SOLIDIFICATION OF WATER IN CONTACT
WITH A CHILL
George Downey
Eric Eddy
MEAE – 4960H02
Prof. Ernesto Gutierrez-Miravete
Due: May 10, 2001
TABLE OF CONTENTS
List of Symbols
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Introduction .
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Background .
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Exact Solution (G.Downey).
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Problem Description
Assumptions
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Scope
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Mathematical Formulation
Governing Equations
Neumann’s Solution
Error Function
Solution methods .
Particular Solution Results
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Numerical Solution (E.Eddy). .
Mathematical Formulation.
Problem Description
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Numerical Solution Results.
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Error Analysis.
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Discussion .
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Conclusions
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References .
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Appendix (1-11).
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20-47
Program Names:
Exact
Bisection.f
Secant.f
Numerical
10steps.f
20steps.f
(coarse mesh algorithm)
(fine mesh algorithm)
1
List of Symbols
tc, K1,2, - thermal conductivity (w/m* oK)
rho,p - density (kg/m3)
cp,c1,2 - specific heat (J/kg*oK)
tfus,T1 - melting/freezing temperature (oK)
hfus - heat of fusion (J/kg)
xtot - total length of slab (m)
tsurf - solid's temperature (oK)
tint,V - liquid's initial temperature (oK)
td,k1,2 - diffusivity (m2/s)
dx - distance between node points on x scale (m)
dt - time between node points on time scale (s)
to - initial temperature (oK)
ho - initial enthalpy (J/kg)
ix - step parameter
nxp1 - parameter
x - position along slab (m)
timet - total time considered (s)
nt - number of time steps
erf(x) – Error function
 - Numerical constant
k - thermal conductivity (w/m* oK)
H - enthalpy (J/kg)
T - temperature (oK)
Cp,s - specific heat of solid (J/kg oK)
Cp,l - specific heat of liquid (J/kg oK)
Tm - melting/freezing temperature (oK)
 s - diffusivity of solid (m2/s)
 l - diffusivity of liquid (m2/s)
 - density (kg/m3)
L - latent heat (J/kg)
t - time (s)
2
Introduction
Substance transformation has been the subject of study since before recorded
history. One of the more easily seen transformations is solidification from a melt. This
can be readily seen when water freezes to ice or forms ice crystals on your window
when it is cold outside. The study of solidification has been of long interest to people
beyond human curiosity. The study of solidification in metal casting dates back as far
5000 B.C. in the near east1. In the early 1800’s, Stefan published the first discussion of
solidification on the subject of polar ice cap formation 2.
The impact of solidification/melting is still finding new applications in the 21st
century. Due to economic factors, there has been continuous development in process
optimization for casting materials such as plastics and metals. Extensive research has
been conducted for the casting of gas turbine blades, which are exposed to a
combination of cyclic loading and high temperatures. To achieve the properties
necessary in this application the processing requires special conditions for directional or
single crystal formation. On a more microscopic scale is the development of
nanophase materials, which show improved property performance in the mechanical
and electrical industry.
The analytical study of substance transformation or more specifically melting and
solidification process as mentioned before dates back to the early 1800’s. Analytical
techniques such as Neumann method have been used in a wide range of applications.
One area that the application of these solution methods is important is the solidification
process of castings. The rate and direction of solidification is involved in the commonly
used design tool, the casting modulus, which is use determine what portion of the
casting will solidify last in a mold.
However there are many cases where the exact solution is not available or
difficult to derive. These more complex problems are best solved numerically. In heat
transfer one of the commonly used numerical methods for solution is the finitedifference method. This method was originally limited due to the manpower required to
perform the calculations by hand. With the advent of computers, a finer mesh size
could be achieved providing greater accuracy. The improved accuracy and ease of use
has evolved from home created programs to standard commercial software.
The heat transfer effect of the mold has a significant effect on the structure and
properties of a material molded. Because this process is so important, analytical
studies and numerical analysis have resulted in the development and application of
software to help predict the solidification rate and temperature distribution of a melt.
The mold surface or any other surface, which provides conductive directional heat flux
away from the melt, is considered a chill. Solidification of water, in contact with a chill,
to form ice is another area that has been studied and is the subject of this paper.
1
2
Class notes from MEAE 6960, Casting and Joining of Materials.
Carslaw & Jaeger, “Conduction of Heat in Solids”, 2nd ed.
3
Background
An understanding of the solidification process is necessary before getting into
the analytical problem and solution method of solidification. Solidification is a change of
state or phase change where by there is an enormous and abrupt decrease in atomic
mobility3. The laws of thermodynamics and kinematics govern this process of changing
state. Before a liquid can form into a solid, it first must release all the latent heat stored
in the liquid. This stored energy is called latent heat of fusion 4 and is an important part
of the phase change process. Once the latent heat is released, the liquid will begin to
nucleate and solidify. Nucleation depends on the cohesive and adhesion forces
between the clusters of atoms. Gibbs-Thomson undercooling explains the energy
associated with interfacial tension, which is an activation barrier to nucleation formation.
As pure materials solidify, a solid/liquid interface is created and propagates until
no more liquid remains. Solidification usually occurs at surfaces or chill locations where
the surface provides a nucleation site or a positive temperature gradient such that the
latent heat generated at the interface is dissipated through the solid. A negative
temperate gradient can exist at the solid/liquid interface if the liquid surrounding the
solid is undercooled, such a condition can exists in the center of a melt 5 but will not be
considered for this project.
EXACT SOLUTION
Problem Description/Mathematical Formulation of Problem
The following are the assumptions used in formulating this problem.
Assumptions

A pure substance material with a specific melting point will be considered rather
than a material which freezes over a range of temperatures.

Constant density - no volume changes during phase transformation.

Chemical interactions in the liquid composition at interface will be neglected.

Constitutional supercooling will be neglected.

Since we are focusing on the solidification of a melt with a chill the temperature
gradients in the liquid will be positive such that the latent heat generated at the
interface is dissipated through the solid.
3
ASM Metals Handbook, Vol 15, Basic Concepts in Crystal Growth and Solidification.
The heat effects accompanying changes in state at constant pressure and no temperature change.
5
Ibid, pg. 115.
4
4
The process of solidifying/melting involves a moving interface, which separates the
solid region from the melt. This interface progressively continues until no liquid
remains. The progression of solidification is not a linear problem but for such problems,
an exact solution can be created. The exact solution to this problem will be formulated
using Neumann's method.
Neumann's Method for Semi-infinite region
For the case of semi-infinite region, a surface x = 0 will be maintained at zero
temperature. For the region x > 0 the temperature will be initially at a constant
temperature V which is greater than the melting point, T1.
X(t) will be the mathematical representation the surface of separation between
the solid and liquid phases as a function of time. x will be any distance from the surface
maintained at zero temperature and for x < X(t) contains solid at temperature v1
conversely x > X (t) contains liquid at temperature v2.
Boundary Conditions
Boundary condition for separation:
When x = X (t) , v1=v2=T1
Boundary condition for latent heat at liquid/solid interface:
K1
Where
v1
v
dX
 K 2  L
x
x
dt
Eq. 1
dX
can be considered the velocity of the moving boundary
dt
The amount of energy released during the solidification process is;
Ldx per unit area, where dx is the progression of surface of separation
This energy is subsequently removed by conduction of the chill to allow solidification to
continue.
Boundary conditions for linear heat flow though the material is given by:
 2 v1 1 v1

 0,
x 2 k1 t
Eq. 2
 2 v 2 1 v 2

0
x 2 k 2 t
Eq. 3
5
Neumann’s solution for solidification of a melt
For region x > 0 initially liquid is at a constant temperature V ;
x = 0 maintained at zero temperature
The solution to Eq. 2 & 3 involves the error function, which is the common solution form
for many simi-infinite heat transfer problems.
erfx 
2
x
e
 
