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

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

Answer 1

[Torsten]

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

[Aiden James]

Thank you sir

Answer 2

[Dyuman Joshi]

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

[Aiden James]

Thank you sir