Transcript Modeling Heat Flow in a Thermos

Modeling Heat Flow in a
Thermos
Michael A. Karls
James E. Schershel
Ball State University
The Coffee Cup Problem
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Freshly poured coffee has a temperature of 80 oC.
Assume that the room temperature is a constant
20 oC, and that after two minutes, the coffee has
cooled to 70 oC .
Find a model for the temperature of the coffee at
any time after the coffee is first poured.
2
Newton’s Law of Cooling
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Newton's law of cooling*: The
rate of temperature change of
an object is proportional to the
difference in temperature
between the object and its
surroundings.
Newton’s law of cooling can
be used to model a cooling
cup of coffee!
*Newton experimentally
observed that the rate of loss
of temperature of a hot body is
proportional to the
temperature itself, but it was
Fourier who actually wrote
down an equation to describe
this process of heat transfer.
3
The Coffee Cup Model

Using Newton's law of cooling, a model for a cooling cup
of coffee is given by the following initial value problem for
t 2 (-1,1):
where
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k>0 is a proportionality constant that describes the rate at
which the coffee cools,
T is the surrounding temperature,
T0 is the initial temperature of the coffee,
u(t) is the temperature of the coffee at any time t.
The solution to (1), (2) is:
4
Verifying the Coffee Cup
Model Experimentally
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Texas Instruments (TI) has developed
an inexpensive calculator-sized data
collection interface, known as a
Calculator-Based Laboratory (CBL).
The CBL provides a link between a TI
calculator and sensors that are used to
collect data.
Some of the sensors available:
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pH
Temperature
Pressure
Force
Motion
Using a CBL with a temperature probe,
a TI-85 calculator, and a program
available from TI’s website
(http://education.ti.com/) , temperature
data can be collected, plotted, and
compared to the solution in (3).
5
A Modified Coffee Cup
Problem

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What if we wish to know the temperature at any
position within the cup of coffee at any particular
time?
One way to approach this problem: Think of the
water in the cup as a cylinder of heat conducting
material that is insulated on the side, top, and
bottom.
Instead of a cup full of coffee, we can think of a
thermos full of coffee!
6
The Thermos Problem


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A thermos is filled with hot coffee at an initial
temperature of 80 oC.
Assume that the room temperature is 20 oC and the
top of the thermos is left open.
Find a model for the temperature of the coffee at
any position in the thermos and at any time after the
coffee is first poured.
7
Heat Flow in a Rod
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Joseph Fourier made an
extensive study of heat flow in
objects such as a long rod of
conducting material with an
insulated lateral surface.
Although we are not
considering a solid, we can
try to use the same ideas to
model the temperature in a
cylindrical column of hot
coffee!
Our goal is to find a model for
the temperature of the coffee
in a thermos at any time and
at any position and show that
this model agrees with actual
data.
Instead of hot coffee, we’ll
look at ice-cold water.
8
The Thermos Model


Assume we have a thermos full of ice-cold water, open at the
top.
Since the water is ice-cold, we’ll assume:

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There is no internal convection.
The contribution of heat from radiation is negligible.
There is no evaporative cooling at the top.
Think of the ice-cold water as a cylinder of conducting material
that is insulated at the bottom and on the side.
Heat is transferred by convection at the top surface of the
water.
Assume that at any time the temperature will be the same at
any point of a cross-sectional slice of water.
9
The Thermos Model (cont.)

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Let u(y,t) be the temperature at
any cross-sectional slice of the
water, for 0 · y · a; t ¸ 0.
Let y = 0 and y = a correspond to
the bottom and top of the thermos,
respectively.
Assume the air surrounding the
thermos is at constant
temperature T0.
Suppose the initial temperature
distribution is given some
sectionally smooth function f(y).
By sectionally smooth, we mean
f'(y) exists and is continuous on
[0,a], except possibly at a finite
number of jumps or removable
discontinuities.
y=0
y=a
10
The Thermos Model (cont.)