 2
d
Eq. 4
0
And erf  1 and erf(-x)= -erfx
Approximations of the erfx for small values of x the series for e is used in Eq. 4 to
create the following series.
2
erfx 
Eq. 5
x

n 2n
  d   1  / n!
2
0
n0
Since this series is uniformly convergent, the integrated terms can be summed term by
term 6.
2  (1) n x 2 n 1
erfx 
Eq. 6

 n  0 (2n  1)n!
The following is a graphical representation of these values of erfx and erfc and match
with existing math tables. It can clearly be seen that the function approaches 1 as the
infinite solution. The solution values obtained for the series can be seen in Appendix 1.
x vs. erf x and erfc x
1.2
erf x & erfc x
1
erf x
0.8
0.6
0.4
erfc x
0.2
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
x value
Figure 1
For large values of x a single integration by parts of the error function is required.
However, during solidification of water to ice the values of , or x in this discussion, are
6
Conduction of Heat in Solids, Appendix II
6
often small. Large values of x should not be the case for this problem 7. The values for
larger x values can also be seen in the error function table in Appendix 1.
Knowing that Eq. 6 satisfies the differential equation, the initial and boundary conditions
are applied.
Using the above Boundary Conditions and the error function, an expression for the
solution to the linear heat flow equations is:
v1  Aerf
x
2(k1t )1 / 2
v 2  V  Berf
and
x
2(k1t )1 / 2
Eq. 7 & 8
Since X is proportional to t1/2 :
X  2 k1t 
1/ 2
 is a numerical constant that must be determined by:
K k 1 / 2 (V  T1 )e  k1 / k 2
e 
L 1 / 2
 2 11/ 2

erf K 1 k 2 T1erfc (k1 / k 2 )1 / 2
c1T1
2
Eq. 9
With  being solved; the v1 and v2 relationships can be rearranged to yield:
v1 
v2  V 
T1
x
erf
erf
2(k1t )1 / 2
(V  T1 )
x
erfc
1/ 2
erfc (k1 / k 2 )
2(k 2 t )1 / 2
Eq. 10
Eq. 11
Where 1 is the temperature of the solid and 2, is the temperature of the liquid portion
of the melt. The temperature can be determined given an initial melt temperature V,
position x and time t. The temperature of the liquid portion of the melt will be evaluated
as a check for the numerical solution of the temperature profile during solidification.
This particular solution will need the initial conditions applied and will be evaluated
numerically.
To solve the exact solution formulated by Eq. 11, values for  must be determined
numerically. Once defined, Eq. 11 will be solved numerically with the following
methods:
7
Heat Transfer pg 185
7
 Bisection method (Appendix 2a)
The bisection method repeats halving subintervals and at each step
locating the half containing the solution. One advantage of the bisection
method is that it is simple and should always converge to a solution.
 Secant method (Appendix 2b)
An improvement over the Bisection method is that the Secant method
converges much faster than functional iteration. This method requires
two initial approximations and is similar to Newton’s method but does not
require the derivative evaluation.
The solutions to these methods will be compared to a numerical result obtained from
Matlab. A comparison of these solution methods will be compared in the results
section. The most accurate solution will be used to calculate the melt temperatures as
a function of time and will be the base line for the solution of the partial differential
equation using the fixed grid method.
Particular Solution Results for Neumann’s Solution
Inputs
The case considered is the solidification or formation of ice expose to a chill. The
parameters and inputs used for this problem are as follows:
Liquid Phase (Water)
 , Density
c1, specific heat
K1, conductivity
k1, diffusivity
V1, Initial Temperature
T1, melting temperature
L, Latent heat of fusion
Units
kg/m3
J/kg*K
W/M*K
m2/s
K
K
J/kg
Value
Solid Phase (ICE)

1000
2060
c2
2.218
K2
0.115E-5
k2
283.10
243.14
335,000
Value
1000 (.917 theo.)
4186
0.6025
0.144E-6
One unique change in the formation of water into ice is the reduction in density. This
density reduction causes ice to float in its liquid form. However, for simplification of this
analysis we will consider the density to remain constant
In order to solve the Neumann’s solution, the numerical value  will need to be solved
with the given initial conditions and material parameters.
K k 1 / 2 (V  T1 )e  k1 / k 2
e 
L 1 / 2
 2 11/ 2

erf K 1 k 2 T1erfc (k1 / k 2 )1 / 2
c1T1
2
Eq. 12
Although this is an analytic solution, it will be solved numerically for values of  . First
the equation is rearranged to solve for zero.
8
e
K k 1 / 2 (V  T1 )e  k1 / k 2 L .5
 2 11/ 2

0
erf K1k2 T1erfc (k1 / k2 )1 / 2
c1T1
2
Eq. 13
The roots of this equation can be obtained using software such as Matlab or numerical
root finding algorithms such as Bisection and Secant solution methods. The solution of
Eq. 13 is solved using Matlab and produces a value of 0.75331 for the given conditions
(see Appendix 3 for inputs and outputs.).
Results
 Matlab Inputs
EDU» syms M K1 K2 k1 k2 V T1 c1 L
EDU» K1=2.218,K2=.6025,k1=.00000115,k2=.000000144,V=283.10,T1=273.15,c1=2060,L=335000
 Equation 13 in Matlab syntax
EDU» f = exp(-M^2)/erf(M)-K2*k1^.5*(V-T1)*exp(-k1*M^2/k2)/(K1*k2^.5*T1*(1-erf(M))*(k1/k2)^.5)M*L*pi^.5/(c1*T1)
 Matlab Solution
EDU» solve(f)
ans =
.75330884038373000039945584175972 (.75331 rounded to 5 places)
TABULATED RESULTS FOR NUMERICAL ALGORITHMS
Method
Endpoints/Initial guess
Tolerance
Solution
% Difference from Matlab
Number of Iterations
Bisection
-1 1
.0001
.75311
.026%
15
Secant
.2 1
.0001
.75316
.020%
6
Matlab
.75331
N/A
1
As seen in the tabulated results above the numerical algorithms agree closely
with the solution obtained using the Matlab software. The Bisection method converged
to a good solution after 15 iterations. The difference between them can be due to
approximation and round off error but also due to the fact that the Bisection method can
discard a good intermediate approximation. As expected Secant's method converged
fastest due the quadratic convergence and was more accurate than the Bisection
method. Newton’s method could have been used which may have provided an even
better solution but solving and working with the derivative is a bit cumbersome and not
necessary for the degree of accuracy obtained by these simpler methods. In general,
the amount of difference between the Bisection and the Secant methods can be
considered negligible and of little impact on the solution accuracy compared to other
factors in this problem.
9
Matlab Validation
Using Matlab to solve for  introduces techniques that are unknown to the user so a
quick check for the solution value would be prudent.
Using the values presented in “Conduction of Heat in Solids” (see Appendix 4 for
conditions and material properties). The value of T1 =1 and V-T1 = 2 were chosen and
evaluated. The results was a  of 0.056 which is close to the 0.053 value presented in
solution table in the Appendix 5.
Evaluating Neumann’s Solution
With the values of  known a spreadsheet can be created to solve Neumann’s solution
for any time, t and position, x can be seen in Appendix 6.
Neumann’s Validation
To check the spreadsheet the special case of the initial temperature at or close to the
melting point will be considered. For this case Eq. 13 is reduced to the following:
e 
 L  .5

0
erf
c1T1
2
Eq. 14
The result of imputing the material constant values along with  and T1 as conditions
should be zero. Below is a few data points showing that these numbers indeed result in
a near zero solution.
T1=1
=2
=3
V-T1=0
.053
-----
1
--.077
---
2
----.093
Error
.000077
.000222
.000592
The error seen here in solving for Eq. 13 and for solving Eq. 12 (see Appendix 6) is
probably due to rounding.
Exact Solution Results
With  solved for with the given initial conditions of this problem (V,=283.1 K and
T1=273.15 K) the spreadsheet will be used to evaluate the temperature profile at the
following points (x = 0.02m, x = 0.04m, x = 0.06m). These points represent positions
near the beginning middle and end of the slab evaluated by the numerical solution.
These points will the base line values and will be mapped against the numerical
solution in the Error Analysis section.
10
Figure 2
According to the exact solution methods these values decline but not to the
solidification temperature of 273.15 K in the timeframe considered. Since this melt
remains a liquid, Eq. 11 dictates the temperature at a position, x for a time, t. However,
if the melt did solidify (i.e. temperature at or below 273.15 K), which would be expected,
Eq. 10 would have to be applied for the times after which the liquid formed a solid. This
is necessary since the heat transfer characteristics are different for ice than for water.
The solid (ice) has a lower conductivity and diffusivity values (see input properties) and
would expect a slower rate of heat transfer from the liquid. The graph does show a
decrease in temperature toward the temperature of the chill. The graph also shows a
decrease in the rate of cooling for positions further way from the chill. Unfortunately,
since the liquid did form a solid a boundary graph could not be constructed because the
melt does not solidify. Had solidification been determined a graph of the boundary
position would be created and used as a comparison to the numerical solution method.
Although the graphed values look reasonable, the results do not seem to be
consistent with the numerical result and intuition. Also when using greater values of t
the melt reduces in temperature but does seem to reach the solidification temperature.
This discrepancy will be discussed more in the Discussion and Conclusion sections.
NUMERICAL SOLUTION
Problem Description / Mathematical Formulation of Problem
The solidification rate of a specified material is determined by locating the
boundary position of the solid/liquid interface as a function of time. This rate is
dependent on the material's properties and initial conditions that were stated earlier. In
11
this analysis, the direction of the solidification is being forced as a constant temperature
is held at one end of the material's volume while the liquid portion is given an initial
temperature. The one-dimensional analysis allows the system to be solved in mesh
fashion numerically by solving for the temperature at any mesh point at specific times.
The temperature of the mesh point can easily be compared to the material's melting
point to determine what phase it is in at that time. Knowing the temperature at two
adjacent mesh points will narrow the location of the phase boundary to a point where
simple interpolation can be effective to further refine the exact location.
It is desirable to combine the heat equations for both the solid and liquid phases
to simplify and minimize the calculations. The combined heat equation appears as:
  T 
H
k

