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Panumat
el 14 de Mzo. de 2013
Comentada: imen baklouti
el 19 de Mayo de 2016
This file .m MathWorks It eror Error using ==> zeros Maximum variable size allowed by the program is exceeded.
Error in ==> ASM1 at 6 QX=zeros(1:length(time)); if true % code
clear
dt = 0.5e-3; % day step size days = 200; % day days of simulation time=(0:dt:days); QX=zeros(1:length(time)); FMA=QX;FMR=QX;SND=QX;SNH=QX;SNO=QX;SS=QX;XBA=QX;XBH=QX;XI=QX;XND=QX;XP=QX;XS=QX;
bA = 0.05; % g SS g COD-1 heterotrophic yield
bH = 0.22; % day-1 heterotrophic decay fp = 0.08; % g non-biodegradable/g biomass iXB= 0.086; % g N/g COD in biomass iXP= 0.06; % ka = 0.08; kh = 3; % day-1 hydrolysis constant KS = 20; % mg COD L-1 half reaction constant for COD removal KOA=0.4; % mg O2 L-1 half reaction constant for O2 in anoxic tank KOH=0.2; KNH=1.0; KNO=0.5; KX =0.03; YA =0.24; % g SS g COD-1 autotrophic yield YH =0.67; % g SS g COD-1 heterotrophic yield ng =0.8; nh =0.4; uA =0.8; % day-1 growth of autotrophic bacteria uH =6.0; % day-1 growth of heterotrophic bacteria
Q0 = 30; % L day-1 influent flow rate QX(1) = 0; % L day-1 daily sludge removal SND0=15; % mg N L-1 soluble organic N SNH0=15; % mg N L-1 NH4-N SNO0=1; % mg N L-1 NO3-N SO =2.0; % mg O2 L-1 DO in aeration tank SS0=200; % mg O2 L-1 soluble COD in influent VH=5; % L Volume of aeration tank XI0=3; % mg L-1 inert material concentration in influent XND0=9; % mg N L-1 particulate N in influent XS0=100; % mg COD L-1 particulate COD in influent XTE=0; % mg SS L-1 SS in effluent
% initial values SND(1) = 5; % mg N L-1 soluble organic N in aeration tank SNH(1) = 1; % mg N L-1 NH4-N in aeration tank SNO(1) =20; % mg N L-1 NO3-N in aeration tank SS(1) = 5; % mg COD L-1 soluble COD in aeration tank XBA(1)=800; % mg COD L-1 autotrophic bacteria in aeration tank XBH(1)=8000; % mg COD L-1 autotrophic bacteria in anoxic tank XI(1)=10; % mg L-1 inert material concentration in anoxic tank XND(1)=2; % mg N L-1 particulate N in aeration tank XP(1)=12; % mg N L-1 particulate N in aeration tank XS(1)=60; % mg N L-1 soluble N in aeration tank
TSS(1)=(XBA(1)+XBH(1)+XS(1)+XP(1))/1.2+XI(1); % mg SS L-1 Total current MLSS TS=TSS(1)*VH;
for i=2:days/dt+1
r1 = uH*SS(i-1)/(KS+SS(i-1))*SO/(KOH+SO)*XBH(i-1); r2 = uH*SS(i-1)/(KS+SS(i-1))*KOH/(KOH+SO)*SNO(i-1)/(KNO+SNO(i-1))*ng*XBH(i-1); r3 = uA*SNH(i-1)/(KNH+SNH(i-1))*SO/(KOA+SO)*XBA(i-1); r4 = bH*XBH(i-1); r5 = bA*XBA(i-1); r6 = ka*SND(i-1)*XBH(i-1); r7 = kh*XS(i-1)/XBH(i-1)/(KX+XS(i-1)/XBH(i-1))*(SO/(KOH+SO)+nh*KOH/(KOH+SO)*SNO(i-1)/(KNO+SNO(i-1)))*XBH(i-1); r8 = r7*XND(i-1)/XS(i-1);
XBA(i)=XBA(i-1)+(r3-r5-QX(i-1)*XBA(i-1)/VH)*dt; XBH(i)=XBH(i-1)+(r1+r2-r4-QX(i-1)*XBH(i-1)/VH)*dt; XS(i)=XS(i-1)+((Q0*XS0-QX(i-1)*XS(i-1))/VH+(1-fp)*(r4+r5)-r7)*dt; XI(i)=XI(i-1)+(Q0*XI0-QX(i-1)*XI(i-1))/VH*dt; XP(i)=XP(i-1)+(fp*(r4+r5)-QX(i-1)*XP(i-1)/VH)*dt; SS(i)=SS(i-1)+((Q0*SS0-Q0*SS(i-1))/VH-1/YH*(r1+r2)+r7)*dt; SNH(i)=SNH(i-1)+((Q0*SNH0-Q0*SNH(i-1))/VH-iXB*(r1+r2)-(iXB+1/YA)*r3+r6)*dt; SNO(i)=SNO(i-1)+((Q0*SNO0-Q0*SNO(i-1))/VH-(1-YH)/2.86/YH*r2+r3/YA)*dt; SND(i)=SND(i-1)+((Q0*SND0-Q0*SND(i-1))/VH-r6+r8)*dt; XND(i)=XND(i-1)+((Q0*XND0-Q0*XND(i-1))/VH+(iXB-fp*iXP)*(r4+r5)-r8)*dt;
TSS(i)=(XBA(i)+XBH(i)+XS(i)+XP(i))/1.2+XI(i);
QX(i)=QX(i-1)+dt*0.001*(TSS(i)-8000);
if QX(i) < 0 QX(i)=0; end if QX(i) > 0.05*Q0 QX(i)=0.05*Q0; end
end
FMA=Q0*(SS0+XS0)./(VH*(XBA+XBH)/1.2); % apparent F/M FMR=Q0*(SS0+XS0)./(VH*(XBA+XBH)/1.2); % real F/M NM=Q0*(SNH0+SNO0+SND0+XND0)./(VH*(XBA+XBH)/1.2);
r=XTE/(XBA(i)+XBH(i)+XS(i)+XI(i)*1.2+XP(i))*1.2; XIE(i)=r*XI(i);
count=1; table(count,:) = [time(1) QX(1) FMA(1) FMR(1) SND(1) SNH(1) SNO(1) SS(1) XBA(1) XBH(1) XI(1) XND(1) XP(1) XS(1)]; for k=1:(i-1) if rem(k*dt, 0.5) == 0 count=count+1; table(count,:) = [time(k) QX(k) FMA(k) FMR(k) SND(k) SNH(k) SNO(k) SS(k) XBA(k) XBH(k) XI(k) XND(k) XP(k) XS(k)]; end end
end
1 comentario
imen baklouti
el 19 de Mayo de 2016
it's correct as asm1 model? the initial values maybe are not good!
Respuesta aceptada
Image Analyst
el 14 de Mzo. de 2013
Before this line that caused the error:
QX=zeros(1:length(time))
out this line:
fprintf('numberOfElements = length(time) = %d', length(time));
and tell me what it says in the command window. By the way, do you know how to use the debugger? Then you might look up zeros() in the help and try it this way:
QX=zeros(1,length(time))
so you're getting a 1D array instead of passing it an array and getting a huge multidimensional array.
4 comentarios
Image Analyst
el 14 de Mzo. de 2013
I don't know. It looks like an array divided by a scalar since XBA(i) and so on are scalars. So it's not dividing by a matrix like it said. Try this
denom = (XBA(i)+XBH(i)+XS(i)+XI(i)*1.2+XP(i));
whos denom
Tell me what it says. It should be a 1x1 double - a scalar, not an array.
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