Lecture 14
Power Flows
Read: Chapter 6.1 – 6.6, 6.8 –6.11
Midterm Exam is on March 8st
Dr. Lei Wu
Department of Electrical and Computer Engineering
EE 333
POWER SYSTEMS ENGINEERING
2
IEEE Power and Energy Society—
Interest meeting--
Formation of a Clarkson Student Chapter
Wednesday March 7
5pm
CAMP 178
IEEE Power and Energy Society (PES) is the primary professional group for power engineers. Being a
member of a PES student chapter is an indication of your interest in power engineering. Members have
access to PES publications, particularly the Power and Energy Magazine, which reports on state of the art
issues in power engineering. A power engineering student chapter at Clarkson will allow us to hold power
engineering oriented meetings and projects, potential activities will be discussed at the meeting.
Outline
� Bus-branch model of power systems
� The power flow problem
� Iterative methods for solving nonlinear equations
� Approximations to the power flow problem
3
Bus-Branch Model of Power Systems
� A power system includes
� Loads
� Generators
� Transmission lines
� Transformers
4
Load Models
� Ultimate goal is to supply loads with electricity at constant
frequency and voltage
� Electrical characteristics of individual loads matter, but usually
they can only be estimated actual loads are constantly changing,
consisting of a large number of individual devices only limited
network observability of load characteristics
� Aggregate models are typically used for analysis
� Two common models
� constant power: Si = Pi + jQi
� constant impedance: Si = |V|2 / Zi
5
Generator Models
� Engineering models depend upon application
� Generators are usually synchronous machines
� For generators we will use two different models:
� a steady-state model, treating the generator as a constant power
source operating at a fixed voltage; this model will be used for
power flow and economic analysis
� a short term model treating the generator as a constant voltage
source behind a possibly time-varying reactance
6
Bus-branch model of power systems
� A power system includes
� Transmission lines
� Transformers
7
� Generators
� Loads
Power Flow Problem
� The most common power system analysis tool is the power flow
(also known sometimes as the load flow)
� Power flow determines how the power flows in a network under balanced
three-phase steady-state conditions.
� Also determines all bus voltages and all currents, as well as equipment
losses can be obtained
� Because of constant power models, power flow is a nonlinear problem
� Power flow is a steady-state analysis tool
� Each bus has 4 variables:
� P
� Q
� V
� δ
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� Bus types
� Reference bus (VΘ bus) : Usually select a bus with a large
generator, , calculate P and Q.
� PQ bus: load buses where P and Q are given, calculate V and δ
� PV bus: Voltage controlled buses ( typically are buses with
generators) where P and V are given, calculate Q and δ.
� Usually Q of a generator is limited by Qmin and Qmax, which are
dependent upon the generator's MW output
� In the calculation procedure
� if Q < Qmin, fix Q = Qmin
� If Q > Qmax, fix Q = Qmax
� "type-switching“: Change the bus type to PQ and continue the
calculation procedure.
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1 0o∠
� Bus types
� Reference bus (VΘ bus) : Usually select a bus with a large
generator, , calculate P and Q.
� PQ bus: load buses where P and Q are given, calculate V and δ
� PV bus: Voltage controlled buses ( typically are buses with
generators) where P and V are given, calculate Q and δ.
10
1 0o∠
Linear VS Nonlinear systems
� f(x) is a linear system as long as
� f(ax+by) = a*f(x) + b*f(y)
� The output is proportional to the input
� The principle of superposition holds
� Linear example f(x) = c*x
� Nonlinear example f(x) = cx2 , f(x) = c*sin(x)
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Linear Power System Elements
� Linear circuit elements, which may be analyzed by
superposition.
� Resistor
� Inductor
� Capacitor
� Independent voltage and current sources
V = R I
V =
1V =
j L I
Ij C
ω
ω
Nonlinear Power System Elements
� Constant power loads and generator injections are nonlinear
and hence systems with these elements can not be analyzed by
superposition
� Nonlinear problems can be very difficult to solve, and usually
require an iterative approach
� Gauss-Seidel (G-S)
� Newton-Raphson (N-R)
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Nonlinear Systems May Have Multiple Solutions or No Solution
� Example 1: x2 - 2 = 0 has real solutions x = ± 2
� Example 2: x2 + 2 = 0 has no real solution
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f(x) = x2 - 2 f(x) = x2 + 2
two solutions where f(x) = 0 no solution f(x) = 0
Gauss Iteration
� With the Gauss method, we need to rewrite the equation in an
implicit form x= h(x).
� To iterate, we first make an initial guess of x, x(0) , and then
iteratively solve x(k+1)= h(x(k)), until we find a fixed point, x, such
that x= h(x).
� Stopping criterion
15
( ) ( ) ( 1) ( )
( ) 22
1
i
If x is a scalar with
If x is a vector with
OR max x
v v v v
nv
iji
i
x x x x
x x
ε
ε
+
=
∞
∆ < ∆ −
∆ < ∆ = ∆
∆ = ∆
∑x
x
≜
Gauss Iteration Example
( 1) ( )
(0)
( ) ( )
Example: Solve - 1 0
1
Let = 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
v
v x v x
+
− =
= +
=
Power Flow Analysis - Gauss
� We know neither complex bus voltages nor complex current injections
� Rather, we know the complex power being consumed by the load, and
the power being injected by generators plus their voltage magnitudes
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bus=I Y V1
N
k kn nn
Y V=
= ∑I
* * *
1
N
k k k k k k kn nn
S P jQ V I V Y V=
= + = = ∑* * *
1
N
k k k k k k kn nn
S P jQ V I V Y V=
= − = = ∑* 1
*1 1 1
N k Nk
kn n kk k kn n kn nn n n kk
SY V Y V Y V Y V
V
−
= = = += = + +∑ ∑ ∑
* 1
*1 1
1 k Nk
k kn n kn nkk n n kk
SV Y V Y V
Y V
−
= = +
= − −
∑ ∑
Three-Bus Power Flow Example –Gauss-Seidel
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20 50 10 20 10 30
10 20 26 52 16 32
10 30 16 32 26 62bus
j j j
Y j j j
j j j
− − + − + = − + − − + − + − + −
V Θ P Q
Bus 1 1.05 0 ? ?
Bus 2 ? (1) ? (0) -4 -2.5
Bus 3 1.04 ? (0) +2 ?
Using the MVA base of 100MVA
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