电力系统导论实验报告 本文关键词：导论，电力系统，实验，报告
电力系统导论实验报告 本文简介：ExperimentIntroductiontopowersystems学生姓名：学生姓名：学学号：号：专业班级：专业班级：实验名称实验名称：：电力系统导论（双语）电力系统导论（双语）20201414年年66月月55日日CONTENTS1、、EXPERIMENT1.1BUSADMITTANCEMAT
电力系统导论实验报告 本文内容：
Experiment
Introduction
to
power
systems
学生姓名：学生姓名：
学学
号：号：
专业班级：专业班级：
实验名称实验名称：：
电力系统导论（双语）电力系统导论（双语）
20201414
年年
6
6
月月
5
5
日日
CONTENTS
1、、EXPERIMENT
1.1
BUS
ADMITTANCE
MATRIX.1-6
2、、EXPERIMENT
2.6
BUS
IMPEDANCE
MATRIX6-13
3、、EXPERIMENT
3.13
GAUSS-SEIDEL
AND
NEWTON
METHOD13-16
4、、PERSENAL
SUMMARY.16
0
Experiment
1
Bus
Admittance
Matrix
1.
Objective
To
write
a
simple
program
in
MATLAB?
for
the
algorithm
of
bus
admittance
matrix.
2.
System
Requirement
Computer
with
MATLAB?
6
or
above
installed.
3.
Procedure
1.0
Launch
the
MATLAB
program.
2.0
Go
to
FILE
NEW
M-file.
3.0
Write
a
function
Y
=
The_Node_Admittance_Matrix(TopoStructureAndBranchPara)
for
the
formation
of
the
bus
admittance
matrix.
1
4.0
TopoStructureAndBranchPara
is
the
transmission
line,cable
and
transformer
input
data
and
contains
five
columns
parameters.
The
first
two
columns
are
the
line
bus
numbers
and
the
remaining
columns
contain
the
line
resistance
and
reactance
in
per-unit
and
transformer
tap
ratio
or
capacitor
of
transmission
line.
5.0
The
function
should
return
the
bus
admittance
matrix.
4.
Exercises
Use
the
written
function,Y
=
The_Node_Admittance_Matrix
(TopoStructureAndBranchPara)
to
obtain
the
Ybus
of
the
following
power
system
network:
Q1.
You
are
required
to
write
the
Ybus
topological
structure
and
parameter
into
a
text
file.
(Hint:
use
the
matlab
text
compiler
to
write
down
the
table
1
data,using
the
comma
to
separate
the
parameters,and
save
it
use
the
name
of
4_Power_System_Data.dbf)
Q2.
You
are
required
to
write
out
the
program
flow
figure
of
forming
a
nodal
admittance
matrix.
Hint.
You
are
required
to
compile
a
program
to
form
the
Ybus
Matrix,the
following
program
is
a
reference
program
to
you.
Figure
:
One-line
diagram
of
power
system
For
example,from
the
textbook
“power
system
analysis”
No.2
edition
3
on
page
61~62
NodalAdmittanceMatrix
=
1.0421
-
8.2429i
-0.5882
+
2.3529i
0
+
3.6667i
-0.4539
+
1.8911i
-0.5882
+
2.3529i
1.0690
-
4.7274i
0
0
Table
1：：Transformer
and
transmissssion
Line
data
From
Bus#To
Bus#R(p.u)X(p.u)B(p.u)or
ratio
KOthers
120.10.4j0.01528
1300.31.1
140.120.5j0.01920
240.080.40J0.01413
2
0
+
3.6667i
0
0
-
3.3333i
0
-0.4539
+
1.8911i
0
0
0.9346
-
4.2616i
5.The
flow
chart
Figure
:
The
flow
chart
of
Forming
Nodal
Admittance
Matrix
The
program
is:
%function
OutPut=The_Node_Admittance_Matrix(handles)
%is
a
subroutine
of
PowerSystemCalculation
function
OutPut=The_Node_Admittance_Matrix(handles)
%the
following
program
is
open
a
data
file
and
get
the
Number
of
%
Node
and
Branch
data
to
form
a
nodal
addmittance
matrix
%the
following
code
is
open
a
file
and
read
the
data
of
power
system
network
[fname,pname]
=
uigetfile(.dbf,Select
the
network
parametre
data-file
);
TopoStructureAndBranchPara=
csvread(fname);
[NumberOfBranch,NumberOfPara]=size(TopoStructureAndBranchPara);
Temporary1=max(TopoStructureAndBranchPara(:,1));
Temporary2=max(TopoStructureAndBranchPara(:,2));
if
Temporary1
Temporary2
NumberOfNode=Temporary1;
else
NumberOfNode=Temporary2;
end
%The
following
program
is
to
form
the
Nodal
Admittance
Matrix
%
and
the
Topologic
structure
and
Branch
Parametres
are
arranged
%
I,J,R,X,C/K,and
pay
attention
to
the
inpedence
of
transformer
is
in
the
%
side
of
Node
J
and
the
ratio
of
transformer
1:K
is
in
the
side
of
Node
I
for
CircleNumber1=1:NumberOfBranch
for
CircleNumber2=1:NumberOfBranch
NodalAdmittanceMatrix(CircleNumber1,CircleNumber2)=0;
end
end
for
CircleNumber=1:NumberOfBranch
if
TopoStructureAndBranchPara(CircleNumber,5)
0.85
NodalAdmittanceMatrix(TopoStructureAndBranchPara(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPara(CircleNumber,1)))=.
