7. Heat Transfer by Radiation Radiation Exchange with Emitting and Absorbing Gases Introduction Heat transfer by radiation needs no medium to be occurred (complete vacuum), but radiation can be happened in case of presence of fluids such that water liquid or gases. The participation of such medium depends on whether this medium absorbs, emits, or scatters radiation. If this medium is completely transparent, it has no effect on radiation passing through it, mediums such that is called nonparticipating medium. Air at ordinary temperatures and pressures is nonparticipating medium. Gases that of monatomic molecules such as Ar and He and symmetric diatomic molecules such as N2 and O2 are transparent to radiation, except at extremely high temperatures. Gases with asymmetric molecules such as H2O, CO2, CO, and SO2, and hydrocarbons HmCn may participate in the radiation process by absorption at moderate temperatures, and by absorption and emission at high temperature such as those encountered in combustion chambers. The propagation of radiation through a medium can be complicated further by presence of aerosols such as dust, ice particles, liquid droplets, and soot (unburned carbon) that scatter radiation. Scattering refers to the change of direction of radiation due to reflection, refraction, and diffraction. Radiation Properties of a Participating Medium Consider a participating medium of thickness L, as shown in figure (1). A spectral radiation beam of intensity Iλ,0 is incident on the medium, which is attenuated as it propagates due to absorption. The decrease in the intensity of radiation as it passes through a layer of thickness dx is proportional to the intensity itself and the thickness dx. According to Beer’s law, one can write: ππΌπ (π₯) = −ππ΄ πΌπ (π₯) ππ₯ Where the constant of proportionality ππ΄ is the spectral absorption coefficient of the medium, whose unit is m-1. Separating the variables and integrating from x = 0 to x = L gives: Figure (1) The attenuation of a radiation beam while passing through an absorbing medium of thickness L πΌπ,πΏ = π −πππΏ πΌπ,0 Where the absorptivity of the medium is assumed to be independent of x. The spectral transmissivity (τλ) of a medium can be defined as: πΌπ,πΏ = π −πππΏ πΌπ,0 The spectral transmissivity of a medium represents the fraction of radiation transmitted by the medium at a given wavelength. For nonscattering medium (nonreflection) medium, radiation is τπ = 1 either absorbed or transmitted. Therefore αλ + τπ = 1, and the spectral absorptivity of a medium of thickness L is: πΌπ = 1 − ππ = 1 − π −πππΏ According to Kirchoff’s law, the spectral emissivity of the medium is: ππ = πΌπ = 1 − ππ = 1 − π −πππΏ Note that the spectral absorptivity, transmissivity, and emissivity of a medium are dimensionless quantities, with values less than or equal to 1. The spectral absorption coefficient of a medium (and thus ππ , πΌπ , πππ ππ ), in general, vary with wavelength, temperature, pressure, and composition. Emissivity and Absorptivity of Gases and Gas Mixtures The spectral absorptivity of, for example, CO2 is given in figure (2) as a function of wavelength. It is clear the band nature of absorption and the strong nongray characteristics. The shape and the width of this absorption bands vary with temperature and pressure, but the magnitude of absorptivity also varies with the thickness of the gas layer. Therefore, the absorptivity values without specified thickness and pressure are meaningless. Figure (2) Spectral absorptivity of CO2 at 830 K and 10 atm for a In spite of the non-gray nature of absorptivity path length of 38.8 cm. of the gas, satisfactory results can be obtained by assuming the gas to be gray, and using an effective total absorptivity and emissivity determined by some averaging process. Radiation properties of an absorbing and emitting gas are usually reported for a mixture of