The initial temperature of the copper sphere is given

C. The initial temperature of the
copper sphere is given at time zero. You may
assume that Bi < 0.1, so the lumped capacitance
model can be used. You may assume k for copper
is constant, and can look up properties at 300K.
The heat transfer is unsteady state, and the
temperature of the sphere is measured by a
thermocouple inserted at the center of the sphere
(r = 0).
Note: You will need to use a spreadsheet to plot
data, but be careful to make the plot fit an
appropriate equation so that you can find h from a
trendline fit to the graph.
Spreadsheet has calculations, and is used to find h. Below
are the calculations for the two AIR experiments.
The blue data are for no fan, the red for with a fan.
The equations give slopes which must match -hA/mc
p
Problem 7
Time
(sec)
Data
Temp, C
(no fan)
Temp, C
(fan)
Temp, C
(water)
0 185 185 185
2 180 176 160.1
4 175.5 168 141
6 171.2 160.5 125
8 167.1 153.6 110.8
10 163.2 147 97.5
12 159.4 140.8 85.2
14 155.7 134.4
75.1
16 152 128.3
66.6
18 149.1 122.4
60.8
20 146.4 116.8
55.5
22 143.8 111.6
51.6
24 141.3 106
48.4
26 138.9 101.3
45.1
28 136.6 97
42.8
30 134.4 93.1
40.9
no fan fan no fan fan no fan fan
T-Tinf T-Tinf (T-Tinf)/(Ti-Tinf) (T-Tinf)/(Ti-Tinf) Time (sec) ln{(T-Tinf)/(Ti-Tinf)} ln{(T-Tinf)/(Ti-Tinf)}
160 160 1 1 0 0 0
155 151 0.96875 0.94375 2 -0.031748698 -0.057893978
150.5 143 0.940625 0.89375 4 -0.061210731 -0.112329185
146.2 135.5 0.91375 0.846875 6 -0.090198268 -0.166202175
142.1 128.6 0.888125 0.80375 8 -0.11864278 -0.218467003
138.2 122 0.86375 0.7625 10 -0.146471904 -0.271152771
134.4 115.8 0.84 0.72375 12 -0.174353387 -0.32330925
130.7 109.4 0.816875 0.68375 14 -0.202269195 -0.380162925
127 103.3 0.79375 0.645625 16 -0.230986729 -0.437536439
124.1 97.4 0.775625 0.60875 18 -0.254086123 -0.496347605
121.4 91.8 0.75875 0.57375 20 -0.276082937 -0.555561518
118.8 86.6 0.7425 0.54125 22 -0.297732408 -0.613874
116.3 81 0.726875 0.50625 24 -0.319000756 -0.680724661
113.9 76.3 0.711875 0.476875 26 -0.339852945 -0.740500877
111.6 72 0.6975 0.45 28 -0.360252765 -0.798507696
109.4 68.1 0.68375 0.425625 30 -0.380162925 -0.854196602
y = -0.0127x - 0.0147
R² = 0.9945
y = -0.0285x + 0.0082
R² = 0.9992
y = -0.0134x
R² = 0.9902
y = -0.0281x
R² = 0.9989
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0 10 20 30 40
ln{(T-Tinf)/(Ti-Tinf)}
Time (sec)
Series1
Series2
Linear (Series1)
Linear (Series2)
Linear (Series1)
Linear (Series2)

Both equations should have been forced to go through the origin, since the theoretical equation has no intercept
when plotted this way. (Note: I converted the data into a linear plot; it is possible to leave with a ln or exp, as long as a
different fit is used).
A = 4 * pi * r 2 = 4 * pi * 0.0125^2 = 0.00196 m 2
m = 4/3 * pi * r 3 * density = 4/3 * pi * (0.0125)^3 * 8933 = 0.0731 kg
density = 8933 kg/m 3
c
p
= 385 J/kg-K
With no fan, hA/mc
p
= 0.0134 s -1 = h* 0.00196m 2 / {0.0731 kg) * (385 J/kg-K)
h (no fan) = 192 W/m
2
K
With the fan, hA/mc
p
= 0.0281 s -1 = h* 0.00196m 2 / {0.0731 kg) * (385 J/kg-K)
h (fan)= 403 W/m 2 K
Similarly, for water, a third line can be added to the graph:
For the water, hA/mc
p
= 0.0801 s -1 = h* 0.00196m 2 / {0.0731 kg) * (385 J/kg-K)
h (water)= 1150 W/m 2 K (higher value than air; as expected)
y = -0.0127x - 0.0147
R² = 0.9945
y = -0.0134x
R² = 0.9902
y = -0.0285x + 0.0082
R² = 0.9992
y = -0.0281x
R² = 0.9989
y = -0.0801x
R² = 0.9971
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
0 5 10 15 20 25 30 35
ln{(T-Tinf)/(Ti-Tinf)}
Time (sec)
Series1
Series2
Water
Linear (Series1)
Linear (Series1)
Linear (Series2)
Linear (Series2)
Linear (Water)

8. A large solid sphere with a 10 inch diameter initially has a uniform temperature of 67 o F until it is placed in an oven at
400 o F (for the purposes of this problem, you can assume that the sphere is surrounded by air on all sides and
does not touch any surface).
A. Draw a qualitative sketch of T(r) from –r to r for a yellow pine sphere for four time points, including t = 0, t =
∞ and two times in between. Assume Bi > 0.1.
B. Draw a similar qualitative sketch of T(r) from –r to r for a 10 inch diameter aluminum sphere, assuming Bi <
0.1.