Thermal radiation presentation

Radiation Heat Transfer including shape function

1. Thermal Radiation
Beijing Institute of Technology
Group –‘9’

2. Presenters
普拉赛(Nepal)
2820170058
比拉(Pakistan)
2820166036

3. Thermal radiation:
•Emitted by matter as a result of vibrational and rotational
movements of molecules, atoms and electrons.
•The energy is transported by electromagnetic waves (or
photons).
•Radiation requires no medium for its propagation,
therefore, can take place also in vacuum.
•All matters emit radiation as long as they have a finite
(greater than absolute zero) temperature

4. Thermal radiation:
•Absorptivity,α,thefractionof
incidentradiationabsorbed.
•Reflectivity,ρ,thefractionof
incidentradiationreflected.
•Transmissivity,τ,thefractionof
incidentradiationtransmitted
Reflected
Radiation
Absorbed
Radiation
Transmitted
Radiation1=++
++=
Q
Q
Q
Q
Q
Q
QQQQ




FromConservationofEnergy,that:
Hence:
α + ρ + τ = 1
Body with α ＝1, —Absolute black-body
Body with ρ ＝1，—Absolute white-body
Body with τ ＝1，—Absolute transparent
body.

5. Black Body radiation
Black body (definition): A hypothetical object that
absorbs all of the radiation that strikes on it. It
also emits radiation (“Energy flux”) at a
maximum rate for its given temperature. For an
ideal black body ,
α ＝1
The Sun is very similar to an
“ideal emitter” (or “Black body”)
(NOTE: the Earth isn’t as ideal as a “black body”

6. Black Body radiation

7. Thermal Radiation
The thermal radiation have two important features:
❖Continuousfrequency spectrumthat depends only on the body's
temperature
❖Radiation is emitted by all parts of a plane surface in all direction
into the hemisphere above the surface, but the directional
distribution of emitted (incident) radiation is not uniform

8. Radiation Intensity
•Due to the directional distribution feature of thermal
radiation, we need a quantity to describes the magnitude of
radiation emitted or incident in specified direction.
•This quantity is radiation intensity denoted by I. It is the
rate of energy emitted ‘dq’ at wavelength λat θand Ф
direction per unit area of emitting surface normal to this
direction per unit solid angle d??????about this direction per
unit wavelength interval dλ
??????λ,θ,Ф=
��
��
����??????∙�??????∙��

9. Radiation Intensity
dA
1 dA
1·cosθ
θ
�ωis a solid angle:
�??????=
��
�
�
�

10. Relation of Intensity to Emissive power
The radiation emitted by blackbody surface is isotropic (intensity I
b
is independent of the direction). Integrating the radiation power
for solid angles covering the whole upper hemisphere results to
relationship between the emissive power E
band the intensity I
b.
??????
�is the spectral heat flux.
??????
�=න
�
�??????
න
�
ൗ
??????
�
??????
�,��,??????,??????���??????���??????�??????�??????=π�
??????

11. •Planck's law describes the
spectral density of
electromagnetic radiation
emitted by a black body in
thermal equilibrium at a given
temperature T.
•The Planck distribution is shown
in the figure as a function of
wavelength for different body
temperatures.
Planck’s law
Black Body radiation

12. Plank’s Law
•The Planck law describes
theoretical spectral
distribution for the emissive
power of a black body.
•�
??????,??????=
�1
??????
5
[exp
??????2
????????????
−1]
•where C
1=3.742x10
8
(W.mm
4
/m
2
) and
C
2=1.439x10
4
(m m.K) are
two constants.
•??????
�,��,??????=
����
�
�
??????
[�??????�
���
�??????
�
??????
−�]
•Where ℎ=6.626×10
−34
�∙??????and
??????
�=1.381×10
−23
�/�are the
universal Plank and
Boltzmann constants,
respectively.
•??????
??????=2.998×10
8
??????/??????is the speed
of light in vacuum, and Tis
the absolute temperature of
the blackbody(K).

13. Wien’s Displacement Law
The blackbody spectral distribution has a maximum and
that the corresponding wavelength �
��??????depends on
temperature.Peak can be found for different temps using
Wien’s Displacement Law:
�
��????????????=�
�
Where the third radiation constants
�
�=������∙??????
Black Body radiation

14. Black Body radiation
Ifspectralintensityisnorelatedto??????,then0≤θ≤π/2
0≤ϕ≤2π
ϕ
dϕ
θ
dθ
dAn
dA1
r
n
I(θ)
Lambert’s Law
�??????=�
0????????????????????????