the gas with non-participating gases rather than the pure gas. The emissivity and absorpitivity of a gas component in a mixture depends primarily on its density, which is a function of temperature and partial pressure of the gas. For single participating gas in a mixture, the emissivity of such gas is function of its partial pressure (Pw for H2O, Pc for CO2), temperature (Tg) and the mean distance traveled by the radiation beam (L). Figure (3-a) and Figure (3-b) show charts of the emissivity of water vapor in air and Carbon dioxide (CO2) in air at total pressure equals to 1 atm and at different temperature (Tg) ; respectively. 2 Figure (3) Emissivities of H2O and CO2 gases in a mixture of non participating gases at a total pressure of 1 atm for a mean beam length of L (1m. atm = 3.28 ft.atm) Emissivity at a total pressure P other than P = 1 atm is determined by multiplying the emissivity value at 1 atm by a pressure correction factor Cw for water vapor and Cc for CO2, (as shown in figure (4) and accordingly: ππ€ = πΆπ€ ππ€,1ππ‘π and ππΆ = πΆπΆ ππΆ,1ππ‘π Figure (4) Correction factors for the emissivities of H2O and CO2 gases at pressures other than 1 atm for use in the relations ππ€ = πΆπ€ ππ€,1ππ‘π and ππΆ = πΆπΆ ππΆ,1ππ‘π 3 In case of the existence of water vapor and CO2 together in the gas mixture, an equivalent emissivity of the gas is given by: ππ = ππΆ + ππ€ − Δπ Where Δπ is the emissivity correction factor, which accounts for the overlap of emission bands. For a gas mixture that contains both CO2 and H2O gases, Δπ is plotted in figure (5). Figure (5) Emissivity correction Δε for use in ππ = ππΆ + ππ€ − Δπ The emissivity of a gas also depends on the mean length an emitted radiation beam travels in a gas before reaching a boundary surface, and thus the shape and size of the gas body involved. The mean beam length (L) for various gas volume shapes are listed in table (1). Table (1) Mean beam length L for various gas volume shapes 4 The absorptivity of a gas containing CO2 and H2O gases for radiation emitted by a source at temperature Ts can be determined similarly from: πΌπ = πΌπΆ + πΌπ€ − Δα Where Δα = Δε and is determined from figure (5) at the source temperature Ts. The absorptivity of CO2 and H2O can be determined from the emissivity charts [figure (3) and (4)] as: ππ 0.65 For CO2: πΌπΆ = πΆπΆ × ( π ) For H2O: πΌπ€ = πΆπ€ × ( π ) π ππ 0.45 π π × ππΆ (ππ , ππΆ πΏ ππ ) π π × ππ€ (ππ , ππ€ πΏ ππ ) π As it is clear the emissivities should be evaluated using Ts instead of Tg, PcLTs/Tg instead of Pc L and PwLTs/Tg instead of Pw L. Also it is noted that the absorpitivity of the gas depends on source temperature Ts as well as the gas temperature Tg. The pressure correction factors Cc and Cw are evaluated using PcL and PwL as in emissivity calculation. When the total emissivity of a gas εg at temperature Tg is known, the emissive power of the gas (radiation emitted by the gas per unit surface area) can be expressed as: πΈπ = ππ π ππ 4 . Then the rate of radiation energy emitted by a gas to a boundary surface of area As becomes: πΜπ,π = ππ π΄π π ππ 4 If the boundary surface is black at temperature Ts, the surface will emit radiation to the gas at a rate of π΄π π ππ 4 without reflection any, and the gas will absorb this radiation at a rate of πΌπ π΄π π ππ 4 , where πΌπ is the absorpitivity of the gas. Then the net rate of radiation heat transfer between the gas and a black surface surrounding it becomes: For black enclosure: πΜπππ‘ = π΄π π(ππ ππ 4 − πΌπ ππ 4 ) If the surface is not black, a correction is made for the case of ππ > 0.7, and the net rate of radiation heat transfer becomes: πΜπππ‘,ππππ¦ = ππ + 1 π +1 Μ πππ‘,πππππ = π ×π π΄π π(ππ ππ 4 − πΌπ ππ 4 ) 2 2 The emissivity of wall surfaces of furnaces and combustion chambers are typically greater than 0.7, and thus the previous relation is suitable to be applied in such cases. 5