A HEAT TRANSFER TEXTBOOK - THIRD EDITION Episode 1 Part 7 pot

Fourier, 1822 4.1 The well-posed problem The heat diffusion equation was derived in Section 2.1and some atten-tion was given to its soluatten-tion.. A well-posed heat conduction problem i

Analysis of Heat Conduction

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some steady one-dimensional

The Analytical Theory of Heat, J Fourier, 1822

4.1 The well-posed problem

The heat diffusion equation was derived in Section 2.1and some

atten-tion was given to its soluatten-tion Before we go further with heat conducatten-tion

problems, we must describe how to state such problems so they can

re-ally be solved This is particularly important in approaching the more

complicated problems of transient and multidimensional heat

conduc-tion that we have avoided up to now

A well-posed heat conduction problem is one in which all the relevant

information needed to obtain a unique solution is stated A well-posed

and hence solvable heat conduction problem will always read as follows:

Find T (x, y, z, t) such that:

some region, R, which might extend to infinity.1

2 T = Ti (x, y, z) at t = 0

This is called an initial condition, or i.c.

(a) Condition 1 above is not imposed at t = 0.

(b) Only one i.c is required However,(c) The i.c is not needed:

i In the steady-state case: ∇ · (k∇T ) + ˙ q = 0.

ii For “periodic” heat transfer, where ˙q or the boundary

con-ditions vary periodically with time, and where we ignorethe starting transient behavior

3 T must also satisfy two boundary conditions, or b.c.’s, for each

co-ordinate The b.c.’s are very often of three common types

(a) Dirichlet conditions, or b.c.’s of the first kind:

T is specified on the boundary of R for t > 0 We saw such

b.c.’s in Examples2.1,2.2, and2.5

(b) Neumann conditions, or b.c.’s of the second kind:

The derivative of T normal to the boundary is specified on the boundary of R for t > 0 Such a condition arises when the heat flux, k(∂T /∂x), is specified on a boundary or when , with the help of insulation, we set ∂T /∂x equal to zero.2

(c) b.c.’s of the third kind:

A derivative of T in a direction normal to a boundary is

propor-tional to the temperature on that boundary Such a conditionmost commonly arises when convection occurs at a boundary,and it is typically expressed as

−k ∂T

∂x



bndry= h(T − T∞ )bndry

when the body lies to the left of the boundary on the

x-coor-dinate We have already used such a b.c in Step 4 of Example2.6, and we have discussed it in Section1.3as well

1(x, y, z) might be any coordinates describing a position
r : T (x, y, z, t) = T (
r , t).

2Although we write ∂T /∂x here, we understand that this might be ∂T /∂z, ∂T /∂r ,

or any other derivative in a direction locally normal to the surface on which the b.c is specified.

Figure 4.1 The transient cooling of a body as it might occur,

subject to boundary conditions of the first, second, and third

kinds

This list of b.c.’s is not complete, by any means, but it includes a great

number of important cases

Figure4.1shows the transient cooling of body from a constant initial

temperature, subject to each of the three b.c.’s described above Notice

that the initial temperature distribution is not subject to the boundary

condition, as pointed out previously under 2(a)

The eight-point procedure that was outlined in Section2.2for solving

the heat diffusion equation was contrived in part to assure that a problem

will meet the preceding requirements and will be well posed

4.2 The general solution

Once the heat conduction problem has been posed properly, the first step

in solving it is to find the general solution of the heat diffusion equation

We have remarked that this is usually the easiest part of the problem

We next consider some examples of general solutions

One-dimensional steady heat conduction

Problem4.1emphasizes the simplicity of finding the general solutions oflinear ordinary differential equations, by asking for a table of all generalsolutions of one-dimensional heat conduction problems We shall workout some of those results to show what is involved We begin the heat

diffusion equation with constant k and ˙ q:

∇2T + q˙

k = 1α

Cylindrical coordinates with a heat source: Tangential conduction.

This time, we look at the heat flow that results in a ring when two pointsare held at different temperatures We now express eqn (2.11) in cylin-drical coordinates with the help of eqn (2.13):

Figure 4.2 One-dimensional heat conduction in a ring.

This result is consistent with the lumped-capacity solution described in

Section1.3 If the Biot number is low and internal resistance is

unimpor-tant, the convective removal of heat from the boundary of a body can be

prorated over the volume of the body and interpreted as

The general solution in this situation was given in eqn (1.21) [A

partic-ular solution was also written in eqn (1.22).]

