b( ) ) ( ) PROBLEM = W / m = 1991 W / m. = W / m 4 m = 3164 W.

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1 PROBLEM.6 KNOWN: rea, temperature, irradiation and spectral absorptivity of a surface. FIND: bsorbed irradiation, emissive power, radiosity and net radiation transfer from the surface. SCHEMTIC: SSUMPTIONS: () Opaque, diffuse surface behavior, () Spectral distribution of solar radiation corresponds to emission from a blackbody at 5800 K. NLYSIS: The absorptivity to solar irradiation is α b ( ) 0 λgλdλ α 0 λeλ 5800 K dλ αs = = = αf( 0.5 m) F G E μ + α ( ). b From Table., λt = 900 μm K: F (0 0.5 μm) = 0.50 λt = 5800 μm K: F (0 μm) = 0.70 λt =,600 μm K: F (0 μm) = 0.9 α s = 0.8( ) + 0.9( 0.9) = 0.9. G = G = W / m = 55 W / m. ( ) Hence, abs αs S The emissivity is ε = b( ) 0 ε λeλ 00 K d λ / Eb = ε F( 0.5 μ m) + ε F ( ). From Table., λt = 00 μm K: F (0 0.5 μm) = 0 λt = 00 μm K: F (0 μm) = 0 λt = 800 μm K F (0 μm) = 0. Hence, ε = ε = 0.9, E = εσ T 8 s = W / m K ( 00 K) = 306 W / m. The radiosity is ( α ) [ ] J = E + ρsgs = E + s GS = W / m = 99 W / m. The net radiation transfer from the surface is ( α ) ( ) qnet = E SGS s = W / m m = 36 W. COMMENTS: Unless 36 W are supplied to the surface by other means (for example, by convection), the surface temperature will decrease with time.

2 PROBLEM.8 KNOWN: Small disk positioned at center of an isothermal, hemispherical enclosure with a small aperture. FIND: Radiant power [μw] leaving the aperture. SCHEMTIC: SSUMPTIONS: () Disk is diffuse-gray, () Enclosure is isothermal and has area much larger than disk, (3) perture area is very small compared to enclosure area, () reas of disk and aperture are small compared to radius squared of the enclosure. NLYSIS: The radiant power leaving the aperture is due to radiation leaving the disk and to irradiation on the aperture from the enclosure. That is, qap = q + G. () The radiation leaving the disk can be written in terms of the radiosity of the disk. For the diffuse disk, q = J cosθ ω π () and with ε = α for the gray behavior, the radiosity is J σ T = εeb( T) + ρg= εσ T + ( ε) 3 (3) where the irradiation G is the emissive power of the black enclosure, E b (T 3 ); G = G = E b (T 3 ). The solid angle ω follows from Eq.., ω = /R. Combining Eqs. (), (3) and () into Eq. () with G = σ T3, the radiant power is q ( ) ap = σ ε T ε T 3 cosθ σ π + + T 3 R 8 W π qap = ( 900K) + ( 0.7)( 300K) ( 0.005m) cos 5 π m K π / ( 0.00m) π 8 + ( 0.00m) W / m K ( 300K) 0.00m ( ) qap = ( ) μw = 79 μw. COMMENTS: Note the relative magnitudes of the three radiation components. lso, recognize that the emissivity of the enclosure ε 3 doesn t enter into the analysis. Why? ()

