Transcript Document

2.7 Radiation
2.7.1 Definitions and laws
Heat transfer by conduction and convection required the existence of a material
medium, either a solid or a fluid. However, it is not required in heat transfer by
radiation.
Radiation can travel through an empty space at the speed of light in the form of
an electromagnetic wave. As shown in the electromagnetic spectrum in Fig. 2.7.1,
thermal radiation covers the range of wavelength from 0.1~100 μm.
2.7.1.1 Absorptivity
Thermal radiation impinging on the surface of an opaque solid is either absorbed
or reflected. The absorptivity is defined as the fraction of the incident radiation that
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is absorbed:
q(a)
  (i )
q
[2.7.1]
Where q(a) is the energy absorbed per unit area per unit time and q(i) is the
energy impinging per unit area per unit time.
Let us define ql(a) and ql(i) such that ql(a)dl and ql(i)dl represent respectively
the absorbed and incident energies per unit area per unit time in the wavelength
range l to l+dl. The monochromatic absorptivity l is defined as :
ql dl ql
 l  (i )  (i )
ql dl ql
(a)
(a)
[2.7.2]
For any real body αλ< 1 and depend on l. A graybody is a hypothetical
one for which αλ < 1 but independent of l and temperature. The limiting case
of αλ = 1 for all l. and temperature is known as a blackbody. In other words,
a blackbody absorbs all the incident radiation.
2.7.1.2 Emissivity
q (e)
The emissivity of a surface is defined as   ( e )
qb
[2.7.3]
Where q(e) and qb(e) are the energies emitted per unit area per unit time by a
real body and a blackbody, respectively. Table 2.7-1 lists the emissivities of some
Material. These averaged value have been used widely even though the emissivity
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actually depends on the wavelength and the angle of emission.
Let us define ql(e) and qbl(e) such that ql(e)dl and qbl(e)dl represent respectively
the energies emitted per unit area per unit time in the wavelength range l to l+dl
of a real body and blackbody. The monochromatic emissivity l is defined as :
ql dl
ql
 l  (e)  (e)
qbl dl qbl
(e)
(e)
[2.7.4]
l=1 for blackbody and <1 for a real body.
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2.7.1.3 Kirchhoff’s law
Consider the body enclosed in the cavity shown in Fig. 2.7.2.
Suppose the body is at the same temperature as the wall of the
cavity, that is, the two are at thermodynamic equilibrium with each
other. Since there is no net heat transfer from the cavity wall to The
body, the energy emitted by the body must be equal to the energy absorbed.
[2.7.5]
q(e) A  q(i ) A
Where q(e) is the energy emitted per unit area of the body per unit time, A the surface
area of the body, q(i) the energy impinging per unit area of the body per unit time, and
a the absorptivity of the body. If the body is a blackbody, Eq.[2.7.5] becomes
qb A  q (i ) A
(e)
[2.7.6]
q (e)
Dividing Eq. [2.7.5] by Eq.[2.7.6], we get

(e)
qb
Substituting Eq. [2.7.7] into Eq.[2.7.3], we have
 
[2.7.7]
[2.7.8]
This is Kirchhoff’s law, which states that for a system in thermodynamic equilibrium
the emissivity and absorptivity are the same. Following a similar procedure, we can
show that
 l  l
[2.7.9] 48
2.7.1.4 Plank’s distribution law
Plank derived the following equation for the energy emitted by a blackbody as a
function of the wavelength and temperature:
[2.7.10]
Where
T = temperature in K
h = Plank’s constant
c = speed of light
k = Boltzmann’s constant
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2.7.1.5 Stefan-Boltzmann law
The Plank distribution law can be integrated over wavelengths from zero to infinity
to determine the total emissive energy of a blackbody:
[2.7.11]
or
qb
( e)
 T 4
[2.7.12]
Where  (Stefan-Boltzmann constant) = 5.676 x 10-8 Wm-2K-4.
2.7.1.6 The solid angle
Consider a hemisphere of radius
r surrounding a differential area
dA1at the center. Fig. 2.7-4a shows
only one-quarter of the hemisphere.
On the hemisphere
dA2  (a b )(c d )
 (r sin d )(rd )
[2.7.13]
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The solid angle that intersects dA2 on the hemisphere, shown in Fig.2.7-4b,
is defined as:
dA2
d  2  sin dd
r
[2.7.14]
Fig. 2.7-5 shows the emission from a
differential surface area dA1. Consider
the direction that is at an angle θ from
the normal to the surface. The projection
of dA1 along the direction is cosθdA1. Let
I be the energy emitted per unit projected
area per unit time per unit solid angle:
I
so that
dQ
(dA1 cos )d
q (e) 
dQ
  I cosd
dA1
where Q is the radiation heat transfer rate.
[2.7.15]
[2.7.16]
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Substituting Eq. [2.7.14] into Eq.[2.7.16] and integrating over the entire hemisphere
q
(e)
 2
 I
 0
  / 2