15. Black Body radiation
Ifspectralintensityisnorelatedto??????,then
Lambert’s Law
�??????=�
0????????????????????????s
r  n

Our skin absorbs more photons
at noon than at evening even if
the intensity of radiation is the
same (neglecting photon
absorption in atmosphere).

16. Black Body radiation
The Stefan-Boltzmann law
Thetotal,hemisphericalemissivepower,E(W/m
2
),istherateat
whichradiationisemittedperunitareaatallpossiblewavelengths
andinallpossibledirections.
??????=න
�
∞
??????
����
??????=න
�
∞
න
�
�??????
න
�
ൗ
??????
�
??????
�,��,??????,??????���??????���??????�??????�??????��
Forablackbody,SubstitutingthePlanckdistributionintoabove
Equation,thetotalemissivepowerofablackbodyE
bmaybe
expressedinanotherway:

17. Black Body radiation
The Stefan-Boltzmann law
The total intensity associated with blackbody emission is ??????
�=
??????
�
??????
??????
�=න
�
∞
�
�
�
??????
[�??????�
�
�
�??????
−�]
��
Performing the integration, it may be shown that
�
??????=????????????
4
The result is termed the Stefan–Boltzmann law. It
can be stated as “The emissive power of a black body
(per unit area per unit time) is directly proportional
to the fourth power of its absolute temperature”.
Wheretheunitis：
W/m
2
.（bmeans
blackbody）
??????is Proportionality constant called Stefan-Boltzmann constant.
??????=5.676 x 10
-8
W/m
2
-K
4
unit is：W/m
2
K
4

18. What we know about blackbody radiation
•the shape of the distribution
•the peak shifts according to Wien's law
•the total power output is described by the Stefan-
Boltzmann law

19. Practical body
Practical body has smaller emissive power than that of black body in the
same temperature. The ratio of the emissive power of the practical body to
the black body is called emissivity, i.e.


bE
E
=)( 4
TE=
So, for any practical body, we have：
A blackbody is said to be a diffuse
emitter since it emits radiation energy
uniformly in all directions.

20. Practical body
Kirchhoff’s law:
Imagine two parallel plates, one, which is a blackbody surface
(
b=1) and the other one is gray (α,).
E
b(??????
1)
E
b(??????
2)
(1-α)E
b(??????
2)
reflected
radiation
radiation
emitted by
gray plate
received
radiation
Resulting flux from left to right
E
b(??????
1)-(1-α)E
b(??????
2)-E
b(??????
2)=0

21. Practical body
For thermal balance，??????
1=??????
1=T
Then ,
E
b
α
=
E
b

The Kirchhoff's law :“The emissivity of a body radiating energy at a temperature,
T, is equal to the absorptivity of the body when receiving energy from a source at
a temperature, T.”
Emissivity equals absorptivity. =α(ideal emitter of radiation is an ideal absorber).
Or, the higher absorptivity，has strong the emissivity.
In reality, (,,T)≠ (,,T)

22. The radiation and View factor
SupposewehavetwosurfacesattemperatureT
1
andT
2,butbotharefiniteinarea,andneither
surfaceiscompletelyenclosedbytheother.
Anexamplemightbethefloorandceilingofa
room.Onlyafractionoftheenergyleavingthe
ceilingstrikesthefloorandviceversa.
Toaccountforthisincompleteexchangeofenergy,
wedefinetheviewfactor,F
1-2.
F
1-2=FractionofenergyleavingA
1reachingA
2.
In the same way , we can estimate F
2-1
??????
2
??????
1
F
1-2
F
2-1

23. The radiation and View factor
Net radiation between the surfaces is：
�
12=�
??????1??????
1�
12−�
??????2??????
2�
21
Under the balance conditions, ??????
1=??????
2�
12=0 ,E
b1=E
b2
Then,
Hence，per unit time the radiation energy emitted by the first surface that
reaches second surface is: E
b1A
1F
1-2
Similarly, the energy from surface 2 that reaches surface 1 is：E
b2A
2F
2-1
From this equation, it can be found that the
geometric factor has no relation to the process.
??????
1�
12=??????
2�
21

