Radiation

Radiation

• Radiation
- Energy Exchange through space at the speed of light
-Radiation may occur due to electron bombardment, electric
discharge etc

• Thermal Radiation
- All substances above absolute zero temperature emit
radiation which is independent of the external agency
- Radiation that is result of temperature only is called
thermal radiation
Emissivity, Reflectivity and Transmittivity
• Radiations move through space in straight lines or beams.
• Only substance in the sight of the radiating body can intercept
radiations
• From the intercepting body:
• Fraction of radiation falling on the body that is reflected is called reflectivity
of the body (r )
• Fraction of radiation falling on the body that is absorbed is called
absorptivity of the body (a)
• Fraction of radiation falling on the body that is transmitted is called
transmittivity of the body ( t)
• A sum of these three fractions is unity r+ a+ t = 1
Specular and Diffused Reflections
• Specular Reflections-
Angle of incident is
same as that of the
angle of reflection. It
presents mirror of the
source to the observer

• Diffused Reflections-
Incident beam is
distributed in all
directions after
reflection
 Thermal Radiations

Thermal radiation is electromagnetic radiation emitted from all matter that is

at a non-zero temperature in the wavelength range from 0.1 μm to 100 μm. It
includes part of the ultraviolet (UV), and all of the visible and infrared (IR)
Emissive power
• Energy emitted by the body per unit area per unit time

• Total Emissive power E

• Total radiant energy emitted by the surface in all directions over the
entire wavelength range per unit time per unit area

• Monochromatic Emissive power Eλ

• Total radiant energy emitted by the surface at a given wavelength per
unit time per unit area in all directions is known as monochromatic
emissive power
BLACK BODY
• Radiation is not heat and when transformed to heat on
ABSORPTION it is no longer radiation.
• The transmitted and reflected radiations usually fall on other
body and eventually converted to heat perhaps after many
successive reflections.
• Maximum possible absorptivity = 1
• Only if a body absorbs all incident radiations does not reflect
or transmit any radiation
• A body that absorbs all incident radiations is called BLACK
BODY
Stefan-Boltzman Law
• The amount of radiant energy emitted per unit time per unit
area of a black body at absolute temperature T is directly
proportional to the fourth power of the temperature.

EB α T4 EB = σ T4
σ = Stefan Boltzman Constant (5.67x10 ̄8 W/ m²K⁴)

Normally a body radiating heat is simultaneously receiving heat

from other body as radiation.
Consider a black surface 1 at temperature T1 is completely
another black surface 2 at temperature T2. The net radiant heat
flux is given by
Qnet = σ (T14 – T24)
Emissivity of non-black surfaces (Ɛ )
• The ratio emissive power of the nonblack body to that of the
black body is known as emissivity of the nonblack body.
Ɛ = E/Eb E= ƐσTb4
Kirchhoff's Law
• The ratio of emissive power of a body to the absorptivity is same for all
bodies and equal to the emissive power of a black body at that
temperature.
• Consider two surfaces, one absolutely black at temp Tb another nonblack
surface at temp T. These two surfaces are arranged parallel to one
another and are so close that the radiations of one falling on another.
The radiant energy E emitted by a nonblack surface impinges on a
E black and completed absorbed.
The radiant energy Eb emitted from the black surface strikes on the
Eb nonblack surface.
If the absorptivity of the nonblack surface is a, it will absorb aEb
radiations and (1-a)Eb will be reflected back for full absorption on the
black surface.
(1-a)Eb
The radiant interchange for the nonblack would be E -aEb
Hence,
E –aEb = 0 E =aEb E/a = Eb E/Eb= a

The relationship can be extended by considering different surfaces

E1/a1 = E2/a2= E3/a3 = E4/a4 = Eb/ab = Eb = f (T)

E1/ Eb = a1 but E1/ Eb = Ɛ1 a1 = Ɛ1

The emissivity and the absorptivity of real surfaces are equal for radiations
with identical temperature and wavelengths
Wien’s Displacement Law
• The product of the absolute temperature and the wavelength at which
emissive power is maximum is constant
λmT = 0.0029 mK

https://physhydro.community.uaf.edu/unit-2/
 Heat Exchange between black bodies
• Heat exchange
• The net energy exchange from A1 to A2 or A2
between two to A1 depend on what portion of radiation by
radiating bodies 1 one will be intercepted by the other
and 2 A2 • Solution to this- Radiation shape factor/view
T2 factor/configuration factor/geometrical
factor
r • “The fraction of radiative energy that diffused
from one surface element and strikes the
other surface with no intervening
A1 reflecctions”
T1 • Fij – Shape factor from surface i to surface j
• F12 – Shape factor from surface 1 to surface 2
• F12 = (Direct radiations from surface 1 incident on surface 2)/
( Total radiations emitting from surface 1)
F12 = Q12 / σ1A1T14
Q12 = F12 σ1A1T14
Similarly, F21 shape factor from surface 2 to surface 1
F21 = Q21 / σ2A2T24
Q21 = F21 σ2A2T24

