Fig. 1 depicts a three-bus system. Suppose the three units are always running, w
ID: 2293919 • Letter: F
Question
Fig. 1 depicts a three-bus system. Suppose the three units are always running, with the following characteristics Unit 1: $40/MWh, POMW, P820MW Unit 2: $50/MWh, PiOMW, P 60OMW Unit 3: $55/MWh, Pi OMW,P610MW The total demand for three hours is: 1000, 1100, 900 MW. The load distribution always has a ratio of: D1:D2:D3-200:350:450, which means the total load will be allocated to D1, D2 and D3 by this ratio. For example, in hour 2, D11100*200/(200 + 350+450) D2 1100*350/(200 + 350+450) D3 1100*450/(200 + 350+450) Loads in other hours can be obtained similarlv Transmission line impedances in per unit and constraints in MW are labeled in the figure. Sbase 100 MW. The transmission line constraints are LI-250 MW, L2-200 MW, L3-200 MW G1 G2 0.1j LI MW Busl Bus2 Di D2 0.125 L2 MW 0.2j L3 MW Bus3 D3 G3 Fig. 1 A three-bus power system.Explanation / Answer
code:
matlab code
% ==========Power flow program using Newton-Raphson Method========= %
%
% ================================================================= %
%%
%------------------Important----------------------%
% load flow equations are non-linear in nature. To find the solution of such equations, it is assumed that
% 1. solution exists thats is (f(x)=0)
% 2. equation are continuous and diffrentiable.
% How to solve these non-linear equations of power (P,Q) ?.
% one can solve these equations by considering assumptions 1 and 2 and using linear
% approximation method (also known as newton-raphson method). also to be
% noted here is that the taylor series expnasion is the basis of the newton-raphson method.
% so if we have one equation whose solution we assume exists and is
% continuous and diffretiable then according to the newton-raphson
% f(x0)=-(x-x0)*f'(x0) ----(1)
% where x0 is our initial guess.Equation 1 here gives the linear approximation of the function f(x) at point x0.
% this equation will always result in straight line (Tagent line of f(x) at the x0)
% thus our solution in this case by manipulating equation 1 will be
% x = x0-f(x0)/f'(x0) ------ (2) (updated value of x0)
% It should be noted that the equation 1 is linear equation , and higher order derivatives (higher than 1) are
% neglected. hence for the solution of f(x) (non- linear equation), an
% iterative method is required to match the solution we are looking for (because we have truncted the non-linear
% equation after first derivative, so obivously the solution will not be exact, so iterations are required to get the
% exact solution).
% Equation 1 can be used in matrix form, in that case, f'(x0), will be our
% Jacobian which is a matrix of a first order partial derivatives of the non-linear
% function (P,Q) evaluated at x0.
%___________________________________________________________________________
%---------------Bascis-----------------%
% When applied to power flow problem, the newton raphson is used to solve
% for a vector of bus voltage magnitudes and angles i.e.
% x = [angle;abs(V)]
% Things to note :
% 1) the voltage and angle at the swing bus is specified and therefore not part of the the solution vector
