Hutton - Fundamentals of Finite Element Analysis (523155), страница 46

Просмтор этого файла доступен только зарегистрированным пользователям. Но у нас супер быстрая регистрация: достаточно только электронной почты!

Текст из файла (страница 46)

The finis surrounded by air at temperature 68 ◦ F . The thermal properties of the fin are as givenin Example 7.4. Use four equal-size four-node rectangular elements to obtain a finiteelement solution for the steady-state temperature distribution in the fin.■ SolutionFigure 7.10b shows four elements with element and global node numbers. Given thenumbering scheme selected, we have constant temperature conditions at global nodes1, 2, and 3 such thatT1 = T2 = T3 = 180 ◦ Fwhile on the other edges, we have convection boundary conditions that require a bit ofanalysis to apply. For element 1 (Figure 7.10c), for instance, convection occurs alongelement edge 1-2 but not along the other three element edges.

Noting that s = −1 and632 in.180 F9342 in.68 F52112(b)551124(a)824(c)487(d)Figure 7.10 Example 7.5:(a) Two-dimensional fin. (b) Finite element model.(c) Element 1 edge convection. (d) Element 2 edgeconvection.7Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.4 Heat Transfer in Two DimensionsN 3 = N 4 = 0 on edge 1-2, Equation 7.43 becomes1−r 1 1 +r 1htk (1)=[ 1 − r 1 + r 0 0 ] a drhS 0 4−10(1 − r) 2 1 − r 2 0 012hta (1 + r) 2 0 0  1−rdr=000 04−1000 0Integrating as indicated gives8 4 0 08250(0.5)hta4800=4k (1)hS =4(3)  0 0 0 0  4(3)(12) 2  00 0 0 000.0579 0.0290 0 0 0.0290 0.0579 0 0 = 000 0004800000000000 0where the units are Btu/(hr-◦ F ).The edge convection force vector for element 1 is, per Equation 7.44,!(1)f hS"  21−r 2 1 50(68)(0.5) 2  2 hT a thT a ta 21+r==a dr =0022 2(12) 2  0   −1 0005.9028  5.9028 =Btu/hr0 0where we again utilize s = −1 , N 3 = N 4 = 0 along the element edge bounded by nodes1 and 2.Next consider element 2.

As depicted in Figure 7.10d, convection occurs along twoelement edges defined by element nodes 1-2 (s = −1) and element nodes 2-3 (r = 1) .For element 2, Equation 7.43 isk (2)hS 1−r 1 ht   1 + r = [ 1 − r 1 + r 0 0 ]a dr0 4 −100 1 1 − s +[ 0 1 − s 1 + s 0 ]b ds 1+s−1 0247Hutton: Fundamentals ofFinite Element Analysis2487. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transferor, after integrating,8hta4k (2)hS =4(3)  0048000000000htb0+0  4(3)  000084004800000and, since a = b , 0.0579 0.0290008 4 0 0 50(0.5) 2  4 16 4 0  =  0.0290 0.1157 0.0290 0  Btu/(hr-◦ F)k (2)hS =200.02900.0579004804(3)(12)00000 0 0 0Likewise, the element edge convection force vector is obtained by integration along thetwo edges as!(2)f hS" 1−r 0 1 1 hTa t 1+r1−s=a dr +b ds 01+s2−1 −1 00  2   5.9028 50(68)(0.5) 2 411.8056 ==Btu/hr2 5.9028 2(12) 2  00Identical procedures applied to the appropriate edges of elements 3 and 4 result ink (3)hS050(0.5) 2 0=4(3)(12) 2  00k (4)hS!!(3)f hS(4)f hS""050(0.5) 2 0=4(3)(12) 2  00 0 0 000008 4 00  =  0 0.0579 0.0290 Btu/(hr-◦ F)4 16 40 0.0290 0.1157 0.0290 0 4 8000.0290 0.057900000 5.9028 =Btu/hr11.8056 5.90280 0 =Btu/hr5.9028 5.90280084 0000=4 080000000  Btu/(hr-◦ F)0 0.0579 0.0290 0 0.0290 0.0579Hutton: Fundamentals ofFinite Element Analysis7.

