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TRANSIENT HEAT CONDUCTION

Therefore, lumped system analysis is not applicable. However, we can still use
it to get a “rough” estimate of the time of death. The exponent b in this case is

hAs h 8 W/m2 C
b



rcpV rcp Lc (996 kg/m3)(4178 J/kg C)(0.0689 m)

2.79 105 s1
We now substitute these values into Eq. 4–4,

T (t )  T 25  20 105 s1)t

ebt ⎯→
e(2.79
Ti  T 37  20

which yields
t
43,860 s
12.2 h
Therefore, as a rough estimate, the person died about 12 h before the body
was found, and thus the time of death is 5 AM.
Discussion This example demonstrates how to obtain “ball park” values using
a simple analysis. A similar analysis is used in practice by incorporating con-
stants to account for deviation from lumped system analysis.

4–2 ■
TRANSIENT HEAT CONDUCTION IN LARGE
PLANE WALLS, LONG CYLINDERS, AND
SPHERES WITH SPATIAL EFFECTS
In Section 4–1, we considered bodies in which the variation of temperature
within the body is negligible; that is, bodies that remain nearly isothermal dur-
ing a process. Relatively small bodies of highly conductive materials approxi-
mate this behavior. In general, however, the temperature within a body changes
from point to point as well as with time. In this section, we consider the varia-
tion of temperature with time and position in one-dimensional problems such as
those associated with a large plane wall, a long cylinder, and a sphere.
Consider a plane wall of thickness 2L, a long cylinder of radius ro, and
a sphere of radius ro initially at a uniform temperature Ti, as shown in
Fig. 4–11. At time t
0, each geometry is placed in a large medium that is at
a constant temperature T and kept in that medium for t  0. Heat transfer
takes place between these bodies and their environments by convection with a
uniform and constant heat transfer coefficient h. Note that all three cases pos-
sess geometric and thermal symmetry: the plane wall is symmetric about its
center plane (x
0), the cylinder is symmetric about its centerline (r
0),
and the sphere is symmetric about its center point (r
0). We neglect radia-
tion heat transfer between these bodies and their surrounding surfaces, or in-
corporate the radiation effect into the convection heat transfer coefficient h.
The variation of the temperature profile with time in the plane wall is
illustrated in Fig. 4–12. When the wall is first exposed to the surrounding
medium at T  Ti at t
0, the entire wall is at its initial temperature Ti. But
the wall temperature at and near the surfaces starts to drop as a result of heat
transfer from the wall to the surrounding medium. This creates a temperature
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CHAPTER 4

T Initially T T Initially T
T = Ti T = Ti T
h h h h
Initially h
T = Ti
0 0 ro
L x r
0 ro
r

FIGURE 4–11
Schematic of the simple
geometries in which heat
(a) A large plane wall (b) A long cylinder (c) A sphere transfer is one-dimensional.

gradient in the wall and initiates heat conduction from the inner parts of the
wall toward its outer surfaces. Note that the temperature at the center of the Ti
wall remains at Ti until t
t2, and that the temperature profile within the wall t = t1 t=0
remains symmetric at all times about the center plane. The temperature profile t = t2
gets flatter and flatter as time passes as a result of heat transfer, and eventually t = t3 t→
becomes uniform at T
T. That is, the wall reaches thermal equilibrium T
with its surroundings. At that point, heat transfer stops since there is no longer
a temperature difference. Similar discussions can be given for the long cylin- 0
L x
der or sphere. h Initially T
T = Ti
h
Nondimensionalized One-Dimensional Transient
Conduction Problem FIGURE 4–12
The formulation of heat conduction problems for the determination of the Transient temperature profiles in a
one-dimensional transient temperature distribution in a plane wall, a cylinder, plane wall exposed to convection
or a sphere results in a partial differential equation whose solution typically from its surfaces for Ti  T.
involves infinite series and transcendental equations, which are inconvenient
to use. But the analytical solution provides valuable insight to the physical
problem, and thus it is important to go through the steps involved. Below we
demonstrate the solution procedure for the case of plane wall.
Consider a plane wall of thickness 2L initially at a uniform temperature of
Ti, as shown in Fig. 4–11a. At time t
0, the wall is immersed in a fluid at
temperature T∞ and is subjected to convection heat transfer from both sides
with a convection coefficient of h. The height and the width of the wall are
large relative to its thickness, and thus heat conduction in the wall can be ap-
proximated to be one-dimensional. Also, there is thermal symmetry about the
midplane passing through x
0, and thus the temperature distribution must be
symmetrical about the midplane. Therefore, the value of temperature at any
x value in L  x  0 at any time t must be equal to the value at x in
0  x  L at the same time. This means we can formulate and solve the heat
conduction problem in the positive half domain 0  x  L, and then apply the
solution to the other half.
Under the conditions of constant thermophysical properties, no heat gener-
ation, thermal symmetry about the midplane, uniform initial temperature, and
constant convection coefficient, the one-dimensional transient heat conduction
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TRANSIENT HEAT CONDUCTION

problem in the half-domain 0  x  L of the plain wall can be expressed as

(see Chapter 2)
2
T 1 T
Differential equation:
(4–10a)
x2 a t
T(0, t) T(L, t)
Boundary conditions:
0 and k
h[T(L, t)  T] (4–10b)
x x

Initial condition: T(x, 0)
Ti (4–10c)

where the property a
k/rcp is the thermal diffusivity of the material.

