Massoud M. Engineering Thermofluids: Thermodynamics, Fluid Mechanics, and Heat Transfer
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572 IVd. Heat Transfer: Thermal Radiation
λ
1
T = 0.4 × (5489 + 273) = 2304.8
µ
m·K, from the table we find the fraction for
F( )0
1
λ
→ = 0.12057
λ
2
T = 0.7 × (5489 + 273) = 4033.4
µ
m·K from the table we find the fraction for
F( )0
2
λ
→ = 0.48675
a) Percentage of energy emitted in the shorter than the visible range is about 12%
b) Percentage of energy emitted in the longer than the visible range is about (1 –
0.48675) ≅ 51%
c) Percentage of energy emitted in the visible range = 0.48675 – 0.12057 ≅ 37%
d) The rate of energy is:
21
)(
λλ
→
∆
b
E = (0.48675 – 0.12057) × [5.67E-8 × (5489 +
273)
4
] = 22.88 MW/m
2
The blackbody represents our sun, which emits energy in the infrared, visible and
ultraviolet as shown above.
Table IVd.2.1. Balckbody radiation functions (Incropera)
3. Real Surfaces 573
Example IVd.2.5. A blackbody is at 2000 K. Find the rate of radiant energy
emission in the cone shown in the figure for
ϕ
= 45
o
(
ϕ
=
π
/4 radian) at wave-
length 1 to 5 µm.
ϕ
Solution: The rate of radiant energy emission is given by Equation IVd.1.2. If in-
tegrated in the limits given:
³
λθϕϕϕλθϕϕϕλ
ππ
λ
ππ
λ
λ
λ
λ
dddIdddIdEE
ee
³³³³³ ³
===
5
1
2
0
4/
0
,
5
1
2
0
4/
0
,
])sincos([])sincos([
2
1
where I
λ
,e
is treated as a constant since a blackbody intensity is only a function of
wavelength. Note that:
cos sin (1 / 4) sin(2 ) ( 2 ) (1 / 4)[ cos( 2 )] 0.25(cos 2 cos 2 )dd
α
ββ
β
α
α
ϕϕϕ ϕ ϕ ϕ
α
β
³³
==−=−
λπλπλθπ
λλ
π
λ
dIdIddIE
eee
³³³³
==−=
5
1
,
5
1
,
5
1
2
0
,
5.0]2[25.0})]2/cos(0[cos{25.0
)]10()50([25.0/25.05.005
5
1
,
5
1
,
5
1
,
→−→====
³³³
FFEdEEEdEdIE
bbbbbe
λλλπ
λλλ
To find the fractions, we need the arguments
λ
1
T and
λ
2
T. These arguments are
calculated as:
λ
1
T = 1 × 2000 = 2000
µ
m·K, from Table IVd.2.1 we find the fraction as
F( )0
1
λ
→ = 0.0667
λ
2
T = 5 × 2000 = 10000
µ
m·K from Table IVd.2.1 we find the fraction as
F( )0
2
λ
→ = 0.9142. The fraction is 0.9142 – 0.0667 = 0.8475
E = 0.25 × 5.67E-8 × 2000 = 0.227E6 W/m
2
.
3. Real Surfaces
Although the introduction of blackbody greatly simplified analysis, blackbody re-
mains a mere concept. Real surfaces must be treated differently. For example,
real surfaces in general are not diffuse emitters and their spectral emission does
not fully conform to the Planck distribution. In this section, we discuss the emis-
sion of radiant energy from real surfaces in the context of a new surface property
known as emissivity. We also discuss real surface response to irradiation. The ir-
radiation will be discussed in the context of three additional surface properties
known as reflectivity, absorptivity, and transmissivity.