Applications in HeatTransferText© The McGraw−HillCompanies, 20047.4 Heat Transfer in Two DimensionsAs no internal heat is generated, the corresponding { f (e)Q } force vector for each element iszero; that is,!" f (e)=Q{N } dA = {0}QAfor each element.On the other hand, each element exhibits convection from its surfaces, so the lateralconvection force vector is(1 − r)(1 − s) 1 1 !"1  (1 + r)(1 − s) (e){N } dA = 2hTaf h = 2hTaab dr ds4  (1 + r)(1 + s) A−1 −1 (1 − r)(1 + s) which evaluates to   44   11.8056 !" 2hT ab  4  2(50)(68)(0.5) 2  4   11.8056 a=f (e)==h244 11.8056 4 4(12)   4411.8056and we note that, since the element is square, the surface convection forces are distributedequally to each of the four element nodes.The global equations for the four-element model can now be assembled by writingthe element-to-global nodal correspondence relations asL (1) = (2) L= (3) L= (4) =L[1452][4785][5896][2563]and adding the edge convection terms to obtain the element stiffness matrices as0.6906 −0.0713 −0.2585 −0.1003 (1)  −0.0713 0.6906 −0.1003 −0.2585 k= −0.2585 −0.1003 0.6327 −0.1003 −0.1003 −0.2585 −0.1003 0.63270.6906 −0.0713 −0.2585 −0.1003 (2)  −0.0713 0.7484 −0.0713 −0.2585 k= −0.2585 −0.0713 0.6906 −0.1003 −0.1003 −0.2585 −0.1003 0.63270.6327 −0.1003 −0.2585 −0.1003 (3)  −0.1003 0.6906 −0.0713 −0.2585 k= −0.2585 −0.0713 0.7484 −0.0713 −0.1003 −0.2585 −0.0713 0.6906249Hutton: Fundamentals ofFinite Element Analysis2507.

Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transfer0.6327 −0.1003 −0.2585 −0.1003 (4)  −0.1003 0.6327 −0.1003 −0.2585 k= −0.2585 −0.1003 0.6906 −0.0713 −0.1003 −0.2585 −0.0713 0.6906Utilizing the direct assembly-superposition method with the element-to-global nodeassignment relations, the global conductance matrix is0.6906 −0.10030 −0.0713[K ] =  −0.25850000−0.10030−0.0713 −0.258500001.2654 −0.1003 −0.2585 −0.2006 −0.2585000−0.1003 0.69060−0.2585 −0.0713000−0.258501.3812 −0.20060−0.0713 −0.25850−0.2006 −0.2585 −0.2006 2.5308 −0.2006 −0.2585 −0.2006 −0.2585 −0.2585 −0.07130−0.2006 1.38120−0.2585 −0.0713 00−0.0713 −0.258500.7484 −0.2585000−0.2585 −0.2006 −0.2585 −0.2585 1.3812 −0.0713 000−0.2585 −0.07130−0.0713 0.7484The nodal temperature vector is180  180  180  T4 {T } =T5T6  T7 T8 T9and we have explicitly incorporated the prescribed temperature boundary conditions.Assembling the global force vector, noting that no internal heat is generated, weobtain{F} =17.7084 + F1  35.4168 + F2 17.7084+F3 35.4168 47.222435.416823.611235.416823.6112Btu/hrwhere we use F1 , F2 , and F3 as general notation to indicate that these are unknown“reaction” forces.

