power simulation lab record new

ANNA UNIVERSITY OF TECHNOLOGY TIRUCHIRAPPALLI

Tiruchirappalli - 620024

DEPARARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

A RECORD OF PRACTICAL WORK

Name :

Reg. No. :

Subject : POWER SYSTEM SIMULATION LABORATORY

Subject code : PS9402

ANNA UNIVERSITY OF TECHNOLOGY TIRUCHIRAPPALLI

Tiruchirappalli - 620024

DEPARARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

BONAFIDE CERTIFICATE

This is to certify that Mr./Ms __________________________with the registration number has satisfactorily completedthe practical course PS9402 – POWER SYSTEM SIMULATIONLABORATORY prescribed for IV semester, M.E. Power SystemEngineering (MBCBS) by Anna University of Technology, Tiruchirappalliduring the academic year 2011 – 2012.

Faculty in-charge Signature of the Coordinator

Submitted for the practical examination held on ____________

Internal Examiner External Examiner

CONTENTS

EX.NO LIST OF EXPERIMENTS PAGE SIGNATURE

1 Load Flow Analysis by Gauss SeidelMethod Using MATLAB

2 Load Flow Analysis by NewtonRaphson Method Using MATLAB

3 Load Flow Analysis by FastDecoupled Method Using MATLAB

4 Small Signal Stability Analysis ofSMIB Using MATLAB

5Small Signal Stability Analysis ofMulti machine system UsingMATLAB

6 Economic Dispatch with generatinglimits and including line losses

7(a)Unit Commitment by dynamicprogramming method UsingMATLAB

7(b)Unit Commitment by priority listmethod Using MATLAB

8 Contingency analysis method

9 Induction motor starting

10Relay co-ordination of radialtransmission/distribution system

LOAD FLOW ANALYSIS BY GAUSS SEIDEL METHOD

AIM :

To determine the power flow analysis using Gauss seidel method

SOFTWARE REQUIRED :

MATLAB

THEORY :

Load flow studies are one of the most important aspects of power system planning andoperation. The load flow gives us the sinusoidal steady state of the entire system - voltages, realand reactive power generated and absorbed and line losses. Since the load is a static quantityand it is the power that flows through transmission lines, the purists prefer to call this PowerFlow studies rather than load flow studies. We shall however stick to the original nomenclatureof load flow.

Fig. 1.1 The simple power system used for load flow studies.

Table 1.1 Line impedance and line charging data of the system of Fig 1.1

Line (bus to bus) Impedance Line charging ( Y /2)1-2 0.02 + j 0.10 j 0.0301-5 0.05 + j 0.25 j 0.0202-3 0.04 + j 0.20 j 0.0252-5 0.05 + j 0.25 j 0.0203-4 0.05 + j 0.25 j 0.0203-5 0.08 + j 0.40 j 0.0104-5 0.10 + j 0.50 j 0.075

Experiment : 1

Date :

Table 1.2 Ybus matrix of the system of Fig. 1.1.

1 2 3 4 51 2.6923 - j

13.4115- 1.9231 + j

9.61540 0 - 0.7692 + j

3.84622 - 1.9231 + j

9.61543.6538 - j18.1942

- 0.9615 + j4.8077

0 - 0.7692 + j3.8462

3 0 - 0.9615 + j4.8077

2.2115 - j11.0027

- 0.7692 + j3.8462

- 0.4808 + j2.4038

4 0 0 - 0.7692 + j3.8462

1.1538 - j 5.6742 - 0.3846 + j1.9231

5 - 0.7692 + j3.8462

- 0.7692 + j3.8462

- 0.4808 + j2.4038

- 0.3846 + j1.9231

2.4038 - j11.8942

Table 1.3 Bus voltages, power generated and load - initial data.

Bus no. Bus voltage Power generated LoadMagnitude (pu) Angle (deg) P (MW) Q (MVAr) P (MW) P (MVAr)

1 1.05 0 - - 0 02 1 0 0 0 96 623 1 0 0 0 35 144 1 0 0 0 16 85 1.02 0 48 - 24 11

The basic power flow equations (1) and (2) are nonlinear. In an n -bus power system, letthe number of P-Q buses be np and the number of P-V (generator) buses be ng such that n = np +ng + 1. Both voltage magnitudes and angles of the P-Q buses and voltage angles of the P-V busesare unknown making a total number of 2np + ng quantities to be determined. Amongst the knownquantities are 2np numbers of real and reactive powers of the P-Q buses, 2ng numbers of realpowers and voltage magnitudes of the P-V buses and voltage magnitude and angle of the slackbus. Therefore there are sufficient numbers of known quantities to obtain a solution of the load

(1)

(2)

flow problem. However, it is rather difficult to obtain a set of closed form equations from (1) and(2). We therefore have to resort to obtain iterative solutions of the load flow problem.

At the beginning of an iterative method, a set of values for the unknown quantities arechosen. These are then updated at each iteration. The process continues till errors between all theknown and actual quantities reduce below a pre-specified value. In the Gauss-Seidel load flowwe denote the initial voltage of the i th bus by Vi

(0) , i = 2, ... , n , This should read as the voltageof the i th bus at the 0th iteration, or initial guess. Similarly this voltage after the first iteration willbe denoted by Vi

(1) . In this Gauss-Seidel load flow the load buses and voltage controlled busesare treated differently. However in both these type of buses we use the complex power equationgiven in (2) for updating the voltages. Knowing the real and reactive power injected at any buswe can expand (2) as

We can rewrite (3) as

In this fashion the voltages of all the buses are updated. We shall outline this procedure withthe help of the system of Fig. 1.1, with the system data given in Tables 1.1 to 1.3. It is to benoted that the real and reactive powers are given respecttively in MW and MVAr. However theyare converted into per unit quantities where a base of 100 MVA is chosen.

Updating Load Bus Voltages

Let us start the procedure with bus-2 of the 5 bus 7 line system given in fig: 1.1. Since this isload bus, both the real and reactive power into this bus is known.

We can therefore write from (4)

(3)

(4)

(5)

From the data given in Table 1.3 we can write

It is to be noted that since the real and reactive power is drawn from this bus, boththese quantities appear in the above equation with a negative sign. With the values of the Y buselements given in Table 1.2 we get V2

1 = 0.9927 < − 2.5959° .

The first iteration voltage of bus-3 is given by

Note that in the above equation since the update for the bus-2 voltage is already available,we used the 1st iteration value of this rather than the initial value. Substituting the numerical datawe get V3

(1) = 0.9883 < − 2. 8258° . Finally the bus-4 voltage is given by

Solving we get V4(1) = 0. 9968 < −3.4849° .

Updating P-V Bus Voltages

It can be seen from Table 1.3 that even though the real power is specified for the P-Vbus-5, its reactive power is unknown. Therefore to update the voltage of this bus, we must firstestimate the reactive power of this bus

And hence we can write the kth iteration values as

(6)

(7)

(8)

(10)

This is computed as 0.0899 per unit. Once the reactive power is estimated, the bus-5voltage is updated as

It is to be noted that even though the power generation in bus-5 is 48 MW, there is alocal load that is consuming half that amount. Therefore the net power injected by this bus is24 MW and consequently the injected power P5, inj in this case is taken as 0.24 per unit. Thevoltage is calculated as V5

(1) = 1.0169 < − 0.8894° . Unfortunately however the magnitude ofthe voltage obtained above is not equal to the magnitude given in Table 1.3.