574 IVd. Heat Transfer: Thermal Radiation
3.1. Characteristics of Real Surfaces, Emissivity
For real surfaces we then define spectral directional emissivity, which implies that
the emissivity from a real surface depends both on direction and on the wave-
length. The spectral directional emissivity,
ε
λ,φ
(
λ
,
θ
,
φ
, T) is defined as the ratio of
the radiation intensity emitted from a real surface having temperature T, at wave-
length
λ
in the direction
θ
and
ϕ
, to the radiation intensity of a blackbody at tem-
perature T and wavelength
λ
:
),(
),,,(
),,,(
,
,
,
TI
TI
T
b
e
λ
ϕθλ
ϕθλε
λ
λ
φλ
=
IVd.3.1
Expectedly, for real surfaces emissivity is always less than unity. Since a black-
body is a diffuse emitter, no directional dependency appears in the denominator of
Equation IVd.3.1. To facilitate analysis, we eliminate directional dependence
from the numerator by finding an average value for
ε
λ
,
φ
over all directions:
³³
³³
³³
³³
==
ππ
ππ
λ
ππ
λ
ππ
λ
λ
θϕϕϕ
θϕϕϕϕθλε
θϕϕϕλ
θϕϕϕϕθλ
λε
2
0
2/
0
2
0
2/
0
,
2
0
2/
0
,
2
0
2/
0
,
]cossin[
]cossin),,,([
]cossin),([
]cossin),,,([
),(
dd
ddT
ddTI
ddTI
T
e
b
e
where we replaced the intensity, ),,,(
,
TI
e
ϕθλ
λ
in the numerator from Equa-
tion IVd.3.1 by substituting for
),(),,,(),,,(
,,,
TITTI
be
λϕθλεϕθλ
λφλλ
= and by
canceling I
λ
,b
(
λ
, T) from both numerator and denominator. For most surfaces,
),,,(
,
T
ϕθλε
ϕλ
is not a strong function of
θ
. Hence, we can reasonably assume
that
),,,(
,
T
ϕθλε
ϕλ
≅ ),,(
,
T
ϕλε
ϕλ
. Making this assumption and carrying out
the integral we get:
³
=
2/
0
,
cossin),,(2),(
π
λλ
ϕϕϕϕλελε
dTT
e
IVd.3.2
We may also define a spectral hemispheric emission as:
),(
),(
),(
,
TE
TE
T
b
λ
λ
λε
λ
λ
λ
=
IVd.3.3
using this definition, we may find the total hemispherical emissive power,
ε
(T) by
using Equations IVd.1.3 :
)(
),(),(
)(
)(
)(
0
,
TE
dTET
TE
TE
T
b
b
b
³
∞
==
λλλε
ε
λλ
IVd.3.4
Emissivity in general varies not only with temperature but also with such surface
conditions as roughness, texture, color, degree of oxidation, and any coating. The
above successive definitions allowed us to express emissivity only in terms of
temperature. Emissivity for various materials and surface coatings are obtained
3. Real Surfaces 575
experimentally as shown in Tables A.IV.8 and A.IV.9 for metallic and nonmetal-
lic surface, respectively.
Example IVd.3.1. The spectral emissivity of a substance is given as 0.1 for 0
≤
λ
≤ 3 µm and 0.9 for
λ
≥ 3 µm. Find the total hemispherical emissivity at 2000 K.
Solution: Substituting numerical values for wavelengths and the corresponding
emissivities, in Equation IVd.3.4, we find:
4
3
,
3
0
,
),(9.0),(1.0
T
dTEdTE
bb
σ
λλλλ
ε
λλ
³³
∞
+
=
By separating terms, we obtain the band emission:
)30(8.09.0)]30(1[9.0)30(1.0
),(9.0),(1.0
4
3
,
3
0
,
→−=→−+→=
+
=
³³
∞
FFF
T
dTEdTE
bb
σ
λλλλ
ε
λλ
For
λ
T = 3 × 2000 = 6000 gives F(0 → 3) = 0.7378
Therefore,
ε
= 0.9 – 0.8 ×0.7378 = 0.31
3.2. Characteristics of Real Surfaces, Absorptivity, Reflectivity,
Transmissivity
Earlier, we defined emissivity with respect to radiant energy emitted by a surface.
