Hutton - Fundamentals of Finite Element Analysis (523155), страница 53
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Applications in HeatTransferText© The McGraw−HillCompanies, 20047.8 Time-Dependent Heat Transfer70T4T ( C)6050T340T2302051015t, sec2025Figure 7.21 Time histories of the nodaltemperatures.where [C] now represents the reduced 3 × 3 capacitance matrix. Utilizing Equation 7.121and multiplying by [C ]−1 yields 0.4988 −0.3521 0.0880 T2 T2 18.0456 T2 = T3− −0.3521 0.5869 −0.3521 T3t + −6.4558 tT 3 T4 i +1T4 iT4 i0.0880 −0.3521 0.49887.7762as the two-point recurrence relation.Owing to the small matrix involved, the recurrence relation was programmed into astandard spreadsheet program using time step t = 0.1 sec. Calculations for nodal temperatures T2 , T3 , and T4 are carried out until a steady state is reached. Time histories ofeach of the nodal temperature are shown in Figure 7.21.
The figure shows that steadystate conditions T2 = 67.5◦ C , T3 = 55 ◦ C , and T4 = 42.5◦ C are attained in about 30 sec.Interestingly, the results also show that the temperatures of nodes 3 and 4 initiallydecrease. Such phenomena are physically unacceptable and associated with use of a consistent capacitance matrix, as is discussed in Chapter 10.7.8.2 Central Difference and BackwardDifference MethodsThe forward difference method discussed previously and used in Example 7.11is but one of three commonly used finite difference methods.
The others are thebackward difference method and the central difference method. Each of these isdiscussed in turn and a single two-point recurrence relation is developed incorporating the three methods.In the backward difference method, the finite approximation to the firstderivative at time t is expressed asT (t ) − T (t − t )Ṫ (t ) ∼=t(7.122)283Hutton: Fundamentals ofFinite Element Analysis2847. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transferso that we, in effect, look back in time to approximate the derivative during theprevious time step. Substituting this relation into Equation 7.116 gives[C]{T (t)} − {T (t − t)}+ [K ]{T (t)} = {FQ (t)} + {Fg (t)}t(7.123)In this method, we evaluate the nodal temperatures at time t based on the state ofthe system at time t − t , so we introduce the notation t = ti , ti−1 = t − t ,i = 1, 2, 3, .
. . . Using the described notation and rearranging, Equation 7.123becomes([C] + [K ]t){T (ti )} = [C]{T (ti−1 )} + FQ (ti )t + Fg (ti )ti = 1, 2, 3, . . .(7.124)If the nodal temperatures are known at time ti−1 , Equation 7.124 can be solvedfor the nodal temperatures at the next time step (it is assumed that the forcingfunctions on the right-hand side are known and can be determined at ti ). Notingthat the time index is relative, Equation 7.125 can also be expressed as([C ] + [K ]t ) {T (ti+1 )}= [C ] {T (ti )} + FQ (ti )t + Fg (ti )ti = 0, 1, 2, . .
.(7.125)If we compare Equation 7.125 with Equation 7.121, we find that the major difference lies in the treatment of the conductance matrix. In the latter case, theeffects of conductance are, in effect, updated during the time step. In the case ofthe forward difference method, Equation 7.121, the conductance effects are heldconstant at the previous time step. We also observe that Equation 7.125 cannot besolved at each time step by “simply” inverting the capacitance matrix. The coefficient matrix on the left-hand side changes at each time step; therefore, moreefficient methods are generally used to solve Equation 7.125.Another approach to approximation of the first derivative is the central difference method. As the name implies, the method is a compromise of sorts betweenforward and backward difference methods. In a central difference scheme, thedependent variable and all forcing functions are evaluated at the center (midpoint)of the time step.
In other words, average values are used. In the context of transientheat transfer, the time derivative of temperature is still as approximated by Equation 7.119 but the other terms in Equation 7.120 are evaluated at the midpoint ofthe time step. Using this approach, Equation 7.120 becomes{T (t + t )} − {T (t )}T (t + t ) + T (t )[C ]+ [K ]t2 FQ (t + t ) + FQ (t )Fg (t + t ) + Fg (t )=+(7.126)22The forcing functions on the right-hand side of Equation 7.126 are either knownfunctions and can be evaluated or “reactions,” which are subsequently computedHutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 2004Referencesvia the constraint equations. The left-hand side of Equation 7.126 is now, however,quite different, in that the unknowns at each step Ti (t + t ) appear in both capacitance and conductance terms.
