Bill's comments in these text boxes
Project 1 : Introduction to MSE 510
Projects
Data Analysis Using Example Diffusion Data for Aluminum Ions
in Aluminum Oxide
Student Name
MSE 510
Due 01/24/2012
Ÿ Project Objectives
1. Familiarize yourself with projects in this course
2. Familiarize yourself with Mathematica
3. Encourage you to review chapter 1 in the textbook
Ÿ Summary
Diffusion data is selected from a peer-reviewed journal article and imported, plotted, and
analyzed using Mathematica. The source of the data is the diffusion studies of Paladino and
Kingery A1E and the following Mathematica functions are utilized:
® The raw data is imported from a .csv file using "Import" and plotted using "ListPlot"
® A non-linear mathematical fit is performed using "NonlinearModelFit"
® "Manipulate" is used to show the effects of changing two relevant parameters on the
fit of the data
® Relevant parameters for diffusion are extracted from the data fit and compare to the
published data
® Statistical error of the non-linear fit is analyzed
® Integration and differentiation are perfomed on the non-linear fit equation
Relevant discussion and comments are included where appropriate.
Add an Abstract no more than 150 words long that explain your findings.
Import, Plot, and Fit Data
Import the data to a list designated as "data" :
2
MSE 510 - Project 1 - Excellent Example.nb
In[1]:=
Out[1]=
data =
Import@"D:\\Bill\\Bsu\\Courses\\MSE 510 ElecOptDielProps\\Projects Exams Quizzes
@MSE 510D\\Projects\\2013 Projects\\Project 1 @Ch 1D\\Student
Projects\\AaronWoodard\\Aaron Woodard diffdata.csv"D
992178., 1.1 ´ 10-10 =, 92103., 4. ´ 10-11 =, 92093., 3.5 ´ 10-11 =, 92073., 2.8 ´ 10-11 =,
92043., 1.9 ´ 10-11 =, 92003., 1.6 ´ 10-11 =, 91993., 1.1 ´ 10-11 =, 91943, 1.9 ´ 10-12 ==
Paladino and Kingery measured the amount of aluminum that diffused across an interface joining two cylinders of
aluminum oxide. A plot of the experimental results is generated showing the diffusion coefficient as a function of
anneal temperature. The diffusion coefficent describes the rate of atomic motion in the material.
In[2]:=
dataplot = ListPlotAdata, Frame ® True,
GridLines ® Automatic, PlotStyle ® 8Red, AbsolutePointSize@5D<,
PlotRange ® 881900, 2200<, 80, 1.4 * 10 ^ - 10<<, FrameLabel ® 9"T HKL", "D Hcm2 •sL"=,
PlotLabel ® "Diffusion Rate vs. T For Al in Al2 O3 "E
Diffusion Rate vs. T For Al in Al2 O3
1.4 ´ 10-10
1.2 ´ 10-10
D Hcm2•sL
1. ´ 10-10
8. ´ 10-11
Out[2]=
6. ´ 10-11
4. ´ 10-11
2. ´ 10-11
0
1900
1950
2000
2050
T HKL
2100
2150
Label your plot. That is, give it a figure
number and caption. e.g., "Figure 1:
Diffusivity of Al in Al2O3 as a function of
temperature. Data taken from reference
[1]."
You can then get rid of the title at the top.
Do this for all your figures.
You can then refer to the figure in the text
2200
by its title (i.e., figure 1).
From the plot, it is apparent that the diffusion rate increases with increasing temperature. It is well known that the
equation for the diffusion coefficient has the form of the Arrhenius rate equation, and is given as
D = Do expI
In[3]:=
-E A
M
kT
where Do is the pre-exponential consant, E A is the activation energy for diffusion, k is the Boltzmann constant, and T is
temperature @2D . A model of this form is now fitted to the data.
