how can i put a limitation rules for below codes?

i have some values of Pch=Pcharging and Pdis=Pdischarging that i mentioned below and i want to write codes for limitation but the codes that i wrote is not working specially for Pch. Could you please and please help me. thanks

i mean;

Pmax= 3600wh
Pmin=0wh

Pch wont be 3750 , must not be bigger than 3600

Pdis wont be -150, must not be smaller than 0

Pch=[2600 2750 2800 2950 3010 3100 3300 3580 3750];
Pdis=[3450 3330 2900 1500 1200 850 300 22 -150];

it these codes right? if not how can i write it?

%%% for Pch:
    a=1;
    b=0;
result_data=[];
while a<=length(Pch);
    b= Pch(a)+b;
    result_data(a+1)=b;
    a=a+1;
    if result_data <=  Pmax ;
        finalresult = result_data;
    elseif result_data > Pmax ;
        break;
    end
    finalresult;   
end
   %%% for Pdis
if Pdis(1,i) >= Pmin;
           Pdis_1=Pdis(1,i);
    elseif Pdis(1,i) < Pmin
     Pdis(Pdis<0)=0;
end 

[Rik]

What should happen with the values that are not allowed? Should they be removed from the vector, or should they be assigned the boundary value?

[Mamad Mamadi]

Because the capacity of the battery is 3600wh then we cannot charge the battery bigger than 3600wh so we dont need the values that bigger than Pmax=3600wh we can remove it.

And also for Pdis is the same action, it cannot be smaller than Pmin=0

Answer 1

[Rik]

With logical indexing it is easy to remove impossible values:

Pmax= 3600;
Pmin=0;

Pch=[2600 2750 2800 2950 3010 3100 3300 3580 3750];
Pdis=[3450 3330 2900 1500 1200 850 300 22 -150];

Pch(Pch>Pmax | Pch<Pmin)=[];
Pdis(Pdis>Pmax | Pdis<Pmin)=[];

[Mamad Mamadi]

hi Rik im so sorry for replying late, would you mind if i ask you your email adress?

[Rik]

A late reply is not a problem. But why would you need my email address? If you have any issue implementing my answer, feel free to comment it here. If you have a new question you can post it and use my profile page to send me a link.

[Mamad Mamadi]

I wanted to send you some matlab codes and describ my problems in private LOL but no matter i will post it here. tnx anyway :)

[Mamad Mamadi]

Hi Rik 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;

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 to solve 'a' as i mentioned below for example ;

when (Pch & Pdis)=0 %at the same time when both Pch and Pdis become zero, then

a=PLoad(i,1) - Ppv_N(i,1)

thanks