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Power Systems I
The Power Flow Solution
l Most common and important tool in power systemanalysis
u also known as the “Load Flow” solution
u used for planning and controlling a system
u assumptions: balanced condition and single phase analysis
l Problem:
u determine the voltage magnitude and phase angle at each bus
u determine the active and reactive power flow in each line
u each bus has four state variables:
n voltage magnitude
n voltage phase angle
n real power injection
n reactive power injection
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Power Systems I
The Power Flow Solution
u Each bus has two of the four state variables defined or given
l Types of buses:
u Slack bus (swing bus)
n voltage magnitude and angle are specified, reference bus
n solution: active and reactive power injections
u Regulated bus (generator bus, P-V bus)
n models generation-station buses
n real power and voltage magnitude are specified
n solution: reactive power injection and voltage angle
u Load bus (P-Q bus)n models load-center buses
n active and reactive powers are specified (negative values for loads)
n solution: voltage magnitude and angle
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Power Systems I
Newton-Raphson PF Solution
l Quadratic convergence
u mathematically superior to Guass-Seidel method
l More efficient for large networks
u number of iterations required for solution is independent of
system size
l The Newton-Raphson equations are cast in natural powersystem form
u solving for voltage magnitude and angle, given real and reactivepower injections
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Power Systems I
Newton-Raphson Method
l A method of successive approximation using Taylor’sexpansion
u Consider the function: f ( x) = c, where x is unknown
u Let x[0] be an initial estimate, then ∆ x[0] is a small deviation fromthe correct solution
u Expand the left-hand side into a Taylor’s series about x [0] yeilds
c x x f =∆+ ]0[]0[
( ) ( ) c xdx
f d x
dx
df x f =+∆
+∆
+ L
2]0[
2
2
21]0[]0[
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Power Systems I
Newton-Raphson Method
u Assuming the error, ∆ x[0], is small, the higher-order terms areneglected, resulting in
u where
u rearranging the equations
( ) ]0[]0[]0[]0[ xdx
df cc x
dx
df x f ∆
≈∆⇒≈∆
+
( )]0[]0[ x f cc −=∆
]0[]0[]1[
]0[
]0[
x x x
dx
df
c x
∆+=
∆
=∆
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Power Systems I
Example
l Find the root of the equation: f (x ) = x 3 - 6x 2 + 9x - 4 = 0
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Power Systems I
Newton-Raphson Method
0 1 2 3 4 5 6-10
0
10
20
30
40
50
x
f(x) = x
3
-6x
2
+9x-4
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Power Systems I
( )∑
∑∑
=
==
+∠−∠=−
=−
+∠==
n
j
jij jijiii
iiii
n
j
jij jij
n
j
jiji
V Y V Q jP
I V Q jP
V Y V Y I
1
*
11
δθδ
δθ
Power Flow Equations
l KCL for current injection
l Real and reactive power injection
l Substituting for I i yields:
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Power Systems I
Power Flow Equations
( )
( )∑
∑
=
=
+−−=
+−=
n
j
jiijij jii
n
j
jiijij jii
Y V V Q
Y V V P
1
1
sin
cos
δδθ
δδθ
l Divide into real and reactive parts
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Power Systems I
Newton-Raphson Formation
( )
( )
( ) ( )( )
=
=
=
+−−=
+−=
∑
∑
=
=
][
][
][
][
][
][
1
][][][][][
1
][][][][][
sin
cos
k
inj
k
injk
k
k
k
sch
inj
sch
inj
n
j
k
j
k
iijij
k
j
k
i
k
i
n
j
k
j
k
iijij
k
j
k
i
k
i
xQ
xP x f V
xQ
Pc
Y V V Q
Y V V P
δ
δδθ
δδθ
l Cast power equations into iterative form
l Matrix function formation of the system of equations
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Power Systems I
Newton-Raphson Formation
( )
( )( )
( )dx
xdf
dx xdf
x f c x x
x x x f c
k
k
k k k
solutionsolution
][
][
][][]1[
]0[ of estimateinitial
−+=
==
+
l General formation of the equation to find a solution
l The iterative equation
l The Jacobian - the first derivative of a set of functions
a matrix of all combinatorial pairs
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Power Systems I
The Jacobian Matrix
( )
∆
∆
∆
∆
=
∆
∆
∆
∆
∆∆
∂∂
∂∂
∂∂
∂∂
=
∆∆
⇒
−
−
∂∂
∂∂
∂∂
∂∂
∂ ∂∂∂∂∂∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
−
−
−
−−
−
−−
−−
−
−−
−
−−
−−
mn
n
V
Q
V
