Rajshahi University of Engineering & Technology
(RUET)
Course name: Power System I Sessional
Course no: EEE3212
Date of experiment: 22-01-18
Date of submission: 30-01-18
Experiment no: 07
Name of the experiment: Experimental study of power flow analysis.
Submitted by-
Name: MD. Naiem-ur-Rahman
Roll no: 141020
Section: A
Session: 2014-15
Experiment No. : 07
Name of the experiment: Experimental study of power flow analysis.
Objectives: The objective of this experiment is to calculate the voltages of the
busses.
Theory: Power flow studies, commonly known as load flow, form an important part
of power system analysis. They are necessary for planning, economic scheduling,
and control of an existing system as well as planning its future expansion. The
problem consists of determining the magnitudes and phase angle of voltages at each
bus and active and reactive power flow in each line.
In solving a power flow problem, the system is assumed to be operating under
balanced conditions and a single-phase model is used. Four quantities are associated
with each bus. These are voltage magnitude |V|, phase angle , real power P, and
reactive power Q. The system buses are generally classified into three types:
Slack bus: One bus, known as slack or swing bus, is taken as reference where the
magnitude and phase angle of the bus voltages are specified. This bus makes up the
difference between the scheduled loads and generated power that are caused by the
losses in the network.
Load buses: At these buses the active and reactive powers are specified. The
magnitude and the phase angle of the bus voltages are unknown. These buses are
called P-Q buses.
Regulated buses: These buses are the generator buses. They are also known as
voltage-controlled buses. At these buses, the real power and voltage magnitude are
specified. The phase angles of the voltages and the reactive power are to be
determined. The limits on the value of the reactive power are also specified. These
buses are called P-V buses.
The power flow equation is usually expressed in terms of the elements of the bus
admittance matrix. Since the off-diagonal elements of the bus admittance matrix Ybua
shown by uppercase letters, are Yij yij , and the diagonal elements are Yii y ij
The equations are as follows:
Pi sch jQisch
YijV j( k )
Vi*(k ) j i
Vi ( k 1) …………………………………………………..(7.1)
Yii
And
n
Pi ( k 1) {Vi *(k ) [Yii YijV j( k ) ]} j i …………………………………………(7.2)
j 1
j i
n
Qi( k 1) {Vi ( k ) [Vi ( k )Yii YijV j( k ) ]} j i ……………………………………..(7.3)
j 1
j i
Example 6.8:
Figure 7.1 shows the one-line diagram of a simple three-bus power system with
generators at buses 1 and 3. The magnitude of voltage at bus 1 is adjusted to 1.05
pu. Voltage magnitude at bus 3 is fixed at 1.04 pu with a real power generation of
200 MW. A load consisting of 400 MW and 250 Mvar is taken from bus 2. Line
impedances are marked in per unit on a 100 MVA base, and the line charging
susceptances are neglected. Obtain the power flow solution by the GAUSS-SEIDAL
method including line flows and line losses.
1 2
0.02+j0.04
400
MW
0.01+j0.03 0.0125+j0.025
250
Mvar
Slack Bus 3
V1=1.05 0
0
|V3|=1.04
200
MW
Fig 7.1: One line diagram of example 6.8
M file:
clc
clear
zdata=[1 2 .02 .04
1 3 .01 .03
2 3 .0125 .025];
nl=zdata(:,1);
nr=zdata(:,2);
R=zdata(:,3);
X=zdata(:,4);
nbr=length(zdata(:,1));
nbus=max(max(nl),max(nr));
Z=R+j.*X;
y=ones(nbr,1)./Z;
Y=zeros(nbus,nbus);
for k=1:nbr;
if nl(k)>0 && nr(k)>0
Y(nl(k),nr(k))=Y(nl(k),nr(k))-y(k);
Y(nr(k),nl(k))=Y(nl(k),nr(k));
end
end
for n=1:nbus
for k=1:nbr
if nl(k)==n | nr(k)==n
Y(n,n)=Y(n,n)+y(k);
end
end
end
BMVA=100;
vi=[1.05 0;
1 0;
1.04 0];
s=[0 0;
400 250;
200 0];
p=s(:,1)/BMVA;
q=s(:,2)/BMVA;
S(2)=-p(2)+i*q(2);
for m=1:7
vi(2)=((S(2)/conj(vi(2)))-Y(1,2)*vi(1)-Y(2,3)*vi(3))/Y(2,2);
Q(3)=-imag(conj(vi(3))*(vi(3)*Y(3,3)+Y(1,3)*vi(1)+Y(2,3)*vi(2)));
P(3)=200/BMVA;
S(3)=P(3)-i*Q(3);
vi(3)=((S(3)/conj(vi(3))-Y(3,1)*vi(1)-Y(3,2)*vi(2)))/Y(3,3);
w=angle(vi(3));
vi(3)=1.04*(cos(w)+i*sin(w));
v3(m)=vi(3);
v2(m)=vi(2);
Q(m)=Q(3);
end
v3
v2
Q
Output:
v3 =
Columns 1 through 4
1.0400 - 0.0052i 1.0400 - 0.0073i 1.0400 - 0.0083i 1.0400 - 0.0088i
Columns 5 through 7
1.0400 - 0.0089i 1.0400 - 0.0090i 1.0400 - 0.0090i
v2 =
Columns 1 through 4
0.9746 - 0.0423i 0.9711 - 0.0434i 0.9707 - 0.0448i 0.9707 - 0.0453i
Columns 5 through 7
0.9706 - 0.0456i 0.9706 - 0.0456i 0.9706 - 0.0457i
Q=
1.1600 1.3881 1.4608 1.4485 1.4563 1.4595 1.4608
Discussion: We compare the output with the calculated result and they matched. So
we can say that our program is right.