We now attempt to nondimensionalize the problem by defining a dimen-
sionless space variable X
x/L and dimensionless temperature u(x, t)

[T(x, t)  T]/[Ti  T]. These are convenient choices since both X and u vary
between 0 and 1. However, there is no clear guidance for the proper form of
the dimensionless time variable and the h/k ratio, so we will let the analysis
indicate them. We note that
2
u u L T, u L2 T u 1 T



and

X (x/L) Ti  T x X2 Ti  T x t Ti  T t

Substituting into Eqs. 4–10a and 4–10b and rearranging give

2
u L2 u u(1, t) hL
2

and
u(1, t) (4–11)
X a t X k

Therefore, the proper form of the dimensionless time is t
at/L2, which is

(a) Original heat conduction problem:
2
called the Fourier number Fo, and we recognize Bi
k/hL as the Biot num-
T 1 T

, T(x, 0)
Ti ber defined in Section 4–1. Then the formulation of the one-dimensional
x2 a t
transient heat conduction problem in a plane wall can be expressed in
T(0, t) T(L, t)

0,  k
h[T(L, t)  T] nondimensional form as
x x
2
T
F(x, L, t, k, , h, Ti) u u
Dimensionless differential equation:
(4–12a)
X2 t
(b) Nondimensionalized problem: u(0, t) u(1, t)
2
u u Dimensionless BC’s:
0 and
Biu(1, t)

, u(X, 0)
1 X X
X2 t
u(0, t) u(1, t) (4–12b)

0,
Biu(1, t)
X X Dimensionless initial condition: u(X, 0)
1 (4–12c)
u
f(X, Bi, t) where
T(x, t)  Ti
FIGURE 4–13 u(X, t)
Dimensionless temperature
Nondimensionalization reduces the T  Ti
number of independent variables in x
X

one-dimensional transient conduction L Dimensionless distance from the center
problems from 8 to 3, offering great hL
convenience in the presentation of Bi
Dimensionless heat transfer coefficient (Biot number)
k
results. at
t

Fo Dimensionless time (Fourier number)
L2
The heat conduction equation in cylindrical or spherical coordinates can be
nondimensionalized in a similar way. Note that nondimensionalization
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CHAPTER 4

reduces the number of independent variables and parameters from 8 to 3—

from x, L, t, k, a, h, Ti, and T to X, Bi, and Fo (Fig. 4–13). That is,
u
f(X, Bi, Fo) (4–13)

This makes it very practical to conduct parametric studies and to present

results in graphical form. Recall that in the case of lumped system analysis,
we had u
f(Bi, Fo) with no space variable.

Exact Solution of One-Dimensional Transient Conduction

Problem*
The non-dimensionalized partial differential equation given in Eqs. 4–12 to-
gether with its boundary and initial conditions can be solved using several an-
alytical and numerical techniques, including the Laplace or other transform
methods, the method of separation of variables, the finite difference method,
and the finite-element method. Here we use the method of separation of vari-
ables developed by J. Fourier in 1820s and is based on expanding an arbitrary
function (including a constant) in terms of Fourier series. The method is ap-
plied by assuming the dependent variable to be a product of a number of func-
tions, each being a function of a single independent variable. This reduces the
partial differential equation to a system of ordinary differential equations,
each being a function of a single independent variable. In the case of transient
conduction in a plain wall, for example, the dependent variable is the solution
function u(X, t), which is expressed as u(X, t)
F(X)G(t), and the applica-
tion of the method results in two ordinary differential equation, one in X and
the other in t.
The method is applicable if (1) the geometry is simple and finite (such as a
rectangular block, a cylinder, or a sphere) so that the boundary surfaces can be
described by simple mathematical functions, and (2) the differential equation
and the boundary and initial conditions in their most simplified form are lin-
ear (no terms that involve products of the dependent variable or its derivatives)
and involve only one nonhomogeneous term (a term without the dependent
variable or its derivatives). If the formulation involves a number of nonhomo-
geneous terms, the problem can be split up into an equal number of simpler
problems each involving only one nonhomogeneous term, and then combin-
ing the solutions by superposition.
Now we demonstrate the use of the method of separation of variables by ap-
plying it to the one-dimensional transient heat conduction problem given in
Eqs. 4–12. First, we express the dimensionless temperature function u(X, t) as
a product of a function of X only and a function of t only as
u(X, t)
F(X)G(t) (4–14)

Substituting Eq. 4–14 into Eq. 4–12a and dividing by the product FG gives
1 d2F 1 dG

(4–15)
F dX2 G dt
Observe that all the terms that depend on X are on the left-hand side of the
equation and all the terms that depend on t are on the right-hand side. That is,

*This section can be skipped if desired without a loss of continuity.

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TRANSIENT HEAT CONDUCTION

the terms that are function of different variables are separated (and thus the
name separation of variables). The left-hand side of this equation is a function
of X only and the right-hand side is a function of only t. Considering that both
X and t can be varied independently, the equality in Eq. 4–15 can hold for any
value of X and t only if Eq. 4–15 is equal to a constant. Further, it must be a
negative constant that we will indicate by l2 since a positive constant will
cause the function G(t) to increase indefinitely with time (to be infinite),
which is unphysical, and a value of zero for the constant means no time de-
pendence, which is again inconsistent with the physical problem. Setting
Eq. 4–15 equal to l2 gives
d2F dG
l2F
0 and l2G
0 (4–16)
dX2 dt

whose general solutions are

C2 sin (lX) and G
C3e l t
2
F
C1 cos (lX) (4–17)

and
u
FG
C3e l t[C1 cos (lX) C2 sin (lX)]
e l t[A cos (lX)
2 2
B sin (lX)]
(4–18)

where A
C1C3 and B
C2C3 are arbitrary constants. Note that we need to
determine only A and B to obtain the solution of the problem.
Applying the boundary conditions in Eq. 4–12b gives
u(0, t)

0 → el t(Al sin 0 Bl cos 0)
0 → B
0 → u
Ael t cos (lX)
2 2

u(1, t)

