72
Analytical Heat Transfer
The above temperature decay solution is plotted in Figure 4.3. Therefore, the
solid temperature can be predicted with time for a given material with certain
geometry under the cooling or heating condition. The material with a smaller
thermal time constant (τ t = (ρVc/hA s )) can quickly reach the environment
temperature.
Now, the question is under what condition the lumped capacitance solution can be used. The answer is that the Biot (Bi) number must be less than
0.1. Therefore, to use 0-D solution, the condition Bi = (hL c /k) < 0.1 must be
satisfied. The Biot number is defined as the ratio of surface convection (h, the
convection heat transfer coefficient from the solid surface; L c , the characteristic
length of the solid material) to solid conduction (k, the thermal conductivity
of the solid material). The smaller Bi number implies a small-sized material with high-conductivity exposure to a low convection cooling or heating
fluid. For the case of smaller Bi, the temperature inside the solid material
changes uniformly (independent of location) with environmental cooling or
heating during the transient. Of course, Bi = 0.1 implies that we may have
10% error by using the lumped solution. The smaller Bi is better for using
the lumped solution. Another point is that the solution can be applied to any
geometry if the condition of Bi < 0.1 is valid. For a given solid geometry,
the characteristic length L c = (volume/surface area) = (V/A s ). For example,
as shown in Figure 4.1, the characteristic length L c = L is for a 2L-thick slab
(plane wall),
1
1
L c = R o for the cylinder and L c = R o for the sphere coordinate.
2
3
4.1.1 Radiation Effect
If we also consider radiation flux q ""
r and internal heat generation q ˙, the energy
balance equation 4.3 can be rewritten as
d(ρVCT) = −hA
""
s (T − T ∞ ) + qV ˙ + A s q
dt
r
where q ""
r
4
= radiation gain from solar flux constant, or q ""
r
radiation loss
T − T 4 ; q ˙ = heat generation = I 2
εσ(
)
=
=
=
sur
R due to electric current and resistance
heating = constant.
If we consider q ""
r = constant and q ˙ = constant, the solution of the above
equation can be obtained by Equation 4.4 by setting
�
� qV + A ""
s q
θ = (T − T
r
)
˙
∞ −
hA s
��
Analytical Heat Transfer
The above temperature decay solution is plotted in Figure 4.3. Therefore, the
solid temperature can be predicted with time for a given material with certain
geometry under the cooling or heating condition. The material with a smaller
thermal time constant (τ t = (ρVc/hA s )) can quickly reach the environment
temperature.
Now, the question is under what condition the lumped capacitance solution can be used. The answer is that the Biot (Bi) number must be less than
0.1. Therefore, to use 0-D solution, the condition Bi = (hL c /k) < 0.1 must be
satisfied. The Biot number is defined as the ratio of surface convection (h, the
convection heat transfer coefficient from the solid surface; L c , the characteristic
length of the solid material) to solid conduction (k, the thermal conductivity
of the solid material). The smaller Bi number implies a small-sized material with high-conductivity exposure to a low convection cooling or heating
fluid. For the case of smaller Bi, the temperature inside the solid material
changes uniformly (independent of location) with environmental cooling or
heating during the transient. Of course, Bi = 0.1 implies that we may have
10% error by using the lumped solution. The smaller Bi is better for using
the lumped solution. Another point is that the solution can be applied to any
geometry if the condition of Bi < 0.1 is valid. For a given solid geometry,
the characteristic length L c = (volume/surface area) = (V/A s ). For example,
as shown in Figure 4.1, the characteristic length L c = L is for a 2L-thick slab
(plane wall),
1
1
L c = R o for the cylinder and L c = R o for the sphere coordinate.
2
3
4.1.1 Radiation Effect
If we also consider radiation flux q ""
r and internal heat generation q ˙, the energy
balance equation 4.3 can be rewritten as
d(ρVCT) = −hA
""
s (T − T ∞ ) + qV ˙ + A s q
dt
r
where q ""
r
4
= radiation gain from solar flux constant, or q ""
r
radiation loss
T − T 4 ; q ˙ = heat generation = I 2
εσ(
)
=
=
=
sur
R due to electric current and resistance
heating = constant.
If we consider q ""
r = constant and q ˙ = constant, the solution of the above
equation can be obtained by Equation 4.4 by setting
�
� qV + A ""
s q
θ = (T − T
r
)
˙
∞ −
hA s
��