Hutton - Fundamentals of Finite Element Analysis (523155), страница 42

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The one-dimensional case of heat transfer with mass transport is alsodeveloped. The three-dimensional case of axial symmetry is developed in detailusing appropriately modified two-dimensional elements and interpolationfunctions. Heat transfer by radiation is not discussed, owing to the nonlinearnature of radiation effects. However, we examine transient heat transfer andinclude an introduction to finite difference techniques for solution of transientproblems.7.2 ONE-DIMENSIONAL CONDUCTION:QUADRATIC ELEMENTChapter 5 introduced the concept of one-dimensional heat conduction via theGalerkin finite element method. In the examples of Chapter 5, linear, two-nodefinite elements are used to illustrate the concepts involved. Given the development of the general interpolation concepts in Chapter 6, we now apply a higherorder (quadratic) element to a previous example to demonstrate that (1) the basicprocedure of element formulation is unchanged, (2) the system assembly procedure is unchanged, and (3) the results are quite similar.222Hutton: Fundamentals ofFinite Element Analysis7.

Applications in HeatTransferText© The McGraw−HillCompanies, 20047.2 One-Dimensional Conduction: Quadratic Element223EXAMPLE 7.1Solve Example 5.4 using two, three-node line elements with equally spaced nodes. Theproblem and numerical data are repeated here as Figure 7.1a.■ SolutionPer Equation 5.62, the element equations arek x A N i (x )xx 2x 2dT 2dN i dT−kAQ N i (x ) dx = 0dx+Axdx x 1dx dxx1i = 1, 3x1where now there are three interpolation functions per element.The interpolation functions for a three-node line element are, per Equations 6.23–6.251N 1 (s) = 2 s −(s − 1)2N 2 (s) = −4s(s − 1)1N 3 (s) = 2s s −2The components of the conductance matrix are then calculated asx 2ki j = k x AdN i dN jdxdx dxi, j = 1, 3x1and the heat generation vector components arex 2f Qi = AQ N i dxi = 1, 3x1and all f Q components are zero in this example, as there is no internal heat source.InsulatedqinAlCu0.5 m0.5 mkal 200 W/m- C kcu 389 W/m- C(a)80 C123E14E2(b)Figure 7.1(a) Geometry and data for Example 7.1, outside diameter = 60 mm,qin = 4000 W/m2.

(b) Two-element model, using quadratic elements.5Hutton: Fundamentals ofFinite Element Analysis2247. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat TransferIn terms of the dimensionless coordinate s = x /L , we have dx = L ds and d/dx =(1/L ) d/ds, so the terms of the conductance matrix are expressed askx Aki j =L1dN i dN jdsds dsi, j = 1, 30The derivatives of the interpolation functions aredN 1= 4s − 3dsdN 2= 4(1 − 2s)dsdN 3= 4s − 1dsTherefore, on substitution for the derivatives,k 11kx A=L1kx A(4s − 3) ds =L0=kx AL1(16s 2 − 24s + 9) ds20316s− 12s 2 + 9s31=07k x A3LVia mathematically identical procedures, the remaining terms of the conductance matrixare found to bek 12 = k 21 = −k 13 = k 31 =k 22 =kx A3L16k x A3Lk 23 = k 32 = −k 33 =8k x A3L8k x A3L7k x A3LA two-element model with node numbers is shown in Figure 7.1b.

Substitutingnumerical values, we obtain, for the aluminum half of the rod (element 1), 7 −8 12.6389 −3.0159 0.3770 (1) 200(/4)(0.006) 2 −8 16 −8  =  −3.0159 6.0319 −3.0159 k=3(0.5)1 −8 70.3770 −3.0159 2.6389Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.2 One-Dimensional Conduction: Quadratic Elementand for the copper portion (element 2), 7 −8 15.1327 −5.8660 0.7332 (2) 389(/4)(0.006) 2 −8 16 −8  =  −5.8660 11.7320 −5.8660 k=3(0.5)1 −8 70.7332 −5.8660 5.1327At the internal nodes of each element, the flux terms are zero, owing to the nature of theinterpolation functions [N 2 (x 1 ) = N 2 (x 2 ) = 0] .

Similarly, at the junction between thetwo elements, the flux must be continuous and the equivalent “forcing” functions arezero. As no internal heat is generated, Q = 0 , that portion of the force vector is zero foreach element. Following the direct assembly procedure, the system conductance matrixis found to be2.6389 −3.0159 0.377000 −3.0159 6.0319 −3.015900 ◦[K ] =  0.3770 −3.0159 7.7716 −5.8660 0.7332  W/ C00−5.8660 11.7320 −5.8660 000.7332 −5.8660 5.1327and we note in particular that “overlap” exists only at the juncture between elements.

