How to perform iterations for lopp

I have imported N and V values from excel, only these two variables are varying. Now i have to write the code for Qo=1000,i have to find N & J values then Qo=2500 find N& J values upto 10000.For N & V i gave 23 inputs .so for Qo=1000 i have to get 23 different N & J values, but here iam getting the values for only one Qo value

Qo=1000:1500:10000; 
nX =length(No);
q = length(Qo);
q1 = q*nX;
Solutions= cell(q1, 2);  
for i=1:q
    for S=1:nX
        X =((bt*Qo(i)*(1+R)*Xo)+(Y*1000*a*V(S)*J*bs))/(bt*((Qo(i)*(1+R))-((V(S)*Y*k*N)/((Kn*1000)+N))+(V(S)*kd)));
        eqn1=(Qo(i)*(1+R)*(No(S)-N)-(V(S)*((a*J*1000)+((X*k*N)/((Kn*1000)+N)))))==0;
        eqn2=(sqrt((Lstar^1.76)+(5.2*(Nstar-Nstar_min)))-(Lstar^0.88))/2.6==jstar^0.88;
        Solutions{S} = vpasolve([eqn1,eqn2],[N,J]);
    end

[Walter Roberson]

you have a double loop but your output is only indexed by one of the two variables

[sunitha]

U r right..in my code only last value of 'Qo' that is 10,000 is considered for the calculation. Can u adjust the code so that I get iterations for all Qo values

Answer 1

[Walter Roberson]

syms N J
Qo=1000:1500:10000; 
nX =length(No);
q = length(Qo);
Solutions= cell(nX, q);  
for i=1:q
    for S=1:nX
        X =((bt*Qo(i)*(1+R)*Xo)+(Y*1000*a*V(S)*J*bs))/(bt*((Qo(i)*(1+R))-((V(S)*Y*k*N)/((Kn*1000)+N))+(V(S)*kd)));
        eqn1=(Qo(i)*(1+R)*(No(S)-N)-(V(S)*((a*J*1000)+((X*k*N)/((Kn*1000)+N)))))==0;
        eqn2=(sqrt((Lstar^1.76)+(5.2*(Nstar-Nstar_min)))-(Lstar^0.88))/2.6==jstar^0.88;
        Solutions{S, i} = vpasolve([eqn1,eqn2],[N,J]);
    end
end

[sunitha]

Thank you sir

Answer 2

[sunitha]

Hello sir,

I am facing the problem in this code.At first it will take No value =400 it will go to loop perform iterations and after again it has to take 500 value. i am getting zero values for second No values.

No=400:100:600;  n=3;  N=4;
 aX=length(No); 
 A=cell(1,aX);
 for i=1:aX
          for j=1:1:3    %%%%% no of reactors 
            Ne(j,i)=No(j,i)*(N/No(j,i))^(j./n);
    No(j+1,i+1)=Ne(j,i);
          end 
    end ;

[Walter Roberson]

You never store into No(2,1) .

You initialize No as a 1 x 3 vector. Then for loops, i = 1, j = 1, you store into No(j+1,i+1) -> No(2,2) . So you never store into column 1 of No. But you use column 1 for all of i = 1, so you end up dividing by 0.

[sunitha]

I am getting the error that index in position 1 exceeds array and I am getting only one value

[Walter Roberson]

At what point in your code do you write to location No(2,1) ?

Answer 3

[sunitha]

fFOR

I have to do 3 iterations,I st iteration it has to take No=400,calculate J1 value and goes to 2nd iteration

for i=1:aX
  N1=No(i);
          for j=1:3    %%%%% no of reactors 
            Ne(i,j)=N1(j)*(N/N1(j))^(j./n);
    N1(j+1)=Ne(i,j);
Ans(i,1)=No(i);
Ans(i,j+1)=Ne(i,j);
eqn1=(sqrt((Lstar^1.76)+(5.2*(Nstar-Nstar_min)))-(Lstar^0.88))/2.6==jstar^0.88;
soln=vpasolve([eqn1],[J]);
J1=double(soln);
          end 
end 

[Walter Roberson]

J1(i,j) = double(soln);

However, notice that

eqn1=(sqrt((Lstar^1.76)+(5.2*(Nstar-Nstar_min)))-(Lstar^0.88))/2.6==jstar^0.88;

does not use Ne or N1 or No or Ans (the variables you assign to in the loops) so you are going to be calculating exactly the same thing every iteration.

Your original code with two equations in N and J had the same problem. Also, there was no obvious connection between the variables such as Lstar and the variables N or J, so the fragments posted before give the impression that the equation is independent of the variable, J, being solved for.

