MEE 5230 Radiation Heat Transfer – D. Pahinkar
Radiation – Basic Concepts
•
•
•
•
•
•
All matter continuously emits radiation
Radiation requires no intervening medium
c
Radiation is emitted by surface at wavelength  and frequency :  =

8
-6
In a vacuum, co = 2.99810 m/s, and 1 m = 10 m
Thermal radiation (portion of UV, visible, IR) occurs at 0.1 <  < 100 m
(For reference, visible radiation occurs at 0.4 <  < 0.7 m)
Page 1
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Spectral Dependence
• Radiation emitted from a surface displays spectral and
directional dependence:
• The amount of radiation emitted by an opaque surface varies
with wavelength, so:
– spectral distribution over all wavelengths, or
– monochromatic/spectral components associated with particular
wavelengths
Page 2
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Directional Dependence
•
•
Emission and incidence of radiation
from/to surface depend on direction:
directional radiation intensity
The amount of radiation emitted from
a surface, dA1 and propagating in a
particular direction, ,  is
represented in terms of a differential
solid angle d associated with the
direction:
dA
d   2n
r
dAn = r 2 sin  d d
d =
dAn
= sin  d d
r2
Page 3
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Radiation Intensity
•
The spectral intensity I,e associated with emission from a surface
element dA1 in the solid angle d about ,  and the wavelength
interval d about :
I  ,e (  , ,  ) 
•
dq
( dA1 cos  )  d  d 
W m 2 sr  m
Radiation flux can be defined in terms of projected area dA1cos
because there are surfaces that can be approximated as having I,e
independent of direction (diffuse surfaces, isotropic radiation)
dq
= I  ,e (  , , ) dA1 cos  d
d
Spectral radiation flux:
dq 
•
dq = I  ,e (  , , ) cos  d
= I  ,e (  , , ) cos  sin d d
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MEE 5230 Radiation Heat Transfer – D. Pahinkar
Radiation Intensity
Spectral hemispherical heat flux:
2  2
q (  ) =   I  ,e (  , , ) cos  sin d d
0
0
Total heat flux in all directions at all wavelengths:

q =  q (  )d 
0
•
Spectral emissive power (W/m2m) corresponds to spectral emission
over all possible directions
2
E (  ) =  0
•
•
/2
0
I  ,e (  , , ) cos  sin d d
Based on actual surface area while I,e is based on the projected area
The total emissive power (W/m2) corresponds to emission over all
directions and wavelengths

E =  0 E (  ) d 
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MEE 5230 Radiation Heat Transfer – D. Pahinkar
Radiation Intensity
•
For many surfaces, diffuse emission is a reasonable approximation
(intensity of emitted radiation independent of direction:
I  ,e (  , , ) = I  ,e (  )
•
This leads to:
E  () =  I ,e ()
•
E =  Ie
( W m ) sr ( W sr m )
2
2
Thus, emissive power (E) is related to intensity of radiation (Ie) through
the above
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MEE 5230 Radiation Heat Transfer – D. Pahinkar
Irradiation (Incident Radiation)
•
•
Could be from emission and reflection
occurring at other surfaces
Could have spectral and directional
dependence based on the spectral intensity
I,i(,,)
2
G ( ) =
(W
•
m 2 sr
)
/2
 
0
I  ,i cos  sin  d  d 
0
Total incident radiation (W/m2), where  is of
the sender:

G =  G  ( )d
0
•
As was the case for emissive power, for
diffuse incident radiation:
G  (  ) = I  ,i (  )
G = I i
Page 7
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Radiosity
• Radiosity is a measure of all radiation leaving a surface,
both emitted and reflected

J =  J  ( )d
0
J  () =  I ,e+ r
& for diffuse surfaces J =  Ie+ r
Page 8
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Example
•
Given a diffuse surface with spectral emission as:
200
E
(W/m2 m)
5
10
15
 (m)
20
Find
(a) Total emissive power of this surface, and
(b) Total intensity of emitted radiation in the normal direction and at 30
deg to normal.
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MEE 5230 Radiation Heat Transfer – D. Pahinkar
Example Continued
(a) Total emissive power

E =  E  ( )d
0
5
10
15
20

0
5
10
15
20
=  0 d +  100 d +  200 d +  100 d +  0 d
E = 2000 W/m 2
b) The total intensity of emitted radiation in the normal direction and at
30o to normal:
E
 Ie = = 637 W/sr  m 2