% 2) the bus voltage magnitudes at PV buses are specified and therefore not
% part of abs (V). Only the bus voltage magnitudes of PQ buses are included
% 3) angle includes all PQ and PV buses
% therefore vector x is an m*1 vector with
% m = 2*n_pq+n_pv ; Where n_pq are number of PQ buses and n_pv are number
% of PV buses
% __________________________________________________________________________
%------------------Start of load flow program--------------%
% Three bus system . Bus 1 = Slack, Bus 2 = PV and bus 3 = PQ
clear all
clc
j = sqrt(-1);
V = zeros(3,1);
S = zeros(3,1);
Mismatch = zeros(3,1);
%------------------ Base values----------------%
kVLL=345;
MVA3Ph=100;
Zbase=kVLL^2/MVA3Ph;
%%%%%%%---------- line parameters in ohms/km----------%
XL_km=0.376;
RL_km= 0.037; B_km=4.5;
%---------Line Susceptances--------%
B13_Micro_Mho=4.5*200; %200 km long
B12_Micro_Mho=4.5*150; %150 km long
B23_Micro_Mho=4.5*150; %150 km long
%---------Line impedances------------%
Z13_ohm=(RL_km+j*XL_km)*200; %200 km long
Z12_ohm=(RL_km+j*XL_km)*150; %150 km long
Z23_ohm=(RL_km+j*XL_km)*150; %150 km long
%------- line impedances in per unit--------%
Z13=Z13_ohm/Zbase;
Z12=Z12_ohm/Zbase;
Z23=Z23_ohm/Zbase;
%-------- susceptances in per unit----------%
B13=B13_Micro_Mho*Zbase*10^-6;
B12=B12_Micro_Mho*Zbase*10^-6;
B23=B23_Micro_Mho*Zbase*10^-6;
%---------- YBUS Creation-------------%
Y(1,1)=1/Z12 + 1/Z13;
Y(1,2)=-1/Z12;
Y(1,3)=-1/Z13;
Y(2,1)=-1/Z12;
Y(2,2)=1/Z12 + 1/Z23;
Y(2,3)=-1/Z23;
Y(3,1)=-1/Z13;
Y(3,2)=-1/Z23;
Y(3,3)=1/Z13 + 1/Z23;
Y; % Print Y=G+jB Admittance Matrix
%----------Conductance Values------------%
G(1,1)=real(Y(1,1));
G(1,2)=real(Y(1,2));
G(1,3)=real(Y(1,3));
G(2,1)=real(Y(2,1));
G(2,2)=real(Y(2,2));
G(2,3)=real(Y(2,3));
G(3,1)=real(Y(3,1));
G(3,2)=real(Y(3,2));
G(3,3)=real(Y(3,3));
%--------Susceptance Values----------%
B(1,1)=imag(Y(1,1));
B(1,2)=imag(Y(1,2));
B(1,3)=imag(Y(1,3));
B(2,1)=imag(Y(2,1));
B(2,2)=imag(Y(2,2));
B(2,3)=imag(Y(2,3));
B(3,1)=imag(Y(3,1));
B(3,2)=imag(Y(3,2));
B(3,3)=imag(Y(3,3));
%--------- Given Specifications in pu (Known)----------%
V1MAG=1.0;
ANG1=0;
V2MAG=1.05;
P2sp=1.0;
P3sp=-0.9;
Q3sp=-0.6;
% -------------Calulate ANG2, V3MAG and ANG3---------------%
% ----Solution Parameters----%
Tolerance= 0.001;
Iter_Max=25;
%----- Initialization-------%
Iter=0;
i=0;
ConvFlag=1;
delANG2=0;
delANG3=0;
delMAG3=0;
%-------------- to be determined (Flat start)-----------%%%%%%%%
ANG2=0;
ANG3=0;
V3MAG=1.0;
%------------------ Start Iteration Process for N-R-----------------%
while( ConvFlag==1 && Iter < Iter_Max)
Iter=Iter+1;
i=i+1;
%%----------- update the Voltage and Angles to calculate new jacobian---%
ANG2=ANG2+delANG2;
ANG3=ANG3+delANG3;
V3MAG=V3MAG+delMAG3;
%------- Creation of Jacobian J--------%
% ------ delP2/delANG2=J(1,1)
J(1,1) = V2MAG*(V1MAG*(B(2,1)*cos(ANG2-ANG1)-G(2,1)*sin(ANG2-ANG1))+V3MAG*(B(2,3)*cos(ANG2-ANG3)-G(2,3)*sin(ANG2-ANG3)));
%%%%--------- delP2/delANG3 = J(1,2)------------%
J(1,2)=V2MAG*V3MAG*(G(2,3)*sin(ANG2-ANG3)-B(2,3)*cos(ANG2-ANG3));
%%%%%%%------ delP2/delV3Mag = J(1,3)--------%
