1
Real-time Power System Operation
Network Modeling
1.0
Introduction
This section reviews some basic concepts which will be used throughout the
course to ease the course understanding. It covers, phasor representation, active
and reactive power derivation, Per Unit systems, network modeling and
development of matrices representing the network.
2.0
Phasor representation
Consider a general sinusoidal function f(t):
f(t)= Fmax cos(ωt+φ)=√ 2
F cos(ωt+φ)
(1)
There are three parameters associated with this function, Fmax, the maximum
value, ω which is the frequency and the φ which is the phase angle. F is called
the RMS value (Root Mean Square).
This function can be written in a different form using Euler's equation:
ejΘ= cos Θ + j sin Θ
(2)
Using this equation, it can be shown that f(t) can be written as:
f(t)= √ 2 Re [ F ej(ωt+φ)]= √ 2 Re [ Fejωtejφ ]= √ 2 Re [ Fpejωt]
(3)
where Fp = Fejφ is called the phasor representation of equation 3.
It can be shown that two sinosuidal functions with the same frequency can be
added up by vector additions of their phasors. The two network theories (
Kirchhof's voltage and current laws) can be applied using phasors as given
below:
Kirchhof's Voltage Law:
The phasor addition of voltage drops in a closed
circuit is equal to zero,
Kirchhof's Current Law:
The total phasor addition of all currents coming to
a node is equal to zero.
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In power system studies, phasors are used for representations of currents,
voltages, real power and reactive power. It can be shown that the real and
reactive power in a single circuit are given by:
P= |V|.| I |. Cos φ
Q= |V|.| I |. Sin φ
(4)
(5)
where φ is the phase angle between the current and the voltage. Equations 4 and
5 can be put in a complex form of :
S= P + J Q= |V| .| I |. Cos φ+ j |V |.| I |. Sin φ = V. I *
(6)
where I * is defined by:
I * =| I | . e-jΘi
(7)
where Θi is the phase angle of the current.
We can also State S in terms of Y or Z as given below:
S= P + J Q= V. I * =Z. | I |2= Y * . | V |2
(8)
For three phase balanced systems, it can be shown that the transmitted real
power is three times the real power per phase.
P3Φ=3 P = 3 | V |. | I |. Cos φ = √ 3. |VL |. | I |. Cos φ
(9)
where VL is the line voltage which is
| VL | =√ 3. | V |
(10)
Similar to equation (8) above the complex power for three phases is defined as:
S3Φ = 3 S= 3 V. I *
3.0
(11)
Per Unit Representation
In power system analysis, electrical quantities such as voltages, currents,
impedances and powers are expresses in per-unit values. Per unit system offers
several advantages:
1-
The data obtained have ranges which are not as diverse as the values
stated in ohms, amperes, kilovolts etc. For example in a power system
network there could be buses with voltage level changing from 12 kV all
the way to 500 KV. This range could reach two decades with values which
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are very different in magnitudes.
Computationally.
2-
This
situation
is
not
efficient
Per unit systems removes the confusion on phase power versus three
phase power, phase voltage versus line voltage and transformer primary
and secondary voltages.
In system studies it is more practical to choose a common base value for voltage
( Vb ) and MVA ( Sb ). Assuming that these values are phase quantities, the two
values can be used to obtain the base values for phase current ( Ib ):
Sb = Vb . Ib
(12)
The per unit values can be defined as:
Vpu= V/ Vb
Ipu= I / Ib
Spu=S / Sb
(13)
(14)
(15)
In terms of he chosen values we can define base impedance and base
admittance Zb and Yb:
Zb = Vb / Ib = Vb2/ Sb
(13)
Yb = 1/ Zb = Sb/ Vb2
(14)
and the per unit values are:
Zpu= Z / Zb = Z . Sb / Vb2
(15)
Ypu= Y / Yb = Y . Vb2/ Sb
(16)
Sometimes it is necessary to convert an impedance from one base system to
another. Using equation 15 one gets:
Zpu1= Z . Sb1 / Vb12
(17)
Zpu2=Z . Sb2 / Vb22
(18)
By eliminating Z, the conversion formula is obtained as:
Zpu2=Zpu1 . Sb2. Vb12 / Sb1. Vb22
4.0
Per Unit Representation for 3 Phase Systems
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(19)
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In three phase systems the base current can be calculated by the following
equation:
I base= S 3ph base/ √ 3 V Lbase= S 3ph base/ 3 V LGbase
(20)
for base values we have the following equations:
VLG base= VL base/ √ 3
(21)
VL pu= VL /VL base
(22)
VLG pu= VLG /VLG base
(23)
Knowing that VL = √ 3 VLG then,
VLG pu= VL pu
(24)
In a similar way, it can be shown that:
S1Ph pu= S =S3ph pu
(25)
If both sides of equation (11), is divided by S 3ph base, we get:
S3ph pu =3 V. I */
S3ph base =3 V. I */ 3 VLGbase Ibase
(26)
or
S3ph pu = VLGpu I*pu
(27)
Please note that in this equation, the factor three has disappeared.
