3867 words - 16 pages

Transient, One-Dimensional Heat Conduction

in a Convectively Cooled Sphere

Gerald Recktenwald∗

March 16, 2006†

1

Overview

This article documents the numerical evaluation of a well-known analytical

model for transient, one-dimensional heat conduction. The physical situation

is depicted in Figure 1. A sphere of uniform material is initially at a uniform

temperature Ti . At time t = 0 the sphere is immersed in a stream of moving

ﬂuid at some diﬀerent temperature T∞ . The external surface of the sphere exchanges heat by convection. The local heat ﬂux from the sphere to the ﬂuid

is

q = h(Ts − T∞ )

(1)

where h is the heat transfer coeﬃcient, and Ts is the local surface temperature.

The problem ...view middle of the document...

g. [1, 2]. A universal solution is obtained in terms of the dimensionless variables

θ∗ =

T − T∞

,

Ti − T∞

r∗ =

r

,

ro

Fo =

αt

.

2

ro

(5)

The dimensionless form of the boundary condition in Equation (3) is

∂θ∗

∂r∗

∗

= Bi θs

(6)

r ∗ =1

where the Biot number is

hro

(7)

k

∗

and θs = (Ts − T∞ )/(Ti − T∞ ). If Bi

1 the internal temperature gradient is

small compared to the scaled diﬀerence between the surface temperature and

the ﬂuid. In that case the temperature in the sphere is spatially uniform.

The analytical solution is the inﬁnite series

Bi =

∞

∗

2

Cn exp(−ζn Fo)

θ=

n=1

where

Cn =

1

sin(ζn r∗ )

ζn r ∗

4 [sin(ζn ) − ζn cos(ζn )]

.

2ζn − sin(2ζn )

Copyright c 2011, Gerald Recktenwald. All rights reserved.

(8)

(9)

3

20

f(ζ)

0

−20

0

5

10

15

20

f(ζ)

0

−20

0

10

20

ζ

30

40

50

Figure 2: Plots of f (ζ ) = 1 − ζ cot(ζ ) − Bi for Bi = 5 showing roots (*) and

singularities (o) of Equation (10). Upper plot shows the range 0 ≤ ζ ≤ 15.

Lower plot shows the range 0 ≤ ζ ≤ 50.

The ζn are the positive roots of

1 − ζn cot(ζn ) = Bi.

(10)

Equations (8) through (10) provide a compact representation of the solution.

Obtaining numerical values from these formulas is a nontrivial eﬀort except in

the case where Fo

1 when a one term approximation is suﬃcient. In general,

at least the ﬁrst few terms terms in the series are needed.

3

Evaluating the Solution in Matlab

In this section, a procedure for evaluating Equation (8) is presented. The Matlab programs, or (for long programs) the function prologues, are listed in the

appendix.

3.1

Finding the Roots of Equation (10)

The ﬁrst task is to ﬁnd the roots, ζn , of Equation (10) for a given Bi. This is a

standard root-ﬁnding problem, with two important complications. First, there

are an inﬁnite number of ζn on the positive real line. To evaluate Equation (8)

a large number of roots must be available and sorted in order of increasing

ζn . A missing root will cause erroneous evaluation of the series. The second

complication is that Equation (10) has a singularity between successive roots.

Copyright c 2011, Gerald Recktenwald. All rights reserved.

3.1

Finding the Roots of Equation (10)

4

Figure 2 shows the function f (ζ ) = 1 − ζ cot(ζ ) − Bi for Bi = 5 in the range

0 ≤ ζ ≤ 50. Note that f (ζ ) = 0 when ζ is a root of Equation (10). The top

half of Figure 2 shows f (ζ ) (for Bi = 5) in the range 0 ≤ ζ ≤ 15. The roots

are identiﬁed with the * symbol, and the singularities are identiﬁed by vertical

dashed lines and the o symbol. The bottom half of Figure 2 shows f (ζ ) over the

larger range 0 < ζ ≤ 50. As ζ increases the spacing between roots approaches a

constant, and the roots are located midway between the singularities. At small

ζ the spacing between the roots is not equal.

The zRoots function in Listing 1 uses the zfun function (Listing 2), the

bracket function (Listing 3), and the built-in fzero function to ﬁnd the roots

of Equation (10). fzero is a very eﬃcient root-ﬁnding...

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