Hutton - Fundamentals of Finite Element Analysis (523155), страница 48
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For example,1 11 1N 1 dr ds =−1 −1−1 −11111 (1 − r ) 2 (1 − s) 2 (1 − r )(1 − s) dr ds = =14422−1−1An identical result is obtained when the other three functions are integrated. The integralcorresponding to convection from the element lateral surfaces is thenI1 = 2h AT(3)1+T(3)2+T(3)3+T(3)44− TaThe first term in the parentheses is the average of the nodal temperatures, and this is ageneral result for the rectangular element. Substituting numerical valuesI1 =2(50)(1) 2144111.982 + 90.966 + 89.041 + 106.507− 684= 21.96 Btu/hrNext, we consider the edge convection terms. Along edge 8-9,I2 =h([N ]{T } − Ta ) d A 8−9A 8−9and, since r = 1 along that edge, d A 8−9 = t b ds, and the integral becomes1I2 = ht b([N r =1 ]{T } − Ta ds)−11= ht b−11[0411−s1+s0] ds {T } − ht bTads−1Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.4 Heat Transfer in Two Dimensions= h(2t b) 0= h A edgeT12(3)20 {T } − h(2t b)Ta12+T(3)32− TaAgain, we observe that the average temperature of the nodes associated with the area ofthe edge appears.
Stated another way, the convection area is allocated equally to the twonodes, and this is another general result for the rectangular element. Inserting numericalvalues,I2 =50(0.5)(1)14490.966 + 89.041− 682= 3.82 Btu/hrBy analogy, the edge convection along edge 6-9 isI3 = h A edgeT(3)3+T(3)42− Ta=50(0.5)(1)14489.041 + 106.507− 682= 5.17 Btu/hrThe total convective heat flow rate for element 3 is thenḢ(3)h= I1 + I2 + I3 = 30.95 Btu/hr7.4.5 Internal Heat GenerationTo this point in the current discussion of heat transfer, only examples havingno internal heat generation ( Q = 0) have been considered. Also, for twodimensional heat transfer, we considered only thin bodies such as fins.
Certainlythese are not the only cases of interest. Consider the situation of a body ofconstant cross section having length much larger than the cross-sectional dimensions, as shown in Figure 7.13a (we use a rectangular cross section for convenience). In addition, an internal heat source is imbedded in the body and runsparallel to the length. Practical examples include a floor slab containing a hotwater or steam pipe for heating and a sidewalk or bridge deck having embeddedheating cables to prevent ice accumulation. The internal heat generation sourcein this situation is known as a line source.Except very near the ends of such a body, heat transfer effects in the z direction can be neglected and the situation treated as a two-dimensional problem,as depicted in Figure 7.13b.
Assuming the pipe or heat cable to be small incomparison to the cross section of the body, the source is treated as acting at asingle point in the cross section. If we model the problem via the finite elementmethod, how do we account for the source in the formulation? Per the first ofEquation 7.38, the nodal force vector corresponding to internal heat generation is (e) Tf Q =Q[N ] t d A =Q{N }t d A(7.55)AA259Hutton: Fundamentals ofFinite Element Analysis2607. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transferyx3yP(x0, y0)Heat source2xz(a)1(b)Figure 7.13Figure 7.14(a) Long, slender body with internal heat source.(b) 2-D representation (unit thickness inz-direction).Concentrated heatsource Q* located atpoint P(x0, y0) in atriangular element.where, as before, t is element thickness.
In this type of problem, it is customaryto take t as unity, so that all computations are per unit length. In accordance withthis convention, the source strength is denoted Q ∗ , having units typicallyexpressed as Btu/(hr-ft2) or W/m2. Equation 7.55 then becomesf(e) Q=∗Q [N ] d A =ATQ ∗ {N } d A(7.56)AThe question is now mathematical: How do we integrate a function applicable ata single point in a two-dimensional domain? Mathematically, the operation isquite simple if the concept of the Dirac delta or unit impulse function is introduced. We choose not to take the strictly mathematical approach, however, in theinterest of using an approach based on logic and all the foregoing informationpresented on interpolation functions.For illustrative purposes, the heat source is assumed to be located at a knownpoint P = (x 0 , y0 ) in the interior of a three-node triangular element, as in Figure 7.14.
If we know the temperature at each of the three nodes of the element,then the temperature at point P is a weighted combination of the nodal temperatures. By this point in the text, the reader is well aware that the weighting factorsare the interpolation functions. If nodal values are interpolated to a specific point,a value at that point should properly be assigned to the nodes via the same interpolation functions evaluated at the point.
