Topic 5 PowerFlow

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Power Systems Analysis ECNG 3012 5-Jan-May 2009

The bus admittance matrix [Ybus]

1 2

3

1 !

1 ! 1 !

1 !

-1 !

2 -1 -1

-1 2 -1

-1 -1 3

1 2 3

1

2

3

Y BUS =

1


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Changes in the system [Ybus]

! Remove a line.!Equivalent to adding a line in parallel to the

one to be removed with y = -yline" Diagonal –yline ii & jj

" Off-diagonal +yline ij & ji

! Add a line.

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! Remove a bus! Assume we are interested only in m buses in

a system with n buses (m


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Ib

= 0 = Y ab

t Va

+ Y bb

Vb

Vb = -Y bb-1 Y ab

t Va

Then:

Ia= Y aa Va + Y ab Vb = Y aa Va + Y ab(-Y bb

-1 Y abt Va)

Ia = [Y aa – Y ab Y bb-1

Y abt

]Va

Ia = [Y BUSEQUIV] Va

[Y BUSEQUIV] = Y aa – Y ab Y bb

-1 Y abt

KRON’S

REDUCTION

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Power Flow

!

Operation point of thesystem.! Steady State.! Given conditions of

generation, load &

configuration: OperationPoint

Buses (V/!)Branches (P, Q)

Solve Equations (Balance generation and load)

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Power Flow Main Objectives

!

To calculate P&Q flow through elements." Observe power flow and check overloads.

" Effects of contingencies.

" Effects of configuration changes.

!

To calculate voltage magnitude and angle on buses." Quality of service." Strategies to operate elements with voltage control ( Taps, Exc.

Generator , Capacitors.)

! To design the optimal operation & distribution of

loads.

! To define operation guidelines.

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Power Flow

V1 V2

V5

V3

V4

SD1

SD4 SD5

SD3

SG2 SG3SG1

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Power Flow Basics

BasicEquations

Power Balance on each node

Power Balance for the system

! Sinput = ! Soutput

SGEN = SLOADS + SLOSSES

S = P + jQ

(Generated real power is calculated with economic dispatch)

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Power Flow

P+"PQ+"Q I

R + jX Vg /! Vc /0

0

P + jQ

" V

Vg

I Vc IR

IX

!

!

" V We want:" V =f(P,Q)

# V = f(P,Q)

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Power Flow

Vg2

= (Vc +" V)2

+ (" V)2

Vg2 = (Vc + RI cos$ + IX sin$ )

2 + (IX cos$ - RI sin$ )2

P = VcI cos $ ; Q = VcI sin $

Vg2 = Vc + RP + XQ

2 + XP - RQ 2

Vc Vc Vc Vc

" V = RP + XQ Vc

" V = XP – RQ

Vc

If R


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Power Flow

! If load P increases # increases

! If load Q increases Vc

reduces

! Real power flow #sending end > #receiving end

! Reactive power flow |Vsending end| >|Vreceiving end|

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Power Flow - Definitions

! At each bus:Vk (Voltage magnitude), #k (Voltage angle),Pk (Total injected real power), Qk(Total injected reactive

power)

! Reduced number of unknowns. Assumptions:

" At most generator buses, the active power PG is controlled (by

speed governor) and the voltage magnitude is controlled (by thevoltage regulator). Treat these as known.

" At most load buses, a reasonable approximation is that the loadactive and reactive power demand PD and QD are known.

" At one generator bus, leave the active power as a variable (tomake up system losses).

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Power Flow – Bus Types

! Load bus (PQ)

" known real (P) and reactive (Q) power injections

! Voltage controlled bus (PV)

" known real (P) power injection and the voltagemagnitude (V)

! Slack bus (swing bus)

"

known voltage magnitude (V) and voltage angle ($)"must have one generator as the slack bus

" takes up the power slack due to losses in the network

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Power Flow – Bus Types

! LOAD (PQ)" Given: Pk, Qk" Unknown: Vk, #k

" Example: Loads, transformer buses

! VOLTAGE CONTROLLED (PV)" Given: Pk, Vk" Unknown: Qk, #k

" Example: Generation buses, reactive powercompensation buses

! SLACK(V #)" Given: Vk, #k" Unknown: Pk, Qk

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Power Flow Equations

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Power Flow – Gauss-Seidel

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Power Flow – Gauss- Seidel

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! Complete set of equations

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! System characteristics!Since both components (V & $) are specified

for the slack bus, there are 2(n - 1) equationswhich must be solved iteratively.

!For the load buses, the real and reactivepowers are known: scheduled" the voltage magnitude and angle must be

estimated."

in per unit, the nominal voltage magnitude is 1 p.u." the angles are generally close together, so an

initial value of 0 degrees is appropriate.

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! For the generator buses, the real power andvoltage magnitude are known" the real power is scheduled."

the reactive power is computed based on theestimated voltage values." the voltage is computed by Gauss-Seidel, only the

angle is kept" the complex voltage is found from the schedule

magnitude and the iterative angle part.

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Select initial voltages

for each bus

Find Maximum Voltage Change

"EMAX = | Ei! – Ei

! -1|MAX OVER i

"EMAX! %Calculate Line Flows,

Losses, Mismatch, etc

Print Results Stop

YesNo

Do for all ii = 1 … N

(i "ref)

% = Specified Voltage

Convergence

Tolerance

Start

Solve for EiNEW

EiNEW

= f(Pi,E j) j = 1 … N

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Power Flow – Gauss-Seidel Example

21

3

4

5

6

Bus Type Given Unknown

1 Load P1,Q1 V1,#1

2 Voltage

controlled

P2,V2 Q1,#2

3 Load P3,Q3 V3,#3

4 Load P4,Q4 V4,#4

5 Load P5,Q5 V5,#5

6 Slack V6,#6 P6,Q6

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Calculate [Y bus

]: Y 11

, Y 12

, Y 13

, Y 21

, Y 22

, Y 26

, Y 31

, Y 33

, Y 34

, Y 43

,

Y 43, Y 44, Y 45, Y 54, Y 55, Y 56, Y 62, Y 64, Y 66

Bus V(0) #(0)

1 1.0 0o

2 Vknown

0o

3 1.0 0o

4 1.0 0o

5 1.0 0o

6 Vknown

0o

Initial voltages (Flat start)

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Bus 1 V1(1) =1 -P1 + jQ1

Y 11 V1(0)*- Y 12 V2(0) - Y 13 V3(0)

Q2(1) = V2(0) [ V1(1){G21 sin(!21) – B21 cos (!21)}

+ V6{G26 sin(!26) – B26 cos (!26)}

V2(1) =1 P2 – jQ2

Y 22 V2(0)*- Y 21 V1(1) - Y 26 V6

Bus 2

!21 = !2 - !1 ; !26 = !2 - !6

Multiply V2(1) such that |V2(1)| = V2 given

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Example – Equations

V3(1) =1 -P3 + jQ3

Y 33 V3(0)*- Y 31 V1(1) - Y 34 V4(0)

Bus 3

V4(1) =1 -P4 + jQ4

Y 44 V4(0)*- Y 43 V3(1) - Y 45 V5(0) - Y 46 V6

Bus 4

V5(1) = 1 -P5 + jQ5

Y 55 V5(0)*- Y 45 V4(1)

Bus 5

Iterate with these equations31


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Power Systems Analysis ECNG 3012 5-Jan May 2009

Gauss – Seidel Method

! Simple , basic

! Slow convergence

! Convergence problems

Acceleration Factor

Vi(k)acc = Vi(k-1) + ! (Vi(k) – Vi(k-1))

1.0" ! "2.0

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Topic 5 PowerFlow