Hutton - Fundamentals of Finite Element Analysis (523155), страница 52

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The consistent capacitance matrix is so called becauseit is formulated on the basis of the same interpolation functions used to describethe spatial distribution of temperature. In our approach, using Galerkin’s method,the consistent matrix is a natural result of the mathematical procedure. An alternate approach produces a so-called lumped capacitance matrix. Whereas the consistent matrix distributes the capacitance throughout the element by virtue of theinterpolation functions, the lumped capacitance matrix ascribes the storagecapacity strictly to the nodes independently.

The difference in the two approaches is discussed in terms of heat transfer and, in more detail, in Chapter 10in the context of structural dynamics.Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.8 Time-Dependent Heat TransferThe model assembly procedure for a transient heat transfer problem is exactly the same as for a steady-state problem, with the notable exception that wemust also assemble a global capacitance matrix. The rules are the same. Elementnodes are assigned to global nodes and the element capacitance matrix terms areadded to the appropriate global positions in the global capacitance matrix, aswith the conductance matrix terms.

Hence, on system assembly, we obtain theglobal equations[C ]{ Ṫ } + [K ]{T } = {FQ } + {Fg }(7.116)where we must recall that the gradient force vector {Fg } is composed of either(1) unknown heat flux values to be determined (unknown reactions) or (2) convection terms to be equilibrated with the flux at a boundary node.7.8.1 Finite Difference Methods for the TransientResponse: Initial ConditionsThe finite element discretization procedure has reduced the one-dimensionaltransient heat transfer problem to algebraic terms in the spatial variable via theinterpolation functions. Yet Equation 7.116 represents a set of ordinary, coupled,first-order differential equations in time.

Consequently, as opposed to the steadystate case, there is not a solution but multiple solutions as the system respondsto time-dependent conditions. The boundary conditions for a transient problemare of the three types discussed for the steady-state case: specified nodal temperatures, specified heat flux, or convection conditions. However, note that theboundary conditions may also be time dependent.

For example, a specified nodaltemperature could increase linearly with time to some specified final value. Inaddition, an internal heat generation source Q may also vary with time.A commonly used approach to obtaining solutions for ordinary differentialequations of the form of Equation 7.116 is the finite difference method. As discussed briefly in Chapter 1, the finite difference method is based on approximating derivatives of a function as incremental changes in the value of the functioncorresponding to finite changes in the value of the independent variable. Recallthat the first derivative of a function f (t ) is defined byf˙ =dff (t + t ) − f (t )= limt→0dtt(7.117)Instead of requiring t to approach zero, we obtain an approximation to thevalue of the derivative by using a small, nonzero value of t to obtainf˙ ∼=f (t + t ) − f (t )t(7.118)and the selected value of t is known as the time step.To apply the procedure to transient heat transfer, we approximate the timederivative of the nodal temperature matrix as{T (t + t )} − {T (t )}{ Ṫ } ∼=(7.119)t279Hutton: Fundamentals ofFinite Element Analysis2807.

Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat TransferSubstituting, Equation 7.116 becomes[C ]{T (t + t )} − {T (t )}+ [K ]{T (t )} = {FQ (t )} + {Fg (t )}t(7.120)Note that, if the nodal temperatures are known at time t and the forcing functionsare evaluated at time t, Equation 7.120 can be solved, algebraically, for the nodaltemperatures at time t + t . Denoting the time at the ith time step as ti = i (t ),i = 0, 1, 2, . . . , we obtain[C ] {T (ti+1 )} = [C ] {T (ti )} − [K ] {T (ti )} t + {FQ (ti )}t + {Fg (ti )}t(7.121)as the system of algebraic equations that can be solved for {T (ti+1 )} .

Formally,the solution is obtained by multiplying Equation 7.121 by the inverse of thecapacitance matrix. For large matrices common to finite element models, inverting the matrix is very inefficient, so other techniques such as Gaussian elimination are more often used. Note, however, that the system of algebraic equationsgiven by Equation 7.121 must be solved only once to obtain an explicit solutionfor the nodal temperatures at time ti+1 .The method just described is known as a forward difference scheme (alsoknown as Euler’s method) and Equation 7.121 is a two-point recurrence relation. If the state of the system (nodal temperatures and forcing functions) isknown at one point in time, Equation 7.121 gives the state at the next point intime.

Solving the system sequentially at increasing values of the independentvariable is often referred to as marching in time. To begin the solution procedure, the state of the system must be known at t = 0 . Therefore, the initial conditions must be specified in addition to the applicable boundary conditions.Recall that the general solution to an ordinary, first-order differential equationcontains one constant of integration.

