10. Modeling and Analysis - Hea= t and Mass Transfer, 2nd Edition

=20 =20

=20

=20 =20 =20 =20

=20 =20 =20 =20 Previous Chapter

9. Radiation Heat Tr= ansfer

=20 =20

=20 =20 =20 =20 Next Chapter

11. Experimental Hea= t Transfer

=20 =20

10 Modeling and Analysis

= 10.1 INTRODUCTION

Modeling and analysis are use= d to optimize the design of a thermal system. The processes that are carrie= d out in a system can be represented in terms of mathematical equations for= the analysis. Modeling a physical system and analyzing the results of the = processes with suitable correlations is known as simulation and it is equiv= alent to actual experiments on the system. Simulation of any physical proce= ss can provide relevant inputs for the design and optimization of a process= or system. Once a model of a system is created, it can be subjected to a l= arge number of operating conditions and the outputs will characterize the b= ehavior of the system. An optimum design condition can be derived from thes= e outputs. It is also important to understand the different types of models= , procedures to develop them, governing equations, material properties, var= iables, and validation of results.

= 10.2 BASICS OF MODELING

A model can be either descrip= tive or predictive and can be used to describe and explain different physic= al concepts. Models can be physically made to show the internal assembly of= components and also to explain the working principles of systems such as h= eat engines, pumps, and other machine tools. These models are known as desc= riptive types. The performance of the systems mentioned above can be predic= ted using predictive models. Using the governing equations that represent t= he processes in a predictive model, the variation of parameters such as tem= perature, pressure, and velocity with respect to time can be obtained. Proc= esses in thermal systems are usually described by partial differential equa= tions, which are time-dependent and multidimensional. In modeling and analy= sis, it is necessary to neglect unimportant factors and also to combine dif= ferent variables to simplify the analysis. Understanding the physical conce= pt is important to create a satisfactory model. Various techniques such as = dimensional analysis are also being used to create models. In some systems,= when it is difficult to simulate the processes for modeling and analysis, = experimental data are used to built the model and analyzed using dimensiona= l analysis. Experimental data are also used for the validation of the resul= ts obtained using model analysis.

= 10.3 PREDICTIVE MODELS

There are four predictive mod= els that are in use for the optimum design of a thermal system. They are: (= i) physical models, (ii) analog models, (iii) mathematical models, and (iv)= numerical models. These models have their own characteristics and applicab= le for specific systems and applications. Models are also classified as ste= ady-state or dynamic, probabilistic or deterministic, distributed or lumped= , and discrete or continuous.

In st= eady-state model, operating variables and properties do not change with tim= e and in the case of dynamic models time-dependent variables are included. = In some cases, both dynamic (in initial stage) and steady-state models (fin= al stage) are used to obtain the required results. In the case of probabili= stic model, uncertainties in the system are represented by probability dist= ributions. Deterministic model predicts the system behavior with good degre= e of acceptance. Distributed models give information about the variation of= parameters at any specified location and the lumped models provide average= values of the parameters of the considered volume. If individual data of p= arameters are considered, then a discrete model is used and for systems wit= h flow of fluids, continuous models are used.

10.3.1= Physical Models

To obtain the required experi= mental results and to study the behavior of a system, a physical model is m= ade similar to the actual system with reduced dimensions as scaled-down mod= els. To study the drag force on a car using wind tunnel and forces acting o= n ships and submarines using water channels, physical models are used.

10.3.2= Analog Models

Analog models use the solutio= n and results of a well-known problem to get the results of another similar= problem. These are used in the fields such as fluid mechanics and heat tra= nsfer. The analogy of conduction heat transfer to electrical circuit is an = example of an analog model. The analog model and the actual system are not = similar in physical appearance but obey the same physical principles. Analo= g models are used only to understand the physical phenomena and have limite= d use in design.

10.3.3= Mathematical Models

Mathematical models represent= the behavior and performance of a system by means of mathematical equation= s. Thermal systems can be designed with the help of the results of the math= ematical models without actually fabricating a prototype.

Governing equations provide = the basis for the procedure of design and optimization of a system. A mathe= matical model of a heat exchanger is described by a few correlations, which= will show the dependence of the heat transfer rate on the temperatures of = the fluids and dimensions of the heat exchanger, forming the basis for opti= mum design. The correlations are either arrived based on physical principle= s in the case of theoretical models or by curve fitting of experimental dat= a in the case of empirical models.

Mathematical models for desi= gn and optimization of a thermal system are usually developed with the foll= owing assumptions:

Modelling of the system is done a= ssuming that the system works in steady-state or transient condition. The t= wo important characteristics for this assumption are the response time (t_R) of the system and variation t= ime (t_v) of the ambient or ope= rating conditions. A system may be treated as being in a steady-state if th= e response time (t_R =E2=86=92 =CE=B1) i= s very large or for conditions with t_R = =E2=89=AA t_v and t_v =E2=89=AA t_R. In some = thermal systems, the process or parameters vary periodically with time and = may also be analyzed using steady-state models. All other systems are treat= ed as transient models.
The systems may be modeled as one-, tw= o-, or three-dimensional based on the required accuracy of prediction.
A lumped mass approximation is assumed= for a model if the temperature, species concentration, or other variables = are uniform within the system.
For simplificat= ion of the analysis, boundary conditions for the model may be approximated = as simple, smooth, and uniform (flat, cylinder, sphere, etc.).
Mathematical models are simplified by = neglecting the effects that are relatively small.
Ideal conditions are also assumed some= times in mathematical modeling.
Material property data are to be used = accurately in mathematical models.
The law of conservation of mass, m= omentum, and energy provide the basis for the governing equations of a mode= l.

= 10.4 GOVERNING EQUATIONS

Heat transfer and fluid flow = processes are governed by the laws of conservation of mass, momentum, and e= nergy with differential and integral equations. Analytical or numerical sol= utions are possible for the governing equations. Though used widely, numeri= cal solution frequently introduces questions of accuracy and stability. In = some specific cases, the governing differential equations of heat transfer = and fluid flow problems involve only one independent variable. Sometimes, t= wo independent variables can be expressed in terms of a single variable tha= t is termed as similarity variable, which considerably simplifies a problem= .