Let’s now consider radiant energy being intercepted by a surface. Such surface
may constitute a medium. Semitransparent medium is a generic term for sub-
stances such as water and glass. In general, some of the incident radiation may be
reflected (shown with subscript r), some may be absorbed (shown with subscript
a) in the medium, and some transmitted (shown with subscript t) away from the
medium. If the substance is opaque, then the incident energy is either reflected or
absorbed (Figure IVd.3.1). The absorbed portion increases the internal energy of
the medium. The reflected energy in the visible spectrum would constitute the
color of a substance. The absorption and reflection of radiant energy occur in a
very thin layer of the surface. From a radiation balance we find:
G
λ
,r
+ G
λ
,a
+ G
λ
,t
= G
λ
Dividing through by G
λ
, we find:
ρ
+
α
+
τ
= 1 IVd.3.6
where
ρ
,
α
, and
τ
are the reflectivity, absorptivity, and transmissivity. In Equa-
tion IVd.3.6, average values are used for these parameters. In general however,
ρ
,
α
, and
τ
are functions of wavelength and direction.
576 IVd. Heat Transfer: Thermal Radiation
I
r
r
a
d
i
a
t
i
o
n
R
e
f
l
e
c
t
i
o
n
Absorption
T
r
a
n
s
m
i
s
s
i
o
n
Semitransparent
Medium
G
λ
G
λ
,r
G
λ
,t
G
λ
,a
I
r
r
a
d
i
a
t
i
o
n
R
e
f
l
e
c
t
i
o
n
Absorption
Opaque
Medium
G
λ
G
λ
,r
G
λ
,a
Figure IVd.3.1. Comparison of semitransparent and opaque surfaces
Absorptivity,
α
λ
,
φ
(
λ
,
θ
,
φ
) determines the fraction of the incident energy ab-
sorbed by the surface:
),,(
),,(
),,(
,
,,
,
ϕθλ
ϕθλ
ϕθλα
λ
λ
φλ
i
ai
I
I
=
IVd.3.7
To find the average surface absorptivity we substitute from Equation IVd.1.5 and
integrate Equation IVd.3.7:
λθϕϕϕϕθλ
λθϕϕϕϕθλϕθλα
α
ππ
λ
ππ
λϕλ
dddI
dddI
G
G
i
ai
a
}]cossin),,([{
}]cossin),,(),,([{
0
2
0
2/
0
,
0
2
0
2/
0
,,,
³³ ³
³³ ³
∞
∞
== IVd.3.8
where
α
in Equation IVd.3.8 is known as the total hemispherical absorptivity.
Reflectivity,
ρ
λ
,
φ
(
λ
,
θ
,
φ
) determines the fraction of the incident energy re-
flected by the surface:
),,(
),,(
),,(
,
,,
,
ϕθλ
ϕθλ
ϕθλρ
λ
λ
φλ
i
ri
I
I
=
IVd.3.9
To find the average surface reflectivity we substitute from Equation IVd.1.5 and
integrate Equation IVd.3.9:
λθϕϕϕϕθλ
λθϕϕϕϕθλϕθλρ
ρ
ππ
λ
ππ
λ
dddI
dddI
G
G
i
ri
r
}]cossin),,([{
}]cossin),,(),,([{
0
2
0
2/
0
,
0
2
0
2/
0
,,
³³ ³
³³ ³
∞
∞
==
where
ρ
in Equation IVd.3.10 is known as the total hemispherical reflectivity.
Shown in Figure IVd.3.2. are two types of surface reflection. While polished sur-
faces are specular, most surfaces are diffuse reflectors.
3. Real Surfaces 577
Incident
Ray
Specular
Reflection
ϕϕ
Polished Surface
Incident
Ray
Diffuse
Reflection
Rough Surface
Figure IVd.3.2. Incident radiation on rough and polished surfaces
Transmissivity,
τ
λ
,
ϕ
(
λ
,
θ
,
ϕ
) is the fraction of the incident energy transmitted.
If averaged,
τ
= G
λ
,t
/G.
