DESCAPACHO ECONOMICO
1.
DESPACHO ECONOMICO DESPRECIANDO PERDIDAS Y SIN CONSIDERAR LIMITES DEL GENERADOR EJERCICIO 1
C1=210+6.3P1+0.007P1^2 C2=200+6.7P2+0.008P2^2 C3=190+6.9P3+0.008P3^2 PD=200
ALGORITMO clear all format short alpha=[200;210;250];beta=[7.0;7.1;7.2];gama=[0.004;0.005;0.006]; lambda=input('ingrese lambda=input('ingrese lambda='); lambda='); PD=200; DP=3;%iniciamos DP=3;%iniciamos con un error igual 3
iter=0; while abs(DP)>=0.0001 while abs(DP)>=0.0001 iter=iter+1; P=(lambda-beta)./(2*gama); DP=PD-sum(P); J=sum(ones(length(gama),1)./(2*gama)); %Jacobiano Dlambda=DP/J; lambda=lambda+Dlambda; disp(['iter disp(['iter lambda Dlambda DP P1 P2 P3']) P3']) disp([iter,lambda,Dlambda,DP, P(1), P(2), P(3)]) end costo_total=sum(alpha+beta.*P+gama.*P.^2)
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RESULTADOS
ingrese lambda=8 iter lambda Dlambda DP P1 P2 P3 1.0000
7.7351 -0.2649 -81.6667 125.0000 90.0000 66.6667
iter lambda Dlambda DP P1 P2 P3 2.0000 7.7351 0.0000 0.0000 91.8919 63.5135 44.5946 costo_total = 2.1411e+003
2.
DESPACHO ECONOMICO DESPRECIANDO PERDIDAS Y CONSIDERANDO LIMITES DEL GENERADOR
EJERCICIO 1 C1=200+7.0P1+0.004P1^2 C2=210+7.1P2+0.005P2^2 C3=250+7.2P3+0.006P3^2 Limites de los generadores
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ALGORITMO clear all PD=500; %1ra iteracion iter=1; lan=6; P1_1=(lan-7)/(2*0.004); P2_1=(lan-7.1)/(2*0.005); P3_1=(lan-7.2)/(2*0.006); DP_1=PD-(P1_1+P2_1+P3_1); Dlan_1=DP_1/(1/(2*0.004)+1/(2*0.005)+1/(2*0.006)); %2da iteracion iter=iter+1 lan=lan+Dlan_1 P1_2=(lan-7)/(2*0.004) P2_2=(lan-7.1)/(2*0.005) P3_2=(lan-7.2)/(2*0.006) DP_2=PD-(P1_2+P2_2+P3_2) Dlan_2=DP_2/(1/(2*0.004)+1/(2*0.005)+1/(2*0.006))
Resultados de la 2da iteración
iter = 2 lan = 8.7081 P1_2 = 213.5135 P2_2 = 160.8108 P3_2 = 125.6757 DP_2 =
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Cuando obtenemos los Resultados de la 2da iteración el DP tiende a cero pero se observa que el segundo generador excede su límite superior. Entonces esta generadora queda estabilizada en su limite superior P2=150MW. Entonces se sigue iterando con 3ra y 4ta iteración clear all PD=500; %1ra iteracion iter=1; lan=6; P1_1=(lan-7)/(2*0.004); P2_1=(lan-7.1)/(2*0.005); P3_1=(lan-7.2)/(2*0.006); DP_1=PD-(P1_1+P2_1+P3_1); Dlan_1=DP_1/(1/(2*0.004)+1/(2*0.005)+1/(2*0.006)); %2da iteracion iter=iter+1; lan=lan+Dlan_1; P1_2=(lan-7)/(2*0.004); P2_2=(lan-7.1)/(2*0.005); P3_2=(lan-7.2)/(2*0.006); DP_2=PD-(P1_2+P2_2+P3_2); Dlan_2=DP_2/(1/(2*0.004)+1/(2*0.005)+1/(2*0.006)); %3ra iteracion iter=iter+1; P2=150; DP_3=PD-(P1_2+P2+P3_2); Dlan_3=DP_3/(1/(2*0.004)+1/(2*0.006)); %4ra iteracion iter=iter+1; lan=lan+Dlan_3; P1_4=(lan-7)/(2*0.004); P3_4=(lan-7.2)/(2*0.006); DP_4=PD-(P1_4+P2+P3_4);
