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McCready\nProfessor and Chair of Chemical and \ Biomolecular Engineering\nUniversity of Notre Dame\nNotre Dame IN 46556\nUSA", FontSize->14], StyleBox["\n\nMark.J.McCready.1@nd.edu\n", FontSize->12], 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, if you use it, this notice remain visible to other users. \nThere \ is no charge for copying and dissemination \n\n", StyleBox["Version: 9/11/05\nMore recent versions of this notebook should \ be available at the web site:\n", FontSize->12], ButtonBox["http://www.nd.edu/~mjm/Antoine.nb", ButtonData:>{ URL[ "http://www.nd.edu/~mjm/longwave.nb"], None}, ButtonStyle->"Hyperlink"] }], "Text", FontSize->10], Cell["\<\ You will often have to use a numerical solution procedure to find \ answer to an equation, or the roots of a polynomial or other complex function \ where you cannot just solve to a single variable answer. Such problems are \ typically solved with an iteration procedure where you start with an initial \ \"guess\" for your answer and then use a formal procedure to alter this \ answer, presumably continually getting closer to the real answer. You stop \ when the answer is \"sufficiently close\". You may have already solve many of these problems on a calculator by \"cut \ and try\" where you are the correction procedure. Another approach is to use \ \"GoalSeek\" in Excel. If you look in numerical analysis texts you will find methods such as \ Regula-Falsi or Newton-Raphson which are commonly used for most of these \ types of problems. In CBE 20258 you will learn more about these. Here is I want you to realize: 1. What is really going on in a interative solution 2. That formal methods exist which usually work well. 3. That you will probably want to use these as opposed to making something \ up yourself!\ \>", "Text"], Cell["Newton-Raphson or \"Newton's Method\"", "Subtitle"], Cell[TextData[{ "The basis for this method is a Taylor series expansion of a function.\n\n\ We can formally write:\n\nf[x] = f[x0] + f'[x0] (x-x0) + f''[x0] ", Cell[BoxData[ \(TraditionalForm\`\((x - x0)\)\^2\)]], " /2 + h.o.t." }], "Text"], Cell[TextData[{ "If you know ", StyleBox["Mathematica", FontSlant->"Italic"], " it shows this:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Series[f[x], {x, x0, 3}]\)], "Input"], Cell[BoxData[ FormBox[ InterpretationBox[ RowBox[{\(f(x0)\), "+", RowBox[{ RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "(", "x0", ")"}], " ", \((x - x0)\)}], "+", RowBox[{\(1\/2\), " ", RowBox[{ SuperscriptBox["f", "\[Prime]\[Prime]", MultilineFunction->None], "(", "x0", ")"}], " ", \(\((x - x0)\)\^2\)}], "+", RowBox[{\(1\/6\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((3)\), Derivative], MultilineFunction->None], "(", "x0", ")"}], " ", \(\((x - x0)\)\^3\)}], "+", InterpretationBox[\(O(\((x - x0)\)\^4)\), SeriesData[ x, x0, {}, 0, 4, 1], Editable->False]}], SeriesData[ x, x0, { f[ x0], Derivative[ 1][ f][ x0], Times[ Rational[ 1, 2], Derivative[ 2][ f][ x0]], Times[ Rational[ 1, 6], Derivative[ 3][ f][ x0]]}, 0, 4, 1], Editable->False], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "We then need to understand several things.