Power The Flow Solution
Most common and important tool in power system
analysis
also known as the Load Flow solution
used for planning and controlling a system
assumptions: balanced condition and single phase analysis
Problem:
determine the voltage magnitude and phase angle at each bus
determine the active and reactive power flow in each line
each bus has four state variables:
voltage magnitude
voltage phase angle
real power injection
reactive power injection
Power Systems I
The Power Flow Solution
Each bus has two of the four state variables defined or given
Types of buses:
Slack bus (swing bus)
Regulated bus (generator bus, P-V bus)
voltage magnitude and angle are specified, reference bus
solution: active and reactive power injections
models generation-station buses
real power and voltage magnitude are specified
solution: reactive power injection and voltage angle
Load bus (P-Q bus)
models load-center buses
active and reactive powers are specified (negative values for loads)
solution: voltage magnitude and angle
Power Systems I
Newton-Raphson PF Solution
Quadratic convergence
More efficient for large networks
mathematically superior to Guass-Seidel method
number of iterations required for solution is independent of
system size
The Newton-Raphson equations are cast in natural power
system form
solving for voltage magnitude and angle, given real and reactive
power injections
Power Systems I
Newton-Raphson Method
A method of successive approximation using Taylors
expansion
Consider the function: f(x) = c, where x is unknown
Let x[0] be an initial estimate, then x[0] is a small deviation from
the correct solution
(
)
f x[ 0 ] + x[ 0] = c
Expand the left-hand side into a Taylors series about x[0] yeilds
()
f x[0]
Power Systems I
2
df [ 0 ] 1 d f [ 0] 2
+ x + 2 2 x
+L = c
dx
dx
(
)
Newton-Raphson Method
Assuming the error, x[0], is small, the higher-order terms are
neglected, resulting in
()
fx
[0]
where
df [ 0 ]
df [ 0]
[0]
+ x c c x
dx
dx
()
c[ 0 ] = c f x[ 0 ]
rearranging the equations
c [ 0 ]
x[ 0] =
df
dx
x[1] = x[ 0 ] + x[ 0]
Power Systems I
Example
Find the root of the equation: f(x) = x3 - 6x2 + 9x - 4 = 0
Power Systems I
Newton-Raphson Method
50
40
30
20
10
3
2
f(x) = x -6x +9x-4
0
-10
0
Power Systems I
1
2
3
x
4
5
6
Power Flow Equations
KCL for current injection
n
n
I i = Yij V j = Yij V j ij + j
j =1
j =1
Real and reactive power injection
Pi j Qi = Vi* I i
Substituting for Ii yields:
n
Pi j Qi = (Vi ) Yij V j ij + j
j =1
Power Systems I
Power Flow Equations
Divide into real and reactive parts
n
Pi = Vi V j Yij cos( ij i + j )
j =1
n
Qi = Vi V j Yij sin ( ij i + j )
j =1
Power Systems I
Newton-Raphson Formation
Cast power equations into iterative form
n
(
Pi[ k ] = Vi[ k ] V j[ k ] Yij cos ij i[ k ] + [j k ]
)
j =1
n
(
Qi[ k ] = Vi[ k ] V j[ k ] Yij sin ij i[ k ] + [j k ]
)
j =1
Matrix function formation of the system of equations
sch
Pinj
c = sch
Qinj
Power Systems I
x
[k ]
[ k ]
= [ k ]
V
()
fx
[k ]
( )
( )
Pinj x[ k ]
=
Qinj x[ k ]
Newton-Raphson Formation
General formation of the equation to find a solution
c = f ( xsolution )
The iterative equation
x
x[ 0 ] = initial estimate of xsolution
[ k +1]
()
()
c f x[ k ]
= x[ k ] +
df x[ k ]
dx
The Jacobian - the first derivative of a set of functions
()
df x[ k ]
a matrix of all combinatorial pairs
dx
Power Systems I
The Jacobian Matrix
P
P
df (x )
= Q
dx
Q
P
1
P
1
1
M
M
P
Pn 1 n1
= Q11
Q1 1
M M
Qnm
Qn m 1
Power Systems I
P
V
V
Q
V
L
P
1
n1
P
1
V1
L
O
M
M
O
L
Pn1
n1
Pn1
n1
Q1
V1
Q1
V1
L
M
M
