# IMPEDANCE MATRIX USING MATLAB

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```					                                                                       Power System Simulation Laboratory

Exp No:1
Date: 31.12.2009

FORMATION OF Y BUS ADMITTANCE MATRIX AND BUS

IMPEDANCE MATRIX USING MATLAB

AIM:

To obtain the bus admittance matrix (Y bus) of the given network using MATLAB

THEORY:

The meeting point of various components in a power system is called bus. The bus or bus bar
is a conductor made of Cu or Al having negligible resistance. Hence the bus bar will have zero
voltage drop when it conducts the rated current. Therefore the buses are considered as points of
constant voltage in a power system.

When the power system is represented by impedance /reactance diagram it can be considered as a
circuit or network. The buses can be treated as nodes and the voltages of all buses (nodes) can be
solved by conventional node analysis technique.

Let N be the number of major or principal nodes in the circuit or network. Since the voltages of
node can be measured only with respect to reference point one of node is considered as reference
node. Now the network will have (N –I) independent voltages. In nodal analysis the independent
voltages are solved by writing Kirchhoff‟s current law(KCL) equations. For N-1 nodes in the circuit.
of writing KCL equations, the voltage sources in the circuit should be converted to equivalent current
sources.

Let V1,V2…..Vn. Node voltages of nodes 1,2…..n respectively.I11,I22 ,I33   …..Inn sum of current
sources connected to nodes 1,2,3…..n respectively.

Y jj = Sum of admittances connected to node.

Y   jk     = Negative of the sum of admittance connected between node j and node k. Now the N no of
nodal equations for n bus system will be in the form shown below

Y11V1+Y12V2+Y13V3+……..Y1nVn =I11

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Y21V1+Y22V2+Y23V3+……...Y2nVn=I22
…      …   …     …    …     …
…      …   …     …    …     …
Yn1V1+Yn2V2+Yn3V3+………..YnnVn=Inn

The above n‟ no of equations can be arranged in the matrix form as shown in below.

I BUS =YBUS VBUS

When I bus is the vector of injected bus current, the current is positive when flowing away from the
bus. Vbus is the vector of bus voltages measured from the reference node. bus is known as bus
admittance matrix. The diagonal elements of each node is the sum of admittance connected to it . It is

n= no of independent buses in the system.

YII                   j≠i

The off diagonal element is equal to the negative of admittance between the nodes. It is known as

Yij=Yji=-Yij

When bus currents are known equation(1) can be solved for the n bus voltages.

Vbus= Y -1bus Ibus.

The inverse of the bus admittance matrix is known as bus impedence matrix(Zbus).

The admittance matrix obtained with one of the buses as reference is non singular other wise the nodal
matrix is singular.

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If the transformer is present in the network, get the values for the transformer line impendence and
also off nominal turns ratio, and also get the line charging admittance is given in the network

OFF NOMINAL TURNS RATIO

When the voltage or turns ratio of the transformer is not used to decide the ratio of base KV then its
voltage or turns ratio.

Consider a transformer with turns ratio a:1. This can be represented as an idel autotransformer in series
with an admittance. Let p-q represents the input and output buses of the transformer. The ideal
autotransformer is shown between P and t buses, while series admittance is shown between t and q.

Itq=Current flowing from t to q

Itq=(Vt-Vq)Ypq               ___________         (1

The terminal current at P

Ip= (Vt-Vp) Ypq/a            __________         (2)

Vt= Vp/a                     __________          (3)

The terminal current at q is similarly

Iq= (Vq-Vt)Ypq                __________(4)

Substituting for VtIq=(Vq-Vp/a) Ypq

=(aVq-Vp) Ypq/a

Consider an equivalent π network mode for the transformer shown below

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For π network

Ip= (Vp-Vq)A+VpB          ______________(6)

Iq=(Vq-Vp)A+VqC            ______________(7)

Let Vp=0 and Vq=1 and substitute in (2) and (6)

Ip= - Ypq/a and Ip= -A

A= Ypq/a

Assuming Ep=0 and Eq =1 in equation (5) and (7)

Iq=Ypq      and           Iq=A+C

= Ypq/a +c

And hence C=(1-1/a) Ypq

Equating the current in equation (1) and (6) and substitute for A from equation

B=1/a(1/a-1) Ypq

Thus we obtain equivalent π model in term of admittance and OFF nominal turn ratio as in figure.

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LINE CHARGING

The line having a length between 50 to 150 Km and voltage s not exceeding 110 KV. In such lines
the line charging capacitance can be lumped either at both ends of the line or at the center of line to
give reasonably good analysis of line

.

If the given value of admittance is Ypq , the line charging admittance value is taken as Ypq/2 Write a
program to formulate bus admittance matrix (y bus) of the given network.

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Line No     Between buses      Impedance       Line         Charging Off     Nominal

1           1-2                0.025+j0.1      J0.1                   __

2           2-3                0.02+j0.08      j0.02                  __

3           3-4                0.05+j0.2            __               __

4           1-4                0.04+j0.16           __               0.92

Given Values

Number of buses = 4
Line impedance between bus 1-2 ,Z12 =0.025+j0.1 Ω
Line impedance between bus2-3, Z23 =0.02+j0.08 Ω
Line impedance between bus 3-4,Z34=0.05+j0.02 Ω
Line impedance between bus1-4,Z14=0.04+j0.16 Ω
Transformer impedance between buses 1 & 4
trz(1,4) =0-1.75j

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Off nominal turns ratio of transformer
a(1,4)=0.92
Generator impedance for bus (1)
g(1) =1+0.5j
lba(12) =j0.01

lca(2,3)   =j0.02

Calculation of YBus Matrix

Y12 = 1/Z12 =1/(0.025+j0.1)
= 2.3529-j9.4118
Y23=1/Z23 = 1/(0.02+j0.08)
= 2.3529-j9.4118
=2.9412-j11.7647
Y34 = 1/Z34 = 1/(0.05+j0.2)
=1.1765-j4.7059
Y14(Initial) = 1/(Z14+trZ14)
= 1/(0.04+j.16-j1.175)
= 0.0518+j0.6285
Y14 = Y14(Initial)/a(1,4)
= 0.0158+j0.6285/.92
=0.172+j0.6832
Y11=Y11+Y14(Initial)*(a(1,4)-1)/a(1,4)
= 0+(0.0158+j0.6285)*(0.92-1)/0.92
= -0.00137-j0.0547
Yg1= 1/(0,5j+1) =0.8-j0.4
Y11= Y11+Y12+Y14+Yg1+hrca12
=0.00149+j0.0594+2.3529-j9.4118+0.0172+j0.6832+(0.8-0.4j)+j.005

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= 3.1716-j9.0642
Y22=Y22+Y12+Y23+hra12+hrca23
=0+(2.3529-j9.4118)+(2.9412-j11.7647)+j0.005+j0.01
=5.2941-j21.1615
Y33=Y33+Y23+Y34+hrca23
=0+(2.9412-j11.7647)+(1.1765-j4.7059)+j0.01
=4.1177-j16.4606
Y44=Y44+Y14+Y34
(0.00137-j0.0547)+(0.0172+j0,6832)+(1.765-j4.7059)
=1.1923-j4.0774
Y12=Y21=-Y12
Y13+Y31=-Y13=0
Y14=Y41=-Y14= -0.0172-j0.6832
Y23=Y32= -Y23= _2.942-j11.7647
Y24=Y42= -Y24=0
Y34=Y43= -Y34= -1.1765+j4.7059

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ALGORITHM

Step 1: Get the total number of buses.

Step 2: Check whether generator or transformer or line charging admittance or a combination of
all these are present in the given network.

Step 3: Get the values of line impedances say from bus p to q for all buses in the network.

Step 4: If transformer is present in the network ,get the values for transformer line impedance and
also off nominal turns ratio.

Step 5: Find the diagonal and off diagonal admittance for the corresponding buses which are
connected to the transformer using following formulas.

Z(p,q)=Zpq+TrZ(p,q)

Ypq=1/Z(p,q)

Ypq=y(p,q)/a(p,q) where a(p,q) is the off nominal turns ratio.

Ypp=Ypp+{y p,q)(1-a(p,q))/[a(p,q)*a(p‟q)]}

Yqq=yqq+[y(p,q)*(a(p,q)-1)/a(p,q)]

Step 6: If generator is present get the value for generator impedance for the corresponding bus.

Y(p,p)=y(p,p)+1/g(p)

Step 7: Get the line charging admittance value given in the network. Connect the given values
into half line charging admittance by given value by two.

Step 8: Form the diagonal & off diagonal elements which can be formatted as bus admittance
matrix(Y bus)

Step 9: Print the resultant Y bus matrix.

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FLOWCHART

START

CHECK WHETHER THE GENERATOR OR TRANSFORMER
OR LINE CHARGING ADMITTANCE COMBINATION IS
PPRESENT IN THE NETWORK

GET THE VALUES OF THE LINE IMPEDENCE FROM BUS P
TO Q FOR ALL BUSES IN THE NETWORK

IF THE IMPEDANCE
TRANSFORMER IS
PRESENT IN THE
NETWORK

GET TRANSFORMER
LINE IMPEDANCE

`

FIND THE DIAGONAL AND OFF

1
1

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1
1

IF GENERATOR
PRESENT IN THE
NETWORK

GET GENERATOR IMPEDANCE

GET THE LINE CHARGING

FORM THE DIAGONAL AND OFF DIAGONAL ELEMENTS
WHICH CAN FORMULATE THE Y BUS

PRINT Y BUS

STOP

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Y BUS

Clc;

Clear all

n=input(„Enter the total number of buses=‟);

for j=1:n

for i=1:n

if(==j)

Z(i,j)=0;

Fprintf(„Enter the generator impedance G(%d):‟j);

G(j)=input(“);

G(i)=G(j);

Else

If (i>j)

Fprintf(„Enter the value of Z(%d,%d):‟j,i);

Z(j,i)=input(“);

Z(i,j)=Z(j,i);

L(j,i)=input(“);

L(i,j)=L(j,i);

fprintf(„Enter the off nominal turns ratio,A(%d,%d):‟j,i);

A(i,j)=input(“);

A(i,j)=A(j,i);

fprintf(„Enter the transformerimpedance,T(%d,%d):‟j,i);

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T(i,j)=input(“);

T(i,j)=T(j,i);

end

end

end

end

end

Z

for p=1:n

for q=1:n

if(p~=q&&Z(p,q)~=0&&A(p,q)~=0)

X(p,q)=1/Z(p,q)+T(p,q);

W(p,q)=X(p,q)/A(p,q);

Y(p,q)= -1*W(p,q);

Else

if(p~=q&&Z(p,q)~=0)

X(p,q)=1/Z(p,q)+T(p,q);

Y(p,q)= -1*X(p,q);

Else

Y(p,q)=0;

end

end

end

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end

for i=1:n

for p=1:n

for q=1:n

if(i==p&&p~=q&&i<q&&A(p,q)~=0&&G(p)~=0)

Y(i,p)=Y(i,p)+(X(p,q)*(1-A(p,q)/A(p,q)))+(1/G(p)+0.5*L(p,q)+W(i,q);

else

if(i==p&&p~=q&&i>q&&A(p,q)~=0G&&(p)~=0)

Y(i,p)=Y(i,p)+(X(p,q)*(A(p,q)-1)/A(p,q)))+(1/G(p)+0.5*L(p,q)+L(p,q);

else

if(i==p&&p~=q)

Y(i,p)=Y(i,p)+(X(p,q)+0.5*L(p,q)+L(p,q);

end

end

end

end

end

Y

Enter the total number of buses=4

Enter the generator impedence G(1):1+5i

Enter the value of Z(1,2):0.025+0.1i

Enter the off-nominal turns ratio,A(1,2):0

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Enter the transformer impedance,T(1,2):0

Enter the value of Z(1,3)

Enter the line charging admittance ,L(1,3):0

Enter the off-nominal turns ratio,A(1,3):0

Enter the transformer impedance,T(1,3):0

Enter the value of Z(1,4):.04+.16i

Enter the off –nominal turns ratio,A(1,4):.92

Enter the transformer impedance,T(1,4):-1.75i

Enter the generator impedance G(2):0

Enter the value of Z(2,3):.02+.08i

Enter the off-nominal turns rotio,A(2,3):0

Enter the transformer impedance,T(2,3):0

Enter the value of Z(2,4):0

Enter the off-nominal turns ratio,A(2,4):0

Enter the transformer impedance,T(2,4):0

Enter the generator impedance G(3):0

Enter the value of Z(3,4):.05+.2i

Enter the off-nominal turns ratio,A(3,4):0

Enter the transformer impedance,T(3,4):0

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Enter the generator impedance G(4):0

Z=

0                  0.0250+0.1000i        0                      0.0400+0.1600i

0.0250+0.1000i     0                             0.0250+0.8000i 0

0                       0.0200+0.0800i       0                          0.0500+0.2000i

0.0400+0.1600i      0                            0.0500+0.2000i     0

Y=

3.1716-9.0642i -2.3529+9.4118i               0                    -0.0172-0.6832i

-2.3529+9.4118i 5.2941-21.1615i              -2.9412+11.7647i      0

0            -2.9412+11.7647i            4.1176-16.4606i           -1.1765+4.7059i       -0.0172-0.6832i
0                -1.1765+4.7059i             1.1923-4.0773i

RESULT

Program to formulate bus admittance matrix is executed and results are verified with that of
sample calculation.

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Exp. No. 2
Date: 4.3.2010
SEQUENCE COMPONENTS OF POWER SYSTEM NETWORK
WITH SINGLE LINE TO GROUND FAULT USING MATLAB

a. DETERMINATION OF FAULT PARAMETERS

AIM:
To determine the fault current and fault voltage for a single line to ground fault on an unloaded
generator

THEORY:

A Fault in a circuit is any failure, which interferes its normal flow of current. The faults are
associated with abnormal change in current, voltage and frequency of the power system. The faults
may cause damage to the equipments if it is allowed to persist for a long time.

Faults are classified into symmetrical and unsymmetrical faults. In symmetrical faults, the fault
currents are equal in all the phases and can be analyzed on per phase basis. In unsymmetrical faults, the
fault currents are unbalanced and so they can be analyzed only using symmetrical components.

The three-phase fault is the only symmetrical fault. All other types of faults are unsymmetrical
faults. The various unsymmetrical faults are

1. line to ground fault
2. Line to line fault
3. Double line to ground fault
4. one or two open conductor faults
The fault condition of the power system can be divided into sub transient, transient and steady
state periods. The currents in the various parts of the system and in the fault are different in these
periods.

Since any unsymmetrical fault causes unbalanced currents to flow in the system, the
unsymmetrical faults are analyzed using symmetrical components.

Let Va, Vb, Vc = three phase unbalanced voltage vectors

Va1, Vb1, Vc1 = Positive sequence components of voltages

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Va2, Vb2, Vc2 = Negative sequence components of voltages

Va0, Vb0, Vc0 = zero sequence components of voltages

Now the unbalanced voltage vectors are related to sequence component voltages by the
following equations

Va0 = Vb0 = Vc0
Vb1 = a2 Va1                Vc1 = aVa1
Vb2 = aVa2                   Vc2 = a2 Va2
a = 1<1200                    a2 = 1<2400
Let Ia, Ib, Ic be a set of three phase unbalanced current vectors.

Let Ia1, Ib1, Ic1 = positive sequence components of current

Ia2, Ib2, Ic2 = negative sequence components of currents

Ia0, Ib0, Ic0 = zero sequence components of currents

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Ia0 = Ib0 = Ic0
Ib1 = a2 Ia1                       Ic1 = aIa1
Ib2 = aIa2                          Ic2 = a2 Ia2
a = 1<1200                         a2 = 1<2400

Problem:
Two 11kV, 20MVA, 3Phase, star connected generators operate in parallel as shown in figure.

The positive, negative and zero sequence reactances of each being respectively, j0.18, j0.15,
j0.1p.u. The star point of one of the generator is isolated and that of other is earthed through a 2-
Ohm resistor. A single line to ground fault occurs at the terminals of one of the generators.
Estimate

1. Fault current

2. Current in grounded resistor

3. Voltage across the grounding resistor

Solution

Let us choose the generator values as the base values

MVAb = 20MVA

kVb    = 11kV

Base impedance, Zb = kVb2 / MVAb

= 112 / 20      = 6.05 Ohm

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P.U value of neutral resistance

= Actual value / Base impedance

= 2/6.05

= 0.3306

The single line diagram and the sequence networks of the given power systems are shown in
figure.

