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(b) Differential element as a control volume.dqxdxdx227Hutton: Fundamentals ofFinite Element Analysis2287. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transfermaterial properties. The convective heat transfer across the surface, denoted q h ,represents the heat flow rate (heat flux) across the surface per unit surface area.To apply the principle of conservation of energy to the control volume, we needonly add the convection term to Equation 5.54 to obtain∂q xdx A dt + q h P dx dtq x A dt + Q A dx dt = U + q x +(7.1)∂xwhere all terms are as previously defined except that P is the peripheral dimension of the differential element and q h is the heat flux due to convection.

The convective heat flux is given by [2]q h = h(T − Ta )(7.2)whereh = convection coefficient, W/(m2-◦ C ), Btu/(hr-ft2-◦ F )T = temperature of surface of the bodyTa = ambient fluid temperatureSubstituting for q h and assuming steady-state conditions such that U = 0 ,Equation 7.1 becomesQA = Adq x+ h P (T − Ta )dx(7.3)which, via Fourier’s law Equation 5.55, becomeskxhPd2 T+Q=(T − Ta )dx 2A(7.4)where we have assumed k x to be constant.While Equation 7.4 represents the one-dimensional formulation of conduction with convection, note that the temperature at any position x along the lengthof the body is not truly constant, owing to convection.

Nevertheless, if the crosssectional area is small relative to the length, the one-dimensional model can giveuseful results if we recognize that the computed temperatures represent averagevalues over a cross section.7.3.1 Finite Element FormulationTo develop the finite element equations, a two-node linear element for whichT (x ) = N 1 (x )T1 + N 2 (x )T2(7.5)is used in conjunction with Galerkin’s method.

For Equation 7.4, the residualequations (in analogy with Equation 5.61) are expressed asx2 hPd2 Tkx 2 + Q −i = 1, 2(T − Ta ) N i (x ) A dx = 0(7.6)dxAx1Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.3 One-Dimensional Conduction with Convectionorx2Ad2 Tk x 2 N i (x ) dx − h Pdxx1x2x2T (x ) N i (x ) dx + Ax1Q N i (x ) dxx1x2+ hPT aN i (x ) dx = 0i = 1, 2(7.7)x1Integrating the first term by parts and rearranging,x2x2dN i dTkx AT (x ) N i (x ) dxdx + h Pdx dxx1x1x2= Ax2Q N i (x ) dx + h P T ax1N i (x ) dx + k x A N i (x )x1xdT 2dx x1i = 1, 2(7.8)Substituting for T (x ) from Equation 7.5 yieldsx2kx AdN idxx2dN 2dN 1N i (x )[N 1 (x )T1 + N 2 (x )T2 ] dxT1 +T2 dx + h Pdxdxx1x1x2= Ax2Q N i (x ) dx + h P T ax1x1xdT 2N i (x ) dx + k x A N i (x )dx x1i = 1, 2(7.9)The two equations represented by Equation 7.9 are conveniently combined intoa matrix form by rewriting Equation 7.5 as T1T (x ) = [N 1 N 2 ]= [N ]{T }(7.10)T2and substituting to obtainx2 kx AdNdxT x2dN{T } dx + h P [N ] T [N ]{T } dxdxx1x1x2= Ax2Q[N ] dx + h P T aTx1xdT 2[N ] dx + k x A [N ]dx x1Tx1Equation 7.11 is in the desired finite element form: (e) (e) (e) (e) k{T } = f Q + f h + f gT(7.11)(7.12)229Hutton: Fundamentals ofFinite Element Analysis2307.

Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transferwhere [k (e) ] is the conductance matrix defined asT x2 x2 (e) dNdNk= kx Adx + h P [N ] T [N ] dxdxdxx1(7.13)x1The first integral is identical to that in Equation 5.66, representing the axial conduction effect, while the second integral accounts for convection.Without loss of generality, we let x 1 = 0 , x 2 = L so that the interpolationfunctions arexN1 = 1 −L(7.14)xN2 =LThe results of the first integral are as given in Equation 5.68, so we need performonly the integrations indicated in the second term (Problem 7.2) to obtain (e) (e) (e) k x A 1 −1hPL 2 1+= kc + khk=(7.15)1 2L −1 16(e)where [k (e)c ] and [k h ] represent the conductive and convective portions of thematrix, respectively.

Note particularly that both portions are symmetric.The forcing function vectors on the right-hand side of Equation 7.12 includethe internal heat generation and boundary flux terms, as in Chapter 5. These aregiven byL Q N 1 dx  (e) 0fQ =A L(7.16) Q N 2 dx 0 dT − dx  (e) qx=0q10f g = kx A=A=A−qx=L−q2dT dx L(7.17)where q1 and q2 are the boundary flux values at nodes 1 and 2, respectively. Inaddition, the forcing function arising from convection is 1 N1 dx  h PT L (e) 1a0f h = h PT a 1=(7.18)12 N2 dx 0where it is evident that the total element convection force is simply allocatedequally to each node, like constant internal heat generation Q.Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.3 One-Dimensional Conduction with Convection1234M3 M2 M1M231M1ConvectionBC1BC2xFigure 7.3 Convection boundary condition at node M + 1 of an M-element,one-dimensional heat transfer finite element model.7.3.2 Boundary ConditionsIn the one-dimensional case of heat transfer under consideration, two boundaryconditions must be specified.