Using this premise, the nodal forces forHutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.5 Heat Transfer with Mass Transportthe triangular element become (assuming Q ∗ to be constant) N1 (x0 , y0 ) (e) f Q = Q∗N (x , y ) dA 2 0 0 N3 (x0 , y0 )A(7.57)For a three-node triangular element, the interpolation functions (from Chapter 6)are simply the area coordinates, so we now have L 1 (x0 , y0 ) L 1 (x0 , y0 ) (e) L 2 (x0 , y0 ) dA = Q ∗ A L 2 (x0 , y0 )f Q = Q∗(7.58)L (x , y )L (x , y )3A00300Now consider the “behavior” of the area coordinates as the position of the interior point P varies in the element. As P approaches node 1, for example, areacoordinate L 1 approaches unity value.
Clearly, if the source is located at node 1,the entire source value should be allocated to that node. A similar argument canbe made for each of the other nodes. Another very important point to observehere is that the total heat generation as allocated to the nodes by Equation 7.58 isequivalent to the source. If we sum the individual nodal contributions given inEquation 7.58, we obtain3since3i=1Q∗(e)i=3)L(e)1+L(e)2+L(e) * ∗3 Q A= Q∗ A(7.59)i=1L i = 1 is known by the definition of area coordinates.i=1The foregoing approach using logic and our knowledge of interpolationfunctions is without mathematical rigor.
If we approach the situation of a linesource mathematically, the result is exactly the same as that given by Equation 7.58 for the triangular element. For any element chosen, the force vectorcorresponding to a line source (keep in mind that, in two-dimensions, this lookslike a point source) the nodal force contributions are (e) ∗{N (x 0 , y0 )} d AfQ = Q(7.60)AThus, a source of internal heat generation is readily allocated to the nodes of afinite element via the interpolation functions of the specific element applied.7.5 HEAT TRANSFER WITH MASS TRANSPORTThe finite element formulations and examples previously presented deal withsolid media in which heat flows as a result of conduction and convection. An additional complication arises when the medium of interest is a flowing fluid.In such a case, heat flows by conduction, convection, and via motion of themedia.
The last effect, referred to as mass transport, is considered here for the261Hutton: Fundamentals ofFinite Element Analysis2627. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat TransferConvectionqhqin qx qmqx qmqoutQ,Uqx dqxdqmdx qm dxdxdxdx(a)(b)Figure 7.15(a) One-dimensional conduction with convection and mass transport. (b) Control volume forenergy balance.one-dimensional case. Figure 7.15a is essentially Figure 7.2a with a major physical difference.
The volume shown in Figure 7.15a represents a flowing fluid (asin a pipe, for example) and heat is transported as a result of the flow. The heatflux associated with mass transport is denoted qm , as indicated in the figure. Theadditional flux term arising from mass transport is given byqm = ṁcT (W or Btu/hr)(7.61)where ṁ is mass flow rate (kg/hr or slug/hr), c is the specific heat of the fluid(W-hr/(kg-◦C) or Btu/(slug-◦F)), and T (x ) is the temperature of the fluid (◦C or◦F). A control volume of length dx of the flow is shown in Figure 7.15b, wherethe flux terms have been expressed as two-term Taylor series as in past derivations.
Applying the principle of conservation of energy (in analogy withEquation 7.1),dq xdx A dtq x A dt + qm dt + Q A dx dt = U + q x +dxdqm+ qm +dx dt + q h P dx dt(7.62)dxConsidering steady-state conditions, U = 0 , using Equations 5.51 and 7.2 andsimplifying yieldsddTdqmhPkx+Q=+(T − Ta )(7.63)dxdxdxAwhere all terms are as previously defined. Substituting for qm into Equation 7.63,we obtainddTd ṁchPkx+Q=T +(T − Ta )(7.64)dxdxdx AAwhich for constant material properties and constant mass flow rate (steady state)becomeskxd2 Tṁc dThP+Q=+(T − Ta )2dxA dxA(7.65)Hutton: Fundamentals ofFinite Element Analysis7.
Applications in HeatTransferText© The McGraw−HillCompanies, 20047.5 Heat Transfer with Mass Transport263With the exception of the mass transport term, Equation 7.65 is identical toEquation 7.4. Consequently, if we apply Galerkin’s finite element method, theprocedure and results are identical to those of Section 7.3, except for additionalstiffness matrix terms arising from mass transport.
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