resultados=([iter,DP_4,lan,P1_4,P2,P3_4]); disp(['iter disp(['iter DP lan P1 P2 P3 ']) ' ]) disp([resultados]) PG=sum(P1_4+P2+P3_4)
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Resultados finales iter 4.0000
DP 0
PG = 500 CT = 4.6085e+003
lan 8.7600
P1 220.0000
P2 150.0000
P3 130.0000
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EJERCICIO 2 C1=200+7.0P1+0.004P1^2 C2=210+7.1P2+0.005P2^2 C3=250+7.2P3+0.006P3^2 Limites de los generadores
PD=500MW SOLUCION OTRA FORMA
Esta vez utilizaremos la función 1er Paso:
fmincon
Creamos 2 archivos “ function”
Archivo 1: funcionobjetivo2 function f=funcionobjetivo2(p) function f=funcionobjetivo2(p) F1=200+7*p(1)+0.004*p(1)^2; F2=210+7.1*p(2)+0.005*p(2)^2; F3=250+7.2*p(3)+0.006*p(3)^2 f=F1+F2+F3 end
Archivo 2: restriccion2 function [c,ceq]=restriccion2(p) function [c,ceq]=restriccion2(p) c=[50-p(1);p(1)-250;100-p(2);p(2)-150;80-p(3);p(3)-150]; pd=500; ceq=[pd-p(1)-p(2)-p(3)]; end
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2do Paso: Creamos 1 archivo
“scrip”
clc; format short p0=[1 1 1];% 1];% punto inicila de interacion options=optimset('algorithm' options=optimset( 'algorithm', ,'active-set' 'active-set'); ); [p,fval]=fmincon(@funcionobjetivo2,p0,[],[],[],[],[],[],@restriccion2,opt ions); disp('solucion:' disp('solucion:') ) P1=p(1) P2=p(2) P3=p(3) PT=P1+P2+P3 Ct=200+7*P1+0.004*P1^2+210+7.1*P2+0.005*P2^2+250+7.2*P3+0.006*P3^2
solucion: P1 = 220.0000
P2 = 150
P3 = 130.0000
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3.
DESPACHO ECONOMICO ECONOMICO CONSIDERANADO PERDIDAS EJERCICIO 4
Las funciones combustible-costo de 3 plantas térmicas en $/h están dadas por:
Las generadoras tienen los siguientes límites.
Las pérdidas de la siguiente expresión
potencia
activa
están
dadas
por
la
Donde los coeficientes de perdidas están expresados en pu sobre una base de 100MVA Determinar el despacho óptimo de generación cuando la carga total del sistema es 150MW