Biu(1,t) → Ael tl sin l
BiAel t cos l → l tan l
Bi
2 2

But tangent is a periodic function with a period of p, and the equation

ltanl
Bi has the root l1 between 0 and p, the root l2 between p and 2p,
the root ln between (n1)p and np, etc. To recognize that the transcendental
equation l tan l
Bi has an infinite number of roots, it is expressed as
ln tan ln
Bi (4–19)

Eq. 4–19 is called the characteristic equation or eigenfunction, and its roots
are called the characteristic values or eigenvalues. The characteristic equa-
tion is implicit in this case, and thus the characteristic values need to be deter-
mined numerically. Then 2 it follows that there are an infinite number of
solutions of the form Ael t cos (lX), and the solution of this linear heat con-
duction problem is a linear combination of them,

u
a Anel nt cos (ln X)
2
(4–20)
n
1

The constants An are determined from the initial condition, Eq. 4–12c,

u(X, 0)
1 → 1
a An cos (ln X) (4–21)
n
1
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CHAPTER 4

This is a Fourier series expansion that expresses a constant in terms of an in-

finite series of cosine functions. Now we multiply both sides of Eq. 4–21 by
cos(lm X), and integrate from X
0 to X
1. The right-hand side involves an
infinite number of integrals of the form
0cos(lm X) cos(ln X)dx. It can be
1

shown that all of these integrals vanish except when n
m, and the coefficient un
An e ln t cos(ln X)
2

An becomes
4 sin ln
An

1 1 2ln sin(2ln)

 cos (l X)dX
A 
4 sin ln
n n cos (ln X)dx → An

2
(4–22) ln tan ln
Bi
0 0
2ln sin (2ln)
For Bi
5, X
1, and t
0.2:

This completes the analysis for the solution of one-dimensional transient n ln An un

heat conduction problem in a plane wall. Solutions in other geometries such
1 1.3138 1.2402 0.22321
as a long cylinder and a sphere can be determined using the same approach.
The results for all three geometries are summarized in Table 4–1. The solution 2 4.0336 0.3442 0.00835
for the plane wall is also applicable for a plane wall of thickness L whose left 3 6.9096 0.1588 0.00001
surface at x
0 is insulated and the right surface at x
L is subjected to con- 4 9.8928 0.876 0.00000
vection since this is precisely the mathematical problem we solved.
The analytical solutions of transient conduction problems typically involve
infinite series, and thus the evaluation of an infinite number of terms to deter- FIGURE 4–14
mine the temperature at a specified location and time. This may look intimi- The term in the series solution of
dating at first, but there is no need to worry. As demonstrated in Fig. 4–14, the transient conduction problems decline
terms in the summation decline rapidly as n and thus ln increases because of rapidly as n and thus ln increases
the exponential decay function e l nt. This is especially the case when the di- because of the exponential decay
2

mensionless time t is large. Therefore, the evaluation of the first few terms of function with the exponent lnt.
the infinite series (in this case just the first term) is usually adequate to deter-
mine the dimensionless temperature u.

Approximate Analytical and Graphical Solutions

The analytical solution obtained above for one-dimensional transient heat
conduction in a plane wall involves infinite series and implicit equations,
which are difficult to evaluate. Therefore, there is clear motivation to simplify

TABLE 4–1
Summary of the solutions for one-dimensional transient conduction in a plane wall of
thickness 2L, a cylinder of radius ro and a sphere of radius ro subjected to convention from
all surfaces.*
Geometry Solution ln’s are the roots of

4 sin ln
e lnt cos (lnx /L)
2
Plane wall u
a ln tan ln
Bi
n
1 2ln sin(2ln)
 2
J1 (ln)
J1 (ln)
u
a ln J 02 (ln) J 12 (ln) e lnt J0 (lnr /ro)
2
Cylinder ln
Bi
n
1 J0 (ln)

4(sin ln  ln cos ln) l2nt sin (ln x / L)
u
a e l  ln cot ln
Bi
Sphere n
1 2ln  sin(2ln) ln x / L

*Here u
(T  Ti)/(T  Ti) is the dimensionless temperature, Bi
hL/k or hro /k is the Biot number, Fo
t
at  L2
or at  ro2 is the Fourier number, and J0 and J1 are the Bessel functions of the first kind whose values are given in
Table 4–3.
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TRANSIENT HEAT CONDUCTION

the analytical solutions and to present the solutions in tabular or graphical

form using simple relations.
The dimensionless quantities defined above for a plane wall can also be
used for a cylinder or sphere by replacing the space variable x by r and the
half-thickness L by the outer radius ro. Note that the characteristic length in
the definition of the Biot number is taken to be the half-thickness L for the
plane wall, and the radius ro for the long cylinder and sphere instead of V/A
used in lumped system analysis.
We mentioned earlier that the terms in the series solutions in Table 4–1 con-
verge rapidly with increasing time, and for t  0.2, keeping the first term and
neglecting all the remaining terms in the series results in an error under 2 per-
cent. We are usually interested in the solution for times with t  0.2, and thus
it is very convenient to express the solution using this one-term approxima-
tion, given as

T(x, t)  T 2
Plane wall: uwall

A1el1 t cos (l1x/L), t  0.2 (4–23)
Ti  T
T(r, t)  T 2
Cylinder: ucyl

A1el1 t J0(l1r/ro), t  0.2 (4–24)
Ti  T
T(r, t)  T 2 sin(l r/r )
Sphere: usph

A1el1 t 1 o
, t  0.2 (4–25)
Ti  T l1r/ro

where the constants A1 and l1 are functions of the Bi number only, and their
values are listed in Table 4–2 against the Bi number for all three geometries.
The function J0 is the zeroth-order Bessel function of the first kind, whose
value can be determined from Table 4–3. Noting that cos (0)
J0(0)
1 and
the limit of (sin x)/x is also 1, these relations simplify to the next ones at the
center of a plane wall, cylinder, or sphere:

