Hutton - Fundamentals of Finite Element Analysis (523155), страница 50
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. . , M(7.76)Inserting the matrix form of Equation 7.71 and rearranging, we have∂ [N ] ∂ N i∂ [N ] ∂ N i∂ [N ] ∂ N ikx+ ky+ kz{T } dV∂x ∂x∂y ∂y∂z ∂zV=Q N i dV − (q x n x + q y n y + q z n z ) N i d Ai = 1, . . . , M (7.77) VAEquation 7.77 represents a system of M algebraic equations in the M unknownnodal temperatures {T}. With the exception that convection effects are not included here, Equation 7.77 is analogous to the two-dimensional case representedby Equation 7.34. In matrix notation, the system of equations for the threedimensional element formulation is ∂ [N ] T ∂ [N ]∂ [N ] T ∂ [N ]∂ [N ] T ∂ [N ]kx+ ky+ kzdV {T }∂x∂x∂y∂y∂z∂zVT=Q[N ] dV − (q x n x + q y n y + q z n z )[N ] T d A(7.78)VAand Equation 7.76 is in the desired form (e) (e) (e) (e) kT= fQ + f q(7.79)Comparing the last two equations, the element conductance (stiffness) matrix is (e) ∂ [N ] T ∂ [N ]∂ [N ] T ∂ [N ]∂ [N ] T ∂ [N ]k=kx+ ky+ kzdV∂x∂x∂y∂y∂z∂zV(7.80)Hutton: Fundamentals ofFinite Element Analysis7.
Applications in HeatTransferText© The McGraw−HillCompanies, 20047.6 Heat Transfer in Three Dimensionsthe element force vector representing internal heat generation is (e) fQ =Q[N ] T dV(7.81)Vand the element nodal force vector associated with heat flux across the elementsurface area is (e) f q = − (q x n x + q y n y + q z n z )[N ] T d A(7.82)A7.6.1 System Assembly and Boundary ConditionsThe procedure for assembling the global equations for a three-dimensionalmodel for heat transfer analysis is identical to that of one- and two-dimensionalproblems. The element type is selected (tetrahedral, brick, quadrilateral solid,for example) based on geometric considerations, primarily. The volume is thendivided into a mesh of elements by first defining nodes (in the global coordinatesystem) throughout the volume then each element by the sequence and numberof nodes required for the element type.
Element-to-global nodal correspondencerelations are then determined for each element, and the global stiffness (conductance) matrix is assembled. Similarly, the global force vector is assembledby adding element contributions at nodes common to two or more elements.The latter procedure is straightforward in the case of internal generation, asgiven by Equation 7.81. However, in the case of the element gradient terms,Equation 7.82, the procedure is best described in terms of the global boundaryconditions.In the case of three-dimensional heat transfer, we have the same three typesof boundary conditions as in two dimensions: (1) specified temperatures,(2) specified heat flux, and (3) convection conditions.
The first case, specifiedtemperatures, is taken into account in the usual manner, by reducing the systemequations by simply substituting the known nodal temperatures into the systemequations. The latter two cases involve only elements that have surfaces (elementfaces) on the outside surface of the global volume. To illustrate, Figure 7.17ashows two brick elements that share a common face in an assembled finite element model. For convenience, we take the common face to be perpendicular tothe x axis.
In Figure 7.17b, the two elements are shown separately with the associated normal vector components identified for the shared faces. For steady-stateheat transfer, the heat flux across the face is the same for each element and, sincethe unit normal vectors are opposite, the gradient force terms cancel. The resultis completely analogous to internal forces in a structural problem via Newton’sthird law of action and reaction.
Therefore, on interelement boundaries (whichare areas for three-dimensional elements), the element force terms defined byEquation 7.82 sum to zero in the global assembly process.What of the element surface areas that are part of the surface area of the volume being modeled? Generally, these outside areas are subjected to convection269Hutton: Fundamentals ofFinite Element Analysis2707. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transferyz12x(a)qxqx12n៝1 (1, 0, 0)៝n2 (1, 0, 0)x(b)Figure 7.17(a) Common face in two 3-D elements. (b) Edge view ofcommon face, illustrating cancellation of conductiongradient terms.conditions.