PrintA"D = ", fitexp = NonlinearModelFitAdata, a * ãb•x , 8a, b<, xEE
H*Calculates the fit model and outputs the resulting equation*L
D = FittedModelB 47.0619 ã-58 346.2•x F
MSE 510 - Project 1 - Excellent Example.nb
In[4]:=
3
plotdf =
PlotAfitexp@xD, 8x, 1900, 2200<, Frame ® True, GridLines ® Automatic, PlotStyle ® Blue,
PlotRange ® 881900, 2200<, 80, 1.4 * 10 ^ - 10<<, FrameLabel ® 9"T HKL", "D Hcm2 •sL"=E;
Show@dataplot, plotdf, PlotLabel ® "Diffusion Rate vs. T For Al in Al2 O3
Blue Line = Fit, Red Dots = Data"D
Diffusion Rate vs. T For Al in Al2 O3
Blue Line = Fit, Red Dots = Data
1.4 ´ 10-10
1.2 ´ 10-10
Label your plot. That is, give it a figure
number and caption. Do this for all your
figures. You can then refer to the figure in
the text by its title (i.e., figure #).
Out[5]=
D Hcm2•sL
1. ´ 10-10
8. ´ 10-11
6. ´ 10-11
4. ´ 10-11
2. ´ 10-11
0
1900
1950
2000
2050
2100
2150
2200
T HKL
The exponential model appears to fit the data well, as expected based on the known diffusion coefficient equation. In
general, the fit appears to be better at higher temperatures and diffusion rates. There is less agreement of the fit with
the observed data at lower temperatures. The accuracy of the fit will be analyzed during the error analysis performed in
a subsequent section.
The constants in the diffusion coefficent equation (excluding the Boltzmann constant) are dependent on the material
properties and conditions under which diffusion takes place. For example, crystal structure, grain structure, atomic
bonding, and defect characteristics affect the value of the pre-exponential constant and activation energy for a particular
situation. Using the built-in "Manipulate" function, the effects of changes in these values on the diffusion rate can be
investigated.
4
MSE 510 - Project 1 - Excellent Example.nb
In[6]:=
Clear@d, g, h, TD H*initialize variables;
g = pre-exponential, h = activation energy*L
k = 8.617 * 10 ^ - 5;H*Boltzmann constant, eV•K*L
-h
d@g__, h_, T_, kD := g * ã k*T
H*define the function for the diffusion coefficient equation*L
ManipulateAShowAPlotAd@g, h, T, kD, 8T, 1900, 2200<,
Frame ® True, GridLines ® Automatic, PlotStyle ® Blue,
PlotRange ® 881900, 2200<, 80, 1.4 * 10 ^ - 10<<, FrameLabel ® 9"T HKL", "D Hcm2 •sL"=,
PlotLabel ® "Diffusion Rate vs. T For Al in Al2 O3 "E, dataplotE,
99g, 47, "Pre-Exponential Hcm2 •sL"=, 0, 2000, 1=,
88h, 5.03, "Activation Energy HeVL"<, 4.0, 8.0, .01<E
Pre-Exponential Hcm2 •sL
Activation Energy HeVL
47
Very nice that you show values
before creating the pdf file
5.03
Diffusion Rate vs. T For Al in Al2 O3
1.4 ´ 10-10
Out[9]=
1.2 ´ 10-10
D Hcm2•sL
1. ´ 10-10
8. ´ 10-11
Label your plot. That is, give it a figure
number and caption. Do this for all your
figures. You can then refer to the figure in
the text by its title (i.e., figure #).
6. ´ 10-11
4. ´ 10-11
2. ´ 10-11
0
1900
1950
2000
2050
2100
2150
2200
T HKL
Relevant diffusion parameters can be extracted from the fit of the data. The resulting equation for the non-linear fit
gives the estimated values of the pre - exponential and activation energy for the material and conditions studied by
Paladino and Kingery. These values and their corresponding units are listed in the table below.
In[10]:=
GridA98"Parameter", "Symbol", "Value", "Unit"<,
8"Activation Energy", "EA ", "5.03", "eV"<,
9"Pre-Exponential", "DO ", "47.06", "cm2 •s"==, Alignment ® Left,
Frame ® AllEH*generates a custom table of values*L
Out[10]=
Nice. Do provide a label and title for your
table. That is, give it a table number and
Pre-Exponential
DO
47.06 cm2 •s
Title. That is, "Table 1: Extracted Diffusions
Parameters" and place it at the top or
By comparison, Paladino and Kingery found values for the
pre-exponential
activation
of 28
4.94,
beginning
of theandtable.
Do energy
this for
all and
your
respectively. By examination, using the manipulation chart
presented
above,
these
values
appear
to
also
produce
tables. You can then refer to the table in a
good fit for the data. In fact, by further manipulation of the chart, there is an infinite number of combinations that will
the text by its title (i.e., Table #).
