EE 369 POWER SYSTEM ANALYSIS Lecture 12 Power Flow Tom Overbye and Ross Baldick 1.

EE 369POWER SYSTEM ANALYSIS

Lecture 12Power Flow

Tom Overbye and Ross Baldick

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Announcements• Homework 9 is: 3.47, 3.49, 3.53, 3.57, 3.61, 6.2, 6.9, 6.13,

6.14, 6.18, 6.19, 6.20; due November 7. (Use infinity norm and epsilon = 0.01 for any problems where norm or stopping criterion not specified.)

• Read Chapter 12, concentrating on sections 12.4 and 12.5.• Homework 10 is 6.23, 6,25, 6.26, 6.28, 6.29, 6.30 (see

figure 6.18 and table 6.9 for system), 6.31, 6.38, 6.42, 6.46, 6.52, 6.54; due November 14.

• Homework 11 is 6.43, 6.48, 6.59, 6.61, 12.19, 12.22, 12.20, 12.24, 12.26, 12.28, 12.29; due Nov. 21.

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Power System Planning

Source: Midwest ISO MTEP08 Report 3

MISO Generation Queue

Source: Midwest ISO MTEP08 Report 4

MISO Conceptual EHV Overlay

Black lines are DC, blue lines are 765kV, red are 500 kV

Source: Midwest ISO MTEP08 Report 5

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considerable more!“Competitive Renewable Energy Zones”

study identified most promising wind sites,Building around $5 billion (original

estimate, now closer to $7 billion) of transmission to support an additional 11 GW of wind.

Will be completed in 2014.

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CREZ Transmission Lines

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NR Application to Power Flow

** * *

1 1

We first need to rewrite complex power equations

as equations with real coefficients (we've seen this earlier):

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

Recall e cos sin

iji i i

ik i k

j

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 k

n

i k ik ik ik ikk

n

i i k ik ik ik ik Gi Dik

n

i i k ik ik ikk

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

)ik Gi DiQ Q

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Newton-Raphson Power FlowIn the Newton-Raphson power flow we use Newton's

method to determine the voltage magnitude and angle at

each bus in the power system that satisfies power balance.

We need to solve the power balance equ

1

1

ations:

( cos sin ) 0

( sin cos ) 0

n

i k ik ik ik ik Gi Dik

n

i k ik ik ik ik Gi Dik

V V G B P P

V V G B Q Q

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Power Flow Variables

1

1

For convenience, write:

( ) ( cos sin )

( ) ( sin cos )

The power balance equations are then:

( ) 0

( ) 0

n

i i k ik ik ik ikk

n

i i k ik ik ik ikk

i Gi Di

i Gi Di

P V V G B

Q V V G B

P P P

Q Q Q

x

x

x

x

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Power Flow Variables

2

n

2

Assume the slack bus is the first bus (with a fixed

voltage angle/magnitude). We then need to determine

the voltage angle/magnitude at the other buses.

We must solve ( ) , where:

n

V

V

f x 0

x

2 2 2

2 2 2

( )

( )( )

( )

( )

G D

n Gn Dn

G D

n Gn Dn

P P P

P P P

Q Q Q

Q Q Q

x

xf x

x

x

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N-R Power Flow Solution

(0)

( )

( 1) ( ) ( ) 1 ( )

The power flow is solved using the same procedure

discussed previously for general equations:

For 0; make an initial guess of ,

While ( ) Do

[ ( )] ( )

1

End

v

v v v v

v

v v

x x

f x

x x J x f x

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Power Flow Jacobian Matrix

1 1 1

1 2 2 2

2 2 2

1 2 2 2

2 2 2 2 2 2

1 2

The most difficult part of the algorithm is determining

and factorizing the Jacobian matrix, ( )

( ) ( ) ( )

( ) ( ) ( )( )

( ) ( )

n

n

n n n

f f fx x x

f f fx x x

f f fx x x

J x

x x x

x x xJ x

x x

2 2

( )n

x

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Power Flow Jacobian Matrix, cont’d

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Jacobian elements are calculated by differentiating

each function, ( ), with respect to each variable.

