PSA final edited NEW.doc

7/28/2019 PSA final edited NEW.doc

1/96

SSM COLLEGE OF ENGINEERING, KOMARAPALAYAM: 638183

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

POWER SYSTEM SIMULATION LABORATORY

Name:______________________________________________

Course & Branch:_____________________________________

Reg. No: _______________________Semester______________

Sub Code & Name:____________________________________


2/96

LIST OF EXPERIMENTS

1. COMPUTATION OF LINE PARAMETERS

2. MODELLING OF TRANSMISSION LINE

3. FORMATION OF BUS ADMITTANCE MATRIX

4. FORMATION OF BUS IMPEDANCE MATRIX

5. LOAD FLOW ANALYSIS- GAUSS SEIDAL METHOD

6. SHORT CIRCUIT ANALYSIS

7. LOAD FLOW ANALYSIS- NEWTON RAPHSON METHOD

8. LOAD FLOW ANALYSIS FAST DECOUPLED LOAD FLOW METHOD

9. LOAD FREQUENCY DYNAMICS OF SINGLE AND TWO AREA SYSTEM

10. TRANSIENT STABILITY ANALYSIS: SMIB SYSTEM

11. TRANSIENT STABILITY ANALYSIS MULTI MACHINE INFINITE BUS

SYSTEM

12. ECONOMIC LOAD DISPATCH WITHOUT LOSSES USING DIRECT

METHOD

13. ECONOMIC LOAD DISPATCH WITH LOSS

14. ELECTROMAGNETIC TRANSIENTS


3/96

INDEX

Ex. No Date Name of the ExperimentPage

No

Marks

AwardedSignature


4/96

INDEX

Ex. No Date Name of the ExperimentPage

No

Marks

AwardedSignature


5/96

Ex. No :

DATE :

FORMATION OF BUS ADMITTANCE MATRIX

AIM:

To write a program in matlab to obtain the bus admittance matrix for the given powersystem network by direct inspection method.

ALGORITHM:

STEP 1: Start the Program.

STEP 2: Get the number of buses and number of lines.

STEP 3: For each line get the line impedance & off line charging impedance.Calculate the line admittance by reciprocating the line impedance.

STEP 4: Initialize the Ybus matrix.

STEP 5: Form the diagonal elements of the Ybus matrix as the sum of then

admittances connected to the bus and off-diagonal elements as the negative of the

admittances.

STEP 6: Print the Ybus matrix.

STEP 7: Stop the program.


6/96

FLOW CHART:

FORMATION OF BUS ADMITTANCE MATRIX


7/96

PROBLEM:

Determine the Ybus matrix for the given four bus system. The datas are given in the table

below.

Bus Code Line impedance p.u Off-line charging admittance p.u

Start

Read the no of lines

(nl) & no of buses(nb)

Read the bus impedance

& off line chargingadmittance

Calculate the Ybus matrix

Initialize the Ybus matrix

Set i = 1

Form the Ybus matrix using the formula

i =i+1

Is

Print the Ybus matrix

Stop

yes

No


8/96

1 - 2 0.2+0.8j 0.02j

1 - 3 0.1+0.4j 0.01j

2 - 3 0.3+0.9j 0.03j

2 - 4 0.25+1j 0.04j

3 - 4 0.2+0.8j 0.02j

PROGRAM:

%Y BUS FORMATION BY INSPECTION METHODclear all;

clc;

nb=input('enter the no. of buses');nl=input('enter the no. of lines');

z=zeros(nb,nb);

y=zeros(nb,nb);yb=zeros(nb,nb);

yo=zeros(nb,nb);

input('frombus tobus impedance line charging admittance');for i=1:nl

fb(i)=input('');tb(i)=input('');

z(fb(i),tb(i))=input('');z(tb(i),fb(i))=z(fb(i),tb(i));

yc=input('');

yo(fb(i))=yo(fb(i))+yc;yo(tb(i))=yo(tb(i))+yc;

end

%LINE ADMITTANCE MATRIX CALCULATION

for i=1:nb

for j=1:nb

if z(i,j)~=0y(i,j)=1/z(i,j);

end

end

end

%YBUS FORMATION

for i=1:nb

yb(i,i)=yb(i,i)+yo(i);for j=1:nb

if i~=j

yb(i,i)=yb(i,i)+y(i,j);

yb(i,j)=-y(i,j);end

end

endyb

OUTPUT:


9/96

enter the no. of buses4

enter the no. of lines5

from bus to bus impedance line charging admittance1

2

0.2+0.8i

0.02i1

30.1+0.4i

0.01i

2

30.3+0.9i

0.03i

24

0.25+1i0.04i3

4

0.2+0.8i0.02i

yb =

0.8824 - 3.4994i -0.2941 + 1.1765i -0.5882 + 2.3529i 0

-0.2941 + 1.1765i 0.8627 - 3.0276i -0.3333 + 1.0000i -0.2353 + 0.9412i

-0.5882 + 2.3529i -0.3333 + 1.0000i 1.2157 - 4.4694i -0.2941 + 1.1765i0 -0.2353 + 0.9412i -0.2941 + 1.1765i 0.5294 - 2.0576i

MODEL CALCULATION:


10/96


11/96

RESULT:

Ex. No:

DATE :


12/96

FORMATION OF BUS IMPEDANCE MATRIX

AIM:

To write a program in matlab to obtain the bus impedance matrix for the given power

system network by direct method.

