Radiation: Processes and Properties
Radiation Exchange Between Surfaces
Chapter 12 & 13
1
Introduction related to Ch 1
•
•
•
•
Consider a solid that is initially at a higher temperature Ts than that
of its surroundings Tsur, but around which there exists a vacuum.
The presence of the vacuum prevents energy loss from the surface
of the solid by conduction or convection. However, our intuition
tells us that the solid will cool and eventually achieve thermal
equilibrium with its surroundings.
This cooling is associated with a reduction in the internal energy
stored by the solid and is a direct consequence of the emission of
thermal radiation from the surface.
In turn, the surface will intercept and absorb radiation coming
from the surroundings. However, if Ts < Tsur the net heat transfer
rate by radiation qrad,net is from the surface, and the surface will
cool until Ts reaches Tsur.
Recall that the case discussed in Ch 1 was if the solid has
temperature of Ts and the surroundings is isothermal surface (Tsur
is fixed):
 =  E b (T s ) − G =  (T s4 −T sur4 )
q rad
(1.7 )
2
Lecture# 1
Main Goals of Ch. 12 & Ch. 13
1.
Identify the main characteristics of thermal radiation.
•
define the irradiation, “define” and “compute” the emissive power
•
describe the characteristics of a blackbody, Sketch the Planck distribution
2.
“compute” the band emission
3.
“define” emissivity and compute it
4.
“define” absorptivity, and compute it
•
Define a diffuse surface
•
Define a gray surface
5.
Define & compute the view factor
6.
Compute net radiative heat exchange between surfaces.
3
General Considerations
General Considerations
• Attention is focused on thermal radiation, whose origins are associated with emission
from matter at an absolute temperature T < 0.
• The mechanism of emission is related to energy released as a result of oscillations and
transitions of the many electrons that comprise matter.
• Radiation may also be intercepted and absorbed by matter, resulting in its increase in
thermal energy.
• The nature of radiation:
- One theory states that the physical manifestations of radiation may be explained by viewing it
as particles (aka photons or quanta).
- In other cases, radiation behaves as an electromagnetic wave.
But, in all cases, radiation can be characterized by a wavelength λ and frequency ν which
are related through the speed at which radiation propagates in the medium of interest,
=c

In vacuum: c = co = 2.998 x 108 m/s
4
Electromagnetic
Spectrum
Thermal radiation is confined to the infrared, visible and
ultraviolet regions of the spectrum (0.1 > λ > 100 µm).
Physicist,
Nuclear engineer
Spectral ~ function of (λ)
Electrical engineer
Directionality ~ function of (ϴ)
(Not included)
5
Radiation Heat Fluxes: Definitions
12.2 Radiation Heat Fluxes and Material Properties
reflected Radiation G ref
=
G
incidint Radiation
absorbed Radiation G abs
=
=
G
incidint Radiation
transmitted Radiation G tr
=
t=
G
incidint Radiation
r=
When an irradiation (G) fall onto a surface, part of this radiation is reflected, part is absorbed,
and part is transmitted.
r +  + t = 1 for any medium.
r +  = 1 for an opaque medium.
6
Directional Considerations
12.3 Directional Considerations and Radiation Intensity
Directional emissivity
• In general, radiation fluxes can be determined only
from knowledge of the directional and spectral nature
of the radiation.
• Radiation emitted by a surface will be in all directions
associated with a hypothetical hemisphere about the
surface
and
is
characterized
by
a
directional
Hemispherical emissivity
distribution.
• The term “Hemispherical” = integrated over all angles
2
E  (  ) = 0
/ 2
0
I  ,e (  , ,  ) cos sin d d 
Directional effects are not
considered in this course
7
The Blackbody
12.4 Blackbody Radiation and Its Intensity
• The Blackbody is defined as a perfect emitter and absorber of radiation.
I.
For a prescribed temperature and wavelength no surface can emit more radiation
than a blackbody (ideal emitter,  = 1).
II.
Also, blackbody absorbs all incident radiation, regardless of wavelength and
direction ( = 1)
• A blackbody is a diffuse emitter which means it emits radiation
uniformly in all direction.
• The radiation energy emitted by a blackbody per unit time and per unit surface
area can be determined from the Stefan-Boltzmann Law: (also called blackbody
emissive power)
E b =  T4
 W m2  where  = 5.67  10−8  2W 4 
 m .K 
8
Planck Distribution
The Spectral (Planck) Distribution of Blackbody Radiation
The spectral distribution of the blackbody emissive power (determined theoretically
and confirmed experimentally) is
Planck distribution over all wavelengths.
C1
E  ,b (  ,T ) =  I  ,b (  ,T ) = 5
 exp (C 2 / T ) − 1
First radiation constant:
C 1 = 3.742 x 108 W   m 4 / m 2
Second radiation constant: C 2 = 1.439 x 104  m  K
•
•
As the temperature increases,
more radiation appears at
shorter wavelengths.
The wavelength at which the
peak emissive power occurs
for a given temperature can
be obtained from Wien’s
displacement law: (dashed line)
( T )max.power = 2898 m.K
•
Plank’s law is valid for a surface
in a vacuum or gas. For other
mediums, it needs to be modified
by replacing C1 by C1/n2, where
n is the index of refraction of the
medium (e.g. n = 1.5 for water).
In contrast, for T < 800 K,
emission is mostly in the
infrared region of the
spectrum and is not visible.
9
Total blackbody emissive power; Eb
• Integrating Plank’s distribution over all wavelengths; gives

