calling a callback function from script

4 visualizzazioni (ultimi 30 giorni)
hello everyone,
I am trying to calculate the steam table data using the XSteam.m script. I have written a callback loop to calculate the density of the steam at perticular temperature and pressure values. the function is as follows.
where a is the final pressure
When i am trying to run the script it gives an error as
Error using zeros
Maximum variable size allowed by the program is exceeded.
Error in Untitled2 (line 5)
y = zeros(1:((a-1)/b));
Please check the script and help me with it.
With regards.
a = 10;
b = 0.01;
T = 200;
y = zeros(1:((a-1)/b));
L = length(y);
for i = 0:L
p(i)= 1+b*i;
y(i) =XSteam('rho_pT',i,T);
end
  1 Commento
Stephen23
Stephen23 il 7 Feb 2020
This code
zeros(1:((a-1)/b))
defines an array with size
zeros(1,2,3,4,5,6,7,8,9,10,11,...,900)
which would require so much memory that it can't even be calculated using double numeric class. Impressive, but unlikely to be very useful. Perhaps you meant:
zeros(1,((a-1)/b));
% ^ comma

Accedi per commentare.

Risposta accettata

JESUS DAVID ARIZA ROYETH
JESUS DAVID ARIZA ROYETH il 7 Feb 2020
Modificato: JESUS DAVID ARIZA ROYETH il 7 Feb 2020
is y = zeros(1,((a-1)/b)); you have ":" instead of a ","
a = 10;
b = 0.01;
T = 200;
y = zeros(1,((a-1)/b));
L = length(y);
for i = 1:L
p(i)= 1+b*i;
y(i) =XSteam('rho_pT',i,T);
end
  3 Commenti
Steven Lord
Steven Lord il 7 Feb 2020
There's no such thing as element 0 in an array in MATLAB. The first element of p is p(1) not p(0).
In this case there's no need to create p inside the for loop.
p = 1 + b*(0:L);
I don't know whether your XSteam function can accept a vector of values as its second input. If it can (and returns a vector the same size as that second input, where each element of the output is the result of operating on the corresponding element of that second input) you could eliminate the loop altogether.
y = XSteam('rho_pT', 0:L, T);
vishnuvardhan naidu tanga
no, the function can not give an vector output. It only gives scalar outputs, and when I try this
p = 1 + b*(0:L);
all the values are turn to zero and i have used p(1).

Accedi per commentare.

Più risposte (0)

Categorie

Scopri di più su MATLAB in Help Center e File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by