This simple system can then be modeled by the following
initial value-boundary value problem for 0<y<a, t>0:

Equation (4) is known as the one-dimensional heat equation
and , c, , and h are the density, specific heat, thermal
conductivity, and convection coefficient, respectively.
11
The Thermos Model (cont.)

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Recall Fourier's law of heat
conduction
where q(y,t) is the heat flow rate.
Using (8), we see that boundary
condition (5) indicates that there is
no heat flow through the bottom of
the thermos.
Similarly, (8) shows that boundary
condition (6) describes the
interaction of the top of the water
with the air via Newton's law of
cooling.
The known initial temperature
distribution is given by (7).
12
The Thermos Model (cont.)

Using the technique of separation of variables or Fourier’s
Method, the solution to (4)-(7) is found to be:
where n is the unique solution to
in the interval

The coefficients Bn are given by:
13
Verifying the Thermos Model
Experimentally


Now that we have a model for our
thermos, we collect temperature
data and test our model
experimentally.
The following materials are
required:
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Four TI-85 calculators with
temperature collection software
Four CBLs with temperature
probes
One thermos
Two twist ties
One rubber band
Ice
Water
Ruler
Freezer
14
Experimental Procedure

The experimental procedure is as follows:
1.
2.
3.
4.
5.
Measure the depth of the interior of the thermos.
Use the two twist-ties to connect three of the
temperature probe cables together, with the
probes arranged so that the distance between
each probe is half of the depth of thermos.
Place the probes in the freezer.
Fill the thermos with ice water and allow it to sit
for roughly two hours to pre-chill.
Then remove the ice from the thermos and topoff the thermos with ice-cold water.
15
Experimental Procedure
(cont.)
6.
7.

Insert the three temperature probes that are
twist-tied together into the thermos and use the
rubber band to keep the probes at the proper
heights: one at the bottom, one in the middle,
and one at the top, just below the surface of the
water. Use the fourth temperature probe to
record the ambient room temperature.
Use the TI-85 calculators and the CBLs to
record the temperature data.
Collect temperature data once every 120
sec for 4 hr.
16
Assigning Values to
Coefficients in Our Model

To compare our model to the experimental data, we
need to specify the parameters in the model.
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Length of the thermos: a = 0.28 m
Ambient room temperature: T0 = 21 oC.
Density of water:  = 1000 kg/m3
Specific heat of water: c = 4186 J/(kg oC)
Thermal conductivity of water:  = 0.602784 J/(m s oC)
Convection coefficient is h = 83.72 J/(m 2 s oC).
The choice for convection coefficient is a guess based
on convection coefficients found in the paper “The virtual
cook: Modeling heat transfer in the kitchen,” Physics
Today 52 (11), pp. 30—36 (1999).
17
Initial Temperature Distribution
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We also need to specify an initial temperature
distribution function f(y).
If the thermos were well-shaken, we would expect
that the temperature would initially be uniform.
Experimentally we find:


The temperature readings initially decrease slightly, then
rise.
The top of the thermos has a higher initial temperature
than the middle and bottom.
18
Initial Temperature Distribution
(cont.)
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
To take this temperature difference into account, we
use a piecewise linear function for f(y).
f(y) is constructed from the points (0,Tb) , (a/2,Tm) ,
and (a,Ta) , where Tb, Tm , and Ta are the
measured initial temperatures at the bottom (y = 0),
middle ( y = a/2), and top ( y = a), respectively.
19
Determining the Number of
Terms in Our Model
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We now use (10) and (11)
to find the values of n
numerically.
These values of n and
the initial temperature
distribution f(y) can then
be substituted into (12) to
find the coefficients, Bn, for
our model.
Graphically comparing the
nth partial sum of (9) at t =
0 to the initial temperature
distribution f(y) , thirty
terms in (9) appear to be
enough for our model.
20
Graphical Comparison of
Series Solution to f(y)
21
Model vs. Experimental
Results
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Unfortunately, as we see in
Fig. 1, the model doesn't
match the experimental results
very well.
How far are we off? One
measure of the error is the
mean of the sum of the
squares for error (MSSE)
which is the average of the
sum of the squares of the
differences between the
measured and model data
values
MSSE for this model:

