Why can't i plot a graph for a against t?
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k_n= 7*10^-11;
n= 2;
i=1;
t(i)=0; %in hours
a = sym('a')
while t<=2.5 %2.5hours
a= 1-exp(-k_n*(t(i))^n); %degree of hydration
dt=1/3600; %1sec interval
t(i+1)=t(i)+dt;
i=i+1;
end
figure(1);
plot(t,a,'r'),xlabel('t [h]'),ylabel('degree of hydration');
hold on;
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Accepted Answer
madhan ravi
on 15 Mar 2019
Edited: madhan ravi
on 16 Mar 2019
k_n= 7*10^-11;
n= 2;
dt=1/3600; %1sec interval
t=0:dt:2.5;
a= 1-exp(-k_n*(t).^n); %degree of hydration
t=0:dt:2.5;
figure(1);
plot(t,a,'r'),
xlabel('t [h]')
ylabel('degree of hydration')
2 Comments
madhan ravi
on 16 Mar 2019
Edited: madhan ravi
on 16 Mar 2019
Refer links:
Also do a course in Matlab website called MATLAB onramp which takes just few hours to complete and is free.
Ok see which was the mistake you made with a small example:
for k=1:5
a=k; % you kept overwriting the variable in each iteration
end
% how it should be done:
a=zeros(1,5); % pre-allocate
for k=1:5
a(k)=k;
% ^^^- this is how you save values in each iteration
end
Note: Your task can be done trivially without loop. Vectorization method is preferrable than a loop.
More Answers (1)
Rik
on 15 Mar 2019
Here are some variations you could try:
k_n= 7*10^-11;
n= 2;
i=1;
dt=1/3600; %1sec interval
t=zeros(1,floor(2.5/dt)); %in hours
a=zeros(size(t));
while t(i)<=2.5 %2.5hours
a(i)= 1-exp(-k_n*(t(i))^n); %degree of hydration
t(i+1)=t(i)+dt;
i=i+1;
end
t(i:end)=[];a(i:end)=[];%remove unneeded parts of preallocated space
figure(1);
plot(t,a,'r'),xlabel('t [h]'),ylabel('degree of hydration');
%or simpler:
k_n= 7*10^-11;
n= 2;
fun=@(t) 1-exp(-k_n.*t.^n);
figure(2)
subplot(1,2,1)
fplot(fun,[0,2.5])
%or explicit:
subplot(1,2,2)
t=0:1/3600:2.5;
a=fun(t);
plot(t,a)
2 Comments
Rik
on 15 Mar 2019
You were overwriting a on every iteration instead of forming a vector as well. You can compare my code with yours to see the changes I've made.
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