Question

A spherical aluminum ball at a uniform temperature of 250°C is dropped into an oil |t [s] Tscat [C] bath at 50°C. The ball's surface temperature is measured using a thermocouple. The 250.3 table below (also provided as a spreadsheet in a separate file) shows the recorded 238.1 10 temperature versus time every ten seconds for 300 seconds. If the Bi #<<1, the 20 230.1 lumped capacitance solution 30 221.0 6h ө(С) — өоехр (- pCD 40 219.0 50 210.2 is valid. Here, 0 is the difference between the ball and free stream temperatures, h is 60 197.9 the heat transfer coefficient. The density (p), specific heat (C), and diameter (D) of the ball are 2700 [kg/m³), 1000 [kJ/(kg-K)] and 1 [cm], respectively. a. Plot 0(t) as a function of time. 70 196.0 80 193.1 90 185.9 100 176.7 b. Transform this function to make it a linear function of time. 110 172.4 Plot the transformed variable as a function of time and use a linear regression C. 120 170.3 to determine the slope of the best fit straight line through the transformed 130 163.0 data. 140 153.5 Use the regression slope to determine the heat transfer coefficient. Was the lumped capacitance solution valid? Use 235 [W/(m-K)] for the d. 150 151.3 e. 160 149.8 thermal conductivity of the aluminum. 170 143.4 180 139.0 190 137.0 200 132.3 210 131.4 220 131.1 230 122.9 240 126.6 250 114.7 260 112.4 270 115.7 280 110.8 290 104.1 300 104.4

Answer 1

017) , Os exb (esim ty By taking log in botu side- ln (024)) = f (8,00) - from the Data Oct Tsurface - 50 How clope of the linear line in -0.004273868092 Here o=2700 Kg/m3 Ć = 1000 nolk.g-k D 21 cm z o.olm Hene lobe Goh 0-0042738 PCD ha PCD x 0.0042738 2 2700X1000 Y 0.61 40.6042138 e 99.232 6

&Matlab code for plotting and finding h clear all close all fall given data t=0:10:300; T_surf=[250.3 238.1 230.1 221.0 219 210.2 197.9 196 193.1 185.9 176.7... 172.4 170.3 163 153.5 151.3 149.8 143.4 139 137 132.3 131.4 131.1 ... 122.9 126.6 114.7 112.4 115.7 110.8 104.1 104.4); th=T_surf-50; figure(1) plot(t,th) xlabel('t') ylabel('theta(t)) title('t vs. theta(t) plot') box on figure (2) plot(t,log(th)) xlabel('t') ylabel('log(theta(t))') title('t vs. log(theta(t)) plot') box on Xplotting best fit line of linear data p=polyfit(t, log(th), 1); pp-polyval(Pit); hold on plot(t,pp) legend ( 'Actual data', 'Linear fit') fprintf('Here slope of the line is %f and intercept is & \n',p(1),P(2)) 물 End of code Here slope of the line is -0.004274 and intercept is 5.282015

t vs. theta(t) plot theta(t) 50 100 150 200 250 tvs. log(theta(t) plot Actual data -Linear it logithetat) 100 150 300

%%Matlab code for plotting and finding h
clear all
close all

%all given data
t=0:10:300;
T_surf=[250.3 238.1 230.1 221.0 219 210.2 197.9 196 193.1 185.9 176.7 ...
    172.4 170.3 163 153.5 151.3 149.8 143.4 139 137 132.3 131.4 131.1 ...
    122.9 126.6 114.7 112.4 115.7 110.8 104.1 104.4];
th=T_surf-50;
figure(1)
plot(t,th)
xlabel('t')
ylabel('theta(t)')
title('t vs. theta(t) plot')
box on

figure(2)
plot(t,log(th))
xlabel('t')
ylabel('log(theta(t))')
title('t vs. log(theta(t)) plot')
box on

%plotting best fit line of linear data
p=polyfit(t,log(th),1);
pp=polyval(p,t);
hold on
plot(t,pp)
legend('Actual data','Linear fit')

fprintf('Here slope of the line is %f and intercept is %f\n',p(1),p(2))

%%%%%%%%%%%%%%%%%% End of Code %%%%%%%%%%%%%%%%