Parameter
Symbol Value Unit
Activation Energy EA
5.03 eV
produce a line that fits the data. In order to determine the correct parameters, the data must be plotted as Ln D vs. 1/T.
This is the general convention by which diffusion data is plotted.
MSE 510 - Project 1 - Excellent Example.nb
5
By comparison, Paladino and Kingery found values for the pre-exponential and activation energy of 28 and 4.94,
respectively. By examination, using the manipulation chart presented above, these values appear to also produce a
good fit for the data. In fact, by further manipulation of the chart, there is an infinite number of combinations that will
produce a line that fits the data. In order to determine the correct parameters, the data must be plotted as Ln D vs. 1/T.
This is the general convention by which diffusion data is plotted.
Import the new data set for Ln D and 1/T :
In[11]:=
Out[11]=
data2 =
Import@"D:\\Bill\\Bsu\\Courses\\MSE 510 ElecOptDielProps\\Projects Exams Quizzes
@MSE 510D\\Projects\\2013 Projects\\Project 1 @Ch 1D\\Student
Projects\\AaronWoodard\\Aaron Woodard diffdata2.csv"D
880.459137, - 22.9<, 80.475511, - 23.9<, 80.477783, - 24.1<, 80.482393, - 24.3<,
80.489476, - 24.7<, 80.499251, - 24.9<, 80.501756, - 25.2<, 80.514668, - 27.<<
By taking the natural logarithm of both sides of the diffusion coefficient equation, the equation becomes
Ln D = Ln Do - E A /kT
which is in the form of a straight line, y = mx + b @3D . Therefore, the pre-exponential and activation energy can be
obtained by a linear fit of the Ln D vs. 1/T data plot. The slope of the fit line is -E A /k and the intercept is Ln Do . Plot
the new data and perform a linear fit:
In[12]:=
dataplot2 = ListPlotAdata2, Frame ® True,
GridLines ® Automatic, PlotStyle ® 8Red, AbsolutePointSize@5D<,
PlotRange ® Automatic, FrameLabel ® 9"1000•T HK-1 L", "Ln D Hcm2 •sL"=,
PlotLabel ® "Diffusion Rate vs. T For Al in Al2 O3 "E;
Print@"D = ", fitline = LinearModelFit@data2, x, xDD
plotfitline =
PlotAfitline@xD, 8x, .45, .52<, Frame ® True, GridLines ® Automatic, PlotStyle ® Blue,
PlotRange ® Automatic, FrameLabel ® 9"1000•T HK-1 L", "Ln D Hcm2 •sL"=E;
Show@dataplot2, plotfitline, PlotLabel ® "Fit = Blue Line; Data = Red Points"D
D = FittedModelB 7.19797 - 65.2783 x F
Fit = Blue Line; Data = Red Points
-23
-24
Ln D Hcm2•sL
Out[15]=
Label your plot. That is, give it a figure
number and caption. Do this for all your
figures. You can then refer to the figure in
the text by its title (i.e., figure #).
-25
-26
-27
0.46
0.47
0.48
0.49
1000•T HK -1L
0.50
0.51
6
MSE 510 - Project 1 - Excellent Example.nb
Fit = Blue Line; Data = Red Points
Slope = -23
EA
k
Ln D Hcm2•sL
-24
In[16]:=
Increasing diffusion
coefficient
-25
-26
Increasing Temperature
-27
0.45
0.46
0.47
0.48
0.49
0.50
0.51
0.52
1000•T HK -1L
Fit = Blue Line; Data = Red Points
Slope = -23
Nice
EA
k
Label your plot. That is, give it a figure
number and caption. Do this for all your
figures. You can then refer to the figure in
the text by its title (i.e., figure #).
Ln D Hcm2•sL
-24
Out[16]=
-25
Increasing diffusion
coefficient
-26
Increasing Temperature
-27
0.45
0.46
0.47
0.48
0.49
0.50
0.51
0.52
1000•T HK -1L
Good observation on the variation of the
data at lower temperatures. What statistical
test can be performed to determine the
validity of removing data outliers?