Partial differential equatio= ns are encountered very frequently in many problems for which numerical tec= hniques are available. The transient heat and fluid flow problems involve t= ime as an independent variable. Based on the geometry considered, one-, two= -, or three-dimensional solutions are used to express the temperature distr= ibution, pressure, and velocity in flow fields and energy transfer rates. T= he form of the analytical or numerical solution obtained may be classified = based on of the highest derivatives that appear in each variable.

A two-dimensional second-ord= er governing equation is more preferred for the heat transfer problems and = is represented in a general form as

where the coefficients A<= /em>, B, C, D, E, F, and G= may be functions of the two independent variables x and = y and the dependent variable =CF=95, where =CF=95 is= used to denote a generalized dependent variable; when thermal applications= are considered, the generalized variable will be replaced by specific vari= ables, such as temperature, density, velocity, or pressure. The governing <= a href=3D"https://learning.oreilly.com/library/view/heat-and-mass/978933251= 5680/xhtml/chapter010.xhtml#eq10.1">equation (10.1) is classified as el= liptic when B²=E2=80=934AC <= ; 0, parabolic when B²=E2=80=934AC<= /em> =3D 0, and hyperbolic when B²=E2= =80=934AC > 0. The solution procedure depends on these mathemat= ical classifications and also on boundary conditions.

10.4.1= Elliptic Governing Equation for Steady-state Conduction Heat Transfer Prob= lems

The conduction heat transfer = problem with steady-state two-dimensional case is governed by the Laplace o= r Poisson equations, which are elliptic, and needs specification of boundar= y conditions at all the boundaries. Iterative and direct methods are availa= ble for solving the algebraic equations that are generated from the numeric= al solutions. The Laplace and the Poisson equations, which generally repres= ent the steady-state problems, are written, respectively, as

The s= teady-state, constant-property, two-dimensional conduction temperature dist= ribution without internal heat generation satisfies the Laplace equation. T= he Poisson equation is used if an internal heat source is present

10.4.2= Parabolic Governing Equations for Transient Conduction Heat Transfer Probl= ems

The transient one-dimensional= heat transfer problem is governed by a parabolic equation that is solved n= umerically by marching in time, starting with an initial condition. The gov= erning equation of a transient two-dimensional conduction problem is parabo= lic in nature with respect to its time dependence and the elliptic behavior= with respect to the spatial coordinates. The temperature distribution is o= btained at each time interval in terms of the distribution at the preceding= interval by marching in time. The simplest parabolic equation for one-dime= nsional transient heat conduction is of the form

where t and x are the time and space coordinates, respectively, and A is the = thermal diffusivity. The boundary conditions at the two ends of a plate at = any instant and an initial condition for temperature are required to solve = this equation. Transient two-dimensional conduction is governed by an equat= ion of the form

where t denotes the= time variable and the internal heat source term G is also include= d. By comparing the highest derivatives in any two of the independent varia= bles, and with the help of Equation (1= 0.1), it is readily verified that = Equation (10.5) is parabolic in time and elliptic in space. An initial = condition and two boundary conditions in each spatial coordinate are requir= ed to solve this equation.

10.4.3= Hyperbolic Governing Equations for Convection Heat Transfer Problems

Hyperbolic equations are the = governing equations for convection dominated flows and they may be solved b= y marching in time or along certain characteristic directions. If more than= two independent variables are to be considered, the governing equations re= tain the characteristics of the three types of equations mentioned above, a= s determined by the highest derivatives in each of the independent variable= s.

Fluid flow problems usually = have a nonlinear term due to the inertia or acceleration component in the m= omentum equation. In addition, the energy equation has a corresponding term= called the convection term, which involves the flow field. For transient t= wo-dimensional fluid flow convection heat transfer problems, the governing = equations are of the form

where = =CF=95 denotes momentum or temperature, u and v = are the velocity components, and A is the diffusivity for momentum= or heat, and G is the pressure gradients in the momentum equation= or to volumetric heating in the energy equation. Equation (10.6) is parabolic in time and elliptic in space.= However, for high-speed flows, the terms on the left side dominate and the= equation becomes hyperbolic in time and space.

The term G in Equation (10.6) will introduce an integ= ral for solid angles of radiation heat transfer problems. An integral diffe= rential equation will result due to this term. Integral equations are also = used in radiation heat transfer problems with emitting and absorbing surfac= es. The integral equations may be converted numerically to algebraic equati= ons. The integral equations are many times replaced by average values of al= gebraic equations for the radiation heat transfer between surfaces. Radiati= on problems are nonlinear when coupled with convection or conduction proces= ses.

10.4.4= Integral Forms

The differential equations ei= ther apply locally or at every point of an appropriate time=E2=80=93space c= ontinuum. Many times global forms of the equations are required and they ma= y be obtained by integrating the differential equations over a suitable reg= ion of time and space. The resulting integral equations may differ in appea= rance from the generalized partial differential equation but the solutions = retain the characteristic nature of the governing partial differential equa= tion. For the momentum and energy equations, the differential and integral = forms represent local and global balances of momentum and heat, respectivel= y, and provide alternative starting points for a numerical solution.

The integral forms require s= pecification of a region of integration and a two-dimensional region is sho= wn in Fig. 10.1. The two-dimen= sional Poisson equation (10.3) can= be integrated over this region as

Figure 10.1 A Planar Region, S, with a Boundary Curve <= em>C

where dS =3D dxdy. T= his equation shows that the integral of the Poisson equation is zero. Vario= us moments of the Poisson equation are also zero and may be formed by addin= g appropriate weighting functions inside the integral and used to create fi= nite element methods.

= 10.5 NUMERICAL METHODS

Numerical models are based on= mathematical solutions that need computer to obtain the behavior of a syst= em for various operating conditions and design parameters. Both analytical = and numerical techniques are used for simple and com= plicated models, respectively. Numerical techniques are used for restructur= ing and discretization of governing algebraic equations, ordinary or partia= l differential equations, integral equations or a combination of these equa= tions suitable to the process, or system under consideration.

Solution to the numerical mo= dels involve selection of a finite element or finite difference method, dis= cretizing mathematical equations suitable for computation, choosing factors= such as grid size and time step and developing a code to obtain the result= s. Numerical models also need material properties and heat transfer coeffic= ients to obtain the solution.