T0  T 2
Center of plane wall (x
0): u0, wall

A1el1 t (4–26)
Ti  T
T0  T 2
Center of cylinder (r
0): u0, cyl

A1el1 t (4–27)
Ti  T
T0  T 2
Center of sphere (r
0): u0, sph

A1el1 t (4–28)
Ti  T

Comparing the two sets of equations above, we notice that the dimensionless
temperatures anywhere in a plane wall, cylinder, and sphere are related to the
center temperature by
ucyl usph

cos a b,
J0 a b ,
uwall l1x l1r sin (l1rro)
and
(4-29)
u0, wall L u0, cyl ro u0, sph l1rro

which shows that time dependence of dimensionless temperature within a

given geometry is the same throughout. That is, if the dimensionless center
temperature u0 drops by 20 percent at a specified time, so does the dimen-
sionless temperature u0 anywhere else in the medium at the same time.
Once the Bi number is known, these relations can be used to determine the
temperature anywhere in the medium. The determination of the constants A1
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CHAPTER 4

TABLE 4–2 TABLE 4–3

Coefficients used in the one-term approximate solution of transient one- The zeroth- and first-order Bessel
dimensional heat conduction in plane walls, cylinders, and spheres (Bi
hL/k functions of the first kind
for a plane wall of thickness 2L, and Bi
hro /k for a cylinder or sphere of h J0(h) J1(h)
radius ro )
0.0 1.0000 0.0000
Plane Wall Cylinder Sphere 0.1 0.9975 0.0499
Bi l1 A1 l1 A1 l1 A1 0.2 0.9900 0.0995
0.3 0.9776 0.1483
0.01 0.0998 1.0017 0.1412 1.0025 0.1730 1.0030
0.4 0.9604 0.1960
0.02 0.1410 1.0033 0.1995 1.0050 0.2445 1.0060
0.04 0.1987 1.0066 0.2814 1.0099 0.3450 1.0120
0.5 0.9385 0.2423
0.06 0.2425 1.0098 0.3438 1.0148 0.4217 1.0179
0.6 0.9120 0.2867
0.08 0.2791 1.0130 0.3960 1.0197 0.4860 1.0239
0.7 0.8812 0.3290
0.1 0.3111 1.0161 0.4417 1.0246 0.5423 1.0298
0.8 0.8463 0.3688
0.2 0.4328 1.0311 0.6170 1.0483 0.7593 1.0592
0.9 0.8075 0.4059
0.3 0.5218 1.0450 0.7465 1.0712 0.9208 1.0880
0.4 0.5932 1.0580 0.8516 1.0931 1.0528 1.1164 1.0 0.7652 0.4400
0.5 0.6533 1.0701 0.9408 1.1143 1.1656 1.1441 1.1 0.7196 0.4709
0.6 0.7051 1.0814 1.0184 1.1345 1.2644 1.1713 1.2 0.6711 0.4983
0.7 0.7506 1.0918 1.0873 1.1539 1.3525 1.1978 1.3 0.6201 0.5220
0.8 0.7910 1.1016 1.1490 1.1724 1.4320 1.2236 1.4 0.5669 0.5419
0.9 0.8274 1.1107 1.2048 1.1902 1.5044 1.2488
1.0 0.8603 1.1191 1.2558 1.2071 1.5708 1.2732 1.5 0.5118 0.5579
2.0 1.0769 1.1785 1.5995 1.3384 2.0288 1.4793 1.6 0.4554 0.5699
3.0 1.1925 1.2102 1.7887 1.4191 2.2889 1.6227 1.7 0.3980 0.5778
4.0 1.2646 1.2287 1.9081 1.4698 2.4556 1.7202 1.8 0.3400 0.5815
5.0 1.3138 1.2403 1.9898 1.5029 2.5704 1.7870 1.9 0.2818 0.5812
6.0 1.3496 1.2479 2.0490 1.5253 2.6537 1.8338
7.0 1.3766 1.2532 2.0937 1.5411 2.7165 1.8673 2.0 0.2239 0.5767
8.0 1.3978 1.2570 2.1286 1.5526 2.7654 1.8920 2.1 0.1666 0.5683
9.0 1.4149 1.2598 2.1566 1.5611 2.8044 1.9106 2.2 0.1104 0.5560
10.0 1.4289 1.2620 2.1795 1.5677 2.8363 1.9249 2.3 0.0555 0.5399
20.0 1.4961 1.2699 2.2880 1.5919 2.9857 1.9781 2.4 0.0025 0.5202
30.0 1.5202 1.2717 2.3261 1.5973 3.0372 1.9898
40.0 1.5325 1.2723 2.3455 1.5993 3.0632 1.9942 2.6 0.0968 0.4708
50.0 1.5400 1.2727 2.3572 1.6002 3.0788 1.9962 2.8 0.1850 0.4097
100.0 1.5552 1.2731 2.3809 1.6015 3.1102 1.9990 3.0 0.2601 0.3391
 1.5708 1.2732 2.4048 1.6021 3.1416 2.0000 3.2 0.3202 0.2613

and l1 usually requires interpolation. For those who prefer reading charts to
interpolating, these relations are plotted and the one-term approximation so-
lutions are presented in graphical form, known as the transient temperature
charts. Note that the charts are sometimes difficult to read, and they are sub-
ject to reading errors. Therefore, the relations above should be preferred to the
charts.
The transient temperature charts in Figs. 4–15, 4–16, and 4–17 for a large
plane wall, long cylinder, and sphere were presented by M. P. Heisler in 1947
and are called Heisler charts. They were supplemented in 1961 with transient
heat transfer charts by H. Gröber. There are three charts associated with each
geometry: the first chart is to determine the temperature T0 at the center of the
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TRANSIENT HEAT CONDUCTION