Example IVd.3.2. An opaque surface is exposed to radiation. Use the spectral
hemispherical absorptivity and irradiation profiles given below to find
α
.
0.0
0.2
α
λ
(
λ
)
0.4
0.6
0.8
1.0
0246 8
10 12 14
16
λ
(
µ
m)
G
λ
(
λ
) kW/m
2
.
µ
m
0
2468
10 12 14
16
λ
(
µ
m)
0.0
0.2
0.4
0.6
0.8
1.0
Solution: To find total absorptivity, we carryout the integrals in Equation
IVd.3.8. The numerator becomes:
λθϕϕϕϕθλϕθλα
ππ
λφλ
dddI
ai
}]cossin),,(),,([{
0
2
0
2/
0
,,,
³³ ³
∞
=
λπλλα
λλ
dI
ai
})]2/1(2)[()({
0
,,
³
∞
and denominator:
λθϕϕϕϕθλ
ππ
λ
dddI
i
}]cossin),,([{
0
2
0
2/
0
,
³³ ³
∞
=
λπλ
λ
dI
ai
})]2/1(2)[({
0
,,
³
∞
Using the definition of absorptivity, Equation IVd.3.8:
³³
∞∞
=
00
)(/)()(
λλλλλαα
λλλ
dGdG
To carryout the numerator and denominator integrals we need to express
α
and G
as functions of
λ
:
For 0
µ
m <
λ
< 2
µ
m:
α
= 0.2 G
λ
= 0 kW/m
2
·
µ
m
For 2
µ
m <
λ
< 4
µ
m:
α
= 0.4 G
λ
= 0 kW/m
2
·
µ
m
578 IVd. Heat Transfer: Thermal Radiation
For 4
µ
m <
λ
< 12
µ
m:
α
= 0.9 G
λ
= (
λ
– 4)/10 kW/m
2
·
µ
m
For 12
µ
m <
λ
:
α
= 0.9 G
λ
= 0.8 kW/m
2
·
µ
m
We now substitute for G
λ
to be able to integrate
Example IVd.3.3. The surface temperature and total hemispherical emissivity in
Example IVd.3.2 are 327 C and 0.85, respectively. Find the surface temperature
after exposure.
Control Volume
Adiabatic Boundary
C
o
n
v
e
c
t
i
o
n
E
m
i
s
s
i
o
n
I
r
r
a
d
i
a
t
i
o
n
R
e
f
l
e
c
t
i
o
n
E
G
G
r
h, T
f
Solution: We find the surface temperature from an energy balance for the control
volume shown in the figure. In steady state, the net energy gain (loss) from the
surface is equal to the rate of energy received by the surface from irradiation (G)
minus the rate of energy removed from the surface by reflection, surface emission,
heat convection, and heat condition in the surface. If the back surface is insulated
and the whole surface can be assumed to have one temperature then there is no
conduction heat transfer:
)(
Cr
qEGGq
′′
++−=
′′
If we ignore the heat loss due to convection and substitute for G
r
=
ρ
G = (1 –
α
)G
and for E =
ε
E
b
we get the net heat gain (loss) as =
′′
q
α
G –
ε
E
b
.
Substituting for G, calculated in Example IVd.3.2 as
³
∞
=
0
)(
λλ
λ
dGG = 7800
W/m
2
, we find:
=
′′
q
α
G –
ε
(
σ
T
4
) = 0.728 × 7800 – 0.85 × [5.67E-8 × (327 + 273)
4
] = –568
W/m
2
.
