lecture 12

EE 369POWER SYSTEM ANALYSIS

Lecture 12Power Flow

Tom Overbye and Ross Baldick

1

Announcements

• 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.

• Midterm 2, Thursday, November 12, covering up to and including material in HW9

• Homework 10 is: 3.49, 3.55, 3.57, 6.2, 6.9, 6.13, 6.14, 6.18, 6.19, 6.20; due 11/19. (Use infinity norm and epsilon = 0.01 for any problems where norm or stopping criterion not specified.)

2

Transmission System Planning

Source: Federal Energy Regulatory Commission 3

ERCOT

•4Source: US Energy Information Administration

ERCOT• Has considerable wind and expecting considerable more!• “Competitive Renewable Energy Zones” study identified

most promising wind sites,• ERCOT ISO planned approximately $5 billion (original

estimate, now closer to $7 billion) of new transmission to support an additional 11 GW of wind:– Used tools such as power flow to identify whether plan could

accommodate wind generation.• Built by transmission companies.• Mostly completed by 2014.

5

CREZ Transmission Lines

6

NR Application to Power Flow

** * *

1 1

We first need to rewrite complex power equationsas 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 kj

V e V

j

7

Real Power Balance Equations* *

1 1

1

1

1

( )

(cos sin )( )

Resolving into the real and imaginary parts:

( cos sin )

( sin

ikn n

ji i i i ik k i k ik ik

k kn

i k ik ik ik ikk

n

i Gi Di i k ik ik ik ikkn

i Gi Di i k ik ikk

S P jQ V Y V V V e G jB

V V j G jB

P P P V V G B

Q Q Q V V G

cos )ik ikB

8

Newton-Raphson Power FlowIn the Newton-Raphson power flow we use Newton'smethod to determine the voltage magnitude and angle ateach 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 Dikn

i k ik ik ik ik Gi Dik

V V G B P P

V V G B Q Q

9

Power Balance Equations

•10

1

1

For convenience, write:

( ) ( cos sin )

( ) ( sin cos )

The power balance equations are then:( ) 0( ) 0

n

i i k ik ik ik ikkn

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 PQ Q Q

x

x

xx

Power Balance Equations• Note that Pi( ) and Qi( ) mean the functions that

expresses flow from bus i into the system in terms of voltage magnitudes and angles,

• While PGi, PDi, QGi, QDi mean the generations and demand at the bus.

• For a system with a slack bus and the rest PQ buses, the power flow problem is to use the power balance equations to solve for the unknown voltage magnitudes and angles in terms of the given bus generations and demands, and solve for the real and reactive injection at the slack bus.

•11

Power Flow Variables

2

n

2

Assume the slack bus is the first bus (with a fixedvoltage 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 PQ Q Q

Q Q Q

x

xf x

x

x

12

N-R Power Flow Solution

(0)

( )

( 1) ( ) ( ) 1 ( )

The power flow is solved using the same procedurediscussed 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

13

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 determiningand factorizing the Jacobian matrix, ( )

( ) ( ) ( )

( ) ( ) ( )( )

( ) ( )

n

n

n n n

f f fx x xf f fx x x

f f fx x x

J x

x x x

x x xJ x

x x

2 2

( )n

x

14

Power Flow Jacobian Matrix, cont’d

1

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

in

i i k ik ik ik ik Gik

f xf 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

f x V V G B

f x V V G B j i

15

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 10busj j

V j j

x Y16

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 Dikn

i k ik ik ik ik Gi Dik

V V G B P P

V V G B Q Q

2 1 22

2 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) 1.0Now calculate the power flow Jacobian

( ) ( )( )

( ) ( )

10 cos 10sin10 sin 10cos 20

P V

Q V V

P Px xV

Q Qx xV

VV V

x

x

J x

18

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 00 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 1.08.82 1.986

( )1.788 8.199

0.2 8.82 1.986 0.212 0.2330.9 1.788 8.199 0.279 0.8586

(

f x

J x

x

f (2) (3)

(3)2

0.0145 0.236)

0.0190 0.85540.0000906

( ) Close enough! 0.8554 13.520.0001175

V

x x

f x

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Two Bus Solved Values

Line Z = 0.1j


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 real and reactive power output

21

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 x 2)

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

2 2 2 2

20.875(10) 1.0

10 cos 10sin 2.5 0( )

0 510 sin 10cos 20

V

V V

J x

22

Low Voltage Solution, cont'd1

(1)

(2) (2) (3)

0 2.5 0 2 0.8Solve

0.25 0 5 0.875 0.0751.462 1.42 0.921

( )0.534 0.2336 0.220

x

f x x x

Line Z = 0.1j


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

23

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

25

Three Bus PV Case Example

Line Z = 0.1j

Line Z = 0.1j Line Z = 0.1j


200 MW 100 MVR

170.0 MW 68.2 MVR

-7.469 Deg

Three 1.000 pu

30 MW 63 MVR

2 2 2

3 3 3

2 2 2

For this three bus case we have( )

( ) ( ) 0( )

D

G

D

P PP P

V Q Q

xx 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 and reactive 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 ofthe bus voltage (i.e., constant power). However, thepower flow can be easily extended to include voltagedependence with both the real and reactive

1

1

load. Thisis done by making and a function of :

( cos sin ) ( ) 0

( sin cos ) ( ) 0

Di Di in

i k ik ik ik ik Gi Di ikn

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 isconstant impedance, with corresponding per unitadmittance of 2.0 1.0 :

( ) 2.0 (10sin ) 2.0 0

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

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 Jacobian10 cos 10sin 4.0

( )10 sin 10cos 20 2.0V VV V V

J x

29

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

30

Voltage Dependent Load, cont'd

Line Z = 0.1j


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.

32

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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lecture 12

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