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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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Changes in the system [Ybus]
! Remove a bus! Assume we are interested only in m buses in
a system with n buses (m
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Changes in the system [Ybus]
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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Power Flow – Gauss-Seidel
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Power Flow – Gauss-Seidel
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Power Flow – Gauss-Seidel
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Power Flow – Gauss-Seidel
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Power Flow – Gauss-Seidel
! Complete set of equations
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Power Flow – Gauss-Seidel
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Power Flow – Gauss-Seidel
! 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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Power Flow – Gauss-Seidel
! 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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Power Flow – Gauss-Seidel
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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Power Flow – Gauss-Seidel Example
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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Power Flow – Gauss-Seidel Example
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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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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