0
cos sin dd
[2.7-17]
This yields the following equation:
q (e)  I
[2.7-18]
Substituting Eq. [2.7-18] into Eq. [2.7-15], we have
d Q
q (e)

cos dA1 d 
[2.7-19]
2.7.2 Radiation between blackbodies
Consider two black surface A1 and A2
shown in Fig. 2.7-6. Define the view factors:
F12= fraction of energy leaving surface 1
that reaches surface 2
F21= fraction of energy leaving surface 2
that reaches surface 1
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As such, the energy leaving surface 1 that reaches surface 2 per unit time is
(e)
Q12  qb1 A1F12
[2.7-20]
Similarly, the energy leaving surface 2 that reaches surface 1 per unit time is
(e)
Q21  qb 2 A2 F21
[2.7-21]
Therefore, the net energy exchange rate is as follows:
(e)
(e)
Q12  Q12  Q21  qb1 A1F12  qb 2 A2 F21
[2.7-22]
If both surfaces are at the same temperature T, there is no net energy exchange
and Q12=0. Then substituting Eq.[2.7-12] into Eq. [2.7-22], we have
Q12  T 4  A1F12  A2 F21   0
[2.7-23]
so that
A1F12  A2 F21
[2.7-24]
Which is called the reciprocity relationship. Substituting Eqs. [2.7-12] and [2.7-24]
into [2.7-22], we get
Q12  A1 F12 T14  T2 4   A2 F21 T14  T2 4 
[2.7-25]
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Let us now proceed to determine the view factors F12 and F21 between the
two black surfaces. From Eq. [2.7-19]
(e)
q
[2.7-26]
d Q12  b1 cos 1dA1d 12

where dΩ12 is the solid angle subtended by dA2 as seen from dA1. Let r12 be the
distance between the centers of dA1 and dA2. Consider the hemisphere of radius
r12 and centered at dA1. The projection of dA2 on the hemisphere is cos2dA2.
Therefore, from Eq.[2.7-14]
d 12 
cos  2 dA2
r12 2
[2.7-27]
Substituting Eq.[2.7-27] into Eq. [2.7-26]
cos 1 cos 2 dA1dA2
 r12 2
[2.7-28]
cos 1 cos 2
dA1dA2
 r12 2
[2.7-29]
(e)
d Q12  qb1
And so
(e)
Q12  qb1 
Similarly it can be shown that
(e)
Q21  qb 2 
cos 1 cos 2
dA1dA2
 r12 2
[2.7-30]
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Substituting Eq. [2.7-29] and Eq. [2.7-30] respectively into Eqs. [2.7-20] and [2.7-21],
We get Eqs. [2.7-31] and [2.7-32]
F12 
1 cos 1 cos  2
dA1dA2
A1 
 r12 2
[2.7-31]
and
F21 
1
A2
cos 1 cos  2
  r122 dA1dA2
[2.7-32]
The integration in Eq. [2.7-31] and [2.7-32] is often difficult and needs to be done
numerically. Analytical equations are available for a number od special cases. Some
examples are given below:
For large (infinite) parallel plates, long (infinite) concentric cylinders and concentric
spheres F12 = 1, as shown in Fig.2.7-7.
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For the two identical, parallel directly opposed rectangles shown in Fig. 2.7-8.
  1  X 2 1  Y 2 1 2