x  x 
t
Eq. 15
Equation 15 is a partial differential equation in terms of temperature, T, time, t,
and enthalpy, H. During a phase change, latent heat must be considered so the
enthalpy results as a step function of temperature.
H  C p , s T  Tm ; T  Tm 
H  C p ,l T  Tm   L; T  Tm 
Eq. 16
This can be more easily seen in Figure 3. The enthalpy vs. temperature
relationship provides a means of numerically relating the two values. The numerical
solution will calculate the enthalpy and determine the respective temperature at that
point on the graph.
Figure 3
12
A fixed grid numerical solution method is taken to solve this solidification
problem. The mesh is divided into N+1 nodes with a spacing of x . Nodes are located
at xi where i=1,2,3,…,N+1. A time mesh is also divided into tj nodes where j=1,2,3,…
The spacing between the time nodes is t .
The finite difference method is used with the fixed grid to calculate the enthalpy
at each node point in the mesh explicitly in time from the previous calculation. The
corresponding temperature is calculated based on the relationship of Eq.16. If many
nodes are used, the calculations can easily become time consuming and laborious. An
algorithm written in Fortran is useful to expedite the process (See Appendix 7).
The algorithm is hard coded with the material constants such as the melting
temperature, heat of fusion, etc. Mesh parameters such as total distance of the slab
and total time considered follows. The algorithm then calculates the step size, time
step and number of steps. An initial temperature and heat of fusion is calculated at
each x node. The first node is forced to be the initial and constant temperature that
was specified as the solid's temperature.
A loop is then started that is based on the number of time steps required. Within
the loop, the enthalpy for each node is calculated based on the explicit finite volume
scheme developed from Eq. 17.
ki   1 , j
2
Ti 1, j  Ti , j
x
 ki  1 , j
2
x
Ti , j  Ti 1, j
x
 i , j
H i , j 1  H i , j
t
Eq. 17
At this point, the position of the enthalpy must be determined based on the step
function of Eq. 16. Based on the position of the enthalpy calculation, h(ix), a
determination of the u(ix) can be made. An excerpt of the algorithm is shown here:
do ix=1,nxp1
if(h(ix).gt.hfus) then
u(ix) = td*(h(ix)-hfus)
t(ix) = tfus+u(ix)/tc
endif
if(h(ix).le.hfus.and.h(ix).ge.0.0) then
u(ix) = 0.0
t(ix) = tfus
endif
if(h(ix).lt.0.0) then
u(ix) = td*h(ix)
t(ix) = tfus+u(ix)/tc
endif
end do
Each of the three "if" statements above corresponds to one of the three lines on
Figure 3. The u(ix) term is calculated based on the slope of the respective line and
forces the t(ix) to be calculated differently.
13
Finally, the algorithm outputs the position and resulting temperature. It then
updates the new initial enthalpy to that which was calculated previously. The process is
repeated for all remaining time steps adhering to the CFL condition and using the
smaller  of liquid or solid:
x 2
Eq. 18
t 
2
A sample output can be found in Appendix 8.
Numerical Solution Results
A mesh system was set up in the algorithm with the following constants for water
as stated earlier. The physical model of the slab consisted of a total length of 0.100
meters divided into eleven nodes. One end of the slab was held at a constant
temperature of 243.15 K while the liquid portion was initially at 283.10 K. Adhering to
Eq. 18, the time interval was calculated in increments of 94.3643 seconds over 39
steps. This produced a final elapsed time of 3680.21seconds. The complete algorithm
output can be found in Appendix 8. The surface model depicting the temperature
distribution of time is shown in Figure 4.
Temperature Distribution Over Time
285
280
275
265
260
255
250
245
Figure 4
14
0.000
3680.210
3491.480
3302.750
3114.020
2925.290
2736.560
2547.840
0.050
2359.110
1981.650
1792.920
2170.380
Time (s)
1604.190
1415.460
1038.010
0.100
1226.740
849.279
471.821
660.550
94.364
240
283.093
Temperature (K)
270
Distance from
Chill (m)
The 3D surface representation of temperature distribution over time offers
significant information about the dynamics of the system. The rate at which each
distance from the chill solidifies can easily be seen. The freezing temperature of
273.15 K appears in the navy blue section of the surface model. As this blue portion
stretches across the entire 3680.21 seconds, the first nine distance intervals pass
through indicating that they have frozen.
The temperature across the body at the first time interval has a very steep slope
but then levels off drastically verses the last time interval where the slope is much more
gradual. The peak temperature always appears at the last node which is furthest away
from the chill. The slope of this line becomes steeper as time passes and could be
expected to approach the initial surface temperature if a longer time period was
observed.
The purpose of this analysis is to locate the solid / liquid boundary position over
a period of time. The mesh method described earlier indicated that the temperature at
specific nodes would be calculated at time intervals. A visual aid is shown in Figure 5
for the last time interval.
Figure 5
Temperature is calculated at each of the 11 nodes at every time interval but the
temperature of fusion may not land on a specific node point. In this case interpolation
between node points on either side of the freezing temperature will be the best estimate
for the actual boundary position. This is most effective and gives the most accurate
estimate when the temperature slope is mild. Referring to Figure 4, the slope is the
least steep at the last time interval.
The freezing temperature was interpolated at each time interval from the raw
data of Appendix 9. To refine the estimate of the actual boundary position as a function
of time, a "Fine Mesh" was established using the same conditions. The algorithm was
15
changed by increasing the parameter, nxp1, from 11 to 21. This essentially doubled
the number of nodes where the temperature would be calculated. In changing the
parameter to 21, the time increment was reduced to 23.5911 seconds and increased
the number of steps to 153. This significantly increased the data to be analyzed. A
comparison between the coarse mesh (10 increments) and fine mesh (20 increments)
was conducted to see if the results differed significantly. These results can be found in
Figures 6 and 7.
Boundary Position vs. Time
Coarse and Fine Mesh
0.100
0.090
0.080
0.070
Position (m)
0.060
Coarse Mesh (10
increments)
0.050
0.040
Fine Mesh (20
