Radiation heat transfer emissivity

In rotary devices, reradiation from the exposed shelf surface to the solids bed is a major design consideration. A treatise on furnaces, including radiative heat-transfer effects, is given by Ellwood and Danatos [Chem. Eng., 73(8), 174 (1966)]. For discussion of radiation heat-transfer computational methods, heat fliixes obtainable, and emissivity values, see Schornshort and Viskanta (ASME Paper 68-H 7-32), Sherman (ASME Paper 56-A-III), and the fohowing subsection. 

Radiation heat transfer. The radiation heat transfer between two parallel planes is reduced by placing a parallel aluminum sheet in the middle of the gap. The surface temperatures are (9j = 40 °C and 62 = 5 °C, respectively the emissivities are ej = e, = 0.8.5. The emissivity of both sides of the aluminum... 

Equations similar to equation 9.158 may be obtained for each of the surfaces in an enclosure, 1 = 1,1 = 2, 1 = 3, 1 = n and the resulting set of simultaneous equations may then be solved for the unknown radiosities, qoi,qm- qun The radiation heat transfer is then obtained from equation 9.140. This approach requires data on the areas and view factors for all pairs of surfaces in the enclosure and the emissivity, reflectivity and the black body emissive power for each surface. Should any surface be well insulated, then, in this case, Qj — 0 and ... 

In Expression (3.96), grad-surf is the radiation heat transferred between S6 and any other visible surface of the electrode. Murthy and Fedorov [65] noted that the surface-to-surface approach (as in Equation 3.96) could lead to some temperature prediction mistakes. According to [65], more accurate results can be expected considering the absorption, emission or scattering in the media. On the present topic, there are still ongoing studies and a common position about the effect of considering the absorption, emission or scattering in the media is still a matter of clarification (see for example [79] and the apparently opposite position of [42] and [65]). 

In this equation, s2 and A2 are the emissivity of body 2 and the effective area of radiation for body 2, respectively. Since radiation heat transfer has to obey the principle of the conservation of energy, the magnitude of the net heat radiated in Eq. (1.8) has to equal the magnitude of the heat radiated in Eq. (1.9). 

SOLUTION Trio rates of radiation heat transfer between a person and the surrounding surfaces at specified temperatures are to be determined in summer and winter. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by convection is not considered. 3 The person is completely surrounded by the interior surfaces of the room. 4 The surrounding surfaces are at a uniform temperature. Properties The emissivity of a person Is r - 0.98 (Table 1-6),... 

The difference between the rates of radiation emitted by the surface and the radiation absorbed is the net radiation heat transfer, If the rate of radiation absorption is greater than the rate of radialion emission, the surface is said to be gaining energy by radiation. Otherwise, the surface is said to be losing energy by radialion. In general, the determination of the net rate of heat transfer by radiation between two surfaces is a complicated matter since it depends on the properties of the surface.s, their orientation relative to each other, and the interaction of the medium between the surfaces with radiation. 

When a surface of emissivity and surface area at a thermodynamic tem-peratwe T, is completely enclosed by a much larger (or black) surface at thermodynamic temperature T separated by a gas (such as air) that does not intervene with radiation, the net rate of radiation heat transfer between these two surfaces is given by (Fig. 1-37)... 

In tliis special case, the emissivity and the surface area of the surrounding sur face do not have any effect on the net radiation heat transfer. 

The roof of a house consists of a 25-cin-lhick concrete slab (k 1.9 W/m C) that is 8-m wide and 10 m long. The emissivity of tlie outer surface of the roof is 0.8, and the convection heal transfer coefficient on (hat surface is estimated to be 18 W/m °C. On a cleat winter nighi, the ambient air is reported to be at I0 C, while the night sky temperature for radiation heat transfer is 170 K. If the inner surface temperature of tberoofisTi = 16"C, determine (he outer surface temperature of the roof and the rate of heat loss through the roof when steady operating conditions are reached. 

C What is effective emissivity for a plane.parallel air space How is it deiermined How is radiation heat transfer through the airspace determined when the effective emissivity is known ... 

Many familiar heal transfer application. involve natural convection as the primary mechanism of heat transfer. Some examples are cooling of electronic equipment such as power transistors, TVs, and DVDs heat transfer from electric baseboard heaters or steam radiators heat transfer from the refrigeration coils and power transmission lines and heat transfer from the bodies of animals and human beings. Natural convection in gases is usually accompanied by radiation of comparable magnitude except for low-emissivity surfaces. 

The emissivity of an ordinary glass surface, for example, is 0.84. Therefore, the effective emissivity of two parallel glass surfaces facing each other is 0.72. Radiation heat transfer between concentric cylinders and spheres is discussed in Chapter 13. 

Minimize radiation heat transfer through the air space. This can be done by reducing the einissivity of glass surfaces by coating them with low-eiTiissivity (or "low-e" for short) material. Recall that the effective emissivity of two parallel plates of emissivities e, and 2 is given by... 

SOLUTION A surface is exposed to solar and sky radiation. The net rate of radiation heat transfer is to be determined for four different combinations of emissivities and solar absorptivities. 