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Top(): 33.7 (oC)2,
Middle(): 0.534 (oC)2,
Bottom(): 0.930 (oC)2.
u
14
12
10
8
6
4
2
t
50
100
150
200
Figure 1
Model: _________
Actual: …………….
22
Model with New Constants vs.
Experimental Results
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Maybe the problem with our model
is our choice of the constant h,
which was chosen fairly arbitrarily.
With the choice of h=920.92 J/(m2
s oC), found by finding the best fit
with integer multiples of our initial
guess of h, we obtain the results
shown in Fig. 2 after recomputing
the values of n and Bn.
Although the model is significantly
closer to the experimental results
for the temperature at the top of
the thermos, the model is still off
by about the same amount for the
middle and bottom of the thermos.
MSSE for this model:



Top(): 1.11 (oC)2,
Middle(): 0.468 (oC)2,
Bottom(): 0.929 (oC)2.
u
14
12
10
8
6
4
2
t
50
100
150
200
Figure 2
Model: _________
Actual: …………….
23
Modified Experimental
Procedure
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Perhaps our assumption that convection plays no
significant role is incorrect.
To reduce the contribution of this pathway for heat
flow, we make the following modification to our
experimental procedure and pack the thermos with
cotton balls to reduce convection currents.
The following additional materials are required for
the revised experimental procedure:
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One bag of 100 cotton balls
One pencil
24
Modified Experimental
Procedure (cont.)
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The modified experimental procedure is as follows:
1. Measure the depth of the interior of the thermos.
2. Use the two twist-ties to connect three of the
temperature probe cables together, with the
probes arranged so that the distance between
each probe is half of the depth of thermos.
3. Place the probes and pencil in the freezer.
4. Fill the thermos with cotton balls, remove the
cotton balls and place them in the freezer.
5. Fill the thermos with ice water and allow it to sit
for roughly two hours to pre-chill.
6. Remove the ice from the thermos and top-off the
thermos with ice-cold water.
25
Modified Experimental
Procedure (cont.)
7.
8.
9.
Take the cotton balls from the freezer and add
the cotton balls one-by-one to the thermos. Use
the chilled pencil to poke them below the
waterline.
Insert the three temperature probes that are
twist-tied together into the thermos and use the
rubber band to keep the probes at the proper
heights: one at the bottom, one in the middle,
and one at the top, just below the surface of the
water. Use the fourth temperature probe to
record the ambient room temperature.
Use the TI-85 calculators and the CBLs to
record the temperature data.
26
Modified Experimental
Procedure (cont.)
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We can now re-test our model using the modified
experimental procedure. The same values of a, ,
and c are used.
The ambient temperature for this data set is slightly
higher, so we take T0 = 24 oC.
Because the cotton, probe leads, and convection
could affect the apparent value of  and our choice
for h was really just a guess, we choose  and h so
that the model matches the data as closely as
possible graphically.
By tripling our initial choice of  and taking multiples
of an initial guess for h =41.86 J/(m2 s oC), we find
that =1.80835 J/(m s oC) and h=1004.64 J/(m2 s
oC) work well.
27
Model With New Constants vs.
Modified Experimental Results
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If we construct a new
piecewise linear f(y) and
compute n and Bn for our new
model data, we obtain the
results shown in Fig. 3 with
thirty terms in the sum for
u(y,t).
Visually, the model seems to
be closer to experimental
results, but the bottom and
middle model values are still
off.
MSSE for this model:
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Top(): 0.396 (oC)2,
Middle(): 0.576 (oC)2,
Bottom:() 0.487 (oC)2.
u
14
12
10
8
6
4
2
t
50
100
150
200
Figure 3
Model: _________
Actual: …………….
28
A Final Modification to Our
Model
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As a final modification of our model, suppose that
we had started with a different initial temperature
distribution, say f(y)=A eB y+C.
This form of f(y) is chosen to model the rapid rise in
initial temperature near the top of the thermos.
To take into account the fact that the temperature
readings initially decrease slightly, then rise, we will
use the points (0, Tmin,b), (a/2,Tmin,m), and (a,Tmin,a),
where Tmin,b, Tmin,m, and Tmin,a are the minimum
temperature readings recorded at these positions.
29
A Final Modification to Our
Model (cont.)
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To find the unknown constants A, B, and C, we solve the
system of equations:
30
Modified Model vs. Modified
Experimental Results
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As before, we compute n and
Bn using the same  and h as
in the previous model. Fig. 4
shows our results.
Of the models we have tried,
this one provides the best
visual fit to the experimental
data.
MSSE for this model:
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Top(): 0.142 (oC)2,
Middle(): 0.0664 (oC)2,
Bottom(): 0.0449 (oC)2.
u
14
12
10
8
6
4
2
t
50
100
150
200
Figure 4
Model: _________
Actual: …………….
31
A Final Test of Our Model
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As a final test of our model, we'll
use it to answer the question: How
long does the thermos keep things
cold?
To do so, we look at the average
value of the temperature of the
water as a function of time, which
is given by the integral
Ave U
Average
Temperature
and Cold
20
15
10
5
1000