J(1,3) = V2MAG*(G(2,3)*cos(ANG2-ANG3)+B(2,3)*sin(ANG2-ANG3));
%%%%%%%%%%%--- delP3/delANG2 = J(2,1)---------%%%
J(2,1) = V3MAG*V2MAG*(G(3,2)*sin(ANG3-ANG2)-B(3,2)*cos(ANG3-ANG2));
%%%%%%%%%%%-------delP3/delANG3 = J(2,2)-----------------%%
J(2,2)= V3MAG*(V1MAG*(B(3,1)*cos(ANG3-ANG1)-G(3,1)*sin(ANG3-ANG1))+V2MAG*(B(3,2)*cos(ANG3-ANG2)-G(3,2)*sin(ANG3-ANG2)));
%%%%%%%%%%%----------delP3/delV3MAG = J(2,3)------%
J(2,3) = 2*G(3,3)*V3MAG+V1MAG*(G(3,1)*cos(ANG3-ANG1)+B(3,1)*sin(ANG3-ANG1)+V2MAG*(G(3,2)*cos(ANG3-ANG2)+B(3,2)*sin(ANG3-ANG2)));
%%%%%%%%%--------------delQ3/delANG2 = J(3,1)--------%
J (3,1)= -V3MAG*V2MAG*(G(3,2)*cos(ANG3-ANG2)+B(3,2)*sin(ANG3-ANG2));
%%%%%%%%%% ----------------delQ3/delANG3 = J(3,2)--------%
J(3,2) = V3MAG*(V1MAG*(B(3,1)*cos(ANG3-ANG1)+G(3,1)*sin(ANG3-ANG1))+V2MAG*(B(3,2)*cos(ANG3-ANG2)+G(3,2)*sin(ANG3-ANG2)));
%%%%%%%%%%-------------delQ3/delV3MAG = J(3,3)------%
J(3,3) = -2*B(3,3)*V3MAG+V1MAG*(G(3,1)*cos(ANG3-ANG1)-B(3,1)*sin(ANG3-ANG1)+V2MAG*(G(3,2)*cos(ANG3-ANG2)-B(3,2)*sin(ANG3-ANG2)));
J = [J(1,1) J(1,2) J(1,3);J(2,1) J(2,2) J(2,3); J(3,1) J(3,2) J(3,3)];
%%%%%%% calculation of updated voltages with angles %%%%%%%%%%%%
V(1) = V1MAG*exp(1i*ANG1);
V(2)= V2MAG*exp(1i*ANG2);
V(3) = V3MAG*exp(1i*ANG3);
%%%%%%%----------- Current injections at each bus based on updated voltages and angles----------%%%%%%%%%
I = Y*V;
%%%%%%%%%----------calculation of P and Q-----------%%%%%%%%%%%
S(1) = V(1)*conj(I(1));
S(2) = V(2)*conj(I(2));
S(3) = V(3)*conj(I(3));
%%%%%%%%%%%------------- Mistmatches--------------%%%%%%%%%%
Mismatch(1) = P2sp-real(S(2));
Mismatch(2) = P3sp-real(S(3));
Mismatch(3) = Q3sp-imag(S(3));
%%%%%%%-------- calculate the deltaANG and deltaVMAG-----------%%%%%%%
del=inv(J)*Mismatch; % for large matrices, this mathod is bad.
% Always use LU factorization to solve this linear equation.
delANG2 = del(1);
delANG3 = del(2);
delMAG3 = del(3);
%%%%%%%%%%%----------- Calculate power flow on transmission line-------%%
P12 = real(V(1))*conj((V(1)-V(2))/Z12);
P13 = real(V(1))*conj((V(1)-V(3))/Z13);
P21 = real(V(2))*conj((V(2)-V(1))/Z12);
P23 = real(V(2))*conj((V(2)-V(3))/Z23);
P31 = real(V(3))*conj((V(3)-V(1))/Z13);
P32 = real(V(3))*conj((V(3)-V(2))/Z23);
Q12 = imag(V(1))*conj((V(1)-V(2))/Z12);
Q13 = imag(V(1))*conj((V(1)-V(3))/Z13);
Q21 = imag(V(2))*conj((V(2)-V(1))/Z12);
Q23 = imag(V(2))*conj((V(2)-V(3))/Z23);
Q31 = imag(V(3))*conj((V(3)-V(1))/Z13);
Q32 = imag(V(3))*conj((V(3)-V(2))/Z23);
P1 = real(S(1));
Q1 = imag(S(1));
P2 = real(S(2));
Q2 = imag(S(2));
P3 = real(S(3));
Q3 = imag(S(3));
%%%%%%%%%%-----------Display the results -----------%%%%%%%%%%%%%%%%
fprintf(' %s %2d %s ', 'Iter ',Iter, ' Mismatch 1 Mismatch 2 Mismatch 3 ')
fprintf(' %7.4f %7.4f %7.4f ', Mismatch(1),Mismatch(2), Mismatch(3) );
if max(abs(Mismatch)> Tolerance)
ConvFlag = 1;
else
ConvFlag = 0;
end
end
J;
%radians into degrees
ANG2DEG=ANG2*180/pi
ANG3DEG=ANG3*180/pi
V3MAG
another method to calculate power
clc;
%this program calculate power flow for:
% n :number of buses
% Ne :number of elements
%using gauss sidel methode with acceleration factor (alpha)
%data of buses and elements is read from 'data.xlsx' excel sheat.