•
Note that Ie is independent of angle for a diffuse emitter
Page 10
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Blackbody Radiation
•
An idealized surface (provides limits on
radiation emission and absorption by
matter)
• For a given temperature, , no surface can
emit more energy (an ideal diffuse emitter)
• Absorbs all radiation of all wavelengths
and direction (ideal absorber)
• Doesn’t exist, but is a good reference
Planck’s Distribution
• Emissive power of a black body as a
function of  at different T (Eb increases
with temperature, and also, varies
continuously with )
E ,b ( ,T) =  I  ,b (  ,T) =
C1

C  
 5 exp  2  − 1
 T  

W m 4
C1 = 2 h c = 3.742 10
m2
2
o
C2 =
8
hco
= 1.439 104 m  K
k
h  Planck’s constant = 6.6256  10−34 J.s
k = Boltzmann’s constant = 1.3805  10−23 J/K
co = 2.998  108 m/s; T  Absolute temperature
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MEE 5230 Radiation Heat Transfer – D. Pahinkar
Stefan-Boltzmann, Wien’s Laws
•
In Planck’s distribution, the peak in emissive power shifts towards
larger  as temperature decreases according to Wien’s Displacement
Law:
 max T = C3 ; C3 = 2897.8 m K
•
•
•
Sun assumed blackbody at ~ 5800 K: peak in visible range
For T < 800K, primarily infrared, not visible
The fractional amount of total blackbody emission appearing at lower
wavelengths increases with increasing T
Integrating Planck’s equation over all wavelengths:
Stefan-Boltzmann Law

Eb =  0 E ,bd  =  T 4
 = 5.67  10−8 W/m 2K 4
• Eb: amount of energy radiated per unit area per unit time
by an ideal radiator
Page 12
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Band Emission
•
The fraction of total blackbody
emission that is in a prescribed
wavelength interval or band 1 <  <
2 is:


 0 2 E ,bd  −  o1 E ,bd 
F( 1 − 2 ) = F( 0− 2 ) − F( 0− 1 ) =
T4
where

F( 0−  )
 E d
= 0  ,b
= f ( T )
T
Page 13
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Radiative Properties of Real Surfaces
Surface Emissivity:
• Emission from a surface compared to the
emission from a blackbody at the same
temperature
Emissivity 
emissive power of surface at some  ,T
emissive power of b.b. at same  ,T
 =
•
E
Eb
(monochromatic / spectral)

Total emissivity:
E
=
=
Eb
•
•
•

  E  b d
0
Eb
= (  F0→1 + - - - )
In many cases,  can be approximated to be
independent of T
Gray body:   ()
  1 always
Page 14
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Representative Emissivities
•
Read examples 12.5 and
12.6 from the book
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MEE 5230 Radiation Heat Transfer – D. Pahinkar
Notes on Emissivities
Example: Aluminum oxide is almost white in the visible range but black
in the IR region (@ 1400 K)
• The apparent color of a surface is a function of reflected light and
absorbed light (snow is white in the visible range, but a blackbody at
long )
Notes:
• Metallic surfaces typically have small 
–
•
•
•
•
•
•
(as low as 0.02 for highly polished gold & silver)
oxidation of surfaces leads to an  in 
non-conductors have relatively high  ( 0.6)
 for conductors  as T  generally
 for nonconductors  or  as T 
 influenced by nature of surface
(fabrication, thermal cycling, reaction w/ environment,..)
Page 16
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Absorption, Reflection and Transmission
Three responses of a semitransparent
medium to irradiation:
• Reflection from the medium ( G ,ref )
• Absorption within the medium ( G ,abs )
• Transmission through the medium
• Radiation balance:
( G ,tr )
G = G ,ref + G ,abs + G ,tr
• The above balance therefore
includes volumetric effects
Are glass and water
• For an opaque body, G,tr = 0
semitransparent or opaque?
G = G ,ref + G ,abs
•  of incident radiation, and material
properties determine whether the
material is semitransparent or
opaque
Page 17
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Absorption and Reflection
•
•
There is no net effect of reflection on the medium, while absorption
raises the internal thermal energy of the medium
Unless an opaque material is at a sufficiently high temperature to emit
visible radiation (e.g., incandescent light, Ts > 1000K), its color is
determined by the spectral dependence of reflection in response to
visible irradiation
– What may be said about reflection for a white surface?
– Black surface?
Page 18
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Reflectivity
Page 19
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Transmissivity
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MEE 5230 Radiation Heat Transfer – D. Pahinkar
Absorptivity
Spectral:
Total:
G  ,abs ( )  based on incident radiation
 (  ) =
G  ( )
G abs
=