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ALGORITMO %C1=200+7*P1+0.008*P1^2; %C2=180+6.3*P2+0.009*P2^2; %C1=140+6.8*P3+0.007*P3^2; %PL=0.000218*P1^2+0.000228*P2^2+0.000179*P3^2; %P1=10:85; %P2=10:80; %P3=10:70; PD=150; %1ra iteracion iter=1; lan=8; P1=(lan-7)/(2*0.008+2*lan*0.000218); P2=(lan-6.3)/(2*0.009+2*lan*0.000228); P3=(lan-6.8)/(2*0.007+2*lan*0.000179); PL=0.000218*P1^2+0.000228*P2^2+0.000179*P3^2; DP=PD+PL-(P1+P2+P3); J=(0.008+0.000218*7)/(2*(0.008+lan*0.000218)^2)+(0.009+0.000228*6.3)/(2*( 0.009+lan*0.000228)^2)+(0.007+0.000179*6.8)/(2*(0.007+lan*0.000179)^2); Dlan=DP/J; %2ra iteracion iter=iter+1; lan=lan+Dlan; P1=(lan-7)/(2*0.008+2*lan*0.000218); P2=(lan-6.3)/(2*0.009+2*lan*0.000228); P3=(lan-6.8)/(2*0.007+2*lan*0.000179); PL=0.000218*P1^2+0.000228*P2^2+0.000179*P3^2; DP=PD+PL-(P1+P2+P3); J=(0.008+0.000218*7)/(2*(0.008+lan*0.000218)^2)+(0.009+0.000228*6.3)/(2*( 0.009+lan*0.000228)^2)+(0.007+0.000179*6.8)/(2*(0.007+lan*0.000179)^2);
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%4ra iteracion iter=iter+1; lan=lan+Dlan; P1=(lan-7)/(2*0.008+2*lan*0.000218); P2=(lan-6.3)/(2*0.009+2*lan*0.000228); P3=(lan-6.8)/(2*0.007+2*lan*0.000179); PL=0.000218*P1^2+0.000228*P2^2+0.000179*P3^2; DP=PD+PL-(P1+P2+P3); J=(0.008+0.000218*7)/(2*(0.008+lan*0.000218)^2)+(0.009+0.000228*6.3)/(2*( 0.009+lan*0.000228)^2)+(0.007+0.000179*6.8)/(2*(0.007+lan*0.000179)^2); Dlan=DP/J; resultados=([iter,DP,lan,P1,P2,P3]); disp(['iter disp(['iter DP lan P1 P2 P3 ']) ' ]) disp([resultados])
CT=200+7*P1+0.008*P1^2+180+6.3*P2+0.009*P2^2+140+6.8*P3+0.007*P3^2
RESULTADOS
iter 4.0000
DP -0.0004
CT = 1.5927e+003
lan 7.6789
P1 35.0908
P2 64.1319
P3 52.4768
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SEGUNDA FORMA ALGORITMO clear all all;clc ;clc format short alpha=[200;180;140]; beta=[7;6.3;6.8]; gama=[0.008;0.009;0.007]; B=[0.000218;0.000228;0.000119]; PD=150; a=[10 85]; b=[10 80]; c=[10 70]; lambda=8; DP=20; iter=0; while abs(DP)>=0.0001 while abs(DP)>=0.0001 iter=iter+1; P=(lambda-beta)./(2*(gama+lambda*B)); if (a(1)<=P(1)<=a(2)&&b(1)<=P(2)<=b(2)&&P(3)<=c(1)) if P(3)=c(1); elseif elseif (a(1)<=P(1)<=a(2)&&b(1)<=P(2)<=b(2)&&P(3)>=c(2)) P(3)=c(2); elseif elseif (a(1)<=P(1)<=a(2)&&c(1)<=P(3)<=c(2)&&P(2)<=b(1)) P(2)=b(1); elseif elseif (a(1)<=P(1)<=a(2)&&c(1)<=P(3)<=c(2)&&P(2)>=b(2)) P(2)=b(2); elseif elseif (b(1)<=P(2)<=b(2)&&c(1)<=P(3)<=c(2)&&P(1)<=a(1)) P(1)=a(1); elseif elseif (b(1)<=P(2)<=b(2)&&c(1)<=P(3)<=c(2)&&P(1)>=a(2)) P(1)=a(2); end
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RESULTADOS
Iter 1.0000
lambda DP 7.6997 -46.9857
P1 51.3136
P2 78.5292
P3 70.0000
Iter 2.0000
lambda 7.6604
DP -6.2109
P1 36.1461
P2 65.0680
P3 56.8249
Iter 3.0000
lambda 7.6596
DP -0.1265
P1 34.1489
P2 63.2963
P3 54.3783
Iter 4.0000
lambda 7.6596
DP -0.0026
P1 34.1082
P2 63.2602
P3 54.3285
Iter 5.0000
lambda 7.6596
DP -0.0001
P1 34.1074
P2 63.2595
P3 54.3275
CT =