As we have one such equation corresponding to each nodal temperature, the value of each nodal temperature must bespecified at time zero. If the initial conditions are so known, the recurrence relation can be used to compute succeeding nodal temperatures.

Prior to discussingother schemes and the ramifications of time step selection, the following simpleexample is presented.EXAMPLE 7.11Figure 7.20a shows a cylindrical rod having diameter of 12 mm and length of 100 mm.The pin is of a material having thermal conductivity 230 W/(m-◦C), specific heat900 J/(kg-◦C), and density 2700 kg/m3. The right-hand end of the rod is held in contactwith a medium at a constant temperature of 30◦C.

At time zero, the entire rod is at a temperature of 30◦C when a heat source is applied to the left end, bringing the temperature ofthe left end immediately to 80◦C and maintaining that temperature indefinitely. Using theforward difference method and four two-node elements, determine both transient andsteady-state temperature distributions in the rod.

No internal heat is generated.Hutton: Fundamentals ofFinite Element Analysis7. Applications in HeatTransferText© The McGraw−HillCompanies, 20047.8 Time-Dependent Heat Transfer1InsulatedT 30 C, t 0T 80 C, t 0d130 C22Figure 7.20(a) Cylindrical rod of Example 7.11. (b) Node and element numbers.■ SolutionIn the solution for this example, we set up the general procedure then present the resultsfor one solution using one time step for the transient portion.

The node numbers andelement numbers are as shown in Figure 7.20b. Since the length and area of each elementare the same, we compute the element capacitance matrix asC900(2700 ) (0.012) 2 (0.025) c A L 2 124==121662.2902 1.1451=J/ ◦ C1.1451 2.290212where we have implicitly performed the integrations indicated in Equation 7.115 andleave the details as an end-of-chapter exercise. Similarly, the element conductancematrix isk(e)200 (0.012) 2 k A 1 −114==−1L −1 10.0250.9408−0.9408=W/ ◦ C−0.94080.94083(b)(a)(e)325 mmL281−11For the one-dimensional case with uniform geometry and material properties, the systemassembly is straightforward and results in the global matrices2.2902 1.1451000 1.1451 4.5804 1.145100 [C] =  01.1451 4.5804 1.14510  001.1451 4.5804 1.1451 0001.1451 2.29020.9408 −0.9408000 −0.9408 1.8816 −0.940800[K ] = 0−0.94081.8816−0.9408000−0.9408 1.8816 −0.9408 000−0.9408 0.9408As no internal heat is generated {FQ } = 0 and, as we have specified boundary temperatures, the flux forcing term is an unknown.

Note that, in the transient case, the flux terms445Hutton: Fundamentals ofFinite Element Analysis2827. Applications in HeatTransferCHAPTER 7Text© The McGraw−HillCompanies, 2004Applications in Heat Transferat the boundaries (the “reactions”) are time dependent and can be computed at each timestep, as will be explained. Hence, the gradient “force vector” isq1 A  0 {Fg } =0 0 −q5 AHaving taken care of the boundary conditions, we now consider the initial conditionsand examine the totality of the conditions on the solution procedure.

It should beclear that, since we have the temperature of two nodes specified, the desired solutionshould provide the temperatures of the other three nodes and, therefore, should be a3 × 3 system. The reduction to the 3 × 3 system is accomplished via the followingobservations:1.2.If T1 = 80 ◦ C = constant, then Ṫ1 = 0 .If T5 = 30 ◦ C = constant, then Ṫ5 = 0 .The equations can be modified accordingly.

In this example, the general equationsbecome   080   q1 A  Ṫ2   T2  0 [C] Ṫ3 + [K ] T3 =0   0 T Ṫ  4  4−q5 A030Consequently, the first and fifth equations become1.1451 Ṫ2 + 0.9408 (80) − 0.9408 T2 = q 1 A1.1451 Ṫ4 − 0.9408 T4 + 0.9408 (30) = −q 5 Arespectively. The three remaining equations are then written as   4.5804 1.145101.8816 −0.94080 Ṫ2  T2  1.1451 4.5804 1.1451  Ṫ3 +  −0.9408 1.8816 −0.9408  T3  T4Ṫ401.1451 4.58040−0.9408 1.8816 75.264 =028.224For this example, the capacitance matrix is inverted (using a spreadsheet program) toobtain[C]−10.2339 −0.0624 0.0156=  −0.0624 0.2495 −0.0624 0.0156 −0.0624 0.2339Hutton: Fundamentals ofFinite Element Analysis7.

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