Hutton - Fundamentals of Finite Element Analysis (523155), страница 47

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Generally, the symmetry is observed geometrically; that is, thephysical domain of interest is symmetric about an axis or plane. Geometric symmetry is not, however, sufficient to ensure that a problem is symmetric. In addition, the boundary conditions and applied loads must be symmetric about theaxis or plane of geometric symmetry as well.

To illustrate, consider Figure 7.12a,depicting a thin rectangular plate having a heat source located at the geometriccenter of the plate. The model is of a heat transfer fin removing heat from a central source (a pipe containing hot fluid, for example) via conduction and convection from the fin. Clearly, the situation depicted is symmetric geometrically. But,is the situation a symmetric problem? The loading is symmetric, since the heatsource is centrally located in the domain.

We also assume that k x = k y so that thematerial properties are symmetric. Hence, we must examine the boundary conditions to determine if symmetry exists. If, for example, as shown in Figure 7.12b,the ambient temperatures external to the fin are uniform around the fin and theconvection coefficients are the same on all surfaces, the problem is symmetricabout both x and y axes and can be solved via the model in Figure 7.12c.

For thissituation, note that the heat from the source is conducted radially and, consequently, across the x axis, the heat flux q y is zero and, across the y axis, the heatflux q x must also be zero. These observations reveal the boundary conditions forthe quarter-symmetry model shown in Figure 7.12d and the internal forcingfunction is taken as Q/4 . On the other hand, let us assume that the upper edge ofthe plate is perfectly insulated, as in Figure 7.12e.

In this case, we do not have253Hutton: Fundamentals ofFinite Element Analysis2547. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transferyh, Tay2bh, TaxQh, Tab2a(a)h, Taa(b)(c)xyqy 0Insulatedyh, Tah, Taqx 02bInsulatedx(d)ah, Ta(e)x(f )Figure 7.12 Illustrations of symmetry dictated by boundary conditions.symmetric conditions about the x axis but symmetry about the y axis exists.

Forthese conditions, we can use the “half-symmetry” model shown in Figure 7.12f,using the symmetry (boundary) condition q x = 0 across x = 0 and apply theinternal heat generation term Q/2 .Symmetry can be used to reduce the size of finite element models significantly. It must be remembered that symmetry is not simply a geometric occurrence.

For symmetry, geometry, loading, material properties, and boundaryconditions must all be symmetric (about an axis, axes, or plane) to reduce themodel.7.4.4 Element ResultantsIn the approach just taken in heat transfer analysis, the primary nodal variablecomputed is temperature. Most often in such analyses, we are more interested inthe amount of heat transferred than the nodal temperatures. (This is analogous tostructural problems: We solve for nodal displacements but are more interested instresses.) In finite element analyses of heat transfer problems, we must back substitute the nodal temperature solution into the “reaction” equations to obtainglobal heat transfer values.

(As in Example 7.5, when we solved the partitionedmatrices for the heat flux values at the constrained nodes.) Similarly, we can backHutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.4 Heat Transfer in Two Dimensionssubstitute the nodal temperatures to obtain estimates of heat transfer propertiesof individual elements as well.The heat flux components for a two-dimensional element, per Fourier’s law,areqqM∂ T (e)∂ N i (e)= −k xT i∂x∂xi=1(e)x= −k x(e)yM∂ T (e)∂ N i (e)= −k y= −k yT i∂y∂yi=1(7.47)where we again denote the total number of element nodes as M.

With the exception of the three-node triangular element, the flux components given by Equation 7.47 are not constant but vary with position in the element. As an example,the components for the four-node rectangular element are readily computedusing the interpolation functions of Equation 6.56, repeated here asN 1 (r, s) =1(1 − r )(1 − s)4N 2 (r, s) =1(1 + r )(1 − s)41N 3 (r, s) = (1 + r )(1 + s)4N 4 (r, s) =(7.48)1(1 − r )(1 + s)4Recalling that∂1 ∂=∂xa ∂rand∂1 ∂=∂yb ∂swe haveqq(e)x(e)y=−4kx ∂ N i (e)T ia i=1 ∂r=−kx (e)(e)(e)(e) (s − 1)T 1 + (1 − s)T 2 + (1 + s)T 3 − (1 + s)T 44a4ky ∂ N i (e)=−T ib i=1 ∂ s=−ky (e)(e)(e)(e) (r − 1)T 1 − (1 + r )T 2 + (1 + r )T 3 + (1 − r )T 44b(7.49)255Hutton: Fundamentals ofFinite Element Analysis2567. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transferand these expressions simplify to) (e)) (e)kx (e)(e) *(e) *qx =−(1 − s) T 2 − T 1 + (1 + s) T 3 − T 44a(7.50)) (e)) (e)ky (e)(e) *(e) *qy =−(1 − r ) T 4 − T 1 + (1 + r ) T 3 − T 24bThe flux components, therefore the temperature gradients, vary linearly in a fournode rectangular element.

