Power Flow Analysis Well known as : Load Flow
Power Flow Analysis
Well known as : Load Flow
The Power Flow Problem
• Power flow analysis is fundamental to the study of power
systems.
• In fact, power flow forms the core of power system analysis.
• power flow study plays a key role in the planning of additions
or expansions to transmission and generation facilities.
• A power flow solution is often the starting point for many other
types of power system analyses.
• In addition, power flow analysis is at the heart of contingency
analysis and the implementation of real-time monitoring
systems.
Problem Statement
For a given power network, with known complex power
loads and some set of specifications or restrictions on
power generations and voltages, solve for any unknown
bus voltages and unspecified generation and finally for
the complex power flow in the network components.
Network Structure
Power Flow Study Steps
1. Determine element values for passive network components.
2. Determine locations and values of all complex power loads.
3. Determine generation specifications and constraints.
4. Develop a mathematical model describing power flow in the network.
5. Solve for the voltage profile of the network.
6. Solve for the power flows and losses in the network.
7. Check for constraint violations.
Formulation of the Bus Admittance Matrix
• The first step in developing the mathematical model
describing the power flow in the network is the
formulation of the bus admittance matrix.
• The bus admittance matrix is an n*n matrix (where n is
the number of buses in the system) constructed from the
admittances of the equivalent circuit elements of the
segments making up the power system.
• Most system segments are represented by a combination
of shunt elements (connected between a bus and the
reference node) and series elements (connected between
two system buses).
Bus Admittance Matrix
Formulation of the bus admittance matrix follows two
simple rules:
1. The admittance of elements connected between node
k and reference is added to the (k, k) entry of the
admittance matrix.
2. The admittance of elements connected between nodes
j and k is added to the (j, j) and (k, k) entries of the
admittance matrix.
• The negative of the admittance is added to the (j, k)
and (k, j) entries of the admittance matrix.
Bus Admittance Matrix
Bus Admittance Matrix
Node-Voltage Equations
Applying KCL at each node yields: Defining the Y’s as
The Y-Bus
The current equations reduced to
In a compact form
Where,
Gauss Power Flow
*
* * *i
1 1
* * * *
1 1
*
*1 1,
*
*1,
We first need to put the equation in the correct form
S
S
S
S1
i i
i
i
n n
i i i ik k i ik kk k
n n
i i i ik k ik kk k
n ni
ik k ii i ik kk k k i
ni
i ik kii k k i
V I V Y V V Y V
V I V Y V V Y V
Y V Y V Y VV
V Y VY V
Difficulties
• Unless the generation equals the load at every bus, the
complex power outputs of the generators cannot be
arbitrarily selected.
• In fact, the complex power output of at least one of the
generators must be calculated last, since it must take up
the unknown “slack” due to the uncalculated network
losses.
• Further, losses cannot be calculated until the voltages are
known.
• Also, it is not possible to solve these equations for the
absolute phase angles of the phasor voltages. This simply
means that the problem can only be solved to some
arbitrary phase angle reference.
Difficulties
• For a 4- bus system, suppose that SG4 is arbitrarily allowed to float
or swing (in order to take up the necessary slack caused by the
losses) and that SG1, SG2, SG3 are specified.
Remedies
• Now, with the loads known, the equations are seen as
four simultaneous nonlinear equations with complex
coefficients in five unknowns. (V1, V2, V3, V4 and SG4).
• Designating bus 4 as the slack bus and specifying the
voltage V4 reduces the problem to four equations in four
unknowns.
Remedies
• The slack bus is chosen as the phase reference for all
phasor calculations, its magnitude is constrained, and
the complex power generation at this bus is free to take
up the slack necessary in order to account for the
system real and reactive power losses.
• Systems of nonlinear equations, cannot (except in rare
cases) be solved by closed-form techniques.
Load Flow Solution
• There are four quantities of interest associated with each
bus:
1. Real Power, P
2. Reactive Power, Q
3. Voltage Magnitude, V
4. Voltage Angle, δ
• At every bus of the system, two of these four quantities will
be specified and the remaining two will be unknowns.
