(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 4.0, MathReader 4.0, or any compatible application. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 396677, 15866]*) (*NotebookOutlinePosition[ 414240, 16482]*) (* CellTagsIndexPosition[ 413546, 16461]*) (*WindowFrame->Normal*) Notebook[{ Cell["Creeping flow past a stationary sphere", "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\nmjm@nd.edu\n\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: 8/8/00\nMore recent versions of \ this notebook should be available at the web site:\n", ButtonBox["http://www.nd.edu/~mjm/creepingsphere.nb", ButtonData:>{ URL[ "http://www.nd.edu/~mjm/creepingsphere.nb"], None}, ButtonStyle->"Hyperlink"] }], "Text"], Cell[TextData[{ "This notebook shows how to solve creeping (intertialess) flow past a \ stationary sphere (Stokes's Problem)\n\n", StyleBox["Reference: S. Middleman (1999) ", "SmallText"], StyleBox["An Introduction to Fluid Dynamics:", "SmallText", FontWeight->"Plain", FontSlant->"Italic"], StyleBox["Principles of Analysis and Design,\nWiley", "SmallText", FontWeight->"Plain"], StyleBox[" ", "SmallText"], StyleBox["pp 166-171", "SmallText", FontWeight->"Plain"], StyleBox[".", "SmallText"] }], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[GraphicsData["Bitmap", "\<\ CF5dJ6E]HGAYHf4PAg9QL6QYHgOol2 00000goo001oo`02Ool01@00Oomoogoo00000goo00D007ooOomoo`0000=oo`05001oogooOol00003 Ool01@00Oomoogoo00000goo00D007ooOomoo`0000=oo`05001oogooOol00003Ool01@00Oomoogoo 00000goo00D007ooOomoo`0000=oo`05001oogooOol00003Ool01@00Oomoogoo00000goo00D007oo Oomoo`0000=oo`05001oogooOol00003Ool00`00Oomoo`02001>Ool009Aoo`8006Yoo`8004eoo`00 Tgoo0P00K7oo0P00C7oo002BOol2001^Ool2001;Ool0095oo`800003Ool007oo009oo`05001oogoo Ool00003Ool01@00Oomoogoo00000goo00D007ooOomoo`0000=oo`05001oogooOol00003Ool01@00 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In all of these situations, there is a \ need to either keep the particles suspended, or, to hasten their settling. \ To understand this problem, we need to find an expression that gives the \ steady -state drag on a particle. There is no simple relation that is valid \ for all Reynolds numbers. However, if the particles is a sphere and the \ Reynolds number sufficiently small, a relation called \"Stokes Law\" is valid \ and it can be obtained through solution of the Navier-Stokes equations. We will derive this relation below and in the process will hopefully learn \ something about the physics of \"creeping\" flows. It is worthwhile to \ emphasize that even if a particle is not a sphere, the relation will give the \ qualitative prediction of the drag and should not be to wrong quantiatively. \ \ \>", "Text"], Cell[TextData[{ "Consider the flow past a stationary sphere where the Reynolds number is \ significantly less than unity. Inertia can be neglected. The far away \ velocity field is a constant straight flow. To make the problem as simple as \ possible, we will use spherical coordinates with the \[Phi] axis oriented \ parallel to the flow. Thus we have \[Phi] symmetry and no ", Cell[BoxData[ $TraditionalForm\`u$]], Cell[BoxData[ $TraditionalForm\`\_\[Phi]$]], ". This means that the ", Cell[BoxData[ $TraditionalForm\`u\_\[Phi]$]], " equation is not needed. If we used a different coordinate system, or \ different orientation, the problem would be more complicated. " }], "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell[CellGroupData[{ Cell["Learning Objectives", "Subtitle"], Cell[CellGroupData[{ Cell["Physical Issues", "Subsubsection"], Cell[TextData[{ "1. We are restricting the problem to the case where inertia forces are \ much weaker than viscous forces. \t\tThus we expect that the interia terms \ of the Navier-Stokes equation should be much smaller than the viscous and \ pressure terms and that thus they can be neglected. For this case the \ Reynolds number is very small. If we make the equations dimensionless all \ terms are no larger in magnitude than about unity. Thus the parameters that \ appear in these equations, and which can have values much different from on, \ determine which terms are needed for the solution. In the nondimensional \ equations, the ", ButtonBox["Reynolds number multiplies the intertia terms and these will \ consequently be neglected in the solution", ButtonData:>"neglect_inertia_terms", ButtonStyle->"Hyperlink"], ". \n\n2. Since the Reynolds number is small, the fluid goes only where \ specifically pushed and it will stop if the forcing is stopped. ", ButtonBox["Thus the geometry of the flow field is determined by the \ boundaries.", ButtonData:>"boundary_determines_flow", ButtonStyle->"Hyperlink"], " \n\n3. Because viscous forces dominate the flow field, the fluid can \ never accelerate above the free stream value even if an obstacle causes the \ fluid to be squeezed. Thus the velocity in the region of the sphere just ", ButtonBox["slows down and then returns to the free stream value", ButtonData:>"fluid_does_not_speed_up", ButtonStyle->"Hyperlink"], ". \n\n4. Both normal stresses and tangential stresses contribute to the \ drag on the sphere. These can be termed ", ButtonBox["form drag", ButtonData:>"form_drag", ButtonStyle->"Hyperlink"], " and ", ButtonBox["skin drag", ButtonData:>"skin_drag", ButtonStyle->"Hyperlink"], ". \n\n5. Consistent with the fluid not accelerating, the ", ButtonBox["pressure never increases above the free stream value", ButtonData:>"pressure_decreases", ButtonStyle->"Hyperlink"], ". The fluid has no inertia that would cause a pressure increase as the \ fluid slows down.\n\n6. The velocity decays slowly (as ", Cell[BoxData[ $TraditionalForm\`1\/r$]], ") and thus the disturbance is felt very far away from the sphere. This \ makes it difficult to do a real experiment, in a reasonable size container, \ that allows that sphere to fall at a speed specified by the drag that is \ predicted from the analysis here. ", ButtonBox["The very high Reynolds number case decays much faster", ButtonData:>"disturbance_felt_faraway", ButtonStyle->"Hyperlink"], ". " }], "Text", CellTags->"physical_objectives"] }, Open ]], Cell[CellGroupData[{ Cell["Mathemematical Issues", "Subsubsection"], Cell[TextData[{ "1. Linear and nonlinear partial differential equations: In general \ nobody can make much progress solving nonlinear partial differential \ equations analytically. Thus, if we are to solve problems involving the \ Navier Stokes equations, there will have to be physically based \ simplifications that allow us to solve linear PDE's or we will be a \ mathematical technique (e.g., a perturbation method) that linearizes the \ equations. In the present problem the vanishing inertia, which allows us to \ neglect the left side of the equations results in a linear problem. \n\n2. \ Solving PDE's: You may not have solved a set of coupled partial differential \ equations before. It is not as hard as it sounds.