how is it possible to solve below matlab codes problem?
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Hi everybody could you please help me in these codes?
i=1:8784 %one year (hours)
if Ppv_N(i,1) > PLoad(i,1);
Pch(1,i)=(Pbat*(1-sigma))+(Ppv_N(i,1)-(PLoad(i,1))/eta_i)*eta_b;
a=Ppv_N(i,1) - PLoad(i,1)
end
else Ppv_N(i,1) < PLoad(i,1)
Pdis(1,i)=(Pbat*(1-sigma))-((PLoad(i,1)/eta_i)-Ppv_N(i,1));
now after finding Pch and Pdis from 1 to 8784 hours, i want to write another code or loop to solve 'b' between 1:8784 , for example ;
when (Pch & Pdis)=0 %at the same time when both Pch and Pdis become zero, then
b=PLoad(i,1) - Ppv_N(i,1)
- which i mean when Pch and Pdis at the same time became zero then use b=PLoad(i,1) - Ppv_N(i,1)
thanks
0 commentaires
Réponses (1)
David Hill
le 14 Jan 2020
I was somewhat confused with your question. Hopefully, this helps:
for i=1:8784 %one year (hours)
if Ppv_N(i,1) > PLoad(i,1);
Pch(1,i)=(Pbat*(1-sigma))+(Ppv_N(i,1)-(PLoad(i,1))/eta_i)*eta_b;
a=Ppv_N(i,1) - PLoad(i,1);
elseif Ppv_N(i,1) < PLoad(i,1)
Pdis(1,i)=(Pbat*(1-sigma))-((PLoad(i,1)/eta_i)-Ppv_N(i,1));
end
end
b=[];
for i=1:8784
if Pch(1,i)==Pdis(1,i)
b=[b,PLoad(i,1) - Ppv_N(i,1)];%not sure if you will have more than one occurrance
end
end
3 commentaires
Voir également
Catégories
En savoir plus sur Loops and Conditional Statements dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!