• Each of the system buses may be classified in accordance
with which of the two quantities are specified
Bus Classifications
Slack Bus — The slack bus for the system is a single bus for which the voltage
magnitude and angle are specified.
• The real and reactive power are unknowns.
• The bus selected as the slack bus must have a source of both real and
reactive power, since the injected power at this bus must “swing” to take
up the “slack” in the solution.
• The best choice for the slack bus (since, in most power systems, many
buses have real and reactive power sources) requires experience with the
particular system under study.
• The behavior of the solution is often influenced by the bus chosen.
Bus Classifications
Load Bus (P-Q Bus) : A load bus is defined as any bus of the system for which the real and reactive power are specified.
• Load buses may contain generators with specified real and reactive power outputs;
• however, it is often convenient to designate any bus with specified injected complex power as a load bus.
Voltage Controlled Bus (P-V Bus) : Any bus for which the voltage magnitude and the injected real power are specified is classified as a voltage controlled (or P-V) bus.
• The injected reactive power is a variable (with specified upper and lower bounds) in the power flow analysis.
• (A P-V bus must have a variable source of reactive power such as a generator.)
Solution Methods
• The solution of the simultaneous nonlinear power flow
equations requires the use of iterative techniques for
even the simplest power systems.
• There are many methods for solving nonlinear
equations, such as:
- Gauss Seidel.
- Newton Raphson.
- Fast Decoupled.
Guess Solution
• It is important to have a good approximation to the
load-flow solution, which is then used as a starting
estimate (or initial guess) in the iterative procedure.
• A fairly simple process can be used to evaluate a good
approximation to the unknown voltages and phase
angles.
• The process is implemented in two stages: the first
calculates the approximate angles, and the second
calculates the approximate voltage magnitudes.
Gauss Iteration Method
Gauss Iteration Example
( 1) ( )
(0)
( ) ( )
Example: Solve - 1 0
1
Let k = 0 and arbitrarily guess x 1 and solve
0 1 5 2.61185
1 2 6 2.61612
2 2.41421 7 2.61744
3 2.55538 8 2.61785
4 2.59805 9 2.61798
v v
v v
x x
x x
k x k x
Stopping Criteria
=
Example
A 100 MW, 50 Mvar load is connected to a generator
through a line with z = 0.02 + j0.06 p.u. and line charging
of 0.05 p.u on each end (100 MVA base). Also, there is a
0.25 p.u. capacitance at bus 2. If the generator voltage is
1.0 p.u., what is V2?
100+j50 0.25 p.u.
Z = 0.02 + j0.06
V= 1 0
Y-Bus
2
2 bus
bus
22
The unknown is the complex load voltage, V .
To determine V we need to know the .
15 15
0.02 0.06
5 14.95 5 15Hence
5 15 5 14.70
( Note - 15 0.05 0.25)
jj
j j
j j
B j j j
Y
Y
Solution
*2
2 *22 1,2
2 *2
(0)2
( ) ( )2 2
1 S
1 -1 0.5( 5 15)(1.0 0)
5 14.70
Guess 1.0 0 (this is known as a flat start)
0 1.000 0.000 3 0.9622 0.0556
1 0.9671 0.0568 4 0.9622 0.0556
2 0
n
ik kk k i
v v
V Y VY V
jV j
j V
V
v V v V
j j
j j
.9624 0.0553j
Solution (cont.)
Gauss-Seidel Iteration
( 1) ( ) ( ) ( )2 12 2 3
( 1) ( 1) ( ) ( )2 13 2 3
( 1) ( 1) ( 1) ( ) ( )2 14 2 3 4
( 1) ( 1) ( 1)( 1) ( )2 1 2 3 4
Immediately use the new voltage estimates:
( , , , , )
( , , , , )
( , , , , )
( , , , ,
v v v vn
v v v vn
v v v v vn
v v vv vn n
V h V V V V
V h V V V V
V h V V V V V
V h V V V V V
)
The Gauss-Seidel works better than the Gauss, and
is actually easier to implement. It is used instead
of Gauss.