\n\n\ti. The general \ procedure is to turn a PDE into an ordinary differential equation -- \ preferably an \t\tequation that you know how to solve. In this case (and in \ many situations) this is done by ", StyleBox["Separation of Variables", FontSlant->"Italic"], ". ", ButtonBox["Separation of variables", ButtonData:>"boundary_determines_flow", ButtonStyle->"Hyperlink"], " involves using a solution form, say F(x,y) that is a function which is \ product of the functions of each of the independent variables, say X(x) Y(y).\ \n\tii. This assumed form of the solution is substituted into the equations \ and if it is going to work, you will either have one of functions appearing \ as an ", ButtonBox["identical factor in every term", ButtonData:>"identical_factor", ButtonStyle->"Hyperlink"], " (the case for this problem) or you will be able to group all of the ", StyleBox["X(x)", FontSlant->"Italic"], " functions together and all of the ", StyleBox["Y(y)", FontSlant->"Italic"], " terms together. Since ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], " can be varied independently, each of these groups of terms should be \ equation to the same constant. \n\t\n3. You may not have solved a set of \ coupled partial ordinary equations before. It is not as hard as it sounds.\n\ \t\n\ti. You solve these by ", ButtonBox["eliminating dependent variables", ButtonData:>"eliminate_dependent_terms", ButtonStyle->"Hyperlink"], " (just as you would for a set of algebraic \t\tequations) and derivatives \ of independent variables in terms of the variable that you have chosen.\n\t\ ii. If a dependent variable, that you need to eliminate (in this case \ Pressure) appears only as a \t\t\tderivative in two similar complex \ equations, it is often possible to ", ButtonBox["take an extra derivative in one or both of the equations to \ create an identical term in each equation", ButtonData:>"eliminate_pressure", ButtonStyle->"Hyperlink"], ". Then the equations can be subtracted to eliminate the variable. \n\n\ 4. Once you have gone through the process, you will have created an ", ButtonBox["Euler differential equation", ButtonData:>"Euler_equation", ButtonStyle->"Hyperlink"], ".\n\n5. An Euler equation can be ", ButtonBox["solved by assuming that the solution is a polynomial", ButtonData:>"How_to_solve_Euler", ButtonStyle->"Hyperlink"], " in the independent \t\tvariable. " }], "Text", CellTags->"mathematical_objectives"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Mathematical Formulation ", "Subtitle"], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " aside" }], "Subsubsection"], Cell["\<\ I have used a mix of input notation to show how it can be done. \ \ \>", "Text"], Cell["\<\ I would enter the dynamic boundary condition from a key pad \ as\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ $dc = \(-\ \[Gamma]$\ D[\[Eta][x, t], {x, 2}]\ - \n\t\t$\[Rho]\_1$ $(\(-D[\[Phi]\_1[x, y, t], t]$ - U1\ D[\[Phi]\_1[x, y, t], x] - g\ \[Eta][x, t])\) + \n\t\t\t$\[Rho]\_2$ $(\(-D[\[Phi]\_2[x, y, t], t]$ - \n\t\t\t\t\tU2\ D[\[Phi]\_2[x, y, t], x] - g\ \[Eta][x, t])\)\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{$-\[Gamma]$, " ", RowBox[{ SuperscriptBox["\[Eta]", TagBox[$(2, 0)$, Derivative], MultilineFunction->None], "(", $x, t$, ")"}]}], "-", RowBox[{$\[Rho]\_1$, " ", RowBox[{"(", RowBox[{$\(-g$\ $\[Eta](x, t)$\), "-", RowBox[{ SubsuperscriptBox["\[Phi]", "1", TagBox[$(0, 0, 1)$, Derivative], MultilineFunction->None], "(", $x, y, t$, ")"}], "-", RowBox[{"U1", " ", RowBox[{ SubsuperscriptBox["\[Phi]", "1", TagBox[$(1, 0, 0)$, Derivative], MultilineFunction->None], "(", $x, y, t$, ")"}]}]}], ")"}]}], "+", RowBox[{$\[Rho]\_2$, " ", RowBox[{"(", RowBox[{$\(-g$\ $\[Eta](x, t)$\), "-", RowBox[{ SubsuperscriptBox["\[Phi]", "2", TagBox[$(0, 0, 1)$, Derivative], MultilineFunction->None], "(", $x, y, t$, ")"}], "-", RowBox[{"U2", " ", RowBox[{ SubsuperscriptBox["\[Phi]", "2", TagBox[$(1, 0, 0)$, Derivative], MultilineFunction->None], "(", $x, y, t$, ")"}]}]}], ")"}]}]}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "In ", StyleBox["Standard form", FontSlant->"Italic"], ", which is a little shorter, you would need the type set window to make \ this preactical. " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ $dc = \(-\((\[Gamma]\ \[PartialD]\_{x, 2}\[Eta][x, t])$\) - \[Rho]\_1\ $(\(-\[PartialD]\_t \[Phi]\_1[x, y, t]$ - U1\ \[PartialD]\_x \[Phi]\_1[x, y, t] - g\ \[Eta][x, t])\) + \[Rho]\_2\ $(\(-\[PartialD]\_t \[Phi]\_2[x, y, t]$ - U2\ \[PartialD]\_x \[Phi]\_2[x, y, t] - g\ \[Eta][x, t])\)\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{$-\[Gamma]$, " ", RowBox[{ SuperscriptBox["\[Eta]", TagBox[$(2, 0)$, Derivative], MultilineFunction->None], "(", $x, t$, ")"}]}], "-", RowBox[{$\[Rho]\_1$, " ", RowBox[{"(", RowBox[{$\(-g$\ $\[Eta](x, t)$\), "-", RowBox[{ SubsuperscriptBox["\[Phi]", "1", TagBox[$(0, 0, 1)$, Derivative], MultilineFunction->None], "(", $x, y, t$, ")"}], "-", RowBox[{"U1", " ", RowBox[{ SubsuperscriptBox["\[Phi]", "1", TagBox[$(1, 0, 0)$, Derivative], MultilineFunction->None], "(", $x, y, t$, ")"}]}]}], ")"}]}], "+", RowBox[{$\[Rho]\_2$, " ", RowBox[{"(", RowBox[{$\(-g$\ $\[Eta](x, t)$\), "-", RowBox[{ SubsuperscriptBox["\[Phi]", "2", TagBox[$(0, 0, 1)$, Derivative], MultilineFunction->None], "(", $x, y, t$, ")"}], "-", RowBox[{"U2", " ", RowBox[{ SubsuperscriptBox["\[Phi]", "2", TagBox[$(1, 0, 0)$, Derivative], MultilineFunction->None], "(", $x, y, t$, ")"}]}]}], ")"}]}]}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "Finally it may be more easily ", StyleBox["read", FontVariations->{"Underline"->True}], " in ", StyleBox["Traditional form", FontSlant->"Italic"], " as" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ FormBox[ RowBox[{"dc", "=", RowBox[{ RowBox[{"-", RowBox[{"(", RowBox[{"\[Gamma]", " ", FractionBox[$\[PartialD]\^2\( \[Eta](x, t)$\), $\[PartialD]x\^2$, MultilineFunction->None]}], ")"}]}], "-", RowBox[{$\[Rho]\_1$, " ", RowBox[{"(", RowBox[{ RowBox[{"-", FractionBox[$\[PartialD]\[Phi]\_1[x, y, t]$, $\[PartialD]t$, MultilineFunction->None]}], "-", RowBox[{"U1", " ", FractionBox[$\[PartialD]\[Phi]\_1[x, y, t]$, $\[PartialD]x$, MultilineFunction->None]}], "-", $g\ \(\[Eta](x, t)$\)}], ")"}]}], "+", RowBox[{$\[Rho]\_2$, " ", RowBox[{"(", RowBox[{ RowBox[{"-", FractionBox[$\[PartialD]\[Phi]\_2[x, y, t]$, $\[PartialD]t$, MultilineFunction->None]}], "-", RowBox[{"U2", " ", FractionBox[$\[PartialD]\[Phi]\_2[x, y, t]$, $\[PartialD]x$, MultilineFunction->None]}], "-", $g\ \(\[Eta](x, t)$\)}], ")"}]}]}]}], TraditionalForm]], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{$-\[Gamma]$, " ", RowBox[{ SuperscriptBox["\[Eta]", TagBox[$(2, 0)$, Derivative], MultilineFunction->None], "(", $x, t$, ")"}]}], "-", RowBox[{$\[Rho]\_1$, " ", RowBox[{"(", RowBox[{$\(-g$\ $\[Eta](x, t)$\), "-", RowBox[{ SubsuperscriptBox["\[Phi]", "1", TagBox[$(0, 0, 1)$, Derivative], MultilineFunction->None], "(", $x, y, t$, ")"}], "-", RowBox[{"U1", " ", RowBox[{ SubsuperscriptBox["\[Phi]", "1", TagBox[$(1, 0, 0)$, Derivative], MultilineFunction->None], "(", $x, y, t$, ")"}]}]}], ")"}]}], "+", RowBox[{$\[Rho]\_2$, " ", RowBox[{"(", RowBox[{$\(-g$\ $\[Eta](x, t)$\), "-", RowBox[{ SubsuperscriptBox["\[Phi]", "2", TagBox[$(0, 0, 1)$, Derivative], MultilineFunction->None], "(", $x, y, t$, ")"}], "-", RowBox[{"U2", " ", RowBox[{ SubsuperscriptBox["\[Phi]", "2", TagBox[$(1, 0, 0)$, Derivative], MultilineFunction->None], "(", $x, y, t$, ")"}]}]}], ")"}]}]}], TraditionalForm]], "Output"] }, Open ]], Cell["\<\ Note that you can convert any \"cryptic\" input expression by \ selecting the cell, going up to the cell menu and selecting \"convert to \ traditional form\". This is \"shift curly t\" on the keypad. \ \>", "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Governing equations", "Subsection"], Cell[CellGroupData[{ Cell[TextData[{ "The momentum equation for velocity in the radial direction, ", Cell[BoxData[ $TraditionalForm\`u\_r$]], " is:" }], "Subsubsection", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ $ureq = \[Rho] \((u\_r[ r, \[Theta]]\ \ \ \[PartialD]\_r u\_r[ r, \[Theta]] + \(u\_\[Theta][ r, \[Theta]]\/r$ \[PartialD]\_\[Theta] u\_r[ r, \[Theta]] - u\_\[Theta][r, \[Theta]]\^2\/r)\) + \[IndentingNewLine]\ \[PartialD]\_r p[ r, \[Theta]] - \[Mu] $(\ \[PartialD]\_r\((r\^2\ \[PartialD]\_r u\ \_r[r, \[Theta]])$\/r\^2 + \[PartialD]\_\[Theta]$(Sin[\[Theta]]\ \ \[PartialD]\_\[Theta] u\_r[r, \[Theta]])$\/$Sin[\[Theta]]\ r\^2$ - $2\ \ u\_r[r, \[Theta]]$\/r\^2 - $2\ \[PartialD]\_\[Theta] u\_\[Theta][r, \ \[Theta]]$\/r\^2 - $2\ u\_\[Theta][r, \[Theta]]\ \ Cot[\[Theta]]$\/r\^2)\)\)], "Input", AspectRatioFixed->True], 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None of \ the other terms have any of these variables. Since the equation is \ dimensionless, all of the terms have a magnitude of about one, or are much \ less than this or exactly zero. Thus the value of the Reynolds number (in \ this case much less than one) determines which terms should be be kept for \ the solution. \n\nFor the present problem we will eliminate the terms \ containing the Reynolds number by taking the limit of \[Mu]->\[Infinity]. \ This is the same as taking the Reynolds number going to 0. 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The obvious answer is that we need to \ reduce them to ODE's that we know how to solve. OK, how do we do this? We might expect that the general form of the solution is S(s) \ \[CapitalTheta](\[Theta]). If this starts to work there are formal expansion \ techniques in terms of orthogonal polynomials that could allow us to proceed \ with no other knowledge of the solution. However, this would be extremely \ tedious and involve use of mathmatics that you probably don't know. Thus we invoke physics. In the low Reynolds number limit, the qualitiative behavior of the flow field \ is very much determined by the boundaries because the fluid has no inertia \ and thus flows only where pushed. In this case the \"boundary\" is that far \ away the flow is in a straight line. Thus we look at the flow field at \ infinity to see its functional dependence. It could be that this will \ provide insight into the shape. From the far field solutions,the likely form of the solution is\ \>", "Text", CellTags->"boundary_determines_flow"], Cell[BoxData[ $\(usr = Cos[\[Theta]]\ F[s];$\)], "Input", AspectRatioFixed->True], Cell[BoxData[ $\(us\[Theta] = Sin[\[Theta]]\ G[s];$\)], "Input", AspectRatioFixed->True], Cell[TextData[ButtonBox["back to objectives(physical)", ButtonData:>"physical_objectives", ButtonStyle->"Hyperlink"]], "Text"], Cell[TextData[ButtonBox["back to objectives(mathematical)", ButtonData:>"mathematical_objectives", ButtonStyle->"Hyperlink"]], "Text"], Cell[TextData[ButtonBox["back to conclusions", ButtonData:>"conclusions", ButtonStyle->"Hyperlink"]], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["\<\ Go through the equations using the substitution to make ODE's\ \>", \ "Subsubsection"], Cell[CellGroupData[{ Cell["\<\ Start with the continuity equation. We make the substitutions and \ we are pleased to see that the \[Theta] dependance of every term is the same. \ So far the \"guessed\" form of the solution is working. \ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"conteqsoln", "=", RowBox[{"conteq2", "/.", RowBox[{"{", RowBox[{ RowBox[{ RowBox[{ SuperscriptBox[$u\&~\_r$, TagBox[$(a1_, a2_)$, Derivative], MultilineFunction->None], "[", $s, \[Theta]$, "]"}], "\[RuleDelayed]", $\[PartialD]\_\({s, a1}, {\[Theta], a2}$usr\ \)}], ",", RowBox[{ RowBox[{ SuperscriptBox[$u\&~\_\[Theta]$, TagBox[$(a1_, a2_)$, Derivative], MultilineFunction->None], "[", $s, \[Theta]$, "]"}], "\[RuleDelayed]", $\[PartialD]\_\({s, a1}, {\[Theta], a2}$us\ \[Theta]\)}], ",", $u\&~\_r[s, \[Theta]] \[Rule] usr$, ",", $u\&~\_\[Theta][s, \[Theta]] \[Rule] us\[Theta]$}], "}"}]}]}]], "Input", CellTags->"identical_factor"], Cell[BoxData[ FormBox[ RowBox[{$\(2\ \(cos(\[Theta])$\ $F(s)$ + 2\ $cos(\[Theta])$\ $G(s)$\)\/s\), "+", RowBox[{$cos(\[Theta])$, " ", RowBox[{ SuperscriptBox["F", "\[Prime]", MultilineFunction->None], "(", "s", ")"}]}]}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[ButtonBox["back to objectives(mathematical)", ButtonData:>"mathematical_objectives", ButtonStyle->"Hyperlink"]], "Text"], Cell["\<\ We will need to substitute for G[s] in terms of F[s] in the \ momentum equations so we can end up with a single dependent variable, F[s] in \ the final equation.