%add the folder path
%exampe: addpath('F:th yearst termpower systems');
%%
Edata=xlsread('MA_PowerFlow.xlsx',1);%reading elements data
Ndata=xlsread('MA_PowerFlow.xlsx',2);%reading nodes power data
%%Ne=size(Edata,2)%no. of elements
%Ne=size(Edata,1);%no. of elements
Ne=max(Edata(:,1));%no. of elements
%n=size(Ndata,1);%no of nodes
Nmax=max(Edata(:,2:3)).';%the max of 1st and 2nd coulomn
n=max(Nmax);%no of nodes
Rs=Edata(:,4);
XLs=Edata(:,5);
Ysh=[Edata(:,1),Edata(:,6)];
%%
%**** declaring Y variabbles***%
Ybus_simple=zeros(n);
Ybus_shunt=zeros(n);
Ybus_mutual=zeros(n);
Ybus_traffratio=zeros(n);
Zpriv=Rs+1j*XLs;%Zprimitive diagonal as vector (no traff ratio+no mutual)
Yprid=diag(1./Zpriv);
%Zprid=diag(Zpriv)
%Yprid=spfun(inv,Zprid);%Apply function to nonzero sparse matrix elements
%%
%check TR exsitance to avoid errors
for k=1:length(Edata(:,7))
if isnan(Edata(k,7))
Edata(k,7)= 1;
end
end
TRpart=[Edata(:,1:3),Edata(:,7)];
for k=(1:Ne)
if TRpart(k,1)==1
else
Yprid(k,k)=Yprid(k,k)/TRpart(k,4);
Ybus_shunt(TRpart(k,2),TRpart(k,2))=Ybus_shunt(TRpart(k,2),TRpart(k,2))+(Yprid(k,k)/TRpart(k,4))*(1/TRpart(k,4)-1);
Ybus_shunt(TRpart(k,3),TRpart(k,3))=Ybus_shunt(TRpart(k,3),TRpart(k,3))+Yprid(k,k)*(-1/TRpart(k,4)+1);
end
end
%%
for k=(1:Ne)
Enodes=Edata(k,2:3);
BB=zeros(n);%building block matrix without [1,-1;-1,1]
BB(Enodes(1,1),Enodes(1,1))=Yprid(k,k);
BB(Enodes(1,1),Enodes(1,2))=-Yprid(k,k);
BB(Enodes(1,2),Enodes(1,2))=Yprid(k,k);
BB(Enodes(1,2),Enodes(1,1))=-Yprid(k,k);
Ybus_simple=Ybus_simple+BB;
end
%%
for c=(1:Ne);
nodes=Edata(c,2:3);
BB=zeros(n);
BB(nodes(1,1),nodes(1,1))=BB(nodes(1,1),nodes(1,1))+(Edata(c,6)/2);
BB(nodes(1,2),nodes(1,2))=BB(nodes(1,2),nodes(1,2))+(Edata(c,6)/2);
Ybus_shunt=Ybus_shunt+BB;
end
%%
%check M exsitance to avoid errors
for k=1:length(Edata(:,8))
if isnan(Edata(k,8))
Edata(k,8)= 0;
end
end
Mpart=Edata(:,8:10);
for k=(1:Ne)
if Mpart(k,1)==0
else
[mm, nn]=Edata(Mpart(k,2),2:3);
[pp, qq]=Edata(Mpart(k,3),2:3);
Ybus_mutual(mm,pp)= Mpart(k,1);
Ybus_mutual(nn,pp)=-Mpart(k,1);
Ybus_mutual(mm,qq)=-Mpart(k,1);
Ybus_mutual(nn,qq)= Mpart(k,1);
end
end
%%
Ybus=Ybus_simple+Ybus_shunt+Ybus_traffratio+Ybus_mutual;
Zbus=inv(Ybus);
%clearing dummy variables
clearvars Nmax Rs XLs Zpriv BB mm nn pp qq Yprid Ybus_simple Ybus_shunt Ybus_mutual Ybus_traffratio c k TRpart Mpart
%%
%***calculation of bus voltage and power***%
error=ones(n,3);%error difference between old&new
e_ind=ones(3*n,1);%error finish indicator
e=1;%while condition
type=Ndata(:,8);
Pdelta=Ndata(:,4)-Ndata(:,6);
Qdelta=Ndata(:,5)-Ndata(:,7);
Vbus=Ndata(:,2).*exp(1j*Ndata(:,3));
Qgen=Ndata(:,5);
Qdel=Ndata(:,7);
%defining Qmax and Qmin for the bus if there are no limits of them.
