Introduction
The purpose of this experiment is to predict the temperature drop of a test
bar as a function of time. The temperature drop of the test bars is predicted to vary
exponentially with time as what is given by the lumped capacitance method.
The prediction of the temperature drop is based on the lumped capacitance
method with the fulfillment of Biot Number. The lumped capacitance method is
the simplest unsteady conduction model. Thus, in order to satisfy the lumped
capacitance method the Biot number must always less than 0.1.
The Biot number can be obtained theoretically as well as experimentally.
The generalized equation for the Biot number is given in the Analytical
Approach section while the Biot number obtained experimentally is predicted to
be the ratio of the change of internal temperature to the change of film temperature
of the test bar.
Also, the time required by the test bar to cool from its initial temperature to
a certain final temperature can be obtained on rearranging the generalized form of
the lumped capacitance method (see also Analytical Approach).
Analytical Approach
In this experiment, a dimensionless parameter known as the Biot number is
introduced. The Biot number plays a very significant role in deriving and verifying the
validity of the lumped capacitance method. It is also a fundamental role in conduction
problems that involve surface convection effects.
The figure below depicts the derivation of the Biot number .
T
Ts,2
Ts,1
T , h
x
L
Schematic of the fundamental derivation of Biot number on a steady-state temperature
distribution in a plane wall with surface convection.
The Biot number is
Ts ,1  Ts ,2
Ts ,2  T

( L kA) hL

 Bi
(1 hA)
k
where L = thickness
k = thermal conductivity of solid
h = convective heat transfer coefficient
A = surface area
In this experiment, some assumptions are made before suing the lumped
capacitance method to predict the temperature drop across the test material. The
assumptions made are:
1.
2.
3.
4.
5.
6.
7.
No radiation
Negligible internal resistance of the material
No heat generation
Temperature of the material is constant
Constant physical properties
Ambient temperature (T) is constant
Area average film coefficient
The test materials used in this experiment are aluminum and brass bars. The
lumped capacitance method is the simplest unsteady conduction model. The Biot number
must be less than or equal to 0.1 in order to satisfy the lumped capacitance method.
However, there is a more general definition for the Biot number which will be used
throughout this experiment. Lc in the following equation is the characteristic length of
the bar used to calculate the Biot number.
Bi 
hLc
k
where Lc 
Volume V

Area
A
Since no heat generation occurs in the test bars, one can assure that the rate of
heat transfer through the bars by conduction is equal to the rate of heat transfer by
convection. Thus a relationship used to predict the Biot number obtained experimentally
is shown below:
q conduction  q convection
Tint
 hAT film
x
Tint
hx

 Bi
T film
k
kA
where Tint = internal temperature of test bars
Tfilm = film temperature of test bars
In order to predict the time required for the test bars to drop to certain
temperature, the following equation, which is a simplified form of the lumped
capacitance method, is needed. Detail derivation of this relationship will be shown in the
Appendix.
t
Vc  Ti  T 
hA
 where
ln
 T  T 
t =
 =
V =
c =
h =
Ti =
T =
T =
time
density of test bar
volume of test bar
specific heat capacity of test bar
average convective heat transfer coefficient
initial temperature of the test bar
ambient temperature
temperature of the test bar
However, in order to determine the convective heat transfer coefficient, h, other
dimensionless parameters were used. These are the Nussult number, Nu and the
Rayleigh number, Ra.
Ra L 
g (Ts  T ) L3c

where
g = local acceleration of gravity
 = expansion coefficient
whereas  and  can be obtained from handbook.
However, the RaL for the vertical side and the top and bottom sides of the bar is needed to
calculate the NuL. The RaL for different orientations differ from their characteristic
length, Lc. For vertical side Lc will be the height of bar whereas for top and bottom sides
the Lc will be the ratio of the area of the bar to its perimeter.
After knowing the RaL, the Nussult number can be calculated. The Nussult
number for the vertical side of the bar takes the form
0.67 Ra 1L/ 4
Nu L  0.68 
[1  (0.492 / Pr) 9 /16 ]4 / 9
for RaL < 109
However, the Nussult number for the upper surface of the cooled bar take the
form of the following
Nu L  0.27 Ra 1L/ 4
for ( 105 < RaL < 1010 )
The Nussult number for the lower surface of the cooled bar is
Nu L  0.54 Ra 1L/ 4
for ( 107 < RaL < 1011 )
With the known Nussult number for different sides of the bar, the convective heat
transfer coefficient for each side can be obtained by using the following equation.
h
k
Nu L
Lc
After calculating the convective coefficients associate with each side of the bar,
the average value is taken to calculate the lumped capacitance.
Introducing the generalized form of the lumped capacitance method:
T (t )  T  (Ti  T )e

