My Lab2012

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EXPT. NO: 01 DATE:

Available Transfer Capability calculation using an existing load flow program

Aim:

To calculate the available transfer capability of an existing power system, a three bus system.

Software Used:

MATLAB 8

Algorithm:STEP 1. Start.

STEP 2. Read the bus data, line data and transformer data.STEP 3. Read the voltage limits Vmin and Vmax.

STEP 4. Compute the Y-bus matrix.

STEP 5. Select an interface for which ATC need to be determined SATC, NSATC.

STEP 6. Set a suitable tolerance and run Newton Raphson Method.STEP 7. Increase the power demand at the load bus and the generator bus of the interface by a

small value.

STEP 8. Run NR method.

STEP 9. Check for voltage limit violations. If yes go to next step else go to step 7.STEP 10. Calculate ATC.

STEP 11. Stop.

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Input:

1-slackbus

2-loadbus

3-Generator bus

P2=400MW

P3=200MWQ2=-250MWV1=1.05pu

V3=1.04pu

Ybus = [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];

Coding:clcclear all%%Calculation of ATC%%%Taking example of a three bus systems%1-Slack bus%2-Load Bus%3-Generator Bus%%Defining Ybus%%Taking base value=100MVA%P2=-400MW,Q2=-250MW,P3=200MW;

y=[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];G=real(y);B=imag(y);theta=angle(y);Y=abs(y);iter=0;iter1=0;Sb=100;P=[0 -4 2];Q=[0 -2.5 0];V=[1.05 1 1.04];d=[0 0 0];

%%Given values and assuming valuesPs=[0 -4 2];

delM=1;e=0.1;

z=2;fora=1:100

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while(z>1)

nbus=3;npv_bus=1;

P_c=zeros(nbus,1);Q_c=zeros(nbus,1);%calculating P and Q

fork=2:nbusform=1:nbus

P_c(k)=P_c(k)+V(k)*V(m)*Y(k,m)*cos(d(k)-d(m)-theta(k,m));end

end

fork=2form=1:nbus

Q_c(k)=Q_c(k)+V(k)*V(m)*Y(k,m)*sin(d(k)-d(m)-theta(k,m));end

endP_c(2)=P_c(2,:);P_c(3)=P_c(3,:);Q_c(2)=Q_c(2,:);

%%Calculation of Jacobian elementsfork=1:3for m=1:3ifk==mH(k,m)=-Q_c(k)-B(k,k)*V(k)^2;L(k,m)=Q_c(k)-B(k,k)*V(k)^2;N(k,m)=P_c(k)+V(k)^2*G(k,k);J(k,m)=P_c(k)-V(k)^2*G(k,k);

elseH1(k,m)=V(k)*V(m)*Y(k,m)*sin(d(k)-d(m)-theta(k,m));L1(k,m)=H1(k,m);J1(k,m)=-V(k)*V(m)*Y(k,m)*cos(d(k)-d(m)-theta(k,m));N1(k,m)=-J1(k,m);

endendend%%Creating Jacobian matrixJac=[H(2,2) H1(2,3) N(2,2)

H1(2,3) H(3,3) N1(2,3)J(2,2) J1(3,2) L(2,2)];

%%Calculating mismatch vectorsdelm=[P(2)-P_c(2);P(3)-P_c(3);Q(2)-Q_c(2)];

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%%Calculating inverse matrix

del=inv(Jac)*delm;d(2)=d(2)+del(1,:);d(3)=d(3)+del(2,:);V(2)=V(2)+del(3,:);

delM=abs(del);ifmax(delM)>e

break

endifiter>=10000

breakenditer=iter+1;end

%%Calculating base line flows

ifa==1Vb2=V(2)*(cosd(2)+j*sind(2));Vb1=V(1)*(cosd(1)+j*sind(1));Vb3=V(3)*(cosd(3)+j*sind(3));Ib12=y(1,2)*(Vb1-Vb2);Ib21=-Ib12;Ib13=y(1,3)*(Vb1-Vb3);Ib31=-Ib13;Ib23=y(2,3)*(Vb2-Vb3);Ib32=-Ib23;S12b=Vb1*conj(Ib12);S21b=Vb2*conj(Ib21);S13b=Vb1*conj(Ib13);S31b=Vb3*conj(Ib31);S23b=Vb2*conj(Ib23);S32b=Vb3*conj(Ib32);

Sl_12b=S12b+S21b;Sl_13b=S13b+S31b;Sl_23b=S23b+S32b;end%%%Increasing demand across loadbus

Q_s=zeros(nbus,1);fork=1:3

form=1:nbus

Q_s(k)=Q_s(k)+V(k)*V(m)*Y(k,m)*sin(d(k)-d(m)-theta(k,m));end

endP(2)=1.05*P(2);Q(2)=1.05*Q(2);P(3)=1.05*P(3);

%%Checking contingencies.....Real Power, Reactive Power and Undervoltages%%and overvoltages in all buses

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%%Taking load bus limit=800Mw max n 50MVAR min.........voltagelimit...0.95