3
NodalAdmittanceMatrix(TopoStructureAndBranchPara(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPara(CircleNumber,1)))+.
TopoStructureAndBranchPara(CircleNumber,5)^2/.
(TopoStructureAndBranchPara(CircleNumber,3)+.
j*TopoStructureAndBranchPara(CircleNumber,4))
;
NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranchPa
ra(CircleNumber,2))=.
NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranchPa
ra(CircleNumber,2))+.
1/((TopoStructureAndBranchPara(CircleNumber,3)+j*TopoStructureAndBranchPara(CircleNumber,4
)));
NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa
ra(CircleNumber,2))=.
NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa
ra(CircleNumber,2)).
-TopoStructureAndBranchPara(CircleNumber,5)/.
((TopoStructureAndBranchPara(CircleNumber,3)+j*TopoStructureAndBranchPara(CircleNumber,4))
);
NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranchPa
ra(CircleNumber,1))=.
NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa
ra(CircleNumber,2));
else
NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa
ra(CircleNumber,1))=.
NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa
ra(CircleNumber,1))+.
+1/(TopoStructureAndBranchPara(CircleNumber,3)+.
Experiment
2
Bus
Impedance
Matrix
4
j*TopoStructureAndBranchPara(CircleNumber,4))+j*TopoStructureAndBranchPara(CircleNumber,5)
;
NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranchPa
ra(CircleNumber,2))=.
NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,2),TopoStructureAndBranchPa
ra(CircleNumber,2))+.
+1/(TopoStructureAndBranchPara(CircleNumber,3)+.
j*TopoStructureAndBranchPara(CircleNumber,4))+j*TopoStructureAndBranchPara(CircleNumber,5)
NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa
ra(
CircleNumber,2))=.
NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa
ra(
CircleNumber,2)).
-1/(TopoStructureAndBranchPara(CircleNumber,3)+.
j*TopoStructureAndBranchPara(CircleNumber,4));
NodalAdmittanceMatrix(TopoStructureAndBranchPara(
CircleNumber,2),TopoStructureAndBranchPara(CircleNumber,1))=.
NodalAdmittanceMatrix(TopoStructureAndBranchPara(CircleNumber,1),TopoStructureAndBranchPa
ra(
CircleNumber,2));
end
end
The
result
is：
NodalAdmittanceMatrix
=
1.0421
-
8.2429i
-0.5882
+
2.3529i
0
+
3.6667i
-0.4539
+
1.8911i
-0.5882
+
2.3529i
1.0690
-
4.7274i
0
0
0
+
3.6667i
0
0
-
3.3333i
0
-0.4539
+
1.8911i
0
0
0.9346
-
4.2616i
5
Experiment
2
Power
Grid
Bus
Impedance
Matrix
1.
Objective
?
To
write
a
simple
program
in
MATLAB?
for
the
algorithm
of
bus
impedance
matrix.
2.
System
Requirement
Computer
with
MATLAB?
6
or
above
installed.
3.
Procedure
1.0
Launch
the
MATLAB
program.
2.0
Go
to
FILE
NEW
M-file.
