Thermodynamics

Chapter 2 | 95In general, both e and a of a surface depend on the temperature and thewavelength of the radiation. Kirchhoff’s law of radiation states that theemissivity and the absorptivity of a surface are equal at the same temperatureand wavelength. In most practical applications, the dependence of e and a onthe temperature and wavelength is ignored, and the average absorptivity of asurface is taken to be equal to its average emissivity. The rate at which a surfaceabsorbs radiation is determined from (Fig. 2–73)Q # abs aQ # incident1W2(2–56)where Q # incident is the rate at which radiation is incident on the surface and a isthe absorptivity of the surface. For opaque (nontransparent) surfaces, the portionof incident radiation that is not absorbed by the surface is reflected back.The difference between the rates of radiation emitted by the surface and theradiation absorbed is the net radiation heat transfer. If the rate of radiationabsorption is greater than the rate of radiation emission, the surface is said tobe gaining energy by radiation. Otherwise, the surface is said to be losingenergy by radiation. In general, the determination of the net rate of heat transferby radiation between two surfaces is a complicated matter since it dependson the properties of the surfaces, their orientation relative to each other, andthe interaction of the medium between the surfaces with radiation. However,in the special case of a relatively small surface of emissivity e and surfacearea A at absolute temperature T s that is completely enclosed by a muchlarger surface at absolute temperature T surr separated by a gas (such as air)that does not intervene with radiation (i.e., the amount of radiation emitted,absorbed, or scattered by the medium is negligible), the net rate of radiationheat transfer between these two surfaces is determined from (Fig. 2–74)Q # rad esA 1T 4 s T 4 surr21W2(2–57)In this special case, the emissivity and the surface area of the surroundingsurface do not have any effect on the net radiation heat transfer.TABLE 2–4Emissivity of some materials at300 KMaterialEmissivityAluminium foil 0.07Anodized aluminum 0.82Polished copper 0.03Polished gold 0.03Polished silver 0.02Polished 0.17stainless steelBlack paint 0.98White paint 0.90White paper 0.92–0.97Asphalt pavement 0.85–0.93Red brick 0.93–0.96Human skin 0.95Wood 0.82–0.92Soil 0.93–0.96Water 0.96Vegetation 0.92–0.96·Q incident· ·Q ref = (1 – α) Q incident· ·Q abs = α Q incidentEXAMPLE 2–19Heat Transfer from a PersonFIGURE 2–73The absorption of radiation incident onan opaque surface of absorptivity a.Consider a person standing in a breezy room at 20°C. Determine the total rateof heat transfer from this person if the exposed surface area and the averageouter surface temperature of the person are 1.6 m 2 and 29°C, respectively,and the convection heat transfer coefficient is 6 W/m 2 · °C (Fig. 2–75).Solution A person is standing in a breezy room. The total rate of heat lossfrom the person is to be determined.Assumptions 1 The emissivity and heat transfer coefficient are constant anduniform. 2 Heat conduction through the feet is negligible. 3 Heat loss byevaporation is disregarded.Analysis The heat transfer between the person and the air in the room willbe by convection (instead of conduction) since it is conceivable that the airin the vicinity of the skin or clothing will warm up and rise as a result ofheat transfer from the body, initiating natural convection currents. It appearsLARGEENCLOSURE,A,T sSMALLBODYT surrQ radFIGURE 2–74Radiation heat transfer between abody and the inner surfaces of a muchlarger enclosure that completelysurrounds it.