Thegradient term at node 1 is computed asf g1 dT = − kx A= q 1 A = 4000(0.06) 2 = 11.3097 Wdx x 14while the heat flux at node 5 is an unknown to be calculated via the system equations.The system equations are given by  T1 11.3097 T  0 −3.0159 6.0319 −3.0159002   0.3770 −3.0159 7.7716 −5.8660 0.7332  T3 =0   T4 000−5.8660 11.7320 −5.8660    80−Aq5000.7332 −5.8660 5.13272.6389−3.01590.377000Prior to solving for the unknown nodal temperatures T1 –T4 , the nonhomogeneous boundary condition T5 = 80 ◦ C must be accounted for properly. In this case, we reduce the system of equations to 4 × 4 by transposing the last term of the third and fourth equationsto the right-hand side to obtain  2.3689 −3.0159 0.37700T1   11.3907  −3.0159 6.0319 −3.0159 T200= 0.3770 −3.0159 7.7716 −5.8660   T3   −0.7332(80)   00−5.8660 11.7320T45.8660(80)11.3097 0= −58.6560 489.2800225Hutton: Fundamentals ofFinite Element Analysis2267.

Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat TransferSolving the equations by Gaussian elimination (Appendix C), the nodal temperatures areT1 = 95.11 ◦ CT2 = 90.14 ◦ CT3 = 85.14 ◦ CT4 = 82.57 ◦ Cand the heat flux at node 5 is calculated using the fifth equation− Aq 5 = 0.7332 T3 − 5.8660 T4 + 5.1327 (80)to obtainq 5 = 4001 .9 W/m2which is observed to be in quite reasonable numerical agreement with the heat input atnode 1.The solution for the nodal temperatures in this example is identical for boththe linear and quadratic interpolation functions. In fact, the solution we obtainedis the exact solution (Problem 7.1) represented by a linear temperature distribution in each half of the bar.

It can be shown [1] that, if an exact solution existsand the interpolation functions used in the finite element formulation include theterms appearing in the exact solution, then the finite element solution corresponds to the exact solution. In this example, the quadratic interpolation functions include the linear terms in addition to the quadratic terms and thus capturethe exact, linear solution. The following example illustrates this feature in termsof the field variable representation.EXAMPLE 7.2For the quadratic field variable representation(x ) = a0 + a1 x + a2 x 2determine the explicit form of the coefficients a0 , a1 , a2 in terms of the nodal variables ifthe three nodes are equally spaced.

Then use the results of Example 7.1 to show a2 = 0for that example.■ SolutionUsing the interpolation functions from Example 7.1, we can write the field variablerepresentation in terms of the dimensionless variable s as(s) = (2s 2 − 3s + 1)1 + 4(s − s 2 )2 + (2s 2 − s)3Collecting coefficients of similar powers of s,(s) = 1 + (42 − 31 − 3 )s + (21 − 42 + 23 )s 2Hutton: Fundamentals ofFinite Element Analysis7.

Applications in HeatTransferText© The McGraw−HillCompanies, 20047.3 One-Dimensional Conduction with ConvectionTherefore,a0 = 1a1 = 42 − 31 − 3a2 = 21 − 42 + 23Using the temperature results of Example 7.1 for the aluminum element, we have1 = T1 = 95.142 = T2 = 90.143 = T3 = 85.14a2 = 2(95.14) − 4(90.14) + 2(85.14) = 0For element 2, representing the copper portion of the bar, the same result is obtained.7.3 ONE-DIMENSIONAL CONDUCTIONWITH CONVECTIONOne-dimensional heat conduction, in which no heat flows from the surface of thebody under consideration (as in Figure 5.8), is not commonly encountered. Amore practical situation exists when the body is surrounded by a fluid mediumand heat flow occurs from the surface to the fluid via convection. Figure 7.2ashows a solid body, which we use to develop a one-dimensional model of heattransfer including both conduction and convection. Note that the representationis the same as in Figure 5.8 with the very important exception that the assumption of an insulated surface is removed.

Instead, the body is assumed to be surrounded by a fluid medium to which heat is transferred by convection. If the fluidis in motion as a result of some external influence (a fan or pump, for example),the convective heat transfer is referred to as forced convection. On the otherhand, if motion of the fluid exists only as a result of the heat transfer taking place,we have natural convection. Figure 7.2b depicts a control volume of differentiallength, which is assumed to have a constant cross-sectional area and uniformTaConvectionqhqinqxqoutQUqx dx(a)(b)Figure 7.2 One-dimensional conduction with surface convection.(a) General model.

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