System with Finite Internal and Surface Resistances
• The previous section dealt with a system in which the internal
resistance was negligible.
• This section deals with the case in which internal and surface
resistances are both significant.
• In this situation, T = T(x,t) for plane wall and T =T(r,t) for
cylindrical and spherical walls.
• In general, the solution of differential equations are complex in
form, often involving series solutions.
• Calculations made from such solutions are presented here in
graphical form (known as Heisler charts and Grober charts)
for three common geometries: the infinite slab or plane wall,
the infinite cylinder, and the sphere.
• We will also consider the solution for a semi-infinite slab.
• Our interest is in predicting the temperature profile with time
and the heat-transfer rate.
System with Finite Internal and Surface Resistances
• Let us first consider a plate of width 2L that extends to infinity
into the page, as shown in the figure.
• Initially, the plate is at a uniform temperature Ti.
• At some time zero, the left and right surfaces are exposed to a
fluid at temperature T∞.
Heisler chart
Heisler chart Grober chart
• Qmax is the initial internal energy of the wall relative to the fluid
temperature T∞. By definition,
Qmax VC (Ti T )
• The parameter Q is the total amount of energy that has passed
through the wall up to any time t of interest.
• The Heisler charts are general solutions for the geometries to
which each applies.
Plane of
Additional Considerations symmetr
y
• A plane of symmetry is treated the same mathematically as an
insulated surface, so the half thickness of the infinite plate
problem (1/2 x 2L = L) equals the actual thickness of the
insulated wall.
Example 4.3
A concrete wall is 24 cm thick and, initially, its temperature is
50°C. One side of the wall is well insulated, while the other side is
exposed to combustion gases at 600°C. The heat transfer
coefficient on this side is 30 W/(m2.K). Determine the time
required for the insulated surface to reach 150°C and the
temperature profile in the wall at that instant. Use the following
properties: k = 1.25 W/(m.K), = 550 kg/m3, C = 890 J/(kg.K).
Solution
• Heat is transferred by convection to one side of the wall and by
conduction through it.
• The concrete wall can be modeled using the infinite plate
solution, making suitable assumptions.
• A plane of symmetry is treated the same mathematically as an
insulated surface, so the characteristic length equals the actual
thickness of the wall in this example.
• The first step in this (or any unsteady) problem is to calculate
the Biot number to determine if a lumped capacitance
equations can be used. For this concrete wall, we have:
V AL
Lc L
As A
where L is thickness of the wall.
hLc 30(0.24)
Bi 5.76
k 1.25
• Because this value is greater than 0.1, the lumped capacity
approach will not give an accurate result.
• The characteristics length in this case is the actual thickness of
the wall
• The reciprocal of the Biot number required to use in the
Heisler chart is
1 1
0.174
Bi 5.67
• The dimensionless temperature is (To - T∞) / ( Ti - T∞) where To
is the temperature at the plate centerline, which, due to
symmetry, is at the surface of the insulated side. We therefore
write
To T 150 600
0..82
Ti T 50 600
• Using these values in the
Heisler chart gives
t
Fo t 2 0.4
L
• Then, we calculate the
time using the Fourier number
L2 Fo
t 0.174
k 1.25
2.55 10 6 m 2 /s
C 550(890)
and the time is
(0.24) 2 (0.4)
t 6
9022 s 2.5 hr
2.55 10
• To determine the temperature profile, we use
Figure 4.6c. At a value of 1/ Bi = 0.174, we
read the following values:
• The wall temperature at the exposed side is 492oC, and the
temperature at the insulated side is 149°C. The temperature
profile is graphed in shown figure.
Infinite Cylinder
• The cylinder must be long enough (at least having a length-to-
radius ratio L/ro ≥ 10).
Heisler chart
• Note that for infinite cylinder the characteristic length used in the
Biot number for the lumped-capacity equation (Lc = ro/2), is
different from the characteristic length used in the Heisler and
Grober charts (Lc = ro).
Infinite Cylinder
Heisler chart Grober chart
• Note that for infinite cylinder the characteristic length used in the
Biot number for the lumped-capacity equation (Lc = ro/2), is
different from the characteristic length used in the Heisler and
Grober charts (Lc = ro).
Sphere Heisler chart
• Note that for sphere the characteristic length used in the Biot
number for the lumped-capacity equation (Lc = ro/3), is different
from the characteristic length used in the Heisler and Grober
charts (Lc = ro).
Sphere
Heisler chart Grober chart
• Note that for sphere the characteristic length used in the Biot
number for the lumped-capacity equation (Lc = ro/3), is different
from the characteristic length used in the Heisler and Grober
charts (Lc = ro).