Single line diagram

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The Thevenin‟s equivalent of the sequence network is as shown in figure

For a single line to ground fault

Ia1 = Ia2 = Ia0

If = Ia = 3Ia1

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Hence, the Thevenin‟s of a sequence networks are connected in series as shown in the figure.
The fault current is calculated by using the prefault voltage Vpf = 1 pu

From the figure we get

0
=

= 0.9741 <-150 pu

To find the fault current

Fault current, If      = Ia

= 3Ia1

= 3 *0.9741<-150

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= 2.9223<-150 pu

Base current,            Ib     =

=

= 1049.7A

Actual value of fault current

= p.u value of fault current * Base current

=2.9223<-150 * 1049.7

= 3067.5<-150A

= 3.0675<-150kA

To find the current through the neutral resistor

The current through the neutral resistor is same as that of the fault current

Current through the neutral resistor

= 2.9823<-150 pu

= 3.0675<-150 kA

To find the voltage across grounding resistor

From the Thevenin‟s equivalent of the zero sequence network we get

The voltage across grounding resistor

=3RnIa0

=3RnIa1

=3*0.3306*0.9741<-150 pu

=0.9661<-150 pu

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Actual value of the voltage across grounding resistor

= pu value of voltage * kVb/√3

Where kVb is the line value

= 0.9661<-150 * 11/√3

= 6.1356<-150 kV

ALGORITHM :

Step 1: Get the ratings of the generator power and voltage in MVA and kV respectively.

Step 2: Get the values of zero (X0) , positive (X1) and negative(X2) reactance‟s respectively

Step 3: Get the values of neutral to ground resistance from the generator (R).

Step 4:

Z b = ( Voltage ratings of the generator)2

(Power ratings of generator)

Step 5: Compute the per unit value of the neutral to ground resistance,

Rn =R/ Zb

Step 6 : Find the Thevenin equivalent reactance (Th)

Step 7 : Assume the prefault voltage

Vpf = 1 p.u

Step 8: Compute the single phase to ground fault current,

Ia1 = Vpf / Th

Step 9: Compute the fault current,

If1 = 3*Ia1

Step 10: Compute the base current,

Ib= (Power rating*106)/(√3*Voltage rating*103)

Step 11: Compute the actual fault current,

If = If1 * Ib

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Step 12 : Compute voltage across the grounding resistor,

Vgr = 3* Rn *Ia1

Step 13: Compute the actual voltage across the grounding resistor ,

Vagr = Vgr * (Voltage rating)/√3

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FLOWCHART

Start

Get the ratings of generator

Get the value of Symmetrical reactance

Get the value of neutral to ground resistance

Compute Base impedance

Compute per unit value of
neutral to ground resistance

Find the Thevenin equivalent reactance

Assume prefault voltage Vpf = 1p.u

Compute fault currents and voltages

Stop

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PROGRAM
clc;
clear all;
P=input('Enter the value of the power of generator in MVA:');
V=input('Enter the value of the voltage of generator in kV:');
X0=input('Enter the value of the zero sequence reactance:i');
X1= input('Enter the value of the positive sequence reactance:i');
X2=input('Enter the value of negative sequence reactance:i');
R=input('Enter the value of nutral to ground resistance from the generator:');
Zb=V^2/P;
Rn=R/Zb;
TX(1)=complex(0,X1/2);
TX(2)=complex(0,X2/2);
TX(3)=complex(3*Rn,X0);
Th=0;
for i=1:3
Th=Th+TX(i);
end
Vpf=1;
Ia1=Vpf/Th;
If1=3*Ia1;
Ib=(P*10^6)/(sqrt(3)*V*10^3);
If=If1*Ib;
fprintf('Actual fault current is');
If
I1=abs(If);
I2=angle(If);
Magnitude=I1
Angle=I2*180/pi
Vgr=3*Rn*Ia1;
Vagr=Vgr*V/sqrt(3);
fprintf('Actual value of voltage across grounding resistor is');
Vagr
Magnitude=abs(Vagr)
Angle=angle(Vagr)*180/pi

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OUTPUT:
Enter the value of the power of generator in MVA: 20

Enter the value of the voltage of generator in kV: 11

Enter the value of the zero sequence reactance: i0.1

Enter the value of the positive sequence reactance: i0.18

Enter the value of negative sequence reactance : i0.15

Enter the value of nutral to ground resistance from the generator: 2

Actual fault current is

If =

2.9638e+003 -7.9195e+002i

Magnitude =

3.0678e+003

Angle =

-14.9604

Actual value of voltage across grounding resistor is

Vagr =

5.9276 - 1.5839i

Magnitude =

6.1356

Angle =

-14.9604

RESULT:
Fault current and fault voltage for a single line to ground fault on an unloaded generator is
determined using MATLAB.

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Exp No: 3

Date: 18.2.2010

MODELING OF SINGLE MACHINE POWER SYSTEM USING

AIM:

To simulate swing equation for a single machine infinite bus system.

A 20 MVA, 50 Hz generator delivers 18 MW over a double circuit line to an infinite bus. The

generator has kinetic energy of 2.52 MJ/MVA at rated speed the generator transient reactance is Xd` =

0.35 pu. Each transmission circuit has R=0 and reactance of 0.2 pu on a 20 MVA base. | E`| = 1.1 pu

and infinite bus voltage V =1.0< 0°. A three phase short circuit occurs at the mid point of one of the

transmission lines. Plot swing curve with fault cleared by simultaneous opening of breakers at both

ends of the line at 2.5 cycles and 6.25 cycles after occurrence of fault. And also plot the swing curve

over a period of 0.5 seconds if the fault is sustained.

THEORY:

Under normal operating conditions the relative position of rotor axis and the resultant magnetic

field is fixed. The angle between the two is known as power angle or torque angle. During any

disturbance, rotor will decelerate or accelerate with respect to the synchronously rotating air gap mmf

and a relative motion begins. The equation describing this relative motion is known as swing equation.

Consider a synchronous generator developing an electromagnetic torque Te and running at the

synchronous speed ωsm. If Tm is the driving mechanical torque under steady state operations with

losses neglected we have Tm=Te.

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A departure from steady state due to a disturbance results in an accelerating (T m>Te) or decelerating

(Tm<Te) torque Ta on the rotor.

If J is the combined moment of the prime mover and the generator neglecting frictional and

damping torques. From law‟s of rotation we have

J d2 Өm / dt2 = Ta = Tm-Te

Where Өm is the angular displacement of the rotor with respect to the stationary reference axis

on the stator.

ωsm is the constant angular velocity

Өm= ωsm.t+δm

δm is the rotor position before disturbance at time t = 0

ωm= d Өm / dt = ωsm + d δm /dt

Rotor acceleration is

d2 Өm / dt2 = d2 δm / dt2

J d2 δm / dt2 = Tm-Te

J ωm d2 δm / dt2 = ωm Tm - ωm Te

Since angular velocity times torque is equal to the power, we right the above equation in terms of

power.

J ωm d2 δm / dt2 = Pm - Pe

The quantity J ωm is called the inertia constant and is denoted by M. It is related to kinetic energy of

the rotating masses, ωk

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ωk=(1/2)J ωm 2 = (1/2) M ωm

M= 2 ωk /m

Although M is called inertia constant, it is not really constant. When the rotor speed deviates from the

synchronous speed.

M = 2 ωk / ωsm

The swing equation in terms of the inertia constant becomes.

M d2 δm /dt2 = Pm - Pe

It is more convenient to write the swing equation in terms of the electrical power angle δ. If P is the

number of poles of a synchronous generator, the electrical power angle δ is related to mechanical

power angle δm by

δ = (p/2) δm

ω = (p/2) ωm

Swing equation in terms of electrical power angle is

(2/p) M d2 δ / dt2 = Pm - Pe

Since power system analysis is done in per unit system, the swing equation is usually expressed in per

unit. Dividing by the base power SB, and substituting for M.

(2/p)( 2ωk / ωsm) d2 δ / dt2 = (Pm / SB )–( Pe / SB )

H is a constant called per unit inertia constant.

H = (Kinetic energy in MJ at rated speed)/ Machine rating in MVA = ωk / SB

The unit of H is seconds.

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(2.2H/ Pωsm) d2 δ / dt2 = Pm (pu) – Pe (pu)

Pm (pu) and Pe (pu) are the per unit mechanical and electrical power. The electrical angular velocity is

related to the mechanical angular velocity

ωsm = (2/p)ωs

(2H/ ωs) d2 δ / dt2 = Pm (pu) – Pe (pu)

In terms of frequency F0 and to simplify the rotation the subscript pu is omitted and the powers are

understood to be in per unit.

(H/ Π F0) d2 δ / dt2 = Pm - Pe

Where δ is in electrical radians. If δ is expressed in electrical degrees, the swing equation becomes,

(H/180) d2 δ / dt2 = Pm - Pe

PROCEDURE:

1. Open Matlab - new-model file and save it

2. Click start button on Matlab- start-Simulink-library browser

3. Select the required blocks and elements from the library and drag it to the model file

4. Connect the blocks and save it.

5. Initialize the integrator 1 block with value of δ0 in rad (0.377 ≈ 21.6°)

6. Set simulation time (0.1 sec) and start simulation

7. Change the threshold values of switch from (2.5/50) to (6.5/50)

8. For sustained condition give another value higher than both

9. Click on the scope so that output is obtained

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Fig No : 1 Single Machine Infinite Bus System

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Fig No. 2 : Swing Curve for 2.5/50

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Fig No. 3 : Swing Curve for 6.5/50

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Fig No. 4 : Swing Curve with sustained oscillations.

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Fig No. 5: Scope output curve for Gain

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RESULT:

Swing curve is generated using simulink.

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Power System Simulation Laboratory

Exp No: 4

Date: 7.1.2010

SHORT CIRCUIT STUDIES OF POWER SYSTEM
USING Mi POWER
AIM:

To perform the short circuit analysis on a given 5- bus system using Mi Power and get the

values of short circuited parameters by using

A) Gauss- Siedel Method

B) Newton Raphson method

THEORY:

When an abnormal condition arises in a power system such as fault, an insulator flashover, or

lightning stroke to the transmission tower, high current flows in the power system. These currents are

sensed through the relays to isolate the faulty section of the power system. Delay in the operation of

the circuit breaker might result in creating instability in the system and also cause burn out of costly

equipments such as transformer, generator etc.

Short circuit study is an important study which provides vital information regarding the

magnitude of fault current through various components of the power system during short circuit. This

helps in proper selection of the circuit breaker and relays. It also helps in relay co-ordination.

In arriving at a mathematical model for short circuit studies, a number of assumptions are made

which simplify the formulation of the problem. Certain valid assumptions are:

1.         The load, line charge capacitances and other shunt connections to the ground are

neglected.

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2.         The generator is represented by a voltage source in series with a reactance which is

taken to be sub-transient or transient reactance.

3.         All the transformers are considered to be at their nominal taps

4.         If the resistance of the transmission lines are sufficiently smaller than the reactance

the resistance are neglected

5.         Pre-fault voltage at all the buses is assumed to be(1+j0) per unit

The mathematical model of the system can be derived as follows:

Fig(1).The figure shows the equivalent of the power system consisting of m generator buses

Voa       : Pre- fault voltage referred to phase “A”

Zi‟       : Generator Transient impedance, i referes to generator bus.

1,2,… m : Generator buses

n         : Total number of bus

o         : Ground bus

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Consider the voltage source Voa is shorted ie.nodes o and o‟ are shorted, the whole network is

reduced to a passive network. We may write the mathematical equation as

[Vbus]   = [Zbus][Ibus]                     …………………………(1)

Now introduce the voltage source between o and o‟. This results in raising all the bus voltage

by Voa. Hence ,we may write

[Vbus]            = [Zbus][Ibus] + bVoa      ………………………(2)

Where b           = [1,1,………1]TSuppose that a three phase fault occurs at pth bus. The voltages

are as shown in Figure (2).
   1

   2





V1(F)                                                                                       P

   n
V2(F)
   o
ZF          VP(F)                                 
FF

Vn(F)

Fig(2)
   P

   n

   o

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Zf              = Fault Impedance

If             = Fault Current

Ip(F)          = Current injected at bus „p‟

It is clear from the above figure (2) that

If        = -Ip(F)          ……………………….(3)

Vf        = Vp(F) =If.Zf    ………………………..(4)

Ii(F) = 0; Vi(F) = unknown for i = 1,2,……….n, I # P.

This follows from the fact that the venin‟s theorem is applied to n –port network shown in the

above figure. Expanding equation 2 for n port description.

V1(F) = Z11I1(F) + ………………..+ Z1PIp(F) +…………..+Z1nIn(F) + Voa

V2(F) = Z21I1(F) + ………………+ Z2PIP(F) + …………….+Z2nIn(F) + Voa

VP(F) = ZP1I1(F) +…………..+ ZPPIP(F) + …………….+ZPnIn(F) + Voa

Vn(F) = Z21I1(F) +……………+Z2PIP(F) + ……………+ ZnnIn(F) + Voa

Using the results 3&4 in the above equation, we get

Zf IF      = -ZPPIF + Voa

IF    =         Voa
ZPP + ZF
Therefore VF       = ZF . Voa       = VP(F)
ZPP + ZF

The other bus voltages are obtained from the rest of the equations in equation (5).

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Vi(F) =    Voa -     ZiP    . Vo
a

ZPP + ZF

Vi(F) = Voa         1-     ZiP      where i= 1,2,………………..n
ZPP + ZF

Knowing the fault current (IF) and all the bus voltages, the line currents and generator for
currents can be computed.

Algorithm:

Step 1: Start

Step 2: Read line data, machine data, and transformer data, Fault impedance, etc.

Step 3: Compute Ybus matrix and calculate modified Ybus matrix.

Step 4: Form(Zbus) by inverting the modified Ybus matrix.

Step 5: Initialize count 1=0.

Step 6: I = I+1. This means fault occurs a bus‟I‟.

Step 7:Compute fault current at faulted bus and bus voltages at all buses.

Step 8: Compute all line currents and generator currents.

Step 9: Check whether 1 is less than the number of buses, if so go to step-6 otherwise go to next
step.

Step 10: Print the results.

Step 11: Stop.

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FLOW CHART

START

data, Fault impedance,etc.

Compute Y bus matrix & modified Y bus matrix

I=0

I=I+1 Fault occurs at bus I

Compute fault current at faulted bus
and bus voltages at all buses.

Compute all line currents and
generator currents

IS                      yes
I<nb?
No

Print the results

STOP

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The figure shows the single line diagram of a 6-bus system with two identical generating units,

five lines and two transformers. Per unit transmission line series impedances and shunt susceptances

are given on 100 MVA base, generator‟s transient impedance and transformer leakage reactances are

given in the accompanying table. If a fault occurs in bus 5-4, find the fault MVA. The data is given

below:

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If a 3 phase to ground fault occurs at bus 5   find the fault MVA. The data is given below.

BUS CODE                IMPEDENCE Zpq             LINE CHARGING

P-q                      Zpq                     Y’pq/2

3-4                  0.00+j0.15                     0

3-5                  0.00+j0.10                     0

3-6                   0.00+j0.2                     0

5-6                  0.00+j0.15                     0

4-6                  0.00+j0.10                     0

Generator details:

G1=G2=100 MVA,11 KV with Xd‟ =10%

Transformer details:

T1=T2=11/110 KV, 100 MVA, leakage reactance=x= 5%

All impedances are on 100 MVA base

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PROCEDURE:

The software Mi power introduces a better interface between the human and the computer for

the easier calculation of power flow problems like fault analysis, load flow analysis,

contingency occurrence etc.

PROCEDURE TO ENTER THE DATA:

Open Power system network editor. Select menu option Data base                 Configure. Click

Browse button. Open dialog box is popped up, where we can browse the desired directory and specify

the name of the database to be associated with the single line diagram . Click the Open button after the

entering the desired database name. Configure Database dialog will appear with path chosen. Click

on OK button in the Configure database dialog, the following configuration information appears.

Uncheck the power system libraries and Standard Relay Libraries.