For a thermal system, mathem= atical modeling is most preferred with analytical solution for best accurac= y and if it is difficult to get then a numerical solution is used. If numer= ical solution is also not possible, then a physical model can be used with = correlations derived from curve fitting the data.

10.5.1= Numerical Solutions =E2=80=93 Finite Difference Methods

The concepts of finite differ= ence methods and constructing the finite difference approximations, namely = direct approximation, Taylor series, and control volume approaches, are dis= cussed in this section. Several numerical concepts that are directly releva= nt to a numerical solution such as the truncation, discretization, and roun= d-off errors in a solution, consistency, convergence, and stability propert= ies of a solution method are also discussed.

10.5.1.1 Direct approximation me= thod

In this method partial deriva= tives are directly replaced by ratios of discrete differences. The method i= s based on calculation of finite discrete differences instead of infinitesi= mal differences. Consider the two-dimensional region sketched in Fig. 10.2.

Figure 10.2 Two-Dimensional Region

The region is covered by gri= d lines with at uniform intervals =CE=94x and =CE=94y and= indexed in two-dimensions by integers i and j. The inter= actions of the grid lines define a set of node points at which the dependen= t variable =CF=95 (x, y) is defined. Let =CF=95<= sub class=3D"EBsub">ij =3D =CF=95(x_i, y_j),a where x_i and y_j indi= cate the (x, y) values at node point (i, j). In = the case of three-dimensional, time-dependent problems, the dependent varia= ble =CF=95(x, y, z, t) is denoted by The indices i, j, k deno= te the spatial locations x, y, and z and the index l<= /em> denotes the time level. The time factor is indexed so that t<= sub class=3D"EBsub">l + 1 =3D t_l + =E2=88=86t, where =E2=88=86t is the t= ime step.
It is possible to approximat= e derivatives of =CF=95 with respect to x in terms of dis= crete differences and =E2=88=82=CF=95/=E2=88=82x= at (i, j) can be approximated by =E2=88=86=CF=95/=EF=80=A0=E2=88=86 x, where =E2=88=86's denote discrete differen= ces. Three approximations for (=E2=88=82=CF=95/=E2=88=82x)_{i, j} c= an be written in terms of differences of values at nodes i =E2=88= =92 l, i, and i + l:

Equations (10.8)=E2=80=93(10.10) are the two-point forward, two-point backward, and three-point= central difference approximations, respectively.

The second derivative =E2=88= =82²=CF=95/=E2=88=82x² at (i, j) can be approximated by

which is the three-point, cen= tral, second difference approximation. The direct approximation approach is= to substitute approximations of the forms of Equations (10.8)=E2=80=93= (10.11) into the appropriate governing partial differential equations. = The substitutions, however, introduce approximate equality signs in the res= ulting difference equations and it is necessary to use a Taylor series appr= oach to remove the approximation.

10.5.1.2 Taylor series approach<= /h4>
The Taylor series expansion i= s used when the function to be expanded is reasonably smooth. The =CF= =95 values at (i + 1, j) and (i =E2=88=92 1= , j) may be expressed in terms of Taylor series expansions about (= i, j) in Fig. 10= .3 as

Figure 10.3 The Control Volume Approach

The last term is the remaind= er that gives the error introduced by truncating the series after a finite = number of terms and the derivative of the remainder is evaluated at some po= int in the interval between i and i + 1, or i an= d i =E2=88=92 1. In finite difference applications, it is generall= y useful to retain the remainder in full derivative form. The approximation= s for the first and second derivatives at (i, j) are

The second term in Equations (10.13)=E2=80=93(10.16) has a finite difference approximation f= or the derivative including the plus a truncation error.

In Equations (10.13)=E2=80=93(10.16), the truncation errors approach zero because either the m= esh is refined (=E2=88=86x =E2=86=92 0) or the coefficient derivat= ives vanish and the coefficient derivatives in the error terms are also of = higher order than the exponents of =E2=88=86x. Equations (10.13)=E2=80=93(10.16) provide four possible approximations for the first = and second derivatives at the grid point (i, j). If three= grid points are in a row, then they represent all the derivative approxima= tions that are possible at the center point. Higher-order approximations re= quire additional grid points and for (=E2=88=82=CF=95/=E2=88=82x)_i,j they may be written as

where the subscript j is suppressed. The first two equations are correct to third-order, forwar= d and backward, four-point differences, and these equations become fourth-o= rder correct if used to evaluate =E2=88=82=CF=95/=E2=88=82x at i + (1/2) or i =E2=88=92 (1/2). Equation (10.19) is correct to fourth-order and is a= five-point central difference approximation for (=E2=88=82=CF=95)= =E2=88=82x)_i,j.

10.5.1.3 Control volume approach= and conservation

The control volume or finite = volume approach is based on the conservation of mass, momentum, or thermal = energy. This approach use numerical balances of a conserved variable over s= mall control volumes and always starts from an integral conservation statem= ent, which applies locally and globally. It will also satisfy the governing= differential equation in an average sense over the control volume and if p= roperly formulated, the numerical solution will also satisfy the conservati= on principle globally.

The first step in the contro= l volume approach is to state the governing conservation principle in integ= ral form and to specify the region of application. Then divide the physical= region as shown in Fig. 10.3 = into a set of nonoverlapping control volumes of size =E2=88=86x by= =E2=88=86y.

For the control volumes, the= integral conservation statement gives the following equations:

If th= e depth in the z direction is assumed as unity, then the integrals= are evaluated over the volume =E2=88=86x =E2=88=86y(1). = The factors i =C2=B1 (1/2) and j =C2=B1 (1/2) in Equation (10.20) indicate the locations = of the faces.

= 10.6 MONTE CARLO METHOD

The Monte Carlo method is a s= tatistical numerical technique that provide a systematic approach to comple= x thermal radiation problem, involving directional and spectral surface pro= perty variations. Radiant energy is assumed to be carried as discrete energ= y packets and the probability technique is applied to these packets to find= their wavelength, direction and their absorption, reflection and scatterin= g magnitudes.

