Hutton - Fundamentals of Finite Element Analysis (523155), страница 45
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Ifthe lateral surfaces do not exhibit convection (i.e., the surfaces are insulated), theconvection terms are removed by setting h = 0 . Note that, in many finite elementsoftware packages, the convection portion of the conductance matrix is not automatically included in element matrix formulation. Instead, lateral surface (aswell as edge) convection effects are specified by applying convection “loads” tothe surfaces as appropriate. The software then modifies the element matrices asrequired.The element forcing functions are described in column matrix (vector)form as (e) TQ[N ] t d A =Q{N } t d AfQ =Aff(e) h(e) gA= 2h Ta=−[N ] d A = 2h Ta{N } d ATAqs n s [N ] T t dS = −S(7.38)Aqs n s {N } t dSSwhere [N ] = {N } is the M × 1 column matrix of interpolation functions.Equations 7.36–7.38 represent the general formulation of a finite element fortwo-dimensional heat conduction with convection from the surfaces.
Note inparticular that these equations are valid for an arbitrary element having M nodesand, therefore, any order of interpolation functions (linear, quadratic, cubic, etc.).In following examples, use of specific element geometries are illustrated.T7.4.2 Boundary ConditionsThe boundary conditions for two-dimensional conduction with convection maybe of three types, as illustrated by Figure 7.8 for a general two-dimensionaldomain. On portion S1 of the boundary, the temperature is prescribed as a knownHutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.4 Heat Transfer in Two Dimensionsh(T Ta)S3T T *S1q*S2Figure 7.8 Types of boundaryconditions for two-dimensionalconduction with convection.constant value TS1 = T ∗ .
In a finite element model of such a domain, every element node located on S1 has known temperature and the corresponding nodalequilibrium equations become “reaction” equations. The reaction “forces” arethe heat fluxes at the nodes on S1 . In using finite element software packages, suchconditions are input data; the user of the software (“FE programmer”) enterssuch data as appropriate at the applicable nodes of the finite element model (inthis case, specified temperatures).The heat flux on portion S2 of the boundary is prescribed as q S2 = q ∗ . Thisis analogous to specified nodal forces in a structural problem.
Hence, for all elements having nodes on S2 , the third of Equation 7.38 gives the correspondingnodal forcing functions as (e) f g = − q ∗ n S2 {N }t dS(7.39)S2Finally, a portion S3 of the boundary illustrates an edge convection condition. In this situation, the heat flux at the boundary must be equilibrated by theconvection loss from S3 .
For all elements having edges on S3 , the convectioncondition is expressed as (e) f g = − q S3 n S3 {N }t dS = − h(T (e) − Ta ){N }t dS(7.40)S3S3Noting that the right-hand side of Equation 7.40 involves the nodal temperatures, we rewrite the equation as (e) Tf g = − h[N ] [N ]{T }t dS3 + h Ta {N }t dS3(7.41)S3S3and observe that, when inserted into Equation 7.36, the first integral term on theright of Equation 7.41 adds stiffness to specific terms of the conductance matrix241Hutton: Fundamentals ofFinite Element Analysis2427. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transferassociated with nodes on S3 .
To generalize, we rewrite Equation 7.41 asf(e) g (e) (e) = − k h S {T } + f h S(7.42)wherek(e) hS=h[N ] T [N ]t dS(7.43)Sis the contribution to the element conductance matrix owing to convection onportion S of the element boundary and (e) f hS =h Ta {N }t dS(7.44)Sis the forcing function associated with convection on S.Incorporating Equation 7.42 into Equation 7.36, we have (e) (e) (e) (e) k (e) {T } = f Q + f h + f g + f h Swhere the element conductance matrix is now given byT T (e) ∂N∂N∂N∂Nk=+ kyt dAkx∂x∂x∂y∂yAT+ 2h[N ] [N ] d A + h [N ] T [N ] t dSA(7.45)(7.46)Swhich now explicitly includes edge convection on portion(s) S of the elementboundary subjected to convection.EXAMPLE 7.443sr1Determine the conductance matrix (excluding edge convection) for a four-node, rectangular element having 0.5 in.
thickness and equal sides of 1 in. The material has thermalproperties k x = k y = 20 Btu/(hr-ft-◦ F ) and h = 50 Btu/(hr-ft2-◦ F ).■ Solution2Figure 7.9 Elementnode numberingfor Example 7.4;the length of eachedge is 1 in.The element with node numbers is as shown in Figure 7.9 and the interpolation functions,Equation 6.56, areN 1 (r, s) =1(1 − r )(1 − s)4N 2 (r, s) =1(1 + r )(1 − s)4Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.4 Heat Transfer in Two DimensionsN 3 (r, s) =1(1 + r )(1 + s)4N 4 (r, s) =1(1 − r )(1 + s)4in terms of the normalized coordinates r and s.