In fact, as will be shown, these terms are the heat flux components atnodes 1, 2, and 3.Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.4 Heat Transfer in Two Dimensions251The global equations for the four-element model are then expressed as0.6906 −0.100300000哸−0.0713 −0.2585 −0.1003 1.2654 −0.1003哸−0.2585 −0.2006 −0.2585000哸 00−0.10030.6906−0.2585−0.0713000 哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹  −0.0713 −0.2585哸 1.3812 −0.200600−0.0713−0.25850哸 −0.2585 −0.2006 −0.2585 −0.2006 2.5308 −0.2006 −0.2585 −0.2006 −0.2585 哸0−0.2585 −0.0713哸 0−0.2006 1.38120−0.2585 −0.0713 00000.7484 −0.25850哸−0.0713 −0.2585000−0.2585−0.2006−0.2585−0.25851.3812−0.0713哸000哸−0.2585 −0.071300−0.071317.7084 + F1  35.4168 + F2 17.7084+F3哹哹哹哹哹 35.4168 =47.2224 35.4168 23.611235.4158 23.6112Taking into account the specified temperatures on nodes 1, 2, and 3, the global equationsfor the unknown temperatures become1.3812 −0.20060 −0.0713 −0.25850  −0.20060−0.0713 −0.25850T94.78084T5 2.5308 −0.2006 −0.2585 −0.2006 −0.2585 176.3904   T−0.2006 1.38120−0.2585 −0.0713 94.78086=T7 −0.258500.7484 −0.2585023.6112 −0.2006 −0.2585 −0.2585 1.3812 −0.0713  T8   35.4168   T9−0.2585 −0.07130−0.0713 0.748423.6112The reader is urged to note that, in arriving at the last result, we partition the global matrixas shown by the dashed lines and apply Equation 3.46a to obtain the equations governingthe “active” degrees of freedom.

That is, the partitioned matrix is of the formK ccK acK caK aaTcTa=FcFawhere the subscript c denotes terms associated with constrained (specified) temperaturesand the subscript a denotes terms associated with active (unknown) temperatures. Hence,this 6 × 6 system represents[K aa ]{Ta } = {Fa } − [K ac ]{Tc }which now properly includes the effects of specified temperatures as forcing functions onthe right-hand side.0.7484180 180  180 哹哹T 4 T5T6 T 7 T8 T9Hutton: Fundamentals ofFinite Element Analysis2527. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat TransferSimultaneous solution of the global equations (in this case, we inverted the globalstiffness matrix using a spreadsheet program) yields the nodal temperatures as  106.507 T4 111.982 T5   106.507 T6◦F=89.041T7T   90.966 8 T  89.0419If we now back substitute the computed nodal temperatures into the first three of theglobal equations, specifically,0.6906T1 − 0.1003T2 − 0.0713T4 − 0.2585T5 = 17.7084 + F1− 0.1003T1 + 1.2654T2 − 0.1003T3 − 0.2585T4 − 0.2006T5 − 0.2585T6 = 35.4168 + F2− 0.1003T2 + 0.6906T3 − 0.2585T5 − 0.0713T6 = 17.7084 + F3we obtain the heat flow values at nodes 1, 2, and 3 as   F1   52.008 = 78.720 Btu/hrF 2 52.008F3Note that, in terms of the matrix partitioning, we are now solving[K cc ]{Tc } + [K ca ]{Ta } = {Fc }to obtain the unknown values in {Fc } .Since there is no convection from the edges defined by nodes 1-2 and 1-3 and thetemperature is specified on these edges, the reaction “forces” represent the heat input(flux) across these edges and should be in balance with the convection loss across the lateral surfaces of the body, and its edges, in a steady-state situation.

This balance is a checkthat can and should be made on the accuracy of a finite element solution of a heat transfer problem and is analogous to checking equilibrium of a structural finite elementsolution.Example 7.5 is illustrated in great detail to point out the systematic procedures for assembling the global matrices and force vectors. The astute readerascertains, in following the solution, that symmetry conditions can be used tosimplify the mathematics of the solution.

As shown in Figure 7.11a, an axis(plane) of symmetry exists through the horizontal center of the plate. Therefore,the problem can be reduced to a two-element model, as shown in Figure 7.11b.Along the edge of symmetry, the y-direction heat flux components are in balance,and this edge can be treated as a perfectly insulated edge. One could then useonly two elements, with the appropriately adjusted boundary conditions to obtainthe same solution as in the example.Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.4 Heat Transfer in Two DimensionsPlane ofsymmetry(a)2T 180F4113625(b)Figure 7.11 Model of Example 7.5,showing (a) the plane of symmetry and(b) a two-element model with adjustedboundary conditions.7.4.3 Symmetry ConditionsAs mentioned previously in connection with Example 7.5, symmetry conditionscan be used to reduce the size of a finite element model (or any other computational model).

Характеристики

Список файлов книги