How to fix a graph with loop

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sophp
sophp el 31 de En. de 2018
Editada: sophp el 31 de En. de 2018
Sorry, I am new to MATLAB, so bear with me. I am trying to plot a graph which shows me the values of Y_B for T = 600:10:850, where k1, k2 and k3 vary with T. Instead, I am given a blank graph with only 599 to 601 on the x axis. Heres my matlab code
MATLAB code
for i = 1:25
T(i) = 600+10i;
k1(i) = 10^7*exp(-12700/T(i));
k2(i) = 5*10^4*exp(-10800/T(i));
k3(i) = 7*10^7*exp(-15000/T(i));
t = 0.2;
Y_B(i) = (k1(i)*t*(k1(i)+k3(i)))/(((k2(i)*t)+1)*(k1(i)+k3(i))*(1+(t*(k1(i)+k3(i)))));
end
plot(T(i),Y_B(i));

Respuesta aceptada

Birdman
Birdman el 31 de En. de 2018
You do not need for loop:
T=600:10:850;
t = 0.2;
k1 = 10^7.*exp(-12700./T);
k2 = 5*10^4.*exp(-10800./T);
k3 = 7*10^7.*exp(-15000./T);
Y_B = (k1.*t.*(k1+k3))./(((k2.*t)+1).*(k1+k3).*(1+(t.*(k1+k3))));
plot(T,Y_B);
  7 comentarios
Jan
Jan el 31 de En. de 2018
@sophp: Remember that [] is the operator for a vertical or horizontal concatenation. While 1:2:20 is a vector, concatenating it with nothing by [1:2:20] replies a vector again. Therefore both produce exactly the same object.
It is not easy to suggest an improvement, because I cannot imagine why you want to assign a vector to the scalar t(i). Most of all the vector is static and does not depend on the loop, so why assigning it in each iteration? What about:
t = 1:2:20;
T = 600:10:850;
for i=1:numel(T)
k1(i) = 1e7 .* exp(-12700 ./ T(i));
k2(i) = 5e4 .* exp(-10800 ./ T(i));
k3(i) = 7e7 .* exp(-15000 ./ T(i));
Y_B = (k1(i) .* t .* (k1(i)+k3(i))) ./ (((k2(i).*t)+1) .* ...
(k1(i)+k3(i)) .* (1+(t(i) .* (k1(i)+k3(i)))));
plot(t, Y_B);
end
I'm not sure if this is what you want. But it is very similar to Bordman's code, which you have accepted already.
sophp
sophp el 31 de En. de 2018
Editada: sophp el 31 de En. de 2018
From this I obtain the attached graph, indicating that T has no effect. I expect there to be different operating curves of t vs Y_B as different values of T are used

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