Qnm
n1
Qn m
V1
O
L
L
O
L
L
P
1
Vnm
1
M
M
Pn1
n 1
Vnm
Q1
V1
Vnm
M M
Qn m
Vn m
Vnm
Jacobian Terms
Real power w.r.t. the voltage angle
Pi
= Vi V j Yij sin ( ij i + j )
i j i
Pi
= Vi V j Yij sin( ij i + j ) i j
j
Real power w.r.t. the voltage magnitude
Pi
= 2 Vi Yii cos ii + V j Yij cos( ij i + j )
Vi
j i
Pi
= Vi Yij cos( ij i + j ) i j
V j
Power Systems I
Jacobian Terms
Reactive power w.r.t. the voltage angle
Qi
= Vi V j Yij cos( ij i + j )
i
j i
Qi
= Vi V j Yij cos( ij i + j ) i j
j
Reactive power w.r.t. the voltage magnitude
Qi
= 2 Vi Yii sin ii + V j Yij sin( ij i + j )
Vi
j i
Qi
= Vi Yij sin( ij i + j ) i j
V j
Power Systems I
Iteration process
Power mismatch or power residuals
difference in schedule to calculated power
Pi
[k ]
= Pi
sch
Pi
[k ]
Qi[ k ] = Qisch Qi[ k ]
New estimates for the voltages
[ k +1]
i
=
[k ]
i
+
[k ]
i
Vi[ k +1] = Vi[ k ] + Vi[ k ]
Power Systems I
Bus Type and the Jacobian Formation
Slack Bus / Swing Bus
one generator bus must be selected and defined as the voltage
and angular reference
Generator Bus
The voltage and angle are known for this bus
The angle is arbitrarily selected as zero degrees
bus is not included in the Jacobian matrix formation
have known terminal voltage and real (actual) power injection
the bus voltage angle and reactive power injection are computed
bus is included in the real power parts of the Jacobian matrix
Load Bus
have known real and reactive power injections
bus is fully included in the Jacobian matrix
Power Systems I
Newton-Raphson Steps
1. Set flat start
For load buses, set voltages equal to the slack bus or 1.00
For generator buses, set the angles equal the slack bus or 0
2. Calculate power mismatch
For load buses, calculate P and Q injections using the known and
estimated system voltages
For generator buses, calculate P injections
Obtain the power mismatches, P and Q
3. Form the Jacobian matrix
Use the various equations for the partial derivatives w.r.t. the
voltage angles and magnitudes
Power Systems I
Newton-Raphson Steps
4. Find the matrix solution (choose a or b)
a. inverse the Jacobian matrix and multiply by the mismatch
power
b. perform gaussian elimination on the Jacobian matrix with the b
vector equal to the mismatch power
compute and V
5. Find new estimates for the voltage magnitude and angle
6. Repeat the process until the mismatch (residuals) are
less than the specified accuracy
Pi[ k ]
Qi[ k ]
Power Systems I
Line Flows and Losses
After solving for bus voltages and angles, power flows
and losses on the network branches are calculated
Transmission lines and transformers are network branches
The direction of positive current flow are defined as follows for a
branch element (demonstrated on a medium length line)
Power flow is defined for each end of the branch
Example: the power leaving bus i and flowing to bus j
Bus i
Iij Vi
IL
Ii0
yi0
Power Systems I
Bus j
yij
Vj Iji
Ij0
yj0
Line Flows and Losses
current and power flows:
i j
ji
I ij = I L + I i 0 = yij (Vi V j ) + yi 0 Vi
I ji = I L + I j 0 = yij (V j Vi ) + y j 0 V j
*
*
*
Sij = Vi I ij = Vi 2 ( yij + yi 0 ) Vi yij V j* S ji = V j I * = V j2 ( yij + y j 0 ) V j yij Vi *
ji
*
power loss:
*
S Loss ij = Sij + S ji
Bus i
Iij Vi
IL
Ii0
yi0
Power Systems I
Bus j
yij
Vj Iji
Ij0
yj0
Example
Using N-R method, find the
phasor voltages at buses 2
and 3
Find the slack bus real
and reactive power
Calculate line flows
j0.02
and line losses
100 MVA base
1
j0.025
j0.04
2
3
138.6 MW
45.2 MVAR
Power Systems I
Slack Bus
V1 = 1.050
256.6 MW
110.2 MVAR