1. To draw the single line diagram:

To draw bus : Click on Bus icon provided on power system tool bar. Draw a bus and give

Bus ID and Bus Name. Click OK

Bus Data

Bus No                      1      2         3      4          5          6

Bus Name                    Bus1 Bus2        Bus3   Bus4       Bus5       Bus6

Nominal Voltage             11     11        110    110        110        110

Area Number                 1      1         1      1          1          1

Zone number                 1      1         1      1          1          1

Contingency weightage       1      1         1      1          1          1

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2.To Draw Transmission Line:

Click on Transmission Line icon provided on power system tool bar.Enter Element ID No

and click OK. Enter structure ref No. and click on Transmission Line Library button. Enter

Transmission Line Library data.

Transmission Line Element Data

Line Number          1           2               3           4             5

Line Name       Line 3-4     Line 3-5      Line 3-6      Line 4-6      Line 5-6

De-Rated          100         100             100         100           100
MVA

No. of circuits       1           1               1           1             1

From Bus No.          3           3               3           4             5

To Bus No.           4           5               6           6             6

Line Length          1           1               1           1             1

From Breaker         5000       5000             5000        5000         5000
Rating

To Breaker         5000       5000             5000        5000         5000
Rating

Structure Ref.        1           2               3           2             1
No.

Transmission Line Library Data

Structure Ref. No.                          1                 2             3

Structure Ref. Name                   Line 3-4 & 5-6    Line 3-5 & 4-6   Line 3-6

Positive sequence resistance                0                 0             0

Positive sequence reactance               0.15               0.1           0.2

Positive sequence susceptance               0                 0             0

Thermal Rating                             100               100           100

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3.To Draw Transformer

Click on Two Winding Transformer icon provided on power system Tool bar. Connect the

transformer in between two buses. Then Element ID dialog will appear. Click OK.

Transformer Element         Data form will be open. Enter the Manufacturer Ref.No. Enter

Transformer Data. Click on Transformer Library button. Enter the data. Save and close

Library screen.

2nd Transformer details

Transformer Number                          2

Transformer Name                            2T2

From Bus Number                             6

To Bus Number                               2

Control Bus Number                          2

Number of units in Parallel                 1

Manufacturer ref.Number                     30

De-Rated MVA                                100

From Breaker Rating                         5000

To Breaker Rating                           350

Nominal Tap Position                        5

4.To Draw Generator : Click on Generator icon provided on power system Tool Bar. Draw

the generator on the bus 1. Element ID dialog will appear. Click OK. Generator Data form

will be opened. Enter the Manufacturer ref.No. Enter Generator Data. Click on Generator

Library button. Enter Generator Library details. Save and close Library screen.

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Name                                       GEN-2

Bus Number                                 2

Manufacturer Ref. Number                   20

Number of Generators in parallel           1

Capability Curve number                    0

De-rated MVA                               100

Specified voltage                          11

Scheduled Power                            80

Reactive Power minimum                     0

Reactive Power maximum                     60

Breaker rating                             350

Type of modeling                           Infinite

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OUTPUT:

I     SHORT CIRCUIT STUDIES
CASE NO : 1          CONTINGENCY : 0 SCHEDULE NO : 0
CONTINGENCY NAME : Base Case
-------------------------------------------------------------------------------
LARGEST BUS NUMBER USED                                 :       6       ACTUAL NUMBER OF BUSES               :       6
NUMBER OF 2 WIND. TRANSFORMERS :                                                   2       NUMBER OF 3 WIND.
TRANSFORMERS : 0
NUMBER OF TRANSMISSION LINES :                                          5
NUMBER OF SERIES REACTORS                                   :       0       NUMBER OF SERIES CAPACITORS              :       0
NUMBER OF BUS COUPLERS                              :           0
NUMBER OF SHUNT REACTORS                                    :       0 NUMBER OF SHUNT CAPACITORS                         :   0
NUMBER OF SHUNT IMPEDANCES                                      :       0    NUMBER OF GENERATORS                :       2
NUMBER OF MOTORS                            :       0
NUMBER OF FILTERS                       :       0
NUMBER OF HVDC CONVERTORS                                       :       0

-------------------------------------------------------------------------------
NUMBER OF ZONES                   : 1
PRINT OPTION                 : 3 (BOTH DATA AND RESULTS PRINT)
PLOT OPTION                  : 0 (NO PLOT FILE GENERATION)
BASE MVA                   : 100.000
NOMINAL SYSTEM FREQUENCY: 50.000
PREFAULT VOLTAGE OPTION: 0 (VOLTAGE OF 1.0 PU IS ASSUMED)
FAULT OPTION                    : 1 (FAULT CONSIDERED AT SELECTED BUSES, ONE AT A
TIME)
FLOW OPTION                   : 0 (NO FAULT CONTRIBUTIONS ARE COMPUTED)

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FAULT TYPE                        : 4 (DOUBLE LINE TO GROUND FAULT FROM Y AND B
PHASES)
POST FAULT VOLT OPTION : 0 (NO COMPUTATION)
-------------------------------------------------------------------------------
TOTAL FAULT BUSES : 2
BUSES :      2     4
-------------------------------------------------------------------------------
FAULT RESISTANCE - PHASE - 0.000000 (PU)
FAULT REACTANCE - PHASE - 0.000000 (PU)
FAULT RESISTANCE - GROUND - 0.000000 (PU)
FAULT REACTANCE - GROUND - 0.000000 (PU)

II
SHORT CIRCUIT STUDIES
CASE NO : 1            CONTINGENCY : 0 SCHEDULE NO : 0
CONTINGENCY NAME : Base Case
-------------------------------------------------------------------------------
LARGEST BUS NUMBER USED                                 :       6       ACTUAL NUMBER OF BUSES               :       6
NUMBER OF 2 WIND. TRANSFORMERS :                                                   2       NUMBER OF 3 WIND.
TRANSFORMERS : 0
NUMBER OF TRANSMISSION LINES :                                          5
NUMBER OF SERIES REACTORS                                   :       0       NUMBER OF SERIES CAPACITORS              :       0
NUMBER OF BUS COUPLERS                              :           0
NUMBER OF SHUNT REACTORS                                    :       0 NUMBER OF SHUNT CAPACITORS                         :       0
NUMBER OF SHUNT IMPEDANCES                                      :       0    NUMBER OF GENERATORS                :       2
NUMBER OF MOTORS                            :       0

NUMBER OF FILTERS                       :       0
NUMBER OF HVDC CONVERTORS                                       :       0
-------------------------------------------------------------------------------
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NUMBER OF ZONES                   : 1
PRINT OPTION                 : 3 (BOTH DATA AND RESULTS PRINT)
PLOT OPTION                  : 0 (NO PLOT FILE GENERATION)
BASE MVA                   : 100.000
NOMINAL SYSTEM FREQUENCY: 50.000
PREFAULT VOLTAGE OPTION : 0 (VOLTAGE OF 1.0 PU IS ASSUMED)
FAULT OPTION                    : 1 (FAULT CONSIDERED AT SELECTED BUSES, ONE AT A
TIME)
FLOW OPTION                   : 0 (NO FAULT CONTRIBUTIONS ARE COMPUTED)
FAULT TYPE                  : 7 (3 PHASE AND SLG, ONE AFTER THE OTHER)
POST FAULT VOLT OPTION : 0 (NO COMPUTATION)
-------------------------------------------------------------------------------
TOTAL FAULT BUSES : 1
BUSES :      3
-------------------------------------------------------------------------------
FAULT RESISTANCE - PHASE - 0.000000 (PU)
FAULT REACTANCE - PHASE - 0.000000 (PU)
FAULT RESISTANCE - GROUND - 0.000000 (PU)
FAULT REACTANCE - GROUND - 0.000000 (PU)
-------------------------------------------------------------------------------
FAULT AT BUS NUMBER                   1 : NAME        Bus1
CURRENT (AMPS/DEGREE)                                     FAULT MVA
SEQUENCE (1,2,0) PHASE (A,B,C)                       SEQUENCE (1,2,0) PHASE (A,B,C)
MAGNITUDE ANGLE MAGNITUDE ANGLE                                      MAGNITUDE              MAGNITUDE
--------- ------- --------- -------      ---------      ---------
3572 -89.73          3572 -89.73              1361               1361
0 -90.00        3572 150.27                0           1361
0 -90.00        3572 30.27                 0           1361
R/X RATIO OF THE SHORT CIRCUIT PATH                             : 0.0048
PEAK ASYMMETRICAL SHORT-CIRCUIT CURRENT :                                    10051 AMPS

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PASCC = k x sqrt(2) x If , k = 1.9897
-------------------------------------------------------------------------------
FAULT AT BUS NUMBER                   2 : NAME            Bus2
CURRENT (AMPS/DEGREE)                                        FAULT MVA
SEQUENCE (1,2,0) PHASE (A,B,C)                           SEQUENCE (1,2,0) PHASE (A,B,C)
MAGNITUDE ANGLE MAGNITUDE ANGLE                                          MAGNITUDE             MAGNITUDE
--------- ------- --------- -------      ---------         ---------
71443 -89.73         71443 -89.73                 1361               1361
0 -90.00       71443 150.27                    0            1361
0 -90.00       71443 30.27                 0               1361
R/X RATIO OF THE SHORT CIRCUIT PATH                               : 0.0048
PEAK ASYMMETRICAL SHORT-CIRCUIT CURRENT : 201026 AMPS
PASCC = k x sqrt(2) x If , k = 1.9897
-------------------------------------------------------------------------------
FAULT AT BUS NUMBER                   3 : NAME            Bus3
CURRENT (AMPS/DEGREE)                                        FAULT MVA
SEQUENCE (1,2,0) PHASE (A,B,C)                           SEQUENCE (1,2,0) PHASE (A,B,C)
MAGNITUDE ANGLE MAGNITUDE ANGLE                                          MAGNITUDE             MAGNITUDE
--------- ------- --------- -------      ---------         ---------
5813 -89.17          5813 -89.17              1108                  1108
0 -90.00        5813 150.83                0               1108
0 -90.00        5813 30.83                 0               1108
R/X RATIO OF THE SHORT CIRCUIT PATH                               : 0.0144
PEAK ASYMMETRICAL SHORT-CIRCUIT CURRENT :                                       16187 AMPS
PASCC = k x sqrt(2) x If , k = 1.9689
-------------------------------------------------------------------------------
FAULT AT BUS NUMBER                   4 : NAME            Bus4
CURRENT (AMPS/DEGREE)                                        FAULT MVA
SEQUENCE (1,2,0) PHASE (A,B,C)                           SEQUENCE (1,2,0) PHASE (A,B,C)

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MAGNITUDE ANGLE MAGNITUDE ANGLE                                          MAGNITUDE              MAGNITUDE
--------- ------- --------- -------      ---------          ---------
3871 -89.47          3871 -89.47                  738                738
0 -90.00        3871 150.53                0                 738
0 -90.00        3871 30.53                 0                738
R/X RATIO OF THE SHORT CIRCUIT PATH                                : 0.0092
PEAK ASYMMETRICAL SHORT-CIRCUIT CURRENT :                                        10840 AMPS
PASCC = k x sqrt(2) x If , k = 1.9801
-------------------------------------------------------------------------------
FAULT AT BUS NUMBER                   5 : NAME             Bus5
CURRENT (AMPS/DEGREE)                                         FAULT MVA
SEQUENCE (1,2,0) PHASE (A,B,C)                        SEQUENCE (1,2,0) PHASE (A,B,C)
MAGNITUDE ANGLE MAGNITUDE ANGLE                                          MAGNITUDE              MAGNITUDE
--------- ------- --------- -------      ---------          ---------
3871 -89.47          3871 -89.47                  738                738
0 -90.00        3871 150.53                0                 738
0 -90.00        3871 30.53                 0                738
R/X RATIO OF THE SHORT CIRCUIT PATH                                : 0.0092
PEAK ASYMMETRICAL SHORT-CIRCUIT CURRENT :                                        10840 AMPS
PASCC = k x sqrt(2) x If , k = 1.9801
-------------------------------------------------------------------------------
FAULT AT BUS NUMBER                   6 : NAME             Bus6
CURRENT (AMPS/DEGREE)                                         FAULT MVA
SEQUENCE (1,2,0) PHASE (A,B,C)                        SEQUENCE (1,2,0) PHASE (A,B,C)
MAGNITUDE ANGLE MAGNITUDE ANGLE                                          MAGNITUDE              MAGNITUDE
--------- ------- --------- -------      ---------          ---------
5813 -89.17          5813 -89.17              1108                   1108
0 -90.00        5813 150.83                0                1108
0 -90.00        5813 30.83                 0                1108

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R/X RATIO OF THE SHORT CIRCUIT PATH                                : 0.0144
PEAK ASYMMETRICAL SHORT-CIRCUIT CURRENT :                                     16187 AMPS
PASCC = k x sqrt(2) x If , k = 1.9689
-------------------------------------------------------------------------------
3 phase fault level
Bus No. Name          BUS kV 3PH-fMVA Fault I
NOMINAL                  kA
------- -------- -------- -------- --------
1     Bus1 220.000 1361.2              3.572
2     Bus2 11.000 1361.2 71.445
3     Bus3 110.000 1107.6              5.814
4     Bus4 110.000        737.5        3.871
5     Bus5 110.000        737.5        3.871
6    Bus6 110.000 1107.6              5.814
-------------------------------------------------------------------------------
-------------------------------------------------------------------------------
FAULT AT BUS NUMBER                    1 : NAME         Bus1
CURRENT (AMPS/DEGREE)                                      FAULT MVA
SEQUENCE (1,2,0) PHASE (A,B,C)                         SEQUENCE (1,2,0) PHASE (A,B,C)
MAGNITUDE ANGLE MAGNITUDE ANGLE                                       MAGNITUDE            MAGNITUDE
--------- ------- --------- -------        ---------     ---------
0 -90.00        0 -90.00                   0           0
0 -90.00        0 180.00                   0           0
0 -90.00        0 180.00                   0           0
-------------------------------------------------------------------------------
FAULT AT BUS NUMBER                    2 : NAME         Bus2
CURRENT (AMPS/DEGREE)                                      FAULT MVA
SEQUENCE (1,2,0) PHASE (A,B,C)                         SEQUENCE (1,2,0) PHASE (A,B,C)
MAGNITUDE ANGLE MAGNITUDE ANGLE                                       MAGNITUDE            MAGNITUDE

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--------- ------- --------- -------      ---------          ---------
1 -90.00          2 -90.00               0                  0
1 -90.00          0 180.00               0                  0
1 -90.00          0 180.00               0                  0
-------------------------------------------------------------------------------
FAULT AT BUS NUMBER                   3 : NAME             Bus3
CURRENT (AMPS/DEGREE)                                         FAULT MVA
SEQUENCE (1,2,0) PHASE (A,B,C)                        SEQUENCE (1,2,0) PHASE (A,B,C)
MAGNITUDE ANGLE MAGNITUDE ANGLE                                            MAGNITUDE            MAGNITUDE
--------- ------- --------- -------      ---------          ---------
2953 -88.55          8860 -88.55                  563              1688
2953 -88.55            0    0.00           563                     0
2953 -88.55            0    0.00           563                     0
-------------------------------------------------------------------------------
FAULT AT BUS NUMBER                   4 : NAME             Bus4
CURRENT (AMPS/DEGREE)                                         FAULT MVA
SEQUENCE (1,2,0) PHASE (A,B,C)                        SEQUENCE (1,2,0) PHASE (A,B,C)
MAGNITUDE ANGLE MAGNITUDE ANGLE                                            MAGNITUDE            MAGNITUDE
--------- ------- --------- -------      ---------          ---------
1295 -89.47          3884 -89.47                  247                  740
1295 -89.47            0    0.00           247                     0
1295 -89.47            0    0.00           247                     0
-------------------------------------------------------------------------------
FAULT AT BUS NUMBER                   5 : NAME             Bus5
CURRENT (AMPS/DEGREE)                                         FAULT MVA
SEQUENCE (1,2,0) PHASE (A,B,C)                        SEQUENCE (1,2,0) PHASE (A,B,C)
MAGNITUDE ANGLE MAGNITUDE ANGLE                                            MAGNITUDE             MAGNITUDE
--------- ------- --------- -------      ---------          ---------
1295 -89.47          3884 -89.47                  247                  740

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1295 -89.47             0   0.00                247               0
1295 -89.47             0   0.00                247               0
-------------------------------------------------------------------------------
FAULT AT BUS NUMBER                    6 : NAME              Bus6
CURRENT (AMPS/DEGREE)                                           FAULT MVA
SEQUENCE (1,2,0) PHASE (A,B,C)                           SEQUENCE (1,2,0) PHASE (A,B,C)
MAGNITUDE ANGLE MAGNITUDE ANGLE                                           MAGNITUDE          MAGNITUDE
--------- ------- --------- -------        ---------          ---------
2953 -88.55          8860 -88.55                    563            1688
2953 -88.55             0   0.00                563               0
2953 -88.55             0   0.00                563               0
-------------------------------------------------------------------------------
Single phase fault level
Bus No. Name          BUS kV 1PH-fMVA Fault I
NOMINAL                  kA
------- -------- -------- -------- --------
1     Bus1 220.000          0.0    0.000
2     Bus2 11.000          0.0     0.002
3     Bus3 110.000 1688.1              8.861
4     Bus4 110.000         739.9       3.884
5     Bus5 110.000         739.9       3.884
6     Bus6 110.000 1688.1              8.861
-------------------------------------------------------------------------------
-------------------------------------------------------------------------------
Positive, negative and zero sequence impedancees on 100.000 MVA base
Bus No. Name          BUS kV 3PH-fMVA SLG-fMVA P-Zdrive N-Zdrive Zdrive
Re/Im        Re/Im Re/Im
------- -------- -------- -------- -------- -------- -------- --------
1     Bus1 220.000 1361.2                 0.0 0.00035 0.00003 0.00000

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0.07347 0.01605 10099.99023
2   Bus2 11.000 1361.2          0.0 0.00035 0.00003 0.00000
0.07347 0.01605 10099.99023
3   Bus3 110.000 1107.6 1688.1 0.00130 0.00141 0.00179
0.09027 0.04597 0.04141
4   Bus4 110.000      737.5    739.9 0.00125 0.00126 0.00127
0.13558 0.09421 0.17563
5   Bus5 110.000      737.5    739.9 0.00125 0.00126 0.00127

RESULT:

The contingency was detected in bus number 3 and the result summary is as shown.

