calling a callback function from script

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vishnuvardhan naidu tanga
vishnuvardhan naidu tanga 2020 年 2 月 7 日
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 件のコメント
Stephen23
Stephen23 2020 年 2 月 7 日
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

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JESUS DAVID ARIZA ROYETH
JESUS DAVID ARIZA ROYETH 2020 年 2 月 7 日
編集済み: JESUS DAVID ARIZA ROYETH 2020 年 2 月 7 日
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 件のコメント
Steven Lord
Steven Lord 2020 年 2 月 7 日
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
vishnuvardhan naidu tanga 2020 年 2 月 7 日
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).

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