Net energy exchange from surface 1 to surface 2 is

Q12(net) = F12 σ1A1T14 - F21 σ2A2T24
Reciprocity theorem
When the surfaces 1 and 2 are maintained at same temperature i.e.
T1= T2 =T, there is no net heat exchange.
Hence , (F12 A1 - F21 A2 ) σbT4 = 0 (As σb=σ1=σ2)
As T and σb are both nonzero quantities
F12 A1 - F21 A2 = 0
F12 A1 = F21 A2-------------------------------Reciprocity theorem
Hence,
Q12 (net) = F12 A1 σb(T14 – T24) Valid only for
= F21 A2 σb(T14 – T24) black surfaces
Shape factor- Perpendicular plates
Shape factor- Parallel Discs
Shape factor- Parallel plates
Two parallel black 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. What is the net
radiant heat exchange between the two plates?
X= 1, Y = 0.5,
D = 0.5
X/D= 1/0.5 =2
Y/D = 0.5/0,5 = 1

The Shape factor F12 = 0.285

Q12 = A1 F12(E1-E2)
= σ A1 F12 (T14-T24)
= (5.67*10-8)(0.5)(0.285) (12734-7734)
= 18333 Watts
=18.333 kW
When the radiating surface exchanges heat with number of
black surfaces, the net heat transfer from the radiating surface
will be:
Q1 (net) = F12 A1 σb(T14 – T24)+ F13 A1 σb(T14 – T34)+ F14 A1 σb(T14 – T44)+
F15A1 σb(T14 – T54)+….+

If the radiations of the surface 1 are intercepted by ‘n’

surfaces then
F12 + F13+ F14 + F15 + F16 + ------- + F1n = 1
• A large black enclosure consists of a box as shown in the figure
below:
20 cm
The bottom surface 1 is at temperature 530 K,
Top the top surface 2 is at 450 K and the vertical
25cm surface 3 is at 475 K . Find the net heat transfer
2
Verticalrate Q12 and Q13. Take shape factors F12 =
3 0.168 and F13 = 0.185.
Q12 (net) = F12 A1 σb(T14 – T24)
= (0.168)(0.25*0.2)(5.67*10-8)(5304-4504)
= 18.03 W
1
Bottom Q13 (net) = F13 A1 σb(T14 – T34)
25 cm
= (0.168)(0.25*0.2)(5.67*10-8)(5304-4754)
= 14.68 W
Q 1 net = 18.03 + 14.68 = 32.71 W
 Heat Exchange between nonblack bodies
Plane at temp T1 Plane at temp T2 Infinite parallel plates
1. Two nonblack surfaces of
E1 equal area at small
distance, all radiations
(1-a2)E1 a2E1 emitted from one surface
a1(1-a2)E1 fall on other
(1-a1)(1-a2)E1 2. The shape factor is unity
3. Surfaces are diffuse and
(1-a1)(1-a2)2E1 a2(1-a1)(1-a2)E1 uniform temperature so
a1(1-a1)(1-a2)2E1 that reflective and
(1-a1)2(1-a2)2E1 emissive properties are
a2(1-a1)2(1-a2)2E1 constant over all the
(1-a1)2(1-a2)3E1 surface
a1(1-a1)2(1-a2)3E1 4. The surfaces are
separated by non-
absorbing medium such as
air
Heat Exchange between nonblack bodies
𝑄1
= E1-a1(1-a2)E1[1+ P+ P2 +P3+--- ]…..P= (1-a1)(1-a2) is less than unity
𝐴1