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Computing this integral
numerically, as we see in Fig. 5,
the model predicts that the
temperature of the thermos will
warm up to 15 oC in about 1.42
days.
Experimentally, the thermos
warms up to about 16 oC in a day
and a half.
2000
3000
4000
5000
6000
7000
t
Figure 5
Model Ave Temperature: _________
“Cold”  Temp < 15 oC : _________
32
Conclusions and Further
Questions
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Using Fourier's model for heat conduction, we have
developed a simple model for the temperature of
ice-cold water in a thermos at any position and any
time.
Based on experimental data, we found that a
modified version of our original model agrees with
our measured data.
One question that arises is why does our last
choice of the initial temperature distribution work so
well?
33
Conclusions and Further
Questions (cont.)
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The main heat transfer mechanism in the thermos
is some combination of conduction and convection.
In practice, when modeling heat transfer due to
conduction or convection, it turns out that Newton's
law of cooling can be used to describe the heat flow
rate.
Therefore, if we assume that for small t, the
temperature in the thermos doesn't change with
time, a natural choice for the initial heat flow rate at
any point in the thermos would be:
34
Conclusions and Further
Questions (cont.)

where Ts is the unknown temperature at the surface of the ice-water
and ĥ is a proportionality constant.
From Fourier's law of heat conduction, we see that for 0<y<a,
which has the general solution
where A, Ts, and ĥ/ are unknown constants. Equation (23) is
exactly the form of our final choice of f(y)!!
35
Conclusions and Further
Questions (cont.)
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This experiment can be safely performed by
students in a classroom setting for the small cost of
four calculators, four CBLs, and a decent thermos
(such as our choice of a Stanley heavy duty
thermos).
Note that the constants h, k, and T0 may need to be
adjusted to match a different experimental
environment.
36
Conclusions and Further
Questions (cont.)
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In addition to the experiment we have outlined, other
questions such as the following could be addressed.
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How does adjusting the constants h and k affect the discrepancy
between the model and the actual data?
Instead of choosing h and k by trial and error, are there
mathematical methods that could be used to minimize the
discrepancy?
What happens if a liquid with different chemical composition than
water, such as cola or lemonade, is put in the thermos? Can h
and k still be chosen to match the model to the measured data?
How well does this model work with hot coffee in the thermos
instead of ice-cold water?
In a model with a hot liquid, are there any other factors that need
to be considered, such as heat lost due to evaporation?
37
Resources
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The MacTutor History of Mathematics
archive: http://www-history.mcs.standrews.ac.uk/index.html
Texas Instruments Educational
Resources: http://education.ti.com/
“Modeling Heat Flow in a Thermos”,
Am. J. Phys. 71 (7), July 2003
38
Supplementary Material
39
Solution of the IVBVP
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IVBVP for Thermos
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To solve (4)-(7), we first
find the steady-state
solution v(y), where v(y) is
the limit of u(y,t) as t ! 1.
Intuitively, we expect that
after a long time, the water
in the thermos should warm
to room temperature and
become independent of
time.
Letting t ! 1 in (4)-(7), we
find that for 0<y<a, v(y)
satisfies:
40
The Steady-State Problem
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The unique solution to (24)-(26) is v(y) ´ T0.
Because u(t,y) ! v(y) as t ! 1, if we let w(y,t) = u(y,t) – v(y),
it follows that w(y,t) ! 0 as t ! 1.
For this reason, we call w(y,t) the transient solution of (1)-(4).
41
The Transient Problem
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Substituting u(y,t)=w(y,t)+v(y) into (1)-(4), we obtain the
transient problem for 0<y<a, t>0:
where g(y) ´ f(y)-T0.
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To solve (27)-(30), we use separation of variables, or Fourier’s
method. With the assumption that w(y,t) = (y)T(t), (27)-(29)
become:
42
Separation of Variables
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Rearranging (31), we see that for all 0<y<a and for all t>0,
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For (35) to hold, both sides must have some common
constant value.
43
Separation of Variables (cont.)
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From (32) and (33) we obtain
boundary conditions in terms
of the product of (y) and T(t).
Equation (32) implies that
either T(t)=0 for all t or '(0)=0.
Because we want a non-trivial
transient solution, we choose
Similarly, (33) implies that
either T(t)=0 for all t, or
44
Separation of Variables (cont.)