For example, if ( ) is the bus real power equation

( ) ( cos sin )

i

i

n

i i k ik ik ik ik Gik

f x

f x i

f x V V G B P P

1

( ) ( sin cos )

( ) ( sin cos ) ( )

Di

ni

i k ik ik ik iki k

k i

ii j ij ij ij ij

j

fx V V G B

fx V V G B j i

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Two Bus Newton-Raphson Example

Line Z = 0.1j

One Two 1.000 pu 1.000 pu

200 MW 100 MVR

0 MW 0 MVR

For the two bus power system shown below, use the Newton-Raphson power flow to determine the voltage magnitude and angle at bus two. Assumethat bus one is the slack and SBase = 100 MVA.

2

2

10 10Unkown: , Also,

10 10bus

j j

V j j

x Y

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Two Bus Example, cont’d

1

1

General power balance equations:

( cos sin ) 0

( sin cos ) 0

For bus two, the power balance equations are

(load real power is 2.0 per unit,

while react

n

i k ik ik ik ik Gi Dik

n

i k ik ik ik ik Gi Dik

V V G B P P

V V G B Q Q

2 1 2

22 1 2 2

ive power is 1.0 per unit):

(10sin ) 2.0 0

( 10cos ) (10) 1.0 0

V V

V V V

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Two Bus Example, cont’d2 2 2

22 2 2 2

2 2

2 2

2 2

2 2

2 2 2

2 2 2 2

( ) 2.0 (10sin ) 2.0

( ) 1.0 ( 10cos ) (10) 1.0

Now calculate the power flow Jacobian

( ) ( )

( )( ) ( )

10 cos 10sin

10 sin 10cos 20

P V

Q V V

P Px x

V

Q Qx x

V

V

V V

x

x

J x

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Two Bus Example, First Iteration(0)2(0)(0)

2

(0) (0)2 2

(0)2(0) (0) (0)

2 2 2

(0) (0) (0)2 2 2(0)

(0) (0)2 2

0For 0, guess . Calculate:

1

(10sin ) 2.0 2.0( )

1.0( 10cos ) (10) 1.0

10 cos 10sin( )

10 sin 10cos

vV

V

V V

V

V

x

f x

J x(0) (0)2 2

1(1)

10 0

0 1020

0 10 0 2.0 0.2Solve

1 0 10 1.0 0.9

V

x

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Two Bus Example, Next Iterations(1)

2

(1)

1(2)

0.9(10sin( 0.2)) 2.0 0.212( )

0.2790.9( 10cos( 0.2)) 0.9 10 1.0

8.82 1.986( )

1.788 8.199

0.2 8.82 1.986 0.212 0.233

0.9 1.788 8.199 0.279 0.8586

(

f x

J x

x

f (2) (3)

(3)2

0.0145 0.236)

0.0190 0.8554

0.0000906( ) Close enough! 0.8554 13.52

0.0001175V

x x

f x

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Two Bus Solved Values

Line Z = 0.1j

One Two 1.000 pu 0.855 pu

200 MW 100 MVR

200.0 MW168.3 MVR

-13.522 Deg

200.0 MW 168.3 MVR

-200.0 MW-100.0 MVR

Once the voltage angle and magnitude at bus 2 are known we can calculate all the other system values,such as the line flows and the generator reactive power output

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Two Bus Case Low Voltage Solution

(0)

(0) (0)2 2

(0)

(0) (0) (02 2 2

This case actually has two solutions! The second

"low voltage" is found by using a low initial guess.

0Set 0, guess . Calculate:

0.25

(10sin ) 2.0( )

( 10cos )

v

V

V V

x

f x2)

(0) (0) (0)2 2 2(0)

(0) (0) (0) (0)2 2 2 2

2

0.875(10) 1.0

10 cos 10sin 2.5 0( )

0 510 sin 10cos 20

V

V V

J x

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Low Voltage Solution, cont'd1

(1)

(2) (2) (3)

0 2.5 0 2 0.8Solve

0.25 0 5 0.875 0.075

1.462 1.42 0.921( )

0.534 0.2336 0.220

x

f x x x

Line Z = 0.1j

One Two 1.000 pu 0.261 pu

200 MW 100 MVR

200.0 MW831.7 MVR

-49.914 Deg

200.0 MW 831.7 MVR

-200.0 MW-100.0 MVR

Low voltage solution

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Two Bus Region of ConvergenceGraph shows the region of convergence for different initialguesses of bus 2 angle (horizontal axis) and magnitude (vertical axis).

