1. ME C3609
AN INTRODUCTION TO RADIATION
HEAT TRANSFER
by
M.N.A. Hawlader

2. RADIATION HEAT TRANSFER
 Lecturer: Dr M.N.A. Hawlader
 Contents:
• Fundamental Concepts
• Laws of blackbody radiation
• Intensity and shape factor
• Radiation exchange between
blackbody and gray surfaces
• Radiation shield

3. RADIATION HEAT TRANSFER
 References
1.Heat Transfer by J.P. Holman, Seventh
Edition, McGraw Hill, Singapore 1992
2. Principles of Heat Transfer by Frank Kreith and
M.S. Bohn, Fourth Edition, Harper and Row,
Singapore 1986.
3. Fundamentals of Heat Transfer by F.P. Incropera
and D.P. Dewitt, John Wiley and Sons, Singapore
1985

4. RADIATION HEAT TRANSFER
Fundamental Concepts
 Conduction and convection: require a
medium to transport energy
Fluid
Conduction +
(a) (b) Conduction/
Convection
convection
Fig.1 Heat transfer in a (a) fin and (b) pipe flow
Radiation : no carrier or medium is required

5. Radiation : no carrier or medium is
required
 Medium: Participating & Non-participating
Q
Q vacuum
Hot
q body
q<Q Q Enclosure
Participating Non-participating
Medium For opaque materials,
absorption, reflection
For gas and semi-transparent materials, and emission takes place
radiation interaction takes place within at the surface
the volume of the medium.

6. RADIATION HEAT TRANSFER
Fundamental Concepts
 Irradiation: radiation incident on a surface, W/m 2
reflected
incident
Absorptance = Gabsorbed/Gincident
=α
Reflectance = Greflected/Gincident diffuse
transmitted
=ρ
Transmittance = Gtrasmitted/Gincident
=τ specular
Radiosity: radiation leaving a surface
due to reflection and emission radiosity

7. Absorptance, Reflectance and Transmittance
For semi transparent material
∴ ρλ + αλ + τλ = 1
If the properties are averaged over the entire
spectrum
ρ+α+τ=1
For opaque medium –
ρλ + αλ = 1
ρ+α=1

8. RADIATION HEAT TRANSFER
Fundamental Concepts
 Nature of radiation
X-rays Ultraviolet
Thermal Microwave
Gama rays Radiation
10-5 10-4 10-3 10-2 10-1 1 10 102 103 104
Figure: Spectrum of electromagnetic radiation

9. RADIATION HEAT TRANSFER
Fundamental Concepts
Thermal Radiation
Spectral Range for thermal radiation:
0.2 µm ≤ λ ≤ 100 µm
0.01 ≤ λ ≤ 0.36 µm  uv
0.36 ≤ λ ≤ 0.76 µm  visible
(0.55 – 0.56µm)  most sensitive
λ > 0.76 µm  Infrared
Thermal radiation and light: Light is the visible part of
thermal radiation

10. RADIATION HEAT TRANSFER
Fundamental Concepts
 Radiation is emitted at different wavelength and
different direction
Spectral distribution Directional distribution
θ
Eλ
wavelength
(a) (b)
Eλ= Monochromatic radiation emission
Figure Radiation emitted by a surface. (a) Spectral distribution.
(b) Directional distribution

11. RADIATION HEAT TRANSFER
Fundamental Concepts
 Plane and Solid angles
r r
dω dα dl
dAn
Solid Angle = ratio of the element, Plane angle = ratio of the element
dAn and the square of arc length and
of the radius, r radius, r
= dAn / r2 = dl/r

12. RADIATION HEAT TRANSFER
Fundamental Concepts
 Radiation Intensity
N
Intensity I of radiation
at any λ is defined as
dAn
the rate at which the
θ
radiant energy is θ
emitted in (θ,φ) dA1
direction per unit dAn X
per unit solid angle,
per unit wavelength. φ φ
W/(m2.sr. µm)
Z (a) (b)
Figure (a) Directional nature of radiation,
(b) spherical co-ordinate system

13. RADIATION HEAT TRANSFER
Fundamental Concepts
 Blackbody Radiation
♣ absorbs all radiation irrespective of
wavelength and direction;
♣ no surface can emit more energy than a
blackbody for a prescribed temp. and
wavelength;
♣ the blackbody is a diffuse emitter.
I λi I λe= I λb
Gλ= E λb
Isothermal blackbody enclosure

