Lecture 15Newton-Raphson Power Flow

Professor Tom OverbyeDepartment of Electrical and

Computer Engineering

ECE 476

POWER SYSTEM ANALYSIS

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Announcements

Homework 7 is 6.12, 6.19, 6.22, 6.45 and 6.50. Due date is October 25

Design Project 2 from the book (page 345 to 348) was due on Nov 15, but I have given you an extension to Nov 29. The Nov 29 date is firm!

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In The News (thanks to Peter Lewis)

Kansas Department of Health and Environment has air quality permits for two new 700 MW coal power plants

– After careful consideration of my responsibility to protect the public health and environment from actual, threatened or potential harm from air pollution, I have decided to deny the Sunflower Electric Power Corporation application for an air quality permit," Roderick Bremby, KDHE secretary, said in a written statement. "I believe it would be irresponsible to ignore emerging information about the contribution of carbon dioxide and other greenhouse gases to climate change and the potential harm to our environment and health if we do nothing."

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Fuel Costs for Electric Generation

Source: EIA Electric Power Annual, 2006 (October 2007)

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Kansas Electric Generation

Here is a summary of Kansas Electric Energy by

Generation Type (% of total)

1990 2005

Coal 69.4% 75.2%

Petroleum 0.2% 2.2%

Natural Gas 7.3% 2.5%

Nuclear 23.0% 19.2%

Hydroelectric 0.0% 0.0%Other Renewables 0.0% 0.9%

Source: EIA State Electricity Profiles, 2005

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Two Bus Case Low Voltage Solution

(0)

2 2(0)2

2 2 2

This case actually has two solutions! The second

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

0Set 0, guess

0.25

Calculate

(10sin ) 2.0f( )

( 10cos ) (10) 1.0

v

V

V V

x

x

2 2 2(0)

2 2 2 2

2

0.875

10 cos 10sin 2.5 0( )

10 sin 10cos 20 0 5

V

V V

J x

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Low Voltage Solution, cont'd

1(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 Convergence

Slide shows the region of convergence for different initialguesses of bus 2 angle (x-axis) and magnitude (y-axis)

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

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PV Buses

Since the voltage magnitude at PV buses is fixed there is no need to explicitly include these voltages in x or write the reactive power balance equations– the reactive power output of the generator varies to

maintain the fixed terminal voltage (within limits)– optionally 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

V ( )

G D

G D

D

P P P

P P P

Q Q

x

x f x x

x

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Solving Large Power Systems

The most difficult computational task is inverting the Jacobian matrix– inverting a full matrix is an order n3 operation, meaning

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

– this amount of computation can be decreased substantially by recognizing that since the 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 Flow

Advantages– fast convergence as long as initial guess is close to solution– large region of convergence

Disadvantages– 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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Decoupled Power Flow Formulation

( ) ( )

( ) ( )( )

( )( ) ( ) ( )

( )2 2 2

( )

( )

General form of the power flow problem

( )( )

( )

where

( )

( )

( )

v v

v vv

vv v v

vD G

v

vn Dn Gn

P P P

P P P

P Pθθ V P x

f xQ xVQ Q

θ V

x

P x

x

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Decoupling Approximation

( ) ( )

( )

( ) ( )( )

( ) ( ) ( )

Usually the off-diagonal matrices, and

are small. Therefore we approximate them as zero:

( )( )

( )

Then the problem

v v

v

v vv

v v v

P QV θ

P0

θ P xθf x

Q Q xV0V

1 1( ) ( )( )( ) ( ) ( )

can be decoupled

( ) ( )v v

vv v v

P Qθ P x V Q x

θ V

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Off-diagonal Jacobian Terms

Justification for Jacobian approximations:

1. Usually r x, therefore

2. Usually is small so sin 0

Therefore

cos sin 0

cos sin 0

ij ij

ij ij

ii ij ij ij ij

j

ii j ij ij ij ij

j

G B

V G B

V V G B

P

V

Qθ

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Fast Decoupled Power Flow

By continuing with our Jacobian approximations we can actually obtain a reasonable approximation that is independent of the voltage magnitudes/angles.

This means the Jacobian need only be built/inverted once.

This approach is known as the fast decoupled power flow (FDPF)

FDPF uses the same mismatch equations as standard power flow so it should have same solution

The FDPF is widely used, particularly when we only need an approximate solution

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FDPF Approximations

ij

( ) ( )( )( ) 1 1

( ) ( )

bus

The FDPF makes the following approximations:

1. G 0

2. 1

3. sin 0 cos 1

Then

( ) ( )

Where is just the imaginary part of the ,

except the slack bus row/co

i

ij ij

v vvv

v v

V

j

P x Q xθ B V B

V VB Y G B

lumn are omitted

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“DC” Power Flow

The “DC” power flow makes the most severe approximations:

– completely ignore reactive power, assume all the voltages are always 1.0 per unit, ignore line conductance

This makes the power flow a linear set of equations, which can be solved directly

1θ B P

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Power System Control

A major problem with power system operation is the limited capacity of the transmission system

– lines/transformers have limits (usually thermal)– no direct way of controlling flow down a transmission line

(e.g., there are no valves to close to limit flow)– open transmission system access associated with industry

restructuring is stressing the system in new ways

We need to indirectly control transmission line flow by changing the generator outputs

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Indirect Transmission Line Control

What we would like to determine is how a change in generation at bus k affects the power flow on a line from bus i to bus j.

