Surface radiosity

For known surface radiosity, Eq. (9.97) gives the radiation flux from a surface with a specified temperature, and Eq. (9.98) gives the temperature of a surface with specified heat flux. Thus, for both cases, the problem is reduced to the evaluation of surface radiosities as follows. 

Uniform Surface Radiosity. Here, first, we limit consideration to gray diffuse surfaces with uniform radiosity. In that case, Eqs. 7.71 and 7.72 apply. Because of the assumption of gray diffuse surfaces, Eq. 7.71 can be rewritten as... 

Evaluation of the AS" s that charac terize an enclosure involves solution of a system of radiation balances on the surfaces. If the assumption is made that all the zones of the enclosure a re gray and emit and reflec t diffusely, then the direct-exchange area ij, as evaluated for the black-siirface pair A and Aj, applies to emission and reflections between them. If at a surface the total leaving-flnx density, emitted phis reflected, is denoted by W (and called by some the radiosity and by others the exitance), radiation balances take the form ... 

If the temperature of a grey surface is known, then the net heat transfer to or from the surface may be determined from the value of the radiosity qo- With regard to signs, the usual convention is that a positive value of g, indicates heat transfer from grey surfaces. 

Radiation arrives at a grey surface of emissivity 0.75 al a constant temperature of 400 K, at the rate of 3 kW/m2. What is the radiosity and the net rate of radiation transfer to the surface What coefficient of heat transfer is required to maintain the surface temperature at 300 K if the rear of the surface is perfectly insulated and the front surface is cooled by convective heat transfer to air at 295 K ... 

For the more complex case of a multi-sided enclosure formed from n surfaces, the radiosities may be obtained from an energy balance for each surface in turn in the enclosure. Thus the energy falling on a typical surface i in an enclosure formed from... 

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 ... 

Irradiation G Total thermal radiation energy incident on a surface per unit time per unit area Irradiation (G), and Radiosity J) are all energy fluxes (i.e., rate... 

Radiosity J Total thermal radiation energy leaving a surface (emitted and reflected) per unit time per unit area of energy transfer per unit area). The three terms, Absorptivity (a), Reflectivity (p), and Transmissivity (x), are all surface properties... 

In addition to the assumptions stated above, we shall also assume that the radiosity and irradiation are uniform over each surface. This assumption is not strictly correct, even for ideal gray diffuse surfaces, but the problems become exceedingly complex when this analytical restriction is not imposed. Sparrow and Cess [10] give a discussion of such problems. The radiosity is the sum of... 

The net energy leaving the surface is the difference between the radiosity and the irradiation ... 

Taking the resistance (1 — e,)/ejAi as zero, we have the network as shown. To calculate the heat flows at each surface we must determine the radiosities J, and J2. The network is solved by setting the sum of the heat currents entering nodes J, and J2 to zero ... 

We define the diffuse radiosity Jn as the total diffuse energy leaving the surface per unit area and per unit time, or... 

Equation (8-77) expresses the diffuse radiation leaving 1 which arrives at 2 and which may contribute to a diffuse radiosity of surface 2. The factor 1 - ps represents the fraction absorbed plus the fraction reflected diffusely. The inclusion of this factor is most important because we are considering only diffuse direct exchange, and thus must leave out the specular-reflection contribution... 

Now consider the diffuse exchange between surfaces 1 and 3. Of the energy leaving 1, the amount which contributes to the diffuse radiosity of surface 3 is... 

The diffuse radiosity <5f a particular surface of the medium is defined by... 

The diffuse radiosity is still defined as in Eq. (8-91), and the net energy exchange with a transmitting surface is given by Eq. (8-93). The analysis of transmitted energy exchange with other surfaces must be handled somewhat differently, however. 

Making use of Eqs. (8-99) and (8-102) gives the complete network for the system as shown in Fig. 8-61. Of course, all the radiation shape factors in the above network are unity, but they have been included for the sake of generality. In this network Jw refers to the diffuse radiosity on the left side of 2, while J 2D is the diffuse radiosity on the right side of this surface. 

The heat transfer at each surface is then evaluated in terms of the radiosities Jj. These parameters are obtained by recalling that the heat transfer can also be expressed as... 

To illustrate the radiation formulation for numerical solution we consider the circular hole 2 cm in diameter and 3 cm deep, as shown in the accompanying figure. The hole is machined in a large block of metal, which is maintained at l000oC and has a surface emissivity of 0.6. The temperature of the large surrounding room is 20°C. A simple approach to this problem would assume the radiosity uniform over the entire heated internal surface. In reality, the radiosity varies over the suiface, and we break it into segments 1 (bottom of the hole), 2. 3, and 4 (sides of the hole) for analysis. 