Consider an enclosure with a= nonparticipating medium inside and whose emissivity, absorptivity, and ref= lectivity are functions of wavelength and direction. The radiative flux emi= tted by a surface i in the enclosure is

where =CE=B5_i is the total emissivity, T_i is the temperature, and A_i is= the area of the surface. So the radiative flux emitted by all the surfaces= is =CF=83_i Q_i A<= sub class=3D"EBsub">i with the total number of energy packets at= M. The energy associated with each packet is

Table 10.1 Boundary treatment and the corres= ponding equations

Each packet has the same ene= rgy content and number of packets for a surface depends on its temperature,= area, and =CE=B5_i. It is required to d= etermine the probability of the energy particle being emitted in a specific= direction and at a wavelength =CE=BB. The direction of emission i= s characterized by the cone angle =CE=B8 and the polar angle = =CF=95 for a spherical coordinate system with the center as the locati= on of emission. The probability function is then to be determined and norma= lized. For a diffusely emitting surface, the probability that the particle = leaves within an angle d=CF=95 is d=CF=95/2=CF=80. Hence, the cumulative probability that it leaves within the angle d= =CF=95 is d=CF=95/2=CF=80. Similarly, the cumulative= probability functions for =CE=B8 and A are to be determi= ned. The probability distribution for =CE=B8 is sin²=CE=B8. The direction (=CE=B8, =CF=95) of the emitted particle is determined by taking a pair of random number= s from a uniformly distributed set that subdivides the range from zero to u= nity. Different methods to get these random numbers are available on comput= ers. These numbers used to assign the probability for the direction of emis= sion. The values of =CE=B8 and =CF=95 are determined from= the probability distributions. Similarly, a third random number determines= the wavelength =CE=BB from the corresponding probability distribu= tion. From the geometry of the enclosure, the point where the particle stri= kes after emission is determined.

To kn= ow whether the particle is absorbed or reflected by the surface on which, i= t impinges, probability method is again used. A random number R is= assigned in the range zero to unity. If R is smaller than the abs= orptivity =CE=B1_=CE=BB, then the partic= le is assumed to be absorbed. Otherwise, it is considered to be reflected. = The absorptivity depends on the angle of impingement and the wavelength. If= the particle is absorbed, then the energy transferred to the surface is fo= und and the next energy packet at the emitting surface i is consid= ered. If R is greater than =CE=B1_=CE=BB, the packet is reflected and once again the probability functions= for the bidirectional reflectivity are involved in the determination of th= e direction. The path of each energy packet is followed until absorption oc= curs. A large number (M_i) of packets sa= tisfactorily represent the radiation emitted at the surface i. The= procedure is repeated to other surfaces of the enclosure. Then the number = of packets (M^=E2=80=B2) absorbed by a s= urface is obtained. The net heat lost by the surface i is

E_i =3D q_i A_i =3DQ_i A_i = =E2=88=92 M=E2=80=B2 =3D e(M_i<= /em> =E2=88=92 M=E2=80=B2).      &nb= sp;   (10.23)

The Monte Carlo method is su= ited for very complex and specialized problems and various special surfaces= and participating media can be treated by it. It can also be used to condu= ction and convection heat transfer problems. Digital computer with a high-c= omputational speed and a large storage capacity is needed to use this metho= d.

10.6.1= Numerical Solution Concepts

The methodology should have c= ertain properties. In most applications, it is not possible to analyze the = complete solution method. Instead a component of method is analyzed and it = should have certain properties.

10.6.1.1 Consistency

The discretization error depe= nds on the grid spacing. The discretization should become exact if the grid= spacing tends to zero. The difference between the discretized equation and= the exact one is called the truncation error. For the method to be consist= ent, the truncation error must become zero.

Some dicretization methods l= ead to truncation errors, which are functions of the mesh spacing. In such = cases, the consistency requirement is only conditionally fulfilled. Even if= the approximations are consistent, it does not necessarily mean that the s= olution of the discretized equation system will become the exact solution o= f the differential equation in the limit of small step size. To satisfy thi= s condition, the solution method has to be stable.

10.6.1.2 Stability

The stable solution should no= t magnify the error that appears in the course of numerical solution proces= s. It is difficult to estimate the stability condition specifically when bo= undary conditions and nonlinearities are present. Due to this reason, the i= nvestigation of stability is done for linear problems with constant coeffic= ients without boundary conditions. The commonly used approach for studying = the stability is the Von-Neumann's method. Many solution schemes require th= at the time step be smaller than a certain limit or that under relaxation b= e used.

10.6.1.3 Convergence

The convergence of the numeri= cal solution is defined as the solution of the discretized equations that t= ends to the exact solution of the differential equation as the grid spacing= tends to zero. A consistent scheme is useless unless the solution method c= onverges. Convergence is usually checked using numerical experiment by repe= ating the calculation on a series of successively refined grids. The soluti= on is said to be grid-independent when the refinement in the grid size does= not affect the solution.

= 10.7 CONCLUSIONS

Modeling a physical system an= d analyzing the results of the processes with suitable correlations are kno= wn as simulation and it is equivalent to actual experiments on the system. = Simulation of any physical process can provide relevant inputs for the desi= gn and optimization of a process or system. Once a model of a system is cre= ated, it can be subjected to a large number of operating conditions and the= outputs will characterize the behavior of the system.

Models can be physically mad= e to show the internal assembly of components and also to explain the worki= ng principles of systems such as heat engines, pumps, and other machine too= ls. Using the governing equations, which represent the processes in a predi= ctive model, the variation of parameters such as temperature, pressure, and= velocity with respect to time can be obtained.

Mathematical models represen= t the behavior and performance of a system with mathematical equations. The= rmal systems can be designed with the help of the results of the mathematic= al models without actually fabricating a prototype. Heat transfer and fluid= flow processes are governed by the laws of conservation of mass, momentum,= and energy with differential and integral equations.

A two-dimensional second-ord= er governing equation is more preferred for the heat transfer problems. The= conduction heat transfer problem with steady-state two-dimensional case is= governed by the Laplace or Poisson equations, which are elliptic and needs= specification of boundary conditions at all the boundaries. The transient = one-dimensional heat transfer problem is governed by a parabolic equation a= nd that is solved numerically by marching in time, starting with an initial= condition. Hyperbolic equations are the governing equations for convection= -dominated flows and they may be solved by marching in time or along certai= n characteristic directions.