Plane at temp T1 Plane at temp T2 Infinite parallel plates

1. Surface 1 emits radiant
E1 energy E1 falls on surface
2
(1-a2)E1 a2E1 2. Surface 2 absorbs a2E1
a1(1-a2)E1 and reminder (1-a2)E1 is
(1-a1)(1-a2)E1 reflected back to surface 1
3. Surface 1 absorbs a1 (1-
(1-a1)(1-a2)2E1 a2(1-a1)(1-a2)E1 a2)E1 and reminder is(1-
a1(1-a1)(1-a2)2E1 a1)(1-a2)E1 reflected back
(1-a1)2(1-a2)2E1 to surface 2

(1-a1)2(1-a2)3E1 a2(1-a1)2(1-a2)2E1
a1(1-a1)2(1-a2)3E1 𝑄1
= E1-a1(1-a2)E1[1+ (1-a1)(1-a2)+(1-
𝐴1
a1)2(1-a2)2 +(1-a1)3(1-a2)3 ]
Q12 = E1-a1(1-a2)E1[1+ P+ P2 +P3+--- ]…..P= (1-a1)(1-a2) is less than unity

For P= is less than unity the series [1+ P+ P2 +P3+--- ] when

expended to infinity give 1/(1-P).
Hence,
𝑄1
= E1-a1(1-a2)E1[1+ P+ P2 +P3+--- ] = [E1-a1(1-a2)E1]/[1-P]
𝐴1

𝑄1
= [E1-a1(1-a2)E1]/[1-(1-a1)(1-a2)]
𝐴1
From Kirchhoff’s emissivity and absorptivity of a surface are equal.
a1=e1 and a2=e2
𝑄1
= E1[1-e1 (1-e2)]/[1-(1-e1)(1-e2)]
𝐴1
𝑄1
= E1[1-e1 (1-e2)]/[1-(1-e1)(1-e2)]
𝐴1
= E1{[1-(e1 (1-e2)/1-(1-e1)(1-e2)]}
𝑄1 1 − 1 − 𝜀1 1 − 𝜀2 − 𝜀1 1 − 𝜀2
= 𝐸1 [ ]
𝐴1 1 − 1 − 𝜀1 1 − 𝜀2

𝑸𝟏 𝟏 − 𝟏 − 𝜺 𝟐 − 𝜺𝟏 + 𝜺𝟏 𝜺𝟐 − 𝜺𝟏 − 𝜺𝟏 𝜺𝟐
= 𝑬𝟏 [ ]
𝑨𝟏 𝟏 − 𝟏 − 𝜺𝟐 − 𝜺𝟏 + 𝜺𝟏 𝜺𝟐

𝑸𝟏 𝟏 − 𝟏 + 𝜺 𝟐 + 𝜺𝟏 − 𝜺𝟏 𝜺𝟐 − 𝜺𝟏 + 𝜺𝟏 𝜺𝟐 ]
= 𝑬𝟏 [ ]
𝑨𝟏 𝟏 − 𝟏 + 𝜺𝟐 + 𝜺𝟏 − 𝜺𝟏 𝜺𝟐 ሻ

𝑸𝟏 𝜺𝟐
= 𝑬𝟏 [ ]
𝑨𝟏 𝜺𝟐 + 𝜺𝟏 − 𝜺𝟏 𝜺𝟐 ሻ
If surface 2 emits radiation of emissive power E2
𝑸𝟏𝟐 = 𝒇𝟏𝟐 𝑨𝟏 𝝈𝒃 𝑻𝟒𝟏 − 𝑻𝟒𝟐
𝑸𝟐 𝜺𝟏 𝒇𝟏𝟐 =
𝜺𝟏 𝜺𝟐
=
𝟏
= 𝑬𝟐 [ ] 𝜺𝟏 + 𝜺𝟐 − 𝜺𝟏 𝜺𝟐 𝟏
+ −𝟏
𝟏
𝑨𝟐 𝜺𝟐 + 𝜺𝟏 − 𝜺𝟏 𝜺𝟐 ሻ 𝜺𝟏 𝜺𝟐

