Radiation transfer between black surfaces

The calculation of the radiation heat transfer between black surfaces is relatively easy because all the radiant energy which strikes a surface is absorbed. The main problem is one of determining the geometric shape factor, but once this is accomplished, the calculation of the heat exchange is very simple. When nonblackbodies are involved, the situation is much more complex, for all the energy striking a surface will not be absorbed part will be reflected back to another heat-transfer surface, and part may be reflected out of the system entirely. The problem can become complicated because the radiant energy can be reflected back and forth between the heat-transfer surfaces several times. The analysis of the problem must take into consideration these multiple reflections if correct conclusions are to be drawn. 

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)... 

A 3-m internal diameter spherical tank made of 2-cm-thlck stainless steel (k = 15 W/m O is used to store iced water at r i == 0°C. Ttie tank is located in a room whose temperature is 7 j - 22°C. The walis of the room are also at 22°C. The outer surface of the tank is black and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. The convection heat transfer coefficients at the inner and the outer surfaces of the tank are h, = 80 W/m °C and I12 = 10 W/m °C, respectively. Determine (a) the rate of heat transfer to the iced water in the lank and (b) the amount of ice at 0 C that melts during a 2d-h period. 

Consider the 5-m x 5-m X S-m cubical furnace shown in Fig. 13-19, whose surfaces closely approximate black surfaces. The base, top, and side surfaces of the furnace are maintained at uniform temperatures of 800 K, 1500 K, and 500 K, respectively. Delermine (a) the net rale of radiation heat transfer be-tvreen the base and the side surfaces, (b) the net rate of radiation heat transfer between the base and the top surface, and (c) the net radiation heat transfer from the base surface. 

SOLUTION The surfaces of a cubical furnace are black and are maintained at uniform temperatures. The net rate of radiation heat transfer between the base and side surfaces, between tfie base and the top surface, and from the base surface are to bo determined. 

If the bounding surface is black at temperature the surface will emit radiation to the gas at a rate of A,crT/ without reflecting any, and the gas will absorb this radiation at a rale of UgA aTj, where is the absorptivity of the gas. Then the net rate of radiation heal transfer between the gas and a black surface suriouiidiiig it becomes... 

The rate of net radiation heal transfer between two black surfaces is determined from... 

The surfaces of a two-surface enclosure exchange heat with one another by thermal radiation. Surface 1 has a tempetature of 400 K, an area of 0.2 m, and a total emissivity of 0.4. Surface 2 is black, has a temperature of 600 K. and an area of 0.3 ntl If the view factor is 0.3, the rate of radiation heal transfer between the two surfaces is ( ) 87W (h) I35W (c) 244W... 

Now reconsider two bodies each with a flat black surface at absolute temperatures T and Tz separated l distance apart in a vacuum, as shown in Fig. 1.15(b). Surface 1 has emissive power Ebi, surface 2 has emissive power Ebz, and the radiation heat transfer between these bodies is... 

Figure US (a) The Stefan-Boltzm aim law, (b) radiation heat transfer between two parallel black surfaces.

The equation of radiative transfer will not be solved here since solutions to some approximations of the equation are well known. In photon radiation, it has served as the framework for photon radiative transfer. It is well known that in the optically thin or ballistic photon limit, one gets the heat flux as q = g T[ - T ) from this equation for radiation between two black surfaces [13]. For the case of phonons, this is known as the Casimir limit. In the optically thick or diffusive limit, the equation reduces to q = -kpVT where kp is the photon thermal conductivity. The same results can be derived for phonon radiative transfer [14,15]. 

The emissivity accounts for the properties of a radiating surface. An ideal radiator, or black body, has a value of e = 1. The rate of radiative heat transfer between two bodies is proportional to the difference between the fourth powers of their temperatures. In most commercial apparatus, not all the radiation from one body reaches the second. Radiation goes out in all directions, and only some of it reaches the intended receiver. The fraction that does is called the area factor or the view factor. Thus,... 

In order to determine the radiation heat-transfer rates between two black surfaces we must determine the general case for the fraction of the total radiant heat that leaves a surface and arrives on a second surface. Using only black surfaces, we consider the case shown in Fig. 4.11-4, in which radiant energy is exchanged between area elements tli4i and dA2 The line r is the distance between the areas and the angles between this line and the normals to the two surfaces are f , and 62 The rate of radiant energy that leaves dA 1... 

In catalytic channels, the flat plate surface temperature in Eq. (3.32) is attained at the channel entry (x O). As the catalytic channel is not amenable to analytical solutions, simulations are provided next for the channel geometry shown in Fig. 3.3. A planar channel is considered in Fig. 3.3, with a length L = 75 mm, height 21) = 1.2 mm, and a wall thickness 5s = 50 pm. A 2D steady model for the gas and solid (described in Section 3.3) is used. The sohd thermal conductivity is k = 6W/m/K referring to FeCr alloy, a common material for catalytic honeycomb reactors in power generation (Carroni et al., 2003). Surface radiation heat transfer was accounted for, with an emissivity = 0.6 for each discretized catalytic surface element, while the inlet and outlet sections were treated as black bodies ( = 1.0). To illustrate differences between the surface temperatures of fuel-lean and fuel-rich hydrogen/air catalytic combustion, computed axial temperature profiles at the gas—wall interface y=h in Fig. 3.3) are shown in Fig. 3.4 for a lean (cp = 0.3) and a rich cp = 6.9) equivalence ratio, p = 1 bar, inlet temperature, and velocity Tj = 300 K and Uin = 10 m/s, respectively. The two selected equivalence ratios have the same adiabatic equilibrium temperature, T d=1189 K. 

Fig. 8.6 Domains used in calculation of radiative heat transfer in a crystal and the gap between the crystal, the crucible, the afterheater and the heater. Left - the gap between the crucible and the afterheater is replaced by a black (radiant, nonreflective) surface. Right - the gap is transparent for radiation exchange between the crystal and the heater.

A plate. 1 m in diameter at 750 K, is to be heated by placing it beneath a hemispherical dome of the same diameter at 1200 K the distance between the plate and the bottom of the dome being 0.5 m, as shown in Figure 9.42. If the surroundings are maintained at 290 K, the surfaces may be regarded as black bodies and heat transfer from the underside of the plate is negligible, what is the net rate of heat transfer by radiation to the plate ... 

Radiation is the transfer of heat between surfaces of solids not in contact with each other and at different temperatures. A so-called black body absorbs all radiation heat, whereas a white body reflects part of it. Radiation (see Figure 5.17c) is described by the equation ... 

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