T0 – T
θ0 =
Ti – T
1.0 k
0.7 hL = 1 Plate
0.5 Bi =
0.4
0.3 100
0.2 80 90
14
60 70
0.1 1.0 10 12
45 50
0.07 9
8 35
0.7

0.05 0. 40
8
0.04 7
0.5

0.03 6 30
0.6

25
0.3

0.02
0.4

18
4

20
0.1

0.01
0.2

16
0.007
0

2.5

0.005
0.05

0.004
1.6 1.4
2 1.8
1.2

0.003
0.002

0.001
0 1 2 3 4 6 8 10 14 18 22 26 30 50 70 100 120 150 300 400 500 600 700
τ = α t/L2

(a) Midplane temperature (from M. P. Heisler, “Temperature Charts for Induction and T Initially T
Constant Temperature Heating,” Trans. ASME 69, 1947, pp. 227–36. Reprinted by permission h T = Ti h
of ASME International.) 0
L x

T – T 2L
Q
θ=
T0 – T Qmax
1.0 x/L = 0.2 1.0
0.9 0.9 Bi = hL/k
0.4
0.8 0.8
0.7 0.7
0.6
0.6 0.6
1
2
5
0.00
0.00
0.00
0.01
0.02
0.05
0.1
0.2

0.5

10

20

50
0.5 0.5
1

5
Bi =

0.4 0.8 0.4

0.3 0.3
0.9
0.2 0.2
0.1 1.0 0.1
Plate 0 Plate
0
0.01 0.1 1.0 10 100 10–5 10– 4 10–3 10–2 10–1 1 10 102 103 104
1 k Bi 2τ = h2α t/k 2
=
Bi hL

(b) Temperature distribution (from M. P. Heisler, (c) Heat transfer (from H. Gröber et al.)
“Temperature Charts for Induction and Constant
Temperature Heating,” Trans. ASME 69, 1947,
pp. 227–36. Reprinted by permission of ASME
International.)

FIGURE 4–15
Transient temperature and heat transfer charts for a plane wall of thickness 2L initially at a uniform temperature Ti
subjected to convection from both sides to an environment at temperature T with a convection coefficient of h.
cen29305_ch04.qxd 11/30/05 3:00 PM Page 233

T0 – T 233
θ0 = CHAPTER 4
Ti – T
1.0
0.7 Cylinder
0.5
0.4 k
5 hr = 1
0.3 o
0.1 Bi =
0.2 4
2. 3 25
5 20
0.1
16 18 90 10
0.07 1. 2 0
8
1.4

0.05 1.6 14 70 80

12
0.04
1.0
1.2

0.03 60

9
10
0.6

50
0.02
0.8

7
8
0.4

40
45
6
0.2

0.01
0.5

30
0

0.007
0.3

35
0.1

0.005
0.004
0.003
0.002

0.001
0 1 2 3 4 6 8 10 14 18 22 26 30 50 70 100 120 140 150 250 350
τ = α t/ro2

(a) Centerline temperature (from M. P. Heisler, “Temperature Charts for Induction and T Initially T
Constant Temperature Heating,” Trans. ASME 69, 1947, pp. 227–36. Reprinted by permission h T = Ti h
of ASME International.)
0 ro r
T – T Q
θ=
T0 – T Qmax
1.0 r/ro = 0.2 1.0
Bi = hro /k
0.9 0.9
0.8 0.4 0.8
0.7 0.7
0.6 0.6 0.6
1
2
5
0.00
0.00
0.00
0.01
0.02
0.05
0.1
0.2
0.5

10

20

50
0.5 0.5
1
2

5
Bi =

0.4 0.4
0.8
0.3 0.3
0.2 0.9 0.2
0.1 1.0 Cylinder 0.1 Cylinder
0 0
0.01 0.1 1.0 10 100 10–5 10– 4 10–3 10–2 10–1 1 10 102 103 104
1 k Bi 2τ = h2α t/k 2
=
Bi hro

(b) Temperature distribution (from M. P. Heisler, (c) Heat transfer (from H. Gröber et al.)
“Temperature Charts for Induction and Constant
Temperature Heating,” Trans. ASME 69, 1947,
pp. 227–36. Reprinted by permission of ASME
International.)
FIGURE 4–16
Transient temperature and heat transfer charts for a long cylinder of radius ro initially at a uniform temperature Ti
subjected to convection from all sides to an environment at temperature T with a convection coefficient of h.
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234
TRANSIENT HEAT CONDUCTION

T0 – T
θ0 =
Ti – T
1.0
0.7 k Sphere
0.5 hr = 1
0.4 o
Bi =
0.3 100
80 90
0.2
12 14 60 70
0.1 3.0

45
50
10
9
2.6 2.8
0.07

35 3
40
7
8
2.4
2.
0.05
2 2

25
5
6
1.