For such convection boundary conditions, the flux conditions ofEquation 7.82 must be in balance with the convection from the area of concern.Mathematically, the condition is expressed as (e) f q = − (q x n x + q y n y + q z n z )[N ] T d A = − qn n[N ] T d AA*)= − h T (e) − Ta [N ] T d AA(7.83)Awhere qn is the flux normal to the surface area A of a specific element face on theglobal boundary and n is the unit outward normal vector to that face. As in twodimensional analysis, the convection term in the rightmost integral of Equation7.83 adds to the stiffness matrix when the expression for T (e) in terms of interpolation functions and nodal temperatures is substituted.
Similarly, the ambienttemperature terms add to the forcing function vector.In most commercial finite element software packages, the three-dimensionalheat transfer elements available do not explicitly consider the gradient force vector represented by Equation 7.82. Instead, such programs compute the system(global) stiffness matrix on the basis of conductance only and rely on the user tospecify the flux or convection boundary conditions (and the specified temperature conditions, of course) as part of the loading (input) data.Hutton: Fundamentals ofFinite Element Analysis7.
Applications in HeatTransferText© The McGraw−HillCompanies, 20047.7 Axisymmetric Heat TransferOwing to the algebraic volume of calculation required, examples of generalthree-dimensional heat transfer are not presented here. A few three-dimensionalproblems are included in the end-of-chapter exercises and are intended to besolved by digital computer techniques.7.7 AXISYMMETRIC HEAT TRANSFERChapter 6 illustrated the approach for utilizing two-dimensional elements andassociated interpolation functions for axisymmetric problems. Here, we illustratethe formulation of finite elements to solve problems in axisymmetric heat transfer. Illustrated in Figure 7.18 is a body of revolution subjected to heat input at itsbase, and the heat input is assumed to be symmetric about the axis of revolution.Think of the situation as a cylindrical vessel heated by a source, such as a gasflame.
This situation could, for example, represent a small crucible for meltingmetal prior to casting.As an axisymmetric problem is three-dimensional, the basic governingequation is Equation 7.70, restated here under the assumption of homogeneity, sothat k x = k y = k z = k , as 2∂ T∂T 2∂ 2Tk+++Q=0(7.84)∂x2∂ y2∂ z2Equation 7.84 is applicable to steady-state conduction only and is expressedin rectangular coordinates. For axisymmetric problems, use of a cylindricalcoordinate system (r, , z ) is much more amenable to formulating the problem.To convert to cylindrical coordinates, the partial derivatives with respect to x andy in Equation 7.84 must be converted mathematically into the corresponding partial derivatives with respect to radial coordinate r and tangential (circumferential)zrqzFigure 7.18 An axisymmetricheat transfer problem.
Allproperties and inputs aresymmetric about the z axis.271Hutton: Fundamentals ofFinite Element Analysis2727. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transfercoordinate . In the following development, we present the general approach butleave the details as an end-of-chapter exercise.The basic relations between the rectangular coordinates x, y and the cylindrical (polar) coordinates r, arex = r cos y = r sin (7.85)r 2 = x 2 + y2ytan =x(7.86)and inversely,Per the chain rule of differentiation, we have∂T∂ T ∂r∂T ∂=+∂x∂r ∂ x∂ ∂x∂T∂ T ∂r∂T ∂=+∂y∂r ∂ y∂ ∂y(7.87)By implicit differentiation of Equation 7.86,2r∂rx∂r= 2x ⇒= = cos ∂x∂xr2r∂r∂ry= 2y ⇒= = sin ∂y∂yrysin ∂1 ∂=− 2 ⇒=−2sec ∂ xx∂xr(7.88)1 ∂1∂cos = ⇒=2sec ∂ yx∂yrso that Equation 7.87 becomes∂T∂Tsin ∂ T= cos −∂x∂rr ∂∂T∂Tcos ∂ T= sin +∂y∂rr ∂For the second partial derivatives, we have∂ 2T∂ ∂T∂ ∂Tsin ∂ ∂ T== cos −∂x2∂x ∂x∂r ∂ xr ∂ ∂x∂ 2T∂ ∂T∂ ∂Tcos ∂ ∂ T== sin +∂ y2∂y ∂y∂r ∂ yr ∂ ∂y(7.89)(7.90)Hutton: Fundamentals ofFinite Element Analysis7.
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