We must therefore force this voltage magnitude to be equal to that specified. This isaccomplished by

This will fix the voltage magnitude to be 1.02 per unit while retaining the phase of −0.8894 ° . The corrected voltage is used in the next iteration.

Convergence of the Algorithm

As can be seen from Table 1.3 that a total number of 4 real and 3 reactive powersare known to us. We must then calculate each of these from (1) and (2) using the values ofthe voltage magnitudes and their angle obtained after each iteration. The power mismatchesare then calculated . The process is assumed to have converged when each of ΔP2 , ΔP3, ΔP4, ΔP5 , ΔQ2 , ΔQ3 and ΔQ4 is below a small pre-specified value. At this point the process isterminated.

Sometimes to accelerate computation in the P-Q buses the voltages obtained is multiplied bya constant. The voltage update of bus- i is then given by

(11)

(12)

(13)

where λ is a constant that is known as the acceleration factor . The value of λ has tobe below 2.0 for the convergence to occur. Table 1.4 lists the values of the bus voltages afterthe 1st iteration and number of iterations required for the algorithm to converge for differentvalues of λ. It can be seen that the algorithm converges in the least number of iterationswhen λ is 1.4 and the maximum number of iterations are required when λ is 2. In fact thealgorithm will start to diverge if larger values of acceleration factor are chosen. The systemdataafter the convergence of the algorithm will be discussed later.

Table 1.4 Gauss-Seidel method: bus voltages after 1 st iteration and number ofiterations required for convergence for different values of λ .

Bus voltages (per unit) after 1st iteration No of

iterations

forconvergence

V2 V3 V4 V5

1 0.9927 2.6 0.98832.83

0.9968 3.48 1.02 0.89

28

2 0.98745.22

0.97668.04

0.991814.02

1.024.39

860

1.8 0.9883 4.7 0.9785 6.8 0.990311.12

1.023.52

54

1.6 0.98934.17

0.98075.67

0.9909 8.65 1.022.74

24

1.4 0.99033.64

0.98314.62

0.9926 6.57 1.022.05

14

1.2 0.99153.11

0.98573.68

0.9947 4.87 1.021.43

19

(12))

PROGRAM

clear alld2r=pi/180;w=100*pi;%line charging admittancesych=j*[0 0.03 0 0 0.02

0.03 0 0.025 0 0.020 0.025 0 0.02 0.010 0 0.02 0 0.0750.02 0.02 0.01 0.075 0];

% the y bus matrix isybus=[2.6923-j*13.4115 -1.9231+j*9.6154 0 0

-0.7692+j*3.8462-1.9231+j*9.6154 3.6538-j*18.1942 -0.9615+j*4.8077 0

-0.7692+j*3.84620 -0.9615+j*4.8077 2.2115-j*11.0027

-0.7692+j*3.8462 -0.4808+j*2.40380 0 -0.7692+j*3.8462 1.1538-j*5.6742 -0.3846+j*1.9231-0.7692+j*3.8462 -0.7692+j*3.8462 -0.4808+j*2.4038 -

0.3846+j*1.9231 2.4038-j*11.8942];g=real(ybus);b=imag(ybus);

% the given parameters and initial condition arep=[0;-0.96;-0.35;-0.16; 0.24];q=[0; -0.62; 0.14; -0.08; -.35];mv=[1.05; 1; 1; 1; 1.02];th=[0; 0; 0; 0];v=[mv(1);mv(2); mv(3); mv(4);mv(5)];acc=input(' enter the acceleration constant:');del=1;indx=0;

% the gauss-seidal iteration stars herewhile del>1e-6

%p-q busesfor i=2:4

tmp1=(p(i)-j*q(i))/conj(v(i));tmp2=0;for k=1:5

if(i==k)tmp2= tmp2+0;

elsetmp2=tmp2+ybus(i,k)*v(k);

end

endvt=(tmp1-tmp2)/ybus(i,i);v(i)=v(i)+acc*(vt-v(i));

end%p-vbusq5=0;

for i=1:5q5=q5+ybus(5,i)*v(i);

endq5=-imag(conj(v(5))*q5);tmp1=(p(5)-j*q5)/conj(v(5));tmp2=0;for k=1:4tmp2=tmp2 +ybus(5,k)*v(k);endvt=(tmp1-tmp2)/ybus(5,5);v(5)=abs(v(5))*vt/abs(vt);

%calculate p and qfor i=1:5

sm=0;for k=1:5

sm=sm+ybus(i,k)*v(k);ends(i)=conj(v(i))*sm;

end

% the mismatchdelp=p-real(s)';delq=q+imag(s)';delpq=[delp(2:5); delq(2:4)];del=max(abs(delpq));indx=indx+1;if indx==1

pauseendend'gs load flow converges in iteration ',indx, pause'final voltage magnitudes are ', abs(v)', pause'final angles inn degree are ' , angle(v)'/d2r, pause' the real poweers in each bus mw are ',( real(s)+[0 0 0 0

0.24])*100, pause' the reactive powers in each bus mvar are ',(-imag(s)+[0

0 0 0 0.11])*100

OUTPUT:

enter the acceleration constant:1.4

gs load flow converges in iteration

indx = 14

final voltage magnitudes are

1.0500 0.9932 1.0162 1.0142 1.0200

final angles inn degree are

0 -5.0750 -7.4101 -7.5119 -3.1811

the real poweers in each bus mw are

126.3701 -96.0000 -35.0000 -16.0000 48.0000

the reactive powers in each bus mvar are

46.2151 -62.0000 14.0000 -8.0000 -3.7719

RESULT :

Thus the load flow solution for the given problem was solved using Gauss-Seidalmethod and verified using MATLAB software

LOAD FLOW ANALYSIS BY NEWTON RAPHSON METHOD

AIM:

To solve load flow problems by using Newton Raphson method using matlabprogram.