The assumption isthat the changein generation isabsorbed by theslack bus

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Power Flow Simulation - Before

One way to determine the impact of a generator change is to compare a before/after power flow.

For example below is a three bus case with an overload

Z for all lines = j0.1

One Two

200 MW 100 MVR

200.0 MW 71.0 MVR

Three 1.000 pu

0 MW 64 MVR

131.9 MW

68.1 MW 68.1 MW

124%

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Power Flow Simulation - After

Z for all lines = j0.1Limit for all lines = 150 MVA

One Two

200 MW 100 MVR

105.0 MW 64.3 MVR

Three1.000 pu

95 MW 64 MVR

101.6 MW

3.4 MW 98.4 MW

92%

100%

Increasing the generation at bus 3 by 95 MW (and hence decreasing it at bus 1 by a corresponding amount), resultsin a 31.3 drop in the MW flow on the line from bus 1 to 2.

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Analytic Calculation of Sensitivities

Calculating control sensitivities by repeat power flow solutions is tedious and would require many power flow solutions. An alternative approach is to analytically calculate these values

The power flow from bus i to bus j is

sin( )

So We just need to get

i j i jij i j

ij ij

i j ijij

ij Gk

V VP

X X

PX P

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Analytic Sensitivities

1

From the fast decoupled power flow we know

( )

So to get the change in due to a change of

generation at bus k, just set ( ) equal to

all zeros except a minus one at position k.

0

1

0

θ B P x

θ

P x

P

Bus k

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Three Bus Sensitivity Example

line

bus

12

3

For the previous three bus case with Z 0.1

20 10 1020 10

10 20 1010 20

10 10 20

Hence for a change of generation at bus 3

20 10 0 0.0333

10 20 1 0.0667

j

j

Y B

3 to 1

3 to 2 2 to 1

0.0667 0Then P 0.667 pu

0.1P 0.333 pu P 0.333 pu

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Balancing Authority Areas

An balancing authority area (use to be called operating areas) has traditionally represented the portion of the interconnected electric grid operated by a single utility

Transmission lines that join two areas are known as tie-lines.

The net power out of an area is the sum of the flow on its tie-lines.

The flow out of an area is equal to

total gen - total load - total losses = tie-flow

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Area Control Error (ACE)

The area control error (ace) is the difference between the actual flow out of an area and the scheduled flow, plus a frequency component

Ideally the ACE should always be zero.Because the load is constantly changing, each utility

must constantly change its generation to “chase” the ACE.

int schedace 10P P f

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Automatic Generation Control

Most utilities use automatic generation control (AGC) to automatically change their generation to keep their ACE close to zero.

Usually the utility control center calculates ACE based upon tie-line flows; then the AGC module sends control signals out to the generators every couple seconds.

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Power Transactions

Power transactions are contracts between generators and loads to do power transactions.

Contracts can be for any amount of time at any price for any amount of power.

Scheduled power transactions are implemented by modifying the value of Psched used in the ACE calculation

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PTDFs

Power transfer distribution factors (PTDFs) show the linear impact of a transfer of power.

PTDFs calculated using the fast decoupled power flow B matrix

1 ( )

Once we know we can derive the change in

the transmission line flows

Except now we modify several elements in ( ),

in portion to how the specified generators would

participate in the pow

θ B P x

θ

P x

er transfer

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Nine Bus PTDF Example

10%

60%

55%

64%

57%

11%

74%

24%

32%

A

G

B

C

D

E

I

F

H

300.0 MW 400.0 MW 300.0 MW

250.0 MW

250.0 MW

200.0 MW

250.0 MW

150.0 MW

150.0 MW

44%

71%

0.00 deg

71.1 MW

92%

Figure shows initial flows for a nine bus power system

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Nine Bus PTDF Example, cont'd

43%

57% 13%

35%

20%

10%

2%

34%

34%

32%

A

G

B

C

D

E

I

F

H

300.0 MW 400.0 MW 300.0 MW

250.0 MW

250.0 MW

200.0 MW

250.0 MW

150.0 MW

150.0 MW

34%

30%

0.00 deg

71.1 MW

Figure now shows percentage PTDF flows from A to I

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Nine Bus PTDF Example, cont'd

6%

6% 12%

61%

12%

6%

19%

21%

21%

A

G

B

C

D

E

I

F

H

300.0 MW 400.0 MW 300.0 MW

250.0 MW

250.0 MW

200.0 MW

250.0 MW

150.0 MW

150.0 MW

20%

18%

0.00 deg

71.1 MW

Figure now shows percentage PTDF flows from G to F

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WE to TVA PTDFs

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Line Outage Distribution Factors (LODFS)

LODFs are used to approximate the change in the flow on one line caused by the outage of a second line– typically they are only used to determine the change in

the MW flow– LODFs are used extensively in real-time operations– LODFs are state-independent but do dependent on the

assumed network topology

,l l k kP LODF P

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Flowgates

The real-time loading of the power grid is accessed via “flowgates”

A flowgate “flow” is the real power flow on one or more transmission element for either base case conditions or a single contingency

– contingent flows are determined using LODFs

Flowgates are used as proxies for other types of limits, such as voltage or stability limits

Flowgates are calculated using a spreadsheet

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NERC Regional Reliability Councils

NERCis theNorthAmericanElectricReliabilityCouncil

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NERC Reliability Coordinators