T (t )  T  (Ti  T )e
 hA 
  t
 mc 

t

mc
hA
On rearranging the above equation yields the form of the equation introduced in the

where
 = thermal time constant =
previous page needed to calculate the time for the temperature drop across the test bar,
i.e.,
t
Vc  Ti  T 
hA

ln
 T  T 
Experimental Program
Since the purpose of this experiment is to predict the temperature drop on
the test bars and then verify the prediction with the results obtained
experimentally, the ambient temperature and the initial temperatures of the bars
are critical. These quantities should be measured as precise as possible for better
precision on the outcome of the experiment. Other quantities that were measured
are masses of the bars and their dimensions (length, width, and height).
The instruments used are General Electric freezer, SRS Stanford Research
Systems (model: SR630), 16 channel thermocouple monitor, Epson Fx-85E
printer, general vernier calipers 0-6”, and OHAUS LS50900 Portable Scale.
The experiment was started by measuring the masses of the bars by means
of the Portogram Balance followed by measuring the lengths, width, and heights
of the two test bars. The two test bars used in this experiment are aluminum and
brass bars. After that, the thermocouple labeled 16 was fixed to measure the
ambient temperature. On the other hand, the thermocouples labeled 3 and 4 were
used to measure the internal temperature of the test bars, i.e., aluminum and brass
bar respectively. Thermocouples labeled 10, 11, and 12 were attached to the
surface of the test bars to measure the film temperature.
In this case,
thermocouple 10 was used to measure the film temperature of the aluminum bar
whereas 11 and 12 were used to measure the brass bar. The starting time of the
experiment was recorded after the above procedures were done.
All the
thermocouples were allowed to reach their steady measurements before the
readings were taken. The test bars were then let to be cooled by putting them in a
freezer.
The thermocouple monitor was connected to a printer which printed the
temperatures read by all thermocouples at every minute. These readings were
needed to compare with the predicted temperature drop by using the Lumped
Capacitance Method.
Results
Since the printer printed the temperatures of all thermocouples at one minute
interval, only 30 data points are used in this section to calculate the result. The table
below shows the specifications of the two test bars used in the experiment.
Table 1
Raw Data For Aluminum and Brass Bars
Variable
Length
Width
Height
Mass
Variable
Length
Width
Height
Mass
Aluminum Bar
Units Uncertain
ty
(inch)
0.01
(inch)
0.01
(inch)
0.01
(g)
0.5
Brass Bar
Units Uncertain
ty
(inch)
0.01
(inch)
0.01
(inch)
0.01
(g)
0.5
1
1.99
6
664
1
1.99
5.99
1662
Table 2 shows that the temperatures of the aluminum and brass bars indicated by
the thermocouple monitor which were printed by the printer.
Since thermocouple 11 and 12 were attached to the brass bar, the average of the
temperatures indicated by these two thermocouples are taken to be used in the
calculations computed in Table 3. With the calculated values in Table 3, the temperature
drop on the test bars as a function of time can be plotted.
Figure 1 shows the temperature drop, (t) versus time, where
 (t )  T  T  (Ti  T )e
 hA 
t