Newton-Raphson Power Flow
The General Form of the Load-Flow Equations
• In Practice, bus powers Si is specified rather than the bus currents Ii .
• As a result, we have
)(||)(1 1
**
ininniin
N
n
N
n
niniiiii VVYVYVIVjQP
ijijijijijijijijij jBGYjYYY sin||cos||||
i
i*
V
SI
Load-Flow Equations
• These are the static power flow equations. Each equation is complex,
and therefore we have 2n real equations. The nodal admittance matrix
current equation can be written in the power form:
)(||)(1 1
**
ininniin
N
n
N
n
niniiiii VVYVYVIVjQP
ijijijijijijijijij jBGYjYYY sin||cos||||
Let,
Load-Flow Equations
• Finally,
N
inn
inininniiiii YVVGVP1
2 )cos(||||
N
inn
inininniiiii YVVBVQ1
2 )sin(||||
o This is known as NR (Newton – Raphson) formulation
Newton-Raphson Algorithm
• The second major power flow solution method is the
Newton-Raphson algorithm.
• Key idea behind Newton-Raphson is to use sequential
linearization
General form of problem: Find an x such that
( ) 0ˆf x
Newton-Raphson Method
( )
( ) ( )
( )( ) ( )
2 ( ) 2( )
2
1. For each guess of , , define ˆ
-ˆ
2. Represent ( ) by a Taylor series about ( )ˆ
( )( ) ( )ˆ
1 ( )higher order terms
2
v
v v
vv v
vv
x x
x x x
f x f x
df xf x f x x
dx
d f xx
dx
Newton-Raphson Method
( )( ) ( )
( )
1( )( ) ( )
3. Approximate ( ) by neglecting all terms ˆ
except the first two
( )( ) 0 ( )ˆ
4. Use this linear approximation to solve for
( )( )
5. Solve for a new estim
vv v
v
vv v
f x
df xf x f x x
dx
x
df xx f x
dx
( 1) ( ) ( )
ate of x̂
v v vx x x
Example
2
1( )( ) ( )
( ) ( ) 2
( )
( 1) ( ) ( )
( 1) ( ) ( ) 2
( )
Use Newton-Raphson to solve ( ) - 2 0
The equation we must iteratively solve is
( )( )
1(( ) - 2)
2
1(( ) - 2)
2
vv v
v v
v
v v v
v v v
v
f x x
df xx f x
dx
x xx
x x x
x x xx
Example Solution
( 1) ( ) ( ) 2
( )
(0)
( ) ( ) ( )
3 3
6
1(( ) - 2)
2
Guess x 1. Iteratively solving we get
v ( )
0 1 1 0.5
1 1.5 0.25 0.08333
2 1.41667 6.953 10 2.454 10
3 1.41422 6.024 10
v v v
v
v v v
x x xx
x f x x
Comments
• When close to the solution the error decreases
quite quickly -- method has quadratic convergence
• Stopping criteria is when f(x(v)) <
• Results are dependent upon the initial guess. What
if we had guessed x(0) = 0, or x (0) = -1?