Experiment
2
Bus
Impedance
Matrix
6
3.0
Write
a
function
Z
=
znbus
(z)
for
the
formation
of
the
bus
impedance
matrix.
4.0
z
is
the
line
input
and
contains
three
columns.
The
first
two
columns
are
the
line
bus
numbers
and
the
remaining
columns
contain
the
line
resistance
in
per-unit.
5.0
The
function
should
return
the
bus
impedance
matrix.
4.
Exercises
Use
the
written
function,Z
=
znbus(z)
to
obtain
the
Ybus
of
the
following
power
system
network:
Example
1
Figure
3:
One-line
diagram
of
power
system
For
example,from
the
textbook
“power
system
analysis”
No.2
edition
3
on
page
61~62
Table
1：：Transformer
and
transmissssion
Line
data
From
Bus#To
Bus#R(p.u)X(p.u)B(p.u)or
ratio
KOthers
120.10.4j0.01528
1300.31.1
7
140.120.5j0.01920
240.080.40J0.01413
Q2.
You
are
required
to
write
the
Zbus
into
a
text
file.
(Hint:
use
the
matlab
text
compiler)
Example
2
For
the
system
shown,form
Zbus
matrix
using
the
building
algorithm
Solution
A
line
list
Experiment
2
Bus
Impedance
Matrix
8
Apply
Kron
reduction
to
eliminate
the
last
row
Hint.
You
are
required
to
compile
a
program
to
form
the
Zbus
Matrix.the
following
program
is
a
reference
program
to
you.
The
program
is:
%function
OutPut=The_Node_impedance_Matrix(handles)
%is
a
subroutine
of
PowerSystemCalculation
function
OutPut=The_Node_impedance_Matrix(handles)
%the
following
program
is
open
a
data
file
and
get
the
Number
of
%
Node
and
Branch
data
to
form
a
nodal
impedance
matrix
%the
following
code
is
open
a
file
and
read
the
data
of
power
system
network
[fname,pname]
=
uigetfile(.dbf,Select
the
network
parametre
data-file
);
Topo_Structure_And_Branch_Para=
csvread(fname);
%get
the
electric
power
system
the
number
of
branch
and
the
parametre
of
%
elements
9
[NumberOfBranch,NumberOfPara]=size(Topo_Structure_And_Branch_Para);
%Temporary1---temporary
variable
1
%Temporary2---temporary
variable
2
Temporary1=max(Topo_Structure_And_Branch_Para(:,1));
Temporary2=max(Topo_Structure_And_Branch_Para(:,2));
if
Temporary1
Temporary2
NumberOfNode=Temporary1;
else
NumberOfNode=Temporary2;
end
%
The
following
program
is
to
form
the
Nodal
impedance
Matrix
%
and
the
Topologic
structure
and
Branch
Parametres
are
arranged
%
I,J,R,X,C/K,and
pay
attention
to
the
inpedence
of
transformer
is
in
the
%
side
of
Node
J
and
the
ratio
of
transformer
1:K
is
in
the
side
of
Node
%
%
set
the
initial
value
of
Nodal
Admittance
Matrix
to
zero
for
CircleNumber1=1:NumberOfNode
for
CircleNumber2=1:NumberOfNode
Nodal_impedance_Matrix(CircleNumber1,CircleNumber2)=0;
end
end
for
CircleNumber=1:NumberOfBranch
if
Topo_Structure_And_Branch_Para(CircleNumber,5)
0.85
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(Topo_Structure_And_Branch_Para(Cir
cleNumber,1),Topo_Structure_And_Branch_Para(CircleNumber,1)))=.
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(Topo_Structure_And_Branch_Para(Cir
cleNumber,1),Topo_Structure_And_Branch_Para(CircleNumber,1)))+Topo_Structure_And_Branch_
Para(CircleNumber,5)^2/(Topo_Structure_And_Branch_Para(CircleNumber,3)+.
j*Topo_Structure_And_Branch_Para(CircleNumber,4))
;
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_And
_Branch_Para(CircleNumber,2))=.
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_And
_Branch_Para(CircleNumber,2))+.
1/((Topo_Structure_And_Branch_Para(CircleNumber,3)+j*Topo_Structure_And_Branch_Para(Circle
Number,4)));
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And
_Branch_Para(CircleNumber,2))=.