14. RADIATION HEAT TRANSFER
Fundamental Concepts
 The Planck Distribution
Radiation intensity for blackbody emission is given by
2
2hco
Iλ,b (λ =
,T) λ [exp(hco / λ ) −1]
5
kT
h = universal Planck constant = 6.6256 x 10 -34 (J.s)
k = Boltzman constant = 1.3805 x 10 -23 (J/K)
co = speed of light = 2.988 x108 (m/s)
T = absolute temp. of the blackbody (K)

15. RADIATION HEAT TRANSFER
Fundamental Concepts
 Planck Distribution Equation
Since the blackbody is a diffuse emitter,
Eλ,b (λ = µIλ,b (λ
,T) ,T)
C1
= λ [exp(C 2 / λ ) −1]
5
T
Where,
C1 = 2πhco2 = 3.742 x 108 (W.µm4/m2)
C2 = (hco/k) = 1.439 x 104 (µm.K)

16. Example
 Emissive power of a blackbody is 1kW/
(m2.µm) at a wavelength of 4 µm. Find the
temperature of the body.
 Solution:
Given- λ = 4 µm
Ebλ(T) = 1kW /(m2.µm)
Find – T, the temperature of the
blackbody.

17. Eλ,b (λ = µIλ,b (λ
,T) ,T)
C1
= λ5 [exp(C 2 / λT ) −1]
Where,
C1 = 2πhco2 = 3.742 x 108 (W.µm4/m2)
C2 = (hco/k) = 1.439 x 104 (µm.K)

18. 8
3.743x10
1000 = 5
4 [exp{1.439 x10 /(4T )} − 1]
4
4
1.439 x10
= ln 366.53
4T
T = 609 K

19. Exercise
There is a small circular opening of 40 mm
in diameter in a large spherical cylinder
whose inner surface is maintained at 527oC.
Find the rate of emission of radiation
through this opening.

20. Outline of solution
 Emissive power at the given temp
Eb(T) =σT4
=(5.67x10-8)(800)4
Q = A Eb(T) , Area of the aperutre

21. RADIATION HEAT TRANSFER
Fundamental Concepts
 Wien’s Displacement Law
The black body spectral distribution is characterised by
maximum and the wavelength associated with this maximum
depends upon temperature. Differentiating the following
equation w.r.t λ
C1
Eλ,b (λ =
,T) λ [exp(C 2 / λ ) −1]
5
T
and setting it to zero, gives
λ maxT = C3 = 2897.6, µ m.K

23. RADIATION HEAT TRANSFER
Fundamental Concepts
 The Stefan- Boltzman Law
Total emissive power
∞
E ( )
E = ∫ λλλ d
0
∞ C1
= ∫ dλ
0
λ ( 2 /λ −
5
[
exp C T ) 1]
=σ4T
σStefan Boltzman constant
=
= 5.67 x 10-8 (W/m2.K4)

24. RADIATION HEAT TRANSFER
Fundamental Concepts
 Band Emission
Fraction of radiation emitted within Eλb(λT)
a wavelength band:
λ
0 λ
∫ λb dλ
E , λ1
F 0- λ= 0
∞
= f(λ
T) Σ Eλb(λT)dλ
∫ λ dλ
0
E ,b
λ2 λ
1
∫ λ dλ ∫ λ dλ
E −E
,b ,b
Fλ λ =
1- 2
0
∞
0
= f(λ
T)
∫ λ dλ
0
E ,b
= F (0-λ - F(0-λ
2) 1)
λ1 λ2

26. Radiation Example
 Example 3
Consider a large isothermal enclosure that is maintained
at a uniform temperature of 2000 K.
1. Calculate the emissive power of the radiation that
emerges from a small aperture on the enclosure
surface.
2. What is the wavelength λ 1 below which 10% of the
emission is concentrated. What is the wavelength
above which 10% of the emission is concentrated.
3. Determine the maximum emissive power and the
wavelength at which this emission occurs.
4. If a small object is placed within the enclosure, what is
the irradiation incident on the object

