Transient Response Characteristics and Lumped System
Analysis of Geometrically Similar Objects
MECH595 – Introduction to Heat Transfer
Professor M. Zenouzi
Prepared by:
Andrew Demedeiros, Ryan Ferguson, Bradford Powers
October 1, 2009
Abstract
This report presents a method for determining whether a thermal system can be assumed a lumped
parameter system or a multi-dimensional parameter system. This is done by comparing the Biot
numbers of the different samples. Three materials are considered in this experiment: 1020 steel, brass
and Lexan. In order to facilitate the calculations the transient thermal response is measured and the
convection coefficient is calculated.
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Contents
Introduction .................................................................................................................................................. 4
Discussion of theory...................................................................................................................................... 4
Procedure...................................................................................................................................................... 5
Materials Tested ........................................................................................................................................... 6
Experimental Data ........................................................................................................................................ 6
Steel Cooling Calculations ............................................................................................................................. 8
Lexan Heating Calculations ........................................................................................................................... 9
Results ......................................................................................................................................................... 10
Discussion of Results ................................................................................................................................... 10
Conclusion ................................................................................................................................................... 11
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Introduction
A Lumped heat capacity system is a system which the temperature changes uniformly
throughout a body. In this experiment 3 materials were tested for lumped system analyses. These
materials were placed in a fluid at an elevated temperature and the inner and outer temperatures were
taken as they were heated and cooled. A lumped heat capacity system will change uniformly resulting in
the same temperature at the surface and in the middle of the body.
Discussion of theory
A system can be determined lumped heat capacity system using graphical and mathematical
analyze. A graphical analysis is done by recording the temperature inside an object and at the surface as
the object is heated and cooled. By overlapping the heating and cooling curves of the inner and outer
temperatures it can be determined that the object is a lumped heat capacity system if the two
temperatures are exactly the same at any time. The mathematical approach uses the heating and
cooling curves to approximate a time constant. This time constant is then related to the density, specific
heat and volume. If the Biot number is lass then 0.1 then the system is a lumped heat capacity system.
The first law of thermodynamics states that at any time t, the thermal energy absorbed by the fluid must
be equals the time rate of change of the internal energy of the lumped mass. This can be expressed
mathematically as:
 dT 
− ρc PV 
 = hA(T − T∞ )
 dt 
(1.)
Realizing T is the unsteady temperature of the lumped mass the initial condition is T(0)=Ti. The equation
then becomes:
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  hA  
s
(T − T∞ )  − ρc V t 
=e
(Ti − T∞ )
(2.)
p
Where T∞ the ambient temperature of the air and Ti is is the initial temperature of the object. The
exponential function e is assumed to be e −1 which allows T to be found at the first time constant, τc. The
time constant can now be used to find the convection coefficient, h.
τc =
ρVc
(3.a)
hAs
Solving for h:
h=
ρVc
τ c As
(3.b)
After finding h the Biot number can be calculated. The Biot number determines if a system is lumped or
not. If the Biot number is less than 0.1 the system is a lumped heat capacity system.
Bi =
hLch
k
(4.)
Procedure
The following procedure was used to complete this experiment.
1. Preheat the steel cylinder. When the cylinder reaches steady state temperature as indicated by
the thermocouples remove the cylinders and place them in free stream for cooling.
2. Monitor the temperature output of both thermocouples on the data logger. Wait until steady
state temperature is reached again.
3. Repeat for Lexan and brass. Print and save the temperature versus time curve.
It should be noted that the steel sample was heated and cooled using air flow. The Lexan and brass
samples where both heated and cooled using a water bath.
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Materials Tested
Three materials were tested during this experiment. Each sample had approximately the same
dimensions. The materials and their dimensions are shown below:
•
1020 Steel (50.8 mm height, 50.8 mm diameter)
•
Lexan (50.8 mm height, 50.8 mm diameter)
•
Naval Brass (50.8 mm height, 50.8 mm diameter)
Two thermocouples were embedded in each sample at different depths. These sensors output their
readings to a computer based data logger.
Experimental Data
The temperature versus time curves for each material was collected during the course of the
experiment by an automated data logger. This data was then used to create the following graphs.
Presented in order is steel (Figure 1), Lexan (Figure 2) and brass (Figure 3).
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120
Temperature (°C)
100
80
60
T2
40
T1
20
0
Time (HH:MM:SS)
Figure 1 ~ Temperature response of steel
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80
Temperature (°C)
70
60
50
T1
40
T2
30
20
10
0
13:55:12 14:02:24 14:09:36 14:16:48 14:24:00 14:31:12 14:38:24 14:45:36 14:52:48
Time (HH:MM:SS)
Figure 2 ~ Temperature response of Lexan
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120
Temperature (°C)
100
80
60
T1
40
T2
20
0
Time (HH:MM:SS)
Figure 3 ~ Temperature response of brass
Steel Cooling Calculations
Consider the following equation,
Setting the power of the exponent to one (1) will return the temperature value at one time constant.
Returning to the collected data it was found that the time which corresponds to a temperature of
was 516 seconds.
Solving the time constant equation for h yields,
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By using the obtained values it was found that the convection coefficient (h) was equal to
.
To assure that the system is a lumped sum system the Biot number was calculated.
Lexan Heating Calculations
Setting the power of the exponent to one will return the temperature value at one time constant.
Returning to the collected data it was found that the time which corresponds to a temperature of
was 101 seconds.
Solving the time constant equation for h yields,
By using the obtained values it was found that the convection coefficient (h) was equal to
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.
To prove that the system is not a lumped sum system the Biot number was calculated.
A Biot number of 5.60 shows that the system is not lumped.
Results
Table 1 below shows the Biot numbers and convection coefficients calculated for each of the
materials. As previously stated, a Biot number greater than 0.1 indicates that the material cannot be
considered a lumped system.
Material
1020 Steel
Lexan
Brass
Biot
Number
Lumped
System
0.008 Yes
5.6 No
0.21 No
Convection Coefficient
55.99 W/m²K
125.19 W/m²K
8755.0 W/m²K
Table 1 ~ Results for each material
Discussion of Results
The results obtained through calculation were used to determine which materials could be
considred lumped systems. The steel sample was the only the specimen tested that was certainly a
lumped system. This can be further verified in Figure 1 where the two temperature readings overlap
constantly.
The Lexan and Brass samples both failed to be considered lumped systems. Their respective Biot
numbers were 5.6 and 0.21 respectively. It was clear from inspection of the Lexan temperature response
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that it was not a lumped system. The brass however was not clear. The numerical value shows that it is
just outside the range needed to be a lumped system.
The convection heat transfer coefficient was then calculated for each sample. The brass
exhibited the highest value. This is seen in the graph in Figure 3 where it can be observed that the
sample heated and cooled quicker than the other materials. The steel had the smallest value. This was
visible in that steel took the longest to both heat and cool.
Conclusion
The ability to determine whether a thermal system is lumped parameter system or a multi-
dimensional transient system is important when analyzing that system. Using the equations and
methods presented here, it was possible to determine which material samples met this requirement.
After analyzing the results of the experiment it was determined that steel was the only genuine lumped
parameter system. Although brass was close it could not safely be considered and Lexan was clearly
unsuitable. Using the same methods it was also possible to calculate the convection heat transfer
coefficient.
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