how is it possible to solve below matlab codes problem?
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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
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Answers (1)
David Hill
on 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
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