24. The radiation and View factor
With the relationship of ??????
1�
12=??????
2�
21, we can calculate the radiation of two black
surfaces:
�
12=(�
??????1−�
??????2)??????
1�
12=(�
??????2−�
??????1)??????
2�
21
�
12=
(�
??????1−�
??????2)
1
??????1??????12
=
????????????1
4
−????????????2
4
????????????
=
(�
??????1−�
??????2)
????????????
Where, �
??????is called radiation resistance for heat transfer between two surfaces.
1/(A
1F
12)
�
??????1 �
??????2
�
??????=
1
A
1F
12

25. View factor
✓Very important coefficient to evaluate the radiation.
Properties
1.Enclosures:
Alsocalledcompletenessofthegeometricfactor
Inorderthatwemightapplyconservationofenergyto
theradiationprocess,wemustaccountforallenergy
leavingasurface.Weimaginethatthesurrounding
surfacesactasanenclosureabouttheheatsourcewhich
receiveallemittedenergy.ForanNsurfacedenclosure,
wecanthenseethat:1
1
,=
=
N
j
jiF

26. View factor
✓Very important coefficient to evaluate the radiation.
Properties
2.ReciprocityRelation:
1 2
A
i F
ij= A
j F
ji
??????.??????.
??????
1�
12=??????
2�
21
The reciprocity theorem allow one to
calculateF
ij if one already know F
jiusing
the areas of two surfaces.

27. View factor
✓Very important coefficient to evaluate the radiation.
Properties
3.AssociativeRule
i
j
k
�
�−(�+�)=�
�−�+�
�−�
And also,
??????
�+��
�+�−�=??????
��
�−�+??????
��
�−�
Thefractionofenergyleavingsurfaceiand
strikingthecombinedsurfacej+kwillequalthe
fractionofenergyemittedfromiandstrikingj
plusthefractionleavingsurfaceiandstrikingk.

28. View factor
✓Very important coefficient to evaluate the radiation.
Properties
4.Thegeometricfactoroftwoinfiniteplates.
�
12=�
21=1X12 X21
�
12�
21
Since, the both plates are facing each
other parallelly,

29. View factor
✓Very important coefficient to evaluate the radiation.
Properties
5.Curvedsurfaces.
�
2,1=
??????
1
??????
2
Anon-concavesurfaceiscompletely
enclosedbyothersurface,theradiation
from1to2isreceivedcompletelyby2，
So，F
1，2＝1，atthesametime，F1
F2
A
1
A
2

30. View factor
✓Very important coefficient to evaluate the radiation.
Properties
Three-Dimensional Geometry: Example()
12
2
2
2
2
1
4
2
1
1
/
/
//
ij j i
j
i
i i j j
F S S r r
R
S
R
R r L R r L


= − −


+
=+
==
Coaxial Parallel Disks

31. Grey Body Radiation
Theabsorptivityofagreybodyislesserthan1.
So,someradiationstrikingonagreybodywillbereflectedback.
Radiosity,J,thetotalradiantenergyleavingabodyperunitareaperunit
time.
Irradiation,G,thetotalradiantenergyincidentonabodyperunitareaper
unittime.(−)
1 2
E1
E1

32. Grey Body Radiation(−)
1 2
E1
E1
Net heat transfer from body：AGJQ )(−=
The net emissive energy is GEGEJ
b
)1( −+=+=
Eliminate G，and note ：=α，then：A
JEAJE
Q
bb
 /)1(
)(
/)1(
)(
−
−
=
−
−
=

33. Grey Body Radiationb J1
(−)F1
ൗ
(1−
1)

1??????
111
1
11
1
1
A
JE
Q
b

−
−
=
Electrical Analogy
Where, ൗ
(1−1)
1�1
, is called surface thermal resistance
Greybodyhasaradiationresistanceonitssurface.
Therefore,itsemissivepowerwilldecrease.

34. Grey Body Radiation
Hence，for any two grey surfaces, the radiation heat transfer can be expressed as：
�
12=
(�
??????1−�
??????2)
1
??????1??????12
+
1−1
1??????1
+
1−2
2??????2
E
b1 E
b2
J1 J2
1−
1

1??????
1
1
??????
1�
12
1−
2

2??????
2
Where,
1
�1�12
is Space radiation resistance
1−
1
1�1
&
1−
2
2�2
is Surface resistances.

