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How I can run following code for different p and t values?

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function [x] = LiBrH2O(t,p)
% Function LiBrH2O calculates the concentration for the corresponding
% Lithium Bromide - Water solution temperature and Saturation Pressure,
% within the range as per as curve-fitting equation.
%
% INPUTS:
% p = Saturation Pressure in kPa
% t = Lithium Bromide - Water solution temperature in degree-Celsius
% OUTPUT:
% x = concentration of Lithium Bromide - Water solution in kJ/kg
t=[82 83 84 85 86 87 88 89]
p=[8 7 6 5 4 3 6 8]
if (t>5 && t<175) %---Checks temperature range
%---Constants
A0 = -2.00755; A1 = 0.16976; A2 = -3.133362*10^(-3); A3 = 1.97668*10^(-5);
B0 = 124.937; B1 = -7.71649; B2 = 0.152286; B3 = -7.95090*10^(-4);
C = 7.05; D = -1596.49; E = -104095.5;
%---Refrigerant Temperature Calculation from Saturation Pressure
T = -2*E/(D + (D*D - 4*E*(C-log10(p)))^(0.5)); %---Temp in Kelvin
t1 = T-273; %---Temp in Degree-Celsius
if (t1>-15 && t1<110)
%---Creating polynomial coefficient
P0 = B0 + t1*A0;
P1 = B1 + t1*A1;
P2 = B2 + t1*A2;
P3 = B3 + t1*A3;
%---Calculation for Concentration
P = [P3 P2 P1 (P0 - t)];
X = roots(P);
[r c] = size(X);
for l=1:r
if (X(l,c)>45 && X(l,c)<70)%---Checks Concentration range
x = X(l,c)
end
end
else
disp(t1);
disp('Error: Refrigerant temperature out of range: t1>-15 && t1<110');
end
else
disp(t);
disp('Error: Solution temperature out of range: t>5 && t<175');
end

Accepted Answer

Torsten
Torsten on 23 Sep 2023
t=[82 83 84 85 86 87 88 89];
p=[8 7 6 5 4 3 6 8];
x = arrayfun(@(t,p)LiBrH2O(t,p),t,p);
x
x = 1×8
58.2131 59.9810 61.8768 63.9662 66.3658 69.3153 63.5742 61.3198
function [x] = LiBrH2O(t,p)
% Function LiBrH2O calculates the concentration for the corresponding
% Lithium Bromide - Water solution temperature and Saturation Pressure,
% within the range as per as curve-fitting equation.
%
% INPUTS:
% p = Saturation Pressure in kPa
% t = Lithium Bromide - Water solution temperature in degree-Celsius
% OUTPUT:
% x = concentration of Lithium Bromide - Water solution in kJ/kg
if (t>5 && t<175) %---Checks temperature range
%---Constants
A0 = -2.00755; A1 = 0.16976; A2 = -3.133362*10^(-3); A3 = 1.97668*10^(-5);
B0 = 124.937; B1 = -7.71649; B2 = 0.152286; B3 = -7.95090*10^(-4);
C = 7.05; D = -1596.49; E = -104095.5;
%---Refrigerant Temperature Calculation from Saturation Pressure
T = -2*E/(D + (D*D - 4*E*(C-log10(p)))^(0.5)); %---Temp in Kelvin
t1 = T-273; %---Temp in Degree-Celsius
if (t1>-15 && t1<110)
%---Creating polynomial coefficient
P0 = B0 + t1*A0;
P1 = B1 + t1*A1;
P2 = B2 + t1*A2;
P3 = B3 + t1*A3;
%---Calculation for Concentration
P = [P3 P2 P1 (P0 - t)];
X = roots(P);
[r c] = size(X);
for l=1:r
if (X(l,c)>45 && X(l,c)<70)%---Checks Concentration range
x = X(l,c);
end
end
else
disp(t1);
disp('Error: Refrigerant temperature out of range: t1>-15 && t1<110');
end
else
disp(t);
disp('Error: Solution temperature out of range: t>5 && t<175');
end
end

More Answers (1)

Dyuman Joshi
Dyuman Joshi on 23 Sep 2023
Edited: Dyuman Joshi on 23 Sep 2023
You can run a for loop through the elements of t and p and get the output. Another option is to use arrayfun, which I have done below.
However, you need to define x for values of other cases - if t is not in the range (5,175) or if t1 is not in the range (-15,110), otherwise the code will give an error.
I have used NaN here, see the added 2 pair of values -
t = [82 83 84 85 86 87 88 89 190 150];
p = [8 7 6 5 4 3 6 8 9 500];
x = arrayfun(@LiBrH2O, t, p)
190 Error: Solution temperature out of range: t>5 && t<175 150.4243 Error: Refrigerant temperature out of range: t1>-15 && t1<110
x = 1×10
58.2131 59.9810 61.8768 63.9662 66.3658 69.3153 63.5742 61.3198 NaN NaN
function [x] = LiBrH2O(t,p)
% Function LiBrH2O calculates the concentration for the corresponding
% Lithium Bromide - Water solution temperature and Saturation Pressure,
% within the range as per as curve-fitting equation.
%
% INPUTS:
% p = Saturation Pressure in kPa
% t = Lithium Bromide - Water solution temperature in degree-Celsius
% OUTPUT:
% x = concentration of Lithium Bromide - Water solution in kJ/kg
x = NaN;
if (t>5 && t<175) %---Checks temperature range
%---Constants
A0 = -2.00755; A1 = 0.16976; A2 = -3.133362*10^(-3); A3 = 1.97668*10^(-5);
B0 = 124.937; B1 = -7.71649; B2 = 0.152286; B3 = -7.95090*10^(-4);
C = 7.05; D = -1596.49; E = -104095.5;
%---Refrigerant Temperature Calculation from Saturation Pressure
T = -2*E/(D + (D*D - 4*E*(C-log10(p)))^(0.5)); %---Temp in Kelvin
t1 = T-273; %---Temp in Degree-Celsius
if (t1>-15 && t1<110)
%---Creating polynomial coefficient
P0 = B0 + t1*A0;
P1 = B1 + t1*A1;
P2 = B2 + t1*A2;
P3 = B3 + t1*A3;
%---Calculation for Concentration
P = [P3 P2 P1 (P0 - t)];
X = roots(P);
[r c] = size(X);
for l=1:r
if (X(l,c)>45 && X(l,c)<70)%---Checks Concentration range
x = X(l,c);
end
end
else
disp(t1);
disp('Error: Refrigerant temperature out of range: t1>-15 && t1<110');
end
else
disp(t);
disp('Error: Solution temperature out of range: t>5 && t<175');
end
end

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