Numerical techniques are use= d for restructuring and descretization of governing algebraic equations, or= dinary or partial differential equations, integral equations, or a combinat= ion of these equations suitable to the process or system under consideratio= n.

The Monte Carlo method is a = statistical numerical technique that provide a systematic approach to compl= ex thermal radiation problem, involving directional and spectral surface pr= operty variations.

The convergence of the numer= ical solution is defined as the solution of the discretized equations that = tends to the exact solution of the differential equation as the grid spacin= g tends to zero.

= EXAMPLE PROBLEMS

10.1. A long bar o= f square cross-section at 50=C2=B0C is exposed to air at 25=C2=B0C or water= , at a different temperature. To find the time-dependent temperatu= res at various points in the interior conduction region of the bar analytic= ally, state the two-dimensional governing equation and boundary co= nditions, assuming material properties to be constant.

The governing equation i= s

where T is the = temperature at any point in the material, =CE=B1 is the thermal diffusivity= and t is the time.

If convective heat trans= fer with a uniform heat transfer coefficient h is assumed at the s= urfaces, the initial and boundary conditions for this equation are:

At t =3D 0:T =3D Tf

At at the surfaces

w= here is the thermal conductivity of the m= aterial, n is the normal at the surface taken to be positive outwa= rd, T_f is the fluid temperature and T_t is the uniform initial temperature in t= he body.

10.2. Discuss the = general numerical approaches to solve the governing equation of Problem 1.<= /em>

Finite difference method
The governing equation will= give the temperatures at a finite number of points in the conduction regio= n of the square rod. The spatial derivatives at a point are replaced by exp= ressions written in terms of the temperatures at the neighbouring locations= and the distances between them. This way, algebraic equations are obtained= for each location. These simultaneous equations are solved to get the temp= eratures at different points. Every grid point represent a finite region in= its neighbourhood. Points on the surfaces are considered with the boundary= conditions and special equations are obtained and solved numerically to ob= tain the time-dependent temperature distribution, starting with the given i= nitial temperature distribution.

Finite element method
In the finite element metho= d, the conduction region of the rod is divided into a number of finite elem= ents of different shapes. The triangular building blocks for two-dimensiona= l problems and the tetra-hedral blocks for three-dimensional ones can be us= ed and for these elements, the variation within the elements is considered = linear. Integral equations are obtained for each element and minimization o= f the integrals is to be carried out to satisfy the conservation principles= for obtaining the distribution of the variables in the region.

10.3. Discuss the = method of finding the boundary condition in terms of heat transfer coeffici= ent from basic concepts.

The heat transfer coeffi= cient h is obtained from the available correlations from a solutio= n of the coupled fluid flow problem, which is governed by the basic princip= les of conservation of mass, momentum and energy. The solution of the energ= y equation linked with the flow field yields the temperature field in the f= luid. The heat transfer at the surfaces is given by

where q is the = heat lost by the material at the surface, (=E2=88=82t/=E2=88=82n)₀ is the temperature gradient in the flu= id evaluated at the surface and K is the thermal conductivity of the fluid.= The heat transfer coefficient h is evaluated from

q =3D h(Ts =E2=88= =92 T_f )

where T_s is the surface temperature. Heat transfer coefficient = is a function of temperature and location on the surface. The convection pr= oblem is to be solved first to determine h, which is then used as = a boundary condition for the conduction problem. Numerical methods are gene= rally used to obtain the solution for the convection problem.

In addition to the condu= ction and convection heat transfer processes, actual problems involve radia= tion heat transfer also. If radiation is included in the heat transfer proc= ess, the boundary condition is written as

where =CE=B5 is= the surface emissivity of the body, =CE=B5 is the Stefan=E2=80=93= Boltzman constant, T_s is the surface te= mperature and T_f is the fluid temperatu= re. Numerical methods are needed to solve nonlinear problems such as that g= iven above.

10.4. Consider the two-dimensional steady heat flow problem of a recta= ngular sheet, perfectly insulated on both its surfaces, with size = 1 m =C3=97 2 m and Dirichlet boundary conditions 0=C2=B0C, 0=C2=B0= C, 0=C2=B0C and 100=C2=B0C on its four edges. The sheet is discretized by t= he size x =3D y =3D 0.5 m. Find the steady-state temperatures at the three = interior grid points (1, 1),(1, 2) and (1, 3) as shown in the figure below.=

The governing equation f= or problem is

The governing equation i= s discretized as

Applying the equation to= the unknown points,

For P_=
11: 0 + 0 + 0 + T₁₂ =E2=88=924T₁₁ =3D 0

    For P₁₂:T₁₃ + = 0 + T₁₁ + 0 =E2=88=924T₁₁ =3D 0

    For P₁₃:100 + 0 + T_1=
2 + 0 =E2=88=924T₁₃ =3D 0

Solving the matrix,

By back-substitution,

=E2=88=924 T₁₁ + T₁₂ =3D 0 : T₁₁ =3D 1.78=C2=B0C

Therefore, T₁₁ =3D 1.78=C2=B0C, T₁₂ = =3D 7.14=C2=B0C, T₁₃ =3D 26.78=C2=B0C

10.5. Consider a s= lab of thickness 0.12 m with a thermal conductivity K =3D 30 W/mK in which = energy is generated at a constant rate of 10⁶ W/= m³. The boundary surface at x =3D 0 is = insulated and the one at x =3D 0.12 m is exposed to the temperature of 150= =C2=B0C as shown in the figure given below. Find the temperature distributi= on inside the slab, assuming one-dimensional heat variation with e= qual spacing.

F= or the internal nodes,

For Node 2:T₃ =E2=88=92 2T₂ + T₁ + 30 =3D 0

For Node 3:T₄ =E2=88=92 2T₃ + T₂ + 30 =3D 0

For Node 4:T₅ =E2=88=92 2T₄ + T₃ + 30 =3D 0

For Node 5:T₅ =3D 150

For the insulated bounda= ry,

For Node 1:2T2=E2=88=92 2T₁= + 30 =3D 0

Solving the matrix,

R3 =E2=86=92 R3 + R2

R4 =E2=86=92 R4 + R3

Therefore, T1 =3D 390=C2=B0C, T_{2<=
/sub> =3D 375=C2=B0C, T₃ =3D 330=C2=B0C=
, T₄ =3D 255=C2=B0C.}

10.6. Consider a s= quare plate as shown in the figure given below. With the left and right fac= es maintained at 100=C2=B0C and top and bottom faces maintained at 500=C2= =B0C Thermal conductivity of the plate is 10W/m K. Find the steady-state te= mperature distribution at the interior nodes.