ALGORITHM:

STEP 1: Start the Program.

STEP 2: Get the number of buses and number of lines.

STEP 3: Add the first element a to form the zbus1 matrix.

STEP 4: Add the second element b to the second diagonal to form the Zbus2 matrix.

STEP 5: Add the third element c to the third diagonal to form the Zbus3 matrix.

STEP 6: Initialize the zbus matrix.

STEP 7: Form the Zbus using the formula

++

++=

)1,1(3

),1(3)1,(3),(3),(

nnZbus

knZbusnjZbuskjZbuskjZbus

STEP 8: Print the Zbus matrix.


FLOW CHART:


13/96

FORMATION OF BUS IMPEDANCE MATRIX

Start

Read the input

data a, b, c, d

Add the first element a to

form the zbus1 matrix

Add the second element b to the second

diagonal to form the Zbus2 matrix

Add the third element c to the third

diagonal to form the Zbus3 matrix

Initialize the zbus matrix

Form the Zbus using the formula

Print the

Zbus matrix

Stop


14/96

PROBLEM:

Determine the Zbus matrix for the given four bus system.

PROGRAM:

clear all;

clc;

a=input('enter the value of a;');b=input('enter the value of b;');

c=input('enter the value of c;');

d=input('enter the value of d;');zbuso=[a];

zbus1=[a a;

a a+b;];

zbus2=[a a a;a a+b a+b;

a a+b a+b+c;];

zbus3=[a a a a;a a+b a+b a+b;

a a+b a+b+c a+b+c;

a a+b a+b+c a+b+c+d;];n=input('enter the no of nodes');

zbus4=zeros(3,3);

for j=1:3for k=1:3

zbus4(j,k)= zbus3(j,k)-(zbus3(j,n+1)*zbus3(n+1,k)/zbus3(n+1,n+1));end

endzbus4


15/96

OUTPUT:

enter the value of a; 0.029+0.117i

enter the value of b; 0.058+0.235ienter the value of c; 0.067+0.095i

enter the value of d; 0.16+0.12ienter the no of nodes 3

zbus4 =

0.0294 + 0.0946i 0.0304 + 0.0496i 0.0229 + 0.0293i

0.0304 + 0.0496i 0.0913 + 0.1492i 0.0689 + 0.0881i

0.0229 + 0.0293i 0.0689 + 0.0881i 0.0998 + 0.1064i

MODEL CALCULATION:


16/96


17/96

RESULT:

Ex. No:


18/96

DATE :

COMPUTATION OF LINE PARAMETERS

AIM:

To write a mat lab program to determine the positive sequence line parameters L and C per

phase per km of a 3 single and double circuit for different conductor arrangement.

ALGORITHM:

STEP 1: Start the program.

STEP 2: Read the conductor type, solid or bundled.

STEP 3: Get the input value of Dab, Dbc, Dca, and GMR, d.

STEP 4: Check the type of conductor. If it is a solid conductor calculate the value of

L and C and goto step 6.

STEP 5: If it is a bundled conductor calculate the value of L and C and goto nextstep.

STEP 6: Print the value of L and C.


FLOW CHART:


19/96

COMPUTATION OF LINE PARAMETERS

Start

Enter the type of

conductor

Enter the value of Dab

,

Dbc

, Dca

, and GMR, d

Check the

type ofconductor

Find the value of L & C Find the value of L & C

Print the values of

L & C

Stop

Solid Bundled


20/96

1. Single phase two wire system:

a D a

L=210-7ln(D/R) H/m

Can=20/ln(D/R) F/m

Where,

L= Inductance of conductor

Can=Capacitance of conductor a w.r.t neutral

D= Distance between the conductors (m)

R=0.7788R

R=Radius of the conductors

0= Absolutepermittivity=8.85410-12

2. Three phase single circuit line-unsymmetrical spacing

Lavg=210-7ln(Deq/R) H/m

Can= 2 o /ln(D/R) F/m

Where,

Lavg=Average inductance of conductor



R=0.7788R


o= Absolutepermittivity=8.85410-12


21/96

3. Three phase Double circuit line-symmetrical spacing

Lavg=210-7ln( /2R) H/m

Can=2 o/7ln( /2R) F/m

Where,

Lavg=Average inductance of conductor



R=0.7788R


o= Absolutepermittivity=8.85410-12


22/96

PROBLEM:

A 3 transposed line proposed of ACSR 1, 43,000 `mil, 47/7 bobo link conductor/ph

with flat horizontal spacing of 11m between space AB and BC. The conductor have diameter of3.625cm and GMR=1.439cm. The line is to be replaced by 3 conductor bundle of ACSR= 4,

77,000cmil, 26/7 hawk of conductor having the same cross area of aluminium as the single

conductor.

