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Net radiation

The method is demonstrated here in terms of an enclosure made of N opaque surfaces, some being fiat, some convex, and some concave (Fig. 9.36). First, complete an enclosure by replacing any hole with a black surface at the ambient temperature as we have done before for solutions by electrical analogy. [Pg.472]

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. [Pg.472]

Consider a surface different from surface i, say surface j. The total radiation from this surface is Bj Aj. The fraction of this radiation intercepted by surface i is B AjFji, which may be rearranged in terms of the reciprocity relation [Eq. (9.11) for surfaces i and j] AjFji = AiF(j to give BjAiFij. Thus, the total radiation leaving all surfaces (including /) and intercepted by surface i is [Pg.472]

for each case, we end up with a set of N algebraic equations in terms of the unknown radiosities, B, B%. Bn. These equations can be solved by using a numerical iteration method as shown in the following example. For convenience, the solution procedure for enclosure radiation problems by the method of net radiation is summarized in Table 9.2 in terms of five steps. [Pg.473]

Express the temperature of surfaces with specified temperature in the absolute scale (Kelvin). [Pg.473]


The net radiation power falling the surface is the difference between the incoming and t)utgoing radiation and the difference between absorption and emission ... [Pg.123]

Taking area 1 as that of the plate, area 2 as the underside of the hemisphere, area 3 as an imaginary cylindrical surface linking the plate and the underside of the dome which represents the black surroundings and area 4 as an imaginary disc sealing the hemisphere and parallel to the plate, then, from equation 9.134, the net radiation to the surface of the plate 1 is given by ... [Pg.455]

Evaporation is influenced not only by the amount of energy available but also by the amount of water available. If unlimited volumes of water are available in the soil and on surface areas, this is referred to as potential evapotranspiration, i.e. the highest possible evapotranspiration under given climatic conditions. The actual evapotranspiration is the evapotranspiration which can effectively be observed. This is always lower than potential evapotranspiration and is dependent on water availability, plant and surface attributes, net radiation, air humidity, and wind speed. In mountains with a great deal of barren soil and large areas of debris and rock cover which are unable to store water in any great quantities, and with fastflowing water over steep terrain, actual evapotranspiration is very often limited and hence much lower than potential evapotranspiration. [Pg.43]

For grey bodies, the emissivity, e, must be included and the net radiation per unit area is then... [Pg.32]

The net radiation between the still and its surroundings was measured with a radiometer. The difference between the net and the total radiation was considered to be radiation from the warm components of the still and reflection of atmospheric and solar radiation from the still. The curve of still radiation in Figure 6 is almost a mirror image of the total radiation curve, still radiation being greatest during the day when the still is the warmest and solar radiation is being reflected from the cover. [Pg.174]

Describing the net radiation from body 2 to the other bodies, the following equations apply ... [Pg.23]

Problem A black element 0.5 cm by 0.5 cm, is at a temperature of 800°C and is near a tube of 2 cm diameter. The opening of the tube may be approximated as a black surface, and is at 400°C. Calculate the net radiation exchange along the connecting path R between the square element and the tube opening. [Pg.205]

We have now considered each of the terms that involve radiation in the energy balance of a leaf (Eqs. 7.1 and 7.2). These quantities comprise the net radiation balance for the leaf ... [Pg.330]

Net radiation = Absorbed solar irradiation + Absorbed DR from... [Pg.330]

Using Equations 7.5 through 7.7, we can express the net radiation balance as... [Pg.330]

Before continuing with our analysis of the energy balance of a leaf, we will examine representative values for each of the terms in the net radiation. [Pg.330]

Table 7-1. Representative Values for the Various Terms in the Net Radiation Balance of an Exposed Leaf ... [Pg.332]

A leaf at 25°C at sea level on a sunny day can have a net radiation balance of 370 W m-2 (see top line of Table 7-1). In the current case, just over half of this energy input by net radiation is dissipated by conduction of heat across the boundary layers on both sides of the leaf (190 W m-2 total), followed by forced convection in the surrounding turbulent air outside the boundary layers. [Pg.345]

For the leaf described in the top line of Table 7-1, a heat loss of 180 W m-2 by evaporation dissipates slightly under half of the net radiation balance (370 W m-2) the rest of the energy input is removed by heat conduction across the boundary layer followed by forced convection (Fig. 7-10a). [Pg.346]


See other pages where Net radiation is mentioned: [Pg.576]    [Pg.578]    [Pg.548]    [Pg.124]    [Pg.1063]    [Pg.447]    [Pg.448]    [Pg.458]    [Pg.471]    [Pg.205]    [Pg.8]    [Pg.349]    [Pg.43]    [Pg.49]    [Pg.783]    [Pg.784]    [Pg.226]    [Pg.277]    [Pg.423]    [Pg.336]    [Pg.461]    [Pg.466]    [Pg.23]    [Pg.23]    [Pg.240]    [Pg.318]    [Pg.319]    [Pg.330]    [Pg.330]    [Pg.331]    [Pg.331]    [Pg.331]    [Pg.332]    [Pg.333]    [Pg.347]    [Pg.349]    [Pg.352]   
See also in sourсe #XX -- [ Pg.318 , Pg.330 , Pg.332 ]




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