Experiment
2
Bus
Impedance
Matrix
10
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And
_Branch_Para(CircleNumber,2)).
-Topo_Structure_And_Branch_Para(CircleNumber,5)/.
((Topo_Structure_And_Branch_Para(CircleNumber,3)+j*Topo_Structure_And_Branch_Para(CircleN
umber,4)));
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_And
_Branch_Para(CircleNumber,1))=.
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And
_Branch_Para(CircleNumber,2));
else
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And
_Branch_Para(CircleNumber,1))=.
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And
_Branch_Para(CircleNumber,1))+.
+1/(Topo_Structure_And_Branch_Para(CircleNumber,3)+.
j*Topo_Structure_And_Branch_Para(CircleNumber,4))+j*Topo_Structure_And_Branch_Para(Circle
Number,5);
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_And
_Branch_Para(CircleNumber,2))=.
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,2),Topo_Structure_And
_Branch_Para(CircleNumber,2))+.
+1/(Topo_Structure_And_Branch_Para(CircleNumber,3)+.
j*Topo_Structure_And_Branch_Para(CircleNumber,4))+j*Topo_Structure_And_Branch_Para(Circle
Number,5)
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And
_Branch_Para(
CircleNumber,2))=.
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And
_Branch_Para(
CircleNumber,2)).
-1/(Topo_Structure_And_Branch_Para(CircleNumber,3)+.
11
j*Topo_Structure_And_Branch_Para(CircleNumber,4));
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(
CircleNumber,2),Topo_Structure_And_Branch_Para(CircleNumber,1))=.
Nodal_impedance_Matrix(Topo_Structure_And_Branch_Para(CircleNumber,1),Topo_Structure_And
_Branch_Para(
CircleNumber,2));
end
end
format
short
Nodal_impedance_Matrix*inv(Nodal_impedance_Matrix)
运行结果运行结果：：
Nodal_impedance_Matrix
=
1.0421e+000
-8.2429e+000i
-5.8824e-001
+2.3529e+000i
0
+3.6667e+000i
0
-5.8824e-001
+2.3529e+000i
5.8824e-001
-2.3377e+000i
0
0
0
+3.6667e+000i
0
0
-3.3333e+000i
0
0
0
0
4.5386e-001
-1.8719e+000i
Nodal_impedance_Matrix
=
1.0421e+000
-8.2429e+000i
-5.8824e-001
+2.3529e+000i
0
+3.6667e+000i
-
4.5386e-001
+1.8911e+000i
-5.8824e-001
+2.3529e+000i
1.0690e+000
-4.7274e+000i
0
0
0
+3.6667e+000i
0
0
-3.3333e+000i
0
-4.5386e-001
+1.8911e+000i
0
0
9.3463e-001
-4.2616e+000i
ans
=
1.0000
+
0.0000i
0.0000
+
0.0000i
0.0000
-
0.0000i
0.0000
-
0.0000i
-0.0000
-
0.0000i
1.0000
-
0.0000i
-0.0000
+
0.0000i
-0.0000
-
0.0000i
Experiment
2
Bus
Impedance
Matrix
12
-0.0000
-
0.0000i
-0.0000
-
0.0000i
1.0000
-
0.0000i
-0.0000
0
-
0.0000i
0
+
0.0000i
0.0000
-
0.0000i
1.0000
+
0.0000i
以上就是对阻抗矩阵的验证，其和其逆相乘为单位对角矩阵
ExperimentExperiment
3
3
Gauss-Seidel
Method
1.
Objective
?
To
write
a
simple
program
in
MATLAB?
for
the
algorithm
to
solution
of
nonlinear
algebraic
equations;
?
Known
as
the
method
of
successive
displacements.
2.
Discussion
The
most
common
methods
for
solving
nonlinear
algebraic
equations
are
Gauss-Seidel,Newtow-Rahpson,and
quasi-Newton-Raphson
methods.
We
start
with
one
dimensional
equations
and
then
generalize
to
n-dimensional
equations.
3.
Mathmatics
model
Consider
the
nonlinear
equation
.The
equation
is
broken
into
two
parts
thus:.
We
0)(?xf)(xgx
?
assume
is
an
initial
“guess“of
the
solution,then
“refine“the
solution
using:
)0(
x
)(
)0()1(
xgx?
