Radiation is a mode of heat transfer that does not require a medium. Thermal radiation is emitted from all objects based on their temperature and is characterized by properties like emissivity. Radiation plays a key role in heating the Earth's surface from the Sun. The amount of radiation emitted by an object increases with its temperature, as described by Stefan-Boltzmann law. Radiation interacts with textiles through absorption, emission, transmission, and scattering depending on fiber properties. Research aims to model radiation transfer through textiles and its role in thermal insulation and heat stress protection.

1. Radiation as a Mode of Heat Transfer in Textiles and Clothing
Dr. Kausik Bal
KHT, TUL

2. Radiation is the energy emitted by matter in the form of electro-magnetic waves.
Thermal radiation is energy emitted by matter that is at a finite temperature above the
absolute zero. Unlike conduction or convection, it does not require any medium of transfer.
What is thermal radiation?

3. Radiation in Nature: The Sun
Coronal Mass Ejection as viewed by the Solar Dynamics Observatory on June 7,
2011. Credit: NASA/SDO
Distance to Earth: 149,600,000 km (light travels from the Sun to Earth in about 8 minutes and 19 seconds.)
Surface temperature: 5,778 K
Mass: 1.98930 kg
Radius: 695,500 km

4. Radiation in Nature: Heating of Earth Surface

5. Radiation in Nature: Heating of Earth Surface

6. Radiation in Nature: Heating of Earth Surface

7. Heating of Earth Surface: Global Warming
http://earthobservatory.nasa.gov/Features/GlobalWarming/page2.php

8. Solar Energy Spectrum
Solar Constant = 1.36 kW/m2 (amount of incoming solar radiation per unit area on a plane perpendicular
to the rays at a distance of 1 astronomical unit [AU]).

9. The speed of radiation
Electromagnetic waves are characterized by their frequency (ν) [Hz] and wavelength
(λ) [m] where
λν =
𝑐0
𝑛
n = index of refraction of the medium
(n = 1 for air)
𝑐0 = 3 × 108 [ 𝑚 𝑠] is the speed of light (or the EM wave) in vacuum

10. Radiation at interface of two media
𝐴𝑏𝑠𝑜𝑟𝑝𝑡𝑖𝑣𝑖𝑡𝑦 = α
𝑅𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑣𝑖𝑡𝑦 = ρ
𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑣𝑖𝑡𝑦 = τ
α + ρ + τ = 1
For opaque surface, τ = 0 and hence, α + ρ = 1
Irradiation = Radiation flux
incident on a surface
(denoted by G)
SUN
Incident solar
radiation 100%
Reflected
radiation 8%
Transmitted
radiation 80%
Absorbed
radiation 12%
Outward transfer of
absorbed radiation 8%
Inward transfer of
absorbed radiation 4%

11. Blackbody
Blackbody is a hypothetical (or theoretical) surface which is a perfect absorber of
electromagnetic radiation, i.e., for the surface of blackbody, absorptivity α = 1.
A blackbody absorbs all the radiation that falls on it, converts it into internal energy
(heat), and then re-radiates this energy into the surroundings. The re-radiated thermal
energy, known as blackbody radiation, has a continuous spectrum governed solely by
the body's temperature.

12. Emissivity of a surface
(Total hemispherical) Emissivity ε 𝑇 =
𝐸 𝑇
𝐸 𝑏 𝑇
is the ratio of the total radiation energy
emitted by the surface at a given temperature over all wavelengths in all directions to the
same emitted by a blackbody at the same temperature.
By definition, the emissivity of a blackbody is maximum and equals to unity.
All real surfaces have emissivity less than unity and are known as grey body. In the
extreme case, a white body is a hypothetical surface which does not absorb any
wavelength of radiation incident upon it at any direction.
Materials Temperature (°C) Typical emissivity
Commercial aluminium sheet 100 0.09
Pure highly polished gold 100 0.02
Brick (Building) 1000 0.45
Concrete 0 - 100 0.94
Smooth glass 0 - 200 0.95
Graphite 0 - 3600 0.7 – 0.8
Human skin 36 0.985
Wood (Oak, sanded) 93 0.82
Opaque plastics (any colour) 25 0.95
www.transmetra.ch