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ $geeofs = Simplify[G[ s] /. \((Solve[conteqsoln \[Equal] 0, G[s]])$[$[1]$]]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{$1\/2$, " ", RowBox[{"(", RowBox[{$\(-2$\ $F(s)$\), "-", RowBox[{"s", " ", RowBox[{ SuperscriptBox["F", "\[Prime]", MultilineFunction->None], "(", "s", ")"}]}]}], ")"}]}], TraditionalForm]], "Output"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Now let's work on the momentum equations,", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"ursoln1", "=", RowBox[{"ureq6", "/.", 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\(cos(\[Theta])$\ $F(s)$\), "-", RowBox[{"2", " ", $cos(\[Theta])$, " ", RowBox[{"(", RowBox[{$\(-2$\ $F(s)$\), "-", RowBox[{"s", " ", RowBox[{ SuperscriptBox["F", "\[Prime]", MultilineFunction->None], "(", "s", ")"}]}]}], ")"}]}]}], $s\^2$]}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"uthsoln1", "=", RowBox[{"utheq6", "/.", RowBox[{"{", RowBox[{ RowBox[{ RowBox[{ SuperscriptBox[$u\&~\_r$, TagBox[$(a1_, a2_)$, Derivative], MultilineFunction->None], "[", $s, \[Theta]$, "]"}], "\[RuleDelayed]", $\[PartialD]\_\({s, a1}, {\[Theta], a2}$usr\ \)}], ",", RowBox[{ RowBox[{ SuperscriptBox[$u\&~\_\[Theta]$, TagBox[$(a1_, a2_)$, Derivative], MultilineFunction->None], "[", $s, \[Theta]$, "]"}], "\[RuleDelayed]", $\[PartialD]\_\({s, a1}, {\[Theta], a2}$us\ \[Theta]\)}], ",", $u\&~\_r[s, \[Theta]] \[Rule] usr$, ",", $u\&~\_\[Theta][s, \[Theta]] \[Rule] us\[Theta]$}], "}"}]}]}]], "Input", CellTags->"pthetaeq"], Cell[BoxData[ FormBox[ RowBox[{"-", RowBox[{$1\/s\^2$, RowBox[{"(", RowBox[{ RowBox[{$sin(\[Theta])$, " ", RowBox[{ SuperscriptBox["G", "\[Prime]\[Prime]", MultilineFunction->None], "(", "s", ")"}], " ", $s\^2$}], "+", RowBox[{"2", " ", $sin(\[Theta])$, " ", RowBox[{ SuperscriptBox["G", "\[Prime]", MultilineFunction->None], "(", "s", ")"}], " ", "s"}], "-", RowBox[{ RowBox[{ SuperscriptBox[$p\&~$, TagBox[$(0, 1)$, Derivative], MultilineFunction->None], "(", $s, \[Theta]$, ")"}], " ", "s"}], "+", $\(cos(\[Theta])$\ $cot(\[Theta])$\ $G(s)$\), "-", $\(csc(\[Theta])$\ $G(s)$\), "-", $2\ \(F(s)$\ $sin(\[Theta])$\), "-", $\(G(s)$\ $sin(\[Theta])$\)}], ")"}]}]}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[ButtonBox["back to objectives(mathematical)", ButtonData:>"mathematical_objectives", ButtonStyle->"Hyperlink"]], "Text"], Cell[TextData[{ "We can now eliminate ", StyleBox["G(s)", FontSlant->"Italic"], " and its derivatives with the results from the continuity equation. " }], "Text", CellTags->"eliminate_dependent_terms"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"uthsoln2", "=", RowBox[{"uthsoln1", "/.", RowBox[{"{", RowBox[{$G[s] \[Rule] geeofs$, ",", RowBox[{ RowBox[{ SuperscriptBox["G", TagBox[$(a1_)$, Derivative], MultilineFunction->None], "[", "s", "]"}], "\[RuleDelayed]", $\[PartialD]\_{s, a1}geeofs$}]}], "}"}]}]}]], "Input"], Cell[BoxData[ FormBox[ RowBox[{"-", RowBox[{$1\/s\^2$, RowBox[{"(", RowBox[{ RowBox[{$1\/2$, " ", $sin(\[Theta])$, " ", RowBox[{"(", RowBox[{ RowBox[{$-4$, " ", RowBox[{ SuperscriptBox["F", "\[Prime]\[Prime]", MultilineFunction->None], "(", "s", ")"}]}], "-", RowBox[{"s", " ", RowBox[{ SuperscriptBox["F", TagBox[$(3)$, Derivative], MultilineFunction->None], "(", "s", ")"}]}]}], ")"}], " ", $s\^2$}], "+", RowBox[{$sin(\[Theta])$, " ", RowBox[{"(", RowBox[{ RowBox[{$-3$, " ", RowBox[{ SuperscriptBox["F", "\[Prime]", MultilineFunction->None], "(", "s", ")"}]}], "-", RowBox[{"s", " ", RowBox[{ SuperscriptBox["F", "\[Prime]\[Prime]", MultilineFunction->None], "(", "s", ")"}]}]}], ")"}], " ", "s"}], "-", RowBox[{ RowBox[{ SuperscriptBox[$p\&~$, TagBox[$(0, 1)$, Derivative], MultilineFunction->None], "(", $s, \[Theta]$, ")"}], " ", "s"}], "-", $2\ \(F(s)$\ $sin(\[Theta])$\), "+", RowBox[{$1\/2$, " ", $cos(\[Theta])$, " ", $cot(\[Theta])$, " ", RowBox[{"(", RowBox[{$\(-2$\ $F(s)$\), "-", RowBox[{"s", " ", RowBox[{ SuperscriptBox["F", "\[Prime]", MultilineFunction->None], "(", "s", ")"}]}]}], ")"}]}], "-", RowBox[{$1\/2$, " ", $csc(\[Theta])$, " ", RowBox[{"(", RowBox[{$\(-2$\ $F(s)$\), "-", RowBox[{"s", " ", RowBox[{ SuperscriptBox["F", "\[Prime]", MultilineFunction->None], "(", "s", ")"}]}]}], ")"}]}], "-", RowBox[{$1\/2$, " ", $sin(\[Theta])$, " ", RowBox[{"(", RowBox[{$\(-2$\ $F(s)$\), "-", RowBox[{"s", " ", RowBox[{ SuperscriptBox["F", "\[Prime]", MultilineFunction->None], "(", "s", ")"}]}]}], ")"}]}]}], ")"}]}]}], TraditionalForm]], "Output"] }, Open ]] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["\<\ Eliminate pressure leaving a single dependent variable, F[s]. This \ increases the degree of the derivatives as a trade.\ \>", "Subsubsection"], Cell[TextData[{ "Now we need to combine these equations to eliminate something. The \ obvious thing is pressure. The way to do it is to take the \[Theta] \ derivative of the ", StyleBox["r", FontSlant->"Italic"], " equation and the ", StyleBox["r", FontSlant->"Italic"], " derivative of the the \[Theta] equation. Then subtract them to eliminate \ what now is the same 2nd derivative term of pressure. " }], "Text", CellTags->"eliminate_pressure"], Cell[TextData[ButtonBox["back to objectives(mathematical)", ButtonData:>"mathematical_objectives", ButtonStyle->"Hyperlink"]], "Text"], Cell[CellGroupData[{ Cell[BoxData[ $ursoln3 = D[\ ExpandAll[\ ursoln2], \[Theta]]$], "Input"], Cell[BoxData[ FormBox[ RowBox[{ FractionBox[ RowBox[{"4", " ", $sin(\[Theta])$, " ", RowBox[{ SuperscriptBox["F", "\[Prime]", MultilineFunction->None], "(", "s", ")"}]}], "s"], "+", RowBox[{$sin(\[Theta])$, " ", RowBox[{ SuperscriptBox["F", "\[Prime]\[Prime]", MultilineFunction->None], "(", "s", ")"}]}], "+", RowBox[{ SuperscriptBox[$p\&~$, TagBox[$(1, 1)$, Derivative], MultilineFunction->None], "(", $s, \[Theta]$, ")"}]}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ $uthsoln3 = D[FullSimplify[\ s\ \ uthsoln2], s]$], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"2", " ", $sin(\[Theta])$, " ", RowBox[{ SuperscriptBox["F", "\[Prime]\[Prime]", MultilineFunction->None], "(", "s", ")"}]}], "+", RowBox[{$1\/2$, " ", $sin(\[Theta])$, " ", RowBox[{"(", RowBox[{ RowBox[{"6", " ", RowBox[{ SuperscriptBox["F", "\[Prime]\[Prime]", MultilineFunction->None], "(", "s", ")"}]}], "+", RowBox[{"s", " ", RowBox[{ SuperscriptBox["F", TagBox[$(3)$, Derivative], MultilineFunction->None], "(", "s", ")"}]}]}], ")"}]}], "+", RowBox[{$1\/2$, " ", "s", " ", $sin(\[Theta])$, " ", RowBox[{"(", RowBox[{ RowBox[{"7", " ", RowBox[{ SuperscriptBox["F", TagBox[$(3)$, Derivative], MultilineFunction->None], "(", "s", ")"}]}], "+", RowBox[{"s", " ", RowBox[{ SuperscriptBox["F", TagBox[$(4)$, Derivative], MultilineFunction->None], "(", "s", ")"}]}]}], ")"}]}], "+", RowBox[{ SuperscriptBox[$p\&~$, TagBox[$(1, 1)$, Derivative], MultilineFunction->None], "(", $s, \[Theta]$, ")"}]}], TraditionalForm]], "Output"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Solve the single fourth order Euler equation.", "Subsubsection"], Cell["\<\ When this is done, the result is an Euler type differential \ equation. An Euler equation has the property that the power of the dependent \ variable multiplying a derivative, matches the power of the derivative. \ \ \>", "Text", CellTags->"Euler_equation"], Cell[CellGroupData[{ Cell[BoxData[ $Feq1 = ExpandAll[\(-2$\ s^2\ $(ursoln3 - uthsoln3)$/ Sin[\[Theta]]]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["F", TagBox[$(4)$, Derivative], MultilineFunction->None], "(", "s", ")"}], " ", $s\^4$}], "+", RowBox[{"8", " ", RowBox[{ SuperscriptBox["F", TagBox[$(3)$, Derivative], MultilineFunction->None], "(", "s", ")"}], " ", $s\^3$}], "+", RowBox[{"8", " ", RowBox[{ SuperscriptBox["F", "\[Prime]\[Prime]", MultilineFunction->None], "(", "s", ")"}], " ", $s\^2$}], "-", RowBox[{"8", " ", RowBox[{ SuperscriptBox["F", "\[Prime]", MultilineFunction->None], "(", "s", ")"}], " ", "s"}]}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[ButtonBox["back to objectives(mathematical)", ButtonData:>"mathematical_objectives", ButtonStyle->"Hyperlink"]], "Text"], Cell[TextData[{ "This solution can be obtained easily by assuming that ", Cell[BoxData[ $TraditionalForm\`F(s)\ = \ s\^n$]] }], "Text", CellTags->"How_to_solve_Euler"], Cell["The solution, which matches Middleman, is:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ $Fofs = F[s] /. \(DSolve[Feq1 \[Equal] 0, F[s], s]$[$[1]$]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox[ TagBox["c", C], "4"], " ", $s\^2$}], "+", SubscriptBox[ TagBox["c", C], "3"], "+", FractionBox[ SubscriptBox[ TagBox["c", C], "2"], "s"], "+", FractionBox[ SubscriptBox[ TagBox["c", C], "1"], $s\^3$]}], TraditionalForm]], "Output"] }, Open ]] }, Closed]] }, Open ]], Cell[CellGroupData[{ Cell["Boundary conditions in terms of F", "Subsection"], Cell[CellGroupData[{ Cell["Radial velocities", "Subsubsection"], Cell["\<\ The 0 radial velocity on the surface, s=1, is F=0. \ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ $fsbceq1 = Fofs /. s \[Rule] 1$], "Input"], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ TagBox["c", C], "1"], "+", SubscriptBox[ TagBox["c", C], "2"], "+", SubscriptBox[ TagBox["c", C], "3"], "+", SubscriptBox[ TagBox["c", C], "4"]}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "To match the free stream radial velocity at s-->\[Infinity],F=1. We can't \ directly substitute in s=\[Infinity], because of the s2 term. Actually, ", Cell[BoxData[ FormBox[ SubscriptBox[ TagBox["c", C], "4"], TraditionalForm]]], " will have to be 0. Thus as s-->\[Infinity], F[s] becomes just ", Cell[BoxData[ FormBox[ SubscriptBox[ TagBox["c", C], "3"], TraditionalForm]]], ". " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ $fsbceq2 = C[3]$], "Input"], Cell[BoxData[ FormBox[ SubscriptBox[ TagBox["c", C], "3"], TraditionalForm]], "Output"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Tangential velocities", "Subsubsection"], Cell["\<\ The 0 tangential velocity on the sphere is G=0 on s=1. Matching \ the free stream far away is G=-1 for s\[Rule]\[Infinity]. In terms of F[s] these are: F'[s]=0 @ s=1, F'[s]=0 @ s\[Rule]\[Infinity].\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ $fsbceq3 = D[Fofs, s] /. s \[Rule] 1$], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{$-3$, " ", SubscriptBox[ TagBox["c", C], "1"]}], "-", SubscriptBox[ TagBox["c", C], "2"], "+", RowBox[{"2", " ", SubscriptBox[ TagBox["c", C], "4"]}]}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "We don't do anything about the other infinity boundary condition because \ we already have ", Cell[BoxData[ FormBox[ SubscriptBox[ TagBox["c", C], "3"], TraditionalForm]]], "=0 there, so that the derivative of ", Cell[BoxData[ FormBox[ SubscriptBox[ TagBox["c", C], "3"], TraditionalForm]]], "is automatically 0. The two far away bounday conditions are fit with the \ same constant because of the applicability of separation of variables \ solution and the angular functional forms. " }], "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["\<\ Use the boundary equations to solve for the constants in the \ solution of the flow field.\ \>", "Subsection"], Cell["We can solve the three equations to get the three constants.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ $cees = Solve[{fsbceq1 \[Equal] 0, \ fsbceq2 \[Equal] 1, fsbceq3 \[Equal] 0}, {C[1], C[2], C[3]}] /. \(\(C[ 4]$$\[Rule]$$0$$\ $\)\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{"{", RowBox[{ RowBox[{ SubscriptBox[ TagBox["c", C], "1"], "\[Rule]", $1\/2$}], ",", RowBox[{ SubscriptBox[ TagBox["c", C], "2"], "\[Rule]", $-\(3\/2$\)}], ",", RowBox[{ SubscriptBox[ TagBox["c", C], "3"], "\[Rule]", "1"}]}], "}"}], "}"}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ $fsans = \((Fofs /. cees[\([1]$])\) /. C[4] \[Rule] 0\)], "Input"], Cell[BoxData[ $TraditionalForm\`1 - 3\/\(2\ s$ + 1\/$2\ s\^3$\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ $gsans = ExpandAll[\((\((geeofs /. {F[s] \[Rule] fsans, \(F'$[s] \[Rule] D[fsans, s]})\) /. cees[$[1]$])\) /. C[4] \[Rule] 0]\)], "Input"], Cell[BoxData[ $TraditionalForm\`\(-1$ + 3\/$4\ s$ + 1\/$4\ s\^3$\)], "Output"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Calculation