Question

I need the Matlab code please.

Problem Statement #2: ds (c) using the To 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 of =-139.3441144 20 34411 + - Ta where mum const! of the 1.243800 x 100 6.642308 x 107 1 = 1 72 8.621949 x 1011 Deva ks = 5.12 car where Oxf = the saturation concentration of dissolved oxy- gen in freshwater at 1 atm (mg L-'); and Ta = absolute temperature (K). Remember that Ta = T + 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 7 as a function of a given oxygen concen- tration. Test your function for oss = 8, 10 and 12 mg/L. Check your results. Aving in a trapezoidal channel at a rate of Lenal must Code for problem #2: Paste code here Solution for problem #2: Give solution/output here

Answer 1

*5.201 T--2 :.01 40; celsius Ta - T + 273.15: ost - (T) exp(-139.34411 + (1.575701 - 10*5) ./ (T + 273.15) ... - (6.64230B . 10^) /(1+ 273.25) .2 ... (1.243800 *10^10) ./(1+ 273.15) .3 - (0.621949 . 10^11) / (1 + 273.15). 4); hold on plot (1,03 (1) f - (T) (( -139.34412 (1.575702 . 10^5) ./(T. 273.15) - (6.64230B . 10^7) ./(T. 273.15).^2 + (1.2438 Splot (1, 1)) x 1 - 0 X -40; 1x new - Xu; **__old - x_1: bisection El for 1 - 1:1:23 16 (x1). X__old-x_ (*_*) <0 the signs should be opposite for error equation XI - ((x ** 1)/2); X our estimated root x__ne -X_: tor error equation 11 (4(x_1) • f(x_x) < 0) else (2 x 2) - Ex_) > 0) x1 - x_:

Ex new = X x__old - X 1: bisection e for i=1:1:23 it fix 1) - XI old - XI: (x ) < 0 the signs should be opposite tor error equation *_ Ex - (*_ **_1)/2): * 1 is our estimated root _new_1 for error equation 17 ((x_1) - (x_x) < 0) else (x1) - (x ) > 0) x_1 - X_: ( (*_*) . f(x ) - 0) end terror_a-abat X__new-*__old)/X_new) approximate percent relative error terror_ - ((zoot_2 - *_*)/(zoot_2)): True percent relative error plot error a, x, Splot error_, *, 'D') blot (T , f (x_r),'S') end