27. Example 3 (contd)
Enclosure T = 2000 K
Eλ, b (T)
10% 10%
λ1 λ λ2
Assumption: aperture area is very small compared to the
surface area of the enclosure

28. Example 3-Solution
 W 
1. E = Eb (T) = σT4 = 5.67 x 10-8  2 4 (2000)4 (K)4
m K 
= 9.07 x 105 W/m 2
2. F0−λ1 = 0.1 from table 1
λ1T ≅ 2200 µmK
λ1 = 1.1 µm
F0−λ 2 = 0.9 λ2T ≅ 9382 µmK
λ2 = 4.69 µm

29. Example 3-Solution(contd)
3. From Wien’s Displacement Law
λmaxT = 2898 µmK
For T = 2000, λmax = 1.45 µm
Maximum emissive power:
[E λ,B ( λ, T ) ] max = C1
λ5 max [ exp( C 2 / λ max T ) − 1]
(
3.742 × 10 8 W ⋅ µm 4 / m 2 )
=   1.439 × 10 4 ( µm ⋅ K )  
(1.45) µm exp
5 5
 2898( µm ⋅ K )  − 1

 
  

= 4.10 x 105 W/m2 µm

30. Example 3-Solution(contd)
 4. The answer is same as (1)
 W 
E = Eb (T) = σT4 = 5.67 x 10-8  2 4 (2000)4 (K)4
m K 
5 2
= 9.07 x 10 W/m

31. Example 4
A diffuse surface at a temperature of 1600 K has the
spectral, hemispherical emissivity shown below:
ε2
0.8
ελ (λ) ε1
0.4
0
0 2 4 6
λ, µm
1. Determine total, hemispherical emissivity
2. Calculate the total emissive power
3. At what wavelength will the spectral emissive power
be a maximum?

32. Example 4-solution
Assumption: Surface is a diffuse emitter.
Equation (37)
∞
∫ ε λ (λ , T ) E λ
0
,b ( λ , T ) dλ
ε (T) = E b (T ) (37)
2 5
∫ ε Eλ
0
1 ,b dλ ∫ε
2
2 Eλ ,b dλ
= Eb +
Eb
= ε1 F(0 - 2µ m) + ε2 F(2 - 5µ m)
or ε = ε1 F(0 - 2µ m) + ε2[F(0-5µ m) – F(0 – 2 m)]

33. Example 4-solution(contd)
λ1T = 2 (µm) x 1600 (K) = 3200 (µm⋅K); F0 – 2 = 0.318
λ2T = 5 (µm) x 1600 (K) = 80, 000 (µm⋅K); F0 – 5 = 0.856
ε = 0.4 x 0.318 + 0.8 (0.856 – 0.318) = 0.558

34. Example 4-solution(contd)
2. E = εEb = εσT4
= 0.558 x 5.67 x 10-8 (W/m2 K4) 16004 (K4)
= 207 kW/m2
2898
3. λmax = = 1.81 µm
1600
Since ε is different for different λ, we have to calculate
emissive power for both

35. Exercise
 The filament of tungsten bulb is heated to a
temperature of 2227oC. Find the fraction of
the enrgy in the visible range. The visible
range of the spectrum may be considered as
0.4 µm ≤ λ ≤ 0.7 µm

36. Example
 Example 7
A flat plate collector with no cover plate has a selective
absorber surface emittance of 0.1 and a solar absorptance
of 0.95. At a given time of the day, the absorber surface
temperature is Ts = 120 oC when the solar irradiation is
750 W/m2, the effective sky temperature is -10 oC and the
ambient air temperature is 30 oC. Assume that the convective
heat transfer coefficient for the calm day condition can be
estimated from the following equation is:
h = 0.22 (Ts -T∞ )1/3 W/m2.K
Calculate the useful heat removal rate from the collector
for these conditions. What is the efficiency of the collector
for these conditions?