4. Gray Surfaces
So far we defined four surface properties: emissivity, absorptivity, reflectivity, and
transmissivity. If we obtain one more relation, we can then find any two proper-
ties if the other two properties are given. The additional relation is between emis-
sivity and reflectivity. Kirchhoff in 1860 demonstrated that if a surface is encom-
passed by a blackbody at temperature T
s
, Figure IVd.4.1(a), at thermal equilibrium
(steady state), the energy received by the surface (GA
1
α
1
where
α
1
is the surface
absorptivity and A
1
is the surface area) must be equal to the energy emitted by the
5. Radiation Exchange Between Surfaces 579
G
T
A
T
s
Blackbody
λ
E
λ
Graybody
Real Surface
T: Constant
Blackbody
φ
ε
φ
Graybody
Real Surface
T: Constant
0
30 60 90
0.5 1.0
Coated
Bare
(a) (b) (c)
Figure IVd.4.1. (a) Emission, (b) total directional emissivity of ideal, gray, and real sur-
faces, and (c) emissivity for various surfaces
surface (E
1
A
1
), GA
1
α
1
= E
1
A
1
. Resulting in E
1
/
α
1
= G. If we now replace surface
1 with surface 2 and let the steady state condition to prevail, we again can write
GA
2
α
2
= E
2
A
2
or E
2
/
α
2
= G. We may repeat this experiment for many surfaces. If
we finally place a blackbody in the enclosure and let it also comes to thermal equi-
librium with the enclosure, we may write E
b
/1 = G. As a result:
E
1
/
α
1
= E
2
/
α
2
= ···· = E
b
Using Equation IVd.3.3, we conclude that
ε
λ
,
φ
(T
s
) =
α
λ
,
φ
(T
s
). In general, we want
to know the circumstance under which we can have
ε
λ
=
α
λ
. It turns out that such
condition exists if either surface or irradiation is diffuse. Condition
ε
=
α
exists, if
the irradiation corresponds to emission from a blackbody or if the surface is gray.
That is to say that a gray surface is an idealized material, having constant emissiv-
ity independent of wavelength. A diffuse gray surface is a surface that has emis-
sivity,
ε
λ,φ
and absorptivity
α
λ,φ
independent of direction (due to diffuse assump-
tion) and wavelength (due to the gray surface assumption). A comparison is made
between a blackbody, a gray surface and a real surface in Figure IVd.4.1(b). Fig-
ure IVd.4.1(c) shows the emissivity for the these surfaces.
5. Radiation Exchange Between Surfaces
So far we dealt with single surfaces emitting radiation and being irradiated. We
also considered surfaces being contained in an enclosure. In general however, sur-
faces exchange radiation with other surfaces with arbitrary geometry and orienta-
tion. This complicates the calculation of the total heat transfer, as we should con-
sider the surface geometry and orientation. To properly account for the amount of
energy exchanged between surfaces, we introduce a parameter known as shape or
view factor, F < 1. In this section we discuss means of calculating the view factor.
We also introduce the concept of radiation exchange between surfaces by a net-
work, which is series – parallel arrangement of the involved surfaces.
580 IVd. Heat Transfer: Thermal Radiation
5.1. The View Factor
Consider two differential surfaces A
i
and A
j
oriented arbitrarily and exchanging
radiation, as shown in Figure IVd.5.1. The view factor F
ij
is defined as the ratio of
the radiant energy leaving surface dA
i
and reaching surface dA
j
to the total radiant
energy leaving surface dA
i
. If the distance between these elemental surfaces is R
and the angle between R and
j
n
K
is
ϕ
j
, then the projected area of dA
i
, normal to R
is dA
i
cos
ϕ
i
.
n
i
n
j
A
j
, T
j
A
i
, T
i
ϕ
j
R
dA
j
dA
i
ϕ
i
n
i
dA
i
dΩ
j-i
dA
j
cos
ϕ
j
Figure IVd.5.1. Exchange of radiation for derivation of the view factor
The amount of radiant energy leaving differential area dA
i
per unit time and
unit area is the intensity of radiation, I
i
of surface dA
i
. Hence, the total radiant en-
ergy leaving the elemental area dA
i
per unit time is I
i
dA
i
. The fraction of this en-
ergy radiated in the direction of dA
j
is given by (I
i
dA
i
)cos
ϕ
i
. The amount of en-
ergy leaving dA
i
in the direction of dA
j
and received by dA
j
is given by (I
i
dA
i
)cos
φ
i
dΩ
j-i
where dΩ
j-i
is the solid angle subtended by dA
j
when viewed from
dA
i
. Since dΩ
j-i
= (dA
j
cos
ϕ
j
)/R
2
then the fraction of energy leaving dA
i
and reach-
ing dA
j
per unit time is obtained from (I
i
dA
i
)cos
ϕ
i
(dA
j
cos
ϕ
j
)/R
2
. This argument is
summarized as follows:
Rate of energy leaving dA
i
per s, per cm
2
, and per sr: I
i
Rate of energy leaving the elemental area dA
i
per s and per sr: I
i
dA
i
.