\n\n1. The earth is almost \ flat. So we will ignore any ", StyleBox["h", FontVariations->{"Underline"->True}], "igher ", StyleBox["o", FontVariations->{"Underline"->True}], "rder ", StyleBox["t", FontVariations->{"Underline"->True}], "erms. So we write simply:\n\nf[x] \[TildeTilde] f[x0] + f'[x0] (x-x0) \n\ \nIn this equation, x is the value of the independent variable that gives the \ correct solution. We don't know this. x0 is the current \"guess\". f[x0] \ is the value of our complex function at x0, f'[x0] is the derivative of f at \ x0. If we also formulate the problem so that f[x] = 0 when we reach the \ answer then we have:\n\nx \[TildeTilde] x0 - f[x0] / f'[x0]\n\nor at least we \ should get closer to the answer by making this correction.\n\nSince you are \ expecting that to have to calculate the values of these functions at least \ several times while you are correcting the answer and getting closer to the \ value of test function being 0, you impliment this in some form of a loop. \ \"While\" works well in ", StyleBox["Mathematica", FontSlant->"Italic"], ". \n\nSo \"While\" |f[x]|> ", Cell[BoxData[ \(TraditionalForm\`10\^\(-3\)\)]], ", You do the following calculations and assignments:\ntest = f[xold], \n\ xnew = xold - f[xold] / f'[xold],\nPrint f[x], xnew, xold,\nxold = xnew\n\n\ and this continues until f[x] gets below ", Cell[BoxData[ \(TraditionalForm\`10\^\(-3\)\)]], "(or what ever you think it should.) \n\nI always print out the \ intermediate results when I am testing the calcuation to see if something is \ going wrong.\n" }], "Text"], Cell["\<\ Now let us implement this for our problem with the Antoine \ equation. You have seen this written as Log[P] = A- B/ (C-T) The form of Log[P] being a function of 1/T is suggested from thermodynamic \ arguments that you will here about next semester. However, tthe full form or \ expression is empirical and there is no physical significance to A, B and \ C's, these are just fitting constants. You see for this problem you could solve for T and get the answer for 1 ATM. \ However, we wish to learn the more general iterative computational \ procedure. Here is how we can implement it.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Clear[aa, bb, cc, T]\)], "Input"], Cell[BoxData[ \(FF[T_]\ := Log[10, pressure]\ - \((\ aa - \ bb\ /\((\ \ cc + \ T)\))\)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(\(\(D[Log[oneATM]\ - \((\ aa - \ bb\ /\((\ \ cc + \ T)\))\), \ T]\)\(\ \)\)\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(-\(bb\/\((cc + T)\)\^2\)\)\)], "Output"] }, Open ]], Cell[BoxData[ \(FFPrime[T_]\ := \ \(-bb\)\/\((cc + T)\)\^2\)], "Input"], Cell["Let's see how to use these expressions", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(FF[T]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(-aa\) + \(log(pressure)\)\/\(log(10)\) + bb\/\(cc + T\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(FFPrime[T]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(-\(bb\/\((cc + T)\)\^2\)\)\)], "Output"] }, Open ]] }, Open ]], Cell["Now try to write some code", "Text"], Cell["For Pentane:", "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(aa\ = \ 6.87632;\)\), "\[IndentingNewLine]", \(\(bb\ = \ 1075.780;\)\), "\[IndentingNewLine]", \(\(cc = \ 233.205;\)\)}], "Input"], Cell["\<\ I like xtest for the expression that we want to be 0. However, it \ cannot start at 0 or the While loop won't run.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(xtest\ = \ \(-1\);\)\)], "Input"], Cell["You need an initial T for a \"guess\"", "Text"], Cell[BoxData[ \(\(tinit\ = \ 0;\)\)], "Input"], Cell["\<\ We need to pick a pressure to find the temperature for. This is in \ mm Hg.