For the 1-in. square element, we have2a = 2b = 1 and d A = dx dy = ab dr ds . The partial derivatives in terms of the nor-malized coordinates, via the chain rule, are∂ Ni∂ N i ∂r1 ∂ Ni==∂x∂r ∂ xa ∂ri = 1, 4∂ Ni∂ Ni ∂ s1 ∂ Ni==∂y∂s ∂yb ∂si = 1, 4Therefore, Equation 7.37 becomesk(e)T T 1 1 ∂N∂N 1∂N∂N 1kxt ab dr ds=+ ky∂r∂r a 2∂s∂ s b2−1 −11+ 2h[N ] T [N ]ab dr ds−1or, on a term by term basis,1 1 ki j =−1 −1∂ Ni ∂ N j 1∂ Ni ∂ N j 1kxt ab dr ds+ ky∂r ∂r a 2∂ s ∂ s b21 1+ 2hN i N j ab dr dsi, j = 1, 4−1 −1or1 1 ki j =kx−1 −1∂ Ni ∂ N j b∂ Ni ∂ N j at dr ds+ ky∂r ∂r a∂s ∂s b1 1+ 2hN i N j ab dr dsi, j = 1, 4−1 −1Assuming that k x and k y are constants, we haveki j = k x tba1 1−1 −11∂ Ni ∂ N jadr ds + k y t∂r ∂rb1 1−1 −11+ 2 habN i N j dr ds−1 −1i, j = 1, 4∂ Ni ∂ N jdr ds∂s ∂s243Hutton: Fundamentals ofFinite Element Analysis2447. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat TransferThe required partial derivatives are∂ N11= (s − 1)∂r4∂ N11= (r − 1)∂s4∂ N21= (1 − s)∂r4∂ N21= − (1 + r )∂s4∂ N31= (1 + s)∂r4∂ N31= (1 + r )∂s41∂ N4= − (1 + s)∂r4∂ N41= (1 − r )∂s4Substituting numerical values (noting that a = b ), we obtain, for example,1 1 k 11 = 20−1 −111(s − 1) 2 +(r − 1) 216161 11(1 − r ) 2 (1 − s) 216+ 2(50)−1 −1Integrating first on r,20(0.5)=16(12)k 11100−161−10.5120.512dr ds2dr ds11(r − 1) 3 (s − 1) 2 r −1 + ds3−10.5122 11(1 − r ) 3 (1 − s) ds3−12−1ork 1120(0.5)=16(12)1 8(s − 1) (2) +32−1100ds +160.5122 1(1 − s) 2−18ds3Then, integrating on s, we obtaink 1120(0.5)=16(12)ork 11 =20(0.5)16(12) 2 112(s − 1) 3(1 − s) 38100 0.58 + s −3316 1233−1−11616+33+100160.512 2 88= 0.6327 Btu/(hr-◦ F )33The analytical integration procedure just used to determine k 11 is not the method used byfinite element software packages; instead, numerical methods are used, primarily theGauss quadrature procedure discussed in Chapter 6.
If we examine the terms in the integrands of the equation defining k i j , we find that the integrands are quadratic functionsHutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.4 Heat Transfer in Two Dimensionsof r and s.
Therefore, the integrals can be evaluated exactly by using two Gauss pointsin r and s. Per Table 6.1, the required Gauss points and weighting factors are r i , s j =± 0.57735 and W i , W j = 1.0 , i, j = 1, 2 . Using the numerical procedure for k 11 , we writek 11b= kx ta1 1−1 −11+ 2hab−1= kx tb1(s − 1) 2 dr ds + k y t16a1 1−1 −11(r − 1) 2 dr ds161(r − 1) 2 (s − 1) 2 dr ds16222 2 b1a1W i W j (s j − 1) 2 + k y tW i W j (r i − 1) 2a i =1 j =1 16b i =1 j =1 16+ 2hab22 1W i W j (1 − r i ) 2 (1 − s j ) 216i =1 j =1and, using the specified integration points and weighting factors, this evaluates tok 11b= kx ta 14a 1+ kyt+ 2hab3b 39It is extremely important to note that the result expressed in the preceding equation is thecorrect value of k 11 for any rectangular element used for the two-dimensional heat conduction analysis discussed in this section.
The integrations need not be repeated for eachelement; only the geometric quantities and the conductance values need be substituted toobtain the value. Indeed, if we substitute the values for this example, we obtaink 11 = 0.6327 Btu/(hr-◦ F )as per the analytical integration procedure.Proceeding with the Gaussian integration procedure (calculation of some of theseterms are to be evaluated as end-of-chapter problems), we findk 11 = k 22 = k 33 = k 44 = 0.6327 Btu/(hr-◦ F )Why are these values equal?The off-diagonal terms (again using the numerical integration procedure) are calculated ask 12 = −0.1003k 13 = −0.2585k 14 = −0.1003k 23 = −0.1003k 24 = −0.2585k 34 = −0.1003245Hutton: Fundamentals ofFinite Element Analysis2467.
Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat TransferBtu/(hr-◦ F ), and the complete element conductance matrix is0.6327 −0.1003 −0.2585 −0.1003 (e) −0.1003 0.6327 −0.1003 −0.2585 ◦k= −0.2585 −0.1003 0.6327 −0.1003 Btu/(hr- F )−0.1003 −0.2585 −0.1003 0.6327EXAMPLE 7.5Figure 7.10a depicts a two-dimensional heating fin. The fin is attached to a pipe on itsleft edge, and the pipe conveys water at a constant temperature of 180 ◦ F .
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