A conductor having diameter of 2.1793cm and GMR of 0.8839cm. The new linebetween also have a flat horizontal configuration but it is to be operated at a high voltage and

therefore the phase spacing is to be increased to 14cm as measured from the center of thebundled. The spacing between the conductors in the bundle is 45cm.

(i) Determine the inductance and capacitance per/ph/km.

(ii) Verified the result using the available program.

PROGRAM:

clear all;

clc;disp('enter 1 for solid or 2 for bundled');

method=input('enter your choice:');dab=input('enter dab:');dbc=input('enter dbc:');

dca=input('enter dca:');

ds=input('enter gmr value:');d=input('enter diameter of conductor:');

dm=power((dab*dbc*dca),1/3);

switch(method)

case(1)l=0.2*log1p(dm/ds);

r=d/2;

c=0.0556/log(dm/r);case(2)

bs=input('enter the bundle spacing:');

dsb=power((ds*power(bs,2)),1/3);l=0.2*log1p(dm/dsb);

r=d/2;

rb=power((r*power(bs,2)),1/3);

c=0.0556/log(dm/rb);otherwise

disp('enter 1 or 2:');

end

disp('the inductance value is');disp(l);

disp('the capacitance value is');disp(c);


23/96

OUTPUT:

enter 1 for solid or 2 for bundled

enter your choice:1

enter dab:11

enter dbc:11enter dca:22

enter gmr value:0.01439enter diameter of conductor:0.03625

the inductance value is

1.3742

the capacitance value is

0.0084

enter 1 for solid or 2 for bundled

enter your choice:2enter dab:14enter dbc:14

enter dca:28

enter gmr value:0.008839enter diameter of conductor:0.021793

enter the bundle spacing:0.4572

the inductance value is

0.9950

the capacitance value is

0.0114


24/96

MODEL CALCULATION:


25/96

RESULT:


26/96

Ex. No:

DATE :

MODELLING OF TRANSMISSION LINE

AIM:

To write a mat lab program to understand the modeling and performance of medium

transmission line.

ALGORITHM:


STEP 2: Get the input data of x1, y1, z1 & d.

STEP 3: Calculate the value of r, x, y, z, a, b, c & d.

STEP 4: Get the input data of vr1, pr, pf.

STEP 5: Calculate the value of vr, irs, s, m, ir, vs, is, vrn1, reg, eff.

STEP 6: Print the output.



27/96

FLOW CHART:

MODELLING OF TRANSMISSION LINE

Start

Read the inputdata of r, x, y, d

Calculate the value of r1,x1, y1, z1, a, b, c, d

Read the input

data of vr1, pr, pf

Calculate the value of vr, irs,

s, m, ir, vrn1, vs, is, reg, eff

Print the outputvalues

Stop


28/96

PROBLEM:

A 230V, 60Hz, 3 transmission line is 160km long per phase resistance is

0.124/km & the reactance is 0.497/km & its shunt admittance is 3.310 -6 sin/km. it delivers

40MW at 220kV with 0.9 lagging pf and medium line mode.(i) Determine the voltage, current, regulation and efficiency.

(ii)Verify the result using available program.

PROGRAM:clear all;

clc;r1=input('enter the resistance value per phase per km:');

x1=input('enter the reactance value per phase per km:');

y1=input('enter the admittance value per phase per km:');

d=input('enter the length of the line:');r=d*r1;

x=d*x1;

y=d*y1;z=r+x*j;

a=1+(y*z/2);b=z;c=y*(1+(y*z/4));

d=a;

vr1=input('enter the receiving end voltage 1-1 in kv:');pr=input('enter the receiving end power:');

pf=input('enter the power factor:');

vr=(vr1*1000/1.732);

irs=(pr*10^6/(1.732*vr1*1000*pf));s=-acos(pf);

m=cos(s)+sin(s)*j;

ir=irs*m;vs=(a*vr)+(b*ir);

is=(c*vr)+(d*ir);

vrn1=abs(vs)/abs(a);reg=(vrn1-abs(vr))/abs(vr)*100;

eff=((pr*10^6*100)/(3*abs(vs)*abs(is)));

disp('Z value is');

disp(z);disp('Y value is');

disp(y);

disp('A value is');

disp(a);disp('B value is');

disp(b);disp('C value is');

disp(c);

disp('D value is');disp(d);

disp('sending end voltage is');

disp(vs);

disp('sending end current is');