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Exp No:5
Date: 6.1.2010

LOAD FLOW ANALYSIS USING GAUSS-SEIDEL METHOD FOR
BOTH P-Q&P-V BUS USING MI-POWER
AIM:

To carry out load flow analysis of the given power system by Gauss-Seidel method.

THEORY:

Load flow analysis is the study conducted to determine the steady state operating

condition of the given power system under given conditions.A large number of numerical

algorithms have been developed and Gauss-Seidel method is one of such algorithms.

PROBLEM FORMULATION:

The performance equation of a power system may be written as

[IBUS] = [YBUS] [VBUS]                              1

Selecting one of the buses as the reference bus,we get (n-1) simultaneous equations .The bus loading
equation can be written as
2
Ii =      Pi - jQi            (i=1,2,3……….n)

Vi*

n

Where Pi =      Re    Σ V *Y V
i       ik   k
3

K=1

n

Qi = -Im       Σ     Vi*YikVk                        4
K=1

The bus voltage can be written in the form of

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n

Vi = 1.0    Ii -   ΣY     ij   Vj                          5
Yii            j=1
j=i

( i=1,2,………..n ) & i≠slack bus

Substituting Ii in the expression for Vi we get

n

Vinew =     1.0       Pi - jQi -        Σ
Yij Vi0                     6
Yii        Vi0*      j=1
j≠i

If the latest available voltages are used in the above expression,we get,

n           n

Vinew   = 1.0      Pi - jQi -       Σ
Yij Vj -   n
Σ
Yij Vi0               7
Yii       Vi0*      j=1         j=i+1

The above equation is the required formula.This equation can be solved for bus voltages in an
iterative manner.During each iteration, we compute all the bus voltages and check for convergence is
carried out by comparison with the voltages obtained at the end of the previous iteration.After the
solution is obtained , the slack bus real and reactive powers , the reactive power generation at other
generator buses and line flows can be calculated

This method is easier when compared with other methods , because the calculations are
very simple.This is used for making a good initial start for Newton-Raphson method.

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ALGORITHM

Step 1: Read the data such as line data,specified powers,specified voltages,Q limits
at the generator buses and tolerance for convergence.

Step 2: Compute Y-bus matrix.

Step 3: Initialise all the bus voltages.

Step 4: Iter=1.

Step 5: Consider i=2,where „i‟ is the bus number.

Step 6: Check whether this is P-Vbus or P-Q bus,go to step8 otherwise go to the next Step.

Step 7:Compute Qi check for Q-limit violation.QGi=Qi+QLi.
If QGi>Qi,max,equate to QGi to Qi max there by converting it into P-Q bus.
Qi=Qimax-Qi, then go to step8.
Similarly if QGi<Qi min equate QGi to Qi min there by converting it into P-Q
bus. Qi=Qi min-Q Li then go to step8.
If QGi is within the upper and lower limits,don‟t change QGiQi =QGi-QLi,then
go to step8.

Step 8: Calculate the new value of the bus voltage using Gauss-Seidel formula.

i-1                  n

Vi         =    1.0     Pi - jQi -   ΣY     ij   Vj n -   ΣY      ij   Vj0
Yii      Vi0*        j=1                 j=i+1

Adjust voltage magnitude of the bus to specified magnitude if Q limits are not violated.

Step 9: If all the buses are considered,go to step10.Otherwise increment the bus no

i=i+1 and go to step6.

Step 10: Check for convergence.If there is no convergence go to step11.Otherwise go

to step12.

Step 11: Update the bus voltages using the formula

Vi new = Vi old + α (Vi new- Vi old )        (i=1,2,3……..n),i≠slack bus

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α is the acceleration factor=1.4.

Increment the iteration counter Iter=Iter+1 and go to step5.

Step 12: Calculate the slack bus power,Q at P-V buses,real and reactive line flows,real and reactive
line losses and print all the results including all the complex bus voltages and all the bus angles.

Step 13:Stop.

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FLOW CHART

START

data,Vspec,Pspec,qspec,q-
limits,tol,acceleration-
factor

Compute Y-bus matrix by inspection method

Initialise all the bus voltages suitably

Set Iteration count Iter=1

c

Bus No i=2
b        i=i+1

No
Does i refer to
e
P-V Bus?

Yes

Calculate Q bus i

QGi=QBus i+QLi

a

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a

e

QGi>Qi max                                        QGi<Qi min
Check for Q
limit violation

QGi=Qimax                                                        QGi=Qimin
Qi=Qi calculated
Qispec=Qi max-QLi                                                Qispec=Qi min-QLi

i-1              N

Vinew = 1 Pi - jQi -
*
Σ
YijVjnew -          Σ
Yij Vjold
Yii Vi old   j=1          j=i+1

No
Is this the last
b
bus?

Yes

b

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d

Check whether
Yes                                              No
|V   i
new
-   Vi |≤Tol
old

i=2 to nb

V inew = V iold +α(V inew - V iold)

i=2 to nb
Calculate all the line flows,BusVoltage
magnitudes,bus angles total line
losses.Reactive power generated at P-V
bus,slack bus power                                 At generator buses where Q limits
of the bus voltage to the specified
Print the results                                                   value.

STOP
V inew = (V inew *V ispec)/ |V inew |

V iold = V inew
i=2 to nb

Iter=Iter+1

c

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ONE LINE DIAGRAM

G

1                                                 4
North                           Lake             3
Main
2                         6

1                                        4                           7
3

5
2
5
South          G                                          Elm

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DATA FOR SAMPLE SYSTEM

IMPEDANCE AND LINE CHARGING FOR THE SAMPLE SYSTEM:

Bus code             Impedance                   Line
From -                (R+jX)                  Charging
To                                            B/2

1-2             0.02+j0.06            0.0+j 0.030

1-3             0.08+j0.24            0.0+ j0.025

2-3             0.06+j0.08                0.0+ j0.02

2-4             0.06+j0.08                0.0+ j0.02

2-5             0.04+j0.12            0.0+ j0.015

3-4             0.01+j0.03            0.0+ j0.010

4-5             0.08+j0.24            0.0+ j0.025

GENERATION ,LOADS AND BUS VOLTAGES FOR SAMPLE SYSTEM:

No             Voltage              MW                       MVAR                MW             MVAR

1            1.06+j0.0               0                         0                  0              0

2            1.00+j0.0              40                        30                  20             10

3            1.00+j0.0               0                         0                  45             15

4            1.00+j0.0               0                         0                  40             5

5            1.00+j0.0               0                         0                  60             10

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PROCEDURE TO ENTER THE DATA FOR PERFORMING STUDIES USING MI-
POWER

Mi-Power-Database-Configuration

Open Power system Network Editor.Select Menu option Database                Configure.Configure
Database

dialog is popped up.Click Browse button. Open dialog box is popped up,where you are going to
browse the desired directory and specify the name of the database to be associated with the single line
diagram.Click open button after entering the desired database name.Configure Database dialog will
appear with path chosen. Click Ok button on the Configure Database dialog.Uncheck the Power
System Libraries and Standard Relay Libraries.For this example these Standard libraries are not
needed ,because all the data is given on pu for power system libraries (liketransformer ,line ,generator
),and relay libraries are required only for relay co-ordination studies.If Libraries are selected, standard
libraries will be loaded along with database.Click Electrical Information tab.Since the impedances
are given on 100MVA base,check the pu status.Enter the Base MVA and Base frequency.Click on
Breaker Ratings button to give Breaker ratings.Click OK button to create the database to return to
Network Editor.

Bus Base Voltage Configuration

In the network editor,Configure the base voltages for the single line diagram.Select Menu option

Configure             Base Voltage.If necessary change the Base Voltages ,colour,Bus width and
click OK.

Procedure to Draw First Element-Bus

Click on Bus icon provided on power system tool bar. Draw a bus and a dialog appears prompting to
give the Bus ID and Bus Name. Click OK. Database manager with corresponding Bus Data form will
appear. Modify the Area number, Zone number and Contigency Weightage data if it is other than the
default values. If this data is not furnished, keep the default values. Usually the minimum and
maximum voltage ratings are ±5% of the rated voltage. If these ratings are other than this modify these
fields. Otherwise keep the default values. Bus description field can be effectively used if the bus name
gives more tha 8 characters. If Bus name is more than 8 characters ,then a short name is given in the

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bus name field and the Bus description field can be used to abbreviate the Bus name. For example let
us say the bus name is Northeast, then bus name can be given as NE and the Bus description field can
be North East. After entering data click Save which invokes Network Editor .Follow the same
procedure for remaining buses.

Following table gives data for other buses

Bus               Bus              Nominal
Number             Name             Voltage(kV)

2            South               220

3             Lake               220

4             Main               220

5             Elm                220

Procedure to Draw Transmission Line

Click on Transmission Line icon provided on power system tool bar.To draw the line click in

between two buses and to connect to the from bus double clicking Left Mouse Button on the From

Bus and join it to another bus by double clicking the mouse button on the To Bus. Element ID dialog

will appear. Enter Element ID number and click OK.Database manager with corresponding
Line\Cable Data form will be open.Enter the details of that line. Enter Structure Ref No.as 1 and
click on Transmission Line Library>>button. Line & cable Library form will appear.Enter
Transmission line library data. After entering data Save and Close.Line\Cable Data form will
appear.Click Save,which invokes Network Editor to update next element.Data for elements given in
the following table.

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TRANSMISSION LINE ELEMENT DATA:

Line              From            To             No.Of                Structure
No               Bus            Bus            circuits               Ref.No

1               1              2                1                        1

2               1              3                1                        2

3               2              3                1                        3

4               2              4                1                        3

5               2              5                1                        4

6               3              4                1                        5

7               4              5                1                        2

TRANSMISSION LINE LIBRARY DATA

Structure        Structure Ref      Resistance      Reactance              Line               Thermal
Ref No           Name                                                      charging B/2       Rating

1           Line 1-2            0.02               0.06                  0.03              100

2            Line 1-            0.08               0.24                  0.025             100
3&4-5

3            Line 2-            0.06               0.18                  0.02              100
3&2-4

4           Line 2-5            0.04               0.12                  0.015             100

5           Line 3-4            0.01               0.03                  0.01              100

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Procedure to Draw Generator:

Click on Generator icon provided on power system tool bar.Connect it to bus 1 by clicking the Left

Mouse button on Bus 1.The Element ID dialog will appear.Enter ID number and click OK.Database

with corresponding Generator Data form will appear.Enter all the details

Generator 1 Element Data

Manufacturer Ref No                  1

No.of units parallel                1

Specified Voltage               233.200

Derated MVA                     100

Scheduled Power                   80

Real Power Min                    0

Real Power Max                   80

Reactive Power Min                  0

Reactive Power Max                  60

Since the specified voltage is given as 1.06pu,click Compute Volt button and give 1.06 value.

Voltage will be calculated and appear in the specified voltage field.

Since generator at bus 1 is mentionas slack bus,only specified voltage will have importance.

Enter Manufacturer Ref. No as 1 and click on Generator Library button.Generator library form will

appear

Generator 1 Library Data

MVA Rating                100

MW Rating                80

KV Rating                220

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Manufacturer              Gen1
Name

After entering data Save and close.In Generator Data form click Save.Network Editor screen will be

invoked.Similarly connect generator 2 at bus 2.Enter its details as given in the following table.

Generator 2 Element Data

Manufacturer Ref No                   2

No.of units parallel                 1

Specified Voltage                  220

Derated MVA                      50

Scheduled Power                    40

Real Power Min                     0

Real Power Max                    40

Reactive Power Min                  30

Reactive Power Max                   30

Generator 1 Library Data

MVA Rating                100

MW Rating                 80

KV Rating                220

Manufacturer              Gen1
Name

Click on Load icon provided on power system tool bar.Connect Load 1 at BUS 2 by clicking the Left

Mouse button on Bus 2.Element ID dialog will appear.Give Id number as 1 and say OK.Load Data

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form will appear .Enter load details.Then click Save button,which invokes Network Editor.Enter the

Load detail as given in the following table.

No            No
1              2           20            10
2              5           60            10
3              3           45            15
4              4           40            5

Select Menu option Solve          Load Flow Analysis.When Study Info button is clicked, following

dialog will open.Select Gauss-Seidel method and enter acceleration factor as1.4&P-Tolerance and

Q-Tolerance as 0.0001. Click OK.

Execute Load Flow Analysis and click on Report in Load Flow Analysis dialog to view Report.