Setting both sides of (35) equal to common constant A0, we
find that (y) and T(t) satisfy

The solutions to (39) are of the form

where  is some real constant.
The solution to (38) depends on the choice of constant A0 and
the boundary conditions (36), ’(0)=0, and (37)
45
Separation of Variables (cont.)
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To have nontrivial transient solutions that decay to zero as t !
1, A0 must be negative.
For convenience, we write A0=-2, with >0. Then (38) has
solutions of the form:
Since (36) implies ’(0) = 0, c2 = 0 must hold.
From the other boundary condition, (37), we see that
which means that either c1 = 0, leading to trivial w(y,t) ´ 0, or 
must satisfy:
46
Separation of Variables (cont.)

One can show that there are infinitely many
solutions to (43), with one in each interval:
where n is a positive integer.
47
Separation of Variables (cont.)

Thus to each positive integer n = 1, 2, 3, …, there
corresponds a solution to (27)-(29):
with n satisfying:
48
Separation of Variables (cont.)

Also, for each positive integer n, there is a
corresponding solution to (39) of the form

It follows that for each positive integer n, we have a
solution to (27)-(29) of the form
49
Separation of Variables (cont.)
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
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Note that the boundary value problem
is a special case of a more general problem known as a
regular Stürm-Liouville problem.
When a function (y) solves this type of problem for a certain
2, we call  an eigenfunction with eigenvalue 2.
The eigenfunctions and eigenvalues for this problem satisfy
the orthogonality condition:
50
Separation of Variables (cont.)

We now know what form solutions to the transient problem (27)-(29)
will take for integers n = 1, 2, 3, …

We still need to find a solution to the transient problem that will solve
initial condition (30), namely w(y,0) = g(y).
By the Principle of Superposition, any finite linear combination of
solutions of the linear homogeneous problem (27)-(29) will also be a
solution.
Let's suppose that a solution to (27)-(29) of the form


exists and see what conditions on the coefficients Bn should hold for
(50) to satisfy (30).
51
Separation of Variables (cont.)
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
Putting (50) into initial condition (30),
Fixing n, multiplying both sides of (51) by n(y) = cos(n y),
and integrating each side from 0 to a, the orthogonality
property (48) implies (formally)
The coefficients for each term of (51) can be determined by
solving (52) for Bn:
52
Separation of Variables (cont.)
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
Formally, the solution to the transient problem (27)-(30) is (50)
where Bn is given by (53) and n is a solution to (43), (44).
Since we assumed w(y,t) = u(y,t) – v(y), It follows that the
solution to the original problem (1)-(4) is
53
Making Things Rigorous

In order to justify our formal solution actually
converges and satisfies the IVPBVP (1)-(4), the
following theorems are useful!