29/96

disp(is);

disp('voltage regulation is');

disp(reg);

disp('efficiency is');disp(eff);


30/96

OUTPUT:

enter the resistance value per phase per km:0.124

enter the reactance value per phase per km:0.497

enter the admittance value per phase per km:0.0000033ienter the length of the line:160

enter the receiving end voltage 1-1 in kv:220

enter the receiving end power:40

enter the power factor:0.9Z value is

19.8400 +79.5200i

Y value is

0 +5.2800e-004i

A value is

0.9790 + 0.0052i

B value is19.8400 +79.5200i

C value is-1.3828e-006 +5.2246e-004i

D value is0.9790 + 0.0052i

sending end voltage is

1.3048e+005 +8.0043e+003i

sending end current is

1.0286e+002 +1.7138e+001i

voltage regulation is

5.1217

efficiency is

97.8083


31/96

MODEL CALCULATION:


32/96

RESULT:


33/96

Ex. No:

DATE :

LOAD FLOW ANALYSIS- GAUSS SEIDAL METHOD

AIM:

To compute the line flows and slack bus power for the given power system using GaussSeidal method.

ALGORITHM:


STEP 2: Read power system data and form Ybus.

STEP 3: Assume the initial voltage 010

jEP += other than slack bus.

STEP 4: Set iteration count 0=k and = maxE (tolerance).

STEP 5: Set the bus count p.

STEP 6: Check for slack bus.

STEP 7: Calculate the value of +=

+

=

+

=n

pq

k

qpq

k

q

p

q

pq

p

ppk

p EYEYV

jQPV

1

11

1*

1

STEP 8: Calculate the change in voltage using the formulak

i

k

i

k

pVVV

=

+1

STEP 9: Increment the bus count p=p+1.

STEP10: Check whether all the buses are included.

STEP11: Evaluate the maximum value of change in voltage.

STEP12: Check for convergence. If it does not exceed increment the iteration.

STEP13: Print the value of line flows and slack bus power.

STEP14: Stop the program.

FLOW CHART:


34/96

LOAD FLOW ANALYSIS- GAUSS SEIDAL METHOD

Start

Read system data& form Ybus

Assume other than slack bus

Set convergence value

Set iteration count k=0

Check

forslack

bus

Calculate

Set bus count p = 1

Calculate change in voltage

Advance bus count p=p+1

A

B

C

Yes

No


35/96

Is

A

Evaluate max value

B

Is


36/96

PROBLEM:

Obtain the load flow solution for the given power system using Gauss- Seidal method.

Bus code V Real Power Reactive

Power

Bus Type

1 1.06+j0 - - Slack

2 - 0.2 -0.2 Load

3 - -0.45 0.15 Load4 - -0.4 0.05 Load

5 - -0.6 0.1 Load

The line impedance and line charging admittance are given below. Acceleration factor is 1.6.

Bus Code Impedance per unit Line Charging

Admittance

per unit

1-2 0.02+0.06i 0.03i

1-3 0.08+0.24i 0.025i2-3 0.06+0.18i 0.02i

2-4 0.06+0.18i 0.02i

2-5 0.04+0.12i 0.015i

3-4 0.01+0.03i 0.01i

4-5 0.08+0.24i 0.025i

PROGRAM:

clear all;clc;

n=input('no of buses:');

k=input('no of lines:');for i=1:k

sb(i)=input('sending bus no:');

eb(i)=input('ending bus no:');z(i)=input('line impedance:');

ys(i)=input('line charging admittance:');

endybus=zeros(n,n);

for j=1:k

l=sb(j);

m=eb(j);

ybus(l,l)=ybus(l,l)+1/z(j)+ys(j);ybus(m,m)=ybus(m,m)+1/z(j)+ys(j);

ybus(l,m)=ybus(l,m)-1/z(j);ybus(m,l)=ybus(m,l)-1/z(j);

end

ybussl=[0 0.2-0.2i -0.45+0.15i -0.4+0.05i -0.6+0.1i];

vold=[1.06+0i 1+0i 1+0i 1+0i 1+0i];

v=vold;