OUTPUT:

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CASE NO : 1          CONTINGENCY : 0 SCHEDULE NO : 0

CONTINGENCY NAME : Base Case                                           RATING CONSIDERED : NOMINAL

-------------------------------------------------------------------------------

VERSION NUMBER : 6.1

LARGEST BUS NUMBER USED                               :        5           ACTUAL NUMBER OF BUSES                            :   5

NUMBER OF 2 WIND. TRANSFORMERS :                                               0   NUMBER OF 3 WIND. TRANSFORMERS :                          0

NUMBER OF TRANSMISSION LINES :                                         7

NUMBER OF SERIES REACTORS                              :           0       NUMBER OF SERIES CAPACITORS                           :       0

NUMBER OF CIRCUIT BREAKERS                                 :       0

NUMBER OF SHUNT REACTORS                                  :        0 NUMBER OF SHUNT CAPACITORS                                      :   0

NUMBER OF SHUNT IMPEDANCES                                     :       0

NUMBER OF GENERATORS                          :            2           NUMBER OF LOADS                       :       4

NUMBER OF LOAD CHARACTERISTICS :                                               0   NUMBER OF UNDER FREQUENCY RELAY:
0

NUMBER OF GEN CAPABILITY CURVES:                                               0   NUMBER OF FILTERS                         :   0

NUMBER OF TIE LINE SCHEDULES :                                         0

NUMBER OF CONVERTORS                          :            0           NUMBER OF DC LINKS                        :       0

-------------------------------------------------------------------------------

LOAD FLOW WITH GAUSS-SEIDEL METHOD                                                      :   5

NUMBER OF ZONES                                                :           1

PRINT OPTION                                      :            3 - BOTH DATA AND RESULTS PRINT

PLOT OPTION                                    :              1 - PLOTTING WITH PU VOLTAGE

NO FREQUENCY DEPENDENT LOAD FLOW, CONTROL OPTION:                                                        0

BASE MVA                                     : 100.000000

NOMINAL SYSTEM FREQUENCY (Hzs)                                                      : 60.000000

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FREQUENCY DEVIATION (Hzs)                                    : 0.000000

FLOWS IN MW AND MVAR, OPTION                                      :0

SLACK BUS                                     :   1

TRANSFORMER TAP CONTROL OPTION                                          :     0

Q CHECKING LIMIT (ENABLED)                                    :   0

REAL POWER TOLERANCE (PU)                                      : 0.00010

REACTIVE POWER TOLERANCE (PU)                                         : 0.00010

MAXIMUM NUMBER OF ITERATIONS                                           : 15

BUS VOLTAGE BELOW WHICH LOAD MODEL IS CHANGED : 0.75000

CIRCUIT BREAKER RESISTANCE (PU)                                   : 0.00000

CIRCUIT BREAKER REACTANCE (PU)                                        : 0.00010

TRANSFORMER R/X RATIO                                      : 0.05000

------------------------------------------------------------------------------

ANNUAL PERCENTAGE INTEREST CHARGES                                            : 15.000

.ANNUAL PERCENT OPERATION & MAINTENANCE CHARGES : 4.000

LIFE OF EQUIPMENT IN YEARS                                   : 20.000

ENERGY UNIT CHARGE (KWHOUR)                                       : 2.500 Rs

COST PER MVAR IN LAKHS                                     : 5.000 Rs

-------------------------------------------------------------------------------

ZONE WISE MULTIPLICATION FACTORS

---- -------- -------- -------- -------- -------- -------- --------

0   1.000     1.000     1.000     1.000      1.000     1.000        1.000

1   1.000     1.000     1.000     1.000      1.000     1.000        1.000

-------------------------------------------------------------------------------

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BUS DATA

BUS NO. AREA ZONE BUS KV VMIN-PU VMAX-PU                                         NAME

------- ---- ---- -------- -------- -------- --------

1    1    1 220.000       0.950     1.050      Bus1

2    1    1 220.000       0.950     1.050      Bus2

3    1    1 220.000       0.950     1.050      Bus3

4    1    1 220.000       0.950     1.050      Bus4

5    1    1 220.000       0.950     1.050      Bus5

-------------------------------------------------------------------------------

TRANSMISSION LINE DATA

STA CKT FROM FROM                     TO TO               LINE PARAMETER                  RATING KMS

NODE NAME*             NODE NAME*                 R(P.U) X(P.U.) B/2(P.U.)          MVA

--- --- ---- -------- ---- -------- --------- --------- --------- ------ -----

3 1      1    Bus1     2     Bus2 0.02000 0.06000 0.03000                   100 1.0

3 1      1    Bus1     3     Bus3 0.08000 0.24000 0.02500                   100 1.0

3 1      4    Bus4     5     Bus5 0.08000 0.24000 0.02500                   100 1.0

3 1      2    Bus2     3     Bus3 0.06000 0.18000 0.02000                   100 1.0

3 1      2    Bus2     4     Bus4 0.06000 0.18000 0.02000                   100 1.0

3 1      2    Bus2     5     Bus5 0.04000 0.12000 0.01500                   100 1.0

3 1      3    Bus3     4     Bus4 0.01000 0.03000 0.01000                   100 1.0

-------------------------------------------------------------------------------

-------------------------------------------------------------------------------

TOTAL LINE CHARGING SUSCEPTANCE                                  :   0.29000

TOTAL LINE CHARGING MVAR AT 1 PU VOLTAGE :                                       29.000

-------------------------------------------------------------------------------

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TOTAL CAPACITIVE SUSCEPTANCE                                : 0.00000 pu - 0.000 MVAR

TOTAL INDUCTIVE SUSCEPTANCE                                 : 0.00000 pu - 0.000 MVAR

-------------------------------------------------------------------------------

GENERATOR DATA

SL.NO* FROM FROM                     REAL        Q-MIN        Q-MAX V-SPEC CAP.                  MVA STAT

NODE NAME*          POWER(MW)              MVAR          MVAR             P.U. CURV RATING

------ ---- -------- --------- --------- --------- --------- ---- ------- ----

1    1   Bus1 80.0000         0.0000 60.0000           1.0600     0 100.00         3

2    2   Bus2 40.0000 30.0000 30.0000                   1.0000     0 50.00         3

-------------------------------------------------------------------------------

SLNO FROM FROM                    REAL REACTIVE               COMP COMPENSATING MVAR VALUE CHAR
F/V

*       NODE NAME*              MW        MVAR         MVAR         MIN          MAX        STEP NO NO

STAT

---- ---- -------- -------- -------- -------- ------- ------- ------- ---- ----

1    2    Bus2 20.000 10.000           0.000 0.000 0.000 0.000                 0    0

3    0

2    5    Bus5 60.000 10.000           0.000 0.000 0.000 0.000                 0    0

3    0

3    3    Bus3 45.000 15.000           0.000 0.000 0.000 0.000                 0    0

3    0

4    4    Bus4 40.000       5.000     0.000 0.000 0.000 0.000                 0    0

3    0

-------------------------------------------------------------------------------

INCLUDING OUT OF SERVICE VALUES

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TOTAL SPECIFIED MW GENERATION                             : 120.00000

TOTAL MIN MVAR LIMIT OF GENERATOR : 30.00000

TOTAL MAX MVAR LIMIT OF GENERATOR : 90.00000

TOTAL SPECIFIED MW LOAD                            : 165.00000 reduced 165.00000

TOTAL SPECIFIED MVAR LOAD                            : 40.00000 reduced 40.00000

TOTAL SPECIFIED MVAR COMPENSATION :                                 0.00000 reduced        0.00000

-------------------------------------------------------------------------------

IN SERVICE VALUES

TOTAL SPECIFIED MW GENERATION                             : 120.00000

TOTAL MIN MVAR LIMIT OF GENERATOR : 30.00000

TOTAL MAX MVAR LIMIT OF GENERATOR : 90.00000

TOTAL SPECIFIED MW LOAD                            : 165.00000 reduced 165.00000

TOTAL SPECIFIED MVAR LOAD                            : 40.00000 reduced 40.00000

TOTAL SPECIFIED MVAR COMPENSATION :                                 0.00000 reduced        0.00000

-------------------------------------------------------------------------------

GENERATOR DATA FOR FREQUENCY DEPENDENT LOAD FLOW

SLNO* FROM FROM                   P-RATE         P-MIN       P-MAX %DROOP PARTICI                    BIAS

NODE NAME*                  MW         MW          MW               FACTOR SETTING

C0        C1        C2

------ ---- -------- -------- --------- --------- --------- --------- ---------

1    1    Bus1 80.000          0.0000 80.0000          4.0000     0.0000       0.0000

0.0000      0.0000     0.0000

2    2    Bus2 40.000          0.0000 40.0000          4.0000     0.0000       0.0000

0.0000      0.0000     0.0000

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-------------------------------------------------------------------------------

Acceleration factor : 1.40

-------------------------------------------------------------------------------

Slack bus angle (degrees) : 0.00

-------------------------------------------------------------------------------

-------------------------------------------------------------------------------

Iteration count =      1 Error = 0.052537 Bus =             2

Iteration count =      2 Error = 0.015724 Bus =             5

Iteration count =      3 Error = 0.007669 Bus =             5

Iteration count =      4 Error = 0.002768 Bus =             2

Iteration count =      5 Error = 0.002594 Bus =            5

Iteration count =      6 Error = 0.001050 Bus =             4

Iteration count =      7 Error = 0.000867 Bus =             3

Iteration count =      8 Error = 0.000394 Bus =             2

Iteration count =      9 Error = 0.000217 Bus =             3

Iteration count = 10 Error = 0.000117 Bus =                    3

Iteration count = 11 Error = 0.000044 Bus =                    2

-------------------------------------------------------------------------------

BUS VOLTAGES AND POWERS

NODE FROM             V-MAG ANGLE                 MW        MVAR           MW       MVAR      MVAR

---- -------- ------ ------ -------- -------- -------- -------- --------

1    Bus1 1.0600 0.00 129.534 -7.469                  0.000      0.000    0.000 #<

2    Bus2 1.0475 -2.81          40.000 30.000         20.000 10.000        0.000

3    Bus3 1.0242 -5.00          0.000     0.000     45.000 15.000         0.000

4    Bus4 1.0236 -5.33          0.000     0.000     40.000       5.000   0.000

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5     Bus5 1.0180 -6.15           0.000     0.000     60.000 10.000            0.000

-------------------------------------------------------------------------------

NUMBER OF BUSES EXCEEDING MINIMUM VOLTAGE LIMIT (@ mark) :                                               0

NUMBER OF BUSES EXCEEDING MAXIMUM VOLTAGE LIMIT (# mark) :                                               1

NUMBER OF GENERATORS EXCEEDING MINIMUM Q LIMIT (< mark) :                                            1

NUMBER OF GENERATORS EXCEEDING MAXIMUM Q LIMIT (> mark) :                                            0

-------------------------------------------------------------------------------

LINE FLOWS AND LINE LOSSES

SLNO CS FROM FROM                    TO TO               FORWARD                   LOSS          %

---- -- ---- -------- ---- -------- -------- -------- -------- -------- -------

1 1     1    Bus1     2     Bus2 88.825 -8.610 1.4093 -2.4345                     84.2#

2 1     1    Bus1     3     Bus3 40.710         1.141 1.1911 -1.8583             38.4^

3 1     4    Bus4     5     Bus5     6.334 -2.280 0.0307 -5.1178                  6.8&

4 1     2    Bus2     3     Bus3 24.690         3.535 0.3513 -3.2385             24.7&

5 1     2    Bus2     4     Bus4 27.936         2.957 0.4413 -2.9660             27.5^

6 1     2    Bus2     5     Bus5 54.824         7.346 1.1253 0.1756              52.8\$

7 1 3        Bus3     4     Bus4 18.901 -5.166 0.0357 -1.9898                     19.1&

-------------------------------------------------------------------------------

! NUMBER OF LINES LOADED BEYOND 125%                                     :   0

@ NUMBER OF LINES LOADED BETWEEN 100% AND 125% :                                             0

# NUMBER OF LINES LOADED BETWEEN 75% AND 100% :                                          1

\$ NUMBER OF LINES LOADED BETWEEN 50% AND 75% :                                         1

^ NUMBER OF LINES LOADED BETWEEN 25% AND 50% :                                       2

& NUMBER OF LINES LOADED BETWEEN 1% AND 25% :                                            3

* NUMBER OF LINES LOADED BETWEEN 0% AND 1% :                                        0

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-------------------------------------------------------------------------------

NEW SYSTEM FREQUENCY FOR ISLAND 1                                : 60.000000 Hzs

-------------------------------------------------------------------------------

Summary of results

TOTAL REAL POWER GENERATION :                          169.534 MW

TOTAL REAL POWER DRAWAL -ve g :                         0.000 MW

TOTAL REACT. POWER GENERATION :                          22.531 MVAR

GENERATION pf                      :    0.991

TOTAL SHUNT REACTOR INJECTION :                          0.000 MW

TOTAL SHUNT REACTOR INJECTION :                          0.000 MVAR

TOTAL SHUNT CAPACIT.INJECTION :                         0.000 MW

TOTAL SHUNT CAPACIT.INJECTION :                         0.000 MVAR

TOTAL REAL POWER LOAD                        :   165.000 MW

TOTAL REAL POWER INJECT,-ve L :                      0.000 MW

TOTAL REACTIVE POWER LOAD                        :    40.000 MVAR

TOTAL COMPENSATION AT LOADS :                            0.000 MVAR

TOTAL HVDC REACTIVE POWER                        :     0.000 MVAR

TOTAL REAL POWER LOSS (AC+DC) : 4.584590 MW ( 4.584590+ 0.000000)

PERCENTAGE REAL LOSS (AC+DC) :                         2.704

TOTAL REACTIVE POWER LOSS                        : -17.429301 MVAR

-------------------------------------------------------------------------------

Zone wise distribution

Description         Zone # 1

---------------- ----------

MW generation          169.5343

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MVAR generation            22.5314

MVAR compensation             0.0000

MW loss              4.5846

MVAR loss             -17.4293

MVAR - inductive           0.0000

MVAR - capacitive          0.0000

-------------------------------------------------------------------------------

Zone wise export(+ve)/import(-ve)

Zone # 1 MW & MVAR

------ -------- --------

1      -----

Area wise distribution

Description         Area # 1

---------------- ----------

MW generation          169.5343

MVAR generation            22.5314

MVAR compensation             0.0000

MW loss              4.5846

MVAR loss             -17.4293

MVAR - inductive           0.0000

MVAR - capacitive 0.0000

Exp No:5 b)

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Date: 6.1.2010

LOAD FLOW ANALYSIS USING NEWTON-RAPHSON METHOD
FOR BOTH P-Q&P-V BUS USING MI-POWER
AIM:

To carry out load flow analysis of the given power system by Newton-Raphson method

THEORY:
Load flow analysis is the study conducted to determine the steady state operating condition of the

given system under given conditions. A large number of numerical algorithms have been developed

and Newton-Raphson method is one of such algorithms.

The Newton-Raphson method of load flow analysis is an iterative method which

approximate the set of non-linear simultaneous equations to a set of linear simultaneous equations.

The load flow equations for Newton-Raphson method are non-linear equations

in terms of real and imaginary part of bus voltages.

n
Pp = Σ {ep (eq Gpq +fq Bpq)+fp(fqGpq-eq Bpq)
q=1

n
Qp= Σ{fp (eqGpq+fqBpq)-ep(fqGpq-eqBpq)}
q=1

|Vp|2=ep2 +fp2

Where ep       = Real part of Vp(Voltage of bus –p)

fq     = Imaginary part of Vp(Voltage of bus –p)

Gpq, Bpq       = Conductance and Susceptance of admittance Ypq

ALGORITHM:

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Step1 : Read the data such as line data ,bus data,specified powers at the generator buses

and tolerance for convergence.

Step2: Compute Y-bus matrix.

Step3: Initialise all the bus voltages.

Step4: Iter=1, Bus count=1.

Step5: Check for slack bus. if it is a slack bus then go to step 13 otherwise go to next step.

Step 6: Calculate the real and reactive power of bus P using the following equation.

n
Ppk = Σ {epk (eqk Gpq +fqk Bpq)+fpk(fq kGpq-eqk Bpq)
q=1

n
Qpk= Σ{fpk (eq kGpq+fq kBpq)-ep k(fq kGpq-eq kBpq)}
q=1

Step 7: Calculate the change in real power.

Change in real power Δkp = Pp spec - Pkp

Step 8 : Check for Generator bus. if it is a Generator bus go to step9 otherwise go to step 12.

Step 9: Check for reactive power limit violation of Generator bus.

Step 10: if the calculated reative power is within specified limit then consider this bus as a generator

Bus.

Step 11: if the reactive power limit is violated, then treat this bus as a load bus.

if Qkp< Qp min then Qp spec = Qp min
if Qkp> Qp max then Qp spec = Qp max

Step 12: Calculate the change in reactive power for load bus.

Step 13: Repeat step 5 to 12 until all residues (change in P , Q and V ) are calculated. For this   the bus

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Increment the bus count by 1 and go to Step 5 until the bus count is n.

Step 14: Determine the largest of the absolute value of the residue (ΔE )

Step 15: Compare ΔE and convergence. If ΔE < convergence then go to step 20 else go to next step.

Step 16: Determine the values of Jacobian matrix

Step 17: Calculate the increment in real and reactive part of voltages.

Step 18: Calculate the new bus voltages.

epk+1=epk +Δepk         ;    p=1,2,3,……… n except slack bus

fpk+1=fpk +Δfpk         ;    p=1,2,3,……… n except slack bus

|Vpk+1|=√(epk+1)2 +(fpk+1)2       and   δpk+1 =tan-1 (epk+1∕ fpk+1)

Therefore , Vpk+1=| Vpk+1| < δpk+1

Step 19: Advance the iteration count, i.e., k=k+1 and go to step 4.

Step 20: Calculate the line flows.

