My Lab2012
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Transcript of My Lab2012
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5/24/2018 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 maximum fault current in all nodes from first to last respectively: 5000
Enter maximum fault current in all nodes from first to last respectively: 3000
Enter maximum fault current in all nodes from first to last respectively: 2000
Enter maximum fault current in all nodes from first to last respectively: 1000
Enter load current in each nodes from first to last respectively: 115
Enter load current in each nodes from first to last respectively: 80
Enter load current in each nodes from first to last respectively: 100
Enter load current in each nodes from first to last respectively: 77
Enter load current in each nodes from first to last respectively: 70
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.