G
  E ,b ( ,5800K)d
For solar radiation:
solar =
0


E ,b ( ,5800K)d
0
(since radiation from the sun approximates that from a b.b.
at 5800 K)
Page 21
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Reflectivity and Transmissivity
Reflectivity:
•
G  ,ref
 ( ) =
G
 again, based on incident radiation, I ,i
Specular and diffuse reflection
(polished and rough surfaces)
Transmissivity:
G trans.
=
G
++=1
Page 22
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Kirchhoff’s Law
Always valid, no restrictions
  , =   ,
If irradiation is diffuse or surface
is diffuse


Gray Surface:
• If , and , are
independent of  and , the
surface is gray and diffuse
 =
=
Page 23
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Example
KNOWN:
Spectral hemispherical emissivity, dimensions and initial temperature of
a tungsten filament.
FIND:
(a) Total hemispherical emissivity, , when filament temperature is Ts =
2900 K; (b) Initial rate of cooling, dTs/dt, assuming the surroundings are
at Tsur = 300 K when the current is switched off; and (c) Time required for
the filament to cool from 2900 to 1300 K
Page 24
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Example Continued
ASSUMPTIONS: (1) Filament temperature is uniform at any time (lumped
capacitance), (2) Negligible heat loss by conduction through the
support posts, (3) Surroundings large compared to the filament, (4)
Spectral emissivity, density and specific heat constant over the
temperature range, (5) Negligible convection.
• Tungsten (2900 K);  = 19,300 kg/m3, Cp  185 J/kg  K
• Total emissivity at Ts = 2900 K using band emission factors from
table,
 = (1/ E b ) 

0
F(0→2m) = 0.72
  E  ,b (T)d = 1F(0→2 m) +  2 (1 − F0→2 m ) (1)
at T = 2 m  2900 K = 5800 m  K
 = 0.45  0.72 + 0.1(1 − 0.72) = 0.352
Page 25
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Example Continued
(b) Energy balance on the filament at the instant of time at which the
current is switched off:
E in − E out
•
dTs
= mC p
dt
4
−q rad = −As  ( Ts4 − Tsur
) = mCp dTs /dt
Find the change in temperature with time where As = DL, m = V, and
V = (D2/4)L
4
4
dTs
DL(Ts4 − Tsur
)
4(Ts4 − Tsur
)
=−
=−
2
dt
Cp (D / 4)L
C p D
(2)
dTs
4  0.352  5.67 10−8 W/m 2  K 4 (2900 4 − 300 4 )K 4
=−
= −1977 K / s
2
dt
19,300 kg/m 185 J/kg  K  0.0008m
Page 26
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Example Continued
(c) Separate variables in (2) and integrate:
4
DCp
t
Tf

dt = − 
0
Ti
 t=
DC p
12
dT
11
1
=  3 − 3
4
T
3  Tf Ti 
1
1
 T 3 − T 3  = 4.94s
 f
i 
Page 27
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Example
KNOWN:
Area, temperature, solar irradiation and spectral absorptivity of a surface.
FIND:
Absorbed irradiation (Gabs), emissive power (E), radiosity (J) and net
radiation transfer from the surface (qnet).
Page 28
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Example Continued
ASSUMPTIONS:
(1) Opaque, diffuse surface, (2) Spectral distribution of solar radiation
corresponds to emission from a blackbody at 5800 K.
•
The absorptivity to solar irradiation is
s
•

=

0
  G  d
G
From Table 12.1,

=

0
  E  ,b (5800 K)d
Eb
= 1F(0.5→1m) +  2 F(2 m→ )
T = 2900 m  K : F(0→0.5m) = 0.250
T = 5800 m  K : F(0→1m) = 0.720
T = 11600 m  K : F(0→2m) = 0.941
•
S = 0.8(0.720 − 0.250) + 0.9 (1 − 0.941) = 0.429
Tsun = 5800 K
Note: F0→∞ = 1
Page 29
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Example Continued
•
Hence,
•
The emissivity is
G abs = SG S = 0.429(1200 W/m 2 ) = 515 W/m 2
=