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FLOWCHART:

START

Read line data, bus data, Pp, Qp, Vp spec, Qp max,
Qp min

Compute Y-bus matrix by inspection method

Initialize all the bus voltages suitably

Calculate ep°&fp° for p=1                          z
Set Iter count, Iter=1,Set bus count p=1

No
Check for
Set bus count,p=1
slack bus

d
Calculate Ppk Qpk & ΔPk

f
No
Check for
b                                    generator
bus

Yes
c

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c

Yes

Check if
Set Qpspec=Qpmin
Qpk < Qpmin

No
Yes
Set Qp,spec=Qp,max                         Check if

Qpk > Qpmax

No

b            Calculate ΔPk              Calculate|ΔVp|2

f

No
Check if
d                     p>n

Yes
1

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1

Determine the largest absolute value of
residue, ΔE

Yes
Check if
ΔE<є

No

Determine the elements of                     Calculate the line flows
and slack bus power
the jacobian matrix(J)
and slack bus power
STOP
Calculate the voltage

increment Δepk and Δfpk

Calculate |Vpk+1| and δpk+1

No
Check for
generator bus

Yes

epk+1=|Vp spec|cosδpk+1
fpk+1=|Vp spec|sinδpk+1

count,k=k+1

z

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ONE LINE DIAGRAM:

G

1                                                         4
North                              Lake                  3        Main
2                              6

1                                        4                                   7
3

5
South                    2
Elm          5
G

DATA FOR SAMPLE SYSTEM

IMPEDANCE AND LINE CHARGING FOR THE SAMPLE SYSTEM:

Bus code                  Impedance               Line
From - To                   (R+jX)              Charging
B/2

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1-2               0.02+j0.06            0.0+j
0.030

1-3               0.08+j0.24             0.0+
j0.025

2-3               0.06+j0.08            0.0+
j0.02

2-4               0.06+j0.08            0.0+
j0.02

2-5               0.04+j0.12             0.0+
j0.015

3-4               0.01+j0.03             0.0+
j0.010

4-5               0.08+j0.24             0.0+
j0.025

GENERATION ,LOADS AND BUS VOLTAGES FOR SAMPLE SYSTEM:

No             Voltage              MW                   MVAR               MW             MVAR

1          1.06+j0.0               0                     0                0               0

2          1.00+j0.0               40                   30                20             10

3          1.00+j0.0               0                     0                45             15

4          1.00+j0.0               0                     0                40              5

5          1.00+j0.0               0                     0                60             10

PROCEDURE TO ENTER THE DATA FOR PERFORMING STUDIES USING MI-
POWER

Mi-Power-Database-Configuration

Open Power system Network Editor.Select Menu option Database                 Configure.Configure
Database

dialog is popped up.Click Browse button. Open dialog box is popped up,where you are going to
browse the desired directory and specify the name of the database to be associated with the single line

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diagram.Click open button after entering the desired database name.Configure Database dialog will
appear with path chosen.Click Ok button on the Configure Database dialog. Uncheck the Power
System Libraries and Standard Relay Libraries.For this example these Standard libraries are not
needed ,because all the data is given on pu for power system libraries (liketransformer ,line ,generator
),and relay libraries are required only for relay co-ordination studies.If Libraries are selected, standard
libraries will be loaded along with database.Click Electrical Information tab.Since the impedances
are given on 100MVA base, check the pu status.Enter the Base MVA and Base frequency.Click on
Breaker Ratings button to give Breaker ratings.Click OK button to create the database to return to
Network Editor.

Bus Base Voltage Configuration

In the network editor,Configure the base voltages for the single line diagram.Select Menu option

Configure              Base Voltage.If necessary change the Base Voltages ,colour,Bus width and
click OK.

Procedure to Draw First Element-Bus

Click on Bus icon provided on power system tool bar.Draw a bus and a dialog appears prompting to
give the Bus ID and Bus Name.Click OK. Database manager with corresponding Bus Data form will
appear. Modify the Area number,Zone number and Contigency Weightage data if it is other than the
default values.If this data is not furnished,keep the default values.Usually the minimum and maximum
voltage ratings are ±5% of the rated voltage.If these ratings are other than this modify these
fields.Otherwise keep the default values. Bus description field can be effectively used if the bus name
gives more tha 8 characters.If Bus name is more than 8 characters ,then a short name is given in the
bus name field and the Bus description field can be used to abbreviate the Bus name.For example let us
say the bus name is Northeast, then bus name can be given as NE and the Bus description field can be
North East. After entering data click Save which invokes Network Editor .Follow the same procedure
for remaining buses.

Following table gives data for other buses

Bus                 Bus                  Nominal
Number              Name                 Voltage(kV)

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2                 South                220

3                 Lake                 220

4                 Main                 220

5                 Elm                  220

Procedure to Draw Transmission Line

Click on Transmission Line icon provided on power system tool bar. To draw the line click in

between two buses and to connect to the from bus double clicking Left Mouse Button on the From

Bus and join it to another bus by double clicking the mouse button on the To Bus. Element ID dialog

will appear. Enter Element ID number and click OK. Database manager with corresponding
Line\Cable Data form will be open. Enter the details of that line.

Enter Structure Ref No. as 1 and click on Transmission Line Library>>button.

Line & cable Library form will appear.Enter Transmission line library data .

After entering data Save and Close. Line\Cable Data form will appear.Click Save, which invokes

Network Editor to update next element.Data for elements given in the following table.

TRANSMISSION LINE ELEMENT DATA:

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Line           From              To             No.Of                 Structure
No             Bus              Bus            circuits               Ref.No

1              1                2                1                      1

2              1                3                1                      2

3              2                3                1                      3

4              2                4                1                      3

5              2                5                1                      4

6              3                4                1                      5

7              4                5                1                      2

TRANSMISSION LINE LIBRARY DATA:

Structure   Structure Ref      Resistance      Reactance        Line              Thermal Rating
Ref No     Name                                                charging
B/2

1              Line 1-2           0.02              0.06              0.03              100

2               Line 1-           0.08              0.24          0.025                 100
3&4-5

3               Line 2-           0.06              0.18              0.02              100
3&2-4

4              Line 2-5           0.04              0.12          0.015                 100

5              Line 3-4           0.01              0.03              0.01              100

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Generator 1 Element Data

Manufacturer Ref No                       1

No.of units parallel                    1

Specified Voltage                233.200

Derated MVA                          100

Scheduled Power                       80

Real Power Min                         0

Real Power Max                        80

Reactive Power Min                       0

Reactive Power Max                      60

Since the specified voltage is given as 1.06pu,click Compute Volt button and give 1.06 value.

Voltage will be calculated and appear in the specified voltage field.

Since generator at bus 1 is mentionas slack bus,only specified voltage will have importance.

Enter Manufacturer Ref. No as 1 and click on Generator Library button.Generator library form will

appear

Generator 1 Library Data

MVA Rating                 100

MW Rating                  80

KV Rating                220

Manufacturer               Gen1
Name

After entering data Save and close.In Generator Data form click Save.Network Editor screen will be

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invoked.Similarly connect generator 2 at bus 2.Enter its details as given in the following table.

Procedure to Draw Generator

Click on Generator icon provided on power system tool bar.Connect it to bus 1 by clicking the Left

Mouse button on Bus 1.The Element ID dialog will appear.Enter ID number and click OK.Database

with corresponding Generator Data form will appear.Enter all the details.

Generator 2 Element Data

Manufacturer Ref No                      2

No.of units parallel                   1

Specified Voltage                  220

Derated MVA                          50

Scheduled Power                       40

Real Power Min                        0

Real Power Max                        40

Reactive Power Min                      30

Reactive Power Max                      30

Generator 1 Library Data

MVA Rating                100

MW Rating                  80

KV Rating                220

Manufacturer               Gen1
Name

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Click on Load icon provided on power system tool bar.Connect Load 1 at BUS 2 by clicking the Left

Mouse button on Bus 2.Element ID dialog will appear.Give Id number as 1 and say OK.Load Data

form will appear .Enter load details.Then click Save button,which invokes Network Editor.Enter the

Load detail as given in the following table.

No             No
1              2             20           10
2              5             60           10
3              3             45           15
4              4             40            5

Select Menu option Solve          Load Flow Analysis.When Study Info button is clicked,following

dialog will open.Select Newton-Raphson method and enter acceleration factor as1.4&P-Tolerance and

Q-Tolerance as 0.01.Click OK.

Execute Load Flow Analysis and click on Report in Load Flow Analysis dialog to view Report.

OUTPUT:
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-------------------------------------------------------------------------------
Date and Time : Mon Mar 15 10:25:39 2010
-------------------------------------------------------------------------------
LOAD FLOW BY NEWTON RAPHSON METHOD
CASE NO :     1      CONTINGENCY      : 0    SCHEDULE NO : 0
CONTINGENCY NAME : Basecase             RATING CONSIDERED : NOMINAL
-------------------------------------------------------------------------------
VERSION NUMBER :       6.1
LARGEST BUS NUMBER USED                 :     5      ACTUAL NUMBER OF BUSES                  :     5
NUMBER OF 2 WIND. TRANSFORMERS :              0      NUMBER OF 3 WIND. TRANSFORMERS :              0
NUMBER OF TRANSMISSION LINES            :     7
NUMBER OF SERIES REACTORS               :     0      NUMBER OF SERIES CAPACITORS             :     0
NUMBER OF CIRCUIT BREAKERS              :     0
NUMBER OF SHUNT REACTORS                :     0      NUMBER OF SHUNT CAPACITORS              :     0
NUMBER OF SHUNT IMPEDANCES              :     0
NUMBER OF GENERATORS                    :     2      NUMBER OF LOADS                         :     4
NUMBER OF LOAD CHARACTERISTICS :              0      NUMBER OF UNDER FREQUENCY RELAY:              0
NUMBER OF GEN CAPABILITY CURVES:              0      NUMBER OF FILTERS                       :     0
NUMBER OF TIE LINE SCHEDULES            :     0
NUMBER OF CONVERTORS                    :     0      NUMBER OF DC LINKS                      :     0
-------------------------------------------------------------------------------

LOAD FLOW WITH NEWTON RAPHSON METHOD                        :       6
NUMBER OF ZONES                                             :       1
PRINT OPTION                                                :       3 - BOTH DATA AND RESULTS
PRINT
PLOT OPTION                                                 :       1 - PLOTTING WITH PU VOLTAGE
NO FREQUENCY DEPENDENT LOAD FLOW, CONTROL OPTION:                   0
BASE MVA                                                    : 100.000000
NOMINAL SYSTEM FREQUENCY (Hzs)                              : 60.000000
FREQUENCY DEVIATION (Hzs)                                   : 0.000000
FLOWS IN MW AND MVAR, OPTION                                : 0
SLACK BUS                                                   :       1
TRANSFORMER TAP CONTROL OPTION                              :       0
Q CHECKING LIMIT       (ENABLED)                            :       0
REAL POWER TOLERANCE (PU)                                   :     0.00010
REACTIVE POWER TOLERANCE (PU)                               :     0.00010
MAXIMUM NUMBER OF ITERATIONS                                :      15
BUS VOLTAGE BELOW WHICH LOAD MODEL IS CHANGED               :     0.75000
CIRCUIT BREAKER RESISTANCE          (PU)                    :     0.00000
CIRCUIT BREAKER REACTANCE           (PU)                    :     0.00010

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TRANSFORMER R/X RATIO                                                         :     0.05000
------------------------------------------------------------------------------

ANNUAL PERCENTAGE INTEREST CHARGES                                            :     15.000
ANNUAL PERCENT OPERATION & MAINTENANCE CHARGES                                :      4.000
LIFE OF EQUIPMENT IN YEARS                                                    :     20.000
ENERGY UNIT CHARGE (KWHOUR)                                                   :      2.500 Rs
COST PER MVAR IN LAKHS                                                        :      5.000 Rs
-------------------------------------------------------------------------------

ZONE WISE MULTIPLICATION FACTORS

---- -------- -------- -------- -------- -------- -------- --------
0           1.000         1.000          1.000          1.000         1.000        1.000          1.000
1         1.000         1.000          1.000          1.000       1.000          1.000       1.000

BUS DATA

BUS NO. AREA ZONE                  BUS KV        VMIN-PU        VMAX-PU           NAME
------- ---- ---- -------- -------- -------- --------
1       1      1     220.000          0.950          1.050           Bus1
2       1      1     220.000          0.950          1.050           Bus2
3       1      1     220.000          0.950          1.050           Bus3
4       1      1     220.000          0.950          1.050           Bus4
5       1      1     220.000          0.950          1.050           Bus5

-------------------------------------------------------------------------------
TRANSMISSION LINE DATA

STA CKT FROM FROM                         TO TO                           LINE PARAMETER                   RATING   KMS
NODE NAME*             NODE NAME*                R(P.U)      X(P.U.) B/2(P.U.)               MVA
--- --- ---- -------- ---- -------- --------- --------- --------- ------ -----
3        1       1          Bus1       2           Bus2      0.02000      0.06000            0.03000      100   1.0
0        1       1          Bus1       3           Bus3      0.08000      0.24000            0.02500      100   1.0
3        1       4          Bus4       5           Bus5      0.08000      0.24000            0.02500      100   1.0
3        1       2          Bus2       3           Bus3      0.06000      0.18000            0.02000      100   1.0
3        1       2          Bus2       4           Bus4      0.06000      0.18000            0.02000      100   1.0
3        1       2          Bus2       5           Bus5      0.04000      0.12000            0.01500      100   1.0

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3           1            3      Bus3       4          Bus4         0.01000        0.03000       0.01000           100        1.0
-------------------------------------------------------------------------------
-------------------------------------------------------------------------------
TOTAL LINE CHARGING SUSCEPTANCE                                             :       0.29000
TOTAL LINE CHARGING MVAR AT 1 PU VOLTAGE :                                          29.000
NUMBER OF LINES OPENED ON BOTH THE ENDS                                                              : 1
TOTAL LINE CHARGING SUSCEPTANCE OF EXISTING LINES                                                    :     0.24000
TOTAL LINE CHARGING MVAR AT 1 PU VOLTAGE OF EXISTING LINES :                                                   24.000
-------------------------------------------------------------------------------
TOTAL CAPACITIVE SUSCEPTANCE                                           :        0.00000 pu -         0.000 MVAR
TOTAL INDUCTIVE SUSCEPTANCE                                            :        0.00000 pu -         0.000 MVAR
-------------------------------------------------------------------------------

GENERATOR DATA

SL.NO* FROM FROM                                   REAL          Q-MIN             Q-MAX        V-SPEC CAP.            MVA STAT
NODE NAME*             POWER(MW)               MVAR            MVAR           P.U. CURV      RATING
------ ---- -------- --------- --------- --------- --------- ---- ------- ----
1            1         Bus1      40.0000         30.0000            30.0000         1.0000     0      50.00            3
2            2         Bus2      40.0000         30.0000            30.0000         1.0000     0      50.00            3

-------------------------------------------------------------------------------

SLNO FROM FROM                               REAL REACTIVE                  COMP COMPENSATING MVAR VALUE CHAR                     F/V
*           NODE NAME*                         MW           MVAR            MVAR          MIN       MAX        STEP         NO     NO
STAT
---- ---- -------- -------- -------- -------- ------- ------- ------- ---- ----
1            2           Bus2      60.000         10.000           0.000      0.000       0.000    0.000             0         0
3         0
2            5           Bus5      45.000         15.000           0.000      0.000       0.000    0.000             0         0
3         0
3            3           Bus3      45.000         15.000           0.000      0.000       0.000    0.000             0         0
3         0
4            4           Bus4      40.000          5.000           0.000      0.000       0.000    0.000             0         0
3         0
-------------------------------------------------------------------------------

INCLUDING OUT OF SERVICE VALUES
TOTAL SPECIFIED MW GENERATION                                      :       80.00000
TOTAL MIN MVAR LIMIT OF GENERATOR                                  :       60.00000

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TOTAL MAX MVAR LIMIT OF GENERATOR                 :        60.00000
TOTAL SPECIFIED MW LOAD                          :        190.00000 reduced      190.00000
TOTAL SPECIFIED MVAR LOAD                        :         45.00000 reduced        45.00000
TOTAL SPECIFIED MVAR COMPENSATION                :          0.00000 reduced           0.00000
-------------------------------------------------------------------------------

IN SERVICE VALUES
TOTAL SPECIFIED MW GENERATION                    :         80.00000
TOTAL MIN MVAR LIMIT OF GENERATOR                :         60.00000
TOTAL MAX MVAR LIMIT OF GENERATOR                :         60.00000
TOTAL SPECIFIED MW LOAD                          :        190.00000 reduced        190.00000
TOTAL SPECIFIED MVAR LOAD                        :         45.00000 reduced        45.00000
TOTAL SPECIFIED MVAR COMPENSATION                :          0.00000 reduced           0.00000
-------------------------------------------------------------------------------

GENERATOR DATA FOR FREQUENCY DEPENDENT LOAD FLOW

SLNO*       FROM FROM           P-RATE          P-MIN          P-MAX      %DROOP       PARTICI          BIAS
NODE NAME*              MW               MW              MW                  FACTOR       SETTING
C0             C1          C2
------ ---- -------- -------- --------- --------- --------- --------- ---------
1      1        Bus1    40.000      0.0000           40.0000      4.0000         0.0000        0.0000
0.0000        0.0000        0.0000
2      2        Bus2    40.000      0.0000           40.0000      4.0000         0.0000        0.0000
0.0000        0.0000        0.0000
-------------------------------------------------------------------------------
Slack bus angle (degrees) :              0.00
-------------------------------------------------------------------------------