QQQ
V Q
V QQQ
V
P
V
PPP
V
P
V
PPP
mn
n
V
V
Q
Q
P
P
V V
QQ
V PP
Q
P
dx xdf
mn
mnmn
n
mnmn
mnn
mn
nn
n
nn
mnn
M
M
LL
MOMMOM
LL
LL
MOMMOM
LL
M
M
1
1
1
1
1
1
111
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
δ
δ
δ
δ
δ
δδ
δδ
δδ
δδ
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Power Systems I
Jacobian Terms
( )
( )
( )
( ) jiY V V
P
Y V Y V
V
P
jiY V V P
Y V V P
jiijiji
j
i
i j
jiijij jiiiii
i
i
jiijij ji
j
i
i j
jiijij ji
i
i
≠+−=∂∂
+−+=∂
∂
≠+−−=∂∂
+−=∂∂
∑
∑
≠
≠
δδθ
δδθθ
δδθδ
δδθδ
cos
coscos2
sin
sin
l Real power w.r.t. the voltage angle
l Real power w.r.t. the voltage magnitude
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Power Systems I
l Reactive power w.r.t. the voltage angle
l Reactive power w.r.t. the voltage magnitude
Jacobian Terms
( )
( )
( )
( ) jiY V V
Q
Y V Y V
V
Q
jiY V V Q
Y V V Q
jiijiji
j
i
i j
jiijij jiiiii
i
i
jiijij ji
j
i
i j
jiijij ji
i
i
≠+−−=∂∂
+−+−=∂
∂
≠+−−=∂∂
+−=∂∂
∑
∑
≠
≠
δδθ
δδθθ
δδθδ
δδθδ
sin
sinsin2
cos
cos
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Power Systems I
Iteration process
l Power mismatch or power residuals
u difference in schedule to calculated power
l New estimates for the voltages
][][]1[
][][]1[
][][
][][
k i
k i
k i
k
i
k
i
k
i
k i
schi
k i
k
i
sch
i
k
i
V V V
QQQ
PPP
∆+=
∆+=
−=∆
−=∆
+
+δδδ
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Power Systems I
Bus Type and the Jacobian Formation
l Slack Bus / Swing Bus
u one generator bus must be selected and defined as the voltageand angular reference
n The voltage and angle are known for this bus
n The angle is arbitrarily selected as zero degrees
n bus is not included in the Jacobian matrix formation
l Generator Bus
n have known terminal voltage and real (actual) power injection
n the bus voltage angle and reactive power injection are computed
n bus is included in the real power parts of the Jacobian matrix
l Load Bus
n have known real and reactive power injections
n bus is fully included in the Jacobian matrix
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Power Systems I
Newton-Raphson Steps
1. Set flat start
u For load buses, set voltages equal to the slack bus or 1.0∠0°u For generator buses, set the angles equal the slack bus or 0°
2. Calculate power mismatch
u For load buses, calculate P and Q injections using the known andestimated system voltages
u For generator buses, calculate P injections
u Obtain the power mismatches, ∆P and ∆Q
3. Form the Jacobian matrix
u Use the various equations for the partial derivatives w.r.t. thevoltage angles and magnitudes
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Power Systems I
Newton-Raphson Steps
4. Find the matrix solution (choose a or b)
u a. inverse the Jacobian matrix and multiply by the mismatchpower
u b. perform gaussian elimination on the Jacobian matrix with the b
vector equal to the mismatch powercompute ∆δ and ∆V
5. Find new estimates for the voltage magnitude and angle
6. Repeat the process until the mismatch (residuals) areless than the specified accuracy
ε
ε
≤∆
≤∆][
][
k
i
k i
Q
P
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Power Systems I
Line Flows and Losses
l After solving for bus voltages and angles, power flowsand losses on the network branches are calculated
u Transmission lines and transformers are network branches
u The direction of positive current flow are defined as follows for a
branch element (demonstrated on a medium length line)u Power flow is defined for each end of the branch
n Example: the power leaving bus i and flowing to bus j
V jV i
y j0 yi0
yijBus i Bus j
I ij I ji I L
I j0 I i0
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Power Systems I
Line Flows and Losses
l current and power flows:
l power loss:
V jV i
y j0 yi0
yijBus i Bus j
I ij I ji I L
I j0 I i0
( )
( )***
0
2*
00
jijiiijiijiij
ii jiiji Lij
V yV y yV I V S
V yV V y I I I
ji
−+==
+−=+=→
( )
( )***
0
2*
00
iij j jij j ji j ji
j ji jij j L ji
V yV y yV I V S
V yV V y I I I
i j
−+==
+−=+−=→
jiijij Loss SSS +=
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Power Systems I
Example
j0.04
3
1
2
138.6 MW45.2 MVAR
256.6 MW110.2 MVAR
Slack Bus
V 1
= 1.05∠0°
j0.02 j0.025
l Using N-R method, find thephasor voltages at buses 2and 3
l Find the slack bus real
and reactive powerl Calculate line flows
and line losses
u 100 MVA base