Typically, this means that, if the finite elementmodel of the problem is composed of M elements, one boundary condition isimposed at node 1 of element 1 and the second boundary condition is imposed atnode 2 of element M. The boundary conditions are of three types:1. Imposed temperature. The temperature at an end node is a known value;this condition occurs when an end of the body is subjected to a constantprocess temperature and heat is removed from the process by the body.2. Imposed heat flux. The heat flow rate into, or out of, an end of the body isspecified; while distinctly possible in a mathematical sense, this type ofboundary condition is not often encountered in practice.3.

Convection through an end node. In this case, the end of the body is incontact with a fluid of known ambient temperature and the conduction fluxat the boundary is removed via convection to the fluid media. Assumingthat this condition applies at node 2 of element M of the finite elementmodel, as in Figure 7.3, the convection boundary condition is expressed asdT kx= −q M+1 = −h(TM+1 − Ta )(7.19)dx M+1indicating that the conduction heat flux at the end node must be carriedaway by convection at that node. The area for convection in Equation 7.19is the cross-sectional area of element M; as this area is common to each ofthe three terms in the equation, the area has been omitted. An explanationof the algebraic signs in Equation 7.19 is appropriate here.

If TM+1 > Ta ,the temperature gradient is negative (given the positive direction of the xaxis as shown); therefore, the flux and convection are positive terms.The following example illustrates application of the one-dimensionalconduction/convection problem.EXAMPLE 7.3Figure 7.4a depicts a cylindrical pin that is one of several in a small heat exchange device.The left end of the pin is subjected to a constant temperature of 180 ◦ F . The right end ofthe pin is in contact with a chilled water bath maintained at constant temperature of 40 ◦ F .Hutton: Fundamentals ofFinite Element Analysis2327. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transfer72 F180 F40 F4 in.(a)121T 180F342354q h(T5 40)(b)Figure 7.4 Example 7.3: (a) Cylindrical pin.(b) Finite element model.The exterior surface of the pin is in contact with moving air at 72 ◦ F .

The physical data aregiven as follows:D = 0.5 in.,kx = 120 Btu/(hr-ft-◦ F ),L = 4 in.,hair = 50 Btu/(hr-ft2-◦ F ),hwater = 100 Btu/(hr-ft2-◦ F )Use four equal-length, two-node elements to obtain a finite element solution for thetemperature distribution across the length of the pin and the heat flow rate through the pin.■ SolutionFigure 7.4b shows the elements, node numbers, and boundary conditions. The boundaryconditions are expressed as followsT1 = 180 ◦ FdT kx= −q 5 = −h(T5 − 40)dx 5At node 1:At node 5:Element geometric data is thenL e = 1 in.,P = (0.5) = 1.5708 in.,A = (/4)(0.5) 2 = 0.1963 in.2The leading coefficients of the conductance matrix terms arekx A=Le120h air PL e=6500.19631441121.5708126= 1.9630 Btu/(hr-◦ F)112= 0.0909 Btu/(hr-◦ F)where conversion from inches to feet is to be noted.Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.3 One-Dimensional Conduction with ConvectionSubstituting into Equation 7.15, the element conductance matrix isk (e) = 1.96301−1−11+ 0.09092112=2.1448−1.8721−1.87212.1448Following the direct assembly procedure, the system conductance matrix is2.1448 −1.8721000 −1.8721 4.2896 −1.872100[K ] = 0−1.8721 4.2896 −1.8721000−1.8721 4.2896 −1.8721 000−1.8721 2.1448As no internal heat is generated, f Q = 0 .

The element convection force components perEquation 7.18 are 11.5708 (72)50 (e) hPT aL 1121219.6375=f h ==Btu/hr119.637522Assembling the contributions of each element at the nodes gives the system convectionforce vector as19.6375  39.2750 {Fh } = 39.2750 Btu/hr 39.2750 19.6375Noting the cancellation of terms at nodal connections, the system gradient vectorbecomes simply  Aq1  Aq1Aq1 00  0  {Fg } ===Btu/hr00000  0  −Aq5−Ah water (T5 − 40)−0.1364T5 + 5.4542and the boundary condition at the pin-water interface has been explicitly incorporated.Note that, as a result of the convection boundary condition, a term containing unknownnodal temperature T5 appears in the gradient vector.

This term is transposed in the finalequations and results in a increase in value of the K 55 term of the system matrix. The finalassembled equations are 2.1448 −1.8721000180   19.6375 + Aq1  −1.8721 4.2896 −1.8721 39.275000  T2  =39.27500−1.8721 4.2896 −1.87210  T3   39.275000−1.8721 4.2896 −1.8721    T4 T525.0917000−1.8721 2.2812233Hutton: Fundamentals ofFinite Element Analysis2347. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat TransferEliminating the first equation while taking care to include the effect of the specified temperature at node 1 on the remaining equations gives  T2   376.2530 4.2896 −1.857100  T3   39.2750  −1.8721 4.2896 −1.87210=0−1.8721 4.2896 −1.8721  T  39.2750  4 00−1.8721 2.2812T525.0917Solving by Gaussian elimination, the nodal temperatures are obtained asT2 = 136.16 ◦ FT3 = 111.02 ◦ FT4 = 97.23 ◦ FT5 = 90.79 ◦ FThe heat flux at node 1 is computed by back substitution of T2 into the first equation:2.1448 (180) − 1.8721 (136.16) = 19.6375 + Aq 1Aq 1 = 111.5156 Btu/hrq1 =111.5156≈ 81,805 Btu/hr-ft20.1963 /144Although the pin length in this example is quite small, use of only four elements represents a coarse element mesh.

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