-------------------------------------------------------------------------------
Iteration count 0 maxp 0.450000 maxq 0.285001
Iteration count 1 maxp 0.043225 maxq 0.015055
Iteration count 2 maxp 0.298492 maxq 1.111672
Iteration count 3 maxp 0.011113 maxq 0.041154
Iteration count 4 maxp 0.000015 maxq 0.000114
Iteration count 5 maxp 0.000000 maxq 0.000001

-------------------------------------------------------------------------------
BUS VOLTAGES AND POWERS

NODE FROM               V-MAG   ANGLE            MW           MVAR          MW          MVAR         MVAR

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---- --------           ------ ------       -------- --------                 -------- -------- --------
1          Bus1   1.0000      0.00      158.767         19.952             0.000           0.000        0.000           <
2          Bus2   0.9588     -5.43       40.000         30.000            60.000          10.000        0.000           >
3          Bus3   0.9154     -9.90        0.000          0.000            45.000          15.000         0.000 @
4          Bus4   0.9176     -9.84        0.000          0.000            40.000           5.000        0.000 @
5          Bus5   0.9206     -9.02        0.000          0.000            45.000          15.000        0.000 @
-------------------------------------------------------------------------------

NUMBER OF BUSES EXCEEDING MINIMUM VOLTAGE LIMIT (@ mark) :                                           3
NUMBER OF BUSES EXCEEDING MAXIMUM VOLTAGE LIMIT (# mark) :                                           0
NUMBER OF GENERATORS EXCEEDING MINIMUM Q LIMIT (< mark)                                     :        1
NUMBER OF GENERATORS EXCEEDING MAXIMUM Q LIMIT (> mark)                                     :        1

-------------------------------------------------------------------------------

LINE FLOWS AND LINE LOSSES

SLNO CS FROM FROM                   TO TO                       FORWARD                              LOSS                 %
---- -- ---- -------- ---- -------- -------- -------- -------- -------- -------
1    1       1       Bus1      2        Bus2     158.767            19.952        5.1467           9.6825      160.0!
2    1       1       Bus1      3        Bus3            LINE IS OPEN
3    1       4       Bus4      5        Bus5         -4.863         -1.597        0.0227           -4.1560          6.0&
4    1       2       Bus2      3        Bus3         41.614          8.889        1.2054           0.1019          44.4^
5    1       2       Bus2      4        Bus4         40.888          7.891        1.1530           -0.0637         43.4^
6    1       2       Bus2      5        Bus5         51.119         13.491        1.2333           1.0496          55.1\$
7    1       3       Bus3      4        Bus4         -4.592         -6.214        0.0060           -1.6621          8.4&
-------------------------------------------------------------------------------

! NUMBER OF LINES LOADED BEYOND                 125%                      :      1
@ NUMBER OF LINES LOADED BETWEEN 100% AND 125% :                                 0
# NUMBER OF LINES LOADED BETWEEN                 75% AND 100% :                  0
\$ NUMBER OF LINES LOADED BETWEEN                 50% AND        75% :            1
^ NUMBER OF LINES LOADED BETWEEN                 25% AND        50% :            2
& NUMBER OF LINES LOADED BETWEEN                     1% AND     25% :            2
* NUMBER OF LINES LOADED BETWEEN                     0% AND         1% :         0
-------------------------------------------------------------------------------

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NEW SYSTEM FREQUENCY FOR ISLAND 1                      :    60.000000 Hzs

-------------------------------------------------------------------------------
Summary of results
TOTAL REAL POWER GENERATION           :      198.767 MW
TOTAL REAL POWER DRAWAL -ve g :                0.000 MW
TOTAL REACT. POWER GENERATION :               49.953 MVAR
GENERATION pf                         :        0.970

TOTAL SHUNT REACTOR INJECTION :                0.000 MW
TOTAL SHUNT REACTOR INJECTION :                0.000 MVAR

TOTAL SHUNT CAPACIT.INJECTION :                0.000 MW
TOTAL SHUNT CAPACIT.INJECTION :                0.000 MVAR

TOTAL REAL POWER LOAD                 :      190.000 MW
TOTAL REAL POWER INJECT,-ve L :                0.000 MW
TOTAL REACTIVE POWER LOAD             :       45.000 MVAR
TOTAL COMPENSATION AT LOADS           :        0.000 MVAR
TOTAL HVDC REACTIVE POWER             :        0.000 MVAR

TOTAL REAL POWER LOSS (AC+DC) :             8.767104 MW (        8.767104+   0.000000)
PERCENTAGE REAL       LOSS (AC+DC) :           4.411
TOTAL REACTIVE POWER        LOSS      :    4.952199 MVAR

-------------------------------------------------------------------------------

Zone wise distribution
Description             Zone # 1
---------------- ----------
MW generation           198.7669

MVAR generation          49.9526

-------------------------------------------------------------------------------

LINE FLOWS AND LINE LOSSES

SLNO CS FROM FROM             TO TO                        FORWARD                LOSS            %

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---- -- ---- -------- ---- -------- -------- -------- -------- -------- -------
1   1     1       Bus1       2         Bus2     158.767         19.952       5.1467     9.6825     160.0!
2   1     1       Bus1       3         Bus3            LINE IS OPEN
3   1     4       Bus4       5         Bus5      -4.863         -1.597       0.0227    -4.1560        6.0&
4   1     2       Bus2       3         Bus3      41.614             8.889    1.2054     0.1019       44.4^
5   1     2       Bus2       4         Bus4      40.888             7.891    1.1530    -0.0637       43.4^
6   1     2       Bus2       5         Bus5      51.119         13.491       1.2333     1.0496       55.1\$
7   1     3       Bus3       4         Bus4      -4.592         -6.214       0.0060    -1.6621        8.4&
-------------------------------------------------------------------------------

! NUMBER OF LINES LOADED BEYOND             125%                   :       1
@ NUMBER OF LINES LOADED BETWEEN 100% AND 125% :                           0
# NUMBER OF LINES LOADED BETWEEN             75% AND 100% :                0
\$ NUMBER OF LINES LOADED BETWEEN             50% AND           75% :       1
^ NUMBER OF LINES LOADED BETWEEN             25% AND           50% :       2
& NUMBER OF LINES LOADED BETWEEN                 1% AND        25% :       2
* NUMBER OF LINES LOADED BETWEEN                 0% AND         1% :       0
-------------------------------------------------------------------------------

NEW SYSTEM FREQUENCY FOR ISLAND 1                          :     60.000000 Hzs

-------------------------------------------------------------------------------
Summary of results
TOTAL REAL POWER GENERATION           :          198.767 MW
TOTAL REAL POWER DRAWAL -ve g :                    0.000 MW
TOTAL REACT. POWER GENERATION :                   49.953 MVAR
GENERATION pf                         :            0.970

TOTAL SHUNT REACTOR INJECTION :                    0.000 MW
TOTAL SHUNT REACTOR INJECTION :                    0.000 MVAR

TOTAL SHUNT CAPACIT.INJECTION :                    0.000 MW
TOTAL SHUNT CAPACIT.INJECTION :                    0.000 MVAR

TOTAL REAL POWER LOAD                 :          190.000 MW
TOTAL REAL POWER INJECT,-ve L :                    0.000 MW
TOTAL REACTIVE POWER LOAD             :           45.000 MVAR
TOTAL COMPENSATION AT LOADS           :            0.000 MVAR
TOTAL HVDC REACTIVE POWER             :            0.000 MVAR

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TOTAL REAL POWER LOSS (AC+DC) :            8.767104 MW (   8.767104+   0.000000)
PERCENTAGE REAL      LOSS (AC+DC) :            4.411
TOTAL REACTIVE POWER        LOSS      :    4.952199 MVAR

-------------------------------------------------------------------------------

Zone wise distribution
Description             Zone # 1
---------------- ----------
MW generation           198.7669

MVAR generation          49.9526

Department of Electrical & Electronics Engineering
Power System Simulation Laboratory

Exp No. 6
Date: 08.02.2010

DC LOAD FLOW ANALYSIS USING MATLAB

AIM :

To determine DC power flow among various buses and DC power generated by slack bus using

MATLAB.

THEORY:

The Newton power flow is the most robust power flow algorithm used in practice. The one

draw back is the fact that the term in the Jacobian matrix must be recalculated during each iteration and

then the entire set of linear equations,

Must also be resolved during each iterations. Such situation where a lot of computations have to be
made a linear approximation of the load flow problem can be made to save computation time, this is
called the DC load flow. The DC Load flow is principally different from the decoupled load flow. In
the DC power flow, the non-linear load flow equations are linearized

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to ease the calculation and to speed up the computation, of the unknown voltages, this means that the

actual model of the power system is altered, then it affects the final solution of the load flow.

In the decoupled load flow, the non linear load flow equation are solved iteratively and

approximation are made to the Jacobin matrix only, therefore only the speed of the convergence is

affected, but the final result remains the same. The DC load flow can also be applied to find the fairly

good approximation of the unknown voltages, that can be used as initial values in a Newton-Raphson /

decoupled load flow solution or calculations.

The decoupled power flow results in

The above equations can be written in matrix form as

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Simplifying eqns (1) and (2) using ΔP - Δθ relationship, Assume rik << Xik this changes - Bik to

Eliminate all shunt reactions to ground.

Eliminate all shunts to ground, which arise from auto transformer.

Simplifying the Δ θ – Δ|E1| relationship of eqn (2)

Omit all effects from phase shift transformer

The resulting equations are

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Where the terms in the matrices are

Assuming a branch from i to k.

In DC power flow, a further simplification of the power flow algorithm involves simply

dropping the Q- V (eqn (4) altogether. This results in a completely linear, non iterative , power flow

algorithm. Assume that |Ei| = 1.0pu.

Eqn (3) becomes

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In dc power flow is only good for calculating MW flows on transmission lines and

transformers. It gives no indication of what happen to voltage magnitudes, or MVAR or MVA

flows. The power flowing on each line using the DC power flow is then

PROBLEM :

Determine the dc power flow among various buses and dc power generated by slack bus using

MATLAB. For the following network.

Fig 1

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Susceptance Matrix

Susceptance Matrix after deleting row and column.

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Slack Bus Power = 35MW

Fig 2.

Slack Bus Power = 35MW

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ALGORITHM:

Step 1 : Read the number of buses.

Step 2 : Read the reactance value between the buses.

Step 3 : Formulate the C matrix by inversing the reactance values.

Step 4 : read the reference bus number as i.

Step 5 : Formulate the B matrix by eliminating ith row and column of C matrix

Step 6 : Calculate [θ] = [B]-1 [P]

Step 7 : Evaluate

Pik = (1/Xik) (θi -θk)

Pi = Σ Pk

k = buses connected to „i‟

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FLOW CHART:

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clear all

n=input('Enter no. of buses =');

x=zeros(n,n);

x

for i=1:n

for j =1:n

if (i<j)

fprintf('Enter the values of X(%d,%d):',i,j);

x(i,j)=input('');

x(j,i)=x(i,j);

if (x(i,j)~=0)

B(i,j)=(-1)/x(i,j);

B(j,i)=(-1)/x(j,i);

end

end

end

end

fprintf('the reactance matrix is \n');

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sum=0;

for i=1:n

for j=1:n

sum=sum+B(i,j);

end

B(i,i)=-1*sum;

sum=0;

end

fprintf('The susceptance matrix is \n');

B

p=input('Enter the reference bus = ');

B(p,:)=[];

B(:,p)=[];

fprintf('The susceptance matrix after deleting row n column is \n');

B

F=zeros(n,1);

for i=1:n

if (i~=p)

fprintf('Enter the value of power(%d):',i);

F(i)=input('');

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end

end

F(p,:)=[];

fprintf('The power at the given bus = ')

F

H=inv(B);

J=H*F;

J

W=zeros(n,1);

for j=1:n-1

W(j,1)=J(j,1);

end

for i=1:p-1

T(i,1)=W(i,1);

end

T(p,1)=0;

for i=p+1:n

T(i,1)=W(i-1,1);

end

T

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E=zeros(n,1);

for i=1:n

for j=1:n

if(i~=j && x(i,j)~=0)

E(i,j)=(1/x(i,j))*(T(i,1)-T(j,1));

end

end

end

E

for i=1:n

for j=1:n

end

E

fprintf('Slack bus power is E(%d,%d)=',p,p)

E(p,p)

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RESULT:

Enter no. of buses =3

x=

0       0    0

0       0    0

0       0    0

Enter the values of X(1,2):.2

Enter the values of X(1,3):.4

Enter the values of X(2,3):.25

the reactance matrix is

x=

0       0.2000 0.4000

0.2000            0 0.2500

0.4000 0.2500            0

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The Susceptance matrix is

B=

7.5000 -5.0000 -2.5000

-5.0000    9.0000 -4.0000

-2.5000 -4.0000 6.5000

Enter the reference bus = 3

The Susceptance matrix after deleting row n column is

B=

7.5000 -5.0000

-5.0000    9.0000

Enter the value of power(1):65

Enter the value of power(2):-100

The power at the given bus =

F=

65

-100

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J=

2.0000

-10.0000

T=

2.0000

-10.0000

0

E=

0 60.0000 5.0000

-60.0000      0 -40.0000

-5.0000 40.0000       0

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The power flow

E=

65.0000 60.0000 5.0000

-60.0000 -100.0000 -40.0000

-5.0000 40.0000 35.0000

Slack bus power is E(3,3)=

ans =

35

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Exp No: 7 a)
Date: 24 .2.2010

SIMULATION AND ANALYSIS OF MAGNETIC CIRCUITS USING

Injection of third harmonic voltage using three coupled windings

AIM:
To inject third or higher order harmonic voltage using three coupled windings to a circuit

THEORY:

Coupled windings or mutual inductance

Description

The Mutual Inductance block can be used to model two- or three-windings inductances with

equal mutual coupling, or to model a generalized multi-windings mutual inductance with balanced or

unbalanced mutual coupling.

If two- or three-windings inductances model with equal mutual coupling is used, specify the

self-resistance and inductance of each winding plus the mutual resistance and inductance. The

electrical model for this block in this case is given below:

Fig(1)

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If a general mutual inductance model is used, specify the number of self windings (not just

limited to 2 or 3 windings) plus the Resistance and Inductance matrices that define the mutual coupling

relationship between the windings (balanced or not).

Type of mutual inductance

Set to Two or Three windings with equal mutual terms to implement a three-phase mutual

inductance with equal mutual coupling between the windings

Winding 1 self impedance

The self-resistance and inductance for winding 1, in ohms (Ω) and henries (H).

Winding 2 self impedance

The self-resistance and inductance for winding 2, in ohms (Ω) and henries (H).

Three windings Mutual Inductance

If selected, implements three coupled windings; otherwise, it implements two coupled windings.

Winding 3 self impedance

The Winding 3 self impedance parameter is not available if the Three windings Mutual

Inductance parameter is not selected. The self-resistance and inductance in ohms (Ω) and henries (H)

for winding 3.

Mutual impedance

The mutual resistance and inductance between windings, in ohms (Ω) and henries (H). The

mutual resistance and inductance corresponds to the magnetizing resistance and inductance on the

standard transformer circuit diagram. If the mutual resistance and reactance are set to [0 0], the block

implements three separate inductances with no mutual coupling.

Generalized mutual inductance:

Type of mutual inductance

Set to Generalized mutual inductance to implement a multi windings mutual inductance with

mutual coupling defined by an inductance and a resistance matrix.

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Inductance matrix L

The inductance matrix, in Henrys, that defines the mutual coupling relationship between the

self windings. It must be a N-by-N symmetrical matrix.

Resistance matrix R

The resistance matrix, in Ohms, that defines the mutual coupling relationship between the

self windings. It must be a N-by-N symmetrical matrix.

Negative values are allowed for the self- and mutual inductances as long as the self-inductances are

different from the mutual inductance.

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Fig(3). Output voltage wave form

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RESULT:
Three coupled winding circuit has been Simulated using Simulink.The third and higher order
harmonics voltage was injected and corresponding output waveform was obtained.

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Exp No: 7 b)
Date: 24.2.2010

LINEAR TRANSFORMER
AIM:
To simulate a distribution transformer network feeding line-to-neutral and line-to-line loads.