35. Grey Body Radiation
Electrical Analogy
Two infinite parallel plates 12
X1,2=1 X2,1=1
F
2-1=1
F
1-2=1
�1
�2
1 F
1-2= F
2-1 =1
�
12=
??????(??????
1
4
−??????
2
4
)??????
1
1

1
+
1

2
−1

36. Grey Body Radiation
Electrical Analogy
A small body in a big space
�1
�2
0 F
1-2 =1
A
1
A
2
�
12=??????
1(??????
1
4
−??????
2
4
)??????
1

37. Grey Body Radiation
Insulation SheetEb1 J1
Q2
1/A3X3,2
Q1
A3,ε3,T3
J31 Eb3 J32 J2 Eb2
1 3 2
1-ε1/ε1A11/A1X1,31-ε31/ε31A31-ε32/ε32A3 1-ε2/ε2A2
Before inserting a sheet, the radiation resistance between
two surfaces only includes two surface resistances and
one geometric resistance. After inserting the sheet, the
resistance will increase two surface resistances and one
geometric resistance. Hence, the total resistance will
increase and the heat transfer will decrease.
�
12=??????(??????
1
4
−??????
2
4
)??????
1 
1=
2=Before Insulation If

38. Grey Body Radiation
Insulation Sheet
Before inserting SheetQ1
Eb1 J1 J2 Eb2
Q2
1-ε1/ε1A11/A1X1,21-ε2/ε2A2
A1
ε1
T1 T2
ε2
A2
E
b1 E
b2
J1 J2
1−
1

1??????
1
1
??????
1�
12
1−
2

2??????
2
Have
•Two surface resistances
•One geometric resistance
�
12=??????(??????
1
4
−??????
2
4
)??????
1

1=
2=If
F
1-2= F
2-1 =1

39. Grey Body Radiation
Insulation Sheet
After inserting SheetEb1 J1
Q2
1/A3X3,2
Q1
A3,ε3,T3
J31 Eb3 J32 J2 Eb2
1 3 2
1-ε1/ε1A11/A1X1,31-ε31/ε31A31-ε32/ε32A3 1-ε2/ε2A2
1
??????
3�
32
1
??????
1�
13
1−
31

31??????
3
1−
32

32??????
3
1−
2

2??????
2
1−
1

1??????
1
Have
•Four surface resistances
•Two geometric resistance
Hence, the effective heat
flow rate decreases and
acts as insulation.

40. Grey Body Radiation
Insulation Sheet
For infinite parallel plates, the geometric factor is
�
13=�
31=�
12=1AAAA ===
321 1
11
)(
21
1
4
2
4
1
2,1
−+
−
=

 ATT
Q
b
Before inserting the
sheet, heat transfer
is:
After inserting the
sheet, heat transfer is:1
11
1
11
)(
232311
4
2
4
1
2,33,12,3,1
−++−+
−
===

 ATT
QQQ
b
Obviously 2,12,3,1QQ  ====
232311
If2,12,3,1
2
1
QQ=
Hence,

41. Grey Body Radiation
Insulation Sheet ====
232311
If2,12,3,1
2
1
QQ=
Hence,
Withthesameway,insertingnsheetsbetweentwoinfiniteparallelplates
withsameemissivity,theheattransferis,2,12,,1
1
1
Q
n
Q
n
+
=
Hence, to increase insulation effect, the low emissivity sheet should be used.shoeldwithoutshieldswith
A
q
nA
q






+
=






1
1

42. Heat transfer coefficient for radiation
✓A heat transfer coefficient for radiation is sometimes defined analogously to
the convection coefficient.
✓??????
??????is defined as a radiation coefficient.
✓We know,TT R
TTTT
R
TT
Q
))(()(
2
2
2
1
2
2
2
1
4
2
4
1
2,1
−+
=
−
=
 )(
))(()(
21
21
2
2
2
1
4
2
4
1
2,1
TT
R
TTTT
R
TT
Q
TT
−
++
=
−
=

Finally, radiation coefficientT
r
AR
TTTT ))((
21
2
2
2
1 ++
=

 )(
212,1 TTAQ
r−=
Hence

43. Heat transfer coefficient for radiation
On the other hand, we know
Then the radiation coefficient is
Again, we have:)(
2112,1 TTAQ
r−= ))()(()(
2121
2
2
2
1121
4
2
4
11212,1
TTTTTTAXTTAXQ −++=−=
−−

�
12 �
12))((
21
2
2
2
121 TTTTX
r ++=
−
�
12

44. Thank you for your
time。。。
谢谢你们