Multiplying by t and rearranging Equation 7.126,we obtaint{T (t + t )}[C ] + [K ]2 Fg (t + t) + Fg (t)tFQ (t + t) + FQ (t)= [C] − [K ]{T (t)} ++222(7.127)Equation 7.127 can be solved for the unknown nodal temperatures at time t + tand the “marching” solution can progress in time until a steady state is reached.The central difference methods is, in general, more accurate than the forward orbackward difference method, in that it does not give preference to either temperatures at t or t + t but, rather, gives equal credence to both.In finite difference methods, the key parameter governing solution accuracyis the selected time step t .
In a fashion similar to the finite element method, inwhich the smaller the elements are, physically, the better is the solution, the finitedifference method converges more rapidly to the true solution as the time stepis decreased. These ideas are amplified in Chapter 10, when we examine thedynamic behavior of structures.7.9 CLOSING REMARKSIn Chapter 7, we expand the application of the finite element method into twoand three-dimensional, as well as axisymmetric, problems in heat transfer. Whilethe majority of the chapter focuses on steady-state conditions, we also presentthe finite difference methods commonly used to examine transient effects. Thebasis of our approach is the Galerkin finite element method, and this text stayswith that procedure, as it is so general in application.
As we proceed into applications in fluid mechanics, solid mechanics, and structural dynamics in the following chapters, the Galerkin method is the basis for the development of manyof the finite element models.REFERENCESHuebner, K. H., and E. A. Thornton. The Finite Element Method for Engineers, 2nded. New York: John Wiley and Sons, 1982.2. Incropera, F. P., and D.
P. DeWitt. Introduction to Heat Transfer, 3rd ed. New York:John Wiley and Sons, 1996.1.285Hutton: Fundamentals ofFinite Element Analysis2867. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat TransferPROBLEMS7.17.27.37.47.5For Example 7.1, determine the exact solution by integrating Equation 5.59 andapplying the boundary conditions to evaluate the constants of integration.Verify the convection-related terms in Equation 7.15 by direct integration.For the data given in Example 7.4, use Gaussian quadrature with four integrationpoints (two on r, two on s) to evaluate the terms of the stiffness matrix.
Do yourresults agree with the values given in the example?Using the computed nodal temperatures and heat flux values calculated inExample 7.5, perform a check calculation on the heat flow balance. That is,determine whether the heat input is in balance with the heat loss due to convection.How does this check indicate the accuracy of the finite element solution?Consider the circular heat transfer pin shown in Figure P7.5. The base of the pinis held at constant temperature of 100◦C (i.e., boiling water). The tip of the pinand its lateral surfaces undergo convection to a fluid at ambient temperature Ta .The convection coefficients for tip and lateral surfaces are equal.
Given k x =380 W/m-◦C, L = 8 cm, h = 2500 W/m2-◦C, d = 2 cm, Ta = 30◦C. Use a twoelement finite element model with linear interpolation functions (i.e., a two-nodeelement) to determine the nodal temperatures and the heat removal rate from thepin. Assume no internal heat generation.h, Tah, Ta100 CLFigure P7.57.67.77.87.9Repeat Problem 7.5 using four elements. Is convergence indicated?The pin of Figure P7.5 represents a heating unit in a water heater. The base of thepin is held at fixed temperature 30◦C. The pin is surrounded by flowing water at55◦C.
Internal heat generation is to be taken as the constant value Q = 25 W/cm3.All other data are as given in Problem 7.5. Use a two-element model to determinethe nodal temperatures and the net heat flow rate from the pin.Solve Problem 7.5 under the assumption that the pin has a square cross section1 cm × 1 cm . How do the results compare in terms of heat removal rate?The efficiency of the pin shown in Figure P7.5 can be defined in several ways.One way is to assume that the maximum heat transfer occurs when the entire pinis at the same temperature as the base (in Problem 7.5, 100◦C), so thatconvection is maximized. We then writeLq max =h P (Tb − Ta ) dx + h A(Tb − Ta )0where Tb represents the base temperature, P is the peripheral dimension, and Ais cross-sectional area at the tip.
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