 mc 
From Figure 1, it is seen that the temperatures on the test bars decreases
exponentially as time goes on.
This satisfied the equation defined by the lumped
capacitance method. Also, as seen from the graph, the temperature drop on the aluminum
bar is faster than the temperature drop on the brass bar because the aluminum bar has a
higher thermal conductivity, k. Also, the curves in Figure 1 can also be accounted for the
larger mass of the brass bar which makes the term
hA
becomes smaller.
mc
The
exponential term would then become smaller. Thus, the curve for the brass bar seems to
have less steep slope compared to the aluminum curve.
Table 4 shows the calculated values for the predicted Biot number, which is the
ratio of the change in internal temperature to the change in film temperature of the test
bar. Figure 2 shows the Biot number versus time for both the aluminum and the brass
bar. It is seen that the Biot number fluctuates with time. This could be caused by the
unsteady heat loss from the bars.
The time required for the test bar to cool from its initial temperature to its final
temperature, say 40F, was calculated based on the lumped capacitance method since the
Biot number obtained from the calculation is less than 0.1.
The Biot number for
aluminum bar is 1.85x10-4 whereas for brass bar is 3.98x10-4. Thus, the time required for
the cooling of aluminum bar takes approximately 71.34 min while brass bar takes 75.14
min. Compare to the experimental results obtained from the thermocouple monitor, the
time taken for the aluminum bar to cool to 40F is shorter than what was calculated. The
results show that the aluminum bar took approximately 56 min to cool to 40F.
However, the time predicted for the cooling of the brass bar is fairly close to the actual
results. The predicted time required by the brass bar to cool to 40F was 75.14 min and
the actual time required was approximately 70 min. This shows that the calculated value
is in 7% agreement with the actual result obtained from the experiment.
The calculated value for brass bar is more accurate probably because there were
two thermocouples connected to it and a more steady temperature for the surface
temperature was read. Also, the cooling rate of the bars could be affected by some
barriers in the refrigerator. This makes the cool air inside the refrigerator could not flow
evenly on the surface of the test bars.
Below shows the interpolated air properties used to calculate the RaL and Nu L .
Detailed calculations are shown in the Appendix.
 air  12394
.
kg m 3
30109
.  265.48
 283.29 K
2
  20.29 x10 6 m 2 s
c p  1006
.
kJ kg  K
  353
. x10  3 K 1
  14.4 x10  6 m 2 s
g  9.81m s 2
Pr  0.7113
k  24.96 x10  3 W m  K
Properties evaluated at
T=
40.00
35.00
0
Temperature Drop, t) ( C)
30.00
25.00
20.00
15.00
10.00
5.00
Temperature Drop on Aluminum Bar
Temperature Drop on Brass Bar
0.00
0
10
20
30
40
50
Time (minute)
Figure 1: Temperature Drop on the Test Bars as a Function of Time
60
0.01800
Predicted Biot Number for Aluminum Bar
Predicted Biot Number for Brass Bar
0.01600
Predicted Biot Number, Bi
0.01400
0.01200
0.01000
0.00800
0.00600
0.00400
0.00200
0.00000
0
10
20
30
40
50
Time (minute)
Figure 2: Predicted Biot Number versus Time for Both Aluminum and Brass Bar
60
Table 2
Temperature of Aluminum and Brass Bar Obtained From the Experiment
Time Center Temperature
Surface Temperature
(min)
(C)
(C)
Aluminum Brass Aluminum Brass
Brass
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
81.3
80
77.7
75
72.6
70.5
68.1
66.2
64
62.2
60.4
58.6
57
55.5
53.9
52.5
51.2
49.8
48.5
47.3
46.2
45.1
44
42.9
41.9
41
83.3
82.4
80.4
78.2
76.4
74.6
72.6
70.8
69
67.6
65.8
64.4
62.9
61.5
60
58.6
57.3
56.1
54.8
53.6
52.7
51.4
50.3
49.4
48.3
47.4
81.1
79.8
77.1
74.8
72.3
70.1
68
65.8
63.8
61.8
60
58.4
56.6
55
53.7
52.3
51
49.6
48.3
47.3
46
44.9
43.7
42.8
41.9
40.8
83.1
82
80
77.9
76.1
74.3
72.3
70.7
68.7
67.2
65.4
64
62.6
60.9
59.7
58.2
57.2
55.7
54.5
53.6
52.3
51.2
50.1
49.1
48.2
47.1
82.9
81.8
79.8
77.7
75.7
73.9
72.1
70.3
68.5
67.1
65.4
63.8
62.4
60.8
59.5
58.1
57
55.5
54.5
53.4
52.1
51
50
49.1
48
47.1
Ambient
Temperature (C)
Brass
(Average)
83
81.9
79.9
77.8
75.9
74.1
72.2
70.5
68.6
67.15
65.4
63.9
62.5
60.85
59.6
58.15
57.1
55.6
54.5
53.5
52.2
51.1
50.05
49.1
48.1
47.1
18.2
31
27.7
26.3
25
24.5
23.8
23.8
23.8
23.6
23.4
23.6
23.2
23.4
22.5
22.7
22.7
22.5
22.1
21.8
21.8
21.2
20.5
21.2
21.1
20.7
Conclusion
From the results calculated, the temperature on the each test bar decreases
exponentially with time. The trend is analogous with the equation given by the
lumped capacitance method. However, the time predicted for the cooling of the
aluminum and brass bars were not too accurate due mainly to the uncertainties
arise in the measurement.
The result also shows that the Biot number fluctuates with time. This is
probably caused by the unsteady heat loss from the surface of the test bars.
Appendix
All properties of the test bars were interpolated from handbook corresponding to
their film temperatures, Tf, where Tf is the average temperature of the surface
temperature of the bars and the ambient temperature.
Tf 
Ts  T
2
As discuss in the Analytical Approach section, in order to predict the time
required for the test bars to drop from their initial temperatures to a certain final
temperature, the following procedures were performed.
First find the Rayleigh number, RaL. Assume that the properties of aluminum and
brass bar have negligible temperature difference and their properties were evaluated at
the film temperature. Average surface temperature of the bars are at 301.09K whereas
the ambient temperature is at 265.48K.
Thus, T f 
30109
.  265.48
 283.29 K
2
For the vertical sides of the bars,
Ra L 
g (Ts  T ) L3c