Again it is seen that the linear fit agrees well with the data at higher temperatures, but not as well at relatively lower
temperatures. The data point corresponding to the lowest temperature could be an outlier; it appears that by omitting
this point the accuracy of the fit could be improved. This point also corresponds to the lowest rate of diffusion measured by Paladino and Kingery. It is possible that the measurement device used was more accurate for higher diffusion
rates.
The resulting equation for the linear fit gives the estimated values of the pre-exponential and activation energy as shown
in the table below.
In[17]:=
GridA98"Parameter", "Symbol", "Value", "Unit"<,
8"Activation Energy", "EA ", 65.2783 * 1000 * k, Do
"eV"<,
provide a label and title for
9"Pre-Exponential", "DO ", ã7.19797 , "cm2 •s"==, Alignment ® Left, Frame ® AllE
Out[17]=
Parameter
Symbol Value
Unit
Activation Energy EA
5.62503 eV
Pre-Exponential
DO
1336.71 cm2 •s
your table.
That is, give it a table number and Title.
Place it at the top or beginning of the table.
Do this for all your tables. You can then
refer to the table in the text by its title (i.e.,
Table #).
The values reported here are not in agreement with those reported by Paladino and Kingery. The difference could be
due to the method of data extraction from the paper--values plotted here were extracted visually from the data chart in
the paper and the error could be significant. Raw data values for the paper were not available.
Your arguments to the reasons that your parameter values are not in agreement with Paladino and Kingery are
In summary,method
the data has
importedthe
and validity
plotted in of
its raw
form.
A non-linearis
equation
was fit the
to theliterature
data, but thefor
fit the
fairly sound. A common
tobeen
examine
your
parameters
to search
was not adequate to obtain estimates of relevant diffusion coefficient equation parameters. The data was then plotted in
work of others that have extracted the values. One way to do so is to see who has referenced the work of
an alternate format, consistent with scientific convention for diffusion data. The resulting linear fit provided estimates
Paladino and Kingery.
In addition, calculating the percent error using the values you extracted and the values
for the pre-exponential and activation energy parameters. An error analysis of the linear fit is performed in the next
given by Paladino and Kingery provide a value of accuracy of the data.
section.
MSE 510 - Project 1 - Excellent Example.nb
The values reported here are not in agreement with those reported by Paladino
and Kingery. The difference could be7
due to the method of data extraction from the paper--values plotted here were extracted visually from the data chart in
the paper and the error could be significant. Raw data values for the paper were not available.
In summary, the data has been imported and plotted in its raw form. A non-linear equation was fit to the data, but the fit
was not adequate to obtain estimates of relevant diffusion coefficient equation parameters. The data was then plotted in
an alternate format, consistent with scientific convention for diffusion data. The resulting linear fit provided estimates
for the pre-exponential and activation energy parameters. An error analysis of the linear fit is performed in the next
section.
Error Analysis
The accuracy of the linear model is now analyzed by examining the statistical parameters associated with the fitting
equation. Recall the linear model used to fit the Ln D vs. 1/T data :
In[18]:=
Print@"D = ", fitline = LinearModelFit@data2, x, xDD
D = FittedModelB 7.19797 - 65.2783 x F
Ÿ Fit Parameters
In[19]:=
Print@"Model Parameter Table = ", fitline@"ParameterTable"DD
Estimate
Model Parameter Table = 1
x
Standard Error t-Statistic P-Value
7.19797 3.7537
-65.2783 7.69559
1.91757 0.103615
-8.48256 0.000146805
The Estimates represent the calculated values for the intercept HLn Do L and the acivation energy H - E A • kL.
The Standard Error represents how precisely the estimated
the parameter
was computed
and estimates the
Dovalue
you ofmean
"Estimated
value"?
standard deviation of the sample mean based on the population mean @4D .
The t Statistic is the ratio of the Estimate to the Standard Error. It is used to calculate the P-Value
@5D .
The P-Value represents the probability of getting a t Statistic greater than the calculated value by chance. A very low
value indicates that the parameter estimate is significantly different from zero; a rule of thumb is that a P-Value below
0.05 is significant @5D .
Good.
Based on the model parameters, the linear fit gives a good estimate of the activation energy for diffusion, since the PValue is very small. The estimate for Ln Do may not be accurate, as the P-Value indicates that there is at least a 10%
chance of obtaining the estimate by chance.