Transient Heat transfer in Semi-Infinite Solid
• A semi-infinite solid is an idealized body that
has a single surface and extends to infinity in all
directions.
• This idealized body is used to indicate that the
temperature change in the part of the body in
which we are interested (the region close to the
surface) is due to heat conduction at this
surface.
• The earth, for example, can be considered to be a semi-infinite
medium in determining the variation of temperature near its
surface.
• Also, a thick wall can be modeled as a semi-infinite medium if
we are interested in the variation of temperature in the region near
one of the surfaces, and the other surfaces is too far to have any
impact on the region of interest during the time of observation.
Transient Heat transfer in Semi-Infinite Solid
• The heat equation for transient conduction in a semi-
infinite solid is given by
2T 1 T
, T = T(x,t)
x 2
t
• The initial and interior boundary conditions are
Initial condition at t = 0, T = Ti for all x
interior boundary condition at x = ∞ , T = Ti for all t
• Closed-form solutions have been obtained for three important
surface conditions (B.C. 1), instantaneously applied at t ≠ 0.
These conditions include:
1. application of a constant surface temperature Ts = T(0,t) Ti,
2. application of a constant heat flux, q o
3. exposure of the surface to a fluid characterized by T Ti and
the convection coefficient h.
• For case (1), T(0,t) = Ts, The differential equation can be solved
subject to the initial and boundary conditions by a number of
methods. The final solution is
T ( x, t ) Ts 2 x /( 2 t )
)d
2
exp(-
Ti Ts 0
T ( x, t ) Ts 2 x /( 2 t )
)d
2
exp(-
Ti Ts 0
where is a dummy variable. The right-hand side
of this equation is called the error function
denoted as
x
erf
2 t
Then
T ( x , t ) Ts x
erf
Ti Ts 2 t
The heat flux at the surface is found as
k (Ts Ti )
q s (t )
t
• The two other cases; Case (2) is the case where, at
the left boundary, the heat added is a constant and
Case (3) is the case where convection at the
surface is taken into account.
• Resultant temperature profiles for these cases are
also shown in the figures.
• The solutions to these problems involve
exponential function and the complementary
error function (erfc) defined as
x x
erfc 1 - erf
2 t 2 t
• For case (2), q s qo the temperature distribution
and any position, x, and any time, t, is given by
2q o (t / )1 / 2 x 2 q o x x
T ( x, t ) Ti exp erfc
k 4t k 2 t
The complementary error function erfc() = 1 - erf()
T
• For case (3), k h[(T T (0, t )] ,
x x 0
Rather than write the solutions here, results
are presented in graphical form
• When the surface resistance is negligible, case (1),
h and Bi Fo h t / k
Example 4.4
A large block of steel (k = 45 W/m.oC, = 1.4 10-5 m2/s] is initially at a uniform
temperature of 35oC. The surface is exposed to a heat flux (1) by suddenly raising
the surface temperature to 250oC and (2) through a constant surface heat flux of 3.2
105 W/m2. Calculate the temperature at a depth of 2.5 cm after a time of 0.5 min
for both these cases.
Schematic
(1) (2)
Assumptions
1. Semi-infinite solid
2. Constant properties
Solution
We can use the solution for the semi-infinite solid
x 0.025
0.61
2 t 2 1.4 105 (30)
1- For sudden change of surface temperature
h and Bi Fo h t / k
We have Ti = 35oC and Ts = 250 oC,
Approximate Solution, From Fig. 4.10
T ( x, t ) Ti
0.4
Ts Ti
T ( x, t ) Ti (Ts Ti )(0.4)
T (0.025,30) 35 (250 35)(0.4) 121o C
and
x T ( x, t ) Ts 121 250
erf 0.6
2 t Ti Ts 35 250
Exact Solution
From Table of complementary error function (erfc) at
x /( 2 t ) 0.61
x
erfc 0.38835
2 t
x x
erf 1 - erfc 1 0.38835 0.61165
2 t 2 t
T ( x, t ) Ts x
erf
Ti Ts 2 t
x
T ( x, t ) Ts (Ti Ts )erf 250 (35 250)(0.61165) 118.5o C
2 t
(2) For the constant heat flux, we make use of Eq. 4.36
2q o (t / )1 / 2 x 2 q o x x
T ( x, t ) Ti exp erfc
k 4 t k 2 t
(2)(3.2 105 )[(1.4 10 5 )(30) / ]1 / 2 (3.2 105 )(0.025)
T ( x, t ) 35 exp[( 0.61) ] 2
(0.38835)
45 45
79.3 o C