increments)
0.030
0.020
0.010
0.000
0
500
1000
1500
2000
2500
3000
3500
4000
Time (s)
Figure 6
Temperature Profile Comparison
at 3585.84 seconds
280
275
Temperature (K)
270
265
Coarse Mesh (10
increments)
260
Fine Mesh (20
increments)
255
250
245
240
0.000
0.020
0.040
0.060
0.080
Distance from Chill (m)
Figure 7
16
0.100
0.120
Figure 6 shows the boundary position as a function of time for the coarse and
fine mesh arrangement. The shape of each line shows the same trend with a steeper
slope at the beginning and end with a much milder center section. This indicates that
the boundary layer grew quickly early and late in the time window. Figure 7 shows the
same correlation between the temperature profiles at the last common time increment
of 3585.84 seconds. Both temperature profiles have a fairly constant slope and then
level off near the end of the slab. The fine mesh shows a slightly higher temperature
which correlates with the later boundary position in Figure 6.
Error Analysis
The bases of the numerical solution is using finite differences between nodes
given a specific mesh size. The exact solution is based off the solution of an exact
equation. Since the basis, for the solutions are different an error is expected. To
evaluate this error a comparison plot of the two methods is generated at positions
0.02m 0.04m and 0.06m.
Exact and Numerical Temperature Profiles
Temperature, K
285
280
275
270
265
260
255
250
0
500
1000
1500
2000
2500
3000
3500
4000
Tim e (seconds)
0.02m Postion 'Exact Solution'
0.04m Position 'Exact Solution'
0.06m Position 'Exact Solution'
0.02m Position 'Numerical Solution'
0.04m Position 'Numerical Solution'
0.06m Position 'Numerical Solution'
Figure 8
A plot of the temperature profiles for the chosen positions can be seen in figure
8. It can be seen that numerical solution is approaching the chill temperature of 248.1
K more rapidly than the Neumann’s or exact solution. The rate of cooling seems to
decrease for positions further away from the chill. The cooling rate trends of the two
methods are similar and the error seems to range with a maximum value of about 20
degrees between the temperature profiles at a given position. The correlation seems to
be consistent but not as close as expected.
17
Discussion
The numerical solution used a slab size and time window that depicts the trends
of interest. We initially wanted to validate the numerical solution based on the exact
method called Neumann’s solution. A further study was conducted to show the
differences between numerical models built with different mesh sizes. Figures 6 and 7
both show that the temperature increases as a function of distance but levels out as the
entire slab cools. This is because of change in temperature is reduced as the boundary
position approaches. The boundary position moves rapidly early in the time window
because it is closest to the chill. The temperature reduces rapidly and the boundary
layer moves quickly. The layer levels off and then increases again as the rest of the
slab cools. The furthest distance from the chill only loses temperature in the direction
of the chill. Since there is no more water to cool, the thermal energy dissipates faster
and freezes quickly.
It seems that the numerical and exact solution methods match more closely for
regions of x closer to the chill but in general, the exact solution indicates a much more
gradual solidification rate than the numerical solution. This could be due to error in the
exact solution method. Although an analytic solution method is used to solve this
solidification model there will be error in the solution due to the numerical method used
to solve for the constant . However, the error is significant enough to indicate that
there is a problem. Investigation of the exact solution showed no apparent flaws and
validation confirmed that the formulated solution method works well compared to known
values presented in “Heat Conduction in Solids”. The error is not known but using a
simple lab experiment would help identify which values we should believe and help in
improving the error or mistake.
Also, it should be mentioned that both the exact and numerical methods use
perfectly constant parameters over all temperature ranges, which is not the case in
reality. Most common available references publish an average value for the constants,
which for the purposes of this report work satisfactorily. The assumption was made that
the density remains constant during the phase change, which from a mass transfer
standpoint would play a roll in the total energy required to be released during the
solidification process.
Conclusions
For the solidification to occur the latent heat of fusion must be extracted from the
liquid to drive the melt to freeze. This heat is removed only from the chill, which
maintains its initial temperature below the freezing point of the water. This positive
thermal gradient ensures that solidification will continue.
The numerical results show that water transforms to ice quickly due to the cold
chill. The entire 0.1m slab was frozen after being exposed to the chill for 3600 seconds.
The boundary velocity moved rapidly at the beginning and end of the time sequence but
appeared to level off to a constant velocity of 0.014 mm/s at the center. During this
evaluation there was not a significant difference in the calculations between the coarse
18
and fine mesh sizes. Although the fine mesh should produce a more accurate result, it
seems more logical to use the simplest method for the approximation.
The results of both methods show a non-linear decrease in temperature for
increased time values. It would have been expected that the numerical solution method
and the Neumann’s exact solution methods match more closely despite the number of
influences involved. Although Neumann’s solution was validated for this problem, the
temperature profiles created are significantly different from the numerical solution and
intuitively would expect a greater decrease in temperature. Had a lab experiment been
conducted both temperature profiles should approach the chill temperature. This
difference could be due to a calculation error that was not found and partly due to the
fact that a numerical solution was used to solve  which is subject to rounding errors.
Unfortunately, it was not possible to solve the exact solution without the assistance of a
numerical solution making it not wholly exact.
A practical experiment should be conducted, as good practice and to further
investigate the differences between the numerical solution and exact solution methods.
