Modeling Heat Flow in a Thermos

1
Modeling Heat Flow in a Thermos

• Michael A. Karls
• James E. Schershel
• Ball State University

2
The Coffee Cup Problem

• 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.

3
Newtons Law of Cooling

• 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.
• Newtons 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.

4
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
• k0 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

5
Verifying the Coffee Cup Model Experimentally

• 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
• pH
• Temperature
• Pressure
• Force
• Motion
• Using a CBL with a temperature probe, a TI-85
  calculator, and a program available from TIs
  website (http//education.ti.com/) , temperature
  data can be collected, plotted, and compared to
  the solution in (3).

6
A Modified Coffee Cup Problem

• 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!

7
The Thermos Problem

• 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.

8
Heat Flow in a Rod

• 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, well look at ice-cold
  water.

9
The Thermos Model

• Assume we have a thermos full of ice-cold water,
  open at the top.
• Since the water is ice-cold, well assume
• 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 lost to the air 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.

10
The Thermos Model (cont.)

• 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 ya 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
11
The Thermos Model (cont.)

• This simple system can then be modeled by the
  following initial value-boundary value problem
  for 00

• 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.

12
The Thermos Model (cont.)

• 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).

13
The Thermos Model (cont.)

• Using the technique of separation of variables or
  Fouriers 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

14
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
• 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

15
Experimental Procedure

• The experimental procedure is as follows
• 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 top-off
  the thermos with ice-cold water.

16
Experimental Procedure (cont.)

• 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.

17
Assigning Values to Coefficients in Our Model

• To compare our model to the experimental data, we
  need to specify the parameters in the model.
• 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. 3036
  (1999).

18
Initial Temperature Distribution

• 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.

19
Initial Temperature Distribution (cont.)

• 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.

20
Determining the Number of Terms in Our Model

• 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.

21
Graphical Comparison of Series Solution to f(y)
22
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.

Figure 1
Model _________ Actual .
23
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 h920.92 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.

Figure 2
Model _________ Actual .
24
Modified Experimental Procedure

• 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
• One bag of 100 cotton balls
• One pencil

25
Modified Experimental Procedure (cont.)

• The modified experimental procedure is as
  follows
• 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 and pencil in the freezer.
• Fill the thermos with cotton balls, remove the
  cotton balls and place them in the freezer.
• Fill the thermos with ice water and allow it to
  sit for roughly two hours to pre-chill.
• Remove the ice from the thermos and top-off the
  thermos with ice-cold water.

26
Modified Experimental Procedure (cont.)

• 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.

27
Modified Experimental Procedure (cont.)

• 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
  h1004.64 J/(m2 s oC) work well.

28
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.

Figure 3
Model _________ Actual .
29
A Final Modification to Our Model

• As a final modification of our model, suppose
  that we had started with a different initial
  temperature distribution, say f(y)A eB yC.
• 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,T),
  (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.

30
A Final Modification to Our Model (cont.)

• To find the unknown constants A, B, and C, we
  solve the system of equations

31
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.

Figure 4
Model _________ Actual .
32
A Final Test of Our Model

• 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
• 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.

Figure 5
Model Ave Temperature _________ Cold ? Temp 15 oC _________
33
Conclusions and Further Questions

• 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?

34
Conclusions and Further Questions (cont.)

• 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

35
Conclusions and Further Questions (cont.)

• where Ts is the unknown temperature at the
  surface of the ice-water and h is a
  proportionality constant.
• From Fourier's law of heat conduction, we see
  that for 0

which has the general solution
where A, Ts, and h/? are unknown constants.
Equation (23) is exactly the form of our final
choice of f(y)!!
36
Conclusions and Further Questions (cont.)

• 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.

37
Conclusions and Further Questions (cont.)

• In addition to the experiment we have outlined,
  other questions such as the following could be
  addressed.
• 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?

38
Solution of the IVBVP

• 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 0v(y) satisfies

• IVBVP for Thermos

39
The Steady-State Problem

• 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).

40
The Transient Problem

• Substituting u(y,t)w(y,t)v(y) into (1)-(4), we
  obtain the transient problem for 00

where g(y) f(y)-T0.

• To solve (27)-(30), we use separation of
  variables, or Fouriers method. With the
  assumption that w(y,t) ?(y)T(t), (27)-(29)
  become

41
Separation of Variables

• Rearranging (31), we see that for all 0for all t0,

• For (35) to hold, both sides must have some
  common constant value.

42
Separation of Variables (cont.)

• 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

43
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)

44
Separation of Variables (cont.)

• 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
45
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.
46
Separation of Variables (cont.)

• Thus to each positive integer n 1, 2, 3, ,
  there corresponds a solution to (27)-(29)

with ?n satisfying
47
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

48
Separation of Variables (cont.)

• 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

49
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).
50
Separation of Variables (cont.)

• 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

51
Separation of Variables (cont.)

• 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

52
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.

53
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
• in which the ?is and ?is are not negative, and
  not all zero. If f is sectionally smooth on the
  interval l
• where
• Furthermore, if the series
• converges, then the series on the RHS of (58)
  converges uniformly on l x r.

54
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.

55
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 ? un(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.

56
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.

Figure 1
Model _________ Actual .
57
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 h920.92 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.

Figure 2
Model _________ Actual .
58
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.

Figure 3
Model _________ Actual .
59
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.

Figure 4
Model _________ Actual .