SOFTWARE REQUIRED:

MATLAB

Experiment : 2

Date :

PROGRAM

v=[1.05;1.0;1.04];d=[0;0;0];ps=[-4;2.0];qs=-2.5;yb=[20-j*50 -10+j*20 -10+j*30

-10+j*20 26-j*52 -16+j*32-10+j*30 -16+j*32 26-j*62];

y=abs(yb); t=angle(yb);iter=0;pwracur=0.00025;dc=10;while max(abs(dc))>pwracur

iter=iter+1p=[v(2)*v(1)*y(2,1)*cos(t(2,1)-

d(2)+d(1))+v(2)^2*y(2,2)*cos(t(2,2))+v(2)*v(3)*y(2,3)*cos(t(2,3)-d(2)+d(3));

v(3)*v(1)*y(3,1)*cos(t(3,1)-d(3)+d(1))+v(3)^2*y(3,3)*cos(t(3,3))+v(3)*v(2)*y(3,2)*cos(t(3,2)-d(3)+d(2))];

q=-v(2)*v(1)*y(2,1)*sin(t(2,1)-d(2)+d(1))-v(2)^2*y(2,2)*sin(t(2,2))-v(2)*v(3)*y(2,3)*sin(t(2,3)-d(2)+d(3));

j(1,1)=v(2)*v(1)*y(2,1)*sin(t(2,1)-d(2)+d(1))+v(2)*v(3)*y(2,3)*sin(t(2,3)-d(2)+d(3));

j(1,2)=-v(2)*v(3)*y(2,3)*sin(t(2,3)-d(2)+d(3));j(1,3)=v(1)*y(2,1)*cos(t(2,1)-

d(2)+d(1))+2*v(2)*y(2,2)*cos(t(2,2))+v(3)*y(2,3)*cos(t(2,3)-d(2)+d(3));

j(2,1)=-v(3)*v(2)*y(3,2)*sin(t(3,2)-d(3)+d(2));j(2,2)=v(3)*v(1)*y(3,1)*sin(t(3,2)-

d(3)+d(1))+v(3)*v(2)*y(3,2)*sin(t(3,2)-d(3)+d(2));j(2,3)=v(3)*y(2,3)*cos(t(3,2)-d(3)+d(2));j(3,1)=v(2)*v(1)*y(2,1)*cos(t(2,1)-

d(2)+d(1))+v(2)*v(3)*y(2,3)*cos(t(2,3)-d(2)+d(3));j(3,2)=-v(2)*v(3)*y(2,3)*cos(t(3,2)-d(2)+d(3));j(3,3)=-v(1)*y(2,1)*sin(t(2,1)-d(2)+d(1))-

2*v(2)*y(2,2)*sin(t(2,2))-v(3)*y(2,3)*sin(t(2,3)-d(2)+d(3));dp=ps-p;dq=qs-q;dc=[dp;dq]jdx=j\dcd(2)=d(2)+dx(1);d(3)=d(3)+dx(2);v(2)=v(2)+dx(3);v,d,delta=180/pi*d;

endp1=v(1)^2*y(1,1)*cos(t(1,1))+v(1)*v(2)*y(1,2)*cos(t(1,2)-d(1)+d(2))+v(1)*v(3)*y(1,3)*cos(t(1,3)-d(1)+d(3))

q1=-v(1)^2*y(1,1)*sin(t(1,1))-v(1)*v(2)*y(1,2)*sin(t(1,2)-d(1)+d(2))-v(1)*v(3)*y(1,3)*sin(t(1,3)-d(1)+d(3))q3=-v(3)*v(1)*y(3,1)*sin(t(3,1)-d(3)+d(1))-v(3)*v(2)*y(3,2)*sin(t(3,2)-d(3)+d(2))-v(3)^2*y(3,3)*sin(t(3,3))

OUTPUT:

iter =1dc = -2.8600

1.4384-0.2200

j =54.2800 -33.2800 24.8600-33.2800 64.1664 -16.6400-27.1400 16.6400 49.7200

dx = -0.0455-0.0080-0.0265

v = 1.05000.97351.0400

d = 0-0.0455-0.0080

iter = 2dc = -0.0992

0.0367-0.0509

j = 51.7246 -31.7678 21.3026-32.9797 63.7409 -15.3834-28.5386 17.3988 48.1036

dx = -0.0016-0.0007-0.0018

v = 1.05000.97171.0400

d = 0-0.0471-0.0087

iter = 3dc = -0.0002

0.0014-0.0001

j = 51.5967 -31.6941 21.1474-32.9337 63.6841 -15.3520-28.5482 17.3966 47.9549

dx =1.0e-004 *0.14650.2778

-0.0428v = 1.0500

0.97171.0400

d = 0-0.0471-0.0087

iter = 4dc = 1.0e-004 *-0.0000

-0.5314-0.0001

j =51.5964 -31.6937 21.1471-32.9337 63.6846 -15.3516-28.5482 17.3969 47.9545

dx = 1.0e-005 *-0.0750-0.1223-0.0003

v = 1.05000.97171.0400

d = 0-0.0471-0.0087

p1 =2.1842q1 =1.4085q3 =1.4618

RESULT:

Thus the load flow solution for the given problem solved using Newton raphsonmethod and verified using MATLAB software.

LOAD FLOW ANALYSIS BY FAST DECOUPLED METHOD

AIM

To develop a computer program in matlab to carryout DC Load Flow Analysis of the

given power system by using MATLAB.

SOFTWARE REQUIRED

MATLAB

THEORY

One of the fast methods of load flow analysis of a power system is DC Load Flow

Analysis. It is based on three important simplifying assumptions. These are,

1. All bus voltage magnitudes are assumed to be equal to 1.0 p. u.

2. All angular differences are small, and hence cos (i - j) = 1.0 p. u and sin (i - j) » (i -

j) expressed in radians.

3.All lines and Transformers are characterized by a high “X/R” ratio. That is, X>>R.

Hence, all resistances are neglected.

The basic steps involved in dc load flow analysis are

1. Compute bus susceptance matrix [B] by considering all buses other than slack bus.

Bii sum of line susceptances connected to bus i.

Bijnegative of line susceptance connected between buses i and j.

2 .Compute bus voltage angles using the relation

[] = [B] -1 * [P]

[nb-1] x 1 [nb-1] x [nb-1] [nb-1] x 1

Experiment : 3

Date :

where

[P] is a vector made of specified bus powers at all buses except the slack bus

3. Compute slack bus power using the relation

P1 = i * B1i

4. Compute generator (slack bus) over load, if any.

5 .Compute line flows.

6. Compute over loads on lines, if any.

Power system transmission lines have a very high X/R ratio. For such a system realpower change ∆p are less sensitive to changes in the voltage magnitude and are mostsensitive to changes in phase angle ∆δ.

Therefore, it is reasonable to set elements J2 and J3 of the jacobian matrix to zero.

i=1

PROGRAM:

V1=1.05;V2=1.0;V3=1.04;d1=0;d2=0;d3=0;Ps2=-4;Ps3=2.0;Qs2=-2.5;YB=[20-j*50 -10+j*20 -10+j*30

-10+j*20 26-j*52 -16+j*32-10+j*30 -16+j*32 26-j*62];

Y=abs(YB);t=angle(YB);B=[-52 32; 32 -62]Binv=inv(B)iter=0;pwracur=0.0003;DC=10;while max(abs(DC)) > pwracur

iter=iter+1;p2=V2*V1*Y(2,1)*cos(t(2,1)-

d2+d1)+V2^2*Y(2,2)*cos(t(2,2))+V2*V3*Y(2,3)*cos(t(2,3)-d2+d3);

p3=V3*V1*Y(3,1)*cos(t(3,1)-d3+d1)+V3^2*Y(3,3)*cos(t(3,3))+V3*V2*Y(3,2)*cos(t(3,2)-d3+d2);

Q2=-V2*V1*Y(2,1)*sin(t(2,1)-d2+d1)-V2^2*Y(2,2)*sin(t(2,2))-V2*V3*Y(2,3)*sin(t(2,3)-d2+d3);