of the pressure field", "Subsection"], Cell[TextData[{ "Both of these, of course, match Middleman. We now need to get the \ pressure field. It would have been easy to have obtained the angular form \ from the ", Cell[BoxData[ $TraditionalForm\`u\_r$]], " equation just by substituting in the angular expressions for ", Cell[BoxData[ $TraditionalForm\`u\_r$]], " and ", Cell[BoxData[ $TraditionalForm\`u\_\[Theta]$]], ". However, we did not do this, so we go back to the last time we saw the \ pressure in the momentum equations. This was the \[Theta] momentum equation. \ " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ $Simplify[uthsoln1/Sin[\[Theta]]]$], "Input"], Cell[BoxData[ FormBox[ FractionBox[ RowBox[{$2\ \(F(s)$\), "+", $2\ \(G(s)$\), "+", RowBox[{"s", " ", RowBox[{"(", RowBox[{ RowBox[{$-2$, " ", RowBox[{ SuperscriptBox["G", "\[Prime]", MultilineFunction->None], "(", "s", ")"}]}], "-", RowBox[{"s", " ", RowBox[{ SuperscriptBox["G", "\[Prime]\[Prime]", MultilineFunction->None], "(", "s", ")"}]}], "+", RowBox[{$csc(\[Theta])$, " ", RowBox[{ SuperscriptBox[$p\&~$, TagBox[$(0, 1)$, Derivative], MultilineFunction->None], "(", $s, \[Theta]$, ")"}]}]}], ")"}]}]}], $s\^2$], TraditionalForm]], "Output"] }, Open ]], Cell["We can then substitute for G[s] and F[s], ", "Text"], Cell[CellGroupData[{ Cell[BoxData[ $psoln1 = Simplify[uthsoln1 /. {F[s] \[Rule] fsans, \(F'$[s] \[Rule] D[fsans, s], G[s] -> gsans, \ $G'$[s] \[Rule] D[gsans, s], \[IndentingNewLine]$G''$[s] \[Rule] D[gsans, {s, 2}]}]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ FractionBox[ RowBox[{ SuperscriptBox[$p\&~$, TagBox[$(0, 1)$, Derivative], MultilineFunction->None], "(", $s, \[Theta]$, ")"}], "s"], "-", $\(3\ \(sin(\[Theta])$\)\/$2\ s\^3$\)}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "It should be apparent that the angular dependence is cos(\[Theta]) \ (because we take a \[Theta] derivative of ", StyleBox["p", FontSlant->"Italic"], " in this equation). We need the solution to match a boundary condition on \ pressure (recall that pressure appears as first derivatives). The obvious \ one for the ", StyleBox["r", FontSlant->"Italic"], " direction is that the pressure much match the far away pressure for the \ undisturbed flow, ", Cell[BoxData[ $p\&~\_\[Infinity]$]], ". " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ $p2soln = p\&~[s, \[Theta]] /. \(DSolve[\ Expand[s\ psoln1] \[Equal] 0, p\&~[s, \[Theta]], {s, \[Theta]}]$[$[1]$]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox[ TagBox["c", C], "1"], "[", "s", "]"}], "-", $\(3\ \(cos(\[Theta])$\)\/$2\ s\^2$\)}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ $p3soln = p2soln /. \(C[1]$[s] -> p\&~\_\[Infinity]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[$p\&~$, InterpretationBox["\[Infinity]", DirectedInfinity[ 1]]], "-", $\(3\ \(cos(\[Theta])$\)\/$2\ s\^2$\)}], TraditionalForm]], "Output"] }, Open ]], Cell["\<\ We can transform this back to dimensional pressure for later \ calculation.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ $pfinalsoln = Expand[\((p3soln /. {s \[Rule] r/R})$\ \[Mu]\ U/R]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ FractionBox[ RowBox[{"U", " ", "\[Mu]", " ", SubscriptBox[$p\&~$, InterpretationBox["\[Infinity]", DirectedInfinity[ 1]]]}], "R"], "-", $\(3\ R\ U\ \[Mu]\ \(cos(\[Theta])$\)\/$2\ r\^2$\)}], TraditionalForm]], "Output"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Velocity profiles", "Subsection"], Cell["\<\ Let's calculate the velocity profiles from the answers that we \ have.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ $ur1 = usr /. F[s] \[Rule] fsans$], "Input"], Cell[BoxData[ $TraditionalForm\`\((1 - 3\/\(2\ s$ + 1\/$2\ s\^3$)\)\ $cos(\[Theta])$\)], "Output"], Cell[CellGroupData[{ Cell[BoxData[ $urfinal = U\ ur1 /. s \[Rule] r/R$], "Input"], Cell[BoxData[ $TraditionalForm\`\((R\^3\/\(2\ r\^3$ - $3\ R$\/$2\ r$ + 1)\)\ U\ $cos(\[Theta])$\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ $uth1 = us\[Theta] /. G[s] \[Rule] gsans$], "Input"], Cell[BoxData[ $TraditionalForm\`\((\(-1$ + 3\/$4\ s$ + 1\/$4\ s\^3$)\)\ $sin(\[Theta])$\)], "Output"], Cell[CellGroupData[{ Cell[BoxData[ $uthfinal = U\ uth1 /. s \[Rule] r/R$], "Input"], Cell[BoxData[ $TraditionalForm\`\((R\^3\/\(4\ r\^3$ + $3\ R$\/$4\ r$ - 1)\)\ U\ $sin(\[Theta])$\)], "Output"] }, Open ]] }, Open ]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Examination of the solution", "Subtitle"], Cell[CellGroupData[{ Cell["Contour plot of the velocity field", "Subsection"], Cell[TextData[{ "At this point it is interesting to look at the velocity field. A way to \ do it is to look at the mean velocity as a function of the distance from the \ sphere and the angular location. This is done by computing a total velocity, \ ", Cell[BoxData[ $TraditionalForm\`vr\^2$]], " + ", Cell[BoxData[ $TraditionalForm\`v\[Theta]\^2$]], ". " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ $ufield1\ = Sqrt[urfinal\^2 + uthfinal\^2]$], "Input", AspectRatioFixed->True], Cell[BoxData[ $TraditionalForm\`\@\(\((R\^3\/\(2\ r\^3$ - $3\ R$\/$2\ r$ + \ 1)\)\^2\ U\^2\ $\(cos\^2$(\[Theta])\) + $(R\^3\/\(4\ r\^3$ + $3\ \ R$\/$4\ r$ - 1)\)\^2\ U\^2\ $\(sin\^2$(\[Theta])\)\)\)], "Output"] }, Open ]], Cell["Use the polar to Cartesian transformation:", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ $ucart = ufield1 /. {r \[Rule] \@\(x\^2 + y\^2$, \[Theta] \[Rule] ArcTan[y\/x]}\)], "Input", AspectRatioFixed->True], Cell[BoxData[ $TraditionalForm\`\@\(\(U\^2\ \((R\^3\/\(2\ \((x\^2 + y\^2)$\^$3/2$\) \ - $3\ R$\/$2\ \@\(x\^2 + y\^2$\) + 1)\)\^2\)\/$y\^2\/x\^2 + 1$ + $U\^2\ \ y\^2\ \((R\^3\/\(4\ \((x\^2 + y\^2)$\^$3/2$\) + $3\ R$\/$4\ \@\(x\^2 \ + y\^2$\) - 1)\)\^2\)\/$x\^2\ \((y\^2\/x\^2 + 1)$\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ $\(plot1 = ContourPlot[ ucart /. {R \[Rule] 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.58601 L .52721 .59121 L .51361 .59424 L .5 .59524 L s % End of Graphics MathPictureEnd \ \>"], "Graphics", Evaluatable->False, AspectRatioFixed->True, ImageSize->{401, 401}, ImageMargins->{{34, 0}, {0, 0}}, ImageRegion->{{0, 1}, {0, 1}}, ImageCacheValid->False] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Drag on the sphere", "Subsection"], Cell["\<\ We now get to the point of the problem, find the drag on the \ sphere. \ \>", "Text"], Cell[CellGroupData[{ Cell["Form drag", "Subsubsection"], Cell[TextData[{ "We need to integrate the normal stress, resolved onto the flow direction \ all over the sphere. The normal stress is -p + 2 \[Mu] ", Cell[BoxData[ $TraditionalForm\`\[PartialD]u\_r[r]\/\[PartialD]r$]], ". It must be resolved onto the flow direction which takes Cos[\[Theta]]. \ The area element for a sphere is ", Cell[BoxData[ $TraditionalForm\`R\^2$]], "Sin[\[Theta]] d\[Theta] d\[Phi]. The d\[Phi] provides a 2 \[Pi] and the \ integral becomes:\n\n2 \[Pi] Integrate[-p + 2 \[Mu] D[u[r],r]Cos[\[Theta]] ", Cell[BoxData[ $TraditionalForm\`R\^2$]], " Sin[\[Theta]] ,{\[Theta],0,\[Pi]}]." }], "Text", CellLabel->"In[81]:="], Cell[CellGroupData[{ Cell[BoxData[ $argument = \((\((\(-p$ + 2\ \[Mu]\ D[u[r], r])\) Cos[\[Theta]]\ R\^2\ Sin[\[Theta]])\) /. {p \[Rule] pfinalsoln, \ D[u[r], r] \[Rule] \ D[urfinal, r]}\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{$R\^2$, " ", $cos(\[Theta])$, " ", $sin(\[Theta])$, " ", RowBox[{"(", RowBox[{$\(3\ R\ U\ \[Mu]\ \(cos(\[Theta])$\)\/$2\ r\^2$\), "+", $2\ \((\(3\ R$\/$2\ r\^2$ - $3\ R\^3$\/$2\ r\^4$)\)\ \ U\ \[Mu]\ $cos(\[Theta])$\), "-", FractionBox[ RowBox[{"U", " ", "\[Mu]", " ", SubscriptBox[$p\&~$, InterpretationBox["\[Infinity]", DirectedInfinity[ 1]]]}], "R"]}], ")"}]}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ $FullSimplify[argument /. r \[Rule] R]$], "Input"], Cell[BoxData[ FormBox[ RowBox[{$1\/4$, " ", "R", " ", "U", " ", "\[Mu]", " ", $sin(2\ \[Theta])$, " ", RowBox[{"(", RowBox[{$3\ \(cos(\[Theta])$\), "-", RowBox[{"2", " ", SubscriptBox[$p\&~$, InterpretationBox["\[Infinity]", DirectedInfinity[ 1]]]}]}], ")"}]}], TraditionalForm]], "Output"] }, Open ]], Cell["\<\ The result of this integration is the \"form drag\", that is drag \ caused by normal stress.\ \>", "Text", CellTags->"form_drag"], Cell[TextData[ButtonBox["back to objectives(physical)", ButtonData:>"physical_objectives", ButtonStyle->"Hyperlink"]], "Text"], Cell[TextData[ButtonBox["back to conclusions", ButtonData:>"conclusions", ButtonStyle->"Hyperlink"]], "Text"], Cell[CellGroupData[{ Cell[BoxData[ $formdrag = 2\ \[Pi]\ Integrate[argument, {\[Theta], 0, \[Pi]}] /. r \[Rule] R$], "Input"], Cell[BoxData[ $TraditionalForm\`2\ \[Pi]\ R\ U\ \[Mu]$], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Skin drag", "Subsubsection"], Cell[TextData[{ "Now we need to integrate the shear stress, resolved onto the flow \ direction all over the sphere. The shear stress is ", Cell[BoxData[ $TraditionalForm\`T\_r\[Theta] = \ \(-\(\[Mu](\ r\ \(\[PartialD]\/\[PartialD]r$ $(u\_\[Theta]\/r)$ + \ \ $1\/r$ \[PartialD]u\_r\/\[PartialD]r)\)\)\)]], ". It must be resolved onto the flow direction which takes Sin[\[Theta]]. \ The area element for a sphere is ", Cell[BoxData[ $TraditionalForm\`R\^2$]], "Sin[\[Theta]] d\[Theta] d\[Phi]. The d\[Phi] provides a 2 \[Pi] and the \ integral becomes:\n\n2 \[Pi] Integrate[-p + 2 \[Mu] D[u[r],r]Cos[\[Theta]] ", Cell[BoxData[ $TraditionalForm\`R\^2$]], " Sin[\[Theta]] ,{\[Theta],0,\[Pi]}]." }], "Text", CellLabel->"In[81]:="], Cell[CellGroupData[{ Cell[BoxData[ $\((\ \(-\[Mu]$ $(\ r\ \[PartialD]\_r\((u\_\[Theta][r, \[Theta]]\/r)$ + \ $1\/r$ \ \[PartialD]\_\[Theta] u\_r[r, \[Theta]])\) Sin[\[Theta]]\ R\^2\ Sin[\[Theta]])\)\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{$-R\^2$, " ", "\[Mu]", " ", $\(sin\^2$(\[Theta])\), " ", RowBox[{"(", RowBox[{ FractionBox[ RowBox[{ SubsuperscriptBox["u", "r", TagBox[$(0, 1)$, Derivative], MultilineFunction->None], "(", $r, \[Theta]$, ")"}], "r"], "+", RowBox[{"r", " ", RowBox[{"(", RowBox[{ FractionBox[ RowBox[{ SubsuperscriptBox["u", "\[Theta]", TagBox[$(1, 0)$, Derivative], MultilineFunction->None], "(", $r, \[Theta]$, ")"}], "r"], "-", $\(\(u\_\[Theta]$(r, \[Theta])\)\/r\^2\)}], ")"}]}]}], ")"}]}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ $argumenttang = \((\ \(-\[Mu]$ $(\ r\ \[PartialD]\_r\((u\_\[Theta][r, \[Theta]]\/r)$ + \ $1\/r$ \ \[PartialD]\_\[Theta] u\_r[r, \[Theta]])\) Sin[\[Theta]]\ R\^2\ Sin[\[Theta]])\) /. {\ D[u\_r[r, \[Theta]], \[Theta]] \[Rule] \ D[urfinal, \[Theta]], \[PartialD]\_r\ u\_\[Theta][r, \[Theta]] \[Rule] D[uthfinal, r], u\_\[Theta][r, \[Theta]] \[Rule] uthfinal}\)], "Input"], Cell[BoxData[ $TraditionalForm\`\(-R\^2$\ \[Mu]\ $\(sin\^2$(\[Theta])\)\ $(r\ \ \((\(\((\(-\(\(3\ R\^3$\/$4\ r\^4$\)\) - $3\ R$\/$4\ r\^2$)\)\ U\ \ $sin(\[Theta])$\)\/r - $\((R\^3\/\(4\ r\^3$ + $3\ R$\/$4\ r$ - 1)\)\ \ U\ $sin(\[Theta])$\)\/r\^2)\) - $\((R\^3\/\(2\ r\^3$ - $3\ R$\/$2\ r$ \ + 1)\)\ U\ $sin(\[Theta])$\)\/r)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ $FullSimplify[argumenttang /. r \[Rule] R]$], "Input"], Cell[BoxData[ $TraditionalForm\`3\/2\ R\ U\ \[Mu]\ \(\(sin\^3$(\[Theta])\)\)], \ "Output"] }, Open ]], Cell["\<\ The result of this integration is the \"skin drag\", that is drag \ caused by tangential stress.\ \>", "Text", CellTags->"skin_drag"], Cell[TextData[ButtonBox["back to objectives(physical)", ButtonData:>"physical_objectives", ButtonStyle->"Hyperlink"]], "Text"], Cell[TextData[ButtonBox["back to conclusions", ButtonData:>"conclusions", ButtonStyle->"Hyperlink"]], "Text"], Cell[CellGroupData[{ Cell[BoxData[ $skindrag = 2\ \[Pi]\ Integrate[argumenttang, {\[Theta], 0, \[Pi]}] /. r \[Rule] R$], "Input"], Cell[BoxData[ $TraditionalForm\`4\ \[Pi]\ R\ U\ \[Mu]$], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ $totaldrag = formdrag + skindrag$], "Input"], Cell[BoxData[ $TraditionalForm\`6\ \[Pi]\ R\ U\ \[Mu]$], "Output"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Contour plot of the pressure field", "Subsection"], Cell["\<\ At this point it is interesting to look at the pressure field. \ \ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ $pfield\ = pfinalsoln$], "Input", AspectRatioFixed->True], 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Note that the contours smoothly change as we \ approach the sphere. The darker, the shading, the lower the pressure. The \ point of this calculation is that the pressure just decreases toward the \ sphere. There is no region of pressure that is higher than the far away \ pressure. You might ask why this is significant. Have you ever noticed how the wind \ blowing against a wall exerts a force on the wall? This is because the \ pressure increases as the fluid is slowed down by the wall. The fluid looses \ inertia as its velocity decreases. Momentum is conserved by increasing the \ pressure. 