0.04

0
3.5
8

18 16
.0

0.03
1.4

20
1.6

0.02
1.0

1.2
0.5

0.7

0.01
5

0.007
0.2
0.35

0.005
0.004
0.1 0.05
0

0.003
0.002

0.001
0 0.5 1.0 1.5 2 2.5 3 4 5 6 7 8 9 10 20 30 40 50 100 150 200 250
τ = α t/ro2

(a) Midpoint temperature (from M. P. Heisler, “Temperature Charts for Induction and T Initially T
Constant Temperature Heating,” Trans. ASME 69, 1947, pp. 227–36. Reprinted by permission h T = Ti h
of ASME International.)
0 ro r

T – T Q
θ=
T0 – T Qmax
1.0 r/ro = 0.2 1.0
Bi = hro /k
0.9 0.9
0.8 0.4 0.8
0.7 0.7
0.6 0.6
1
2
5

0.6
0.00
0.00
0.00
0.01
0.02
0.05
0.1
0.2
0.5

10

20

50

0.5 0.5
1
2

5
Bi =

0.4 0.4
0.3 0.8 0.3
0.2 0.9 0.2
0.1 1.0 Sphere 0.1 Sphere
0 0
0.01 0.1 1.0 10 100 10–5 10– 4 10–3 10–2 10–1 1 10 102 103 104
1 = k Bi 2τ = h2α t/k 2
Bi hro

(b) Temperature distribution (from M. P. Heisler, (c) Heat transfer (from H. Gröber et al.)
“Temperature Charts for Induction and Constant
Temperature Heating,” Trans. ASME 69, 1947,

FIGURE 4–17
Transient temperature and heat transfer charts for a sphere of radius ro initially at a uniform temperature Ti subjected to
convection from all sides to an environment at temperature T with a convection coefficient of h.
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235
CHAPTER 4

geometry at a given time t. The second chart is to determine the temperature

at other locations at the same time in terms of T0. The third chart is to deter-
mine the total amount of heat transfer up to the time t. These plots are valid
for t  0.2.
Note that the case 1/Bi
k/hL
0 corresponds to h → , which corre- Ts Ts

sponds to the case of specified surface temperature T. That is, the case in T T
h Ts ≠ T h
which the surfaces of the body are suddenly brought to the temperature T
at t
0 and kept at T at all times can be handled by setting h to infinity
(Fig. 4–18).
The temperature of the body changes from the initial temperature Ti to the (a) Finite convection coefficient
temperature of the surroundings T at the end of the transient heat conduction
process. Thus, the maximum amount of heat that a body can gain (or lose if
Ti  T) is simply the change in the energy content of the body. That is,
Qmax
mcp(T  Ti )
rVcp(T  Ti ) (kJ) (4–30)

where m is the mass, V is the volume, r is the density, and cp is the specific
heat of the body. Thus, Qmax represents the amount of heat transfer for t → .
The amount of heat transfer Q at a finite time t is obviously less than this max- T T
Ts Ts
imum, and it can be expressed as the sum of the internal energy changes h→ h→
throughout the entire geometry as Ts = T

Q
 rc [T(x, t)  T ]dV
V
p i (4-31)
(b) Infinite convection coefficient

where T(x, t) is the temperature distribution in the medium at time t. Assum- FIGURE 4–18
ing constant properties, the ratio of Q/Qmax becomes The specified surface
temperature corresponds to the case
Q



V V


v rcp[T(x, t)  Ti]dV 1 (1  u)dV
(4-32)
of convection to an environment at
T with a convection coefficient h
Qmax rcp(T  Ti)V
that is infinite.
Using the appropriate nondimensional temperature relations based on the one-
term approximation for the plane wall, cylinder, and sphere, and performing
the indicated integrations, we obtain the following relations for the fraction of
heat transfer in those geometries:

a b
1  u0, wall
Q sin l1
Plane wall: (4–33)
Qmax wall l1

a b
1  2u0, cyl
Q J1(l1)
Cylinder: (4–34)
Qmax cyl l1
sin l1  l1 cos l1
a b
1  3u0, sph
Q
Sphere: (4–35)
Qmax sph l31

These Q/Qmax ratio relations based on the one-term approximation are also
plotted in Figures 4–15c, 4–16c, and 4–17c, against the variables Bi and
h2at/k2 for the large plane wall, long cylinder, and sphere, respectively. Note
that once the fraction of heat transfer Q/Qmax has been determined from these
charts or equations for the given t, the actual amount of heat transfer by that
time can be evaluated by multiplying this fraction by Qmax. A negative sign for
Qmax indicates that the body is rejecting heat (Fig. 4–19).
The use of the Heisler/Gröber charts and the one-term solutions already
discussed is limited to the conditions specified at the beginning of this section:
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236
TRANSIENT HEAT CONDUCTION
. the body is initially at a uniform temperature, the temperature of the medium
Qmax
t=0 surrounding the body and the convection heat transfer coefficient are constant
and uniform, and there is no heat generation in the body.
T = Ti
We discussed the physical significance of the Biot number earlier and indi-
T = T
m, cp cated that it is a measure of the relative magnitudes of the two heat transfer
h
mechanisms: convection at the surface and conduction through the solid.
T A small value of Bi indicates that the inner resistance of the body to heat con-
duction is small relative to the resistance to convection between the surface
(a) Maximum heat transfer (t → ) and the fluid. As a result, the temperature distribution within the solid be-
. comes fairly uniform, and lumped system analysis becomes applicable. Recall
t=0
Q that when Bi  0.1, the error in assuming the temperature within the body to
be uniform is negligible.
T = Ti To understand the physical significance of the Fourier number t, we express
m, cp
T = T(r, t) it as (Fig. 4–20)
h
T The rate at which heat is conducted
2
at kL (1/L) T across L of a body of volume L3
t
2

(4–36)
Bi = . . . Q L rcp L /t T
3 The rate at which heat is stored
2α t —— = . . . in a body of volume L3
h—— Qmax
= Bi2τ = . . .
k2
(Gröber chart) Therefore, the Fourier number is a measure of heat conducted through a body
(b) Actual heat transfer for time t relative to heat stored. Thus, a large value of the Fourier number indicates
FIGURE 4–19 faster propagation of heat through a body.
The fraction of total heat transfer Perhaps you are wondering about what constitutes an infinitely large plate
Q/Qmax up to a specified time t is or an infinitely long cylinder. After all, nothing in this world is infinite. A plate
determined using the Gröber charts. whose thickness is small relative to the other dimensions can be modeled as
an infinitely large plate, except very near the outer edges. But the edge effects
on large bodies are usually negligible, and thus a large plane wall such as the
L wall of a house can be modeled as an infinitely large wall for heat transfer pur-
L
L poses. Similarly, a long cylinder whose diameter is small relative to its length
can be analyzed as an infinitely long cylinder. The use of the transient tem-
·
Q
· Qconducted perature charts and the one-term solutions is illustrated in Examples 4–3, 4–4,
and 4–5.