You stated above that Standard Error is based on precision and the P-value
Ÿ Coefficients of Determination
In[20]:=
is based on the Standard Error. In this sentence, you mention accuracy.
between "accurate" and "precise"?
PrintA"R2 = ", fitline@"RSquared"DE
What is the difference
PrintA"Adjusted R2 = ", fitline@"AdjustedRSquared"DE
R2 = 0.923031
Adjusted R2 = 0.910203
The Coefficient of Determination represents how well the model fits the data. The value varies between 0 and 1, with 1
being a perfect fit, meaning that the model can accurately describe every data point. The Adjusted Coefficient of
Determination also represents how well the model fits the data, but also takes into account the degrees of freedom.
Here, the high values indicate that the linear model is a relatively good fit for the data. This agrees with the theoretical
Arrhenius behavior of the diffusion coefficient with temperature, which indicates that Ln D vs. 1/T should be a linear
relationship. As noted earlier, the data point corresponding to the lowest anneal temperature could be the reason that
R^2 is not higher @6D .
8
MSE 510 - Project 1 - Excellent Example.nb
The Coefficient of Determination represents how well the model fits the data. The value varies between 0 and 1, with 1
being a perfect fit, meaning that the model can accurately describe every data point. The Adjusted Coefficient of
Determination also represents how well the model fits the data, but also takes into account the degrees of freedom.
Here, the high values indicate that the linear model is a relatively good fit for the data. This agrees with the theoretical
Arrhenius behavior of the diffusion coefficient with temperature, which indicates that Ln D vs. 1/T should be a linear
relationship. As noted earlier, the data point corresponding to the lowest anneal temperature could be the reason that
R^2 is not higher @6D .
Ÿ Residuals
What are the degrees of freedom? Please define.
The residuals represent the difference between the value predicted by the model and the value of the corresponding
obeserved data point.
In[22]:=
Print@"Residuals = ", fitline@"FitResiduals"DD
ListPlot@fitline@"FitResiduals"D, PlotRange ® Automatic, Filling ® Axis,
Frame ® True, PlotLabel ® "Ln D vs. 1•T Linear Fit Residuals"D
Residuals = 8- 0.126285, - 0.0574176, - 0.109105,
- 0.0081723, 0.0541939, 0.492289, 0.355811, - 0.601315<
Ln D vs. 1•T Linear Fit Residuals
0.4
Label your plot. That is, give it a figure
number and caption. Do this for all your
figures. You can then refer to the figure in
the text by its title (i.e., figure #).
0.2
Out[23]=
0.0
-0.2
-0.4
-0.6
0
2
4
Good. Unknown factors could be other competing
kinetic mechanisms including other diffusion
mechanism
or reactions
6
8
From the residual plot, the values least well described by the model are the three data points that correspond to measurements taken at the lowest anneal temperatures. This supports the theory that the measurement device used may not be
sensitive enough to detect these lower amounts of diffused particles. It is also possible that, as temperature increases, it
becomes the dominant mechanism controlling the diffusion rate. At lower temperatures, there could be unknown
factors controlling the rate of diffusion.
Ÿ Confidence Intervals
A confidence interval represents a range where the true value of the estimated parameter lies according to a given
probability. The confidence intervals for the parameters estimated by the linear fit are presented below.
MSE 510 - Project 1 - Excellent Example.nb
In[24]:=
9
Print@"80% Confidence Interval = ",
fitline@"ParameterConfidenceIntervals", ConfidenceLevel ® .8DD
Print@"85% Confidence Interval = ",
fitline@"ParameterConfidenceIntervals", ConfidenceLevel ® .85DD
Print@"90% Confidence Interval = ",
fitline@"ParameterConfidenceIntervals", ConfidenceLevel ® .90DD
Print@"95% Confidence Interval = ",
fitline@"ParameterConfidenceIntervals", ConfidenceLevel ® .95DD
Print@"99% Confidence Interval = ",
fitline@"ParameterConfidenceIntervals", ConfidenceLevel ® .99DD
ListPlot@Table@fitline@"ParameterConfidenceIntervals", ConfidenceLevel ® pD,
8p, 8.8, .85, .9, .95, .99<<D, Frame ® True, GridLines ® Automatic,
PlotLabel ® "Confidence Intervals: 80%, 85%, 90%, 95%, 99%"D
80% Confidence Interval = 881.79356, 12.6024<, 8- 76.3581, - 54.1985<<
85% Confidence Interval = 881.00372, 13.3922<, 8- 77.9774, - 52.5793<<
90% Confidence Interval = 88- 0.0961403, 14.4921<, 8- 80.2322, - 50.3244<<
95% Confidence Interval = 88- 1.987, 16.3829<, 8- 84.1087, - 46.4479<<
99% Confidence Interval = 88- 6.71859, 21.1145<, 8- 93.8091, - 36.7475<<
Confidence Intervals: 80%, 85%, 90%, 95%, 99%
Label your plot. That is, give it a figure
number and caption. Do this for all your
figures. You can then refer to the figure in
the text by its title (i.e., figure #).