The net heat flow from surface 1 to surface 2 per unit time sis given by:
𝑄12 𝑄1 𝑄2
= − but A1 = A2 𝑄12 = 𝑄1 - 𝑄2
𝐴1 𝐴1 𝐴2
𝑄12 𝐸1 𝜀2 − 𝐸2 𝜀1 𝑄12 𝜀1 𝜎𝑏 𝑇14 𝜀2 − 𝜀2 𝜎𝑏 𝑇24 𝜀1
= 𝐴 = 𝐴
𝐴1 𝜀1 + 𝜀2 − 𝜀1 𝜀2 𝐴1 𝜀1 + 𝜀2 − 𝜀1 𝜀2
From Stefan- Boltzmann law, for nonblack surfaces
𝜺𝟏 𝜺𝟐
𝐸1 = 𝜀1 𝜎𝑏 𝑇14 𝑸𝟏𝟐 = 𝑨𝝈𝒃 (𝑻𝟒𝟏 − 𝑻𝟒𝟐 ሻ
𝜺𝟏 + 𝜺𝟐 − 𝜺𝟏 𝜺𝟐
𝐸2 = 𝜀2 𝜎𝑏 𝑇24
Interchange factor f12
𝟏
For parallel plates 𝒇𝟏𝟐 = 𝑸𝟏𝟐 = 𝒇𝟏𝟐 𝑨𝟏 𝝈𝒃 𝑻𝟒𝟏 − 𝑻𝟒𝟐
𝟏 𝟏
+ −𝟏
𝜺𝟏 𝜺𝟐

𝟏
For Concentric cylinders 𝒇𝟏𝟐 =
𝟏 𝑨𝟏 𝟏
+ −𝟏
𝜺𝟏 𝑨𝟐 𝜺𝟐

𝟏
For Concentric spheres 𝒇𝟏𝟐 = 𝟏 𝑨𝟏 𝟏
+ −𝟏
𝜺𝟏 𝑨𝟐 𝜺𝟐
Example1
• Calculate the net radiant heat interchange per square meter of very
large planes at temperatures 703 K and 513 K respectively. Assume
that the emissivities of the hot and cold planes are 0.85 and 0.75
respectively.
𝝈𝒃 (𝑻𝟒𝟏 −𝑻𝟒𝟐 ሻ
• (𝑸/𝑨ሻ𝟏𝟐 = 𝟏 𝟏
+ −𝟏
𝜺𝟏 𝜺𝟐

5.67 × 10−8 7034 − 5134

=6571 Watts
1 1
+ −1
0.85 0.75
 Example 2
• Calculate the loss of heat by radiation from a steel tube of diameter
70 mm and 3 m long at temperature of 500 K if the tube is located in
a square brick conduit of 0.3 m side at 300 K. Assume emissivity of
steel and brick as 0.79 and 0.93 respectively. 𝟏
𝒇𝟏𝟐 =
𝟏 𝑨𝟏 𝟏
+ −𝟏
𝜺𝟏 𝑨𝟐 𝜺𝟐
𝐴1 𝜎𝑏 𝑇14 − 𝑇24
𝑄12 = =1590 Watts
1 𝐴1 1
+ −1
𝜀1 𝐴2 𝜀2
Electrical Network approach for radiation heat exchange

Radiocity:
Total radiant energy leaving the surface per unit time per unit
area. (Original emittance + reflected portion of radiations incident on it)

Irradiation:
Total radiant energy incident upon the surface per unit time per unit
area (some of it may be reflected to become a part of radiocity)
• A non black opaque surface
Total radiant energy leaving the surface J =
Original emittance (E) of the surface + the energy reflected back (rG)
J=E+ rG
J= 𝜺Eb + rG = 𝜺Eb + (1-a)G For Opaque Surface

Now according to Kirchoff's Law a = 𝜺

J = e Eb + (1 - e )G J - e Eb
G=
1- e
The net rate of energy leaving the surface is given by the difference between
radiocity and the irradiation
𝑸 𝑱 − 𝜺𝑬𝒃
( ሻ =𝑱−
𝑨 𝒏𝒆𝒕 𝟏−𝜺
Radiation Shield
• Reduce the overall heat transfer between two radiating
surfaces

• Radiation shield between radiating surfaces

• Opaque partition

• Very low absorptivity and high reflectivity

Consider heat exchange between two infinite parallel
planes without radiation shield

(Q12)net

With no radiation shield the net heat transfer between the infinite
parallel plates is given by

