x x

x

(0)

( ) ( )

t. Matlab code: x=x0; x=1+sqrt(x).

Start at = 0, arbitrarily guess 1 and iterate:

0 1 5 2.611851 2 6 2.616122 2.41421 7 2.617443 2.55538 8 2.617854 2.59805 9 2.61798

v v

x

x x

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Stopping Criteria

( ) ( ) ( 1) ( )

A key problem to address is when to stop the iteration. With the Gauss iteration we stop when

with

If is a scalar this is clear, but if is a vector weneed to generalize t

v v v vx x x x

x x

( )

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1

he absolute value by using a norm

Two common norms are the Euclidean & infinity

( ) max

v

n

i i ii

x

x x

x x32

Gauss Power Flow

** * *

1 1

* * *

1*

*1 1,

We first need to put the equation in the appropriate form,with power flow expressed in the form ( ) :

i

i

n n

i i i i ik k i ik kk kn

i i i ik kk

ni

ik k ii i ikk k

V h V

S V I V Y V V Y V

S V I V Y V

S Y V Y V YV

*

*1,

S1 .i

n

kk i

ni

i ik kii k k i

V

V Y VY V

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Gauss Power Flow*

*1,

S1We define ( ) by: ( ) ,

Collect the entries ( ) together to form the vector ( ).Then we have expressed the power flow equations inthe form: ( ). (There are o

i

ni

i i ik kii k k i

i

h V h V Y VY V

h V h V

V h V

(0)

(1) (0)

(2) (1)

(

ther ways we can expressthe power flow equations in this form.)

Start with an initial guess and then update according to:

( ),

( ),...

Continue until stopping criterion satisfied:

V

V h V

V h V

V

1) ( ) .V

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