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fprintf(', Busno. || Bus Type || Voltage(pu) ||Angle(deg.) || Power(MW) || Reactive Power(MVAR)\n');fprintf('\n 1 Slack'); fprintf('%26f',V(1), d(1)*180/pi,P_s(1,:)*Sb , Q_s(1,:)*Sb');fprintf('\n 2 Load'); fprintf('%26.5f', V(2), d(2)*180/pi,P(2)*Sb ,Q(2)*Sb);fprintf('\n 3 Generator'); fprintf('%25f',V(3),d(3)*180/pi,P(3)*Sb,Q_s(3,:)*Sb);

fprintf('\n\n ATC=');fprintf('%f',atc*Sb);fprintf('MW');

Output:

ATC CALCULATION USING NEWTON RAPHSON METHOD

Busno. || Bus Type || Voltage(pu) || Angle(deg.) || Power(MW) || Reactive Power(MVAR)

1 Slack 1.049840 0.000000 289.966700 178.325645

2 Load 0.94835 -3.59353 -536.03826 -335.02391

3 Generator 1.038575 -0.769215 268.019128 196.387072

ATC=160.595871MW>>

Conclusion:

Hence, ATC for a three bus system was calculated using newton raphson solved loadflow analysis

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EXPT. NO: 02 DATE:

Computation of harmonic indices generated by a rectifier feeding a R-L load

Aim:

To perform the harmonics analysis of a full bridge rectifier circuit feeding a RL load

Software used:

MATLAB 8, SIMULINK .

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Input:

R=1ohm

L=1mF

Vrms=230V

Simulink model:

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Output:

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Conclusion:

Hence, we can observe that RL load causes distortion in voltage and current in a balanced

supply. THD for current is greater than voltage.

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EXPT. NO: 04 DATE:

Co-ordination of over-current and distance relays for radial line protection

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.

Software Used:

MATLAB 8 .

Formulas Used:

Top= 0.14*TMS/(PSM^0.02-1) PSM = Irelay / Plug setting

Algorithm:

STEP 1. Start.

STEP 2. Prepare single line diagram and collect all equipment details.

STEP 4. Select a bus to have a fault

STEP 3. Select suitable CT Ratio, relay type.STEP 4. Perform the manual calculation and determine the TMS values for the different relays

employed in the system.STEP 5. Plot the variation of relay operating time with change in PSM.

STEP6. Display the resultsSTEP7. Stop.

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Input:

Enter number of nodes: 5

Enter node on which fault occured: 4

Enter maximum fault current in all nodes from first to last respectively: 6000





Enter load current in each nodes from first to last respectively: 115





Coding:

clear all

clcn=input('Enter number of nodes: ');r=input('Enter node on which fault occured: ');fori=1:nz_f(i)=input('Enter maximum fault current in all nodes from first to lastrespectively: ');

endfori=1:r+1z_l(i)=input('Enter load current in each nodes from first to lastrespectively: ');

end

PS=5;%Assuming plug setting of 5A%Calculating CT ratioforj=1:n

temp=0;ifj==nB(j)=z_l(:,n);

endfor i=j+1:n;

B(j)=temp+z_l(i);

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temp=B(j);end

ende=1;i=r;c=1;whilec0ifi==r;TSM(i,e)=0.1;%For tail end relay TMS=1 alwaysz(i)=tc(B,i); %Finding CT ratioPSM(i,e)=z_f(:,i)/(z(i)*PS);%Finding PSMTR(i,e)=0.14*TSM(i,e)/(PSM(i,e)^0.02-1);%Operating time of relayelse

TR(i,e)=TR(i+1,e)+0.5+(TR(i+1,e)+0.5)*0.1;z(i)=tc(B,i);PSM(i,e)=z_f(:,r)/(z(i)*PS);TSM(i,e)=TR(i,e)*(PSM(i,e)^0.02-1)/0.14;endi=i-1;

ende=e+1;c=c+1;

endj=r;fprintf('\n\n\n Co-ordination of overcurrent relays in radial feeder');fprintf('\n\n Number of nodes');fprintf(' %d',n);fprintf('\nFault location node');fprintf(' %d',j);fprintf('\n\n Relay no Operating time(seconds) Plug setting multiplier TimeMultiplier setting\n');whilej>0

fprintf('\n %d',j);fprintf(' %f', TR(j,1), PSM(j,1),TSM(j,1));fprintf('\n');j=j-1;end

fori=1:rplot(PSM(i,:),TR(i,:));hold on;

xlabel('Plug setting Multiplier'); ylabel('operating time of relay(s)');title('Overcurrent relay co-ordination','fontsize',15);endaxis([0 max(PSM(:,1)+20) 0 max(TR(:,1))+0.5]);

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Output:

Co-ordination of overcurrent relays in radial feeder

Number of nodes 5

Fault location node 4

Relay no Operating time(seconds) Plug setting multiplier Time Multiplier setting

4 0.226736 20.000000 0.100000

3 0.799409 10.000000 0.269107

2 1.429350 5.000000 0.33398

1 2.122285 4.000000 0.426183

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Conclusion:

Hence, overcurrent relay co-ordination in a 5 bus system was successfully done.

My Lab2012

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