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00<006>oHkl02V>o00<006>oHkl02V>o00<006>oHkl02V>o00<006>oHkl02V>o00<006>oHkl02V>o 00<006>oHkl02V>o00<006>oHkl02V>o00<006>oHkl02V>o00<006>oHkl02F>o00<006>oHkl02V>o 00<006>oHkl02V>o00<006>oHkl02V>o00<006>oHkl02V>o00<006>oHkl02V>o00<006>oHkl02V>o 00<006>oHkl02V>o00<006>oHkl02V>o00<006>oHkl00f>o0@000F>o000?Hkoo000A0001Hkl00?mS _b5S_`00\ \>"], ImageRangeCache->{{{0, 287}, {287, 0}} -> {-5.84543, -5.78983, 0.0386771, \ 0.0386771}}] }, Open ]], Cell[TextData[{ "The code is as the shading gets darker, the velocity is slower. We see \ that coming in the y=0 axis from x=-Infinity, the velocity is decreasing to \ agree with the: no flow through the surface boundary condition. Looking at \ the x=0 axis, coming in from y=\[Infinity], the velocity is increasing \ reflecting the deflection of fluid around the sphere. This results in a ", StyleBox["Bernoulli", FontSlant->"Italic"], " effect with local fluid acceleration. Below we check some numbers to \ verify these statements. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ $N[vcartinvis /. {x \[Rule] .01, y \[Rule] 1}]$], "Input", AspectRatioFixed->True], Cell[BoxData[ $TraditionalForm\`2.249550078739689`$], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ $N[vcartinvis /. {x \[Rule] 1, y \[Rule] \(- .001$}]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ $TraditionalForm\`2.2499977499999997`^-6$], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ $N[vcartinvis /. {x \[Rule] 100, y \[Rule] 100}]$], "Input", AspectRatioFixed->True], Cell[BoxData[ $TraditionalForm\`0.9999998232233829`$], "Output"] }, Open ]], Cell["\<\ The same question can be asked about the pressure field. We will \ find that the pressure is always at or below the far away value for Re=0. \ Substitute for the pressure field equation, \ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Pressure field for high Reynolds number:", "Subsubsection"], Cell["\<\ In contrast, if a high velocity fluid is impinging on a surface, \ there will be a pressure increase. 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.52721 .59121 L .51361 .59424 L .5 .59524 L s % End of Graphics MathPictureEnd \ \>"], "Graphics", ImageSize->{361, 361}, ImageMargins->{{54, 0}, {0, 0}}, ImageRegion->{{0, 1}, {0, 1}}, ImageCacheValid->False] }, Open ]], Cell["\<\ So that the pressure is highest at y=0, x=1,-1 (i.e. the stagnation \ point) and decreases out to infinity. The pressure is lowest, consistent \ with the Bernoulli effect, at y=1,-1,x=0, and increases out to y=infinity. \ \ \>", "Text", Evaluatable->False, AspectRatioFixed->True] }, Open ]], Cell[TextData[{ "A stream function, which is a function that is always tangent to the \ velocity field, is another way to view the solution of this problem. The Re \ = 0 solution can be used to write the stream function which is, -1/2f[r] ", Cell[BoxData[ $TraditionalForm\`r\^2$]], " ", Cell[BoxData[ $TraditionalForm\`sin[\[Theta]]\^2$]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ $re0stream = 1\/2\ fsans\ r\^2\ Sin[\[Theta]]\^2 /. {s \[Rule] r/R}$], "Input", AspectRatioFixed->True], Cell[BoxData[ $TraditionalForm\`1\/2\ r\^2\ \((R\^3\/\(2\ r\^3$ - $3\ R$\/$2\ r$ \ + 1)\)\ $\(sin\^2$(\[Theta])\)\)], "Output"] }, Open ]], Cell["\<\ We can make a plot of this if we transform from polar to Cartesian \ coordinates\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ $stream1 = re0stream /. {r \[Rule] \@\(x\^2 + y\^2$, \[Theta] \[Rule] ArcTan[y\/x], R \[Rule] 1}\)], "Input", AspectRatioFixed->True], Cell[BoxData[ $TraditionalForm\`\(y\^2\ \((x\^2 + y\^2)$\ $(1 - 3\/\(2\ \@\(x\^2 + y\ \^2$\) + 1\/$2\ \((x\^2 + y\^2)$\^$3/2$\))\)\)\/$2\ x\^2\ \((y\^2\/x\^2 \ + 1)$\)\)], "Output"] }, Open ]], Cell[BoxData[ $\(plot5 = ContourPlot[stream1, {x, \(-5$, 5}, {y, $-5$, 5}, Contours \[Rule] 15, ContourShading \[Rule] False, DisplayFunction 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Creeping flow is a term used for a flow that have effectively no \ inertia. In this case the ", ButtonBox["inertia terms are neglected and the solution is obtained from \ the resulting linear equations", ButtonData:>"neglect_inertia_terms", ButtonStyle->"Hyperlink"], ". The Reynolds number is very much smaller than unity.\n \n2. The \ solution technique involves using a solution form that is ", ButtonBox["deduced from boundaries", ButtonData:>"boundary_determines_flow", ButtonStyle->"Hyperlink"], " of the flow field, faraway from the sphere. \n\n3. Because viscous \ forces dominate the flow field, the fluid can never accelerate above the free \ stream value even if an obstacle causes the fluid to be squeezed. Thus the \ velocity in the region of the sphere just ", ButtonBox["slows down and then returns to the free stream value", ButtonData:>"fluid_does_not_speed_up", ButtonStyle->"Hyperlink"], ". \n\n4. Both normal stresses and tangential stresses contribute to the \ drag on the sphere. These can be termed ", ButtonBox["form drag", ButtonData:>"form_drag", ButtonStyle->"Hyperlink"], " and ", ButtonBox["skin drag", ButtonData:>"skin_drag", ButtonStyle->"Hyperlink"], ". Note that both of these are ", StyleBox["linear", FontWeight->"Bold"], " in the velocity (consistent with the linear governing equations) and \ fluid viscosity. The density, and thus the Reynolds number, does not appear. \ \n\n5. Consistent with the fluid not accelerating, the ", ButtonBox["pressure never increases above the free stream value", ButtonData:>"pressure_decreases", ButtonStyle->"Hyperlink"], ". The fluid has no inertia that would cause a pressure increase as the \ fluid slows down.\n\n6. The velocity decays slowly (as ", Cell[BoxData[ $TraditionalForm\`1\/r$]], ") and thus the disturbance is felt very far away from the sphere. This \ makes it difficult to do a real experiment, in a reasonable size container, \ that allows that sphere to fall at a speed specified by the drag that is \ predicted from the analysis here. ", ButtonBox["The very high Reynolds number case decays much faster", ButtonData:>"disturbance_felt_faraway", ButtonStyle->"Hyperlink"], ". 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