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Radiation heat transfer and clothing comfort

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Radiation heat transfer and clothing comfort

  1. 1. Radiation as a Mode of Heat Transfer in Textiles and Clothing Dr. Kausik Bal KHT, TUL
  2. 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. 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. 4. Radiation in Nature: Heating of Earth Surface
  5. 5. Radiation in Nature: Heating of Earth Surface
  6. 6. Radiation in Nature: Heating of Earth Surface
  7. 7. Heating of Earth Surface: Global Warming http://earthobservatory.nasa.gov/Features/GlobalWarming/page2.php
  8. 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. 9. The speed of radiation Electromagnetic waves are characterized by their frequency (Ξ½) [Hz] and wavelength (Ξ») [m] where λν = 𝑐0 𝑛 n = index of refraction of the medium (n = 1 for air) 𝑐0 = 3 Γ— 108 [ π‘š 𝑠] is the speed of light (or the EM wave) in vacuum
  10. 10. Radiation at interface of two media π΄π‘π‘ π‘œπ‘Ÿπ‘π‘‘π‘–π‘£π‘–π‘‘π‘¦ = Ξ± 𝑅𝑒𝑓𝑙𝑒𝑐𝑑𝑖𝑣𝑖𝑑𝑦 = ρ π‘‡π‘Ÿπ‘Žπ‘›π‘ π‘šπ‘–π‘ π‘ π‘–π‘£π‘–π‘‘π‘¦ = Ο„ Ξ± + ρ + Ο„ = 1 For opaque surface, Ο„ = 0 and hence, Ξ± + 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. 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 Ξ± = 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. 12. Emissivity of a surface (Total hemispherical) Emissivity Ξ΅ 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 (Β°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. 13. Stefan – Boltzmann Law 𝐸 𝑏 𝑇 = σ𝑇4 [ π‘Š π‘š2 ] Οƒ = 5.670 Γ— 10βˆ’8 [ π‘Š π‘š2 𝐾4 ] is Stefan-Boltzmann constant Example: 1. What is the radiation flux emitted by human skin? (Take Ξ΅ = 0.95) Solution: Skin temperature = 305 [K], hence, the radiative heat flux is: 𝐸 𝑇 = 305 = 0.95 Γ— 5.670 Γ— 10βˆ’8 Γ— 3054 = 466 π‘Š π‘š2 2. Calculate the radiation flux from a wall with Ξ΅ = 0.64 which is at 20Β°C. Solution: Wall temperature = 293 [K], hence the radiative heat flux is: 𝐸 𝑇 = 293 = 0.5 Γ— 5.670 Γ— 10βˆ’8 Γ— 2934 = 267 π‘Š π‘š2 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 Ξ΅, the total radiation flux emitted is E 𝑇 = Ρσ𝑇4.
  14. 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. Ξ΅ 𝑇 = Ξ± 𝑇 T T (Ξ΅, Ξ±) E(T)
  15. 15. Wien’s Displacement Law Ξ» π‘šπ‘Žπ‘₯ 𝑇 = πΆπ‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ π‘Šπ‘–π‘’π‘›β€² 𝑠 π‘‘π‘–π‘ π‘π‘™π‘Žπ‘π‘’π‘šπ‘’π‘›π‘‘ π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ = 2897768.5 π‘›π‘š. 𝐾
  16. 16. Planck’s Law of blackbody radiation 𝐸 𝑏λ Ξ», 𝑇 = 𝐢1 Ξ»5 𝑒 𝐢2 λ𝑇 βˆ’ 1 π‘Š π‘š2 . ΞΌπ‘š 𝐢1 = 2πœ‹β„Žπ‘0 2 = 3.742 Γ— 108 π‘ŠΞΌπ‘š4 π‘š2 𝐢2 = β„Žπ‘0 π‘˜ = 1.439 Γ— 104 ΞΌπ‘š. 𝐾 β„Ž = 6.6256 Γ— 10βˆ’34 𝐽. 𝑠 π‘˜ = 1.3805 Γ— 10βˆ’23 𝐽 𝐾 Planck’s constant Boltzmann’s constant
  17. 17. Radiation Geometry I: Solid angle 2D 3D 𝑑ω = 𝑑𝑆 π‘Ÿ2 = sin πœƒ π‘‘πœƒπ‘‘Ο† dS Ο† ΞΈ r Unit of solid angle: sr (steradian)
  18. 18. Intensity of radiation The Radiation Intensity 𝐼𝑒 ΞΈ, Ο† is defined as the rate at which radiation energy 𝑑𝑄 𝑒 is emitted in the ΞΈ, Ο† direction per unit area normal to this direction and per unit solid angle about this direction. 𝐼𝑒 πœƒ, Ο† = 𝑑𝑄 𝑒 𝑑𝐴 cos πœƒ sin πœƒπ‘‘πœƒπ‘‘Ο† π‘Š π‘š2 . π‘ π‘Ÿ
  19. 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 = β„Žπ‘’π‘šπ‘–π‘ π‘β„Žπ‘’π‘Ÿπ‘’ 𝑑𝐸 = Ο†=0 2πœ‹ πœƒ=0 πœ‹ 2 𝐼𝑒 πœƒ, πœ‘ cos πœƒ sin πœƒ π‘‘πœƒπ‘‘πœ‘ π‘Š π‘š2 In case of diffusely emitting surface, 𝐸 = πœ‹πΌπ‘’ π‘Š π‘š2 Therefore, in case of a blackbody, the following is valid: 𝐼 𝑏 𝑇 = 𝐸 𝑏 𝑇 πœ‹ = πœŽπ‘‡4 πœ‹ π‘Š π‘š2 . π‘ π‘Ÿ