·
Qstored
EXAMPLE 4–3 Boiling Eggs
·
Fourier number: τ = α
—–t =Q conducted
———— An ordinary egg can be approximated as a 5-cm-diameter sphere (Fig. 4–21).
L2 ·
Qstored The egg is initially at a uniform temperature of 5 C and is dropped into boil-
FIGURE 4–20 ing water at 95 C. Taking the convection heat transfer coefficient to be
Fourier number at time t can be h
1200 W/m2 · C, determine how long it will take for the center of the egg
to reach 70 C.
viewed as the ratio of the rate of heat
conducted to the rate of heat stored
at that time. SOLUTION An egg is cooked in boiling water. The cooking time of the egg is
to be determined.
Assumptions 1 The egg is spherical in shape with a radius of ro
2.5 cm.
2 Heat conduction in the egg is one-dimensional because of thermal symmetry
about the midpoint. 3 The thermal properties of the egg and the heat transfer
coefficient are constant. 4 The Fourier number is t  0.2 so that the one-term
approximate solutions are applicable.
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237
CHAPTER 4

Properties The water content of eggs is about 74 percent, and thus the ther-
mal conductivity and diffusivity of eggs can be approximated by those of water Egg
at the average temperature of (5 70)/2
37.5 C; k
0.627 W/m · C and
a
k/rcp
0.151 106 m2/s (Table A–9). Ti = 5°C
Analysis The temperature within the egg varies with radial distance as well as
time, and the temperature at a specified location at a given time can be deter-
mined from the Heisler charts or the one-term solutions. Here we use the latter h = 1200 W/ m 2·°C
to demonstrate their use. The Biot number for this problem is T = 95°C
hro (1200 W/m2 C)(0.025 m)
Bi


47.8 FIGURE 4–21
k 0.627 W/m C Schematic for Example 4–3.
which is much greater than 0.1, and thus the lumped system analysis is not
applicable. The coefficients l1 and A1 for a sphere corresponding to this Bi are,
from Table 4–2,

l1
3.0754, A1
1.9958
Substituting these and other values into Eq. 4–28 and solving for t gives
T0  T 2 70  95

1.9958e(3.0754) t ⎯→ t
0.209
2

A1el1 t ⎯→
Ti  T 5  95

which is greater than 0.2, and thus the one-term solution is applicable with an
error of less than 2 percent. Then the cooking time is determined from the de-
finition of the Fourier number to be
tro2 (0.209)(0.025 m)2
t
a

865 s  14.4 min
0.151 106 m2/s
Therefore, it will take about 15 min for the center of the egg to be heated from
5 C to 70 C.
Discussion Note that the Biot number in lumped system analysis was defined
differently as Bi
hLc /k
h(r o /3)/k. However, either definition can be used in
determining the applicability of the lumped system analysis unless Bi  0.1.

EXAMPLE 4–4 Heating of Brass Plates in an Oven

T = 500°C
In a production facility, large brass plates of 4-cm thickness that are initially at h = 120 W/m 2·°C
a uniform temperature of 20 C are heated by passing them through an oven
that is maintained at 500 C (Fig. 4–22). The plates remain in the oven for a
period of 7 min. Taking the combined convection and radiation heat transfer 2L = 4 cm
coefficient to be h
120 W/m2 · C, determine the surface temperature of the Brass
plates when they come out of the oven. plate

Ti = 20°C
SOLUTION Large brass plates are heated in an oven. The surface temperature
of the plates leaving the oven is to be determined.
Assumptions 1 Heat conduction in the plate is one-dimensional since the
FIGURE 4–22
plate is large relative to its thickness and there is thermal symmetry about the Schematic for Example 4–4.
center plane. 2 The thermal properties of the plate and the heat transfer coef-
ficient are constant. 3 The Fourier number is t  0.2 so that the one-term ap-
proximate solutions are applicable.
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238
TRANSIENT HEAT CONDUCTION

Properties The properties of brass at room temperature are k
110 W/m · C,

r
8530 kg/m3, cp
380 J/kg · C, and a
33.9 106 m2/s (Table A–3).
More accurate results are obtained by using properties at average temperature.
Analysis The temperature at a specified location at a given time can be de-
termined from the Heisler charts or one-term solutions. Here we use the charts
to demonstrate their use. Noting that the half-thickness of the plate is L