Multi-Variable Newton-Raphson
1 1
2 2
Next we generalize to the case where is an n-
dimension vector, and ( ) is an n-dimension function
( )
( )( )
( )
Again define the solution so ( ) 0 andˆ ˆ
n n
x f
x f
x f
x
f x
x
xx f x
x
x f x
x ˆ x x
Multi-Variable Case, cont’d
i
1 11 1 1 2
1 2
1
n nn n 1 2
1 2
n
The Taylor series expansion is written for each f ( )
f ( ) f ( )f ( ) f ( )ˆ
f ( )higher order terms
f ( ) f ( )f ( ) f ( )ˆ
f ( )higher order terms
nn
nn
x xx x
xx
x xx x
xx
x
x xx x
x
x xx x
x
Multi-Variable Case, cont’d
1 1 1
1 21 1
2 2 22 2
1 2
1 2
This can be written more compactly in matrix form
( ) ( ) ( )
( )( ) ( ) ( )
( )( )ˆ
( )( ) ( ) ( )
n
n
nn n n
n
f f f
x x xf x
f f ff x
x x x
ff f f
x x x
x x x
xx x x
xf x
xx x x
higher order terms
nx
Jacobian Matrix
1 1 1
1 2
2 2 2
1 2
1 2
The n by n matrix of partial derivatives is known
as the Jacobian matrix, ( )
( ) ( ) ( )
( ) ( ) ( )
( )
( ) ( ) ( )
n
n
n n n
n
f f f
x x x
f f f
x x x
f f f
x x x
J x
x x x
x x x
J x
x x x
Multi-Variable N-R Procedure
1
( 1) ( ) ( )
( 1) ( ) ( ) 1 ( )
( )
Derivation of N-R method is similar to the scalar case
( ) ( ) ( ) higher order termsˆ
( ) 0 ( ) ( )ˆ
( ) ( )
( ) ( )
Iterate until ( )
v v v
v v v v
v
f x f x J x x
f x f x J x x
x J x f x
x x x
x x J x f x
f x
Example
1
2
2 21 1 2
2 22 1 2 1 2
1 1
1 2
2 2
1 2
xSolve for = such that ( ) 0 where
x
f ( ) 2 8 0
f ( ) 4 0
First symbolically determine the Jacobian
f ( ) f ( )
( ) =f ( ) f ( )
x x
x x x x
x x
x x
x f x
x
x
x x
J xx x
Solution
1 2
1 2 1 2
11 1 2 1
2 1 2 1 2 2
(0)
1(1)
4 2( ) =
2 2
Then
4 2 ( )
2 2 ( )
1Arbitrarily guess
1
1 4 2 5 2.1
1 3 1 3 1.3
x x
x x x x
x x x f
x x x x x f
J x
x
x
x
x
Solution, cont’d
1(2)
(2)
2.1 8.40 2.60 2.51 1.8284
1.3 5.50 0.50 1.45 1.2122
Each iteration we check ( ) to see if it is below our
specified tolerance
0.1556( )
0.0900
If = 0.2 then we wou
x
f x
f x
ld be done. Otherwise we'd
continue iterating.
NR Application to Power Flow
*
* * *i
1 1
We first need to rewrite complex power equations
as equations with real coefficients
S
These can be derived by defining
Recal
i
n n
i i i ik k i ik kk k
ik ik ik
ji i i i
ik i k
V I V Y V V Y V
Y G jB
V V e V
jl e cos sinj
=
=
=
Power Balance Equations
* *i
1 1
1
i1
i1
S ( )
(cos sin )( )
Resolving into the real and imaginary parts
P ( cos sin )
Q ( sin cos
ikn n
ji i i ik k i k ik ik
k k
n
i k ik ik ik ikk
n
i k ik ik ik ik Gi Dik
n
i k ik ik ik ik
P jQ V Y V V V e G jB
V V j G jB
V V G B P P
V V G B
)k Gi DiQ Q
NR Power Flow
i1
In the Newton-Raphson power flow we use Newton's
method to determine the voltage magnitude and angle
at each bus in the power system.
We need to solve the power balance equations
P ( cosn
i k ik ikk
V V G
i1
sin )
Q ( sin cos )
ik ik Gi Di
n
i k ik ik ik ik Gi Dik
B P P
V V G B Q Q
Power Flow Variables
2 2 2
n
2
Assume the slack bus is the first bus (with a fixed
voltage angle/magnitude). We then need to determine
the voltage angle/magnitude at the other buses.
