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Notebook[{
Cell["Buckingham Pi method (dimensional analysis)", "Title",
  Evaluatable->False,
  AspectRatioFixed->True],

Cell[TextData[{
  "This notebook has been written in ",
  StyleBox["Mathematica ",
    FontSlant->"Italic"],
  "by \n\nMark J. McCready\nProfessor and Chair of Chemical Engineering\n\
University of Notre Dame\nNotre Dame IN 46556\nUSA\n\n",
  ButtonBox["mjm@nd.edu",
    ButtonData:>{
      URL[ "mailto:mjm@nd.edu"], None},
    ButtonStyle->"Hyperlink"],
  "\n",
  ButtonBox["http://www.nd.edu/~mjm/",
    ButtonData:>{
      URL[ "http://www.nd.edu/~mjm/"], None},
    ButtonStyle->"Hyperlink"],
  "\n\nIt is copyrighted to the extent allowed by whatever laws pertain to \
the World Wide Web and the Internet.\n\nI would hope that as a professional \
courtesy that this notice remain visible to other users. \nThere is no charge \
for copying and dissemination   \n\nVersion:  8/24/00\nMore recent versions \
of this notebook should be available at the web site:\n",
  ButtonBox["http://www.nd.edu/~mjm/",
    ButtonData:>{
      URL[ "http://www.nd.edu/~mjm/dimensionless.nb"], None},
    ButtonStyle->"Hyperlink"],
  StyleBox[ButtonBox["dimensional.analysis",
    ButtonData:>{
      URL[ "http://www.nd.edu/~mjm/dimensionless.nb"], None},
    ButtonStyle->"Hyperlink"],
    FontSlant->"Italic"],
  ButtonBox[".nb",
    ButtonData:>{
      URL[ "http://www.nd.edu/~mjm/dimensionless.nb"], None},
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Cell[CellGroupData[{

Cell["Physical Motivation", "Subtitle",
  Evaluatable->False,
  AspectRatioFixed->True],

Cell[CellGroupData[{

Cell["Why do dimensional analysis?", "Section"],

Cell[TextData[{
  StyleBox["We do not know how to attach significance to the numerical value \
of something unless it is in comparison to another thing that has the same \
dimensions.  For example, 1 is much less that 10^4, but if it is 1 ton of \
sand versus 10^6 grains of sand, we change our mind.  Thus we need to have \
the same dimensions (and units) for our comparison.    \n\nA generalization \
of this kind of comparison is that we need to measure the effects of two \
different things that act similarly.  If we are determining the \"filling\" \
effect of our favorite foods,  we might compare a 15 inch thin crust \
pepperoni pizza to 1 lb of spaghetti with meat sauce.  They are different \
foods, but they both can satisfy a significant appetite.\n  \nA systematic \
way of building into our problems these comparisons of like quantities and \
perhaps more importantly, generalizations that involve comparisons of say \
forces associated with gravity compared to forces associated with a moving \
fluid, say inertia is called dimensional analysis.  The idea is that by \
making quantities dimensionless, we will be making these comparisons between \
different \"things\" that have the ",
    Evaluatable->False,
    AspectRatioFixed->True],
  StyleBox["same",
    Evaluatable->False,
    AspectRatioFixed->True,
    FontVariations->{"Underline"->True}],
  StyleBox[" effect.\n\nA problem that we care about  is pipeflow.  Flow in a \
pipe can be caused by a pressure difference, \[CapitalDelta]p.  Pressure has \
the same dimensions as \[Rho] v",
    Evaluatable->False,
    AspectRatioFixed->True],
  StyleBox["2",
    Evaluatable->False,
    AspectRatioFixed->True,
    FontVariations->{"CompatibilityType"->"Superscript"}],
  StyleBox[", which is a measure of the force associated with the inertia of \
the fluid.  Thus the ratio \[CapitalDelta]p/\[Rho] v",
    Evaluatable->False,
    AspectRatioFixed->True],
  StyleBox["2",
    Evaluatable->False,
    AspectRatioFixed->True,
    FontVariations->{"CompatibilityType"->"Superscript"}],
  StyleBox[" measures force due to pressure to force associated with fluid \
inertia.  Engineers think about their systems in terms of these comparisons \
so they will be important.  The particular ratio that we are describing is \
giving us an idea how difficult it is to pump the fluid in question.          \
",
    Evaluatable->False,
    AspectRatioFixed->True]
}], "Text",
  Evaluatable->False,
  AspectRatioFixed->True]
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Cell[CellGroupData[{

Cell["Examination of simulated data", "Section",
  Evaluatable->False,
  AspectRatioFixed->True],

Cell["Let is create some simulated data and plot it:", "Text"],

Cell[CellGroupData[{

Cell["We need secret forumulas", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(dp = 
      2*\ 0.079\ \((d\ v\ \[Rho]/\[Mu])\)^\((\(- .25\))\)\ \[Rho]\ v^2\ \ /
          d\)], "Input",
  CellLabel->"In[41]:="],