Rate of energy radiated in the direction of dA
j
per s and per sr:
(I
i
dA
i
)cos
φ
i
Rate of energy leaving dA
i
and received by dA
j
per s:
(I
i
dA
i
)cos
φ
i
dΩ
j-i
Rate of energy leaving dA
i
and received by dA
j
per s:
(I
i
dA
i
)cos
φ
i
(dA
j
cos
φ
j
)/R
2
Rate of energy leaving dA
i
and received by dA
j
per s:
(J
i
dA
i
)cos
φ
i
(dA
j
cos
φ
j
)/
π
R
2
where in the last expression we have assumed that the surface emits and reflects
diffusely and substituted for I
e+r
= J/
π
, from the definition of total radiosity. This
now represents hemispheric emission. Hence, the total rate of radiant energy leav-
ing surface i and intercepted by surface j is:
2
[cos cos / ]
ij ij
ij i
Ai Aj
QJ RdAdA
ϕϕπ
=
³³
IVd.5.1
Expressing
ij
Q
= F
ij
A
i
J
i
, by comparing to Equation IVd.5.1, we conclude that:
5. Radiation Exchange Between Surfaces 581
2
[cos cos / ]
ij i i j i j
Ai Aj
F
A R dAdA
ϕϕπ
=
³³
IVd.5.2
Similar arguments can be made for the radiant energy emitted from surface A
j
and
intercepted by surface A
i
. Since the net energy exchange between these surfaces is
the same we conclude that F
ij
A
i
= F
ji
A
j
. This is known as the reciprocity relation.
Note that in this derivation we assumed that the surrounding is not participating in
the radiation exchange. Having the view factors, the net rate of heat transfer by
radiation is:
iji
ji
ij
FA
JJ
Q
/1
−
=
IVd.5.3
where Equation IVd.5.3 is written similar to Equation IV3.6 for the reasons dis-
cussed in Section 5.4.
Example IVd.5.1. A differential area dA
1
is located parallel to and on the center-
line of a finite disk of diameter D (radius of c = D/2). Find the view factor for this
arrangement.
dA
1
R
ϕ
ϕ
D
r
dA
2
dr
H
Solution: Since surface dA
1
is a differential surface, we choose the shaded area at
an arbitrary 0 < r < c as the elemental area of surface A
2
. Clearly, if dA
2
= 2
π
rdr is
integrated in the above range, we would obtain A
2
=
π
D
2
/4. In this problem, cos
ϕ
1
= cos
ϕ
2
= r/R = H/(r
2
+ H
2
)
1/2
. Substituting values in Equation IVd.5.2:
µ
¶
´
µ
¶
´
=
i
j
A
A
ji
ji
dAdA
R
dAF ]
coscos
[
2
112
π
ϕϕ
=
µ
¶
´
+
+
c
rdr
rH
rHH
dA
0
22
222
1
)2(
)(
)/(
π
π
=
c
2
/(H
2
+ c
2
) dA
1
Hence, F
12
= c
2
/(H
2
+ c
2
).
Hamilton and Howell carried out the double integral of Equation IVd.5.2 for
variety of surface orientations as shown in Figures IV5.5.2(a) through (d). In
Equation IVd.5.2, F
ij
is known as the shape, configuration, or view factor. Ta-
ble IVd.5.1 includes the equations from which the above plots are obtained. These
are also available on the accompanying CD-ROM. The user specifies dimensions
and the program finds the F factor.