DP2=Ps2-p2;DP2V=DP2/V2;DP3=Ps3-p3;DP3V=DP3/V3;DQ2=Qs2-Q2;DQ2V=DQ2/V2;DC=[DP2;DP3;DQ2];Dd=-Binv*[DP2V;DP3V];DV=-1/B(1,1)*DQ2V;d2=d2+Dd(1);d3=d3+Dd(2);V2=V2+DV;angle2=180/pi*d2;angle3=180/pi*d3;R=[iter d2 d3 V2 DP2 DP3 DQ2];disp(R)

endQ3=-V3*V1*Y(3,1)*sin(t(3,1)-d3+d1)-V3^2*Y(3,3)*sin(t(3,3))-V3*V2*Y(3,2)*sin(t(3,2)-d3+d2);P1=V1^2*Y(1,1)*cos(t(1,1))+V1*V2*Y(1,2)*cos(t(1,2)-d2+d2)+V1*V3*Y(1,3)*cos(t(1,3)-d1+d3);Q1=-V1^2*Y(1,1)*sin(t(1,1))-V1*V2*Y(1,2)*sin(t(1,2)-d1+d2)-V1*V3*Y(1,3)*sin(t(1,3)-d1+d3);S1=P1+j*Q1Q3

OUTPUT:

B = -52 3232 -62

Binv = -0.0282 -0.0145-0.0145 -0.0236

1.0000 -0.0605 -0.0089 0.9958 -2.8600 1.4384-0.2200

2.0000 -0.0565 -0.0080 0.9653 0.1759 -0.0710-1.5790

3.0000 -0.0442 -0.0087 0.9657 0.6403 -0.45700.0219

4.0000 -0.0448 -0.0090 0.9730 -0.0214 0.00120.3652

5.0000 -0.0477 -0.0087 0.9731 -0.1534 0.11290.0067

6.0000 -0.0476 -0.0086 0.9714 0.0005 0.0026-0.0861

7.0000 -0.0469 -0.0087 0.9713 0.0360 -0.0262-0.0041

8.0000 -0.0469 -0.0087 0.9717 0.0009 -0.00140.0201

9.0000 -0.0471 -0.0087 0.9718 -0.0084 0.00610.0016

10.0000 -0.0471 -0.0087 0.9717 -0.0005 0.0005-0.0047

11.0000 -0.0471 -0.0087 0.9717 0.0020 -0.0014-0.0005

12.0000 -0.0471 -0.0087 0.9717 0.0002 -0.00020.0011

13.0000 -0.0471 -0.0087 0.9717 -0.0005 0.00030.0002

14.0000 -0.0471 -0.0087 0.9717 -0.0001 0.0000-0.0003S1 =

2.1843 + 1.4085iQ3 =1.4617

RESULT:

Thus the Load Flow Analysis of the given power system solved by fast decoupled

method and verified using MATLAB.

SMALL SIGNAL STABILITY ANALYSIS-SMIB

AIM:-

To determine the small signal stability analysis for single machine infinite bussystem.

SOFTWRE REQUIRED:

MATLAB

THEORY:

Power system stability may be broadly defined as that property of the power system

that enables it to remain in a state of operating equilibrium under normal operating condition

and to regain an acceptable state of equilibrium after being subjected to a disturbance.

Power system stability may be broadly classified as (i) rotor angle stability and (ii)

voltage stability. Rotor angle stability is the ability of interconnected synchronous machines

of a power system to remain in synchronism.

Rotor angle stability can further be classified in to transient stability and small signal

stability depending on the type of disturbance. Transient stability is the rotor angle stability

study of a system following large disturbances.

Small signal (or small disturbance) stability is the ability of the power system to

maintain synchronism under small disturbances. The disturbances are considered

sufficiently small for linearization of system equations to be permissible for purpose of

analysis. Instability that may result can be of two forms.

I. Steady increase in rotor angle due to lack of sufficient synchronizing torque.

II. Rotor oscillations of increasing amplitude due to lack of sufficient damping

torque.

There are four modes of oscillations causing small signal instability in a power system.

They are:

Local Modes or Machine System Modes are associated with the swinging of

units at a generating station with respect to rest of the power system. The

frequency range of oscillation is (0.8-2)Hz.

Inter Area Modes are associated with swinging of many machines in one part

of the system against machines in other parts. The frequency range for inter

area modes is (0.2-0.8)Hz.

Experiment : 4

Date :

Control Modes are associated with generating units and other controls.

Torsional Modes are associated with the turbine-generator shaft system rotational

components

PROGRAM:

p=0.9;q=0.3;f=60;xd=0.3*i;xtr=0.15*i;x1=0.5*i;x2=0.93*i;Et=1;Eb=0.995;

w0=2*pi*f;

s=p+q*i;

It=s'/Et';

FL=input('ENTER THE FAULTED LINE NO:');

if FL==1

xt=xd+xtr+x2;

else

xt=xd+xtr+x1;

end

E=Et+It*xd;

D=angle(E)-angle(Eb);

ks=(abs(E)*abs(Eb)*cos(D))/abs(xt);

H=input('ENTER THE VALUE OF INTERTIA CONSTANT , H:');

kd=input('ENTER THE Kd VALUE :');

A=[(-kd/(2*H)) (-ks/(2*H));w0 0];

lambda=eig(A)

w=ks*w0/(2*H);

wn=sqrt(w)/(2*pi)

z=0.5*kd/sqrt(2*ks*H*w0)

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

[VD]=eig(A);

g=[(-0.0019+0.0168*i) (-0.0019-0.0168*i); 1 1];

L=inv(g);

[P]=[L(1,1)*g(1,1) L(1,2)*g(2,1);L(1,2)*g(2,1) L(2,2)*g(2,2)]

theta=acos(z)

D0=5*pi/180;

t=0:.01:3;

Dd=D0/sqrt(1-z^2)*exp(-z*wn*t).*sin(wd*t+theta);

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

plot(t,d),grid

xlabel('t sec'),ylabel('delta degree')

OUTPUT:

Case 1:

ENTER THE FAULTED LINE NO: 1ENTER THE VALUE OF INERTIA CONSTANT: 3.5ENTER THE Kd VALUE: 0

lamda = 0 + 6.5058i0 - 6.5058i

wn = 1.0354

z = 0

wd = 1.0354

P = 0.5000 + 0.0565i 0.5000 - 0.0565i0.5000 - 0.0565i 0.5000 + 0.0565i

theta =1.5708

Case 2:

ENTER THE FAULTED LINE NO:1ENTER THE VALUE OF INTERTIA CONSTANT , H:3.5ENTER THE Kd VALUE :10

lambda = -0.7143 + 6.4665i-0.7143 - 6.4665i

wn = 1.0354

z = 0.1098

wd =1.0292

P = 0.5000 + 0.0565i 0.5000 - 0.0565i0.5000 - 0.0565i 0.5000 + 0.0565i

theta = 1.4608

Case 3:

ENTER THE FAULTED LINE NO:1ENTER THE VALUE OF INTERTIA CONSTANT , H:3.5ENTER THE Kd VALUE :-10

lambda = 0.7143 + 6.4665i0.7143 - 6.4665i

wn =1.0354

z = -0.1098

wd = 1.0292

P = 0.5000 + 0.0565i 0.5000 - 0.0565i0.5000 - 0.0565i 0.5000 + 0.0565i

theta = 1.6808

RESULT:-

Thus the small signal stability analysis of single machine infinite bus system wasdetermined and it was verified by using MATLAB.