for k=1:length(Ndata(:,1))
if isnan(Ndata(k,9))
Ndata(k,9)= inf;
end
end
for k=1:length(Ndata(:,1))
if isnan(Ndata(k,10))
Ndata(k,10)= -inf;
end
end
Qmax=Ndata(:,9);
Qmin=Ndata(:,10);
disp('enter the acceleration factor (1~2),for no acceleration enter 0');
a=input('');
%%
%itterations ... 3:)
if n==size(Ndata,1); %check of data rows=nodes
while e~=0
for L=1:n;
for c=1:3;
if error(L,c)<1e-5;
e_ind((c-1)*n+L)=0;%here we put all errors in one vector
end
end
end
for t=1:n
switch type(t)
case 1%slack bus
error(t,:)=[0 0 0];
case 2%load bus
YVsum=0;
for k=1:n
YVsum=YVsum+Ybus(t,k)*abs(Vbus(k));%sum (V*Y)=Ibus
end
YVsum=YVsum-Ybus(t,t)*abs(Vbus(t));
Vbus_old=Vbus(t);
Vbus(t)=(((Pdelta(t)-1j*Qdelta(t))/abs(Vbus(t)))+YVsum)/Ybus(t,t);
%Vbus(t)=Vbus_old+a*(Vbus(t)-Vbus_old);
error(t,:)=[abs(Vbus(t))-abs(Vbus_old) 0 0];%angle(Vbus(t))-angle(Vbus_old) 0];
case 3
YVsum=0;
for k=1:n
YVsum=YVsum+Ybus(t,k)*abs(Vbus(k));
end
Qdelta_old= Qdelta(t);
Qdelta(t)=-imag(conj(Vbus(t))*YVsum);
Qgen(t)=Qdelta(t)+Qdel(t);
if Qgen(t)>Qmax(t)
Qdelta(t)=Qmax(t)-Qdel(t);
YVsum=YVsum-Ybus(t,t)*abs(Vbus(t));
Vbus_old=Vbus(t);
Vbus(t)=(((Pdelta(t)-1j*Qdelta(t))/abs(Vbus(t)))+YVsum)/Ybus(t,t);
%Vbus(t)=Vbus_old+a(Vbus(t)-Vbus_old);
error(t,:)=[abs(Vbus(t))-abs(Vbus_old) 0 0]; %angle(Vbus(t))-angle(Vbus_old) 0];
elseif Qgen<Qmin(t)
Qdelta(t)=Qmin(t)-Qdel(t);
YVsum=YVsum-Ybus(t,t)*abs(Vbus(t));
Vbus_old=Vbus(t);
Vbus(t)=(((Pdelta(t)-1j*Qdelta(t))/abs(Vbus(t)))+YVsum)/Ybus(t,t);
%Vbus(t)=Vbus_old+a(Vbus(t)-Vbus_old);
error(t,:)=[abs(Vbus(t))-abs(Vbus_old) 0 0];%angle(Vbus(t))-angle(Vbus_old) 0];
else
Qgen(t)=Qdelta(t)+Qdel(t);
error(t,:)=[Qdelta(t)-Qdelta_old 0 0];
end
otherwise
disp('error of bus type (bus no.):');
disp(t);
end
end
e=sum(e_ind);
end
disp (Vbus);
else disp ('number of buses does not match nodes input');
end
clearvars e e_ind c t k type a L Qdelta_old Vbus_old YVsum nodes Enodes
%%
%symmetric fault calculations
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