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(pressure = 380\)], "Input"], Cell[BoxData[ \(TraditionalForm\`380\)], "Output"] }, Open ]], Cell["I will use \"tt\" as the temperature at any time in the loop", "Text"], Cell["\<\ We always need to keep track of the current temperature and the \ previous one. So I set the previous one.\ \>", "Text"], Cell[BoxData[ \(\(ttold\ = \ tinit;\)\)], "Input"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(While[Abs[xtest] > 1\/10\^5\ , \ xtest = \ \ \ FF[ttold]; tt = ttold - FF[ttold]\/FFPrime[ttold]; \n Print["\", \ xtest, "\< T= \>", tt, "\< P= \>", \ 10^\ \ \ \ \((\ aa - \ bb\ /\((\ \ cc + \ tt)\))\)]; ttold\ = \ tt]\)], "Input"], Cell[BoxData[ FormBox[ InterpretationBox[\("xtest= "\[InvisibleSpace]0.3164864734848014`\ \[InvisibleSpace]" T= "\[InvisibleSpace]15.999536533652268`\[InvisibleSpace]" \ P= "\[InvisibleSpace]362.6305563834596`\), SequenceForm[ "xtest= ", 0.31648647348480141, " T= ", 15.999536533652268, " P= ", 362.63055638345958], Editable->False], TraditionalForm]], "Print"], Cell[BoxData[ FormBox[ InterpretationBox[\("xtest= "\[InvisibleSpace]0.02031920030574197`\ \[InvisibleSpace]" T= "\[InvisibleSpace]17.17252848886147`\[InvisibleSpace]" \ P= "\[InvisibleSpace]379.91671668823307`\), SequenceForm[ "xtest= ", 0.020319200305741969, " T= ", 17.17252848886147, " P= ", 379.91671668823307], Editable->False], TraditionalForm]], "Print"], Cell[BoxData[ FormBox[ InterpretationBox[\("xtest= "\[InvisibleSpace]0.00009519328127760573`\ \[InvisibleSpace]" T= "\[InvisibleSpace]17.17807568506827`\[InvisibleSpace]" \ P= "\[InvisibleSpace]379.9999981546699`\), SequenceForm[ "xtest= ", 9.5193281277605735*^-05, " T= ", 17.178075685068269, " P= ", 379.99999815466992], Editable->False], TraditionalForm]], "Print"], Cell[BoxData[ FormBox[ InterpretationBox[\("xtest= "\[InvisibleSpace]2.1089907598081936`*^-9\ \[InvisibleSpace]" T= "\[InvisibleSpace]17.17807580797089`\[InvisibleSpace]" \ P= "\[InvisibleSpace]379.99999999999994`\), SequenceForm[ "xtest= ", 2.1089907598081936*^-09, " T= ", 17.178075807970892, " P= ", 379.99999999999994], Editable->False], TraditionalForm]], "Print"] }, Open ]], Cell["\<\ Here is the analogous Matlab code. I have not bothered to define functions, \ but you could. aa = 6.87632 bb = 1075.780 cc = 233.205 xtest = -1 % tinit = 0 % pressure = 380 ttold = tinit % \twhile abs(xtest) > 1/10^5 , xtest = log10(pressure) - ( aa - bb /(cc + ttold)) tt = ttold - (log10( pressure) - ( aa - bb /( cc + ttold)))/(-bb / (cc + \ ttold) ^2 ) xtest, tt, 10^( aa - bb /( cc + ttold)) ttold = tt \tend end \ \>", "Text"], Cell["Regula-Falsi", "Subtitle"], Cell["\<\ In this implimentation, I use a fake derivative. That is I write \ the f' as the difference between the function evaluation at two different \ temperatures. The convergence is slightly slower and I need more initial \ variables. However this is useful if you cannot get an analytical derivative.