37/96

vacc=vold;

delvmax=10;

itrcount=1;

while(delvmax>.001)for i=2:5

sum=0;

for k=1:5

if(i~=k)sum=sum+(ybus(i,k)*vacc(k));

endend

v(i)=(sl(i)/conj(vacc(i))-sum)/ybus(i,i);

vacc(i)=vold(i)+(1.4*(v(i)-vold(i)));

enddelvmax=max(abs(vacc-vold));

vold=vacc;

itrcount=itrcount+1;end

ybusv


38/96

OUTPUT:

no of buses:5

no of lines:7sending bus no:1

ending bus no:2

line impedance:.02+.06i

line charging admittance:.03isending bus no:1

ending bus no:3line impedance:.08+.24i

line charging admittance:.025i

sending bus no:2


line charging admittance:.02i

sending bus no:2ending bus no:4

line impedance:.06+.18iline charging admittance:.02isending bus no:2

ending bus no:5

line impedance:.04+.12iline charging admittance:.015i

sending bus no:3

ending bus no:4

line impedance:.01+.03iline charging admittance:.01i

sending bus no:4


line charging admittance:0.025i

ybus =

Columns 1 through 4

6.2500 -18.6950i -5.0000 +15.0000i -1.2500 + 3.7500i 0

-5.0000 +15.0000i 10.8333 -32.4150i -1.6667 + 5.0000i -1.6667 + 5.0000i

-1.2500 + 3.7500i -1.6667 + 5.0000i 12.9167 -38.6950i -10.0000 +30.0000i

0 -1.6667 + 5.0000i -10.0000 +30.0000i 12.9167 -38.6950i0 -2.5000 + 7.5000i 0 -1.2500 + 3.7500i

Column 5

0-2.5000 + 7.5000i

0

-1.2500 + 3.7500i

3.7500 -11.2100i


39/96

ybus =

Columns 1 through 4

6.2500 -18.6950i -5.0000 +15.0000i -1.2500 + 3.7500i 0

-5.0000 +15.0000i 10.8333 -32.4150i -1.6667 + 5.0000i -1.6667 + 5.0000i-1.2500 + 3.7500i -1.6667 + 5.0000i 12.9167 -38.6950i -10.0000 +30.0000i

0 -1.6667 + 5.0000i -10.0000 +30.0000i 12.9167 -38.6950i0 -2.5000 + 7.5000i 0 -1.2500 + 3.7500i

Column 5

0

-2.5000 + 7.5000i

0-1.2500 + 3.7500i

3.7500 -11.2100i

v =

Columns 1 through 4

1.0600 1.0469 - 0.0511i 1.0213 - 0.0889i 1.0200 - 0.0949i

Column 5

1.0128 - 0.1090i


40/96

MODEL CALCULATION:


41/96

RESULT:


42/96

Ex. No:

DATE :

SHORT CIRCUIT ANALYSIS

AIM:

To analysis and perform the short circuit of given network by using MATLABprogram.

ALGORITHM:

Step 1: Get the impedance value, number of buses and transient reactance of bus.

Step 2: Get the impedance of respedtive buses and calculate the Ybus matrix.

Step 3: Compute the diagonal and off-diagonal

Adm(i)=1/(imp(i))Ybus(k1,k1)=Ybus(k1,k1)+adm(i)Ybus(k2,k2)=Ybus(k2,k2)+adm(i)

Ybus(k1,k2)=-adm(i)

Ybus(k2,k1)=Ybus(k1,k2)

Step 4: Form Zbus matrix by using Ybus matrix since Zbus=1/Ybus .

Step 5: Calculate the post fault bus voltage by using formula.

Vf(i)=Vo(i)-(Z(1,r)+Vo(r))/[Z(r,r)+Zf]

Step 6: Calculate the fault current.

Step 7: Short circuit current by using formula

Iss(m,m)=(Vf(m)-Vf(n))/(imp(i))

Iss(n,n)=Iss(m,n)

Step 8: Finally short circuit current matrix can be obtained.


43/96

FLOWCHART:

Determine the pre fault voltage of

All busses and current in all lines

Through a load flow study

Vbus=V1,V2,.Vn

Assume rth bus is failed through

fault impedance Zf

Draw network of the system with eneratorsreplaced by transient(or)subtransient

reactance with the emf shorted

Print all the

results

Excite the passive network

Under fault bcondition voltages at rth busVrf=Vr0+Vr0=Vr0-Zrr.If

Stop

Start

Read All the bus

voltage,Transformer and

generators data.


44/96

PROGRAM:

%SHORT CIRCUIT ANALYSIS

%formation of Y bus

s=input('enter the no of impedance values:');b=input('enter the no of buses:');

for i=1:b

trans(i)=input('transient reactance of bus:');

if trans(i)==0;tr(i)=0;

elsetr(i)=1/trans(i);

end

end

for i=1:ssb(i)=input('starting the number:');

rb(i)=input('receiving bus number:');

imp(i)=input('impedance of bus:');end

Ybus=diag(0,(b-1));for i=1:s

k1=sb(i);

k2=rb(i);

adm(i)=1/imp(i);Ybus(k1,k1)=Ybus(k1,k1)+adm(i);

Ybus(k2,k2)=Ybus(k2,k2)+adm(i);

Ybus(k1,k2)=adm(i);

Ybus(k2,k1)=Ybus(k1,k2);end

for i=1:b

Ybus(i,i)=Ybus(i,i)+tr(i);end

Ybus

%computation of Z busZbus=inv(Ybus);