0
 E b (400 K)d/E b = 1F(0.5→1m) + 2 F(2→ )
From Table 12.1,
T = 200 m  K : F(0→0.5m) = 0
Ts = 400 K
T = 400 m  K : F(0→1m) = 0
T = 800 m  K : F(0→2m) = 0
Hence,  = 2 = 0.9,
E = Ts4 = 0.9  5.67 10−8 W/m 2  K 4 (400 K) 4 = 1306 W/m 2
Page 30
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Example Continued
The radiosity is
J = E + s G s = E + (1 − s )G s = [1306 + 0.571  1200]W/m 2 = 1991 W/m 2
The net radiation transfer from the surface is
q net = (E − s G s )As = (1306 − 515)W/m 2  4m 2 = 3164 W
COMMENTS:
Unless 3164 W are supplied to the surface by other means (for example,
by convection or a heating strip), the surface temperature will
decrease with time.
Page 31
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Example
KNOWN:
Temperature, absorptivity, transmissivity, radiosity and convection
conditions for a semitransparent plate, under uniform irradiation from
both sides.
FIND:
Plate irradiation G and total hemispherical emissivity 
measured
(includes
transmitted,
reflected &
emitted)
Page 32
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Example Continued
ASSUMPTIONS:
(1) Steady-state conditions, (2) Uniform surface conditions.
• From an energy balance on the plate
•
Therefore,
E in = E out
2G = 2qconv + 2J
G = 40 W/m 2 K (350 − 300)K + 5000 W/m 2 = 7000 W/m 2
•
•
From the definition of J,
J = E + G + G = E + (1 − )G = E b + (1 − )G
Therefore
J − (1 − )G (5000 W/m 2 ) − 0.6(7000 W/m 2 )
=
=
= 0.94
4
−8
2
4
4
T
5.67  10 W/m  K (350 K)
Page 33
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Example Continued
•
Hence,    and the surface is not gray for the prescribed conditions.
Note:
• The emissivity may also be determined by expressing the plate
energy balance as
2G = 2qconv + 2E
•
Hence,
  T4 =  G − h ( T − T )
0.4(7000 W/m 2 ) − 40 W/m 2  K(50 K)
=
= 0.94
−8
2
4
4
5.67  10 W/m  K (350 K)
Page 34
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Environmental Radiation
Solar radiation:
• Approximated as blackbody at 5800K.
• Direct and diffuse components due to scattering and absorption
• Extra-terrestrial solar irradiation Gs,o depends on:
– Latitude
– Time of day
– Time of year
•
Solar radiation flux:
qS = f x Sc
Sc → solar constant or heat flux ( 1353 W/m 2 )
when earth is at its mean distance from sun
f → correction factor accounting for eccentricity
of the earth's orbit ( 0.97 < f <1.03 )
Page 35
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Solar Radiation
•
Extraterrestrial irradiation of
a surface whose normal is at
a zenith angle  relative to
the sun’s rays is
GS ,o = f x Sc x cos
Interaction of solar radiation
with earth’s atmosphere:
• Absorption by aerosols over
the entire spectrum
• Absorption by gases (CO2,
H2O (v), O3) in discrete
wavelength bands
• Scattering by gas molecules
and aerosols
Page 36
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Solar Radiation
Effect of Atmosphere on Spectral Distribution of Solar Radiation:
• Attenuation over entire spectrum but more pronounced in spectral
bands associated with polar molecules
• Concentration of all radiation in the spectral region 0.3 <  3 m, peak
at  0.5 m
Page 37
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Solar Radiation
Effect of Atmosphere on Directional Distribution of
Solar Radiation:
• Rayleigh scattering is approximately uniform in
all directions (isotropic scattering), while Mie
scattering is primarily in the direction of the sun’s
rays (forward peaked).
• Directional distribution of radiation at the earth’s
surface has two components
• Direct radiation: Unscattered and in the
direction  of the sun’s rays
• Diffuse radiation: Scattered radiation strongly
peaked in the forward direction.
• Calculation of solar irradiation for a horizontal
surface often presumes scattered component is
isotropic
 cos +  I dir
GS = GS ,dir + GS ,dif = qdir
Page 38
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Terrestrial Radiation
Emission by Earth’s Surface:
E =  T
4
Typical emissivities:
Sand/Soil:
Water/Ice:
Vegetation:
Snow:
Concrete/Asphalt:





 0.90
 0.95
 0.92
 0.82
 0.85
•
Emission is typically from surfaces with temperatures in the range of
250 < T < 320K and hence concentrated in the spectral region 4 <  <
40 m, with peak emission at 10 m
Atmospheric emission:
• Largely due to emission from CO2 and H2O (v) and concentrated in
spectral regions 5 <  < 8 m, and  > 13 m
Page 39
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Terrestrial Radiation
•
Earth irradiation due to atmospheric emission is often approximated
by a blackbody emissive power of the form:
4
Gatm =  Tsky
•
•
•
Tsky → effective sky temperature
230K(cold,clear day)< Tsky  285K(warm, overcast day)
Can water in the natural environment freeze if the ambient air
temperature exceeds 273K?
If so, what environmental conditions (wind and sky) favor ice
formation?
Surface
Collection/Rejection of heat:
Snow
0.29
Human skin
0.64
White paint
0.22
Black paint
1.0
Evaporated Al
film
3.0
Rejection
Collection
Page 40
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Greenhouse Effect
Page 41
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Carbon Cycle
Page 42
MEE 5230 Radiation Heat Transfer – D. Pahinkar
Sea Level Effects
Page 43