流体力学与传热 ：4-6- RADIATION HEAT TRANSFER

1、,4.6. RADIATION HEAT TRANSFER,Radiation, which may be considered to be energy streaming through space at the speed of light, may originate in various ways.,Some types of material will emit radiation when they are treated by external agencies.,All substances at temperatures above absolute zero emit r

2、adiation that is independent of external agencies.,Radiation that is the result of temperature only is called thermal radiation.,Fundamental facts concerning radiation,Radiation moves through space in straight lines, or beams, and only substances in sight of a radiating body can intercept radiation

3、from that body.,Radiation as such is not heat, and when transformed into heat on absorption, it is no longer radiation.,The fraction that is absorbed is called absorptivity .,The fraction that is transmitted is called transmissivity .,The fraction of the radiation falling on a body that is reflected

4、 is called reflectivity .,The maximum possible absorptivity is unity. A body which absorbs all incident radiation is called a black body.,The sum of these fractions must be unity, or,Emission of radiation,The radiation emitted by any given mass of substance is independent of other material in sight

5、of , or in contact with, the mass.,The net energy gained or lost by a body is the difference between the energy emitted by the body and that absorbed by it from the radiation reaching it from other bodies.,When bodies at different temperatures are placed in sight of one another inside an enclosure,

6、the hotter bodies loss energy by emission of radiation faster than they receive energy by absorption of radiation from the cooler bodies. Temperatures of hotter bodies decrease.,Wavelength of radiation,Known electromagnetic radiations cover an enormous range of wavelengths, from the short cosmic ray

7、s to long wave broadcasting wave.,Although radiation of any wavelength is, in principle, convertible into heat on absorption by matter, the portion of the electromagnetic spectrum that is of importance in heat flow lies in the wavelength range between 0.5 and 50 m.,Visible light covers a wavelength

8、range of about 0.38 to 0.78 m,At temperature above about 5000C ，heat radiation in the visible spectrum become significant.,The higher the temperature of the radiating body, the shorter the predominant wavelength of the thermal radiation emitted by it.,Emissive power,The monochromatic energy emitted

9、by a radiating surface depends on the temperature of the surface and on the wavelength of the radiation.,At constant surface temperature, a curve can be plotted showing the rate of energy emission as a function of the wavelength.,The monochromatic radiation emitted in this manner from unit area in u

10、nit time, divided by the wavelength, is called the monochromatic radiating power W.,For the entire spectrum of the radiation from a surface, the total radiating power W is the sum of all the monochromatic radiations from the surface, or , mathematically,4.6-2,Blackbody radiation,A blackbody has the

11、maximum attainable emissive power at any given temperature. The ratio of the total emissive power W of a body to that of a blackbody WB is by definition the emissivity of the body, thus,4.6-3,Emissivities of solids,Emissivity usually increases with temperature. Emissivities of polished metals are lo

12、w, in the range 0.03 to 0.08. Emissivities of most oxidized metals range from 0.6 to 0.85, those of nonmetals from 0.65 to 0.95.,Practical source of blackbody radiation No actual substance is a blackbody, although some materials, such as certain grades of carbon black, do approach blackness.,The dis

13、tribution of energy in the spectrum of a blackbody is known accurately. It is given by Plancks law,4.6-6,Laws of blackbody radiation,Plancks law can be shown to be consistent with the Stefan-Boltzmann law by substituting Wb, from Eq 4.6-6 into Eq 4.6-2 and integrating.,A basic relationship for black

14、body radiation is the Stefan-Boltzmann law, which states that the total emissive power of a blackbody is proportional to the fourth power of the absolute temperature, or,Where is a universal constant,Absorption of radiation by opaque solids,Kirchhoffs law states, at temperature equilibrium, the rati

15、o of the total radiating power of any body to the absorptivity of that body depends only upon the temperature of the body.,Thus, consider any two bodies in temperature equilibrium with common surroundings. Kirchhoffs law states that,4.6-9,Thus,By definition, the emissivity of the second body 2 is,If

16、 the first body is blackbody, 1=1, and,4.6-11,4.6-10,Thus, when any body is at temperature equilibrium with its surroundings, its emissivity and absorptivity are equal. Kirchholff law applies whether or not the two surfaces are at same temperature.,Radiation between surfaces,The total radiation from

17、 a unit area of an opaque body of area A1, emissivity1, and absolute temperature T1 is,4.6-13,Qualitatively, the interception of radiation from an area element of a surface by another surface of finite size can be visualized in terms of the angle of vision.,The factor F is called the view factor or

18、angle factor; it depends upon the geometry of the two surface,4.6-14,The equation for two bodies radiating each other can be written in the form,If surface A1 is chosen for A, Eq(4.6-14) can be written,4.6-15,In general, for gray surfaces, Eq(4.6-15)and Eq(4.6-16) can be written,(4.6-26),F12 and F21

19、 are the overall interchange factor and are functions of 1 and 2.,Two large parallel planes,4.6-27,4.6-28,One gray surface completely surrounded by another,problem,A chamber for heat-curing large aluminum sheets, lacquered black on both sides, operates by passing the sheets vertically between two st

20、eel plates 150 mm apart. One of the plates is at 300C, and the other, exposed to the atmosphere, is at 25C.(a) What is the temperature of the lacquered sheet? (b) What is the heat transferred between the walls when equilibrium has been reached? Neglect convection effects. Emissivity of steel is 0.56

21、; emissivity of lacquered sheets is 1.0. Solution (a) Let subscript 1 refer to hot plate, 2 to lacquered sheets, and 3 to cold plate:,1, 3 = 0.56 2 = 1.0 T1 = 573K T3 = 298K From Eq. (4.6-27),Since A1 = A2 ,T2 = 490.4K = 217.4C,(b) From Eq. (4.6-26) the heat flux is,= 5.672 0.56(5.734 4.9044) = 1587W/m2,Check:,= 5.672 0.56(4.9044 2.984) = 1587W/m2,Note: If the lacquered sheet is removed, q13 = 3174 W/m2,Problem A shell-and-tube heat exchanger cons