SMALL SIGNAL STABILITY ANALYSIS OF MULTI MACHINE SYSTEM

AIM:-

To determine the small signal stability analysis for multi machine system.

SOFTWRE REQUIRED:

MATLAB

ALGORITHM:

Step1: Start the program.

Step2: Form the Ybus explicitly Including are represented as constant admittances.

Step3: Reduce the Ybus by eliminating all the passive nodes.

Step4: Transform thr reduced set of equcations to the individual machine co-ordinates.

Step5: Linearise the transformed set of eqns.

Step6: Express the incremental change in algberic variables of the incremental change in state

variables.

Step7:Apply eigen values technique for stability assessment.

Experiment : 5

Date :

PROGRAM:

clear all;clc;%'multimachine system'Mva=160;Xdt=0.49;Xt=0.8;Ke=379.2;W0=(2*60*3.14);Pf1=0.85;Pf2=0.85;P1=0.5;P2=0.5;Vt=1.0+j*0.0;q1=0.5*(tan(acos(Pf1)));q2=0.5*(tan(acos(Pf2)));i1=(P1-j*q1)/(conj(Vt));i2=(P2-j*q2)/(conj(Vt));eq1=Vt+j*Xdt*i1;eq2=Vt+j*Xdt*i2;d10=angle(eq1);d20=angle(eq2);il=i1+i2;Vl=Vt-j*Xt*i1;yl=il/Vl;Z=[j*1.29 inf -j*1.29;

inf j*1.29 -j*1.29;-j*1.29 -j*1.29 (1/(yl+(-1.5504*j)))];

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

y(i,j)=1/Z(i,j);end

endy

eq=[eq1+j*0 eq2+j*0];ygg=y(1:2,1:2);ygng=y(1:2,3:3);yngg=y(3:3,1:2);yngng=y(3:3,3:3);yggr=ygg-(ygng*(inv(yngng))*yngg);d12=d10-d20;m=[yggr(1,1) yggr(1,2)*exp(-j*d12);

yggr(2,1)*exp(j*d12) yggr(2,2)]b=imag(m)t13=((b(2,1))*abs(eq1)*abs(eq2));t23=-((b(2,1))*abs(eq1)*abs(eq2));h=Ke/(2*Mva);dw12=-(t13-t23)/(2*h)a=[0 dw12;

W0 0]lamda=eig(a)wn=imag(lamda(1))fn=(wn/(2*3.14))

OUTPUT:

y = 0 - 0.7752i 0 0 + 0.7752i

0 0 - 0.7752i 0 + 0.7752i

0 + 0.7752 0 + 0.7752i 1.3781 - 1.6415i

m = 0.1803 - 0.5605i 0.1803 + 0.2147i

0.1803 + 0.2147i 0.1803 - 0.5605i

b = -0.5605 0.2147

0.2147 -0.5605

dw12 = -0.2513

a = 0 -0.2513

376.8000 0

lamda = 0 + 9.7309i

0 - 9.7309i

wn = 9.7309

fn =1.5495

Result:

Thus the small signal stability analysis of multi machine system is completed, the modified [A]

matrix for the system was generated, using which eigen values and natural frequency of oscillation

were found.

ECONOMIC LOAD DISPATCH WITH GENERATING LIMITS INCLUDINGLOSSES

AIM :

To understand the fundamentals of economic dispatch and solve the problem using

classical method with generating limits and including line losses.

SOFTWARE REQUIRED :

MATLAB

THEORY :

Purpose of the economic dispatch or optimal dispatch is to reduce the fuel costs for

the power system .By economic load scheduling, we mean to find the generation of the

different generators or plants, so that total fuel cost is minimum and at the same time the

total demand and losses at any instant must be met by the total generation. The economic

dispatch problem involves the solution of two different problems, i.e.. Unit commitment and

online dispatch.

There are two methods used to find the economic dispatch.

1) Base load method;

Where the most efficient unit is loaded to its maximum capability, than the second most

efficient unit is loaded etc.

2) Best point method (incremental method);

Where units are successively loaded to their lowest heat rate point beginning with most

efficient unit and working down to the least efficient unit.

Experiment : 6

Date :

PROGRAM

cost= [200,7.0,0.008;180,6.3,0.009;140,6.8,0.007];mwlimits=[10,85;10,80;10,70];pdt=150;B=[0.000218,0.000093,0.000028;0.000093,0.00028,0.000017;0.000028,0.000017,0.000179];beta1=cost(1,2); beta2=cost(2,2);beta3=cost(3,2);gamma1=cost(1,3); gamma2=cost(2,3);gamma3=cost(3,3);basemva=100;lamda=8;for iter=1:3

P(1)=(lamda-beta1)/(2*(gamma1+(lamda*B(1,1))));P(2)=(lamda-beta2)/(2*(gamma2+(lamda*B(2,2))));P(3)=(lamda-beta3)/(2*(gamma3+(lamda*B(3,3))));PL1=0;for i=1:3for j=1:3PL=(P(i)*B(i,j))*(P(j));PL1=PL1+PL;

endenddelp=pdt+PL1-(P(1)+P(2)+P(3));delpi1=((gamma1+(B(1,1)*beta1)))/(2*((gamma1+(lamda*B(1,1)))^2));delpi2=((gamma2+(B(2,2)*beta2)))/(2*((gamma2+(lamda*B(2,2)))^2));delpi3=((gamma3+(B(3,3)*beta3)))/(2*((gamma3+(lamda*B(3,3)))^2));delpi=delpi1+delpi2+delpi3;dellamda=delp/delpi;nwlamda=lamda+dellamda;lamda=nwlamda;end

Tcost=(cost(1,1)+beta1*(P(1))+gamma1*(P(1))^2)+(cost(2,1)+beta2*(P(2))+gamma2*(P(2))^2)+(cost(3,1)+beta3*(P(3))+gamma3*(P(3))^2);disp('#########################################');disp('The Result is ............');fprintf('Incremental cost of delivered power(system lamda) =');fprintf('%4g ' , nwlamda); fprintf('\n');disp('optimal Dispatch of generation : ');disp(P(1));disp(P(2));disp(P(3));fprintf('Total system loss = '); fprintf('%4g ', PL1);fprintf('MW');fprintf('\n');fprintf('Total Generation Cost= '); fprintf('%4g ', Tcost);fprintf('$/h'); fprintf('\n');

OUTPUT:#########################################The Result is ............Incremental cost of delivered power (system lamda) =7.69954optimal Dispatch of generation :

36.152462.738253.6989

Total system loss = 2.54832 MWTotal Generation Cost = 1599.54 $/h

RESULT:

Thus the Economic load dispatch for the given problem was solved was using classicalmethod with generating limits and including line losses and verified using MATLAB software.