E b ( T ) =  E  ,b (  ,T ) d 
0

C1
= 5
d
exp
C
/

T
−
1

(
)
0  
2


= T
4
• Where Eb is the total blackbody emissive power.
• The Stefan-Boltzmann law gives the total radiation
emitted by a blackbody at all wavelengths from 0 to ∞.
However sometimes, we are interested in finding the
amount of radiation emitted over specified wavelength
Note that E ,b corresponds to
any value on the curve,
whereas Eb corresponds to the
area under the entire curve for
a specified temperature
interval or band in order to account for the spectral effect.
10
Lecture# 2
Band Emission from 0 to 
To avoid numerical integration, the fraction of the total emission from a blackbody in
the wavelength band from 0 to  is determined by the ratio of the shaded section to
the total area under the curve. Hence
F(0→ )

=

0
E  ,b d 
T 4
Note, thr ability to determine Iλ,b and its relation to the maximum intensity from the 3rd and 4th columns.
E  ,b =  I  ,b
11
Example on Band Emission
The temperature of the filament of a light bulb is 2500 K. Assuming the filament to be a blackbody,
•
Determine the fraction of the radiant energy emitted by the filament that falls in the visible range.
•
Determine the wavelength at which the emission of radiation from the filament peaks.
Assume the visible range of the electromagnetic spectrum extends from 0.4 to 0.76 μm.
which means only about 5% of the radiation emitted by the filament of
the light bulb falls in the visible range. The remaining 95% appears in
the infrared region or the “invisible light”.
(  )max.power =
2898  m.K = 1.159 m
2500K
12
Emissivity
Surface Emissivity
• A blackbody (ideal emitter & absorber ) is the reference in describing the emission
and absorption characteristics of real surfaces.
• The emissivity of a real surface depends on directional and spectral features of the
emitted radiation, ε = function (θ, ϕ, λ, T)
• The total, hemispherical emissivity (a directional and spectral average):

E ( T )  0   ( T ) E  ,b ( T ) d 
 (T ) 
=
Eb (T)
 T4
Emissivity is strongly related to temperature
Diffuse surface: is a surface which its properties are independent of direction.
Gray surface: is a surface which its properties are independent of wavelength.
13
Example 12.6: A diffuse surface at 1600 K has the spectral,
hemispherical emissivity shown as follows.
•
•
Determine the total, hemispherical emissivity and
the total emissive power.
At what wavelength will the spectral emissive power
be a maximum?
This will be λmax if the surface emitted as a blackbody
or ε is not varying with λ:
14
Abs, Ref & Trans
Response to Surface Irradiation: Absorption, Reflection
and Transmission
➢Irradiation, G
➢Reflection, Gref = ρG
➢Absorption, Gabs = αG
➢Transmission, Gtr = τG
• 1st law of thermodynamics :
G = G ref + G abs + G tr Hence; 1 = ρ + α + τ
• For opaque (nontransparent) material, Gtr = τ = 0
G = G ref + G abs Hence; 1 = ρ + α
These are for total hemispherical properties (over all direction and all frequencies),
however, we can rewrite these equations for the spectral irradiation as follows;
Gλ in [W/m2.μm]
G  = rG  +  G  + t G 
ρλ (T,λ) spectral reflectivity
α λ (T,λ) spectral absorptivity
τ λ (T,λ) spectral transmittivity
15
Absorptivity

Total, hemispherical
  (  )G  (  ) d 
G
  abs = o 
G
0 G  (  )d 
Reflectivity
Transmissivity

 r (  )G  (  ) d 
G
r  ref = 0 
G
 0 G  (  )d 
 G ( )d 
G
t  tr = 0  ,tr
G
0 G ( )d 