20
10
0
-10
Out[29]=
-20
-30
-40
-50
-80
-60
-40
-20
0
What are the title &
units of the axes?
Can you define
them?
The confidence intervals at all levels are relatively large. The range of values in which the true values may lie would
have a significant effect on the calculation of Do and E A . Using the 95% confidence interval as an example, the high
and low true values of Do and E A are shown in the following table:
10
MSE 510 - Project 1 - Excellent Example.nb
In[30]:=
GridA98"Parameter", "High Value", "Low Value", "Unit"<,
8"EA ", 84.1087 * 1000 * k, 46.4479 * 1000 * k, "eV"<,
9"DO ", ã16.3829 , ã-1.987 , "cm2 •s"==, Alignment ® Left, Frame ® AllE
Out[30]=
Parameter High Value
EA
7.24765
DO
Low Value Unit
4.00242
eV
1.30318 ´ 107 0.137106
cm2 •s
Do provide a label and title for your table. That is, give it
a table number and Title. Place it at the top or beginning
of the table. Do this for all your tables. You can then
refer to the table in the text by its title (i.e., Table #).
The high range of values indicates that the estimates of the parameters are poor. To obtain more accurate estimates,
more data points could be measured or a more accurate measurement device could be used. It is also possible that
there is an unknown factor controlling the rate of diffusion that influences the behavior of the diffusion coefficent in
response to a temperature change. In that case, the diffusion coefficient might not completely obey the Arrhenius
relationship.
Good.
Integration and Differentiation
Since the data plotted as Ln D vs. 1/T fit a straight line, differentiation of the fit line will be a constant. However, as
noted above, the resulting constant is a useful parameter for diffusion--the activation energy. The value is actually
-E A • H1000 *kL, so E A is found by multiplying by Boltzmann's constant and by 1000 (since T was plotted as 1000/T).
The activation energy found for the analyzed diffusion data was presented earlier, however its calculation by differentiation is demonstrated here. Since the second derivative would be zero, it is not considered.
Calculate the activation energy by differentiation of the linear fit :
In[31]:=
Print@"The activation energy for diffusion is = ", D@fitline@xD, xD * - 1000 * k, " eV."D
The activation energy for diffusion is = 5.62503 eV.
The derivative can also be plotted. In this case, it is simply a straight line at 5.625 eV, as expected.
MSE 510 - Project 1 - Excellent Example.nb
11
In[32]:=
Needs@"PlotLegends`"D
dfitline@x_D = fitline '@xD;
Plot@dfitline@xD * - 1000 * k, 8x, 1900, 2200<, PlotRange ® 881900, 2200<, 80, 10<<,
Frame ® True, GridLines ® Automatic, PlotStyle ® Blue,
FrameLabel ® 8"T HKL", "dD•dT HeVL"<, PlotLegend ® 8"Activation Energy"<,
PlotLabel ® "Derivative of Diffusion Rate vs. T For Al in Al2 O3 "D
General::obspkg : PlotLegends` is now obsolete. The legacy version being loaded may
conflict with current Mathematica functionality. See the Compatibility Guide for updating information.
Derivative of Diffusion Rate vs. T For Al in Al2 O3
10
Label your plot. That is, give it a figure
number and caption. Do this for all your
figures. You can then refer to the figure in
the text by its title (i.e., figure #).
dD•dT HeVL
8
6
Out[34]=
4
2
0
1900
1950
Activation Energy
2000
2050
T HKL
2100
2150
2200
Good. I like the
legend.
There is no value in performing integration on the diffusion data, since it will not yield useful information. However,
the built-in Mathematica integration function is used for demonstration purposes.