Zbus

%computation of post fault bus voltage

Z=Zbus;Zf=input('enter fault impedance of bus:');

for i=1:b

vo(i)=1;

endr=input('enter bus fault number:');

disp('post bus fault voltage');for i=1:b

vf(i)=vo(i)-(Z(i,r)*vo(r))/(Z(r,r)+Zf);

endvf

%calculation of fault current

disp('fault current:');

If=vo(r)/(Z(r,r)+Zf);


45/96

If

%calculation of short circuit current

for i=1:s

m=sb(i);n=rb(i);

Iss(m,n)=((vf(m)-(vf(n))))/(imp(i));

Iss(n,m)=Iss(m,n);

enddisp('short circuit line current');

Iss


46/96

MODEL CALCULATION:


47/96

RESULT:


48/96

Ex. No:

DATE :

LOAD FLOW ANALYSIS- NEWTON RAPHSON

METHOD

AIM:

To write a mat lab program to obtain the load flow solution for the Newton Raphsonmethod for the given power system network.

ALGORITHM:


STEP 2: Read the input data and form Ybus.

STEP 3: Set iteration count.

STEP 4: Calculate the real and reactive power except slack bus.

STEP 5: Calculatecalact

calact

QQQ

PPP

==

STEP 6: Check P, Q < (tolerance). If yes than calculate line flow and slack bus

power, then go to step 10, otherwise assemeble jacobian matrix.

STEP 7: Find the inverse of jacobian andK

iV andK

iQ

STEP 8: FindK

i

K

i

K

iVVV +=+1

STEP 9: Increment the iteration count.

STEP10:Stop the program.


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FLOW CHART:LOAD FLOW ANALYSIS- NEWTON RAPHSON METHOD

Start

Read input data

& form Ybus

Set iteration count k=0

Set bus count p=1

Is

p0.001)


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for j=2:3

p(j)=0;

q(j)=0;

for k=1:3p(j)=p(j)+(e(j)*(e(k)*g(j,k)+f(k)*b(j,k))+f(j)*(f(k)*g(j,k)-e(k)*b(j,k)));

q(j)=q(j)-(f(j)*(e(k)*g(j,k)+f(k)*b(j,k))-e(j)*(f(k)*g(j,k)-e(k)*b(j,k)));

end

enddelp=ps-p;

delq=qs-q;for j=2:3

sumpe=0;

sumpf=0;

sumqe=0;sumqf=0;

for k=1:3

if (k~=j)sumpe=sumpe+(e(k)*g(j,k)+f(k)*b(j,k));

sumpf=sumpf+(f(k)*g(j,k)-e(k)*b(j,k));sumqe=sumqe+(f(k)*g(j,k)-e(k)*b(j,k));sumqf=sumqf+(e(k)*g(j,k)+f(k)*b(j,k));

end

if (k~=1) & (k~=j)delpe(j,k)=e(j)*g(j,k)-f(j)*b(j,k);

delpf(j,k)=e(j)*b(j,k)+f(j)*g(j,k);

delqe(j,k)=e(j)*b(j,k)-f(j)*g(j,k);

delqf(j,k)=-e(j)*g(j,k)+(f(j)*b(j,k));end

end

delpe(j,j)=(2*e(j)*g(j,j))+sumpe;delpf(j,j)=(2*f(j)*g(j,j))+sumpf;

delqe(j,j)=(2*e(j)*b(j,j))+sumqe;

delqf(j,j)=(2*f(j)*b(j,j))+sumqf;end

for j=1:2

for k=1:2

delpem(j,k)=delpe(j+1,k+1);delpfm(j,k)=delpf(j+1,k+1);

delqem(j,k)=delqe(j+1,k+1);

delqfm(j,k)=delqf(j+1,k+1);

enddelpm(j)=delp(j+1);

delqm(j)=delq(j+1);end

jacob=[delpem delpfm;delqem delqfm];

dels=[delpm delqm];con=max(dels);

c=inv(jacob)*dels';

for j=2:3

e(j)=e(j)+c(j-1);


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f(j)=f(j)+c(j-1);

end

end

pq

delpm

delqm


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MODEL CALCULATION:


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RESULT:


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Ex. No:

DATE :

LOAD FLOW ANALYSIS FAST DECOUPLED LOAD

FLOW METHODAIM:

To obtain the load flow solution using Fast Decoupled Load flow method for the given

power system.

ALGORITHM:


STEP 2: Read the system data and form the Ybus.

STEP 3: Assume voltage profile as 1+j0.

STEP 4: Set convergence value, iteration count and bus count.

STEP 5: CalculatePP & QP.

STEP 6: Calculate KppKp pspecp =

STEP 7: Test for Convergence KP < P .