13. Stefan – Boltzmann Law
𝐸 𝑏 𝑇 = σ𝑇4
[ 𝑊 𝑚2
]
σ = 5.670 × 10−8
[ 𝑊 𝑚2
𝐾4
] is Stefan-Boltzmann constant
Example:
1. What is the radiation flux emitted by human skin? (Take ε = 0.95)
Solution: Skin temperature = 305 [K], hence, the radiative heat flux is:
𝐸 𝑇 = 305 = 0.95 × 5.670 × 10−8 × 3054 = 466 𝑊 𝑚2
2. Calculate the radiation flux from a wall with ε = 0.64 which is at 20°C.
Solution: Wall temperature = 293 [K], hence the radiative heat flux is:
𝐸 𝑇 = 293 = 0.5 × 5.670 × 10−8
× 2934
= 267 𝑊 𝑚2
The total radiation flux emitted by a blackbody at temperature T is a function of its
temperature only
Therefore, for a real surface (grey body with surface emissivity ε, the total radiation flux
emitted is E 𝑇 = εσ𝑇4.

14. Kirchhoff’s Law
The total hemispherical emissivity of a surface at temperature T is equal to its total
hemispherical absorptivity for radiation coming from a blackbody at the same
temperature.
ε 𝑇 = α 𝑇
T
T
(ε, α)
E(T)

15. Wien’s Displacement Law
λ 𝑚𝑎𝑥 𝑇 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑊𝑖𝑒𝑛′ 𝑠 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 2897768.5 𝑛𝑚. 𝐾

16. Planck’s Law of blackbody radiation
𝐸 𝑏λ λ, 𝑇 =
𝐶1
λ5 𝑒
𝐶2
λ𝑇 − 1
𝑊 𝑚2 . μ𝑚
𝐶1 = 2𝜋ℎ𝑐0
2
= 3.742 × 108 𝑊μ𝑚4 𝑚2
𝐶2 =
ℎ𝑐0
𝑘 = 1.439 × 104
μ𝑚. 𝐾
ℎ = 6.6256 × 10−34
𝐽. 𝑠
𝑘 = 1.3805 × 10−23
𝐽 𝐾
Planck’s constant
Boltzmann’s constant

17. Radiation Geometry I: Solid angle
2D 3D
𝑑ω =
𝑑𝑆
𝑟2
= sin 𝜃 𝑑𝜃𝑑φ
dS
φ
θ
r
Unit of solid angle: sr (steradian)

18. Intensity of radiation
The Radiation Intensity 𝐼𝑒 θ, φ is defined as the rate at which radiation energy 𝑑𝑄 𝑒
is emitted in the θ, φ direction per unit area normal to this direction and per unit
solid angle about this direction.
𝐼𝑒 𝜃, φ =
𝑑𝑄 𝑒
𝑑𝐴 cos 𝜃 sin 𝜃𝑑𝜃𝑑φ
𝑊 𝑚2
. 𝑠𝑟

19. Emissive power
The radiation flux for emitted radiation is the emissive power E, i.e., the rate at which
radiation energy is emitted per unit area of the emitting surface.
𝐸 =
ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒
𝑑𝐸 =
φ=0
2𝜋
𝜃=0
𝜋
2
𝐼𝑒 𝜃, 𝜑 cos 𝜃 sin 𝜃 𝑑𝜃𝑑𝜑 𝑊 𝑚2
In case of diffusely emitting surface, 𝐸 = 𝜋𝐼𝑒 𝑊 𝑚2
Therefore, in case of a blackbody, the following is valid:
𝐼 𝑏 𝑇 =
𝐸 𝑏 𝑇
𝜋
=
𝜎𝑇4
𝜋
𝑊 𝑚2 . 𝑠𝑟

20. Irradiation
The intensity of incident radiation 𝐼𝑖 𝜃, 𝜑 is defined as the rate at which radiation energy 𝑑𝐺 is
incident from the 𝜃, 𝜑 direction per unit area of the receiving surface normal to this direction and
per unit solid angle about this direction. When incident radiation is diffused, 𝐼𝑖 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.
𝐺 =
ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒
𝑑𝐺 =
φ=0
2𝜋
𝜃=0
𝜋
2
𝐼𝑖 𝜃, 𝜑 cos 𝜃 sin 𝜃 𝑑𝜃𝑑𝜑 𝑊 𝑚2
The radiation flux incident on a surface from all directions is called irradiation (G).
For diffusely incident radiation, 𝐺 = 𝜋𝐼𝑒 𝑊 𝑚2