The models of this report are limited in application because the heat is only dissipated
from the chill surface. This is not the case for most everyday situations and would only
be possible only with perfect thermal insulation on all other surfaces.
References
http://www.rh.edu/~ernesto/computer_programs/freezing/sl.f
http://www.rh.edu/~ernesto/C_S2000/cht/Notes/cht11.html
Mills, Anthony F. Heat Transfer. Richard D Irwin, Inc. 1992. p. 185
Carslaw, H.S. and Jaeger, J.C. Conduction of Heat in Solids. 2nd edition.Oxford
University Press. 1959. p. 282-291, 482, 483, 485, 497
19
APPENDIX 1
n
Inputs
0.4
2/sqr(pi)=
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
1.1283792
erf(x)
Summed n "values"
0.451352
0.427280
0.428435
0.428391
0.428392
0.428392
0.428392
0.428392
0.428392
0.428392
0.428392
0.428392
0.428392
0.428392
0.428392
0.428392
0.428392
0.428392
0.428392
0.428392
n value
0.451352
-0.024072
0.001155
-4.4E-05
1.37E-06
-3.59E-08
8.09E-10
-1.6E-11
2.83E-13
-4.5E-15
6.51E-17
-8.65E-19
1.06E-20
-1.21E-22
1.29E-24
-1.28E-26
1.21E-28
-1.07E-30
9E-33
-7.19E-35
x
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
x vs. erf x and erfc x
1.2
1
erf x & erfc x
erf(x)
erf x
0.8
0.6
0.4
erfc x
0.2
0
0
0.2
0.4
0.6
0.8
1
X value
20
1.2
1.4
1.6
erfx
0
0.1125
0.2227
0.3286
0.428392
0.5205
0.603856
0.677801
0.7421
0.7969
0.8427
0.880205
0.910314
0.934008
0.952285
0.966105
erfc
1
0.8875
0.7773
0.6714
0.5716
0.4795
0.3961
0.3222
0.2579
0.2031
0.1573
0.1198
0.0897
0.066
0.0477
0.0339
APPENDIX 2a
BISECTION METHOD
C***********************************************************************
C
BISECTION ALGORITHM 2.1
*
C
*
C***********************************************************************
C TO FIND A SOLUTION TO F(X)=0 GIVEN THE CONTINOUS FUNCTION
C F ON THE INTERVAL , WHERE F(A) AND F(B) HAVE
C OPPOSITE SIGNS:
C
C INPUT: ENDPOINTS A,B; TOLERANCE TOL;
C
MAXIMUM INTERATIONS N0.
C
C OUTPUT: APPROXIMATE SOLUTION P OR A
C
MESSAGE THAT THE ALGORITHM FAILS.
C
CHARACTER NAME1*14,AA*1
INTEGER OUP,FLAG
LOGICAL OK
REAL A,B,FA,FB,X,TOL
INTEGER N0
C DEFINE F
f(x)= EXP(-x**2)/ERF(x)-.000335415*EXP(-7.986111111*x**2)/
&(.229902658.229902658*ERF(x)*6.782329983*
&1.055240177*x)
C f(x)= exp(-x**2)/erf(x)-.001326457*exp(-7.986111111*x**2)/
C&(.204652504-.204652504*erf(x))*6.782233-1.185436374*x
c
OPEN(UNIT=5,FILE='CON',ACCESS='SEQUENTIAL')
c
OPEN(UNIT=6,FILE='CON',ACCESS='SEQUENTIAL')
WRITE(6,*) 'This is the Bisection Method.'
WRITE(6,*) 'Has the function F been created in the program? '
WRITE(6,*) 'Enter Y or N '
WRITE(6,*) ' '
READ(5,*) AA
IF(( AA .EQ. 'Y' ) .OR. ( AA .EQ. 'y' )) THEN
OK = .FALSE.
10
IF (OK) GOTO 11
WRITE(6,*) 'Input endpoints A < B separated by blank '
WRITE(6,*) ' '
READ(5,*) A, B
IF (A.GT.B) THEN
X=A
A=B
B=X
ENDIF
IF (A.EQ.B) THEN
WRITE(6,*) 'A cannot equal B '
WRITE(6,*) ' '
ELSE
FA = F( A )
FB = F( B )
IF ( FA * FB .GT. 0.0 ) THEN
WRITE(6,*) 'F(A) and F(B) have same sign '
WRITE(6,*) ' '
21
11
12
13
14
15
ELSE
OK = .TRUE.
ENDIF
ENDIF
GOTO 10
OK = .FALSE.
IF (OK) GOTO 13
WRITE(6,*) 'Input tolerance '
WRITE(6,*) ' '
READ(5,*) TOL
IF (TOL.LE.0.0) THEN
WRITE(6,*) 'Tolerance must be positive '
WRITE(6,*) ' '
ELSE
OK = .TRUE.
ENDIF
GOTO 12
OK = .FALSE.
IF (OK) GOTO 15
WRITE(6,*) 'Input maximum number of iterations '
WRITE(6,*) '- no decimal point '
WRITE(6,*) ' '
READ(5,*) N0
IF ( N0 .LE. 0 ) THEN
WRITE(6,*) 'Must be positive integer '
WRITE(6,*) ' '
ELSE
OK = .TRUE.
ENDIF
GOTO 14
CONTINUE
ELSE
WRITE(6,*) 'The program will end so that the function F '
WRITE(6,*) 'can be created '
OK = .FALSE.
ENDIF
IF (.NOT.OK) GOTO 40
WRITE(6,*) 'Select output destination: '
WRITE(6,*) '1. Screen '
WRITE(6,*) '2. Text file '
WRITE(6,*) 'Enter 1 or 2 '
WRITE(6,*) ' '
READ(5,*) FLAG
IF ( FLAG .EQ. 2 ) THEN
WRITE(6,*) 'Input the file name in the form - '
WRITE(6,*) 'drive:name.ext'
WRITE(6,*) 'with the name contained within quotes'
WRITE(6,*) 'as example: ''A:OUTPUT.DTA'' '
WRITE(6,*) ' '
READ(5,*) NAME1
OUP = 3
OPEN(UNIT=OUP,FILE=NAME1,STATUS='NEW')
ELSE
OUP = 6
ENDIF
WRITE(6,*) 'Select amount of output '
22
WRITE(6,*) '1. Answer only '
WRITE(6,*) '2. All intermediate approximations '
WRITE(6,*) 'Enter 1 or 2 '
WRITE(6,*) ' '
READ(5,*) FLAG
WRITE(OUP,*) 'BISECTION METHOD'
IF (FLAG.EQ.2) THEN
WRITE(OUP,004)
004
FORMAT(3X,'I',15X,'P',12X,'F(P)')
ENDIF
C STEP 1
I=1
C STEP 2
016 IF (I.GT.N0) GOTO 020
C
STEP 3
C
COMPUTE P(I)
P=A+(B-A)/2
FP=F(P)
IF (FLAG.EQ.2) THEN
WRITE(OUP,005) I,P,FP
005
FORMAT(1X,I3,2X,E15.8,2X,E15.8)
ENDIF
C
STEP 4
IF( ABS(FP).LE.1.0E-20 .OR. (B-A)/2 .LT. TOL) THEN
C
PROCEDURE COMPLETED SUCCESSFULLY
WRITE(OUP,002) P, I, TOL
GOTO 040
ENDIF
C
STEP 5
I=I+1
C
STEP 6
C
COMPUTE A(I) AND B(I)
IF( FA*FP .GT. 0) THEN
A=P
FA=FP
ELSE
B=P
FB=FP
ENDIF
GOTO 016
020 CONTINUE
C STEP 7
C PROCEDURE COMPLETED UNSUCCESSFULLY
IF(OUP.NE.6) WRITE(6,3) N0,P,TOL
WRITE(OUP,3) N0,P,TOL
040 CLOSE(UNIT=5)
CLOSE(UNIT=OUP)
IF (OUP.NE.6) CLOSE(UNIT=6)
STOP
002 FORMAT(1X,'THE APPROXIMATE SOLUTION IS',/,1X
*,E15.8,1X,'AFTER',1X,I2,1X,'ITERATIONS, WITH TOLERANCE'
* ,1X,E15.8)
003 FORMAT(1X,'ITERATION NUMBER',1X,I3,1X,'GAVE APPROXIMATION',
*/,E15.8,1X,'NOT WITHIN TOLERANCE',1X,E15.8)
END
23
INPUT AND OUTPUT FOR BISECTION METHOD
This is the Bisection Method.
Has the function F been created in the program?
Enter Y or N
y
Input endpoints A < B separated by blank
-1 1
Input tolerance
.0001
Input maximum number of iterations
- no decimal point
20
Select output destination:
1. Screen
2. Text file
Enter 1 or 2
1
Select amount of output
1. Answer only
2. All intermediate approximations
Enter 1 or 2
2
BISECTION METHOD
I
P
F(P)
1 0.00000000E+00 Infinity
2 0.50000000E+00 0.96583283E+00
3 0.75000000E+00 0.93932748E-02
4 0.87500000E+00 -0.33032596E+00
5 0.81250000E+00 -0.16806513E+00
6 0.78125000E+00 -0.81423879E-01
7 0.76562500E+00 -0.36564291E-01
8 0.75781250E+00 -0.13726234E-01
9 0.75390625E+00 -0.22022128E-02
10 0.75195312E+00 0.35865903E-02
11 0.75292969E+00 0.68992376E-03
12 0.75341797E+00 -0.75668097E-03
13 0.75317383E+00 -0.33497810E-04
14 0.75305176E+00 0.32824278E-03
15 0.75311279E+00 0.14734268E-03
THE APPROXIMATE SOLUTION IS
0.75311279E+00 AFTER 15 ITERATIONS, WITH TOLERANCE 0.99999997E-04
24
APPENDIX 2b
SECANT’S METHOD
C***********************************************************************
C
*
C
SECANT ALGORITHM 2.4
*
C
*
C***********************************************************************
C
C
C
C TO FIND A SOLUTION TO THE EQUATION F(X)=0
C GIVEN INITIAL APPROXIMATIONS PO AND P1:
C
C INPUT: INITIAL APPROXIMATIONS P0,P1; TOLERANCE TOL;
C
MAXIMUM NUMBER OF ITERATIONS N0.
C