STEP 8: Normalize the mismatch by dividing with bus voltageKV

kpKp

i

ip

=

STEP 9: Solve K using [ ][ ]=

'BV

p

STEP10: Calculatekp

kp

kp +=+1

STEP11: Calculatek

pspecp

k

p QQQ = ,

STEP12: Test for Convergencek

pQ


56/96

FLOW CHART:

Fast decoupled load flow method

Start

Read system data and form

Ybus matrix

Assume voltage profile as 1+j0

Set convergence value

Set iteration count k = 0

Set bus count p = 0

Calculate PP

& QP

Calculate

Test forConvergen

ce

Pi,max

Set Pi = Pi,max

IfP

i P

i,

max

Pi= P

i, max

A

Yes

No

Yes

No

B

C

D


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i = i+ 1

Ifi =

N

Is<

Calculate

and

Is

P <

DLiDPPPP =

Is

PD

< 0 = +

= -

Print generation &

cost of generation

Stop

A

B

No

Yes

k = k+1

C

D

No

Yes

No

Yes

No

Yes


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PROBLEM:

For a 3 bus system the loss co-efficients are

000090121.000

000194971.0005963568.00

000375082.0000049448.0008383183.0

The incremental costs of the 3 units are

97.700964.0

85.700388.0

48.600256.0

3

3

3

2

2

2

1

1

1

+=

+=

+=

PdP

dF

PdP

dF

PdP

dF

Find the optimal generation for =Rs.20/MWhr. Also compute the minimum loss and receivedpower.

PROGRAM:

clear all;

clc;

B=[0.008383183 -0.000049448 0.000375082;0 0.005963568 0.000194971;0 0 0.000090121];b=[6.48 7.85 7.97];

c=[0.00256 0.00388 0.00964];

Pd=input('enter the demand value:');Pl=0;

Pg=0;

L=20;P(2)=0;

P(3)=0;

while(round(Pd+Pl)~=round(Pg))

for n=1:3y=0;

for m=1:3

if (m~=n)y=y+(2*P(m)*B(m,n));

end

endx=b(n)/L;

z=(c(n)/L)+(2*B(n,n));

P(n)=(1-x-y)/z;

endPg=P(1)+P(2)+P(3);

Pl=0;

for i=1:3for j=1:3

Pl=Pl+(P(i)*B(i,j)*P(j));

end


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end

if(round(Pd+Pl)>round(Pg))

L=L+0.001;

elseif (round(Pd+Pl)


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enter the demand value:850

P =

40.7574 51.7197 879.0636

Pl =

121.7179

L =

20.7940

Pd =

850


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MODEL CALCULATION:


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RESULT:

Ex. No:


79/96

DATE :

LOAD FREQUENCY DYNAMICS OF SINGLE AND TWO

AREA SYSTEM

AIM:

To obtain the load frequency dynamics of single area and two area system.

SINGLE AREA SYSTEM:

PROBLEM:

Turbine time constant for a given system is T t=0.5sec governor time constant TS=0.2sec,generator inertia constant h=5sec, governor speed regulation R=0.05p.u, damping co-efficient

B=0.8, sudden load change of PL=0.2p, integrator controller gain KS=7. Construct the simulink

block diagram and obtain the frequency deviation response for given system AGC in single areasystem.

BLOCK DIAGRAM:


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FREQUENCY RESPONSE:

TIME

POWER RESPONSE:

F

R

E

Q

UE

N

C

Y

P

O

W

E

R


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TWO AREA SYSTEMS:

PROBLEM:

A two area system connected by the line has parameters on 100MVA common base

Area 1 2

Speed regulation, R1 0.05 0.0625Frequency sensitive

Load co-efficient, D1

0.6 0.9

Integrator constant, H1 5 4

Governor timeconstant , TG1

0.2sec 0.3sec

Turbine time constant,Tt1

0.5sec 0.6sec

The units are operated in parallel at nominal frequency of 60Hz, the synchronous coefficient is

computed from initial operation is given to be DS=2sec. a load change of 187.5Mw occurs in

area 1.Construct the simulink block diagram and obtain the frequency deviation response.


83/96

BLOCK DIAGRAM:


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FREQUENCY RESPONSE:

TIME

POWER RESPONSE:

TIME

F

RE

Q

U

E

N

C

Y

P

O

W

E

R


85/96

MODEL CALCULATION:


86/96

RESULT:


87/96

Ex. No:

DATE :

TRANSIENT STABILITY ANALYSIS: SMIB SYSTEM

AIM:

To obtain the transient stability response in SMIB system and to make analysis of the

system.

ALGORITHM:

STEP1: Start the program.

STEP2: Get the input data such as terminal voltage, supply voltage, inertia constant,

transient reactance, frequency.

STEP3: Find the frequency response and load angle response of SMIB system using

( )

tsimer

trer

trOnO

trOO

n

n

22

22

11

sin1

+=

+

+=

STEP4: Plot the frequency response and load angle curve.