21. Radiosity
The rate at which radiation energy leaves a unit area of a surface in all directions is termed as
Radiosity (J).
𝐽 =
φ=0
2𝜋
𝜃=0
𝜋
2
𝐼𝑒+𝑟 𝜃, 𝜑 cos 𝜃 sin 𝜃 𝑑𝜃𝑑𝜑 𝑊 𝑚2
G
J
E
𝐺 𝑟𝑒𝑓

22. Radiation Geometry II: View factor
Surface 3
Surface 2
Surface 1
Point source
𝑑𝜔 21
𝐴 1
𝐴 2
𝑛 1
𝑛 2
𝑑𝐴 1
𝑑𝐴 2
𝜃 1
𝜃 2
𝑟
𝐹𝑖𝑗 =
1
𝐴𝑖 𝐴 𝑖 𝐴 𝑗
cos 𝜃𝑖 cos 𝜃𝑗
𝜋𝑟2 𝑑𝐴𝑖 𝑑𝐴𝑖

23. Exposure to radiation: Buildings and vehicles

24. Exposure to radiation: Outdoors

25. Radiation heat transfer
𝑇1 𝑇2
𝑄1
𝑄2
𝐴1 𝐴2 Net radiation transfer from surface 1 to 2 (both black) is:
𝑄12 = 𝐴1 𝐹12 𝜎 𝑇1
4
− 𝑇2
4
𝑊
Net radiation transfer from non-black surface i is:
𝑄𝑖 =
𝐴𝑖 𝜀𝑖
1 − 𝜀𝑖
𝐸 𝑏𝑖 − 𝐽𝑖
Electrical analogy: 𝑄𝑖 =
𝐸 𝑏𝑖−𝐽 𝑖
𝑅 𝑖
where, 𝑅𝑖 =
1−𝜀𝑖
𝐴 𝑖 𝜀𝑖
is Surface Resistance
When the two surfaces are diffuse, opaque and grey, net
radiation heat transfer from surface i to surface j:
𝑄𝑖𝑗 =
𝐽 𝑖−𝐽 𝑗
𝑅 𝑖𝑗
where 𝑅𝑖𝑗 =
1
𝐴 𝑖 𝐹 𝑖𝑗
is Space Resistance
𝐸 𝑏𝑖
𝑅𝑖
𝐽𝑖
𝑅𝑗
𝐽𝑗
𝐸 𝑏𝑗
𝑅𝑖𝑗
𝑄𝑖𝑗

26. http://www.kostic.niu.edu/352/_352-posted/Heat_4e_Chap13-Radiation_HT_lecture-PDF.pdf
Radiation shielding

27. Reflecting surfaces and coating
Surface Absorptivity
Aluminum, dull/rough polished 0.4 - 0.65
Aluminum. polished 0.1 - 0-40
Asbestos Cement, old 0.83
Black matt 0.95
Chromium plate 0.20
Iron, galvanised old 0.89 - 0.92
Grey paint 0.95
Light gren paint 0.95
Limestone 0.33 - 0.53
Red clay brick 0.94
White paint 0.89
For opaque materials, practically there is no transmission 𝜏 = 0 of radiation incident
on its surface. Hence, in such cases, ρ = 1 − 𝛼

28. Scattering
It is the process in which electromagnetic radiation or particles are deflected or
diffused. Such deflection can be due to the presence of other particle (s) or due to
localized non-uniformities of the medium.
Generally speaking, in case of waves (e.g. EM waves), the interaction with a matter
may cause two types of reflections from the surface where the wave is incident, one is
specular reflection and another is diffused reflection. The second type is a common
example of scattering.
In case of light (EM wave) scattering from a small particles, scattering is categorized in
three domains based on a dimensionless parameter.
Rayleigh Scattering: 𝛼 ≪ 1
Mie Scattering: 𝛼 ≈ 1
Geometric Scattering: 𝛼 ≫ 1
D Here, 𝛼 =
𝜋𝐷
λ

29. Thermal radiation and textiles
 Radiation, emitted by a hot surface may pass through the straight
pores (holes) across the textile.
 Radiation, while passing through a textile, may be scattered by
the solid fibres or yarn.
 Radiation incident on fibres, yarns or fabric surface may partly
be absorbed.
 The fibres, yarns or fabric itself may emit radiation as a grey
body which depends on its temperature and emissivity.
 Some fibres may allow the radiation to be transmitted through
them by refraction
 In case of special fibres (e.g., metallic) or in case of textiles with
reflective coating (e.g. metallic coating), a significant amount of
incident radiation may be reflected back by specular reflection.

