Post on 14-Apr-2018
7/27/2019 12B-Chapter 12, Sec 12-1.4 - 12.7 Black
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Radiation: Processes and Properties
Surface Radiative Properties
Chapter 12
Sections 12.4 through 12.7
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Emissivity
Surface Emissivity
Radiation emitted by a surface may be determined by introducing a property
(the emissivity) that contrasts its emission with the ideal behavior of a blackbody
at the same temperature.
The definition of the emissivity depends upon ones interest in resolving
directional and/or spectral features of the emitted radiation, in contrast
to averages over all directions (hemispherical and/or wavelengths (total).
The spectral, directional emissivity:
,
,
,
, , ,, , ,
,
e
b
I TT
I T
The spectral, hemispherical emissivity (a directional average):
2 2
0 0
2 2
0 0
/
,
/
, ,
, , , cos sin,,
, , cos sin
e
b b
I T d dE TT
E T I T d d
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Emissivity (cont)
The total, hemispherical emissivity (a directional and spectral average):
0 b
b b
T E T d E TT
E T E T
,, , ,
To a reasonable approximation, the hemispherical emissivity is equal to
the normal emissivity.
n
Representative values of the total, normal emissivity:
Note:
Low emissivity of polished metals and increasing emissivity for unpolished
and oxidized surfaces.
Comparatively large emissivities of nonconductors.
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Emissivity (cont)
Representative spectral variations:
Note decreasing with increasing for metals and different behavior for nonmetals.,n
Representative temperature variations:
nWhy does increase with increasing for tungsten and not for aluminum oxide?
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Abs, Ref & Trans
Response to Surface Irradiation: Absorption, Reflection
and Transmission
There may be three responses of a semitransparentmedium to irradiation:
Reflection from the medium refG, .
Absorption within the medium absG, .
Transmission through the medium trG, .
Radiation balance
ref abs tr G G G G , , ,
In contrast to the foregoing volumetric effects, the response of an opaque material
to irradiation is governed by surface phenomena and 0, .trG
, ,ref tr G G G
The wavelength of the incident radiation, as well as the nature of the material,
determine whether the material is semitransparent or opaque.
Are glass and water semitransparent or opaque?
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Abs, Ref & Trans (cont)
Unless an opaque material is at a sufficiently high temperature to emit visible
radiation, its coloris determined by the spectral dependence of reflection in
response to visible irradiation.
What may be said about reflection for a white surface? Ablacksurface?
Why are leaves green?
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Absorptivity
Absorptivity of an Opaque Material The spectral, directional absorptivity:
, ,
,
,
, ,, ,
, ,
i abs
i
I
I
The spectral, hemispherical absorptivity:
2 2
0 0
2 2
0 0
/
, , ,
/
,
, , , , cos sin
, , cos sin
abs i
i
G I d d
G I d d
To what does the foregoing result simplify, if the irradiation is diffuse?
If the surface is diffuse?
The total, hemispherical absorptivity:
0
oabsG dG
G G d
If the irradiation corresponds to emission from a blackbody, how may the
above expression be rewritten?
Assuming negligible temperature dependence,
The absorptivity is approximately independent of the surface temperature,
but if the irradiation corresponds to emission from a blackbody, why does
depend on the temperature of the blackbody?
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Reflectivity
Reflectivity of an Opaque Material
The spectral, directional reflectivity: Assuming negligible temperature
dependence:
, ,
,
,
, ,, ,, ,
i ref
i
II
The spectral, hemispherical reflectivity:
22
0 0
/
, , ,
,
, , , , cos sin
, ,
ref i
i
G I d d
G I
To what does the foregoing result simplify if the irradiation is diffuse?
If the surface is diffuse?
The total, hemispherical reflectivity:
0
0
abs G dGG G d
Limiting conditions of diffuse and
spectral reflection. Polished and rough surfaces.
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Reflectivity (cont)
Note strong dependence of and 1 on .
Is snow a highly reflective substance? White paint?
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TransmissivityTransmissivity
The spectral, hemispherical transmissivity: Assuming negligible temperature
dependence,
,trG
G
Note shift from semitransparent to opaque conditions at large and small wavelengths. The total, hemispherical transmissivity:
0
0
trtrG dG
G G d
,
For a semitransparent medium,
11
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Kirchhoffs Law
Kirchhoffs law equates the total, hemispherical emissivity of a surface to its
total, hemispherical absorptivity:
Kirchhoffs Law
However, conditions associated with its derivation are highly restrictive:
Irradiation of the surface corresponds to emission from a blackbody at the
same temperature as the surface.
However, Kirchhoffs law may be applied to the spectral, directional properties
without restriction:
, ,
Why are there no restrictions on use of the foregoing equation?