𝑄12 𝑛𝑒𝑡 = 𝐹𝑔 𝐴1𝜎𝑏 𝑇14 − 𝑇24

12

T1, E1 T2, E2
 𝟏
(𝑭𝒈 ሻ =
𝟏𝟐 𝟏 − 𝜺𝟏 Τ𝜺𝟏 + 𝟏Τ𝑭𝟏𝟐 + 𝟏 − 𝜺𝟐 Τ𝜺𝟐 𝑨𝟏 Τ𝑨𝟐

For infinitely parallel planes F12 = 1 and A1= A2=A

𝟏 𝟏
(𝑭𝒈 ሻ = =
𝟏𝟐 𝟏Τ𝜺𝟏 − 𝟏 + 𝟏 + 𝟏Τ𝜺𝟐 − 𝟏 𝟏Τ𝜺𝟏 + 𝟏Τ𝜺𝟐 − 𝟏

𝟒 𝟒
𝑨𝟏𝝈𝒃 𝑻𝟏 − 𝑻𝟐
𝑄12 𝑛𝑒𝑡 = 𝐹𝑔 𝐴1𝜎𝑏 𝑇14 − 𝑇24 =
12 𝟏Τ𝜺𝟏 + 𝟏Τ𝜺𝟐 − 𝟏
 Placement of radiation shield between these two parallel planes would neither remove nor
add the heat. Under steady state conditions the shield plane attain uniform temperature T3

Equivalent electrical circuit:

(Q13)net (Q32)net

Under steady state conditions (Q13)net=(Q32)net

𝑨𝝈𝒃 𝑻𝟒𝟏 −𝑻𝟒𝟑 𝑨𝟏𝝈𝒃 𝑻𝟒𝟑 −𝑻𝟒𝟐

=
𝟏Τ𝜺𝟏 +𝟏Τ𝜺𝟑 −𝟏 𝟏Τ𝜺𝟐 +𝟏Τ𝜺𝟑 −𝟏
T1, e1 T3, e3 T2, e2
[(𝟏Τ𝜺𝟐 +𝟏Τ𝜺𝟑 −𝟏ሻ(𝑻𝟒𝟏 −𝑻𝟒𝟑 ሻ]−[(𝟏Τ𝜺𝟏 +𝟏Τ𝜺𝟑 −𝟏ሻ 𝑻𝟒𝟑 −𝑻𝟒𝟐 ]
(𝟏Τ𝜺𝟏 +𝟏Τ𝜺𝟑 −𝟏ሻ (𝟏Τ𝜺𝟐 +𝟏Τ𝜺𝟑 −𝟏ሻ
𝑻𝟒𝟏 + 𝑻𝟒𝟐
𝐹𝑜𝑟 𝜀1 = 𝜀2 = 𝜀3 = 𝜀 𝑻𝟒𝟑 =
Example 3
• Consider large parallel plates one at temperature 750 K with emissivity
0.75 and other at 500 K with emissivity 0.5. An aluminium radiation
shield with an emissivity 0.05 ( Both sides) is place between the plates.
(i) Sketch radiation network for the system with and without radiation
shield
(ii) Calculate the percent reduction in the heat transfer rate as a result of
the radiation shield
(i) Radiation network for the system without radiation shield

(Q12)net

T1, E1 T2, E2
(i) Radiation network for the system with radiation shield

(Q13)net (Q32)net

T1, e1 T3, e3 T2, e2

(ii) Rate of heat transfer
(a) Without radiation shield

𝑄12 𝐸𝑏1 −𝐸𝑏2 𝜎 𝑇14 −𝑇24

= σ𝑅
= ……. For parallel plates F12 = 1
𝐴 𝑛𝑒𝑡 ൫1−𝜀1 ൗ𝜀1 ሻ+𝐹12 +൫1−𝜀2 Τ𝜀2 ሻ

𝑄12 𝜎 𝑇14 −𝑇24

= (1Τ𝜀1 ሻ+(1Τ𝜀2 ሻ−1
= 6169.92 W/m2
𝐴 𝑛𝑒𝑡

(a) With radiation shield

𝑄12 𝐸𝑏1 − 𝐸𝑏2 σ𝑹 =
𝟏 𝟏 𝟏 𝟏
= + −𝟏 + + − 𝟏 = 41.333
𝜺𝟏 𝜺𝟑 𝜺𝟑 𝜺𝟐
𝐴 𝑛𝑒𝑡 σ𝑅

𝑄12 𝟔𝟏𝟔𝟗. 𝟗𝟐 − 𝟑𝟒𝟖. 𝟑𝟎

= 348.30W/m2 𝑷𝒆𝒓𝒆𝒄𝒆𝒏𝒕 𝒔𝒂𝒗𝒊𝒏𝒈 = ∗ 𝟏𝟎𝟎
𝐴 𝑛𝑒𝑡 𝟔𝟏𝟔𝟗. 𝟗𝟐
= 𝟗𝟒. 𝟑𝟓
Thank you !!!