In[35]:=
Out[35]=
intfitline@x_D = Integrate@dfitline@xD, xD
H*define a function that equals the integral of the derivative of the fitline*L
- 65.2783 x
12
MSE 510 - Project 1 - Excellent Example.nb
In[36]:=
PlotAintfitline@xD, 8x, .45, .52<, Frame ® True, GridLines ® Automatic,
PlotStyle ® Blue, PlotRange ® Automatic, PlotLegend ® 8"Original Linear
Fit Line
HIntercept = 0L"<, PlotLabel ® "Integration Example",
FrameLabel ® 9"1000•T HK-1 L", "Ln D Hcm2 •sL"=E
Integration Example
Label your plot. That is, give it a figure
number and caption. Do this for all your
figures. You can then refer to the figure in
the text by its title (i.e., figure #).
-30
Ln D Hcm2•sL
-31
-32
Out[36]=
-33
-34
0.45
0.46
Original Linear
Fit Line
HIntercept = 0L
0.47
0.48
0.49
0.50
0.51
0.52
1000•T HK L
-1
Since the integral of the derivative of the fitline was calculated, the fitline is also the result. As expected, the integration
does not give the intercept; however the slope is the same as that of the linear fit line. This indicates that the integration
function was used properly.
Applications
This type of analysis of diffusion data is not limited to the diffusion of aluminum in aluminum oxide. It can be applied
to the diffusion of any type of atom in any material and will show whether a particular diffusion process can be characterized by an Arrhenius relationship to temperature. Specifically for aluminum diffusion in aluminum oxide, one example
application is in characterization of the oxidation rate of aluminum alloys. By performing this type of analysis, the ion
diffusion rate can be determined over different operating temperatures, which assists in the determination of the rate of
oxide growth. Aluminum oxide is typically a protective oxide layer, so the data can be used to determine both the
temperature and length of time at which the alloy should be annealed in order to achieve the desired oxide thickness, or
the rate at which the alloy will oxidize in service.
In order to further characterize the process, the diffusion of oxygen in aluminum oxide could be investigated. Based on
a comparison of the respective diffusion rates of oxygen and aluminum in the oxide, the ion which controls the overall
oxide formation rate can be determined. If this is known, the rate of formation of oxide can be further controlled by
inhibiting or enhancing the rate of diffusion of the rate limiting ion. Even further experiments to determine the mechanism by which diffusion occurs can allow for extensive control over the oxide formation rate. For example, diffusion
through a single crystal of aluminum can be compared to that through polycrystalline aluminum to determine the effects
of the presence of grain boundaries, which can be conduits by which diffusion occurs.
In summary, the analysis presented here can be repeated across different material types and structures to characterize
the rate of formation of any type of alloy, ceramic, polymer or phase for a given situation.
MSE 510 - Project 1 - Excellent Example.nb
References
13
Nice reference list.
[1] Paladino, A.E.; Kingery, W.D. Aluminum ion Diffusion in Aluminum Oxide. J.Chem.Phys. Vol 37, 5. 1962.
[2] Kasap, S.O. Principles of Electronic Materials and Devices. 3rd Ed. McGraw Hill, 2006.
[3] Atkins, P.; de Paula, J. Physical Chemistry. 7th Ed. Oxford University Press, 2002.
[4] Press,W.H.;Flannery,B.P.;Teukolsky,S.A.;and Vetterling,W.T. Numerical Recipes in FORTRAN:The Art of
Scientific Computing. 2nd Ed. Cambridge University Press, 1992.
[5] Bivariate Scatterplot and Fitting. JMP7: Statististical Discovery. Computer Software. SAS Institute, 2007.
[6] Interpreting Regression Results, OriginLabs - Origin Software - http://www.originlab.com/.
Comments:
Excellent work. Very thorough. Writing style very clear. Nice citing of references.
Improvements:
- I encourage you to move beyond the depth of the material in the textbook. For the first project, your analysis is fine. But
I'm looking for a greater level of critical and creative thinking. Mathematica is a tool to be used to analyze the data.
Leverage it to analyze the data in a critical and creative way and to reveal interesting and/or peculiar phenomena.
- Clarity between precision and accuracy
- Determining if data is/are outlier data
- Compare your results to other work in the literature
- Can extend the discussion to the diffusion mechanisms in Al2O2.