THEORY:

Distribution Transformer or Linear Transformer
Implement two-or three-winding linear transformer

75KVA,14.4KV/120/120,50 Hz , R1=27.648Ω, L1=0.22 H, R2 =R3=55.296Ω, L2=L3=0.

Description

The linear transformer block model shown consists of three coupled windings wound on the
same core.

Fig(4). The linear transformer block model

The model takes into account the winding resistances (R1 R2 R3) and the leakage inductances
(L1 L2 L3), as well as the magnetizing characteristics of the core, which is modeled by a linear (Rm
Lm) branch.

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Modeling an Ideal Transformer
To implement an ideal transformer model, set the winding resistances and inductances to zero,
and the magnetization resistance and inductance (Rm Lm) to infinity.

Units
Specify the units used to enter the parameters of the Linear Transformer Block. Select pu to use
per unit. Select SI to use SI units. Changing the Units parameters from pu to SI, or from SI to pu, will
automatically convert the parameters displayed in the mask of the block.

Nominal Power and Frequency
The nominal power rating Pn in volt-amperes (VA) and frequency fn, in hertz (Hz), of the
transformer.

Winding 1 Parameters
The nominal voltage V1, in volts RMS, resistance, in pu or ohms, and leakage inductance, in
pu or henries. The pu values are based on the nominal power Pn and on VI.Set the winding resistances
and inductances to zero to implement an ideal winding.

Winding 2 Parameters
The nominal voltage V2, in volts RMS, resistance,in pu or ohms,and leakage inductance,in pu
or henries.The pu values are based on the nominal power Pn and on V2.Set the winding resistances and
inductances to zero to implement an ideal winding.

Three winding transformer
If selected, implements a linear transformer with three windings; otherwise, it implements a
two-winding transformer.

Winding 3 Parameters
The Winding 3 parameters parameter is not available if the three windings transformer
parameter is not selected.

The nominal voltage in volts RMS (Vrms), resistance, in pu or ohms, and leakage inductance,in
pu or henries.The pu values are based on the nominal power Pn and on V3.Set the winding resistances
and inductances to zero to implement an ideal winding.

Magnetization resistance and reactance
The resistance and inductance simulating the core active and reactive losses. When selected ,
the pu values are based on the nominal power Pn and on VI.For example, to specify 0.2% of active and
reactive core losses, at nominal voltage, use Rm=500 pu and Lm=500 pu.

Rm must have a finite value when the inductance of winding 1 is greater than zero.

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Fig(6).Input voltage wave form

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Fig(7).Input voltage wave form

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Fig(8).Output current wave form

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Fig(9).Output voltage wave form

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RESULT:
Linear transformer network feed up line to neutral and line to line loads was simulated and
corresponding output waveform are obtained.

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Exp No: 8
Date : 04.02.2010

SIMULATION AND MEASUREMENTS OF THREE PHASE
AIM:

To

   Explore the powerlib library in the SimPowerSystems Software

   Learn how to build a simple circuit from the powerlib library

   Interconnect Simulink blocks with the given circuit

The circuit below represents an equivalent power system feeding a 300 km transmission line. The

line is compensated by a shunt indicator at its receiving end. To simplify matters, only one of the three

phases is represented. The parameters shown in the figure are typical of a 735 kV power system.

CIRCUIT TO BE MODELED:

Fig 1. Circuit to be modeled

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PROCEDURE:

    Open the SimPowerSystems main library by entering the following command at the MATLAB

prompt.

powerlib

This command displays a Simulink window showing icons of different block libraries.

    Open these libraries to produce the windows containing the blocks to be copied into the circuit. Each

component is represented by a special icon having one or several inputs and outputs corresponding to

the different terminals of the component:

    Open the Electrical Sources library and copy the AC Voltage Source block into the circuit window.

    Open the AC Voltage Source dialog box by double-clicking the icon and enter the Amplitude, Phase

and Frequency parameters according to the given values.

Note that the amplitude to be specified for a sinusoidal source is its peak value (424.4e3*sqrt(2) volts in this

case).

    Change the name of this block from AC Voltage Source to Vs.

    Copy the parallel RLC branch block, which can be found in the Elements library of powerlib, set its

parameters and name it Z_eq.

    The resistance Rs_eq of the circuit can be obtained from the Parallel RLC Branch block. Duplicate the

Parallel RLC Branch block. Select R for the Branch Type parameter and the set the R parameter

according to the value given. Enter the values as infinity (inf) and zero (0) for L and C.

Once the dialog box is closed, notice that the L and C components have disappeared so that the icon now

shows a single resistor. Name this block Rs_eq.

    The model of a line with uniformly distributed R, L and C parameters normally consists of a delay equal

to the wave propagation time along the line. This model cannot be simulated as a linear system because

a delay corresponds to an infinite number of states. However, a good approximation of the line with a

finite number of states can be obtained by cascading several PI circuits, each representing a small

section of the line.

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A PI section consists of a series R-L branch and two shunt C branches. The model accuracy depends on the

number of PI sections used for the model. Copy the PI Section Line block and set its parameters.

    It is more convenient to use a Series RLC Load block that allows to specify directly inactive and

reactive powers absorbed by the shunt reactor.

Copy the Series RLC Load block, which can be found in the Elements library of powerlib. Name this block

110 Mvar. Set its parameters as follows:

Vn 424.4e3 V

fn    60Hz

P     110e6/300 w (quality factor = 300)

QL 110e6 vars

Qe    0

Note that, as no reactive capacitive power is specified, the capacitor disappears on the block icon when the

dialog box is closed.

    A Voltage Measurement block and current measurement block is used to measure the voltage and

current at different nodes. This block is found in the Measurement library of powerlib.

    To observe the voltage and current a display system is needed. This can be any device found in the

Open the Sinks library and copy the Scope block. If the scope were connected directly at the output of the

voltage measurement, it would display the voltage in volts.

    From the Simulation menu, select Start. A Powergui block is automatically added to the model from

which the values of voltage and current are obtained.

    Open the Scope blocks and observe the parameters at different nodes.

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Fig 2. Simulation block diagram

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RESULT:

Vrms 1          424400<00V

Vrms 2          428722.97<-0.120V

Irms 1          261.83 <-89.930A

Irms 2          284.15<89.710A

A 300 km transmission line has been simulated and the corresponding voltage and current has been

measured.

Department of Electrical & Electronics Engineering
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Exp No: 9
Date: 03.03.2010

MODELING OF AUTOMATIC GENERATION CONTROL FOR A

AIM:

To construct a SIMULINK block diagram of a two area system network and obtain the

frequency deviation response and power deviation response.

THEORY:

As the system load changes continuously the generation is adjusted automatically to restore the

frequency to nominal value. This scheme is known as automatic generation control. The operation

objectives of the LFC are to maintain reasonably uniform frequency, to divide the load between

generators, and to control the tie- line interchange schedules. The change in frequency and tie-line real

power are sensed, which is a measure of the change in rotor angle, δ i.e., the rotor error Δδ to be

corrected. The error signal, i.e., Δf and ΔPtie, are amplified, mixed, and transformed into a real power

command signal ΔPv, which is sent to the prime mover to call for an increment in the torque.

The prime mover, therefore, brings change in the generation output by an amount ΔPg which

will change the values of Δf and ΔPtie within the specified tolerance. The first step of analysis of

design of control system is mathematical modeling of the system. Proper assumptions and

approximations are made to linearize the mathematical equations describing the system and transfer

function model obtained for the following components.

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GENERATOR MODEL:

Applying swing equation of a synchronous machine with small perturbation is given by

2H d2Δδ = ΔPm –ΔPe
ws dt2
Or in terms of small deviation in speed

With speed expressed in per unit, without explicit per unit notation, we have

Taking Laplace transform of above equation, we obtain

The above relation is shown in block diagram form as below.

Fig 1.Generator Model

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The load on a power system consists of a variety of electrical devices. For resistive loads, such

as lighting and heating loads, the electrical power is independent of frequency. Motor loads are

sensitive to changes in frequency. How sensitive it is to frequency depends on the composite of the

speed - load characteristics of all the driven devises. The speed - load characteristic of a composite

load is approximated by ΔPe = ΔPL + DΔw

where ΔPL is the non frequency - sensitive load change, and DΔw is the frequency - sensitive load

change. D is expressed as percent change in load divided by percent change in frequency.

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PRIME MOVER MODEL:

The model for turbine relates changes in mechanical power output Δ Pm to changes in steam

valve position ΔPv. Different types of turbine vary widely in characteristics. The simplest prime mover

model for non reheat steam turbine can be approximated with a single time constant τT resulting in

following transfer function

.

the block diagram for a sample turbine is show below

Fig 3. Prime mover model

The time constant τT is in range of 0.2 to 2.0 seconds.

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GOVERNOR MODEL:

The speed governor mechanism act as comparator whose output ΔPg is the difference between

reference set power ΔPref and power          as given             from      the       governor     speed

characteristics. R is the speed regulation of governor (5-6 percent from zero to full load).

Or in s – domain

ΔPg (s) =ΔPref (s) -    ΔΩ (s)

The command ΔPg is transformed through the hydraulic amplifier to the steam valve position

command ΔPv. Assuming a linear relationship and considering a simple time constant τg, we have the

following S-domain relation.

Fig 4. Governor model

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AGC IN THE TWO AREA SYSTEM:

Consider two areas represented by an equivalent generating unit interconnected by a lossless tie

line with reactance Xtie. Each area is represented by a voltage source behind an equivalent reactance as

in following figure.

Fig 5. Reactance diagram

During normal operations real power transferred over tie line is given by

P12 = |E||E2| Sin δ12 .................................(1)
X12
X12 = X1+Xtie + X2

δ12 =δ1-δ2

Equation 1 can be linearised for a small deviation in the tie line flow ΔP12 from nominal value.

ie,

=Ps δ12............................................... (2)

Ps = Slope of the power angle curve at the initial operating angle δ120= δ 10- δ 20

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........................... (3)

= Ps δ12

Tie line power deviation then takes on the form

ΔP12 = Ps (Δδ1- Δδ2) ...................................... (4)

The tie line flow appears as a load increase in one area and the load decrease in the other area,

depending on the direction of flow. The direction of the flow is dictated by phase angle difference if

Δδ1>Δδ2 the power flows from area 1 to area 2.

A block diagram representation for the two area LFC containing only primary loop is shown in the

following figure.

Fig 6. Block diagram representation for the two area LFC containing only primary loop

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Consider a load change ΔPl1 in area 1. In steady state both areas have same steady state

frequency deviation

ie, Δw = Δw1 = Δw2

and ΔPm1 - ΔP12 - ΔP11 = ΔwD1 ........................(5)

ΔPm2 + ΔP12 = ΔwD2.......................... (6)

The change in mechanical power is determined by governor speed characteristics, given by

ΔPml = -       .................................. (7)

ΔPm2 = .........    ............................ (8)

Substituting (7) & (8) in (5) & (6) and solving for Δw, we have

Where,

B1, B2 are frequency bias factors.

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ΔP12 = Change in the tie line power is given by

PROBLEM:

A two - area system connected by a tie line has the following parameters on a 1000 - MVA common

base.

Table 1

The units are operating in parallel at the nominal frequency of 60Hz. The synchronizing power

coefficient is computed from the initial operating condition and is given to be Ps=2.0 per unit. A load

change of 187.5 MW occurs in area 1. Construct the SIMULINK block diagram and obtain the

frequency deviation response and power deviation step response.

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Fig 7.Simulation block diagram
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Fig 8. Power deviation response

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Fig 9. Power deviation response

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Fig 10.Frequency deviation response

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RESULT:
The block diagram of a two area system network was constructed using Simulink and the frequency
deviation response and power deviation response was obtained

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Exp No: 10

Date: 5.3.2010

MODELING AND SIMULATION OF NON-CONVENTIONAL
ENERGY SYSTEMS USING MATLAB
AIM:
To study the generic model of Double-Fed Induction Generator (DFIG) Driven by a

Wind Turbine.

THEORY:
1. Turbine response to a change in wind speed

Open the “Wind Speed” step block specifying the wind speed. Initially, wind speed is set at

8m/s, then at t=5s, wind speed increases suddenly at 14m/s. Start simulation and observe the signals on

the “Wind Turbine” scope monitoring the wind turbine voltage, current, generated active and reactive

powers, DC bus voltage and turbine speed. At t=5s, the generated active power starts increasing

smoothly(together with the turbine speed) to reach its rated value of 9 MW in approximately 15s.Over

that time frame the turbine speed will have increased from 0.8 pu to 1.21 pu. Initially, the pitch angle

of the turbine blades is zero degree and the turbine operating point follows the red curve of the turbine

power characteristics up to point D. Then the pitch angle is increased from 0 deg to 0.76 deg in order

to limit the mechanical power. Observe also the voltage and the generated reactive power. The reactive

power is controlled to maintain a 1 pu voltage. At nominal power, the wind turbine absorbs 0.68 MVar

(generated Q = -0.68 MVar) to control voltage at 1pu. If we change the mode of operation to “Var

regulation” with the “Generated reactive power Qref” set to zero, we will observe that voltage

increases to 1.021 pu when the wind turbine generates its nominal power at unity power factor.

2. Simulation of voltage sag on the 120 kV system

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We will now observe the impact of a voltage sag resulting from a remote fault on the 120-kV

system. First in the wind speed step block, disable the wind speed step by changing the Final value

from 14 to 8 m/s. Then open the 120 kV voltage source menu. In the parameter “Time variation of”,

select “Amplitude”. A 0.15 pu voltage drop lasting 0.5s is programmed to occur at t =5s. Make sure

that the control mode is still in Var regulation with Qref=0. Start simulation and open the “Grid”

scope. Observe the plant voltage and current as well as the motor speed. Note that the wind farm

produces 1.87 voltage an d current as well as the motor speed. Note that the wind farm produces 1.87

MW. At t=5s, the voltage falls below 0.9 pu and at t=5.22s, the protection system trips the plant

because an under voltage lasting more than 0.2 s has been detected (look at the protection settings and

status in the “PLANT” subsystem). The plant current falls to zero and motor speed decreases

gradually, while the wind farm continues generating at a power level of 1.87 MW. After the plant has

tripped, 1.25 MW of power (P_B25 measured at bus B25) is exported to the grid.

Now, change the wind turbine control mode to “Voltage regulation” and repeat the test. We will notice

that the plant does not trip anymore. This is because the voltage support provided by the 5 Mvar

reactive power generated by the wind turbines during the voltage sag keeps the plant voltage above 0.9

pu protection threshold. The plant voltage during sag is now 0.93 pu.

3. Simulation of a fault on the 25 kV line system.

Finally, we will now observe impact of a single phase to ground fault occurring on the 25 kV

line at B25 bus. First disable the 120 kV voltage step. Now open the “Fault” block menu and select

“Phase A Fault”. Check that the fault is programmed to apply a 9-cycle single phase to ground fault at t

=5 s.

We should observe that when the wind tubine is in “Voltage regulation” mode, the positive sequence

voltage at wind turbine terminals(V1_B575) drops to 0.8 pu during the fault which is above the under

voltage protection threshold(0.75 pu for t>0.1 sec). the wind farm therefore stays in service .However,

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if the “Var regulation “ mode is used with Q ref =0, the voltage drops under 0.7 pu and the under

voltage protection trips the wind point. We can now observe that the turbine speed increases. At t=40s

the pitch angle starts to increase in order to limit the speed.

PROCEDURE:

1. Open matlab - new-model file and save it

2. Click start button on matlab- start-simulink-library browser-sim power systems

3. Select the required blocks and elements from the library and drag it to the model file

4. Connect the blocks and save it

5. Set simulation time ( 2 sec ) and start simulation

6. Click on the scope so that output is obtained.

Department of Electrical & Electronics Engineering
Power System Simulation Laboratory

Fig No: 1 Model of Double-Fed Induction Generator driven by a Wind Turbine

Department of Electrical & Electronics Engineering
Power System Simulation Laboratory

Fig no. 2 : Scope output curves of Grid

Department of Electrical & Electronics Engineering
Power System Simulation Laboratory

Fig no. 3 : Scope output curves of Wind Turbine

Department of Electrical & Electronics Engineering
Power System Simulation Laboratory

RESULT
The generic model of Double-Fed Induction Generator (DFIG) Driven by a Wind Turbine is studied.

Department of Electrical & Electronics Engineering

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