how is it possible to solve below matlab codes problem?

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

Answer 1

[David Hill]

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

[Mamad Mamadi]

Thank you very much for replying, sorry for my bad english ;(

i mean;

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));
if Pch & Pdis <=0
b=PLoad(i,1) - Ppv_N(i,1)

if the results of Pch and Pdis will become;

Pch= [10 15 20 0 50 20 30 0 0];

Pdis=[11 18 15 0 20 35 15 0 0];

then use this ''b=PLoad(i,1) - Ppv_N(i,1)'' only when at time 4 ,8,9 that the result for both Pch and Pdis are 0.

It might be anytime between 1:8784 for example at time 500 or 5000 wherever Pch and Pdis become 0 at the same time, system must run ''b=PLoad(i,1) - Ppv_N(i,1)' .

[David Hill]

Is that not that what I did?

[Mamad Mamadi]

i ran your codes but didnt work it showed an error.

Attempted to access Pch(1,1); index out of bounds

because size(Pch)=[0,0].

i dont know how to solve this error.