Question

matlab please

TABLE P15.5 Dissolved oxygen concentration in water as a function of temperature (°C) and chloride concentration (O/L). Dissolved Oxygen (mg/L) for Temperature (C) and Concentration of Chloride (g/L) C=Og/L c=10 g/l c=20 g/L 14.6 12.9 11.4 12.8 11.3 10.3 11.3 10.1 8.96 10.1 9.03 8.08 9.09 8.17 7.35 8.26 7.46 6.73 6.85 6.20 1.56 the percent relative error for your prediction. Explain pos sible causes for the discrepancy. 15.7 As compared with the models from Probs. 15.5 and 15.6, a somewhat more sophisticated model that accounts for the effect of both temperature and chloride on dis- solved oxygen saturation can be hypothesized as being of the form o=S;(T) +SCC) That is, a third-order polynomial in temperature and a lin car relationship in chloride is assumed to yield superior re- sults. Use the general linear least-squares approach to fit this model to the data in Table P15.5. Use the resulting equation to estimate the dissolved oxygen concentration for a chloride concentration of 15 g/L at T = 12 °C. Note that the true value is 9.09 mg/L. Compute the percent relative error for your prediction

Answer 1

% temparature

T=[0 5 10 15 20 25 30];

% oxygen concentration for chloride concentration 0 g/L
C0=[14.6 12.8 11.3 10.1 9.09 8.26 7.56];

% oxygen concentration for chloride concentration 10 g/L
C10=[12.9 11.3 10.1 9.03 8.17 7.46 6.85];

% oxygen concentration for chloride concentration 20 g/L
C20=[11.4 10.3 8.96 8.08 7.35 6.73 6.20];

% set all chloride concentration
C=[linspace(0,0,length(T)) linspace(10,10,length(T)) linspace(20,20,length(T))];

% set time
T=[T T T];

% set all oxygen concentration
O=[C0 C10 C20];

% define a fitting function for O = f3(T) + f1(C)

myfun =fittype(@(a,b,c,d,e,cValue,TValue) a*TValue.^3+b*TValue.^2+c*TValue+d+e*cValue, 'ind', {'cValue' 'TValue'},'dep','z');

% plynomial fitting
P=fit([T',C'],O',myfun);

% display polynomial curve
disp(P);

% get value of oxygen concentration
v=P(15,12);

% relative error
error=abs(v-9.09);

fprintf('\nRelative error is %.4f\n',error);

General model: P(cValue, TValue) = a*TValue. ^3+b*TValue. ^2+c*TValue+d+e*cValue efficients (with 95% confidence bounds) : = 0.004065 (-2.436e+12, 2.436e+12) -0.1212 (-7.307e+13, 7.307e+13) 0.6923 (-4.871e+14, 4.871e+14) 13.55 (12.98, 14.12) -0.2012 (-0.2239, -0.1786) OUUUUU Relative error is 0.6758