( )
( )
G
n
P P
V
V
x
x f x
2
2 2 2
( )
( )
( )
D
n Gn Dn
G D
n Gn Dn
P
P P P
Q Q Q
Q Q Q
x
x
x
N-R Power Flow Solution
( )
( )
( 1) ( ) ( ) 1 ( )
The power flow is solved using the same procedure
discussed last time:
Set 0; make an initial guess of ,
While ( ) Do
( ) ( )
1
End While
v
v
v v v v
v
v v
x x
f x
x x J x f x
Power Flow Jacobian Matrix
1 1 1
1 2
2 2 2
1 2
1 2
The most difficult part of the algorithm is determining
and inverting the n by n Jacobian matrix, ( )
( ) ( ) ( )
( ) ( ) ( )
( )
( ) ( ) ( )
n
n
n n n
n
f f f
x x x
f f f
x x x
f f f
x x x
J x
x x x
x x x
J x
x x x
Power Flow Jacobian Matrix,
i
i
i1
Jacobian elements are calculated by differentiating
each function, f ( ), with respect to each variable.
For example, if f ( ) is the bus i real power equation
f ( ) ( cos sin )n
i k ik ik ik ik Gik
x V V G B P P
x
x
i
1
i
f ( )( sin cos )
f ( )( sin cos ) ( )
Di
n
i k ik ik ik iki k
k i
i j ik ik ik ikj
xV V G B
xV V G B j i
Two Bus, Example
Line Z = 0.1j
One Two 1.000 pu 1.000 pu
200 MW
100 MVR
0 MW
0 MVR
For the two bus power system shown below, use the Newton-
Raphson power flow to determine the voltage magnitude and angle
at bus two. Assume that bus one is the slack and SBase = 100 MVA.
2
2
10 10
10 10bus
j j
V j j
x Y
Two Bus Example, cont’d
i1
i1
2 2 1 2
22 2 1 2 2
General power balance equations
P ( cos sin )
Q ( sin cos )
Bus two power balance equations
P (10sin ) 2.0 0
( 10cos ) (10) 1.0 0
n
i k ik ik ik ik Gi Dik
n
i k ik ik ik ik Gi Dik
V V G B P P
V V G B Q Q
V V
Q V V V
Two Bus Example, cont’d
2 2 2
22 2 2 2
2 2
2 2
2 2
2 2
2 2 2
2 2 2 2
P ( ) (10sin ) 2.0 0
( ) ( 10cos ) (10) 1.0 0
Now calculate the power flow Jacobian
P ( ) P ( )
( )Q ( ) Q ( )
10 cos 10sin
10 sin 10cos 20
V
Q V V
VJ
V
V
V V
x
x
x x
xx x
First Iteration
(0)
2 2(0)
22 2 2
2 2 2(0)
2 2 2 2
(1)
0Set 0, guess
1
Calculate
(10sin ) 2.0 2.0f( )
1.0( 10cos ) (10) 1.0
10 cos 10sin 10 0( )
10 sin 10cos 20 0 10
0 10 0Solve
1 0 10
v
V
V V
V
V V
x
x
J x
x
12.0 0.2
1.0 0.9
Next Iterations
(1)
2
(1)
1(2)
0.9(10sin( 0.2)) 2.0 0.212f( )
0.2790.9( 10cos( 0.2)) 0.9 10 1.0
8.82 1.986( )
1.788 8.199
0.2 8.82 1.986 0.212 0.233
0.9 1.788 8.199 0.279 0.8586
f(
x
J x
x
(2) (3)
(3)2
0.0145 0.236)
0.0190 0.8554
0.0000906f( ) Done! V 0.8554 13.52
0.0001175
x x
x
Two Bus Solved Values
Line Z = 0.1j
One Two 1.000 pu 0.855 pu
200 MW
100 MVR
200.0 MW
168.3 MVR
-13.522 Deg
200.0 MW 168.3 MVR
-200.0 MW-100.0 MVR
Once the voltage angle and magnitude at bus 2 are known we can
calculate all the other system values, such as the line flows and the
generator reactive power output
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
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 MVR170.0 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
N-R 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