Theorem 1: A convergence theorem for generalized
Fourier series expansions in terms of eigenfunctions.
Theorem 2: Weierstrass M-Test for uniform convergence
of an infinite series.
Theorem 3: Uniform convergence and differentiation of an
infinite series.
54
A Generalized Fourier Series
Convergence Theorem

Theorem 1 (From Powers, Boundary Value Problems): Let 1, 2, … be eigenfunctions of
the regular Stürm-Liouville problem on l<x<r:
in which the i’s and i’s are not negative, and not all zero. If f is sectionally smooth on
the interval l<x<r, then
where
Furthermore, if the series
converges, then the series on the RHS of (58) converges uniformly on l· x · r.
55
Weierstrass M-Test

Theorem 2 (From Rudin, Principles of
Mathematical Analysis): Suppose {fn} is a
sequence of functions defined on a set E, and
suppose
Then  fn converges uniformly on E if  Mn
converges.
56
Uniform Convergence and
Differentiation

Theorem 3 (From Olmstead, Advanced Calculus):
If  un(x) is a series of differentiable functions on
[a,b], convergent at one point x02 [a,b], and if the
derived series  u’n(x) converges uniformly on [a,b],
then the original series converges uniformly on [a,b]
to a differentiable function whose derivative is
represented on [a,b] by the derived series.
57
Model vs. Experimental
Results



Unfortunately, as we see in
Fig. 1, the model doesn't
match the experimental results
very well.
How far are we off? One
measure of the error is the
mean of the sum of the
squares for error (MSSE)
which is the average of the
sum of the squares of the
differences between the
measured and model data
values
MSSE for this model:



Top: 33.7 (oC)2,
Middle: 0.534 (oC)2,
Bottom: 0.930 (oC)2.
U
Model
1
14
12
10
8
6
4
2
50
100
150
200
t
Figure 1
Model: _________
Actual: …………….
58
Model with New Constants vs.
Experimental Results




Maybe the problem with our model
is our choice of the constant h,
which was chosen fairly arbitrarily.
With the choice of h=920.92 J/(m2
s oC), found by finding the best fit
with integer multiples of our initial
guess of h, we obtain the results
shown in Fig. 2 after recomputing
the values of n and Bn.
Although the model is significantly
closer to the experimental results
for the temperature at the top of
the thermos, the model is still off
by about the same amount for the
middle and bottom of the thermos.
MSSE for this model:



Top: 1.11 (oC)2,
Middle: 0.468 (oC)2,
Bottom: 0.929 (oC)2.
U
Model
2
14
12
10
8
6
4
2
50
100
150
200
t
Figure 2
Model: _________
Actual: …………….
59
Model With New Constants vs.
Modified Experimental Results



If we construct a new
piecewise linear f(y) and
compute n and Bn for our new
model data, we obtain the
results shown in Fig. 3 with
thirty terms in the sum for
u(y,t).
Visually, the model seems to
be closer to experimental
results, but the bottom and
middle model values are still
off.
MSSE for this model:



Top: 0.396 (oC)2,
Middle: 0.576 (oC)2,
Bottom: 0.487 (oC)2.
U
Model
3
12
10
8
6
4
2
50
100
150
200
t
Figure 3
Model: _________
Actual: …………….
60
Modified Model vs. Modified
Experimental Results



As before, we compute n and
Bn using the same  and h as
in the previous model. Fig. 4
shows our results.
Of the models we have tried,
this one provides the best
visual fit to the experimental
data.
MSSE for this model:



Top: 0.142 (oC)2,
Middle: 0.0664 (oC)2,
Bottom: 0.0449 (oC)2.
U
Model
4
12
10
8
6
4
2
50
100
150
200
t
Figure 4
Model: _________
Actual: …………….
61