Cell[BoxData[
    \(TraditionalForm\`\(0.158`\ v\^2\ \[Rho]\)\/\(d\ \((\(d\ v\ \[Rho]\)\/\
\[Mu])\)\^0.25`\)\)], "Output",
  CellLabel->"Out[41]="]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(re = \ \((d\ v\ \[Rho]/\[Mu])\)\)], "Input",
  CellLabel->"In[6]:="],

Cell[BoxData[
    \(TraditionalForm\`\(d\ v\ \[Rho]\)\/\[Mu]\)], "Output",
  CellLabel->"Out[6]="]
}, Open  ]]
}, Closed]],

Cell[CellGroupData[{

Cell["We need to load a special graphics package", "Text"],

Cell[BoxData[
    \(<< Graphics`MultipleListPlot`\)], "Input",
  CellLabel->"In[33]:="]
}, Closed]],

Cell[CellGroupData[{

Cell["Here are the calculated data.", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(data1 = 
        Table[{v*Random[Real, { .94, 1.04}], 
              dp} /. {\[Mu] \[Rule]  .01\ \ , \ \[Rho] \[Rule] 1\ \ , \ 
              d \[Rule] 2.54\ \ , v \[Rule] i\ \ }, \[IndentingNewLine]{i, 0, 
            100, 4}]\ ;\)\)], "Input",
  CellLabel->"In[52]:="],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(ListPlot[data1, 
        AxesLabel \[Rule] {"\<v, cm/s\>", \
\*"\"\<\[CapitalDelta]P/L,dyne/\!\(cm\^3\)\>\""}];\)\)], "Input",
  CellLabel->"In[53]:="],

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  "We see that the pressure drop increases with velocity (duh).  We might \
even realize that it is roughly as ",
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  "\n\nDo we need to take data for every situation?  \nCan we collapse the \
data onto one curve? \n\nWe can resolve these issues by doing dimensional \
analysis.  "
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Cell["Procedure for doing dimensional analysis.", "Subtitle"],

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Cell["\<\
How to do pipe flow.  Note that this is a different approach than \
Middleman.  Of course it yields the same answer.\
\>", "Subsubsection",
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  StyleBox["The variables associated with pipe flow are \[CapitalDelta]p, D, \
L, v,\[Mu], \[Rho].  They have dimensions:\n\n\[CapitalDelta]p [=]  m/(l \
\[Theta]",
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\[Theta])\n\n\[Rho] [=] m/l",
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  StyleBox["Note that the dimensional time is given by \[Theta].  There are 3 \
fundamental dimensions associated with these variables and there will be 6 - \
3 = 3 dimensionless groups.  This is surmised because we are doing this as \
part of linear algebra and we need to solve some linear algebraic equations.  \
The only equations that seem obvious are that the groupings of variables are \
dimensionless ( i.e., some sum=0).  There are only three equations but 6 \
variables.  \n\nA dimensionless group must be of the form\n\n\
\[CapitalDelta]p",
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  StyleBox["   \n\nwhere a, b, c, d, e, f satisfy equations that make the \
group dimensionless.  This gives for mass\n\nmass:   a + e + f=0,\n\nlength:\t\
-a +b +c +d -e -3f = 0\n\ntime:\t-2a -d -e  = 0",
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can solve this system for 3 of the variables in terms of the other three.  \
This will give us three dimensionless groups.  "], "Text",
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  StyleBox["It turns out that I need to put the coefficients into equations \
to solve them so I use a \"dot\" product and set it equal to 0.  The \
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I choose the three to solve for by checking some choices and then \
deciding.   \
\>", "Text",
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  StyleBox["Now reconstruct our dimensionless group\n\n\[CapitalDelta]p",
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  StyleBox["Now rearrange this into:\n\n(\[CapitalDelta]p/\[Rho]v",
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  StyleBox[" \n\nEach of the groups that we have made is dimensionless.  This \
has to be true because we know that the entire thing is dimensionless and the \
solution is true for any values of a, c and e.  (Check what would happen if \
each group had some dimension and the values of a, c, and e were changed)  \
Thus we have found three groups and we have an answer.\n\nOf course, L has \
the same dimensions as D so how do we know that a good group is\n\[Mu]/\[Rho] \
v D  as opposed to \[Mu]/\[Rho] v L ??  The answer is physical intuition and \
possible experiment.  \n\nThe way we actually end up using this result is to \
define\n\nReynolds number \n\nRe = \[Rho] v D/\[Mu]\n\nand friction factor\n\n\
f = \[CapitalDelta]p D/2 L \[Rho]v",
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we need only two groups.  For most problems this does not happen and you will \
need all three.  L/D is separately important in this problem if the flow \
distance has not been long enough for the velocity profile to reach a \
configuration that does not change with distance.  \n\nThe reason that it \
happened this time is that \[CapitalDelta]p is inherently proportional to L.  \
If you consider two identical lengths of pipe that have the same Re, the \
pressure drop is the same.  If they are attached end to end, then the \
\[CapitalDelta]P must be two times this value.  Since \[CapitalDelta]P is \
really proportional to L, we can conveniently think of the new variable, \
\[CapitalDelta]p/L which I'll call dpdx.  Now if we redo the analysis we have \
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Redo the analysis with \[CapitalDelta]P/L as a single \
variable\
\>", "Subsubsection"],

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