30. Research on the thermal radiation in textiles
 Theoretical prediction of radiation through woven (and/or knitted fabric) in the
light of the fabric structure.
 Theoretical prediction of radiation through nonwovens and random fibrous
assemblies.
 Development of measurement techniques with fabrics in single and multiple
layers.
 Development of measurement techniques with clothing.
 Empirical and semi-empirical modelling of insulation from thermal radiation in
respect of protection from heat stress.
 Empirical and semi-empirical modelling of radiation transfer and shielding in case
of UV protection.
 Empirical analysis of structure – property relations to find total effective thermal
resistance.

31. Interaction of thermal radiation with fibres and yarns
Considering the typical diameter of textile fibres which has a range 10−6 [𝑚] - 10−4 [𝑚],
and the wavelength of thermal radiation being between 10−7
[𝑚] and 10−3
[𝑚].
 Therefore, fibres can cause scattering of thermal radiation and such scattering is
often considered to be in the Mie Scattering regime.
 Yarns have typical diameters in the range 10−5
[𝑚] - 10−3
[𝑚], and therefore
such yarns as a solid material can also cause scattering of thermal radiation and
such scattering is also often considered to be in the Mie Scattering regime.
 Some researchers have developed models of radiation
heat transfer in fibrous materials such as nonwovens
assuming that there is no scattering.
1. B. Farnworth, Mechanism of heat flow through clothing insulation, Textile Research Journal, Vol. 53 (12), 1983.
2. X. Wan; J. Fan, Heat transfer through fibrous assemblies incorporating reflective interlayers, International Journal of heat & Mass Transfer, Vol.
55, 2012.
3. D. Bhattacharjee & V. K. Kothari, A theoretical model to predict the thermal resistance of plain woven fabrics, Indian Journal of Fibre & Textile
research, Vol. 30 (3), 2005.

32. Modelling thermal radiation transfer through fabrics
In situation where the total heat transfer by conduction through fabric is much higher
than the heat transfer by radiation, the total thermal conductivity (or resistance) can be
considered as a linear sum of the individual components due to conduction and radiation.
λ 𝑒𝑓𝑓 = λ 𝑐𝑜𝑛𝑑 + λ 𝑟𝑎𝑑
In such cases, it is assumed that it is possible to express the radiative heat flux in terms
of the temperature gradient at steady state which resembles Fourier’s law of thermal
conductivity.
𝑞 𝑟𝑎𝑑 = λ 𝑟𝑎𝑑 𝑇1 − 𝑇2
Where λ 𝑟𝑎𝑑 = 4𝜎𝑇 𝑚
3
𝜀1
−1 + 𝜀2
−1 − 1 and 𝑇 𝑚 = 𝑇1 + 𝑇2 2
In case of nonwovens or similar low density fabrics, the radiation is given as
λ 𝑟𝑎𝑑 =
4ℎ𝜎𝑇 𝑚
3
2
𝜀
− 1
𝑒
0.188ℎ ν−1 𝜇
𝑟√𝜋
ℎ = thickness
ν = (idealized) portion of fibres
oriented vertically
𝜇 = filling coefficient of the fabric
1. M. Boguslawska-Baczek; L. Hes, Determination of heat transfer by radiation in textile fabrics by means of method with known emissivity of
plates, Journal of Industrial Textiles, 2013.

33. Radiation heat transfer through clothing
• Clothing acts as a barrier to radiation heat transfer
between skin and environment.
• The insulation or protection provided by the clothing can
reduce heat stress and discomfort and can even be a life
saver when the clothed human is exposed to very intense
thermal radiation.
 Intense solar radiation (dry deserts and snow-capped
mountain peaks)
 Fire-fighting
 Furnace-work
 Space-travel
Very limited models exist for the radiation heat transfer through clothing, some empirical
and some semi-analytical and almost all approximate.
1. E. A. D. Hartog; G. Havenith, Analytical study of the heat loss attenuation by clothing on thermal manikins under radiative heat loads,
International Journal of occupational Safety and Ergonomics, Vol. 16 92), 2010.

34. Protection Vs. Comfort: Clothing for radiative environments
The requirements of protection and comfort are
often contradictory. It may be obvious to give
more weightage to protection in case of short
duration use, but comfort becomes more
important for longer duration of continuous use
and performance.

35. Thank you for your attention.
For further discussion, please contact by email: kb.iitd@gmail.com