  20. 20. Irradiation The intensity of incident radiation 𝐼𝑖 πœƒ, πœ‘ is defined as the rate at which radiation energy 𝑑𝐺 is incident from the πœƒ, πœ‘ 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 = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘. 𝐺 = β„Žπ‘’π‘šπ‘–π‘ π‘β„Žπ‘’π‘Ÿπ‘’ 𝑑𝐺 = Ο†=0 2πœ‹ πœƒ=0 πœ‹ 2 𝐼𝑖 πœƒ, πœ‘ cos πœƒ sin πœƒ π‘‘πœƒπ‘‘πœ‘ π‘Š π‘š2 The radiation flux incident on a surface from all directions is called irradiation (G). For diffusely incident radiation, 𝐺 = πœ‹πΌπ‘’ π‘Š π‘š2
  21. 21. Radiosity The rate at which radiation energy leaves a unit area of a surface in all directions is termed as Radiosity (J). 𝐽 = Ο†=0 2πœ‹ πœƒ=0 πœ‹ 2 𝐼𝑒+π‘Ÿ πœƒ, πœ‘ cos πœƒ sin πœƒ π‘‘πœƒπ‘‘πœ‘ π‘Š π‘š2 G J E 𝐺 π‘Ÿπ‘’π‘“
  22. 22. Radiation Geometry II: View factor Surface 3 Surface 2 Surface 1 Point source π‘‘πœ” 21 𝐴 1 𝐴 2 𝑛 1 𝑛 2 𝑑𝐴 1 𝑑𝐴 2 πœƒ 1 πœƒ 2 π‘Ÿ 𝐹𝑖𝑗 = 1 𝐴𝑖 𝐴 𝑖 𝐴 𝑗 cos πœƒπ‘– cos πœƒπ‘— πœ‹π‘Ÿ2 𝑑𝐴𝑖 𝑑𝐴𝑖
  23. 23. Exposure to radiation: Buildings and vehicles
  24. 24. Exposure to radiation: Outdoors
  25. 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 βˆ’ 𝑇2 4 π‘Š Net radiation transfer from non-black surface i is: 𝑄𝑖 = 𝐴𝑖 πœ€π‘– 1 βˆ’ πœ€π‘– 𝐸 𝑏𝑖 βˆ’ 𝐽𝑖 Electrical analogy: 𝑄𝑖 = 𝐸 π‘π‘–βˆ’π½ 𝑖 𝑅 𝑖 where, 𝑅𝑖 = 1βˆ’πœ€π‘– 𝐴 𝑖 πœ€π‘– is Surface Resistance When the two surfaces are diffuse, opaque and grey, net radiation heat transfer from surface i to surface j: 𝑄𝑖𝑗 = 𝐽 π‘–βˆ’π½ 𝑗 𝑅 𝑖𝑗 where 𝑅𝑖𝑗 = 1 𝐴 𝑖 𝐹 𝑖𝑗 is Space Resistance 𝐸 𝑏𝑖 𝑅𝑖 𝐽𝑖 𝑅𝑗 𝐽𝑗 𝐸 𝑏𝑗 𝑅𝑖𝑗 𝑄𝑖𝑗
  26. 26. http://www.kostic.niu.edu/352/_352-posted/Heat_4e_Chap13-Radiation_HT_lecture-PDF.pdf Radiation shielding
  27. 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 βˆ’ 𝛼
  28. 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 β‰ͺ 1 Mie Scattering: 𝛼 β‰ˆ 1 Geometric Scattering: 𝛼 ≫ 1 D Here, 𝛼 = πœ‹π· Ξ»
  29. 29. Thermal radiation and textiles οƒ˜ Radiation, emitted by a hot surface may pass through the straight pores (holes) across the textile. οƒ˜ Radiation, while passing through a textile, may be scattered by the solid fibres or yarn. οƒ˜ Radiation incident on fibres, yarns or fabric surface may partly be absorbed. οƒ˜ The fibres, yarns or fabric itself may emit radiation as a grey body which depends on its temperature and emissivity. οƒ˜ Some fibres may allow the radiation to be transmitted through them by refraction οƒ˜ 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. 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. 31. Interaction of thermal radiation with fibres and yarns Considering the typical diameter of textile fibres which has a range 10βˆ’6 [π‘š] - 10βˆ’4 [π‘š], and the wavelength of thermal radiation being between 10βˆ’7 [π‘š] and 10βˆ’3 [π‘š].  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 10βˆ’5 [π‘š] - 10βˆ’3 [π‘š], 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. 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. Ξ» 𝑒𝑓𝑓 = Ξ» π‘π‘œπ‘›π‘‘ + Ξ» π‘Ÿπ‘Žπ‘‘ 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. π‘ž π‘Ÿπ‘Žπ‘‘ = Ξ» π‘Ÿπ‘Žπ‘‘ 𝑇1 βˆ’ 𝑇2 Where Ξ» π‘Ÿπ‘Žπ‘‘ = 4πœŽπ‘‡ π‘š 3 πœ€1 βˆ’1 + πœ€2 βˆ’1 βˆ’ 1 and 𝑇 π‘š = 𝑇1 + 𝑇2 2 In case of nonwovens or similar low density fabrics, the radiation is given as Ξ» π‘Ÿπ‘Žπ‘‘ = 4β„ŽπœŽπ‘‡ π‘š 3 2 πœ€ βˆ’ 1 𝑒 0.188β„Ž Ξ½βˆ’1 πœ‡ π‘Ÿβˆšπœ‹ β„Ž = thickness Ξ½ = (idealized) portion of fibres oriented vertically πœ‡ = 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. 33. Radiation heat transfer through clothing β€’ Clothing acts as a barrier to radiation heat transfer between skin and environment. β€’ 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. 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. 35. Thank you for your attention. For further discussion, please contact by email: kb.iitd@gmail.com

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