The source temperature
where G came from.
strongly dependent
-------
-------
Dependency
almost independent on surface
temperature, λ
on G direction and also on Gref
direction (bidirectional)
Strong dependent on λ
Remarks
For example, α of the concrete roof
is about 0.6 for solar radiation with
source temperature of 5762 K but
0.9 for radiation originating from
the surroundings with temperature
300 K
No surface is perfectly diffuse or
specular. However;
- Rough surface ~ diffuse
- Polished (mirror) surface ~ specular
The assumption of diffuse reflection is
mostly used in engineering application
e.g. glass or water, are
Semitransparent ~ short λ
Opaque ~ longer λ.
systems.
Note that the transmissivity
of glass is affected by its iron
content. These factors have an
important behavior on the
selection of cover plate
materials for
- solar collector applications,
- design of windows for
energy conservation,
- optical components of
infrared imaging.
16
Kirchhoff’s Law
Kirchhoff’s Law
• Kirchhoff’s law equates the total, hemispherical emissivity of a surface to its
total, hemispherical absorptivity:
 =
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.
• But, Kirchhoff’s law may be applied to the spectral,
directional properties without restriction:
  , =   ,
✓ The Kirchhoff’s law makes the radiation analysis easier (ε = α), especially for
opaque surfaces where ρ = 1 – α.
17
Problem 12.57:
The spectral, hemispherical absorptivity of an opaque surface and the spectral distribution of
radiation incident on the surface are as shown.
Use:

G
  G d
  abs = o   
G
 0 G d 
(a) What is the total, hemispherical absorptivity of the surface?
18
(b) If it is assumed that ελ = αλ and that the surface is at 1000 K, what is its total,
hemispherical emissivity?
(c) What is the net radiant heat flux to the surface?
19
Solar Radiation (Extra info)
•
The solar energy reaching the edge of the earth’s atmosphere is called the solar constant:
Gs = 1353 W / m2
Owing to the ellipticity of the earth’s orbit, the actual solar constant changes throughout the year within +/- 3.4%. This
variation is relatively small; thus Gs is assumed to be a constant.
•
The effective surface temperature of the sun can be estimated from the solar constant (by treating the sun as a
blackbody).
•
The solar radiation undergoes considerable attenuation as it passes through the atmosphere as a result of absorption
and scattering:
•
Absorption by the oxygen occurs in a narrow band about λ = 0.76 μm.
•
The ozone layer absorbs ultraviolet radiation at wavelengths below λ = 0.3 μm almost completely and radiation in
the range of 0.3 – 0.4 μm considerably.
•
Absorption in the infrared region is dominated by water vapor and carbon dioxide. Dust/pollutant particles also
absorb radiation at various wavelengths.
•
As a result the solar radiation reaching the earth’s surface is about 950 W/m2 on a clear day and much less on a
cloudy day, in the wavelength band 0.3 to 2.5 μm.
20
21
Lecture #3
The View Factor (aka shape or configuration factor)
• Radiation heat transfer between surfaces depends on:
(1) the orientation of the surfaces relative to each other
(2) and, their radiation properties and temperatures.
• So, it is a geometrical parameter that accounts for the
effects of orientation on radiation between surfaces.
• View factor is defined as the fraction of radiation
leaving a surface i and incident on surface j
Fi → j =
radiation reaching j from i
radiation leaving i
• Calculating view factors between surfaces are usually
a very complex and difficult to perform.
Fi → j =
This method could involve complex
integration to be performed
F j →i =
1
Ai
1
Aj

Ai A J

Ai A J
cos i cos  j
R 2
cos i cos  j
R 2
dAi dA j
(13.1)
dAi dA j
22
Why do we need to evaluate Fij?
To calculate the net radiative exchange between two surfaces that can be
approximated as blackbodies;
(
)(
q ij = Radiation leaving surface i − Radiation leaving surface j
and directly strikes surface j
and directly strikes surface i
q ij =
Ai Fij E b ,i
−
A j F ji E b , j
)
How?
I. Analytical method:
By performing the double integral of eq. 13.1,
Table 13.1, 13.2 and Figs. 13.4, 13.5 and 13.6 are
produced for several common geometries.
II. With some relations & rules:
- Reciprocity Rule
- Summation Rule
- Symmetry Rule
- Superposition Rule
23
24
View Factor Relations
I- Reciprocity Rule:
The view factor Fij is not equal to Fji unless the areas of the two surfaces (areas) are equal. It can be
shown that:
A i Fij = A j F ji
II- Summation Rule:
In radiation analysis, we usually form an enclosure. The conservation of energy principle requires that
the entire radiation leaving any surface i within the enclosure be intercepted by all surfaces of
enclosure. Therefore,
N
F
j =1
ij
=1
III- Symmetry Rule:
Two (or more) surfaces that possess symmetry about a third surface will have
identical F from that surface.
N
From summation:
F
j =1
From symmetry:
ij
= F11 + F12 + F13 + F14 + F15 = 1
 F12 = F13 = F14 = F15 = 0.25
25
F12 = 1
Example 13.2
F12 =