2  
X
Y
2
1
2
1
1
1 
  X 1  Y tan
F12 
 Y 1  X tan
 X tan X  Y tan Y 
ln 
2
2
 XY   1  X 2  Y 2 
1

Y
1

X


 

[2.7-33]
where X and Y are defined in the figure.
For the two parallel concentric circular disks shown in Fig. 2.7-9,
2



R
1
F12   X  X 2  4  2  
2
 R1  

where X, R1, and R2 are defined in the figure.
[2.7-34]
For the sphere and the disk shown in Fig. 2.7-10,
1
1 

F12  1 
2
2
1  R2 

[2.7-35]
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Example
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2.7.3 Radiation between graybodies
In radiation heat transfer between blackbodies, all the radiant energy that strikes
a surface is absorbed. In radiation heat transfer between nonblackbodies, all the
energy striking a surface will not be absorbed; part will be reflected. Since the
energy emitted and the energy reflected by a nonblack surface both contribute to
the total energy leaving the surface (J), we can write
J  q(e)  rq(i )   qb(e)  rq(i )
[2.7-36]
Where J is called radiosity, is the total energy leaving a surface per unit area per
unit time and r, called the reflectivity, is the fraction of the incident energy reflected.
From the definition for J, we see that
J - q(i) = net energy leaving a surface per unit area per unit time
and
q(i) - J= net energy received at a surface per unit area per unit time
For an opaque material the incident energy is either reflected or absorbed:
r +  =1
[2.7-37] 60
If Kirchhoff’s law can be applied, that is,  =  according to Eq. [2.7-8],
Eq.[2.7-37] becomes
r+=1
[2.7-38]
From Eqs. [2.7-36] and [2.7-38]
J  q(i )  J 
J   qb( e)
r
(1   ) J  J   qb( e) qb( e)  J


1 
(1   1)
[2.7-39]
Let us consider radiation heat transfer between two gray surface A1 and A2. Since
the net energy transfer from A1 to A2 (Q12) equals either the net energy leaving A1
or the net energy received at A2, we can write, with the help of Eq. [2.7-39]:
and
(e)


(i )
q

J


Q12  A1  J1  q1   A1  b1 1 
 1 1   1 




(e)


J

q
Q12  A2  q2 (i )  J 2   A2  2 b 2 
 1  2   1 


[2.7-40]
[2.7-41]
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Similar to Eqs. [2.7-20] and [2.7-21] for two black surfaces, we can write the
following expressions for two gray surfaces in terms of the view factors:
Q12  J1 A1F12
[2.7-42]
Q21  J 2 A2 F21
[2.7-43]
and
Since A1F12 =A2F21 according to Eq. [2.7-24]
Q12  Q12  Q21  A1F12 ( J1  J 2 )
[2.7-44]
From Eqs. [2.7-40], [2.7-44], and [2.7-41]
 1 1   1 
qb1  J1  Q12 

A
1


 1 
J1  J 2  Q12 

A
F
 1 12 
(e)
 1  2   1 
J 2  qb 2  Q12 

A
2


[2.7-45]
[2.7-46]
(e)
[2.7-47]
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Adding these equations, we have
 1 1   1
1  2   1 
1
qb1  qb 2  Q12 



A
A
F
A
1
1 12
2


(e)
(e)
[2.7-48]
Substituting Eq. [2.7-12] and rearranging, we obtain
Q12  A1 f12 T14  T2 4   A1 f 21 T14  T2 4 
where
1 1   1  1  1  2  1
1
1


A1 f12 A2 f 21
A1
A1F12
A2
[2.7-49]
[2.7-50]
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Example 2.7-3 Radiation shields
Given:
Hot surface T1, 1, cold surface T2, 2, radiation shield T3, 3
Surfaces 1 & 2: two infinite parallel gray surface with the area of A
Find: The radiation heat transfer between two parallel gray surfaces
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Example: Two very large parallel planes with emissivities 0.3 and 0.8 exchange heat.
Find the percentage reduction in heat transfer when a polished-aluminum radiation
shield (=0.04) is placed between them.
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