T= he governing equation is

The discretized equation= is

=E2=88=86x =3D = =E2=88=86y =3D Constant =3D 0.5

T_{i +}_{l, j} + T_{i =E2=88=92 l, j} + T_{i, j +l} + T_{i<=
/em>, j =E2=88=92 l} =E2=88=92 4T_ij<=
/sub> =3D 0

For Node 1:100 + T₂ + T₃ = =E2=88=92 4T₁ + 500 =3D 0

    =             &nb= sp;   T₂ + T3 =E2=88=92 4T₁= =3D =E2=88=92600

For Node 2:100 =E2=88=92= 4T₂ + T₄₌ + T₁ + 500 =3D 0

    =             &nb= sp;   =E2=88=924T₂ + T₄ + T_{1 =3D =E2=88=92600}

For Node3:100 =E2=88=92 = 4T₃ + T_{4<=
/sub> + T₁ + 500 =3D 0}

    =             &nb= sp;   =E2=88=924T₃ + T₄ + T_{1 =3D =E2=88=92600}

For Node 4:100 + T₂ + T₃ = =E2=88=92 4T₄ + 500 =3D 0

    =             &nb= sp;   T₂ + T3 =E2=88=92 4T₄= =3D =E2=88=92600

By solving the matrix

T_{1 =3D 301.67=C2=B0C}

T_{2 =3D 300.83=C2=B0C}

T_{3 =3D 305.85=C2=B0C}

T_{4 =3D 301.67=C2=B0C}

10.7. A block made= up of cast iron (100 cm=C3=975 cm=C3=972 cm) as shown in= figure is taken out from the mould and cooled in air. The casting taken ou= t is initially at 200=C2=B0C. Find out the time required to bring the casti= ng to room temperature.

The properties of cast i= ron are

   = =3D 7272 kg/m³

=3D 52 W/mK

  C_p =3D 420 kJ/kgK

   =3D 1.702=C3=9710^=E2=88=925

We can make the followin= g assumptions to solve this problem:

The system is not in a steady-sta= te.

The heat flow is one-dimensional. (Sin= ce the other dimensions are small compared with the length.)

The material pr= operties remain constant.

Convection heat transfer is only a= t the edge nodes.

The governing equation f= or conduction is

The discretized equation= is

The governing equation f= or convection is

According to Courent's c= riteria,

assume

Solving equations

10.8. A large indu= strial furnace, as shown in the figure, is supported on a= long column of fireclay brick, which is 1 m=C3=971 m on = a side. During steady-state operation, three surfaces of the colum= n are maintained at 500 K, while the bottom surface is insulated. = Using the grid size =E2=88=86x =3D =E2=88=86=EF=80=A0y = =3D 0.25 m, determine the two-dimensional temperature distribution= in the column.

We make the following as= sumptions:

It is a two-dimensional, steady-s= tate problem.

The properties remain constant.

There is no heat generation.

The geometry and boundary equation= s are symmetric.

The governing equation i= s

T_{i +}_{l, j} + T_{i =E2=88=92 l, j} + T_{i, j +l} + T_{i<=
/em>, j =E2=88=92 l} =E2=88=92 4T_ij<=
/sub> =3D 0

By solving the matrix,

  T₁ =3D 355.99 K     T₅ =3D 438.44 K

      = ;T₂ =3D 23.68 K     = ;    T₆ =3D 402.58 = K

   T₃ =3D 400.28 K     T₇ =3D 450.91 K

   T₄ =3D 306.71 K     T₈ =3D 426.74 K

10.9. Using AnSYS<= /em>, plot the temperature distribution in the material packed in betwe= en at a cross-section of a composite cylindrical arrangement as shown in th= e figure. Two stainless steel cylinders of diameters 0.02 and 0.06 m are pl= aced co-axially, with asbestos between them. The inner surface tem= perature of the inner cylinder is 600=C2=B0C and the outer surface temperat= ure of the outer cylinder is 100=C2=B0C. Thermal conductivity of stainless = steel and asbestos are 0.19 and 0.02W/m K, respectively.

Preprocessing: Define the problem:

The element type is Plane 5= 5, four noded quadrilateral, two-dimensional thermal element Preprocessor= =E2=80=94Element type=E2=80=94Add/Edit/Delete=E2=80=94Add/Thermal mass/soli= d quad 4 node 55 Material is specified and thermal conductivity values also= specified create 2D concentric circular area

For the accuracy of results, = select size control in meshing, mesh tool option also select the Mapped 3 o= r 4-sided mesh density controls

Preprocessor=E2=80=94Meshing= =E2=80=94Mesh tool

Solution Phase: Assign loads and solve:

Apply the boundary condition= s

Temperature 600=C2=B0C is maintained at all points and materia= l is considered as a Stainless Steel (Inner surface).

Temperature 100=C2=B0C is maintained at all points and material is = considered as a stainless steel (Outer surface).

Solve the problem Solve=E2=80=94Current LS

Post Processor

Plot the temperature distribution Solution=E2=80=94General
=

post processor=E2=80=94Read Results=E2=80=94First set

Plot Results=E2=80=94Contour plot=E2=80=94Nodal Solution

List the temperatures at each of these nodes. Temperature distr= ibution is shown in the figure given below.

10.10. Determine t= he temperature distribution of a rectangular aluminium rod of 75 mm length = and 3 mm width. The thermal conduction of the material is 200 W/mK and the = temperature at one smaller side is 100=C2=B0C and the remaining three sides= are at 50=C2=B0C. Find the temperature in the midplane of the rod. (Softwa= re used AnSYS.)

The physical model and b= oundary conditions are shown in the figure below.