0.02 m, from Fig. 4–15 we have

1 k 110 W/m C



45.8 T 0  T
∂
Bi hL (120 W/m2 C)(0.02 m)

0.46
at (33.9 10 6
m2
/s)(7 60 s) Ti  T 
t
2

35.6
L (0.02 m)2

Also,

1 k


45.8
Bi hL T  T
∂
0.99
x L T0  T


1
L L
Therefore,
T  T T  T T0  T


0.46 0.99
0.455
Ti  T T0  T Ti  T

and

T
T 0.455(Ti  T)
500 0.455(20  500)
282 C

Therefore, the surface temperature of the plates will be 282 C when they leave
the oven.
Discussion We notice that the Biot number in this case is Bi
1/45.8
0.022,
which is much less than 0.1. Therefore, we expect the lumped system analysis
to be applicable. This is also evident from (T  T)/(T0  T)
0.99, which
indicates that the temperatures at the center and the surface of the plate rela-
tive to the surrounding temperature are within 1 percent of each other. Noting
that the error involved in reading the Heisler charts is typically a few percent,
the lumped system analysis in this case may yield just as accurate results with
less effort.
The heat transfer surface area of the plate is 2A, where A is the face area of
the plate (the plate transfers heat through both of its surfaces), and the volume
of the plate is V
(2L)A, where L is the half-thickness of the plate. The expo-
nent b used in the lumped system analysis is

hAs h(2A) h
b



rcpV rcp (2LA) rcp L
120 W/m2 C


0.00185 s1
(8530 kg/m3)(380 J/kg C)(0.02 m)

Then the temperature of the plate at t
7 min
420 s is determined from

T (t )  T T (t )  500 1

ebt ⎯→
e(0.00185 s )(420 s)
Ti  T 20  500
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CHAPTER 4

It yields

T (t )
279 C

which is practically identical to the result obtained above using the Heisler
charts. Therefore, we can use lumped system analysis with confidence when
the Biot number is sufficiently small.

T = 200°C
EXAMPLE 4–5 Cooling of a Long h = 80 W/ m2 ·°C
Stainless Steel Cylindrical Shaft
Stainless steel
A long 20-cm-diameter cylindrical shaft made of stainless steel 304 comes out shaft
of an oven at a uniform temperature of 600 C (Fig. 4–23). The shaft is then al- D = 20 cm
lowed to cool slowly in an environment chamber at 200 C with an average heat Ti = 600°C
transfer coefficient of h
80 W/m2 · C. Determine the temperature at the
center of the shaft 45 min after the start of the cooling process. Also, deter- FIGURE 4–23
mine the heat transfer per unit length of the shaft during this time period. Schematic for Example 4–5.

SOLUTION A long cylindrical shaft is allowed to cool slowly. The center tem-
perature and the heat transfer per unit length are to be determined.
Assumptions 1 Heat conduction in the shaft is one-dimensional since it is
long and it has thermal symmetry about the centerline. 2 The thermal proper-
ties of the shaft and the heat transfer coefficient are constant. 3 The Fourier
number is t  0.2 so that the one-term approximate solutions are applicable.
Properties The properties of stainless steel 304 at room temperature
are k
14.9 W/m · C, r
7900 kg/m3, cp
477 J/kg · C, and
a
3.95 106 m2/s (Table A–3). More accurate results can be obtained by
using properties at average temperature.
Analysis The temperature within the shaft may vary with the radial distance r
as well as time, and the temperature at a specified location at a given time can
be determined from the Heisler charts. Noting that the radius of the shaft is
ro
0.1 m, from Fig. 4–16 we have
1 k 14.9 W/m C



1.86
T0  T
∂
2
Bi hro (80 W/m C)(0.1 m)

0.40
at (3.95 106 m2/s)(45 60 s) Ti  T
t


1.07
ro2 (0.1 m)2
and
T0
T 0.4(Ti  T)
200 0.4(600  200)
360 C
Therefore, the center temperature of the shaft drops from 600 C to 360 C in
45 min.
To determine the actual heat transfer, we first need to calculate the maxi-
mum heat that can be transferred from the cylinder, which is the sensible
energy of the cylinder relative to its environment. Taking L
1 m,

m
rV
rpro2 L
(7900 kg/m3)p(0.1 m)2(1 m)
248.2 kg

Qmax
mcp(T  Ti)
(248.2 kg)(0.477 kJ/kg · C)(600  200) C

47,350 kJ
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TRANSIENT HEAT CONDUCTION

The dimensionless heat transfer ratio is determined from Fig. 4–16c for a long
cylinder to be
1 1
Bi


0.537
1/Bi 1.86 Q
t
0.62
h 2at Qmax
Bi2t
(0.537)2(1.07)
0.309
k2

Therefore,

Q
0.62Qmax
0.62 (47,350 kJ)
29,360 kJ

which is the total heat transfer from the shaft during the first 45 min of
the cooling.
Alternative solution We could also solve this problem using the one-term
solution relation instead of the transient charts. First we find the Biot number
hro (80 W/m2 C)(0.1 m)
Bi


0.537
k 14.9 W/m C
The coefficients l1 and A1 for a cylinder corresponding to this Bi are deter-
mined from Table 4–2 to be

l1
0.970, A1
1.122
Substituting these values into Eq. 4–27 gives
T0  T 2
u0

A1el1 t
1.122e(0.970) (1.07)
0.41
2

Ti  T

and thus

T0
T 0.41(Ti  T)
200 0.41(600  200)
364 C

The value of J1(l1) for l1
0.970 is determined from Table 4–3 to be
0.430. Then the fractional heat transfer is determined from Eq. 4–34 to be

Q J1(l1) 0.430

1  2u0
12 0.41
0.636
Qmax l1 0.970

and thus
Q
0.636Qmax
0.636 (47,350 kJ)
30,120 kJ
Discussion The slight difference between the two results is due to the reading

error of the charts.

 Plane
surface
T

h x
0 4–3 ■
TRANSIENT HEAT CONDUCTION
 IN SEMI-INFINITE SOLIDS
 A semi-infinite solid is an idealized body that has a single plane surface and
FIGURE 4–24 extends to infinity in all directions, as shown in Figure 4–24. This idealized
Schematic of a semi-infinite body. body is used to indicate that the temperature change in the part of the body in