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.

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- Fall '10
- Brazel
- Heat, Heat Transfer