UNIT COMMITMENT BY DYNAMIC PROGRAMMING

AIM:

To obtain the unit commitment problem by dynamic programming

and verify using MATLAB.

SOFTWARE REQUIRED :

MATLAB

THEORY:

Unit Commitment, further abbreviated as UC, refers to the strategic choice to be

made in order to determine which of the available power plants should be considered to

supply electricity. UC is not the same as dispatching. Dispatching consists of fitting a given

set of power plants into a certain electric demand. UC appoints the set of plants from which

dispatching can choose. The difference between both issues is time. In dispatching

decisions, there is no time to rapidly activate a power plant because the inertia of most

plants will

not allow this. UC therefore prepares a set of plants and stipulates in which time period they

have to be on-line and ready for dispatching.

UC chooses plants taking into account a wide variety of parameters, technological

aspects (such as minimal operation point, minimum up and down time and transient

behavior) as well as economical considerations (such as start-up costs and operational costs)

and social elements (such as availability of staff and work-schemes). UC optimization

enables utilities to minimize electricity generation costs.

2. APPLICATIONS OF UNIT COMMITMENT

For utilities, UC is a problem that is to be evaluated in a time period of one day up to one

week. The power system these utilities need to optimize is usually limited to ten to fifty

power plants.

3. EXISTING METHODS

Many strategies have already been developed to tackle the UC economic optimization.

Experiment : 7(a)

Date :

Brute Force Method: The most evident method is what we call brute force in which

all possible combinations of power plants to provide a given demand are calculated.

The possibilities conflicting with the boundary conditions are struck off the list.

Finally, the most economic of all remaining possibilities is withheld. This method is

not only scientifically clumsy but will also amount in the largest possible calculation

time.

DP Method: Dynamic programming (DP) is a name used for methods in which a-

priori impossible or improbable possibilities are left out. This Method starts from a

previously determined optimal UC planning and gradually adds power plants to

obtain optimal solutions for higher demand.

Decomposition Method: In this method the main problem is decomposed into

several sub-problems that are easier to solve. In order to take into account

uncertainties combine the DP with fuzzy logic. The neural networks can be used to

enable the model to learn from previously made decisions. · It is possible to

decompose UC into a master problem and sub-problems that can be solved

separately. The master problem is optimized (minimal cost), linking the sub-

problems by Lagrange multipliers.

Priority Listing Method: A very simple method is based on Priority Listing in

which power plants are logically ranked. Originally, the plants were ranked

according to full load cost. All plants are initially activated. Then they are shut down

one at a time to check whether or not the overall costs are reduced.

Next to these conservative methods, also some unconventional methods like genetic

algorithms can be used. This is a stochastic adaptive search based on "survival of the

fittest".

PROGRAM:

b=[0.77,23.5;1.60,26.5;2.00,30.0;2.50,32.0];load=9;for i=1:3for y=1:load

F1(y) = ((1/2)*b(i,1)*(y)^2+(b(i,2)*y));endfor x=1:loadF2(x) = ((1/2)*b((i+1),1)*(x)^2+(b((i+1),2)*x));end

h(1)=0+F1(9);h(2)=F2(1)+F1(8);h(3)=F2(2)+F1(7);h(4)=F2(3)+F1(6);h(5)=F2(4)+F1(5);h(6)=F2(5)+F1(4);h(7)=F2(6)+F1(3);h(8)=F2(7)+F1(2);h(9)=F2(8)+F1(1);h(10)=F2(9)+0;

s=[h(1) h(2) h(3) h(4) h(5) h(6) h(7) h(8) h(9) h(10)];n=min(s);f(i)=n;for m=1:9

if(h(m)==n)k(i)=m;

endendp1=k(i)-1;L1=10-k(i);disp('#########################################');

fprintf('%4g ',i);fprintf('st');fprintf('Unitfor');fprintf('%4g',L1);fprintf('MW and');fprintf('%4g','economical');fprintf('\n ');

endq=[f(1) f(2) f(3)];z=min(q);

if(f(1)==z)r=k(1);i=1;

else if(f(2)==z)r=k(2);i=2;

else if(f(3)==z)r=k(3);i=3;

endend

end

L1=r-1;P1=10-r;disp('#########################################');

fprintf('%4g ' , z);fprintf('\n ');fprintf('%4g ' , i);fprintf('Unit for ' );fprintf('%4g ' ,P1);fprintf('MW and ');fprintf('%4g ' , i+1);fprintf('Unit for ');fprintf('%4g ' , L1);fprintf('Mw are economical');fprintf('\n');

OUTPUT

#########################################1 st Unit for 7MW and 101 99 111 110 111 109 105 99

97 108#########################################

2 nd Unit for 6MW and 101 99 111 110 111 109 105 9997 108#########################################

3 rd Unit for 5MW and 101 99 111 110 111 109 105 9997 108#########################################

239.5651 Unit for 7 MW and 2 Unit for 2 Mw are

economical

RESULT :

Thus the given unit commitment problem solved by dynamic programmingmethod and verified using MATLAB.

UNIT COMMITMENT BY PRIORITY LIST METHOD

AIM:

To obtain the unit commitment problem by priority list method

and verify using MATLAB.

SOFTWARE REQUIRED :

MATLAB

THEORY:

Unit Commitment, further abbreviated as UC, refers to the strategic choice to be

made in order to determine which of the available power plants should be considered to

supply electricity. UC is not the same as dispatching. Dispatching consists of fitting a given

set of power plants into a certain electric demand. UC appoints the set of plants from which

dispatching can choose. The difference between both issues is time. In dispatching

decisions, there is no time to rapidly activate a power plant because the inertia of most

plants will

not allow this. UC therefore prepares a set of plants and stipulates in which time period they

have to be on-line and ready for dispatching.

UC chooses plants taking into account a wide variety of parameters, technological

aspects (such as minimal operation point, minimum up and down time and transient

behavior) as well as economical considerations (such as start-up costs and operational costs)

and social elements (such as availability of staff and work-schemes). UC optimization

enables utilities to minimize electricity generation costs.

2. APPLICATIONS OF UNIT COMMITMENT

For utilities, UC is a problem that is to be evaluated in a time period of one day up to one

week. The power system these utilities need to optimize is usually limited to ten to fifty

power plants.

Experiment : 7(b)

Date :

3. EXISTING METHODS

Many strategies have already been developed to tackle the UC economic optimization.

Brute Force Method: The most evident method is what we call brute force in which

all possible combinations of power plants to provide a given demand are calculated.

The possibilities conflicting with the boundary conditions are struck off the list.

Finally, the most economic of all remaining possibilities is withheld. This method is

not only scientifically clumsy but will also amount in the largest possible calculation

time.

DP Method: Dynamic programming (DP) is a name used for methods in which a-

priori impossible or improbable possibilities are left out. This Method starts from a

previously determined optimal UC planning and gradually adds power plants to

obtain optimal solutions for higher demand.