However, recall that, for a C 0 formulation, the gradients are not, in general, continuous across element boundaries. Consequently, theelement flux components associated with an individual element are customarilytaken to be the values calculated at the centroid of the element. For the rectangular element, the centroid is located at (r, s) = (0, 0) , so the centroidal values aresimplyk x ) (e)(e)(e)(e)(e) *qx =−T2 +T3 −T1 −T44a(7.51)k y ) (e)(e)(e)(e)(e) *qy =−T3 +T4 −T1 −T24bThe centroidal values calculated per Equation 7.51, in general, are quite accuratefor a fine mesh of elements. Some finite element software packages compute thevalues at the integration points (the Gauss points) and average those values foran element value to be applied at the element centroid.

In either case, the computed values are needed to determine solution convergence and should bechecked at every stage of a finite element analysis.EXAMPLE 7.6Calculate the centroidal heat flux components for elements 2 and 3 of Example 7.5.■ SolutionFrom Example 7.4, we have a = b = 0.5 in ., k x = k y = 20 Btu/(hr-ft-◦ F) , and fromExample 7.5, the nodal temperature vector is  T1   180   T2 180 T1803  T4  106.507 {T } = T5 = 111.982 ◦ F  106.507   T6 T89.0417T90.9668  T989.041For element 2, the element-global nodal correspondence relation can be written asT(2)1T(2)2T(2)3T(2) 4= [T4T7= [106.507T8T5 ]89.04190.966111.982]Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.4 Heat Transfer in Two Dimensions257Substituting numerical values into Equation 7.49,q(2)x=−12(20)(89.041 + 90.966 − 106.507 − 111.982) = 4617 .84 Btu/(hr-ft 2 )4(0.5)q(2)y=−12(20)(90.966 + 111.982 − 106.507 − 89.041) = −888.00 Btu/(hr-ft 2 )4(0.5)and, owing to the symmetry conditions, we haveq(3)x= 4617 .84 Btu/(hr-ft 2 )q(3)y= 888.00 Btu/(hr-ft 2 )as may be verified by direct calculation.

Recall that these values are calculated at thelocation of the element centroid.The element resultants representing convection effects can also be readilycomputed once the nodal temperature solution is known. The convection resultants are of particular interest, since these represent the primary source of heatremoval (or absorption) from a solid body. The convective heat flux, per Equation 7.2, isq x = h(T − Ta ) Btu/(hr-ft2 ) or W/m 2(7.52)where all terms are as previously defined. Hence, the total convective heat flowrate from a surface area A isḢh =h(T − Ta ) d A(7.53)AFor an individual element, the heat flow rate isḢ(e)h=h(T (e) − Ta ) d A =Ah([N ]{T } − Ta ) d A(7.54)AThe area of integration in Equation 7.54 includes all portions of the element surface subjected to convection conditions.

In the case of a two-dimensional element,the area may include lateral surfaces (that is, convection perpendicular to the planeof the element) as well as the area of element edges located on a free boundary.EXAMPLE 7.7Determine the total heat flow rate of convection for element 3 of Example 7.5.■ SolutionFirst we note that, for element 3, the element-to-global correspondence relation for nodaltemperatures isT(3)1T(3)2T(3)3T(3) 4= [T5T8T9T6 ]Hutton: Fundamentals ofFinite Element Analysis2587.

Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat TransferSecond, element 3 is subjected to convection on both lateral surfaces as well as the twoedges defined by nodes 8-9 and 6-9. Consequently, three integrations are required asfollows:(e)Ḣ h = 2h([N ]{T } − Ta ) d A (e) +A (e)A 8−9+h([N ]{T } − Ta ) d A 8−9h([N ]{T } − Ta ) d A 6−9A 6−9(e)where A is element area in the xy plane and the multiplier in the first term (2) accountsfor both lateral surfaces.Transforming the first integral to normalized coordinates results in1 12h A4−1 −11 11[N ] dr ds {T } − 2habTa([N ] {T } − Ta ) dr ds = 2habI1 = 2hab=1 1−1 −1dr ds−1[N ] dr ds {T } − 2hAT a−1 −1Therefore, we need integrate the interpolation functions only over the area of the element,as all other terms are known constants.

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