Answer 4

[sunitha]

i am using calculated N value in finding J variable and V value . Nstar is a function of N. Here i have a doubt i want to give No values intitial No ,after every iteration whatever i got the value that has to be taken into No. How can i No values here in finding V??

for i=1:n

Ne(j,i)=N1(i)*(N/N1(i))^(i./n);

Se(j,i)=transpose(Ne(j,i));

N1(i+1)=Ne(j,i);

Nstar(j,i)=Se(j,i)/(1000*Kn)

jstar=((sqrt((Lstar^1.76)+(5.2*(Nstar(j,i)-Nstar_min)))-(Lstar^0.88))/2.6)^1.136;

J(j,i)=jstar*(sqrt(Kn*k*Xf*Df));

Ans(j,1)=No(j);

Ans(j,i+1)=Ne(j,i);

J1(j,i)=transpose(J(j,i));

So=transpose(No); %%This is the input values in finding the V value

A=(a*J1(j,i)*1000)*((bt*kd*((Kn*1000)+Se(j,i)))-(Y*k*Se(j,i)*kd));

B=bt*Qo*(1+R)*((k*Se(j,i))*(Y*(No-Se(j,i))+ Xo)+(((Kn*1000)+Se(i,j))*((a*J1(j,i)*1000)-(kd*(No(i,j)-Se(i,j))))));

C=(Qo^2)*((1+R)^2)*(Se(i,j)-No(i,j))*((Kn*1000)+Se(i,j))*bt;

V(j,i)=(sqrt((B.^2)-(4*A*C))-B)/(2*A);

end

end

[Walter Roberson]

Why are you transpose() of a scalar? If you need to take complex conjugate then use conj() instead of transpose() or use ctranspose()

[Walter Roberson]

So=transpose(No);   %%This is the input values in finding the V value

No it is not. You never use So after you assign to it.

[sunitha]

Bze I got the result J variable in 1*3 format ..in order to convert to 3*1 format ..I use transpose. All other values it is in 3*1 format that's y I changed to 3*1 to perform operations

[sunitha]

I can change the No to So in the equation. How can I define the No value in the equation

[Walter Roberson]

Ans(j,1)=No(j);

That tells us No is a vector. It was a vector before in previous posts but you changed your code and did not show assigning to it, so we could not assume that it is still a vector in your current code.

               B=bt*Qo*(1+R)*((k*Se(j,i))*(Y*(No-Se(j,i))+ Xo)+(((Kn*1000)+Se(i,j))*((a*J1(j,i)*1000)-(kd*(No(i,j)-Se(i,j))))));

The first access to No is not indexed so it uses the whole vector and so B will be a vector of results.

The second reference to No is indexed with two variables, implying that No is two dimensional instead of a vector, which is likely a problem compared to using it as a vector for Ans.

V(j,i)=(sqrt((B.^2)-(4*A*C))-B)/(2*A);

B is a vector or possibly an array so the right hand side is vector or array but you are storing it in a scalar location.

Answer 5

[sunitha]

Hello sir,

when running the program i am getting the error and i couldnt understand the error .can u please tell how to rectify this error??

[Walter Roberson]

eqn1=(jstar*Lstar)+((0.5*jstar^2)+(jstar((1+((jstar/3.4)^1.19)^-0.61))))/tanh(jstar/Nstar_min)==Nstar;
                                   ^^^^^^

In MATLAB, NAME followed by ( can mean one of several things:

1. NAME is a variable that has been assigned a function handle, and the function is to be invoked passing in whatever list is inside the (); or
2. NAME is not a variable, but is the name of a function that is "visible" in the scope, and the function is to be invoked passing in whatever list is inside the (); or
3. NAME is not a variable, but is the name of a method for the dominent object class encountered inside the () and the class method is to be called for the object class; or
4. NAME is not a variable, but is the name of a function that the user expects to be in scope, but is not, and the call is in error
5. NAME is a variable and what follows inside the () is a list of indices that are comprised of positive integers, or logical values, or character vectors (but not string objects), which are to act to index the variable; or
6. NAME is a variable and what follows in the () includes things that are not positive integers or logical values or character vectors, and the call is an invalid indexing attempt

Your jstor is not the name of a function or a variable designating a function handle: it is the name of a scalar expression, so jstor can only be followed immediately by () for indexing. However, (1+((jstar/3.4)^1.19)^-0.61) is not a positive integer, so the indexing attempt fails.

There are absolutely no circumstances in MATLAB in which NAME followed immediately by () is ever treated as implied multiplication. Never. If you intend multiplication you must always use .* (element-by-element multiplication) or * (algebraic matrix multiplication -- inner product)