Decomposition Method: In this method the main problem is decomposed into

several sub-problems that are easier to solve. In order to take into account

uncertainties combine the DP with fuzzy logic. The neural networks can be used to

enable the model to learn from previously made decisions. · It is possible to

decompose UC into a master problem and sub-problems that can be solved

separately. The master problem is optimized (minimal cost), linking the sub-

problems by Lagrange multipliers.

Priority Listing Method: A very simple method is based on Priority Listing in

which power plants are logically ranked. Originally, the plants were ranked

according to full load cost. All plants are initially activated. Then they are shut down

one at a time to check whether or not the overall costs are reduced.

Next to these conservative methods, also some unconventional methods like genetic

algorithms can be used. This is a stochastic adaptive search based on "survival of the

fittest".

PROGRAM:

hrate=[ 510 7.2 0.0014310 7.85 0.0019478 7.97 0.00482];

limit=[150 600100 40050 200];

cost=[1.11.01.2];

k1=cost(1,1);k2=cost(2,1);k3=cost(3,1);PD=550;

FLAPC1=(k1*(hrate(1,1)+hrate(1,2)*limit(1,2)+hrate(1,3)*(limit(1,2)^2)))/(limit(1,2));



m={FLAPC1,FLAPC2,FLAPC3};f=max(FLAPC1,FLAPC2);h=min(FLAPC1,FLAPC2);n=min(h,FLAPC3);max1=limit(1,2)+limit(2,2)+limit(3,2);max2=limit(1,2)+limit(2,2)max3=limit(2,2);min1=limit(1,1)+limit(2,1)+limit(3,1);min2=limit(1,1)+limit(2,1)min3=limit(2,1);disp('################################');

fprintf('Unit Commitment');fprintf('\n');fprintf('Combination');fprintf('Min');fprintf('Max');fprintf('\n');fprintf('2+1+3');fprintf('%4g',min1);fprintf('%4g',max1);fprintf('\n');fprintf('2+1');fprintf('%4g',max2);fprintf('\n');fprintf('2');fprintf('%4g',min3);fprintf('%4g',max3);fprintf('\n');fprintf('All three units would be held on until load

reached');fprintf('%4g',max2);fprintf('MW,unit 2 and 1 would be held on until the

load reached');fprintf('%4g',max3);fprintf('MW');fprintf('\n');fprintf('for demand');fprintf('%4g',PD);fprintf('MW unit 2 and 1 would be operated');fprintf('\n');

OUTPUT:

############################################

Unit Commitment

Combination Min Max

2+1+3 300 1200

2+1 250 1000

2 100 400

All three units would be held on until load reached 1000 Mw ,

unit 2 and 1 would be held on until the load reached 400 MW

For Demand 550 Mw unit 2 and 1 would be operated

RESULT :

Thus the given unit commitment problem solved by priority list method andverified using MATLAB.

CONTINGENCY ANALYSIS METHOD

AIM:

To solve the problem by contingency analysis method and verify usingMATLAB.

SOFTWARE REQUIRED:

MATLAB

THEORY:

An important factor in the operation of a power system is the power system security.

Power system involves practices and design to keep the system operating even when the

components like transmission line, transformer or generation fail. Since the specific time at

which the components fail is not known, the system should be operated in such way that, it

will not be left in dangerous condition, even when any contingency occurs contingently

study involves the study of the effect of different credible outages considered one at a time

Experiment : 8

Date :

PROGRAM:

clc;nb=2;nl=3;sb=[1 1 2];eb=[2 3 3];y=1.0./[ 0.6 0.8 0.4 ];ybus=zeros(nb,nb);for i=1

l=sb(i);m=eb(i);ybus(l,l)=ybus(1,1)+y(i)+y(i+1);ybus(m,m)=ybus(m,m)+y(i)+y(i+2);ybus(l,m)=-y(i);ybus(m,l)=ybus(l,m);

endybuszbus=inv(ybus)% generation shift factorsp=1;q=2;m=1;n=3;i=1;fprintf('GENERATION SHIFT FACTOR');a=[ 0.6 0.8 0.4 ];gsf1=(zbus(p,p)-zbus(q,p))/a(i);zbus(n,p)=0;gsf2=(zbus(p,p)-zbus(n,p))/a(i+1);zbus(p,n)=0;gsf3=(zbus(p,q)-zbus(m,n))/a(i+2);gsf1gsf2gsf3% line outage factorsfprintf('LINE OUTAGE DISTRIBUTION FACTOR when outage of line1-3');

zbus(n,p)=0;zbus(p,n)=0;zbus(q,n)=0;zbus(n,q)=0;zbus(n,m)=0;zbus(n,n)=0;L1=(zbus(p,m)-zbus(p,n))-(zbus(q,m)-zbus(q,n));L2=a(i+1)-((zbus(m,m)-zbus(m,n))-(zbus(n,m)-zbus(n,n)));L3=(a(i+1)/a(i));Line1=(L1*L3)/L2;Line1p=2;q=3;m=1;n=3;L1=(zbus(p,m)-zbus(p,n))-(zbus(q,m)-zbus(n,n));L2=a(i+1)-((zbus(m,m)-zbus(m,n))-(zbus(n,m)-zbus(n,n)));L3=(a(i+1)/a(i+2));Line2=(L1*L3)/L2;Line2OUTPUT:

ybus =

2.9167 -1.6667

-1.6667 4.1667

zbus =

0.4444 0.1778

0.1778 0.3111

GENERATION SHIFT FACTOR

gsf1 = 0.4444

gsf2 = 0.5556

gsf3 = 0.4444

LINE OUTAGE DISTRIBUTION FACTOR when outage of line 1-3

Line1 = 1.0000

Line2 = 1.0000

RESULT:

Thus the given problem solved by contingency analysis method and verifiedusing MATLAB.

INDUCTION MOTOR STARTING

AIM :

To perform induction motor starting connected to a system given and generate thecharacteristics curves.

SOFTWARES USED:

MATLAB

ALGORITHM:

Step 1: Start

Step 2: Power factor and current ratio for different percentage speeds.

Step 3: Read the induction motor data-Full Load Amperes (FLA), RPM, frequency,J=63.87, r1,

r2, X1&X2.

Step 4: Read the MVA on 3 phase AC (3HP) and system base(Ssys).

Step 5: Calculate Zs=1/ (S3ph/Ssys).

Step 6: Read the transmission line impedance Ztr.

Step 7: Calculate the synchronous speed Ws=2*pi*rpm/60.

Step 8: Consider steady state model of induction machine, read V1, V2, V3.

Step 9: Set =V3.

Step 10: Motor MVA on (S motor) =3* *FLA.

Step 11: Initialize =0,N=0,Cange in speed =0.

Step 12: Select the set of power factor, current ratio percentage speed.

Step 13: Change in speed =percentage speed (previous)-percentage speed (current).

Step 14: Slips= (100-percentage speed)/100.

Step 15: Calculate power factor angle from power factor ( ).

Step 16: Calculate induction motor equivalent impedance

Zm=(P.F)+j*sin ( )*1/I ratio*Ssys/S motor.

Experiment : 9

Date :

Step 17: Calculate motor current I=V1/(Zs+Ztr+Zm) .