C OUTPUT: APPROXIMATE SOLUTION P OR MESSAGE THAT THE
C
ALGORITHM FAILS.
C
REAL P0,P1,P,FP,Q0,Q1,TOL
INTEGER I,N0,FLAG,OUP
CHARACTER NAME1*30,AA*1
LOGICAL OK
C DEFINE FUNCTION F
F(X)=exp(-x**2)/erf(x)-.000335415*exp(-7.986111111*x**2)/
&(.229902658-.229902658*erf(x))*6.782233-1.055240177*x
cF(X)=exp(-x**2)/erf(x)-.001326457*exp(-7.986111111*x**2)/
c&(.204652504-.204652504*erf(x))*6.782233-1.185436374*x
c
OPEN(UNIT=5,FILE='CON',ACCESS='SEQUENTIAL')
c
OPEN(UNIT=6,FILE='CON',ACCESS='SEQUENTIAL')
WRITE(6,*) 'This is the Secant Method.'
WRITE(6,*) 'Has the function F been created in the program? '
WRITE(6,*) 'Enter Y or N '
WRITE(6,*) ' '
READ(5,*) AA
IF(( AA .EQ. 'Y' ) .OR. ( AA .EQ. 'y' )) THEN
OK = .FALSE.
10
IF (OK) GOTO 11
WRITE(6,*) 'Input initial approximations P0 and P1'
WRITE(6,*) 'separated by blank '
WRITE(6,*) ' '
READ(5,*) P0,P1
IF (P0.EQ.P1) THEN
WRITE(6,*) 'P0 cannot equal P1 '
ELSE
OK = .TRUE.
ENDIF
GOTO 10
11
OK = .FALSE.
12
IF (OK) GOTO 13
WRITE(6,*) 'Input tolerance '
WRITE(6,*) ' '
25
READ(5,*) TOL
IF (TOL.LE.0.0) THEN
WRITE(6,*) 'Tolerance must be positive '
WRITE(6,*) ' '
ELSE
OK = .TRUE.
ENDIF
GOTO 12
13
OK = .FALSE.
14
IF (OK) GOTO 15
WRITE(6,*) 'Input maximum number of iterations '
WRITE(6,*) '- no decimal point '
WRITE(6,*) ' '
READ(5,*) N0
IF ( N0 .LE. 0 ) THEN
WRITE(6,*) 'Must be positive integer '
WRITE(6,*) ' '
ELSE
OK = .TRUE.
ENDIF
GOTO 14
15
CONTINUE
ELSE
WRITE(6,*) 'The program will end so that the function F '
WRITE(6,*) 'can be created '
OK = .FALSE.
ENDIF
IF (.NOT.OK) GOTO 040
WRITE(6,*) 'Select output destination: '
WRITE(6,*) '1. Screen '
WRITE(6,*) '2. Text file '
WRITE(6,*) 'Enter 1 or 2 '
WRITE(6,*) ' '
READ(5,*) FLAG
IF ( FLAG .EQ. 2 ) THEN
WRITE(6,*) 'Input the file name in the form - '
WRITE(6,*) 'drive:name.ext'
WRITE(6,*) 'with the name contained within quotes'
WRITE(6,*) 'as example: ''A:OUTPUT.DTA'' '
WRITE(6,*) ' '
READ(5,*) NAME1
OUP = 3
OPEN(UNIT=OUP,FILE=NAME1,STATUS='NEW')
ELSE
OUP = 6
ENDIF
WRITE(6,*) 'Select amount of output '
WRITE(6,*) '1. Answer only '
WRITE(6,*) '2. All intermediate approximations '
WRITE(6,*) 'Enter 1 or 2 '
WRITE(6,*) ' '
READ(5,*) FLAG
WRITE(OUP,*) 'SECANT METHOD'
IF (FLAG.EQ.2) THEN
WRITE(OUP,4)
4
FORMAT(3X,'I',14X,'P',13X,'F(P)')
26
ENDIF
STEP 1
I=2
Q0=F(P0)
Q1=F(P1)
C STEP 2
016 IF ( I .GT. N0 ) GOTO 020
C
STEP 3
C
COMPUTE P(I)
P=P1-Q1*(P1-P0)/(Q1-Q0)
FP=F(P)
IF (FLAG.EQ.2) THEN
WRITE(OUP,5) I,P,FP
5
FORMAT(1X,I3,2X,E15.8,2X,E15.8)
ENDIF
C
STEP 4
IF( ABS(P-P1) .LT. TOL ) THEN
WRITE(OUP,2) P,I,TOL
GOTO 040
END IF
C
STEP 5
I=I+1
C
STEP 6
C
UPDATE P0, Q0, P1, Q1
P0=P1
Q0=Q1
P1=P
Q1=FP
GOTO 016
020 CONTINUE
C STEP 7
C PROCEDURE COMPLETED UNSUCCESSFULLY
IF(OUP.NE.6) WRITE(6,3) N0,P,TOL
WRITE(OUP,3) N0,P,TOL
040 CLOSE(UNIT=5)
CLOSE(UNIT=OUP)
IF(OUP.NE.6) CLOSE(UNIT=6)
STOP
1 FORMAT(3(E15.8,1X),I2)
2 FORMAT(1X,'THE APPROXIMATE SOLUTION IS',1X,E15.8,1X,
*'AFTER',1X,I2,1X,'ITERATIONS',/,' WITH TOLERANCE',1X,E15.8)
3 FORMAT(1X,'ITERATION NUMBER',1X,I3,1X,'GAVE APPROXIMATION',
*/,E15.8,1X,'NOT WITHIN TOLERANCE',1X,E15.8)
END
C
27
INPUT AND OUTPUT FOR SECANT’S METHOD
This is the Secant Method.
Has the function F been created in the program?
Enter Y or N
y
Input initial approximations P0 and P1
separated by blank
.2 1
Input tolerance
.0001
Input maximum number of iterations
- no decimal point
10
Select output destination:
1. Screen
2. Text file
Enter 1 or 2
1
Select amount of output
1. Answer only
2. All intermediate approximations
Enter 1 or 2
2
SECANT METHOD
I
P
F(P)
2 0.87477899E+00 -0.44401836E+00
3 0.69246149E+00 0.96710443E-01
4 0.72506934E+00 -0.10933518E-01
5 0.72175735E+00 -0.25588274E-03
6 0.72167796E+00 0.65565109E-06
THE APPROXIMATE SOLUTION IS 0.72167796E+00 AFTER 6 ITERATIONS
WITH TOLERANCE 0.99999997E-04
28
APPENDIX 3
NUMERICAL ALGORITHUM
parameter(nxp1=11)
dimension h(nxp1),u(nxp1),t(nxp1)
dimension ho(nxp1),uo(nxp1),to(nxp1)
c
c data
c
tc = 2.218
rho = 1000
cp = 4186
tfus = 273.15
hfus = 335000
c
xtot = .1
timet = 3600
tsurf = 243.15
tinit = 283.15
c
c preliminaries
c
td = tc/(rho*cp)
dx = xtot/float(nxp1-1)
dx2 = dx*dx
dt = dx2/(2.0*td)
nt = int(timet/dt) + 1
c
write(6,*) dx,dt,nt
c
c initialize
c
do ix=1,nxp1
to(ix) = tinit
uo(ix) = tc*(to(ix) - tfus)
ho(ix) = hfus + uo(ix)/td
if(ix.eq.1) then
to(ix)= tsurf
uo(ix) = tc*(to(ix) - tfus)
ho(ix) = uo(ix)/td
endif
end do
c
c time step
c
time = 0.0
do it = 1,nt
time = time + dt
do ix = 2,nxp1-1
h(ix) = ho(ix) + dt*(uo(ix+1)-2*uo(ix)+uo(ix-1))/dx2
end do
h(1) = ho(1)
h(nxp1) = h(nxp1-1)
c
c compute u then t
c
29
do ix=1,nxp1
if(h(ix).gt.hfus) then
u(ix) = td*(h(ix)-hfus)
t(ix) = tfus+u(ix)/tc
endif
if(h(ix).le.hfus.and.h(ix).ge.0.0) then
u(ix) = 0.0
t(ix) = tfus
endif
if(h(ix).lt.0.0) then
u(ix) = td*h(ix)
t(ix) = tfus+u(ix)/tc
endif
end do
c
c output
c
write(6,*) time
x = 0.0
do ix=1,nxp1
write(6,*) x,t(ix)
x = x + dx
end do
c
c update
c
x = 0.0
do ix=1,nxp1
ho(ix) = h(ix)
uo(ix) = u(ix)
write(66,*) x,t(ix),(t(ix)-to(ix))/dt
to(ix) = t(ix)
x = x + dx
end do
c
end do
c
c output
c
x = 0.0
do ix=1,nxp1
write(67,*) x,t(ix)
x = x + dx
end do
c
stop
end
30
APPENDIX 4
1.00000E-02 94.3643 39
94.3643
0. 243.150
1.00000E-02 263.230
2.00000E-02 283.150
3.00000E-02 283.150
4.00000E-02 283.150
5.00000E-02 283.150
6.00000E-02 283.150
7.00000E-02 283.150
8.00000E-02 283.150
9.00000E-02 283.150
1.00000E-01 283.150
188.729
0. 243.150
1.00000E-02 263.150
2.00000E-02 273.190
3.00000E-02 283.150
4.00000E-02 283.150
5.00000E-02 283.150
6.00000E-02 283.150
7.00000E-02 283.150
8.00000E-02 283.150
9.00000E-02 283.150
1.00000E-01 283.150
283.093
0. 243.150
1.00000E-02 258.170
2.00000E-02 273.150
3.00000E-02 278.170
4.00000E-02 283.150
5.00000E-02 283.150
6.00000E-02 283.150
7.00000E-02 283.150
8.00000E-02 283.150
9.00000E-02 283.150
1.00000E-01 283.150
377.457
0. 243.150
1.00000E-02 258.150
2.00000E-02 268.250
3.00000E-02 278.150
4.00000E-02 280.660
5.00000E-02 283.150
6.00000E-02 283.150
7.00000E-02 283.150
8.00000E-02 283.150
9.00000E-02 283.150
1.00000E-01 283.150
471.821
0. 243.150
1.00000E-02 255.700
2.00000E-02 268.150
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
566.186
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
660.550
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
754.914
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
849.279
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
31
274.455
280.650
281.905
283.150
283.150
283.150
283.150
283.150
255.650
265.078
274.400
278.180
281.900
282.527
283.150
283.150
283.150
283.150
254.114
265.025
271.709
278.150
280.354
282.525
282.839
283.150
283.150
283.150
254.087
262.911
271.587