Preprocessing: Define the problem

The Element type is Plane 55, four noded quadrilateral, 2-D th= ermal element Preprocessor=E2=80=94Element type=E2=80=94Add/Edit/Delete=E2= =80=94Add/Thermal mass/Solid/Quad 4 node 55

Material is specified as a Aluminium with thermal conductivity of 2= 00 W/mK

Create 2d rectangular area

For the accuracy of results select size control in meshing, mes= h tool option also select the Mapped 3- or 4-sided mesh density controls Pr= eprocessor=E2=80=94Meshing=E2=80=94Mesh tool solution Phase:Assign loads an= d solve

Solution Phase: Assign loads and solve:

Apply the boundary conditions

One end is fixed and temperature value is specified as 100=C2=B0C R= emaining 3 ends the heat flux value is specified as 50=C2=B0C

Solve the problem Solve=E2=80=94Current LS

Post Processor

Plot the temperature distribution. Solution=E2=80=94General po= st processor=E2=80=94Read Results=E2=80=94First set Plot Results=E2=80=94Co= ntour plot=E2=80=94Nodal Solution

List the temperatures at each of these nodes. Temperature distr= ibution is shown in the figure below.

Temperature distribution

10.11. Water flows= through a pipe of diameter 2.54 cm and length 3 m. The entry temperature a= nd velocity of water are 60=C2=B0C and 2 cm/s, respectively. If th= e inner wall surface temperature is maintained at 80=C2=B0C, find = the outlet temperature of water. Use any CFD software.

The physical model and b= oundary conditions are shown in the figure given below.

Methodology<= /p>

Using the coordinate point, 2D mo= del is created in Gambit

The Quad/ Map mesh is created with the= spacing of 0.025 Mesh=E2=80=94Face=E2=80=94Quad/Map=E2=80=94Select spacing= =E2=80=94Apply

Total no of Nodes are 4207 and Element= s are 7202

The Inlet and Outlet boundary conditio= ns are defined Select=E2=80=94solver=E2=80=94Fluent 5/6 Zones=E2=80=94bound= ary type=E2=80=94Inlet/Outlet

Water is defined as fluid domain

The conditions are applied in Fluent s= olver

The results are visualized using p= ost-processor as shown in the figures below Grid=E2=80=94Display=E2=80=94 C= ontour

Result: Outle= t temperature is 347 K

= EXERCISE PROBLEMS

Derive an expression in finite-di= fference form to find the temperature of the corner grid point of a two-dim= ensional solid under unsteady-state conditions.

Consider the one-dimensional fin with = a base temperature T_b surrounded by a f= luid temperature of T_=E2=88=9E with a h= eat transfer coefficient h. Divide the fin into four segments and = formulate the problem in terms of finite differences to calculate the tempe= ratures at the four points under steady-state conditions.

If the temperature distribution, in a = cube as T =3D axyz + bx² + cz and K_x =3D K_y =3D K, K_z =3D 2k assuming no heat is generated, find =E2=88=82T/=E2= =88=82t.

The internal heat generation within a = plain slab is given by q =3D ce^{=E2=
=88=92ax} where c and a are constants. The left side= of the slab is exposed to convection with a fluid of temperature T₁ and heat transfer coefficient is h. If the ot= her side is insulated, how will you formulate the problem mathematically?

For steady-state heat conduction throu= gh a very long steel tube, formulate the boundary conditions mathematically= . (Ans. The boundary conditions are r =3D r_i, T =3D T_i. and = r =3D r₀, T =3D T₀)

A metal slab with temperature Ti is suddenly immersed in a bath of coolant ma= intained at temperature T_f. The convect= ive heat transfer coefficient is hW/m²K= . Formulate the boundary conditions mathematically. (Ans. = Initial conditions: T =3D T_i at t =3D 0 Boundary condition:

Heat is generated within a hollow sphe= re. The inner surface is insulated and outer surface is exposed to convecti= on to a fluid at temperature T_f. The he= at transfer coefficient =E2=80=98h=E2=80=99. Formulate the problem= mathematically.

In steady-state one-dimensional heat c= onduction through a medium, what are the two cases under which the temperat= ure distribution will not be linear?

q=E2=80=B2=E2=80=B2=E2= =80=B2 =E2=89=A0 0

When q=E2=80=B2=E2=80=B2= =E2=80=B2 =3D 0, but K is a function of x, T (no= n-uniform thermal conductivity).

A metallic plate, 29.9 mm thick is mai= ntained at 670 K on one side and 370 K on the other. Find the heat transfer= through the plate? Assume thermal conductivity K =3D 369.9 W/mK. = Find the temperature distribution in the plate by numerical and analytical = methods and compare. (Ans. 3.7 MW/m^2<=
/sup>)

A hot plate of = length 0.749 m, width 0.499 m and thickness 1.99 cm is placed in air stream= at 292 K. A total of 300 W is lost from the plate surface by radiation. Ta= king the convective heat transfer coefficient as 25 W/m²K and thermal conductivity as 43 W/mK. Calculate the inside tempera= ture of the plate.

10.11 The thickness of a wall is 0.59 = m and it is initially at an uniform temperature of 310 K. Its surface tempe= ratures are suddenly raised to 420 K and 620 K and kept constant at these v= alues for 14 hr 50 min. Estimate the temperature at a depth of 0.149 m from= the surface maintained at 420 K. The thermal diffusivity of the material o= f the wall can be assumed to be 0.519 =C3=97 10^=E2=88=
=926 m²/s (Ans. 400 K)
10.12 A fin is made of brass (K =3D 110 W/m²K) of length 2 cm with a 0.15 cm = =C3=97 35 cm cross-section. The base temperature of the fin is 1008=C2=B0C.= The surrounding fluid temperature is 308 =C2=B0C, and the heat transfer co= efficient is 14 W/m²K. Find the temperature dist= ribution along the fin and the rate of heat transfer to the surrounding flu= id using a finite difference method and prove the same with the conventiona= l method.

10.13 A wooden plate is initially at 3= 08=C2=B0C. It is suddenly exposed to air at 08=C2=B0C. The plate is 15 cm i= n thickness and the other dimensions are very large. The heat transfer coef= ficient on both sides of the plate are 15 W/m²K = and the properties of the plate material are K =3D 0.11 W/mK, = =CF=81 =3D 418 kg/m³ and c =3D 272= 5 J/kgK. After two hours of heating, find the temperatures of the surfaces = and the midplane of the plate

by dividing the thickness of the = plate into three divisions and applying a finite-difference method, and
prove the result by the convention= al analytical heat transfer method.

