Hutton - Fundamentals of Finite Element Analysis (523155), страница 49

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Rather than repeat the derivation of known terms, we develop only the additional terms. If Equation 7.65is substituted into the residual equations for a two-node linear element (Equation 7.6), the additional terms arex2dTṁcN i dxi = 1, 2(7.66)dxx1Substituting for T via Equation 7.5, this becomesx2dN 1dN 2ṁcT1 +T2 N i dxdxdxi = 1, 2(7.67)x1Therefore, the additional stiffness matrix resulting from mass transport isdNdNx2 N1 1 N1 2dxdx dx[kṁ ] = ṁc (7.68)dN1dN2x1N2N2dxdxEXAMPLE 7.8Explicitly evaluate the stiffness matrix given by Equation 7.68 for the two-node element.■ SolutionThe interpolation functions areN1 = 1 −N2 =xLxLand the required derivatives aredN 11=−dxLdN 21=dxLUtilizing the change of variable s = x /L , Equation 7.68 becomes[kṁ ] =ṁcL1 0−(1 − s)−s1−2(1 − s)L ds = ṁc 1s−2and note that the matrix is not symmetric.12 = ṁc −1 12 −1 112Hutton: Fundamentals ofFinite Element Analysis2647.

Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat TransferUsing the result of Example 7.8, the stiffness matrix for a one-dimensionalheat transfer element with conduction, convection, and mass transport is givenby (e) k x A 1 −1hPL 2 1ṁc −1 1++k=1 2L −1 162 −1 1 (e) (e) (e) = k c + k h + k ṁ(7.69)where the conduction and convection terms are identical to those given in Equation 7.15. Note that the forcing functions and boundary conditions for the onedimensional problem with mass transport are the same as given in Section 7.3,Equations 7.16 through 7.19.EXAMPLE 7.9Figure 7.16a shows a thin-walled tube that is part of an oil cooler. Engine oil enters thetube at the left end at temperature 50 ◦ C with a flow rate of 0.2 kg/min.

The tube is surrounded by air flowing at a constant temperature of 15 ◦ C . The thermal properties of theoil are as follows:Thermal conductivity: k x = 0.156 W/(m- ◦ C)Specific heat: c = 0.523 W-hr/(kg-◦ C)The convection coefficient between the thin wall and the flowing air is h =300 W/(m 2 -◦ C) . The tube wall thickness is such that conduction effects in the wall are tobe neglected; that is, the wall temperature is constant through its thickness and the sameas the temperature of the oil in contact with the wall at any position along the length ofthe tube. Using four two-node finite elements, obtain an approximate solution for the temperature distribution along the length of the tube and determine the heat removal rate viaconvection.■ SolutionThe finite element model is shown schematically in Figure 7.16b, using equal lengthelements L = 25 cm = 0.025 m . The cross-sectional area is A = (/4)(20/1000 ) 2 =Air, 15 CT 50 Cmⴢ20 mm100 cm(a)ⴢm121324354(b)Figure 7.16(a) Oil cooler tube of Example 7.9.

(b) Element and node numbers for afour-element model.Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.5 Heat Transfer With Mass Transport3.14(10 −4 ) m 2 . And the peripheral dimension (circumference) of each element isP = (20/1000 ) = 6.28(10 −2 ) m .

The stiffness matrix for each element (note that allelements are identical) is computed via Equation 7.69 as follows:(e) k ckx A 1=L −1=k(e) h−1112ṁc −1=2 −111(e) k ṁk (e) =0.156(3.14)(10 −4 ) 1=−10.025−1.95941.95941.9594−1.9594hPL 216=−2.9810−3.0595=(10 −3 )300(6.28)(10 −2 )(0.025) 216(0.2)(60)(0.523) −1=−123.21653.2950−111112==0.1570.0785−3.138−3.1380.07850.1573.1383.138At this point, note that the mass transport effects dominate the stiffness matrix and we anticipate that very little heat is dissipated, as most of the heat is carried away with the flow.Also observe that, owing to the relative magnitudes, the conduction effects have beenneglected.Assembling the global stiffness matrix via the now familiar procedure, we obtain−2.9810 3.2165000 −3.05950.3143.216500 [K ] = 0−3.05950.3143.2165000−3.05950.3143.2165 000−3.0595 3.2950The convection-driven forcing function for each element per Equation 7.18 isf(e) hhPT aL=2 300(6.28)(10 −2 )(15)(0.025) 113.5325==113.53252As there is no internal heat generation, the per-element contribution of Equation 7.16 iszero.