Separation of variables: A general solution of multidimensional problems

Suppose that the physical situation permits us to throw out all but one ofthe spatial derivatives in a heat diffusion equation Suppose, for example,that we wish to predict the transient cooling in a slab as a function ofthe location within it If there is no heat generation, the heat diffusionequation is

A common trick is to ask: “Can we find a solution in the form of a product

of functions of t and x: T = T (t) · X(x)?” To find the answer, we

substitute this in eqn (4.5) and get

X  T = 1

where each prime denotes one differentiation of a function with respect

to its argument Thus T  = dT/dt and X  = d2X/dx2 Rearrangingeqn (4.6), we get

This is an interesting result in that the left-hand side depends only

upon x and the right-hand side depends only upon t Thus, we set both

sides equal to the same constant, which we call−λ2, instead of, say, λ,

for reasons that will be clear in a moment:

X(x) = A sin λx + B cos λx for λ ≠ 0

T (t) = Ce −αλ2t for λ≠ 0

where we use capital letters to denote constants of integration [In

ei-ther case, these solutions can be verified by substituting them back into

eqn (4.8).] Thus the general solution of eqn (4.5) can indeed be written

in the form of a product, and that product is

T = XT = e −αλ2t (D sin λx + E cos λx) for λ ≠ 0

The usefulness of this result depends on whether or not it can be fit

to the b.c.’s and the i.c In this case, we made the functionX(t) take the

form of sines and cosines (instead of exponential functions) by placing

a minus sign in front of λ2 The sines and cosines make it possible to fit

the b.c.’s using Fourier series methods These general methods are not

developed in this book; however, a complete Fourier series solution is

presented for one problem in Section5.3

The preceding simple methods for obtaining general solutions of

lin-ear partial d.e.’s is called the method of separation of variables It can be

applied to all kinds of linear d.e.’s Consider, for example,

two-dimen-sional steady heat conduction without heat sources:

The general solution is

T = (E sin λx + F cos λx)(e −λy + Ge λy ) for λ≠ 0

Figure 4.3 A two-dimensional slab maintained at a constant

temperature on the sides and subjected to a sinusoidal tion of temperature on one face

varia-Example 4.1

A long slab is cooled to 0◦C on both sides and a blowtorch is turned

on the top edge, giving an approximately sinusoidal temperature tribution along the top, as shown in Fig 4.3 Find the temperaturedistribution within the slab

dis-Solution. The general solution is given by eqn (4.13) We musttherefore identify the appropriate b.c.’s and then fit the general solu-tion to it Those b.c.’s are:

on the top surface : T (x, 0) = A sin π x

The only way that this can be true for all x is if G = 0 Substitute

eqn (4.13), with G = 0, into the second b.c.:

(O + F)e −λy = 0

so F also equals 0 Substitute eqn (4.13) with G = F = 0, into the first

b.c.:

E(sin λx) = A sin π x

L

It follows that A = E and λ = π/L Then eqn (4.13) becomes the

particular solution that satisfies the b.c.’s:

Thus, the sinusoidal variation of temperature at the top of the slab is

attenuated exponentially at lower positions in the slab At a position

of y = 2L below the top, T will be 0.0019 A sin πx/L The

tempera-ture distribution in the x-direction will still be sinusoidal, but it will

have less than 1/500 of the amplitude at y = 0.

Consider some important features of this and other solutions:

• The b.c at y = 0 is a special one that works very well with this

particular general solution If we had tried to fit the equation to

a general temperature distribution, T (x, y = 0) = fn(x), it would

not have been obvious how to proceed Actually, this is the kind

of problem that Fourier solved with the help of his Fourier series

method We discuss this matter in more detail in Chapter5

• Not all forms of general solutions lend themselves to a particular

set of boundary and/or initial conditions In this example, we made

the process look simple, but more often than not, it is in fitting a

general solution to a set of boundary conditions that we get stuck.

• Normally, on formulating a problem, we must approximate real

be-havior in stating the b.c.’s It is advisable to consider what kind of

assumption will put the b.c.’s in a form compatible with the

gen-eral solution The temperature distribution imposed on the slab

by the blowtorch in Example 4.1might just as well have been

ap-proximated as a parabola But as small as the difference between a

parabola and a sine function might be, the latter b.c was far easier

to accommodate

• The twin issues of existence and uniqueness of solutions require

a comment here: It has been established that solutions to all

well-posed heat diffusion problems are unique Furthermore, we know

from our experience that if we describe a physical process correctly,

a unique outcome exists Therefore, we are normally safe to leavethese issues to a mathematician—at least in the sort of problems

we discuss here

• Given that a unique solution exists, we accept any solution as

cor-rect since we have carved it to fit the boundary conditions In thissense, the solution of differential equations is often more of an in-centive than a formal operation The person who does it best isoften the person who has done it before and so has a large assort-ment of tricks up his or her sleeve

4.3 Dimensional analysis

Introduction

Most universities place the first course in heat transfer after an tion to fluid mechanics: and most fluid mechanics courses include some

introduc-dimensional analysis This is normally treated using the familiar method

of indices, which is seemingly straightforward to teach but is

cumber-some and cumber-sometimes misleading to use It is rather well presented in[4.1]

The method we develop here is far simpler to use than the method

of indices, and it does much to protect us from the common errors we

might fall into We refer to it as the method of functional replacement.