Note that radiation heat transfer between a room and its windows is proportional to the emissivity of the glass surface facing the room, and can be expressed as... 

The direction of the net radiation heat transfer depends on the relative magnitudes of 7, (the radiosity) and (, (the emissive power of a blackbody at the teinpeiature of the surface). It is from the surface if > 7,- and to the surface if 7f > ft),-. A negative value for ft indicates that heat transfer is to the surface. All of this radiation energy gained must be removed from the other side of the surface through some mechanism if the surface temperature is to remain constant. 

Consider an enclosure consisting of two opaque surfaces at specified temperatures r, and T2, as shown in Fig. 13-24, and try to determine llie net rate of radiation heat transfer between the two surfaces with the network method. Surfaces 1 and 2 have emissivities c, and and surface areas /1 and A2 and are maintained at uniform temperatures T, and T, respectively. There are only two surfaces in the enclosure, and thus we can write... 

Consider a cylindrical furnace with r = W = 1 m, as shosvn in Fig. 13-27. The lop (surface 1) and the base (surface 2) of the furnace have emissivities ei = 0.8 and 2 = 0.4, respectively, and are maintained at uniform temperatures T 700 K and T2 - 500 K. The side surface closely approximates a blackbody and is maintained at a temperature of = 400 K. Determine the net rate of radiation heat transfer at each surface during steady operation and explain hovr these surfaces can be maintained at specified temperatures. 

Now consider a radiation shield placed between these two plates, as shown in Fig. 13-30. Let the emissivities of the shield facing plates 1 and 2 be 3 j and E, 2, respectively. Note that the emissivity of different surfaces of the shield may be different. The radiation network of this geometry is constructed, as usual, by drawing a surface resistance associated with each surface and connecting these surface resistances with space resistances, as shown in the figure. The resistances are connected in series, and thus the rate of radiation heat transfer is... 

Therefore, when all emissivities are equal, 1 shield reduces the rale of radiation heat transfer to one-half, 9 shields reduce it to one-ienlh, and 19 shields reduce it to onc-twenlielh (or 5 percent) of whal it was when there were no shields. 

A thin aluminum sheet with an emissivity of 0.1 on both sides is placed between two very large parallel plates that are maintained at uniform temperatures Ti = 800 K and T2 = 500 K and have emissivities ci = 0.2 and et 0.7, respectively, as shown in Fig. 13-32. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates and compare the.result to that without the shield. 

Radiation heat transfer between two surfaces can be reduced greatly by inserting belsveen the two siirface.s thin, high-reflectivity (low emissivity) sheets of material called radiation shields. Radiation heat transfer between two large parallel plates separated by N radiation shields is... 

Two very large parallel plates are maintained at uniform temperatures of Ti - 600 K and Tj = 400 K and have emissivities e, = 0.5 and = 0.9, respectively. Determine the net tale of radiation heat transfer between the two surfaces per unit area of the plates. 

A furnace is shaped like a long semicylindrical duct of diameter D = 5 in. The base and the dome of the furnace have emissivities of 0.5 and 0.9 and are maintained at unifonn tern peralures of 300 and 1000 K. respectively. Determine the net rate of radiation heat transfer from the dome to the base surface per unit length during steady operation. 

T vo concentric spheres of diameters D, = 0.3 m and Dj = 0.4 ni are maintained at uniform temperatures 7 1 = 700 K and Tj = -500 K and have emissivities 6 = 0.5 and 62 = 0.7, respectively, Deiemune the nel rate of radiation heat transfer between the two spheres. Also, determine the convection heat transfer coefficient at the ouler surface if both the surrounding medium and Ihe surrounding surfaces are at SO C, Assume Ihe emissivity of the outer surface is 0.35. 

Two very large parallel plates are maintained at uniform temperatures of T, 1000 K and Tj = 800 K and have emissiviiies of e, = Cj = 0.5, respectively. It is desired to reduce the net rate of radiation heat transfer betweert the two plates to one-fifih by placing thin aluminum sheets with an emissivity of 0.1 on both sides between the plates. Defermine ihe number of sheets that need to be inserted. 

Two parallel disks of diameter D = 1 m separated by E = 0.6 m are located directly on top of each other. The disks are separated by a radiation shield whose emissivity is 0.15. Both disks arc black and are maintained at temperatures of 650 K and 400 K, respectively. The environment that the disks are in can be considered to be a blackbody at 300 K. Determine the net rate of radiation heat transfer through the shield under steady conditions. Ansmt 268 W... 

Two coaxial cylindei.s of diameters D = O.IO 111 and O3 = 0.30 in and emissivities e, = 0.7 and 63 = 0.4 aie maintained at uniform temperatures of T = 750 K and = 500 K, respectively. Now a coaxial radiation shield of diameter D3 = 0.20 m and emissivity 63 = 0.2 is placed between tlie two cylinders. Determine the net rate of radiation heat transfer between the two cylinders per unit length of the cylinders and compare the result with that without the shield. 

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