This
process
is
repeated
thus
)(
)1()2(
xgx?
and
on
the
iteration
we
have:
th
n)(
)1()(?
?
nn
xgx
.
If
this
process
is
convergent,then
the
successive
solutions
approach
a
value
which
is
declared
as
the
solution.
Thus
if
at
some
step
we
have:
1?k
13
???
?)()1(kk
xx
where
e
is
the
desired
“accuracy“,then
we
claim
the
solution
has
been
found
to
the
accuracy
?
specified.
4.
System
Requirement
Computer
with
MATLAB?
6
or
above
installed.
5.
Procedure
1.0
Launch
the
MATLAB
program.
2.0
Go
to
FILE
NEW
M-file.
3.0
Write
a
function
program
of
Gauss
Seidel
Method.
6.
Exercises
Example:
Using
the
Gauss-Seidel
method
to
obtain
the
roots
of
the
equation:
0496)(
23
?????xxxxf
First
the
equation
is
expressed
in
a
different
form
thus
??)(46
9
1
23
xgxxx?????
Experiment
2
Bus
Impedance
Matrix
14
And
the
iteration
can
proceed.
Take
a
good
look
at
the
shape
of
the
iterations!
Below
is
the
program
showing
the
process
graphically
(later
showing
how
to
do
it
iteratively).
7.The
flow
chart
of
Gauss
Seidel
method
(Omitted)
8.Reference
Program
and
result.
程序是：程序是：
clear
all
clc
x0=0.5;
n=0;
while
(abs(x0^3-6*x0^2+9*x0-4)0.00001)
x0=-(x0^3-6*x0^2-4)/9;
y=x0;
n=n+1;
end
结果是结果是：n=1627
y=x0=0.99818
clear
all
clc
x0=2.5;
n=0;
while
(abs(x0^3-6*x0^2+9*x0-4)0.00001)
x0=-(x0^3-6*x0^2-4)/9;
y=x0;
n=n+1;
end
结果是结果是：n=7
y=x0=4
仿照高斯--赛德尔法，我们可以写出简单的牛顿法的程序,如下：
牛顿法解方程
x0=0.5;
n=0;
while
(abs((x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9))0.00001)
dx0=-(x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9);
x0=x0+dx0;
n=n+1;
end
结果是：dx0=1.7684e-005
n=15
x0=0.99998
y=
-0.875
x0=0.5;
15
n=0;
while
(abs(x0^3-6*x0^2+9*x0-4)0.00001)
dx0=-(x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9);
x0=x0+dx0;
n=n+1;
end
结果是:
dx0=0.0011305
n=9
x0=0.99887
y=
-0.875
x0=3.5;
n=0;
while
(abs(x0^3-6*x0^2+9*x0-4)0.00001)
dx0=-(x0^3-6*x0^2+9*x0-4)/(3*x0^2-12*x0+9);
x0=x0+dx0;
n=n+1;
end
结果是：dx0=
-2.5283e-006
n=5
x0=4
y=
-0.875
Personal
Summary：：
The
experiment
of
bilingual
class
is
over,here
is
my
personal
summary.
In
my
opinion,first
and
foremost,I
had
to
acknowledge
that
I
have
elementary
know
the
base
using
of
MATLAB,during
approximately
ten
hours’
hard
working.Although
I
have
spent
ten
hours
or
less
on
learning
this
software,I
merely
grasp
the
knowledge
which
is
just
like
the
tip
of
the
iceberg.In
terms
with
the
application
of
this
software,we
just
do
some
simple
steps.For
instance,the
node
admittance
matrix
of
node
impedance
matrix
in
date
input
software,the
experimental
program
input
last
run
results,the
corresponding
node
admittance
matrix
of
node
impedance
matrix
is
obtained.
Besides,I
would
say
I
haven’t
master
the
method
of
how
to
edit
a
program.This
is
a
pity.Nevertheless,I
think
it
is
just
a
program
of
time
.I’ll
pay
more
time
on
fulfill
a
deeper
study.
Last
but
not
least,I
am
really
appreciate
for
teacher’s
patient
teaching
and
conducting.Thank
you
very
much!
Because
the
experimental
time
coincides
with
exam
review
time,so
I
didn
t
understand
a
lot
of
knowledges
in
the
experimental
thoroughly,but
I
still
hope
the
teacher
can
forgive
me.