Step 18: Calculate V3=I*Zm* .

Step 19: Calculate motor Torque(Tm) = (3*[mag (V3/ )] * /s)

Ws*( + /s + ( +

Step 20: Accelerating Torque (Tacc) =Motor Torque(Tm)-Load Torque (TL).

Step 21: Calculate T=JWWs/Tacc.

Step 22: Increment by t.

Step 23: Increment by speed by .

Step 24: Check if all percentage speed sets are covered else goto step 12.

Step 25: Plot Torque Vs Slip &Torque Vs .

Step 26: Stop.

PROGRAM:

clc;clear all;v1=1.026+j*0.0;BaseMVA=1e7;MVA=125e6;zs=BaseMVA/MVA;zc=0.0;zst=0.0;ztr=0.0001+j*0.0001;J=63.87;R1=0.029;R2=0.022;X1=0.226;X2=0.226;Ns=1800;ws=(2*pi*Ns/60);% speed PF ang Iratio

data=[ 0.0 6.0 9.0 0.0;10.0 5.9 18.0 0.0;20.0 5.8 20.0 0.0;30.0 5.7 36.0 0.0;40.0 5.6 50.0 0.0;50.0 5.5 60.0 0.0;60.0 5.5 71.0 0.0;70.0 5.4 81.0 0.0;80.0 5.2 86.0 0.0;90.0 4.0 88.0 0.0;96.5 3.2 89.0 0.0;98.5 1.7 91.0 0.0;];

n=length(data);%PF=data(:,2)%ang=data(:,3)%Iratio=data(:,4)t=0;y=0;for i=1:n

%for k=1:4%Zm=(1/PF(i))*((ang/100)+sin(acos(ang/100))*j)

Zm=(1/data(i,2))*((data(i,3)/100)+sin(acos(data(i,3)/100))*j;Zl=Zm*10/(2.098*3);I=v1/(zs+ztr+Zl);v3=abs(I*Zl*2300/1.732);w=data(i,1)*ws/100;s=(ws-w)/ws;R=(R1+R2/s)*(R1+R2/s);X=(X1+X2)*(X1+X2);Tm=((3*v3*v3*R2)/s)/(ws*(R+X));Tacc=Tm-(data(i,4)/100);del=w-y;delt=(J*del)/Tacc;

t=t+delt;t1(i)=t;w1(i)=w;I1(i)=I;Tm1(i)=Tm;s1(i)=s;y=w;%end

endt1;I1;Tm1;subplot(2,2,1);plot(t1,Tm1);title('Torque Vs time');subplot(2,2,2);plot(s1,Tm1);title('Torque Vs slip');subplot(2,2,3);plot(t1,I1);title('current vs time');subplot(2,2,4);

OUTPUT:

RESULT:

Thus the performance of induction motor starting was observed and the characteristics curveswere drawn.

RELAY CO-ORDINATION OF RADIAL TRASMISSON/DISTRIBUTION SYSTEM

AIM :

To co-ordinate the over-current protection of the given radial system using IDMT relays by

proper selection of their Time Multiplier Setting (TMS) and Plug Setting Multiplier (PSM), such that

there is sufficient time-of-operation discrimination and sequential relay operation.

.SOFTWARES USED:

MATLAB

ALGORITHM:

1. Start.2. Perform the manual calculation and determine the TMS values for the different

relays employed in the system.3. Perform short circuit analysis in ETAP.4. Plot the variation of relay operating time with change in PSM.5. Stop.

EXERCISETo perform relay coordination of the radial distribution system given below i.eto suitably place the relays and determine the T.M.S (Time Multiplier Setting)and P.S.M (Plug Setting Multiplier

Fig 10.1 Single line diagram of radial distribution system

Experiment : 10

Date :

Calculation:

Given Data:

% Impedance, Pos. Seq., 100 MVAb

R X Z

T1 2W XFMR Bus1 Bus2 5.14 79.61 79.78

T1 2W XFMR Bus2 Bus3 74.41 430.86 437.24

Bus1 Bus2 X1=0.7961 pu

Bus2 Bus3 X2=4.3086 pu

Xth = X1+X2 = 5.1047

Short Circuit Current Isc = 1/5.1047 = 0.1959 pu

For fault at 415V side

At 415V bus:

Ibase = 100MVA/(3^0.5*415V) = 139124.6 A

Isc = 0.1959*139124.6 A = 27.25KA

At 11KV bus:

Isc = 27.25KA*415V/11KV = 1028A

Relay R1 at 415V side

Full load current Ifl = 1MVA/(3^0.5*415V) = 1391.24A

Allowing an Overload of 25% Relay Pick up current Ip = 1391.24*1.25 = (1739.05A) = 1800A

Select the CT ratio to be 2000/1A

Plug setting PS = 1800/2000 = 0.9

Plug Setting Multiplier PSM = Fault current/Pick up current =27.5KA/1800A = 15.14

For IDMT Over Current relay top = 0.14(TMS)/(PSM^0.02-1)

Fixing top = 0.1 sec

Time Multiplier Setting TMS = top*(PSM^0.02-1)/0.14 = 0.0399

Relay R2 at 11KV side

Full load current Ifl = 1MVA/(3^0.5*11KV) = 52.49A

Allowing an Overload of 25% Relay Pick up current Ip = 52.49*1.25 = 65.61A


Plug setting PS = 65.61/80 = 0.82

Plug Setting Multiplier PSM = Fault current/Pick up current =1028/65.61 = 15.67



Fault at 11KV bus

Xth = 0.7961 pu

Isc = 1/0.7961 = 1.2561 pu

At 11KV bus:

Ibase = 100MVA/(3^0.5*11KV) = 5248.79 A

Isc = 1.2561*5248.79 = 6.59KA

At 132KV bus:

Isc = 6.59KA*11KV/132KV = 549.42 A

Relay R2 CT ratio = 80/1A,PS = 0.82, TMS = 0.121

PSM = 6.59KA/65.61A = 100.49

top = 0.14(TMS)/(PSM^0.02-1) = 0.175 sec





Plug setting PS = 656.10/800 = 0.82

Plug Setting Multiplier PSM = Fault current/Pick up current =6.59KA/656A = 10.05







Plug setting PS = 54.67/60 = 0.911

Plug Setting Multiplier PSM = Fault current/Pick up current =549.42/54.67 = 10.05



PROGRAM:

'FOR RELAY R4'TMS4=0.044;psm=1;for i=1:10

top=(0.14*TMS4)/((psm^0.02)-1);top4(i)=top;psm4(i)=psm;psm=psm+10;

end'FOR RELAY R3'TMS3=0.0129;psm=1;for i=1:10


end'FOR RELAY R2'TMS2=0.178;psm=1;for i=1:10


end

'FOR RELAY R1'TMS1=0.221;psm=1;for i=1:10


endplot(psm4,top4,'-',psm4,top3,'-',psm4,top2,'-',psm4,top1,'-')

OUTPUT:

RESULT:

The relay co-ordination on the given radial system is done and the inverse-time characteristicsof each relay are plotted.

power simulation lab record new

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