6
F12 = F13 = 0.5
F21 = 0.71
F13 = 0.172
F12 = 0.828
F21 = 0.207
26
Example
Consider the rectangular surfaces shown in the
Figure. Determine the view factors
(a) F13,
(b) F 3→(1+2)
(c) F23
Answer:
(a) F13
= 0.24
(b) F 3→(1+2) = 0.29
(c) F23
•
= 0.05
Here, we use the superposition Rule
27
Blackbody Radiation Exchange
(
)(
q ij = Radiation leaving surface i − Radiation leaving surface j
and directly strikes surface j
and directly strikes surface i
q ij =
Ai Fij E b ,i
−
A j F ji E b , j
)
σT4
• Net radiation transfer from surface i due to exchange with all (N) surfaces of an
enclosure (all black surfaces):
q i =  A i Fij  (T i 4 −T j4 )
N
j =1
• This is if we have blackbodies within the enclosure.
• How about real surfaces?
• Reflection… Reflection … Reflection make the radiation exchange between nonblack surfaces complicated!.
28
Radiation Exchange between Surfaces (Opaque, Diffuse, Gray)
within an Enclosure (Isothermal and no Transmission, τ = 0)
( i = i
= 1 − ri )
• Radiosity J is the total radiation energy outflowing from a surface, per unit area per unit
time. It is the summation of the reflected and the emitted radiation.
• For a surface i that is gray and opaque (εi = αi & αi + ρi = 1), the Radiosity can be
expressed as Ji = εiEbi + ρiGi
= εiEbi + (1 - εi)Gi
• Solve for Gi and substituted in
q i = Ai ( J i − G i )
• For blackbody (εi = 1);
Ji = Ebi = σT4
• Surface radiative resistance (Ohm’s law analogy) of the form:
qi
( E − J i ) , where R
= bi
Ri
i
1− i
=
 i Ai
I surface resistance
qi
29
Lecture# 4
Net Radiation between Two Surfaces
Consider two diffuse, gray, and opaque surfaces of arbitrary shape maintained at uniform
temperatures. The net rate of radiation heat transfer from surface i to surface j can be
expressed
q ij = A i Fij J i − A j F ji J j
Applying reciprocity;
q ij = A i Fij ( J i − J j )
Thermal resistance analogy: (considered as a space resistance) q ij
E b1
J1
1 − 1
R1 =
1A1
q12 = −q 21
R12 = 1
A1F12
J2
1− 2
R2 =
 2A 2
Eb 2
J
(
=
i
−J j )
R ij
q1 = q12 =
, where R ij =
1
A i Fij
 (T14 −T 24 )
1 − 1
1− 2
+ 1 +
1A1 A1F12  2 A 2
II surfaces and space resistances
30
Two-Surface Enclosure
31
Method of Solving Radiation Problem
• In radiation problems, either the temperature or the net rate of heat transfer must be given
for each of the surfaces to obtain a unique solution for the unknown surface temperature and
heat transfer rates.
• We use the network method which is based on the electrical network analogy.
• The following steps should be taken:
1. Form an enclosure; consider dummy surface(s) for openings, room, and approximate
these openings as blackbody of area, temperature, and properties,
2. Draw a surface resistance associated with each surface of the enclosure.
3. Connect the surface resistances with space resistances.
4. Solve the radiations problem (radiosities) by treating it as an electrical network
problem.
32
Example from old exams: Consider coaxial, parallel, black disks separated by 0.20 m. The
lower disk of diameter 0.40 m is maintained at 500 K and the surroundings (surface 3) are
at 300 K.
Ai Fij = Aj Fji
-
Determine the view factors F12, F13, and F23
R1 = 0.5
R2 = 1
S = 0.9
F12 = 0.469
F11 = 0
F13 = 1 − 0.469 = 0.531
-
F11 + F12 + F13 = 1
A1 = 0.03142m 2
( R = r / L,
A1 = 0.1257m 2
i
i
F21 = 0.117
S =1+
F22 = 0
F23 = 0.883
Fij =
R j = rj / L )
1 + R 2j
Ri2
1
S − [ S 2 − 4(rj / ri )2 ]1/2 )
(
2
What temperature will the upper disk of diameter 0.20 m achieve if electrical power of 17.5 W is
supplied to the heater on the back side of the disk?
 T1 = 456K
33
Extra Example Two parallel plates 0.5 by 1.0 m are spaced 0.5 m apart. One plate is maintained at 1000°C
and the other at 500°C. The emissivities of the plates are 0.2 and 0.5, respectively. The plates are located in a
very large room, the walls of which are maintained at 27°C. The plates exchange heat with each other and
with the room, but only the plate surfaces facing each other are to be considered in the analysis.
Find the net heat transfer rate to each plate and the room; neglect other modes of heat transfer, i.e.,
conduction and convection.
34
35
Thank you for being a
student in my class during
term 182 and wish you
successful future!
36