Finally, we must examine the boundary conditions. At node 1, the temperature isspecified but the heat flux q1 = F1 is unknown; at node 5 (the exit), the flux is also unknown. Unlike previous examples, where a convection boundary condition existed, herewe assume that the heat removed at node 5 is strictly a result of mass transport. Physically, this means we define the problem such that heat transfer ends at node 5 and theheat remaining in the flow at this node (the exit) is carried away to some other process.Consequently, we do not consider either a conduction or convection boundary conditionat node 5. Instead, we compute the temperature at node 5 then the heat removed at thisnode via the mass transport relation.

In terms of the finite element model, this means265Hutton: Fundamentals ofFinite Element Analysis2667. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transferthat we do not consider the heat flow through node 5 as an unknown (reaction force).With this in mind, we assemble the global force vector from the element force vectorsto obtain3.5325 + F1 7.065{F} =7.0657.0653.5325The assembled system (global) equations are then −2.9801 3.2165000 T1  −3.0595 0.3143.216500  T2 [K ]{T } = 0−3.05950.3143.21650  T3 00−3.05950.3143.2165   T4 T5000−3.0595 3.29503.5325 + F1 7.065=7.0657.0653.5325Applying the known condition at node 1, T = 50 ◦ C , the reduced system equationsbecome  T2 0.3143.216500160.04  −3.05950.3143.21650  T3 = 7.065 0−3.05950.3143.2165   T4   7.065 T500−3.0595 3.29503.5325which yields the solution for the nodal temperatures as T2 T 3T  4T5=47.448  45.124 42.923 40.928◦CAs conduction effects have been seen to be negligible, the input rate is computed asq in = q m 1 = ṁcT1 = 0.2(60)(0.523)(50) = 3138 Wwhile, at node 5, the output rate isq m 5 = ṁcT5 = 0.2(60)(0.523)(40.928) = 2568 .6 WThe results show that only about 18 percent of input heat is removed, so the cooler is notvery efficient.Hutton: Fundamentals ofFinite Element Analysis7.

Applications in HeatTransferText© The McGraw−HillCompanies, 20047.6 Heat Transfer in Three Dimensions7.6 HEAT TRANSFER IN THREE DIMENSIONSAs the procedure has been established, the governing equation for heat transferin three dimensions is not derived in detail here. Instead, we simply present theequation as∂∂∂∂T∂T∂Tkx+ky+kz+Q=0(7.70)∂x∂x∂y∂y∂z∂zand note that only conduction effects are included and steady-state conditions areassumed.

In the three-dimensional case, convection effects are treated most efficiently as boundary conditions, as is discussed.The domain to which Equation 7.70 applies is represented by a mesh of finiteelements in which the temperature distribution is discretized asMT (x , y, z) =N i (x , y, z)Ti = [N ]{T }(7.71)i=1where M is the number of nodes per element. Application of the Galerkin methodto Equation 7.70 results in M residual equations: ∂∂T∂∂T∂∂Tkx+ky+kz+ Q N i dV = 0∂x∂x∂y∂y∂z∂zVi = 1, . .

. , M(7.72)where, as usual, V is element volume.In a manner analogous to Section 7.4 for the two-dimensional case, thederivative terms can be written as∂∂T∂∂T∂ T ∂ NikxNi =kxNi − k x∂x∂x∂x∂x∂x ∂x∂∂∂T∂T∂ T ∂ NikyNi =kyNi − k y(7.73)∂y∂y∂y∂y∂y ∂y∂∂T∂∂T∂ T ∂ NikzNi =kzNi − kz∂z∂z∂z∂z∂z ∂zand the residual equations become∂∂T∂∂T∂∂TkxNi +kyNi +kzNiQ Ni dVdV +∂x∂x∂y∂y∂z∂zVV ∂ T ∂ Ni∂ T ∂ Ni∂ T ∂ Ni=+ ky+ kzdVi = 1, .

. . , M (7.74)kx∂x ∂x∂y ∂y∂z ∂z VThe integral on the left side of Equation 7.74 contains a perfect differentialin three dimensions and can be replaced by an integral over the surface of thevolume using Green’s theorem in three dimensions: If F(x, y, z), G(x, y, z), and267Hutton: Fundamentals ofFinite Element Analysis2687. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat TransferH(x, y, z) are functions defined in a region of xyz space (the element volume inour context), then ∂F∂G∂H++dV = (Fn x + Gn y + Hn z ) d A(7.75)∂x∂y∂zVAwhere A is the surface area of the volume and n x , n y , n z are the Cartesian components of the outward unit normal vector of the surface area. This theorem is thethree-dimensional counterpart of integration by parts discussed earlier in thischapter.Invoking Fourier’s law and comparing Equation 7.75 to the first term ofEquation 7.74, we have− (q x n x + q y n y + q z n z ) N i d A +Q N i dVAV =kxV∂ T ∂ Ni∂ T ∂ Ni∂ T ∂ Ni+ ky+ kz∂x ∂x∂y ∂y∂z ∂zdVi = 1, .

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