37. Example
 Example 7 - contd
Tsky = -10 oC
T∞ = 30 oC
ε=0.1, G= 750 W/m2
α =0.95
qu

38. Example 7 - contd
Assumption:
a) Steady-state conditions
b) Bottom of collector well insulated
c) Absorber surface is diffuse
1. Energy balance on the absorber
Ein – Eout = 0
Or αsGs + αsGsky – qcon – E – qu = 0
Gsky = σTsky4

39. Sky radiation is concentrated in approximately the same
spectral region as that of the surface emission and as such it
may be reasonable to assume.
αsky ≅ ε = 0.1 4
3
qcon = h (Ts – T∞) = 0.22 (Ts - T∞)
and E = εσTs4
Hence, qu = αsGs + αsky Gsky* - qcon – E*
= 0.95 (750 W/m2) + 0.1 x 5.67 x 10-8 (W/m2K4)
x (393)4K4- 0.22(393-303)1.33 – 0.1x5.67x108
263 (W/m2K4)x3934K4
qu = 712.5 (W/m2) + 27.127 (W/m2) – 87.4 (W/m2) –
135.25 (W/m2)
= 516.9 W/m2
q u 516.9
2. Collector efficiency, η= Gs
=
750
= 0.689
qu = αsGs - εσ (Ts4 – Tsky4) – h (Ts - T∞)
4
= αsGs - εσ (Ts4 – Tsky4) - 0.22 (Ts - T∞) 3

40. Radiation Exchange between Surfaces
 Assumption:surfaces are separated by non-
participating medium which neither emits,
absorb or scatter radiation.
a) black surfaces,
b) diffuse-gray surfaces, and
N
c) enclosures.
The view factor: The view factor Fij is defined as the fraction
of the radiation leaving surface i, which is intercepted by A , T
j j
surface j.::
Reciprocity relation: Fij Ai = Fji Aj :
A1F12=A2F21 R
Summation Rule: Σ J=1 N Fij = 1
1 2 F11 + F12 =1
1
Ai, Ti

41. The view factor: Example
 The view factor F1-3 between the base and
the top surface of the cylinder shown in
Figure can be found from charts. Develop
expressions for the view factors F1-2 and
F2-1 between the base the lateral cylindrical
surface in terms of F1-3.
3
2 H
1
R

43. Solution
From the summation rule
F1-1 + F1-2 + F1-3 = 1
Since F1-1 = 0, the above equation reduces to the following
form
F1-2= 1-F1-3
From the reciprocity relation,
A1F1-2 =A2F2-1
Hence, F2-1= (A1/A2) F1-2
(Find the values of the shape factors, if R = 40 mm and H =
100 mm)

44. Blackbody Radiation Exchange

45. Blackbody Radiation Exchange
Consider radiation exchange between two black surfaces –
qi.→j = (AiJi) Fij (74)
where qi→j = rate at which radiation leaves surface i
and is intercepted by surface j
For a black surface, radiosity = emissive power
.
. . qi→j = Ai Ebi Fij (75)
Similarly, qj→i = Aj Ebj Fji (76)
.
. . net radiative exchange between the two surfaces,
qij = qi→j – qj→i (77)
= Ai Ebi Fij - Aj Ebj Fji
= Ai Fijσ (Ti4 – Tj4) (78)
(since Ai Fji = Aj Fji)

46. Example 10
A furnace cavity, which is in the form of a cylinder of
diameter D = 75 mm and length L = 150 mm, is open at
one end to surroundings that are at a temperature of 27oC.
The sides and bottom may be approximated as blackbodies,
heated electrically, and are maintained at temperatures of
T1 = 1350oC and T2 = 1650oC ,respectively. The sides and
bottom are considered to be well insulated.
Side, T1
Heater wire
L
Insulation
Bottom, T2
How much power is required to maintain the furnace at the
prescribed conditions?

47. Example 10 - contd
A q
T3 = Ts Tsurr = 300 K
3
= Ts
Solution:
Assumption: T1 = 1623 K
L = 0.15m
1. Interior surfaces behave A2, T2 = 1923 K
as blackbodies
D = 0.075m
2. Negligible heat transfer by convection
3. Outer surface of the furnace is adiabatic

48. Power required = Heat losses from the furnace
Heat loss is due to radiation (other losses have been
neglected) from the hypothetical surface A3. The
surroundings are large; the heat transfer from the furnace to
the surroundings may be approximated by treating the
surface A3 to be at T3 = Tsur .
Heat balance, q = q13 + q23
= A1 F13σ (T14 – T34) + A2 F23σ (T24 – T34)
0.0375 L 0.15
rj /L = 0.15
= 0.25; =
ri 0.0375
= 4, From Fig 26
F23 = 0.06