5.0
Matrix Algebra
Matrix algebra is increasingly used in the formulation and solution of complex
engineering problems. Here a brief overview is presented to facilitate the
formulation of the application programs discussed in the following chapters.
A matrix is defined as a rectangular array of numbers, called elements, arranged
in a systematic manner with m rows and n columns. The elements can be real or
complex numbers represented as ai j where I designates the row and J
designates the columns as given below:
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A few different types of matrices are defined below:
Vector: A matrix with a single row or column is called a vector
Transpose of a Matrix: If the rows and columns of an m*n matrix are
interchanges, the resultant n*m matrix is the transpose and is designated by At .
Conjugate of a Matrix: If all the elements of a matrix are replaced by their
conjugates(replace the element a+jb by a-jb), the resultant matrix is the
conjugate and is designated by A*.
Two important matrix operations are described below:
Matrix Additions: Matrices of the same dimension can be added in the form of:
A+B=C where all elements of cij=aij+bij.
Matrix Multiplications: Matrix A (n*m) can be multiplied by B (m*q) resulting in a
matric C(n*q) where the alements of C are:
q
c ∑ a ik .b kj
=
ij
k =1
where I=1,.....m and j=1,.....n.
6.0
Steady State Component Modeling
In steady state analysis, the following components are the most common
elements which need to be modeled:
1- Transmission lines
2- Transformers and phase shifters
3- generators
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4- Shunts and Condensers
5- Loads
The modeling of these components is briefly described below.
6.1
Transmission Lines
Transmission lines are represented by a PI equivalent shown in Figure 1.
Figure 1: PI Equivalent of a Transmission line
The parameters for the PI equivalent are given by:
Y 12 =
1
Z 12
(30)
⎛ R + JWL ⎞
Z 12 = ⎜
⎟ Sinh (Ψd )
⎝ G + JWL ⎠
2
⎛ G + JWC ⎞
Y 12 = ⎜
⎟
⎝ R + JWL ⎠
1/ 2
(
tanh Ψd
(31)
2
)
where
R+JWL
G+JWC
d
Ψ
6.2
is impedance per mile
is shunt admittance per mile
is line length in miles and
is [(R+JWL)(G+JWC)]1/2
Transformers and Phase Shifters
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Transformers and phase shifters can be both modeled by using a complex ratio t
(Magnitude and angle) between the high and low voltage as given in Figure 2. If
the angle is zero then a transformer is modeled as opposed to a constant
magnitude which represents a phase shifter.
Figure 2: Transformer Model
To obtain the transformer PI equivalent similar to the one given in Figure 1, the
values for primary and secondary voltages and currents are related to each
other using the following equations:
I1 / t = (YL + YS )
I 2 = −t
V1t − YLV 2
YLV1 + YLV2
(33)
(34)
Comparing these equations with the ones obtained using the PI equivalent of
Figure 1, the following values can be obtained for the PI equivalent.
Y1 2 = YL t
YS 12 = t (t − 1)YL + T 2YS
YS 21 = (1 − t )YL
6.3
(35)
(36)
(37)
Generators
Only the boundary values of the generator terminal are enough for the generator
representation. These include the values of P and specified terminal voltage and
the maximum and the minimum values of reactive power i.e. Qmax and Qmin.
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6.4
Shunts and Condensers
Shunts and condensers produce reactive power. Shunts are represented by
their impedances. Condensers are represented as generators with zero power.
6.5
Loads
In steady state analysis loads are normally modeled by constant values for P and
Q assuming that they are not voltage dependent. In more elaborate
representations, however, loads can be represented as a function of V.
7.0 Network Equations
The network equations are developed by relating voltage and current at each
node giving:
[I]= [Y] [V]
where I and V are vectors containing voltages and currents at n nodes.
The admittance matrix Y is developed by :
Yii= Sum of all admittances connected to bus i
Yij= - Admittances between i and j
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9
Assignment 1
1. Add and multiply the two matrices A and B given below:
2
3 ⎞
⎛1
⎜
⎟
A=⎜ 0 1
4 ⎟
⎜ −1 −1 − 2⎟
⎝
⎠
1 ⎞
⎛ −1 0
⎜
⎟
B =⎜ 0 −2 1 ⎟
⎜5
0 − 3 ⎟⎠
⎝
2. For the three bus system given below,
a - Obtain the real values for P, Q, V and Z assuming that MVA base is 50
and voltage base is 120 KV
b - Obtain the Y matrix
© Ebrahim Vaahedi 2008