Question

1) A spherical aluminum ball at a uniform temperature of 250°C is dropped into an oil t(s) Tscat t(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 10 238.1 temperature versus time every ten seconds for 300 seconds. If the Bi #<<1, the lumped capacitance solution 20 230.1 30 221.0 6h e(t) = 0,exp( 219.0 40 PCD is valid. Here, 0 is the difference between the ball and free stream temperatures, h is the heat transfer coefficient. The density (p), specific heat (C), and diameter (D) of the 50 210.2 60 197.9 70 196.0 ball are 2700 (kg/m³), 1000 (kJ/(kg-K)] and 1 [cm], respectively. 193.1 80 90 185.9 Plot 0(t) as a function of time. a. 100 176.7 b. Transform this function to make it a linear function of time. c. Plot the transformed variable as a function of time and use a linear regression 172.4 110 120 170.3 to determine the slope of the best fit straight line through the transformed 130 163.0 data. 140 153.5 d. Use the regression slope to determine the heat transfer coefficient. e. Was the lumped capacitance solution valid? Use 235 (W/(m-K)] for the thermal conductivity of the aluminum. 151.3 150 160 149.8 143.4 170 180 139.0 190 137.0 200 132.3 210 131.4 220 131.1 230 240 122.9 126.6 250 114.7 260 112.4 270 115.7 280 110.8 104.1 290 300 104.4

Answer 1

I am using Matlab for this problem.

(a) Plot for the given Data

eeeeee Temprature 0 50 100 150 Time(s) 200 250 300

(b) Conversion of the given Equation to the linear form:

6 forms conversion of the function to unear o(t) = 0o e-(ec) rog (ab) = logat logbo applying log on both sides =) 20g Olt) = logo. - 6h Lt) PCD on maunanging. Lt) +1098. => 109, BLE) c = mx + I whoe m=-ch

(c) Linear Graph is just obtained using the log of Temperature value. I Have included the Raw data with a fitted curve. It can

2.48 • data fitted curve 2.35 2.3 2.25 2.2 2.15 2.1 TO 50 100 150 200 250 300

be noticed from part (b) why I took just log_10 for temprature.

These are the values for Linear fitted plot.

Linear model Poly1:
val(x) = p1*x + p2
Coefficients
p1 = -0.001246  
p2 = 2.38  

I have used inbuilt Matlab package but it can also be done using traditional formula.

(d) Now If we got the Slope then we can easily calculate the value of 'h' i.e. 560.789 (I have included the code in the answer so you can check how I computed it.)

(e) Now simply check the ratio for D(diameter) and h(Thermal heat conductivity) as given in the question for Aluminium. It should be less than 1. I chose Diameter because it is whole lenght of the sphere.

Code:

time = 0:10:300;

temp = [250.3 238.1 230.1 221.0 219.0 210.2 197.9 196.0 193.1 185.9 176.7 172.4 170.3 163.0 153.5 151.3 149.8 143.4 139.0 137.0 132.3 131.4 131.1 122.9 126.6 114.7 112.4 115.7 110.8 104.1 104.4];

plot(time,temp,'r-o')

plot(time,log10(temp)) % Linear Graph is just obtained using log of Temprature value

grid

hold on

xlabel('Time(s)')

ylabel('Temprature')

xyfit = fit(time',log10(temp)','poly1')

plot(xyfit,time,log10(temp))

hold off

rho = 2700; %kg/m^3

C = 1000; %kJ/(kg-k)

D = 1; %cm

h = -(xyfit.p1)*rho*C*D/6 %p1 is slope of the fitted curve

% I have Used negative sign to cancel out negative sign in slope