The importance of dimensional analysis to heat transfer can be madeclearer by recalling Example2.6, which (like most problems in PartI) in-volved several variables Theses variables included the dependent vari-

able of temperature, (T ∞ − Ti);3 the major independent variable, which

was the radius, r ; and five system parameters, r i , ro, h, k, and (T ∞ − Ti ).

By reorganizing the solution into dimensionless groups [eqn (2.24)], wereduced the total number of variables to only four:

This solution offered a number of advantages over the dimensional

solution For one thing, it permitted us to plot all conceivable solutions

for a particular shape of cylinder, (r o/r i ), in a single figure, Fig. 2.13

For another, it allowed us to study the simultaneous roles ofh, k and r o

in defining the character of the solution By combining them as a Biot

number, we were able to say—even before we had solved the problem—

whether or not external convection really had to be considered

The nondimensionalization made it possible for us to consider,

simul-taneously, the behavior of all similar systems of heat conduction through

cylinders Thus a large, highly conducting cylinder might be similar in

its behavior to a small cylinder with a lower thermal conductivity

Finally, we shall discover that, by nondimensionalizing a problem

be-fore we solve it, we can often greatly simplify the process of solving it.

Our next aim is to map out a method for nondimensionalization

prob-lems before we have solved then, or, indeed, before we have even written

the equations that must be solved The key to the method is a result

called the Buckingham pi-theorem.

The Buckingham pi-theorem

The attention of scientific workers was apparently drawn very strongly

toward the question of similarity at about the beginning of World War I

Buckingham first organized previous thinking and developed his famous

theorem in 1914 in the Physical Review [4.2], and he expanded upon the

idea in the Transactions of the ASME one year later [4.3] Lord Rayleigh

almost simultaneously discussed the problem with great clarity in 1915

[4.4] To understand Buckingham’s theorem, we must first overcome one

conceptual hurdle, which, if it is clear to the student, will make everything

that follows extremely simple Let us explain that hurdle first

Suppose that y depends on r , x, z and so on:

y = y(r , x, z, )

We can take any one variable—say, x—and arbitrarily multiply it (or it

raised to a power) by any other variables in the equation, without altering

the truth of the functional equation, like this:

y

x = y x x2r , x, xz

To see that this is true, consider an arbitrary equation:

y = y(r , x, z) = r (sin x)e −z

This need only be rearranged to put it in terms of the desired modified

variables and x itself (y/x, x2r , x, and xz):

y

x = x x23r (sin x) exp

− xz x

We can do any such multiplying or dividing of powers of any variable

we wish without invalidating any functional equation that we choose towrite This simple fact is at the heart of the important example thatfollows:

Example 4.2

Consider the heat exchanger problem described in Fig.3.15 The known,” or dependent variable, in the problem is either of the exittemperatures Without any knowledge of heat exchanger analysis, wecan write the functional equation on the basis of our physical under-standing of the problem:

where the dimensions of each term are noted under the quotation

We want to know how many dimensionless groups the variables ineqn (4.14) should reduce to To determine this number, we use theidea explained above—that is, that we can arbitrarily pick one vari-able from the equation and divide or multiply it into other variables.Then—one at a time—we select a variable that has one of the dimen-sions We divide or multiply it by the other variables in the equationthat have that dimension in such a way as to eliminate the dimensionfrom them

We do this first with the variable (T hin − Tcin), which has the

The interesting thing about the equation in this form is that the only

remaining term in it with the units of K is (T hin − Tcin) No such

term can exist in the equation because it is impossible to achieve

dimensional homogeneity without another term in K to balance it

Therefore, we must remove it

Now the equation has only two dimensions in it—W and m2 Next, we

multiply U (T hin−Tcin) by A to get rid of m2in the second-to-last term

Accordingly, the term A (m2) can no longer stay in the equation, and

Next, we divide the first and third terms on the right by the second

This leaves only Cmin(Thin−Tcin), with the dimensions of W That term

must then be removed, and we are left with the completely

Equation (4.15) has exactly the same functional form as eqn (3.21),

which we obtained by direct analysis

Notice that we removed one variable from eqn (4.14) for each

di-mension in which the variables are expressed If there are n variables—

including the dependent variable—expressed in m dimensions, we then

expect to be able to express the equation in (n − m) dimensionless

groups, or pi-groups, as Buckingham called them.

This fact is expressed by the Buckingham pi-theorem, which we state

formally in the following way:

The interesting thing about the equation in this... exittemperatures Without any knowledge of heat exchanger analysis, wecan write the functional equation on the basis of our physical under-standing of the problem:

where the dimensions of each