Red regionconvergesto the highvoltage solution,while the yellow regionconvergesto the lowvoltage solution

Maximum of 15

iterations24

PV BusesSince the voltage magnitude at PV buses is fixed there

is no need to explicitly include these voltages in x nor write the reactive power balance equations:– the reactive power output of the generator varies to

maintain the fixed terminal voltage (within limits), so we can just set the reactive power product to whatever is needed.

– An alternative is these variations/equations can be included by just writing the explicit voltage constraint for the generator bus:

|Vi | – Vi setpoint = 0

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Three Bus PV Case Example

Line Z = 0.1j

Line Z = 0.1j Line Z = 0.1j

One Two 1.000 pu 0.941 pu

200 MW 100 MVR

170.0 MW 68.2 MVR

-7.469 Deg

Three 1.000 pu

30 MW 63 MVR

2 2 2 2

3 3 3 3

2 2 2

For this three bus case we have

( )

( ) ( ) 0

( )

G D

G D

D

P P P

P P P

V Q Q

x

x f x x

x

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PV Buses

• With Newton-Raphson, PV buses means that there are less unknown variables we need to calculate explicitly and less equations we need to satisfy explicitly.

• Reactive power balance is satisfied implicitly by choosing reactive power production to be whatever is needed, once we have a solved case (like real power at the slack bus).

• Contrast to Gauss iterations where PV buses complicated the algorithm.

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Modeling Voltage Dependent LoadSo far we've assumed that the load is independent of

the bus voltage (i.e., constant power). However, the

power flow can be easily extended to include voltage

dependence with both the real and reactive

1

1

load. This

is done by making and a function of :

( cos sin ) ( ) 0

( sin cos ) ( ) 0

Di Di i

n

i k ik ik ik ik Gi Di ik

n

i k ik ik ik ik Gi Di ik

P Q V

V V G B P P V

V V G B Q Q V

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Voltage Dependent Load Example

2 22 2 2 2 2

2 2 22 2 2 2 2 2

In previous two bus example now assume the load is

constant impedance, with corresponding per unit

admittance of 2.0 1.0 :

( ) 2.0 (10sin ) 2.0 0

( ) 1.0 ( 10cos ) (10) 1.0 0

Now

j

P V V V

Q V V V V

x

x

2 2 2 2

2 2 2 2 2

calculate the power flow Jacobian

10 cos 10sin 4.0( )

10 sin 10cos 20 2.0

V V

V V V

J x

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Voltage Dependent Load, cont'd(0)2(0)(0)

2

2(0) (0) (0)2 2 2(0)

2 2(0) (0) (0) (0)2 2 2 2

(0)

(1)

0Again for 0, guess . Calculate:

1

(10sin ) 2.0 2.0( )

1.0( 10cos ) (10) 1.0

10 4( )

0 12

0Solve

1

vV

V V

V V V

x

f x

J x

x110 4 2.0 0.1667

0 12 1.0 0.9167

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Voltage Dependent Load, cont'd

Line Z = 0.1j

One Two 1.000 pu 0.894 pu

160 MW 80 MVR

160.0 MW120.0 MVR

-10.304 Deg

160.0 MW 120.0 MVR

-160.0 MW -80.0 MVR

With constant impedance load the MW/MVAr load atbus 2 varies with the square of the bus 2 voltage magnitude. This if the voltage level is less than 1.0,the load is lower than 200/100 MW/MVAr.

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In practice, load is the sum of constant power, constant impedance, and, in some cases, constant current load terms: “ZIP” load.

Solving Large Power SystemsMost difficult computational task is inverting the

Jacobian matrix (or solving the update equation):– factorizing a full matrix is an order n3 operation, meaning

the amount of computation increases with the cube of the size of the problem.

– this amount of computation can be decreased substantially by recognizing that since Ybus is a sparse matrix, the Jacobian is also a sparse matrix.

– using sparse matrix methods results in a computational order of about n1.5.

– this is a substantial savings when solving systems with tens of thousands of buses.

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Newton-Raphson Power FlowAdvantages

– fast convergence as long as initial guess is close to solution

– large region of convergenceDisadvantages

– each iteration takes much longer than a Gauss-Seidel iteration

– more complicated to code, particularly when implementing sparse matrix algorithms

Newton-Raphson algorithm is very common in power flow analysis.

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