276.031
280.337
281.596
282.837
282.994
283.150
283.150
253.031
262.837
269.471
275.962
278.814
281.587
282.295
282.994
283.072
1.00000E-01
943.643
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
1038.01
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
1132.37
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
1226.74
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
1321.10
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
283.072
252.994
261.251
269.400
274.143
278.775
280.555
282.291
282.684
283.033
283.033
252.200
261.197
267.697
274.087
277.349
280.533
281.619
282.662
282.858
282.858
252.173
259.949
267.642
272.603
277.310
279.484
281.597
282.239
282.760
282.760
251.549
259.908
266.276
272.476
276.043
279.454
280.861
282.179
282.499
282.499
251.529
258.912
266.192
271.159
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
1415.46
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
1509.83
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
1604.19
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
1698.56
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
1792.92
0. 243.150
275.965
278.452
280.816
281.680
282.339
282.339
251.031
258.860
265.036
271.078
274.806
278.391
280.066
281.578
282.010
282.010
251.005
258.034
264.969
269.921
274.735
277.436
279.984
281.038
281.794
281.794
250.592
257.987
263.977
269.852
273.678
277.359
279.237
280.889
281.416
281.416
250.569
257.285
263.920
268.828
273.606
276.458
279.124
280.326
281.152
281.152
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
1887.29
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
1981.65
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
2076.01
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
2170.38
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
32
250.217
257.244
263.056
268.763
272.723
276.365
278.392
280.138
280.739
280.739
250.197
256.637
263.003
267.890
272.564
275.557
278.252
279.566
280.439
280.439
249.893
256.600
262.263
267.784
271.724
275.408
277.562
279.345
280.002
280.002
249.875
256.078
262.192
266.993
271.596
274.643
277.376
278.782
279.674
279.674
249.614
256.034
261.536
266.894
270.818
274.486
276.712
278.525
9.00000E-02
1.00000E-01
2264.74
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
2359.11
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
2453.47
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
2547.84
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
2642.20
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
279.228
279.228
249.592
255.575
261.464
266.177
270.690
273.765
276.506
277.970
278.876
278.876
249.362
255.528
260.876
266.077
269.971
273.598
275.868
277.691
278.423
278.423
249.339
255.119
260.802
265.423
269.837
272.999
275.644
277.145
278.057
278.057
249.135
255.071
260.271
265.320
269.211
272.741
275.072
276.851
277.601
277.601
249.110
254.703
260.195
264.741
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
2736.56
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
2830.93
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
2925.29
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
3019.66
0. 243.150
1.00000E-02
2.00000E-02
269.030
272.142
274.796
276.337
277.226
277.226
248.926
254.653
259.722
264.613
268.442
271.913
274.239
276.011
276.781
276.781
248.901
254.324
259.633
264.082
268.263
271.340
273.962
275.510
276.396
276.396
248.737
254.267
259.203
263.948
267.711
271.112
273.425
275.179
275.953
275.953
248.709
253.970
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
3114.02
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
3208.39
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
3302.75
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
3397.11
0. 243.150
33
259.107
263.457
267.530
270.568
273.150
274.689
275.566
275.566
248.560
253.908
258.714
263.319
267.013
270.340
272.705
274.358
275.128
275.128
248.529
253.637
258.613
262.863
266.829
269.859
272.349
273.916
274.743
274.743
248.393
253.571
258.250
262.721
266.361
269.589
271.887
273.546
274.330
274.330
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
3491.48
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
3585.84
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
3680.21
0. 243.150
1.00000E-02
2.00000E-02
3.00000E-02
4.00000E-02
5.00000E-02
6.00000E-02
7.00000E-02
8.00000E-02
9.00000E-02
1.00000E-01
248.361
253.322
258.146
262.305
266.155
269.124
271.568
273.150
273.938
273.938
248.236
253.253
257.814
262.151
265.715
268.861
271.137
272.791
273.544
273.544
248.202
253.025
257.702
261.764
265.506
268.426
270.826
272.340
273.168
273.168
248.087
252.952
257.394
261.604
265.095
268.166
270.383
271.997
272.834
272.834
APPENDIX 5
Distance from
Chill (m)
0.000
0.010
0.020
0.030
0.040
0.050
0.060
0.070
0.080
0.090
0.100
Time (seconds)
94.3643 188.729
243.150 243.150
263.230 263.150
283.150 273.190
283.150 283.150
283.150 283.150
283.150 283.150
283.150 283.150
283.150 283.150
283.150 283.150
283.150 283.150
283.150 283.150
283.093
243.150
258.170
273.150
278.170
283.150
283.150
283.150
283.150
283.150
283.150
283.150
377.457
243.150
258.150
268.250
278.150
280.660
283.150
283.150
283.150
283.150
283.150
283.150
471.821
243.150
255.700
268.150
274.455
280.650
281.905
283.150
283.150
283.150
283.150
283.150
566.186
243.150
255.650
265.078
274.400
278.180
281.900
282.527
283.150
283.150
283.150
283.150
660.55
243.150
254.114
265.025
271.709
278.150
280.354
282.525
282.839
283.150
283.150
283.150
754.914
243.150
254.087
262.911
271.587
276.031
280.337
281.596
282.837
282.994
283.150
283.150
849.279
243.150
253.031
262.837
269.471
275.962
278.814
281.587
282.295
282.994
283.072
283.072
943.643
243.150
252.994
261.251
269.400
274.143
278.775
280.555
282.291
282.684
283.033
283.033
1038.01
243.150
252.200
261.197
267.697
274.087
277.349
280.533
281.619
282.662
282.858
282.858
1132.37
243.150
252.173
259.949
267.642
272.603
277.310
279.484
281.597
282.239
282.760
282.760
1226.74
243.150
251.549
259.908
266.276
272.476
276.043
279.454
280.861
282.179
282.499
282.499
Distance from
Chill (m)
0.000
0.010
0.020
0.030
0.040
0.050
0.060
0.070
0.080
0.090
0.100
Time (seconds)
1321.1 1415.46
243.150 243.150
251.529 251.031
258.912 258.860
266.192 265.036
271.159 271.078
275.965 274.806
278.452 278.391
280.816 280.066
281.680 281.578
282.339 282.010
282.339 282.010
1509.83
243.150
251.005
258.034
264.969
269.921
274.735
277.436
279.984
281.038
281.794
281.794
1604.19
243.150
250.592
257.987
263.977
269.852
273.678
277.359
279.237
280.889
281.416
281.416
1698.56
243.150
250.569
257.285
263.920
268.828
273.606