49. From summation rule, F21 + F22 + F23 = 1
F21 = 1 - F23 = 0.94 since F22 = 0
From reciprocity theorem,
A1F12 = A2F21
A2
F12 = F21
A1
= [{0.25 π(0.075 m)2}/{
π(0.075mx0.15m)}]x0.94
= 0.118
From symmetry, F13 = F12

50. Therefore, q
-8 4 4
= π x 0.075 x 0.15 x 0.118 x 5.67 x 10 [1623 -300 ]
+0.25 π (0.075)2 x 0.06 x 5.67 x 10-8 [19234 – 300 4]
= 1639 + 205 W
= 1844 W

51. Radiation exchange between diffuse-gray
surfaces in an enclosure
 Black surfaces are ideal surface, difficult to achieve in
reality.
 In real surfaces, complication arises due to multiple
reflection, with partial absorption occurring each time.
Likely assumptions:
1. Isothermal surface characterized by
uniform radiosity and irradiation.
2. opaque – diffuse surface
3. Non-participating medium.
Problem: To determine net radiative heat flux from each
surface.
(Ti associated with each surface is known.)

52. Radiation exchange between diffuse-gray
surfaces in an enclosure (contd)
Tj, Aj, εj
T1, A1, ε1
Ji qi Gi
Ti, Ai, εi
Fig. 29 Radiation exchange in an enclosure consisting
of gray surfaces and non-participating medium

53. Net radiation exchange at a surface
qi
Ji,Ai Gi,Ai
Qi = Ai (Ji – Gi) (79)
Ai
qi is the net rate at which radiation
leaves surface i.
Radiosity: ρ iGi,Ai Gi,Ai
Ei,Ai
α iGi,Ai
Ji = Ei + ρ iGi (80)
qi = Ai (Ei – α iGi) (81)
since, ρi = 1 - α i for an opaque surface
= 1 - ε i for opaque, diffuse gray surface
.
. . Ji = ε i Ebi + (1 - ε i) Gi (82)
J i - ei Fbi
or Gi= 1- ε
i

54. Substituting in (79) Ji
qi
Ebi
= (83)
qi
where, Ebi – Ji = driving potential
= surface radiative resistance

55. Tj, Aj, εj
T1, A1, ε1
Ji qi Gi
Ti, Ai, εi
Fig. 29 Radiation exchange in an enclosure consisting
of gray surfaces and non-participating medium

56. Radiation exchange between surface
For an enclosure, the irradiation at surface I,
N
Ai Gi = ∑ F ji A j J j
j =1
N
= ∑F A J
j =1
ij i j (from reciprocity theorem)
N
.
. . Gi = ∑F J
j =1
ij j
Substituting into (79) gives
 N 
qi = Ai  J i − ∑ Fij J j 
 
 j =1 

57. N
using ∑F
j =1
ij =1
 N N 
qi =  ∑ Fij J i − ∑ Fij J j 
Ai  
 j =1 j =1 
hence, qi = ∑ A F ( J − J j ) = ∑ qij
N N
i ij i
j =1 j =1 (84)

58. where qij =net radiation exchange
between i & j J1 qi1
(AiFi1)-1
qi Ebi Ji J2
-1
(AiFi2) qi2
1−εi (AiFi3)-1
(AiFin)-1 J3
Aiεi
Jn qi3
qin
Combining equations (83) & (84)
N J −J
Ebi − J i
=∑ i j
(85)
(1 − ε i ) / ε i Ai j =1 ( Ai Fij ) −1
Note:1. Equation (85) is useful when the surface
temperature Ti (and hence Ebi ) is known.
1. Equation (84) is useful when the net radiation
transfer rate, qi , is known.

