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Metode Numerik Perpindahan PAnas

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Dasar dasar analisa numerik untuk analisa perpindahan panas
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METODE NUMERIK PADA PERPINDAHAN PANAS
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  • METODENUMERIK PADAPERPINDAHAN

    PANAS

  • 2WHY NUMERICAL METHODS?In Chapter 2, we solved various heatconduction problems in variousgeometries in a systematic but highlymathematical manner by(1) deriving the governing differentialequation by performing an energybalance on a differential volumeelement,(2) expressing the boundary conditionsin the proper mathematical form, and(3) solving the differential equation andapplying the boundary conditions todetermine the integration constants.

  • 31 LimitationsAnalytical solution methods are limited tohighly simplified problems in simplegeometries.The geometry must be such that its entiresurface can be described mathematically in acoordinate system by setting the variablesequal to constants.That is, it must fit into a coordinate systemperfectly with nothing sticking out or in.Even in simple geometries, heat transferproblems cannot be solved analytically if thethermal conditions are not sufficientlysimple.Analytical solutions are limited to problemsthat are simple or can be simplified withreasonable approximations.

  • 42 Better ModelingWhen attempting to get an analyticalsolution to a physical problem, there isalways the tendency to oversimplify theproblem to make the mathematical modelsufficiently simple to warrant an analyticalsolution.Therefore, it is common practice to ignoreany effects that cause mathematicalcomplications, e.g., nonlinearities (such astemperature dependence of thermalconductivity and the radiation boundaryconditions) in the differential equation or theboundary conditions.A mathematical model intended for anumerical solution is likely to represent theactual problem better.The numerical solution of engineeringproblems has now become the norm ratherthan the exception even when analyticalsolutions are available.

  • 53 FlexibilityEngineering problems often require extensive parametric studies tounderstand the influence of some variables on the solution in order tochoose the right set of variables and to answer some what-ifquestions.This is an iterative process that is extremely tedious and time-consuming if done by hand.Computers and numerical methods are ideally suited for suchcalculations, and a wide range of related problems can be solved byminor modifications in the code or input variables.Today it is almost unthinkable to perform any significant optimizationstudies in engineering without the power and flexibility of computersand numerical methods.

  • 64 ComplicationsSome problems can be solved analytically,but the solution procedure is so complexand the resulting solution expressions socomplicated that it is not worth all thateffort.With the exception of steady one-dimensional or transient lumped systemproblems, all heat conduction problemsresult in partial differential equations.Solving such equations usually requiresmathematical sophistication beyond thatacquired at the undergraduate level, suchas orthogonality, eigenvalues, Fourier andLaplace transforms, Bessel and Legendrefunctions, and infinite series.In such cases, the evaluation of thesolution, which often involves double ortriple summations of infinite series at aspecified point, is a challenge in itself.

  • 75 Human Nature Analytical solutions are necessarybecause insight to the physicalphenomena and engineeringwisdom is gained primarilythrough analysis.The feel that engineers developduring the analysis of simple butfundamental problems serves as aninvaluable tool when interpreting ahuge pile of results obtained from acomputer when solving a complexproblem.A simple analysis by hand for alimiting case can be used to check ifthe results are in the proper range.In this chapter, you will learn howto formulate and solve heattransfer problems numericallyusing one or more approaches.

  • 8FINITE DIFFERENCE FORMULATIONOF DIFFERENTIAL EQUATIONSThe numerical methods for solvingdifferential equations are based on replacingthe differential equations by algebraicequations.In the case of the popular finite differencemethod, this is done by replacing thederivatives by differences.

    Reasonably accurate results can be obtainedby replacing differential quantities bysufficiently small differences.

    AN EXAMPLE

  • 9Finite differenceform of the firstderivative

    Taylor series expansion of the function fabout the point x,

    The smaller the x, thesmaller the error, and thusthe more accurate theapproximation.

  • 10

    Consider steady one-dimensional heat conduction in a plane wall of thicknessL with heat generation.

    Finite difference representationof the second derivative at ageneral internal node m.

    No heat generation

  • 11

    Finite difference formulation for steady two-dimensional heat conduction in a region withheat generation and constant thermalconductivity in rectangular coordinates can beexpressed as

  • 12

    ONE-DIMENSIONAL STEADY HEATCONDUCTION

    In this section we develop the finitedifference formulation of heat conductionin a plane wall using the energy balanceapproach and discuss how to solve theresulting equations.The energy balance method is based onsubdividing the medium into a sufficientnumber of volume elements and thenapplying an energy balance on eachelement.

  • 13

    This equation is applicable to each of theM 1 interior nodes, and its applicationgives M 1 equations for thedetermination of temperatures at M + 1nodes.The two additional equations needed tosolve for the M + 1 unknown nodaltemperatures are obtained by applyingthe energy balance on the two elementsat the boundaries (unless, of course, theboundary temperatures are specified).

  • 14

  • 15

    Boundary ConditionsBoundary conditions most commonly encountered in practice are thespecified temperature, specified heat flux, convection, and radiationboundary conditions, and here we develop the finite differenceformulations for them for the case of steady one-dimensional heatconduction in a plane wall of thickness L as an example.The node number at the left surface at x = 0 is 0, and at the right surfaceat x = L it is M. Note that the width of the volume element for eitherboundary node is x/2.

    Specified temperature boundary conditions:


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