The large room acts like a blackbody at 20°C, so for analysis purposes we can assume the hole is covered by an imaginary black surface S at 20°C. We shall set the problem up for a numerical solution for the radiosities and then calculate the heat-transfer rates. After that, we shall examine an insulated-surface case for this same geometry. 

It is of interest to compare this heat transfer with the value we would obtain by assuming uniform radiosity on the hot surface. We would then have a two-body problem with... 

Thus, the simple assumption of uniform radiosity gives a heat transfer which is 3.9 percent above the value obtained by breaking the hot surface into four parts for the calculation. This indicates that the uniform-radiosity assumption we have been using is a rather good one for engineering calculations. 

It is of interest to compare the heat transfer calculated above with that obtained by assuming surfaces 2, 3, and 4 uniform in temperature and radiosity. Equation (8-41) applies for this case ... 

In this case the assumption of uniform radiosity at the insulated surface gives an overall heat transfer with surface 1 (bottom of hole) that is 18.7 percent too high. 

With such a small difference between the solutions we may conclude that the extra complexity of choosing each surface at a different radiosity is probably not worth the effort, particularly when one recognizes the uncertainties which are present in the surface emissivities. This points out that our assumptions of uniform irradiation and radiosity, though strictly not correct, give answers which are quite satisfactory. 

Two 10 by 30 cm rectangular plates are spaced 10 cm apart and connected by four insulated and re-radiating walls. The plate temperatures are uniform at 1000 and 300°C, and their emissivities are 0.6 and 0.4, respectively. Using the numerical method, determine the net heat transfer under the assumptions that (a) the four re-radiating surfaces act as one surface and have uniform radiosity and (b) the four re-radiating surfaces have radiosities determined from the radiant balance with all other surfaces. Assume that the 1000 and 300°C surfaces have uniform radiosity. Also calculate the temperatures for the re-radiating surfaces for each case above. 

Related Calculations. If the six surfaces are not black but gray (in the radiation sense), it is nominally necessary to set up and solve six simultaneous equations in six unknowns. In practice, however, the network can be simplified by combining two or more surfaces (the two smaller end walls, for instance) into one node. Once this is done and the configuration factors are calculated, the next step is to construct a radiosity network (since each surface is assumed diffuse, all energy leaving it is equally distributed directionally and can therefore be taken as the radiosity of the surface rather than its emissive power). Then, using standard mathematical network-solution techniques, create and solve an equivalent network with direct connections between nodes representing the surfaces. For details, see Oppenheim [8],... 

Matrix characterization of the radiative energy balance at each surface zone is facilitated via definition of three M X 1 vectors the radiative surface fluxes Q = [ i], with units of watts and the vectors H = [if,] and W = [Wi] both having units of W/m2. The arrays H and W define the incident and leaving flux densities, respectively, at each surface zone. The variable W is also referred to in the literature as the radiosity or exitance. Since W = el-E + pIH, the radiative flux at each surface zone is also defined in terms of E, II, and W by three equivalent matrix relations, namely,... 

An Exact Solution to the Integral Equations—The Hohlraum Exact solutions of the fundamental integral equations for radiative transfer are available for only a few simple cases. One of these is the evaluation of the emittance from a small aperture, of area Ai, in the surface of an isothermal spherical cavity of radius R. In German, this geometry is termed a hohlraum or hollow space. For this special case the radiosity W is constant over the inner surface of the cavity. It then follows that the ratio W/E is given by... 

W = total leaving-flux density (also radiosity). a = absorptivity or absorptance ai2, absorptance of surface 1 for radiation from surface 2. 

Surfaces emit radiation as well as rellecting it, and thus the radiation leaving a surface consists of emitted and reflected components, as shown in Fig. 12 21. The calculation of radiation heat transfer between surfaces involves the total radiation energy streaming away from a surface, with no regard for its origin. Thus, we need to define a quantity that represents the rate at which radiation energy leaves a unit area of a surface in all directions. This quantity is called the radiosity J, and is expressed as... 

FIGURE 12-21 The three kinds of radiation flux where is the sum of the emitted and reflected intensities. For a surface that W/m ) emi.ssive power, irradiation, is both a diffuse emitter and a diffuse reflector, /,+r constant, and the ra- radiosity. 

C Foi a surface, how is radiosity defined For diffusely emitting and reflecting surfaces, how is radiosity related to the intensities of emitted and reflected radiation ... 

The total rate at which radiation leaves dA (via emission and reflection) in all directions is the radiosity (which is J, = tt/,) times the surface area,... 

To make a simple radiation analysis possible, it is common to assume the surfaces of an enclosure to be opaque, diffuse, and gray. That is, the siirface.s arc nontransparent, they are diffuse emitters and diffuse reflectors, and their radiation properties are independent of wavelength. Also, each surface of the enclosure is isothennal, and both the incoming and outgoing radiation are uniform over each surface. But first vve review the concept of radiosity introduced in Chap. 12. 

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