A hot metal plate is used to heat air = inside a room. The room air temperature is at 26=C2=B0C. The temperature of= the metal plate is maintained at various temperatures between 80=C2=B0C an= d 800=C2=B0C. Emissivity of the metal surface is 0.295 and the plate does n= ot receive heat from any other source. Write a computer program and plot th= e heat supplied by the plate to room air per square metre of area by convec= tion and radiation heat transfer modes. At what temperature they are equal = in magnitude.
Hint: Fluid Grashof number wi= th

L =3D 1 m

Q_conv= =3D hAT

Q_{rad<=
/sub> =3D =E2=88=91AT⁴.}

(Ans. 380=C2= =B0C)

A heat transfer analysis for temperatu= re distribution in a plane wall provides temperature at several points at a= particular time as given below:

Use a statistical method and = fit a straight line for this data.

A rectangular metal plate of length 90= cm and height 10 cm is maintained under steady-state conditions with heat = flowing from one end at 800=C2=B0C to the other end at 200=C2=B0C. Find the= temperature of the plate at 1/3 and 2/3 of its length using one-dimensiona= l analysis. (Ans. 600=C2=B0C, 400=C2=B0C)

A square plate of size 30 =C3=97 30 cm= is divided into nine equal segments. The left and top sides of the plate a= re maintained at 500=C2=B0C and the bottom and right sides are maintained a= t 100=C2=B0C. Using two-dimensional, steady-state analysis, find the temper= atures at the nodal points.

Ans.  &n= bsp;

Model a composite wall of a house and = find out the insulation thickness required to reduce the heat loss by about= three-fourth of that without insulation. Assume the following factors to s= olve the problem.
Wall thickness: the brick lay= er is 100 mm thick and the cement layer is 40 mm thick. Thermal conductivit= ies of brick and cement layers are 0.7 and 0.5 W/m=C2=B0C, respectively and= that of the insulation material is 0.07 W/m=C2=B0C.

(Hint:= Use any software : AnSYS, CFX, CFD, FLUENT. Create a square/ rectangle mod= el with wall thickness as specified and with an assumed height (as minimum = as possible to reduce the area of analysis). Specify the properties of the = materials (thermal conductivity) and boundary conditions. As the temperatur= es are not specified we can assume suitably (Eg: 45=C2=B0C and 25=C2=B0C as= the outer and inner tem peratures). Heat transfer analysis is to be car ri= ed out using 1D steady-state conduction heat transfer method.)(Ans.= 60 mm)

Model the case of a cylindrical fin ex= tended from a base with a temperature of 600 K. The length and diameter of = the fin are 100 mm and 20 mm, respectively. The fin is exposed to an enviro= nment with a convection coefficient of 30 W/m²K.= Find the temperatures at the axis of the fin at 5 cm and 10 cm for stainle= ss steel (K =3D 17 W/mK) and copper K =3D 385 W/mK). Show that at both loca= tions the temperature of copper fin is higher than that of the stainless st= eel. (Hint: Use any software : AnSYS, CFX, CFD, FLUENT=20
Create the model of the fin a= s rectangular (2D) with length 200 mm and height 20 mm. Specify the boundar= y conditions as follows: temperature at one 20 mm side is 600 K. The other = three sides are to be provided with the convection heat transfer coefficien= t of 30W/m²=C2=B0C

Heat transfer analysis is to = be carried out using the software as a two-dimensional steady-state model o= nce with material property K =3D 17 W/mK for steel and again with = K =3D 385 W/ mK for copper with all other conditions remaining sam= e. Get the temperature distribution and read the values of the temperatures= at the specified locations.)

A metal rod of length 99 mm and di= ameter 49 mm is initially at a temperature of 475 K. The thermal conductivi= ty of the rod may be assumed to be 215 W/mK. The rod is suddenly subjected = to an environment of 345 K with a convection heat transfer coefficient of 5= 30 W/m²K. Find the temperature of the rod at 6mm= from one end and at half the radius from the edge after 60 s.
(Hint: create a two-dimension= al model as a rectangle with length 99 mm and side 49 mm. Specify the bound= ary conditions with convection heat transfer coefficient of 530 W/m²K all around the boundary. Specify the initial temperat= ure of the rod as 475 K uniformly throughout the material. Also specify the= thermal conductivity of the material as 215 W/mK. Heat transfer analysis i= s to be carried out as two-dimensional transient heat transfer model. Run t= he analysis for 60 s and obtain the temperature distribution. Find the temp= erature at the required location.)

Highlight

?

Delete Note

Save Note

Cancel

Highlight
Close

Are you sure you want to permanently delete this note?

No, I changed my mind Yes, delete it

=09
Copy
=09
Add Highlight<= /a>

Add Note
=09

=20 =20 =20 =20 Previous Chapter
9. Radiation Heat Tr= ansfer
=20 =20

=20 =20 =20 =20 Next Chapter
11. Experimental Hea= t Transfer
=20 =20
=20

Table= of Contents for =20 Heat and Mass Transfer, 2nd Edition =20

10

Modeling and Analysis

= 10.1 INTRODUCTION

= 10.2 BASICS OF MODELING

= 10.3 PREDICTIVE MODELS

10.3.1= Physical Models

10.3.2= Analog Models

10.3.3= Mathematical Models

= 10.4 GOVERNING EQUATIONS

10.4.1= Elliptic Governing Equation for Steady-state Conduction Heat Transfer Prob= lems

10.4.2= Parabolic Governing Equations for Transient Conduction Heat Transfer Probl= ems

10.4.3= Hyperbolic Governing Equations for Convection Heat Transfer Problems

10.4.4= Integral Forms

= 10.5 NUMERICAL METHODS

10.5.1= Numerical Solutions =E2=80=93 Finite Difference Methods

10.5.1.1 Direct approximation me= thod

10.5.1.3 Control volume approach= and conservation

= 10.6 MONTE CARLO METHOD

10.6.1= Numerical Solution Concepts

10.6.1.1 Consistency

10.6.1.2 Stability

10.6.1.3 Convergence

= 10.7 CONCLUSIONS

= EXAMPLE PROBLEMS

= EXERCISE PROBLEMS

Highlight

Highlight

Table= of Contents for
=20 Heat and Mass Transfer, 2nd Edition =20