\ \>", \ "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Told = \ 0;\)\), "\[IndentingNewLine]", \(\(Tolder\ = \ 15;\)\), "\[IndentingNewLine]", \(\(xtest\ = \ \(-1\);\)\), "\[IndentingNewLine]", \(\(xtestold\ = \ \(-5\);\)\)}], "Input"], Cell[BoxData[ \(\(T\ = \ Told;\)\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(While[Abs[xtest] - 1\/10\^5 \[GreaterEqual] 0, \ xtest = \ Log[10, 760\ ] - \ \ \((aa - \ bb\ /\((\ \ cc + \ T)\))\); T = Told - \(xtest\ \((Told - Tolder)\)\)\/\(xtest - xtestold\); \ \ \n Print["\", \ xtest, "\< T= \>", T, "\< P= \>", \ 10^\((\((aa - \ bb\ /\((\ \ cc + \ T)\))\))\)]; Tolder = Told; Told = T; xtestold = xtest; If[T > 58, \ T = T - 37];]\)], "Input"], Cell[BoxData[ FormBox[ InterpretationBox[\("xtest="\[InvisibleSpace]0.6175164691487827`\ \[InvisibleSpace]" T= "\[InvisibleSpace]1.6489042957154547`\[InvisibleSpace]" \ P= "\[InvisibleSpace]197.55365049462242`\), SequenceForm[ "xtest=", 0.61751646914878267, " T= ", 1.6489042957154547, " P= ", 197.55365049462242], Editable->False], TraditionalForm]], "Print"], Cell[BoxData[ FormBox[ InterpretationBox[\("xtest="\[InvisibleSpace]0.5851285330797125`\ \[InvisibleSpace]" T= "\[InvisibleSpace]31.438420666355825`\[InvisibleSpace]" \ P= "\[InvisibleSpace]647.5942961111158`\), SequenceForm[ "xtest=", 0.58512853307971247, " T= ", 31.438420666355825, " P= ", 647.5942961111158], Editable->False], TraditionalForm]], "Print"], Cell[BoxData[ FormBox[ InterpretationBox[\("xtest="\[InvisibleSpace]0.06951057736680655`\ \[InvisibleSpace]" T= "\[InvisibleSpace]35.454352344953335`\[InvisibleSpace]" \ P= "\[InvisibleSpace]744.8469873438407`\), SequenceForm[ "xtest=", 0.069510577366806547, " T= ", 35.454352344953335, " P= ", 744.84698734384074], Editable->False], TraditionalForm]], "Print"], Cell[BoxData[ FormBox[ InterpretationBox[\("xtest="\[InvisibleSpace]0.00874652675046006`\ \[InvisibleSpace]" T= "\[InvisibleSpace]36.03241541840855`\[InvisibleSpace]" \ P= "\[InvisibleSpace]759.7388768519867`\), SequenceForm[ "xtest=", 0.0087465267504600597, " T= ", 36.032415418408547, " P= ", 759.73887685198667], Editable->False], TraditionalForm]], "Print"], Cell[BoxData[ FormBox[ InterpretationBox[\("xtest="\[InvisibleSpace]0.00014924187980236425`\ \[InvisibleSpace]" T= "\[InvisibleSpace]36.042450123670115`\[InvisibleSpace]" \ P= "\[InvisibleSpace]759.9994295497587`\), SequenceForm[ "xtest=", 0.00014924187980236425, " T= ", 36.042450123670115, " P= ", 759.9994295497587], Editable->False], TraditionalForm]], "Print"], Cell[BoxData[ FormBox[ InterpretationBox[\("xtest="\[InvisibleSpace]3.2597826926661355`*^-7\ \[InvisibleSpace]" T= "\[InvisibleSpace]36.04247208973163`\[InvisibleSpace]" \ P= "\[InvisibleSpace]759.9999999786924`\), SequenceForm[ "xtest=", 3.2597826926661355*^-07, " T= ", 36.042472089731632, " P= ", 759.99999997869236], Editable->False], TraditionalForm]], "Print"] }, Open ]], Cell["\<\ I can make a log-log plot for the vapor pressures of the \ substances.\ \>", "Text"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " needs to load another package to do the various log plots." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(<< Graphics`Graphics`\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(\(LogLogPlot[ Evaluate[10^\((aa - \ bb\ /\((\ \ cc + \ tt - 273)\))\)], {tt, \ 100, 350}];\)\)], "Input"], Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica %%AspectRatio: .61803 MathPictureStart /Mabs { Mgmatrix idtransform Mtmatrix dtransform } bind def /Mabsadd { Mabs 3 -1 roll add 3 1 roll add exch } bind def %% Graphics %%IncludeResource: font Courier %%IncludeFont: Courier /Courier findfont 10 scalefont setfont % Scaling calculations -3.47715 1.75048 0.464008 0.0408734 [ [.02381 -0.0125 -11 -12 ] [.02381 -0.0125 11 0 ] [.33205 -0.0125 -11 -12 ] [.33205 -0.0125 11 0 ] [.55076 -0.0125 -11 -12 ] [.55076 -0.0125 11 0 ] [.859 -0.0125 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