59. The two-surface Enclosure:
A2,T2,
1 − ε2
1 − ε1 (A1F12)-1 A 2ε 2
q1 A1ε1 J2 Eb2
Eb1 q1 J1 q12 q1 -q2 q12
A1, T1, ε1
Since there are only two surfaces,
Net radiation transfer from surface 1, q1 = net radiation transfer to surface 2, - q2
q1 = - q2 = q12
E bi - J1 J 2 - E b2
q1 = ; -q2 =
1- ε1 1- ε 2
ε1A1 ε2A2
J - J
q12 = 1 2
1
A1F 12

60. 1− ε   1 − ε1 
Eb1 - J1 = q1  
 Aε 
 A ε  = q12
  1 1 

 1 1
 1 
J1 - J2 = q12 
AF  
 1 12 
1− ε 2  1− ε 2 
J2 - Eb2 = -q2   
 A ε  = q12
  Aε
 2 2


 2 2 
1 − ε1 1 1− ε2
Adding Eb1 - Eb2 = q12 [ + + ]
A1ε 1 A1 F12 A2ε 2
(
= σ T1 − T2
4 4
)

61. (86)
Eqn (86) may also be written as:

62. r1 r2
Concentric sphere Concentric Finite Infinite parallel
Cylinders plane
A1 r12 A1 r
= 2 ; F12 = 1 = 1 ; F12 = 1 A1 = A2 = A; F12 = 1
A 2 r2 A 2 r2

63. Radiation Shields
Radiation shield, constructed from low emittance
materials can be used to reduce heat transfer
.
between two surfaces.
q12
1 − ε31 1 − ε32 1 − ε2
1 − ε1 -1 -1
(A1F13) A3ε31 A3ε32 (A3F32) A 2ε2
q1 A1ε1 J1 Eb3 Eb2
J31 q13 q32
Eb1 J32 J2
q1 ε32 -q2
ε31
A1, ε1, T1 A2, ε2, T2
A3T3

64.  Total Resistance
 Rtotal = R1+R2+R3+R4+R5+R6
Where,
 R1= (1-ε1)/A1ε1
 R2= (A1F13)-1
 R3= (1-ε31)/A3ε31
 R4= (1-ε32)/A3ε32
 R5= (A3F32)-1
 R6= (1-ε2)/A2ε2

66. For large parallel plates
σA1 (T14 − T2 4 )
q12 =
1 1 1 − ε 31 1 − ε 32
+ + +
ε1 ε 2 ε 31 ε 32
when ε31 and ε32 are small, the resistance become very
large.
Special case:
ε1 = ε2 = ε31 = ε32
One radiation shield reduces the radiation heat transfer by
50%

68. Example 12
A cryogenic fluid flows through a long tube of diameter D 1
= 20 mm, the outer surface of which is diffuse-gray with ε1
= 0.03 and T1 = 77 K. This tube is concentric with a larger
tube of diameter D2 = 50 mm, the inner surface being
diffuse-gray with ε2 = 0.05 and T2 = 300 K. The space
between the surfaces is evacuated.
1. Calculate the heat gain by cryogenic fluid per unit length
of the tube.
2. If a thin radiation shield of diameter D = 35 mm and
emittance, ε3 = 0.02 (both sides) is inserted midway
between the inner and outer surfaces, calculate the
change in percent of heat gain per unit length of the tube.

69. Example 12-solution
Solution Network: (without shield)
D2 = 50 mm 1 − ε2
T2 = 300 K
1 − ε1 (A1F12)-1 A 2ε2
Radiation Shield Eb1 A1ε1 J2 Eb2
D3 = 35 mm ε2 = 0.05
ε3 = 0.02 J1
Network: (with shield)
Eb1 J1 J31 Eb3 J32 J2 Eb2
D1 = 20 mm
T1 = 77 K R1 R2 R3 R4 R5 R6
ε1 = 0.02

70. Example 12-solution(contd)
 Total Resistance
 Rtotal = R1+R2+R3+R4+R5+R6
Where,
 R1= (1-ε1)/A1ε1
 R2= (A1F13)-1
 R3= (1-ε31)/A3ε31
 R4= (1-ε32)/A3ε32
 R5= (A3F32)-1
 R6= (1-ε2)/A2ε2

71. Example 12-solution(contd)
1. Heat gain by cryogenic fluid, q
Heat gain per unit length of the tube

72. Example 12-solution(contd)
=
=

73. Example 12-solution(contd)
2. When a radiation shield is placed in between the two concentric
tubes, heat gained by the cryogenic given by
where,
=
=

74. Example 12-solution(contd)
Hence :
q' = =
= -0.25,
Percentage reduction in heat gain
w = with shield
w o = without shield
= 49.8 %