Hutton - Fundamentals of Finite Element Analysis (523155), страница 51
Текст из файла (страница 51)
Applications in HeatTransferText© The McGraw−HillCompanies, 20047.7 Axisymmetric Heat TransferApplying Equation 7.89 to the operations indicated in Equation 7.90 yields (withappropriate use of trigonometric identities)∂ 2T∂ 2T1 ∂T∂ 2T1 ∂ 2T+=++∂x2∂ y2∂r 2r ∂rr 2 ∂ 2(7.91)where the derivation represents a general change of coordinates. To relate to anaxisymmetric problem, recall that there is no dependence on the tangential coordinate . Consequently, when Equations 7.84 and 7.91 are combined, the governing equation for axisymmetric heat transfer is 2∂ T1 ∂T∂ 2Tk+++Q=0(7.92)∂r 2r ∂r∂ z2and, of course, note the absence of the tangential coordinate.7.7.1 Finite Element FormulationPer the general procedure, the total volume of the axisymmetric domain is discretized into finite elements.
In each element, the temperature distribution isexpressed in terms of the nodal temperatures and interpolation functions asT (e) =MN i (r, z)T(e)i(7.93)i=1where, as usual, M is the number of element nodes. Note particularly that theinterpolation functions vary with radial coordinate r and axial coordinate z.Application of Galerkin’s method using Equations 7.92 and 7.93 yields theresidual equations 2∂ T1 ∂T∂ 2Tk+++QN i r dr d dz = 0i = 1, . .
. , M∂r 2r ∂r∂ z2V(7.94)Observing that, for the axisymmetric case, the integrand is independent of thetangential coordinate , Equation 7.94 becomes 2∂ T1 ∂T∂ 2T2k+++ Q N i r dr dz = 0i = 1, . . . , M∂r 2r ∂r∂ z2A(e)(7.95)where A (e) is the area of the element in the rz plane. The first two terms of theintegrand can be combined to obtain 1 ∂∂T∂ 2T2kr++ Q N i r dr dz = 0i = 1, .
. . , Mr ∂r∂r∂ z2A(e)(7.96)273Hutton: Fundamentals ofFinite Element Analysis2747. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat TransferObserving that r is independent of z, Equation 7.96 becomes ∂∂T∂∂Tr+r+ Qr N i dr dz = 02k∂r∂r∂z∂zA(e)i = 1, . . . , M(7.97)As in previous developments, we invoke the chain rule of differentiation as, forexample,∂∂T∂∂∂T∂ T ∂ Ni∂Tr Ni= Nir+r⇒ Nir∂r∂r∂r∂r∂r ∂r∂r∂r∂∂ T ∂ Ni∂T=r Ni−ri = 1, .
. . , M(7.98)∂r∂r∂r ∂rNoting that Equation 7.98 is also applicable to the z variable, the residual equations represented by Equation 7.97 can be written as ∂∂∂T∂T2kr Ni+r Nidr dz + 2Q N i r dr dz∂r∂r∂z∂zA(e)A( E) ∂ T ∂ Ni∂ T ∂ Ni= 2k+r dr dzi = 1, . . . , M(7.99)∂r ∂r∂z ∂zA(e)The first integrand on the left side of Equation 7.99 is a perfect differential in twodimensions, and the Green-Gauss theorem can be applied to obtain ∂T∂T2knr + kn z r N i dS + 2Q N i r dr dz∂r∂z(e)S (e) A ∂ T ∂ Ni∂ T ∂ Ni= 2k+r dr dzi = 1, . . .
, M(7.100)∂r ∂r∂z ∂zA(e)where S is the boundary (periphery) of the element and n r and n z are the radialand axial components of the outward unit vector normal to the boundary. Applying Fourier’s law in cylindrical coordinates,∂T∂r∂Tq z = −k∂zqr = −k(7.101)and noting the analogy with Equation 7.33, we rewrite Equation 7.100 as ∂ T ∂ Ni∂ T ∂ Ni2k+r dr dz∂r ∂r∂z ∂zA(e)= 2Q N i r dr dz − 2 qs n s N i r dSi = 1, . . . , M(7.102)A(e)S (e)Hutton: Fundamentals ofFinite Element Analysis7.
Applications in HeatTransferText© The McGraw−HillCompanies, 20047.7 Axisymmetric Heat Transfer275The common term 2 could be omitted, but we leave it as a reminder of thethree-dimensional nature of an axisymmetric problem. In particular, note thatthis term, in conjunction with r in the integrand of the last integral on the righthand side of Equation 7.102, reinforces the fact that element boundaries areactually surfaces of revolution.Noting that Equation 7.102 represents a system of M equations, the form ofthe system is that of (e) (e) (e) (e) kT= fQ + f g(7.103)where [k (e) ] is the element conductance matrix having individual terms definedby ∂ Ni ∂ N j∂ Ni ∂ N jk i j = 2k+r dr dzi, j = 1, .
. . , M(7.104)∂r ∂r∂z ∂zA(e)and {T (e) } is the column matrix (vector) of element nodal temperatures per Equation 7.93. The element forcing functions include the internal heat generationterms given by (e) f Q = 2Q {N } r dr dz(7.105)A(e)and the boundary gradient (flux) components (e) f g = −2 qs n s {N } r dS(7.106)S (e)As has been discussed for other cases, on boundaries common to two elements,the flux terms are self-canceling in the model assembly procedure.
