Question

Need the solution in Matlab

MU UUNIUI) in the tex the same initial guesses as in (b). (5.9) The saturation concentration of dissolved oxygen in freshwater can be calculated with the equation 1.575701 x 105 In 03 = -139.34411 + 5 6.642308 x 107 1.243800 x 1010 + 72 8.621949 x 1011 where Oss = the saturation concentration of dissolved oxy- gen in freshwater at 1 atm (mg L-'); and T, = absolute temperature (K). Remember that Ta = 1 + 273.15, where T = temperature (°C). According to this equation, saturation decreases with increasing temperature. For typical natural waters in temperate climates, the equation can be used to de- termine that oxygen concentration ranges from 14.621 mg/L at 0°C to 6.413 mg/L. at 40 °C. Given a value of oxygen con- centration, this formula and the bisection method can be used to solve for temperature in °C. (a) If the initial guesses are set as 0 and 40°C, how many bisection iterations would be required to determine tem- perature to an absolute error of 0.05 °C? (b) Based on (a), develop and test a bisection M-file function to determine T as a function of a given oxygen concen- tration. Test your function for 0 = 8, 10 and 12 mg/L. Check your results. mwing in a trapezoidal channel at a rate of

Answer 1

solution

a) For the initial guess set 0 and 40°C The number of iterations are usedn element To achieve a given error (tolerance), e It is given by n log2 (Axo Ed) where eo is the initial size Axo = xu - xl where Axo xu -xl = 40-0 Axo = 40°C absolute error= 0.05 Hence n log2 40 0.05) n log2 (800) =9.64 Therefore Approximately n 10 bisection iterations are required To determine temperature of absolute error will be less than 0.05 C

b) Smatlab code %bisection.m clc clear all; close all; Supper and lower limit value xu 0 xl- 40 for osf 8:2:12 &to find the approximte temperature function T f-@ (T) (exp ((-139.34411)(1.575701 10^5). / (T+273.15)) - ((6.642308 10^7)./ ( (T+273.15) .^2)) + ( (1.24380010^10) ./ ( (T+273.15 ).^3)) (8. 621949 10^11)./ ( (T+273.15) .^4)))-osf); lower and upper limit interval x (1) xu; y (1) xl; %calculate interval iteration xr (1) xutxl) /2 %loop executes if (f (x (1)) *f (xr (1)))<0 Bupdate the upper limit x (2) x (1) y (2) xr (1;

else x (2) xr (1); y (2) y (1) Send loop end %test the solution solution=zeros (1,100) loop executes for i-2:100 iterate the interval xr (i) (x (i)y (i)). /2; %check upper and lower limit condition if (f (x (i)) *f (xr (i)))<0 x (i+1) x (i) ;y (i+1)=xr (i); else x (i+1)xr (i) ;y (i+1) =y (i); end Sabsolute error if (xr (i)-xr (i-1))<=0.05 solution (i) =xr (i); end Send loop end disp ('the number of iteration required to get an absolute num2str (osf)]); error 0.05 is disp ('answer=') display (solution (100)); end

//output

Result Executing the program.... Soctave -qf-no-window-system main.m the number of iteration required to get an absolute error 0.05 is 8 answer 26.780 the number of iteration required to get an absolute error 0.05 is 10 answer 15.388 the number of iteration required to get an absolute error 0.05 is 12 answer 7.4652

//copyable code

b)

%matlab code

%bisection.m

clc;

clear all;

close all;

%upper and lower limit value

xu=0;

xl=40;

for osf=8:2:12

%to find the approximte temperature function T

f=@(T) (exp((-139.34411)+((1.575701*10^5)./(T+273.15))-((6.642308*10^7)./((T+273.15).^2))+((1.243800*10^10)./((T+273.15).^3))-((8.621949*10^11)./((T+273.15).^4)))-osf);

%lower and upper limit interval

x(1)=xu;

y(1)=xl;

%calculate interval iteration

xr(1)=(xu+xl)/2;

%loop executes

if (f(x(1))*f(xr(1)))<0

%update the upper limit

x(2)=x(1);

y(2)=xr(1);

else

x(2)=xr(1);

y(2)=y(1);

%end loop

end

%test the solution

solution=zeros(1,100);

%loop executes

for i=2:100

%iterate the interval

xr(i)=(x(i)+y(i))./2;

%check upper and lower limit condition

if (f(x(i))*f(xr(i)))<0

x(i+1)=x(i);y(i+1)=xr(i);

else

x(i+1)=xr(i);y(i+1)=y(i);

end

%absolute error

if (xr(i)-xr(i-1))<=0.05

solution(i)=xr(i);

end

%end loop

end

disp(['the number of iteration required to get an absolute error 0.05 is ' num2str(osf)]);

disp('answer=')

display(solution(100));

end