3 PROBLEM.77 KNOWN: Diffuse-gray sphere is placed in large oven with known wall temperature and experiences convection process. FIND: (a) Net heat transfer rate to the sphere when its temperature is 300 K, (b) Steady-state temperature of the sphere, (c) Time required for the sphere, initially at 300 K, to come within 0 K of the steady-state temperature, and (d) Elapsed time of part (c) as a function of the convection coefficient for 0 h 5 W/ m K for emissivities 0., 0. and 0.8. SCHEMTIC: SSUMPTIONS: () Sphere surface is diffuse-gray, () Sphere area is much smaller than the oven wall area, (3) Sphere surface is isothermal. PROPERTIES: Sphere (Given) : α = m /s, k = 85 W/m K. NLYSIS: (a) From an energy balance on the sphere find qnet = qin qout qnet = αgs + qconv Es qnet = ασ To s + hs ( T Ts ) εσ Ts s. () Note that the irradiation to the sphere is the emissive power of a blackbody at the temperature of the oven walls. This follows since the oven walls are isothermal and have a much larger area than the sphere area. Substituting numerical values, noting that α = ε since the surface is diffuse-gray and that s = πd, find 8 qnet = W m K ( 600K) + 5 W m K ( ) K a f c h W m K 300K π 30 0 m = [ + ] =. () qnet W 9.8 W (b) For steady-state conditions, q net in the energy balance of Eq. () will be zero, 0= ασ Tos + hs( T Tss) εσ Tsss Substitute numerical values and find the steady-state temperature as Tss = 538.K ()

4 PROBLEM 3. KNOWN: Various geometric shapes involving two areas and. FIND: Shape factors, F and F, for each configuration. SSUMPTIONS: Surfaces are diffuse. NLYSIS: The analysis is not to make use of tables or charts. The approach involves use of the reciprocity relation, Eq. 3.3, and summation rule, Eq. 3.. Recognize that reciprocity applies to two surfaces; summation applies to an enclosure. Certain shape factors will be identified by inspection. Note L is the length normal to page. (a) Long duct (L): By inspection, F =.0 RL F = F =.0 = = 0. ( 3/) πrl 3π (b) Small sphere,, under concentric hemisphere,, where = Summation rule F + F + F3 = But F = F 3 by symmetry, hence F = 0.50 F = F = 0.5 = 0.5. (c) Long duct (L): (d) Long inclined plates (L): By inspection, F =.0 RL F = F =.0 = = πrl π Summation rule, F = F = 0.6 = Summation rule, F + F + F3 = But F = F 3 by symmetry, hence F = L F = F = 0.5 = / 0 L ( ) (e) Sphere lying on infinite plane Summation rule, F + F + F 3 = But F = F 3 by symmetry, hence F = 0.5 F = F 0 since. Continued..

5 PROBLEM 3. (Cont.) (f) Hemisphere over a disc of diameter D/; find also F and F 3. By inspection, F =.0 Summation rule for surface 3 is written as F3 + F3 + F33 =. Hence, F3 =.0. F 3 3 = F3 πd π ( D/) D π F 3 = /.0 = π D πd F = F = / = Summation rule for, F + F + F3 = or Note that by inspection you can deduce F = 0.5 (g) Long open channel (L): F = F F3 = = 0.5. Summation rule for F + F + F3 = 0 but F = F 3 by symmetry, hence F = L F = F = = 0.50 = ( π )/ L π COMMENTS: () Note that the summation rule is applied to an enclosure. To complete the enclosure, it was necessary in several cases to define a third surface which was shown by dashed lines. () Recognize that the solutions follow a systematic procedure; in many instances it is possible to deduce a shape factor by inspection.

6 PROBLEM 3.8 KNOWN: rrangement of plane parallel rectangles. FIND: Show that the view factor between and can be expressed as F =, F, F 3 F b g b gb3, g where all F ij on the right-hand side of the equation can be evaluated from Fig. 3. (see Table 3.) for aligned parallel rectangles. SCHEMTIC: SSUMPTIONS: Diffuse surfaces with uniform radiosity. NLYSIS: Using the additive rule where the parenthesis denote a composite surface, F = F3 + F + F3 + F (,) (,)(,3) () where the asterisk () denotes that the F ij can be evaluated using the relation of Figure 3.. Now, find suitable relation for F 3. By symmetry, F3 = F () and from reciprocity between and, F = F (3) Multiply Eq. () by and substitute Eq. (3), with =, F F 3 = = F = F () Substituting for F 3 from Eq. () into Eq. (), and rearranging, F = (,) F(,)(,3) F3 F

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