3 02 01 01 10 20 30 40 lnt tinfti tinf time sec

-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)

The blue data are for no fan, the red for with a fan.
The equations give slopes which must match -hA/mc
p
Both equations should have been forced to go through the origin, since the theoretical equation has no intercept
when plotted this way, thus I use the values -0.0134 for no fan and -0.0281 for with a fan. (Note: I converted the data
into a linear plot; it is possible to solve and plot then curve fit with a ln or exp, as long as an appropriate curve fit
(trendline) is used).
A = 4 * pi * r
2
= 4 * pi * 0.0155^2 = 0.00302 m
2
m = 4/3 * pi * r
3
* density = 4/3 * pi * (0.0155)^3 * 2702 kg/m
3
= 0.0421 kg
density = 2702 kg/m
3
c
p
= 903 J/kg-K
With no fan, hA/mc
p
= 0.0134 s
-1
= h* 0.00302m
2
/ {0.0421 kg) * (903 J/kg-K)
h (no fan) = 169 W/m
2
K
With the fan, hA/mc
p
= 0.0281 s
-1
= h* 0.00302m
2
/ {0.0421 kg) * (903 J/kg-K)
h (fan)= 354 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* 0.00302m
2
/ {0.0421 kg) * (903 J/kg-K)
h (water)= 1008 W/m
2
K (higher value than air; as expected)
Validate Biot number:
Bi = h L
c
/ k
use h’s above
L
c
= r/3 = 0.0155 m / 3 = 0.0052 m
k = 237 W/m-K
air (no fan): Bi = 169 * 0.0052 / 237 = 0.0037 (VALID)
air (fan) Bi = 354 * 0.0052 / 237 = 0.0078 (VALID)
water Bi = 1008 * 0.0052 / 237 = 0.022 (Still Valid, but the better/higher the convection term, it is possible that lumped
capacitance would no longer be valid)
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)