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Diffuse/Gray Surfaces
Diffuse/Gray Surfaces With 2 2
0 0
2 2
0 0
/
,
/
cos sin
cos sin
d d
d d
and 2 20 0
2 2
0 0
/
, ,
/
,
cos sin
cos sin
i
i
I d d
I d d
Under what conditions may we equate to ?
With
0 b
b
E d
E T
,
and 0 G d
G
Under what conditions may we equate to ?
Conditions associated with
assuming a gray surface:
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Problem: Surface Emissivity and Absorptivity
Problem 12.49: Determination of the solar absorptivity and total emissivity
of a diffuse surface from knowledge of the spectral
distribution of and the surface temperature.
KNOWN: Spectral, hemispherical absorptivity of an opaque surface.
FIND: (a) Solar absorptivity, (b) Total, hemispherical emissivity for T s = 340K.
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Problem: Surface Emissivity and Absorptivity (cont)
SCHEMATIC:
ASSUMPTIONS: (1) Surface is opaque, (2) = , (3) Solar spectrum has G = G,Sproportional to E,b (, 5800K).
ANALYSIS: (a) The solar absorptivity may be expressed as
S ,b ,b0 0E ,5800K d / E ,5800K d .
The integral can be written in three parts using F(0 ) terms.
S 1 2 30 0.3 0 1.5 0 0.3 0 1.5F F F 1 F .
From Table 12.1,
T = 0.3 5800 = 1740 mK F(0 0.3 m) = 0.0335T = 1.5 5800 = 8700 mK F(0 1.5 m) = 0.8805.
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Problem: Surface Emissivity and Absorptivity (cont)
Hence,
S 0 0.0355 0.9 0.8805 0.0335 0.1 1 0.8805 0.774.
(b) The total, hemispherical emissivity for the surface at 340K may be expressed as
,b b0E ,340K d / E 340K .
With = , the integral can be written in terms of the F(0 ) function. However, it isreadily recognized that since
0 1.5 m,340K F 0.000 at T 1.5 340 510 m K
there is negligible emission below 1.5 m.
It follows that
0.1
COMMENTS: The assumption = is satisfied if the surface is irradiated diffusely or ifthe surface itself is diffuse. Note that for this surface under the specified conditions of solar
irradiation and surface temperature, S. Such a surface isspectrally selective.
P bl E B l f I di t d S f
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Problem: Energy Balance for an Irradiated Surface
Problem 12.90: Determination of the emissivity and absorptivity of a coated
vertical plate exposed to solar-simulation lamps and the magnitude
of the irradiation required to maintain a prescribed plate
temperature.
KNOWN: Vertical plate of height L 2 m suspended in quiescent air. Exposed surface withdiffuse coating of prescribed spectral absorptivity distribution subjected to simulated solar
irradiation, GS,. Plate steady-state temperature Ts 400 K
P bl E B l f I di t d S f ( t)
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Problem: Energy Balance for an Irradiated Surface (cont)
FIND: Plate emissivity, , plate absorptivity, , and plate irradiation, G.
ASSUMPTIONS: (1) Steady-state conditions, (2) Ambient air is extensive, quiescent, (3)
Spectral distribution of the simulated solar irradiation, GS, , proportional to that of a blackbody
at 5800 K, (4) Coating is opaque, diffuse, and (5) Plate is perfectly insulated on the edges and the
back side, and (6) Plate is isothermal.
PROPERTIES: Table A.4, Air (Tf 350 K, 1 atm): 20.92 10-6 m2s, k = 0.030 W/mK,
= 29.90 10-6 m2/s, Pr 0.700.
SCHEMATIC:
P bl E B l f I di t d S f ( t)
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Problem: Energy Balance for an Irradiated Surface (cont)
ANALYSIS: (a) Perform an energy balance on the plate as shown in the schematic on a per unitplate width basis,
in outE E 0
4s sG T h T T L 0
where and are determined from knowledge of and h is estimated from an appropriatecorrelation.
Plate total emissivity: Expressing the emissivity in terms of the band emission factor, F (0 - T),
1 s 1 s1 20 T 0 TF 1 F
0.9 0 0.1 1 0 0.1
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Problem: Energy Balance for an Irradiated Surface (cont)
Plate absorptivity: With the spectral distribution of simulated solar irradiation proportional to
emission from a blackbody at 5800 K,
1 s 1 s1 20 T 0 TF 1 F
0.9 0.7202 0.1 1 0.7202 0.676
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Problem: Energy Balance for an Irradiated Surface (cont)
Irradiation on the Plate: Substituting numerical values into Eq. (1),
4 48 2
0.676G 0.1 5.67 10 W m K 400 K 25.66W m K 400 300 K 0
2G 1052 W m