Therefore,Equation 7.106 is applicable to element boundaries on a free surface. For such surfaces, the boundary conditions are of one of the three types delineated in previoussections: specified temperature, specified heat flux, or convection conditions.EXAMPLE 7.10Calculate the terms of the conductance matrix for an axisymmetric element based on thethree-node plane triangular element.■ SolutionThe element and nodal coordinates are as shown in Figure 7.19. From the discussions inChapter 6, if we are to derive the interpolation functions from basic principles, we firstexpress the temperature variation throughout the element asT (r, z) = a0 + a1 r + a2 z = N 1 (r, z)T1 + N 2 (r, z)T2 + N 3 (r, z)T3apply the nodal conditions, and solve for the constants.
Rearranging the results in termsof nodal temperatures then reveals the interpolation functions. However, the results areexactly the same as those of Chapter 6, if we simply replace x and y with r and z, so thatHutton: Fundamentals ofFinite Element Analysis2767. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transferz3(r3, z3)2(r2, z2)1(r1, z1)Figure 7.19 Cross section ofra three-node axisymmetricelement.
Recall that theelement is a body of revolution.the interpolation functions are of the form1(b1 + c1 r + d1 z)2 A (e)1N 2 (r, z) =(b2 + c2 r + d2 z)2 A (e)1N 3 (r, z) =(b3 + c3 r + d3 z)2 A (e)N 1 (r, z) =whereb1 = r 2 z 3 − r 3 z 2b2 = r 3 z 1 − r 1 z 3b3 = r 1 z 2 − r 2 z 1c1 = z 2 − z 3c2 = z 3 − z 1c3 = z 1 − z 2d1 = r 3 − r 1d2 = r 1 − r 3d3 = r 2 − r 1(e)and A is the area of the element in the rz plane.Since the interpolation functions are linear, the partial derivatives are constants, soEquation 7.102 becomes ∂ Ni ∂ N j∂ Ni ∂ N jk i j = 2kr dr dz+∂r ∂r∂z ∂z(e)Ak(cc+dd)rdrdzi, j = 1, 3= )i j*2 i j2 A (e)(e)ARecalling from elementary statics that the integral in the last equation represents the firstmoment of the area of the element about the z axis, we haver dr dz = r̄ A (e)A (e)where r̄ is the radial coordinate of the element centroid.
The components of the conductance matrix are thenki j =k r̄(ci c j + di d j )2 A (e)i, j = 1, 3and the symmetry of the conductance matrix is evident.Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.8 Time-Dependent Heat Transfer7.8 TIME-DEPENDENT HEAT TRANSFERThe treatment of finite element analysis of heat transfer has, to this point, been limited to cases of steady-state conditions.
No time dependence is included in suchanalyses, as we have assumed conditions such that a steady state is reached, andthe transient conditions are not of interest. Certainly, transient, time-dependenteffects are often quite important, and such effects determine whether a steady stateis achieved and what that steady state will be. To illustrate time-dependent heattransfer in the context of finite element analysis, the one-dimensional case isdiscussed here.The case of one-dimensional conduction without convection is detailed inChapter 5. The governing equation, by consideration of energy balance in a control volume, Equation 5.54, is∂Tq x A dt + Q A dx dt = U + q x +dx A dt(7.107)∂xwhere the temperature distribution T (x , t ) is now assumed to be dependent onboth position and time.
Further, the change in internal energy U is not zero.Rather, the increase in internal energy during a small time interval is describedby Equation 5.56, and the differential equation governing the temperature distribution, Equation 5.58, iskx∂ 2T∂T+ Q = c2∂x∂t(7.108)where c and denote material specific heat and density, respectively, and t istime.
For time-dependent conduction, the governing equation is a second-orderpartial differential equation with constant coefficients.Application of the finite element method for solution of Equation 7.108proceeds by dividing the problem domain into finite-length, one-dimensionalelements and discretizing the temperature distribution within each element asT (x , t ) = N 1 (x )T1 (t ) + N 2 (x )T2 (t ) = [N (x )] {T (t )}(7.109)which is the same as Equation 5.60 with the notable exception that the nodal temperatures are functions of time.Let us now apply Galerkin’s finite element method to Equation 7.108 toobtain the residual equationsx2 x1∂ 2T∂Tk x 2 + Q − c∂x∂tN i (x ) A dxi = 1, 2(7.110)Noting that∂ 2T∂Ni 2 =∂x∂x∂TNi∂x−∂ Ni ∂ T∂x ∂x(7.111)277Hutton: Fundamentals ofFinite Element Analysis2787. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transferthe residual equations can be rearranged and expressed asx2x2∂ Ni ∂ T∂TkxA dz + cN i A dx∂x ∂x∂tx1x1x2=x2Q N i A dx +x1kxx1∂∂xNi∂T∂xi = 1, 2A dx(7.112)Comparing Equation 7.112 to Equations 5.63 and 5.64, we observe that the firstintegral on the left includes the conductance matrix, the first integral on the rightis the forcing function associated with internal heat generation, and the secondintegral on the right represents the gradient boundary conditions.
Utilizing Equation 7.109, Equation 7.112 can be written in detailed matrix form as x2 Ṫ1N1cA[N 1 N 2 ] dxN2Ṫ2x1dN 1 x2 dx dN 1 dN 2T1 (t )dx= { f Q } + { f g } (7.113)+ kx A T2 (t )dN 2 dxdxx1dxwhere the dot denotes differentiation with respect to time. Note that the derivatives of the interpolation functions have now been expressed as ordinary derivatives, as appropriate. Equation 7.113 is most often expressed as (e) (e) (e) (e) (e) (e) CṪ+ kT= fQ + f g(7.114)where [C (e) ] is the element capacitance matrix defined byC(e)x2 = cAx1N1[N 1N2x2N 2 ] dx = c A[N ] T [N ] dx(7.115)x1and, as implied by the name, indicates the capacity of the element for heat storage. The capacitance matrix defined by Equation 7.115 is known as the consistent capacitance matrix.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