276.458
279.124
280.326
281.152
281.152
1792.92
243.150
250.217
257.244
263.056
268.763
272.723
276.365
278.392
280.138
280.739
280.739
1887.29
243.150
250.197
256.637
263.003
267.890
272.564
275.557
278.252
279.566
280.439
280.439
1981.65
243.150
249.893
256.600
262.263
267.784
271.724
275.408
277.562
279.345
280.002
280.002
2076.01
243.150
249.875
256.078
262.192
266.993
271.596
274.643
277.376
278.782
279.674
279.674
2170.38
243.150
249.614
256.034
261.536
266.894
270.818
274.486
276.712
278.525
279.228
279.228
2264.74
243.150
249.592
255.575
261.464
266.177
270.690
273.765
276.506
277.970
278.876
278.876
2359.11
243.150
249.362
255.528
260.876
266.077
269.971
273.598
275.868
277.691
278.423
278.423
2453.47
243.150
249.339
255.119
260.802
265.423
269.837
272.999
275.644
277.145
278.057
278.057
Distance from
Chill (m)
0.000
0.010
0.020
0.030
0.040
0.050
0.060
0.070
0.080
0.090
0.100
Time (seconds)
2547.84 2642.2
243.150 243.150
249.135 249.110
255.071 254.703
260.271 260.195
265.320 264.741
269.211 269.030
272.741 272.142
275.072 274.796
276.851 276.337
277.601 277.226
277.601 277.226
2736.56
243.150
248.926
254.653
259.722
264.613
268.442
271.913
274.239
276.011
276.781
276.781
2830.93
243.150
248.901
254.324
259.633
264.082
268.263
271.340
273.962
275.510
276.396
276.396
2925.29
243.150
248.737
254.267
259.203
263.948
267.711
271.112
273.425
275.179
275.953
275.953
3019.66
243.150
248.709
253.970
259.107
263.457
267.530
270.568
273.150
274.689
275.566
275.566
3114.02
243.150
248.560
253.908
258.714
263.319
267.013
270.340
272.705
274.358
275.128
275.128
3208.39
243.150
248.529
253.637
258.613
262.863
266.829
269.859
272.349
273.916
274.743
274.743
3302.75
243.150
248.393
253.571
258.250
262.721
266.361
269.589
271.887
273.546
274.330
274.330
3397.11
243.150
248.361
253.322
258.146
262.305
266.155
269.124
271.568
273.150
273.938
273.938
3491.48
243.150
248.236
253.253
257.814
262.151
265.715
268.861
271.137
272.791
273.544
273.544
3585.84
243.150
248.202
253.025
257.702
261.764
265.506
268.426
270.826
272.340
273.168
273.168
3680.21
243.150
248.087
252.952
257.394
261.604
265.095
268.166
270.383
271.997
272.834
272.834
40
Time
Boundary
(seconds) Position (m)
94.3643
0.01500
188.7290
0.02000
283.0930
0.02000
377.4570
0.02490
471.8210
0.02790
566.1860
0.02870
660.5500
0.03220
754.9140
0.03350
849.2790
0.03570
943.6430
0.03790
1038.0100
0.03850
1132.3700
0.04120
1226.7400
0.04190
1321.1000
0.04410
1415.4600
0.04560
1509.8300
0.04670
1604.1900
0.04860
1698.5600
0.04900
1792.9200
0.05120
1887.2900
0.05200
1981.6500
0.05390
2076.0100
0.05510
2170.3800
0.05640
2264.7400
0.05800
2359.1100
0.05880
2453.4700
0.06060
2547.8400
0.06180
2642.2000
0.06380
2736.5600
0.06530
2830.9300
0.06690
2925.2900
0.06880
3019.6600
0.07000
3114.0200
0.07270
3208.3900
0.07510
3302.7500
0.07760
3397.1100
0.08000
3491.4800
0.08480
3585.8400
0.08980
3680.2100
APPENDIX 6
SOLVING TRANSENDATAL EQUATION FOR  USING MATLAB
EDU» syms M K1 K2 k1 k2 V T1 c1 L
EDU» K1=2.218,K2=.6025,k1=.00000115,k2=.000000144,V=283.10,T1=273.15,c1=2060,L=335000
K1 =
2.2180
K2 =
0.6025
k1 =
1.1500e-006
k2 =
1.4400e-007
V=
283.1000
T1 =
273.1500
c1 =
2060
L=
335000
EDU» f=exp(-M^2)/erf(M)-K2*k1^.5*(V-T1)*exp(-k1*M^2/k2)/(K1*k2^.5*T1*(1-erf(M))*(k1/k2)^.5)M*L*pi^.5/(c1*T1)
f=
exp(-M^2)/erf(M)-5558915281605303/16573246628723425280*exp(43445771642400736/5440166188265831*M^2)/(4141558102980793/180143985094819844141558102980793/18014398509481984*erf(M))*46^(1/2)167131971145454779375/158382873170342313984*M
EDU» solve(f)
ans =
.75330884038373000039945584175972
41
APPENDIX 7
42
43
APPENDIX 8
VALIDATION OF UNITS AND CHECK FOR MATLAB AS SOLUTION METHOD
Book Units
EDU» K1=.0053,K2=.0014,k1=.0115,k2=.00144,V=2,T1=1,c1=.502,L=73.6
K1 =
0.0053
K2 =
0.0014
k1 =
0.0115
k2 =
0.0014
V=
2
T1 =
1
c1 =
0.5020
L=
73.6000
EDU» exp(-M^2)/erf(M)-K2*K1^0.5*(V-T1)*exp(-K1*M^2/K2)/K1*K2^0.5*T1*ERFC(M)*(K1/K2)^0.5M*L*3.141592^.5/(c1*T1)
??? Error using ==> erfc
X must be real.
EDU» exp(-M^2)/erf(M)-K2*K1^0.5*(V-T1)*exp(-K1*M^2/K2)/K1*K2^0.5*T1*
(1- ERF(M))*(K1/K2)^0.5-M*L*3.141592^.5/(c1*T1)
MATLAB does not recognize erfc as a
variable and requires that 1-erf be used.
ans =
exp(-M^2)/erf(M)-188012053538455575/13687484102692487299072*exp(-53/14*M^2)*14^(1/2)*(1erf(M))*742^(1/2)-1147473115367467925/4415638697148416*M
44
f = exp(-M^2)/erf(M)-2027084135185505/147573952589676412928*exp(-53/14*M^2)*14^(1/2)*(1erf(M))*742^(1/2)-4571606037320589/17592186044416*M
EDU» solve(f)
ans =
.58334615615361223033256158534803e-1
USING SI UNITS
EDU» K1=2.218,K2=.6025,k1=.00000115,k2=.000000144,V=2,T1=1,c1=2060,L=335000
K1 =
2.2180
K2 =
0.6025
k1 =
1.1500e-006
k2 =
1.4400e-007
V=
2
T1 =
1
c1 =
2060
L=
335000
EDU» f=exp(-M^2)/erf(M)-K2*K1^0.5*(V-T1)*exp(-K1*M^2/K2)/K1*K2^0.5*T1*(1-ERF(M))*(K1/K2)^0.5M*L*3.141592^.5/(c1*T1)
f=
exp(-M^2)/erf(M)-40410813823442875/1203672568807685095424*exp(-4436/1205*M^2)*241^(1/2)*(1erf(M))*1336345^(1/2)-16713195376004424125/57983845202395136*M
EDU» solve(f)
ans =
45
.56499945720744219974065388857484e-1
Back Solving using opposite value from other solution method
EDU» K1=2.218,K2=.6025,k1=.00000115,k2=.000000144,V=2,T1=1,c1=2060,L=335000,M=-.0583346
K1 =
2.2180
K2 =
0.6025
k1 =
1.1500e-006
k2 =
1.4400e-007
V=
2
T1 =
1
c1 =
2060
L=
335000
M=
0.0583
EDU» f=exp(-M^2)/erf(M)-K2*K1^0.5*(V-T1)*exp(-K1*M^2/K2)/K1*K2^0.5*T1*(ERFC(M))*(K1/K2)^0.5M*L*3.141592^.5/(c1*T1)
f=
1.0225
46
APPENDIX 9
SYMBOL
PROPERTY
UNITS
p
c
L
K
k
Tm
V
Density
Specific Heat
Latent Heat
Thermal Cond.
Dissusivity
Melting Temp.
Initial Melt Temp
kg/m3
J/kg*oK
J/kg
W/m*K
m2/s
o
K
o
K
Eq 18
SOLID
ICE
p=
c1=
L=
K1=
k1=
T1=
V=
917
2,060
2.218
0.00000115
273.15
LIQUID
WATER
c2=
K2=
k2=
1,000
4,186
335,000
0.6025
0.000000144
273.15
M*e(M^2)*erf(M)=c1*T1/L*PI^.5
leftside
rightside
Delta=
Eq 14
Lambda M (function of T1,T1-V1)
V1
T1
x
t
0.947722
0.947652
-0.000071
0.753308
283.10
273.15
0.06
2076.01
Eq14 solved for 0 =
-0.000430714
A
B
e^(-M^2)/erf(M) -K2*k1^0.5*(V1-T1)*e^(-k1*M^2/k2)/K1*k2^0.5*T1*ERFC(M)*(k1/k2)^0.5
A - First term
B - Middle term
C - 3rd term
0.794861496
0.000371343
0.794920867
-0.000430714
SUM of parts A,B and C
-0.567472943
Eq. 16 (as a function of Lambda,temperature of liquid,postion and time)
v2 (liquid)
v1 (solid)
282.93
235.43
47
0.713276922
0.286723078
Postion (cm)
Time (s)
=EXP(-2*2)
=EXP(-(2^2))
=EXP(-2^2)
Carefull!!
=0
matches with Eq.14 (large eq.)
eq14 with middle term elimiated (reduces to eq18)
Leftside
0.794861496
rightside
0.794920867
Delta
-5.93717E-05
-Lamda2
C
-M*L*PI^.5/(c1*T1)
erf(lamda)
erc(lamnda)
0.018315639
0.018315639
54.59815003
not correct