(9.81m / s 2 )(353
. x10 3 K 1 )(300.54  265.48 K )(01524
.
m) 3
Ra L 
(14.4 x10  6 m 2 / s)(20.29 x10 6 m 2 / s)
Ra L  14.71x10 6
For the upper and the lower surface of the bars,
(9.81m / s 2 )(353
. x10 3 K 1 )(300.54  265.48 K )(0.008453m) 3
Ra L 
(14.4 x10 6 m2 / s)(20.29 x10 6 m2 / s)
Ra L  4100
All the properties given above were interpolated from handbook at T = 283.29K.
This is followed by calculating the Nussult number with the known RaL.
Also, with the known Nussult number, the convective coefficient can be obtained.
For the vertical side,
0.67 Ra 1L/ 4
[1  (0.492 / Pr) 9 /16 ]4 / 9
0.67(14.71x10 6 ) 1/ 4
Nu L  0.68 
[1  (0.492 / 0.7113) 9 /16 ]4 / 9
Nu L  0.68 
Nu L  32.534
Thus,
h vertical
Nu L k 32.534(24.9632 x10 3 W / m  K )


 5.329W / m2  K
L
01524
.
m
For the upper surface of the bar,
Nu L  0.27 Ra 1L/ 4  0.27(4100) 1/ 4  2.1605
Thus, h top
Nu L k 2.1605(24.9632 x10 3 W / m  K )


 6.3804W / m2  K
L
0.008453m
For the lower surface of the bar,
Nu L  0.54 Ra 1L/ 4  0.54(4100) 1/ 4  4.32
Thus, h bottom
Nu L k 4.32(24.9632 x10 3 W / m  K )


 12.7577W / m2  K
L
0.008453m
The total surface area of the bars is 0.02581m2.
To calculate the average convective coefficient,
h average
h average
0.02323m 2
0.00129m 2
0.00129m 2
2
2
2

2 5.329W / m  K 
2 6.3804W / m  K 
2 12.7577W / m  K
0.02581m
0.02581m
0.02581m
2
 5.7528W / m  K
Take the brass bar as sample calculation.
The measured temperature of the test bar as a function of time, T(t) is derived from the
lumped capacitance method shown below.
The lumped capacitance method is derived first by performing an energy balance on the
bar.
E  E in  E out  E gen
dT
 0  hAT  T   0
dx
dT
 hA

dt
T  T
mc
T
t  hA
dT
Ti T  T  0 mc dt
mc
T  T  (Ti  T )e
 hA 

t
 mc 
However, h is used instead of h because the average convective coefficient for each side
of the bar is needed to take into consideration when calculating T(t).
Thus, by lumped capacitance method
T (t )  T  (Ti  T )e
 hA 
t

 mc 
T (t )  265.48  (30165
.  265.48)e
 ( 5.7528W / m2  K )( 0.02581m2 ) 


(1.662 kg )( 380 J / kg  K )


4
T (t )  265.48  3617
. e  ( 2.351x10 ) t
If (t) = ( T-T ), then the temperature drop of the brass bar is
 (t )  3617
. e  ( 2.351x10
4
)t
On the other hand, the time required for the brass bar to cool from its initial temperature
of 83F(301.65K) to a certain final temperature, say 40F(277.59K), can be computed as
follow:
t
mc  T1  T 

ln
hA  T  T 
.  265.48 
 30165
t  (2.351x10  4 ) 1 ln

 277.59  265.48 
t  4508s  7514
. min  125
. hrs
To predict the Biot number for the bars, the following relationship is needed.
q conduction  q convection
Tint
 hAT film
x
Tint
hx

 Bi
T film
k
kA
The predicted Biot number for the aluminum bar in the first trial is
T
 Ts 813
.  811
.
Bi  center

 0.00318
Ts  T
811
.  18.2
While the predicted Biot number for the brass bar is
83.3  83.0
Bi 
 0.00463
83.0  18.2
The same procedures are repeated for the rest of the trials.