STEP5: Stop the program.


88/96

FLOW CHART:

TRANSIENT STABILITY ANALYSIS

Start

Read the input data such as

terminal voltage, supplyvoltage, inertia constant,transient reactance, frequency

Find the frequency response and load angle using

Plot the frequency response and loadangle curve

Print the result

Stop


89/96

PROBLEM:

A 60Hz synchronous generator having inertia constant H=9.94ms/MVA and a transientreactance of Xd

=0.3 p.u is to an through a purely reactive circuit as in fig. Reactance are

marked on the diagram an a common system base. The generator is delivering real power 0.6

p.u, 0.8 pf lagging to the infinite bus at a voltage V=1 p.u.

Assume the pu damping power coefficients as D=0.138. consider a small disturbance

of =1 with 0.1745 radian. For example the breaker open and then quickly closed. Obtain therotor angle and frequency.

PROGRAM:

TRANSIENT STABILITYclear all;

clc;

e=1.35;v=1;

h=9.94;

x=0.65;

pm=0.6;d=0.138;

fr=60;

pmax=e*v/x;do=asin(pm/pmax)

ps=pmax*cos(do)

wn=sqrt((pi*60)/(h*ps))z=d/2*sqrt((pi*60)/(h*ps))

wd=wn*sqrt(1-z^2)

fd=wd/(2*pi)tau=1/(z*wn)

th=acos(z)

Ddo=10*pi/180

t=0:0.01:3;Dd=(Ddo/sqrt(1-(z^2)))*exp(-z*wn*t).*sin(wd*t+th);

d=(do+Dd)*180/pi;

Dw=-wn*Ddo/sqrt(1-z^2)*exp(-z*wn*t).*sin(wd*t);f=fr+Dw/(2*pi);

subplot(2,1,1);

plot(t,d,'m'),grid;xlabel('t,sec');

ylabel('del,pu');

subplot(2,1,2);

plot(t,f),grid;xlabel('t,sec');


90/96

ylabel('freq,pu');

OUTPUT

do =

0.2931

ps =

1.9884

wn =

3.0882

z =

0.2131

wd =

3.0173

fd =

0.4802

tau =

1.5196

th =

1.3561

Ddo =

0.1745je


91/96

MODEL CALCULATION:


92/96

RESULT:


93/96

Ex. No:

DATE :

ELECTROMAGNETIC TRANSIENTS

AIM:

To study the effect in electromagnetic transients in power supply

THEORY:

Transient phenomena is a periodic function of time and does not-last longer than the operator

for which they lost in very insurant compared with operator time of system.

They are very important. The power system can be considerably made up of lines. Reduces

elements of RF and the circuits is normally organized and carrier load until OC fault occurs, the

fault corresponds to closing the switch. The redistribution is accomplished in general by ac

transient period during which the resultant to relic that these redistribution of I and V cannot

take place instantly for the following reason.

Transient in single circuits with DC source

. R-only

As soon as the switch is closed the current is determined by using ohms law.

I=V/R

No transients will be there

. L-onlyWhen switch is closed the current is given by

I(s)=V(S)/z(s)={VI/s2L}=(V/l)*t

It is shown that when a pure inductance is switched on to a DC source the current at t=0 and

increases linearly for infinity.

.C-only

When switch is closed current is circuit is given by

I(s)=V(s)/Z(s)={V/S}C s=VC

Therefore to have transients in an electrical system the following requirements should be used.

Either inductance or capacitance of both should be present.

1. Fundamental frequency

2. Natural frequency


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3. There is also another component called thermionic due to unbalanced current. Natural

frequency occurs after the occurrence of certain fault which is added to fundamental

frequency which consist transient voltage.

4. Transient which are in form of energy storage magnetic or electric is consider is called

single energy transient. If those are both in magnetic and electric energies it is called

double energy transients. The electromagnetic energy standby an inductance L is (1/2)

LI^2 where I is instantaneous value of current assuming L to be constantly changing in

current which an inductance is allowed.

5. There are only to two components of which store energy and redistribution of energy

following in a circuit change takes a finite time.

1. I cant change instantaneously through induction

2. V across the capacitor cant change instantaneous

3. How of conservation of energy must hold.Which an impulse of strength where s is closed

|Vm*cos/ L S^2+^2 + Vm*s.sin/ L S^2+^2|

V= Vm sin(t + )

I(S)= Vm * cos / (s+)(s2+2)

Now

L-I I(s) = (Vm/(L2+2))( cos ){e-at+(a/)sint-cost}+sin {a cost +sint- e-t }

It can be simplified to

{Vm/(R2+ 2 L2)V}{sin (+0-0)-( sin(0-0) e-at)}

The first term in above can is steady state sinusoidal vibration and second term in the transient

part after infinite line.


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RESULT:


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