Problem One Answer the following questions concerning fundamental radiative heat transfer. (2 points each) Part Question Your Answer
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1 Problem One Answer the following questions concerning fundamental radiative heat transfer. ( points each) Part Question Your Answer A Do all forms of matter emit radiation? Yes B Does the transport of thermal radiation require matter? No C What form of energy is responsible for the emission of thermal radiation? (Potential, Kinetic, Wind, Internal, Chemical, or Work) Internal D What intensive property of any given system is responsible for driving radiation? Temperature E What is the equation for defining wavelength? F What is the term describing how radiation varies with wavelength? Spectral G In addition to the answer for part F, what other feature complicates the nature of thermal radiation? Directionality H What term /concept is also called radiative flux and encompasses radiation incident from all directions? Irradiation I According to the Planck distribution, how does radiation vary with wavelength? Continuously J Since radiative heat transfer has many terms, constants and concepts, what table in the text can you always refer to if you have a question? Table 1.3 Page 777
2 PROBLEM 1.4 KNOWN: Furnace with prescribed aperture and emissive power. FIND: (a) Position of gauge such that irradiation is G = 1000 W/m, (b) Irradiation when gauge is tilted θ d = 0 o, and (c) Compute and plot the gage irradiation, G, as a function of the separation distance, L, for the range 100 L 300 mm and tilt angles of θ d = 0, 0, and 60 o. SCHEMATIC: ASSUMPTIONS: (1) Furnace aperture emits diffusely, () A d << L. ANALYSIS: (a) The irradiation on the detector area is defined as the power incident on the surface per unit area of the surface. That is G = qf d Ad qf d = IeAf cosθ f ωd f (1,) where q f d is the radiant power which leaves A f and is intercepted by A d. From Eqs. 1. and 1.7, ωd f is the solid angle subtended by surface Ad with respect to A f, ωd f = Ad cosθd L. (3) Noting that since the aperture emits diffusely, I e = E/π (see Eq. 1.1), and hence G = E π Af cosθf Ad cosθd L Ad (4) ( ) ( ) Solving for L and substituting for the condition θ f = 0 o and θ d = 0 o, L = Ecos θf cos θd A f π G. (5) 1/ 5 π 3 L = W m (0 10 ) m π 1000 W m = 193 mm 4. < (b) When θ d = 0 o, q f d will be reduced by a factor of cos θ d since ω d-f is reduced by a factor cos θ d. Hence, G = 1000 W/m cos θ d = 1000W/m cos 0 o = 940 W/m. < (c) Using the IHT workspace with Eq. (4), G is computed and plotted as a function of L for selected θ d. Note that G decreases inversely as L. As expected, G decreases with increasing θ d and in the limit, approaches zero as θ d approaches 90 o Irradiation, G (W/m^) thetad = 0 deg thetad = 0 deg thetad = 60 deg Separation distance, L (mm)
3 PROBLEM 1.5 KNOWN: Radiation from a diffuse radiant source A 1 with intensity I 1 = W/m sr is incident on a mirror Am, which reflects radiation onto the radiation detector A. FIND: (a) Radiant power incident on A m due to emission from the source, A 1, q 1 m (mw), (b) Intensity of radiant power leaving the perfectly reflecting, diffuse mirror A m, I m (W/m sr), and (c) Radiant power incident on the detector A due to the reflected radiation leaving A m, q m (μw), (d) Plot the radiant power q m as a function of the lateral separation distance y o for the range 0 y o 0. m; explain features of the resulting curve. SCHEMATIC: ASSUMPTIONS: (1) Surface A 1 emits diffusely, () Surface A m does not emit, but reflects perfectly and diffusely, and (3) Surface areas are much smaller than the square of their separation distances. ANALYSIS: (a) The radiant power leaving A 1 that is incident on A m is q1 m = I1 A1 cosθ1 Δ ω m-1 where ω m-1 is the solid angle A m subtends with respect to A 1, Eq. 1., da θ Δω n A m cos m m-1 = r x o + yo m cos 45 = 10 4 = m sr with θ m = 90 θ1 and θ1 = 45, q W / m sr m 1 m cos = sr = 60 mw < (b) The intensity of radiation leaving A m, after perfect and diffuse reflection, is b W Im = q1 m Am W/m / g / π = = 955. sr π 10-4 m (c) The radiant power leaving A m due to reflected radiation leaving A m is qm = q = Im Am cos θ m Δ ω m where Δω -m is the solid angle that A subtends with respect to A m, Eq. 1., Continued..
4 PROBLEM 1.5 (Cont.) da θ Δω n A m = cos r Lo xo + yo b g = m cos 45 = m sr with θ = 90 - θ m qm = q = 95.5 W/m sr 10-4 m cos sr = 47.8 μ W < (d) Using the foregoing equations in the IHT workspace, q is calculated and plotted as a function of yo for the range 0 y o 0. m. 100 Emitted power from A1 reflected from Am onto A 80 q (uw) yo (m ) From the relations, note that q is dependent upon the geometric arrangement of the surfaces in the following manner. For small values of y o, that is, when θ 1 0, the cos θ 1 term is at a maximum, near unity. But, the solid angles Δω m-1 and Δω -m are very small. As y o increases, the cos θ 1 term doesn t diminish as much as the solid angles increase, causing q to increase. A maximum in the power is reached as the cos θ 1 term decreases and the solid angles increase. The maximum radiant power occurs when y o = m which corresponds to θ 1 = 30.
5 KNOWN: Various surface temperatures. PROBLEM 1. FIND: (a) Wavelength corresponding to maximum emission for each surface, (b) Fraction of solar emission in UV, VIS and IR portions of the spectrum. ASSUMPTIONS: (1) Spectral distribution of emission from each surface is approximately that of a blackbody, () The sun emits as a blackbody at 5800 K. ANALYSIS: (a) From Wien s law, Eq. 1.5, the wavelength of maximum emission for blackbody radiation is C3 898 μm K λ max = =. T T For the prescribed surfaces Hot Cool Surface Sun Tungsten metal Skin metal (5800K) (500K) (1500K) (305K) (60K) λ max (μm) < (b) From Fig. 1.3, the spectral regions associated with each portion of the spectrum are Spectrum Wavelength limits, μm UV VIS IR For T = 5800K and each of the wavelength limits, from Table 1.1 find: λ(μm) λt(μm K) F (0 λ) Hence, the fraction of the solar emission in each portion of the spectrum is: F UV = = 0.15 < F VIS = = < F IR = = < COMMENTS: (1) Spectral concentration of surface radiation depends strongly on surface temperature. () Much of the UV solar radiation is absorbed in the earth s atmosphere.
6 PROBLEM 1.30 KNOWN: Spectral distribution of emissivity for zirconia and tungsten filaments. Filament temperature. FIND: (a) Total emissivity of zirconia, (b) Total emissivity of tungsten and comparative power requirement, (c) Efficiency of the two filaments. SCHEMATIC: ASSUMPTIONS: (1) Negligible reflection of radiation from bulb back to filament, () Equivalent surface areas for the two filaments, (3) Negligible radiation emission from bulb to filament. ANALYSIS: (a) From Eq. (1.36), the emissivity of the zirconia is ε = o ε = ε + ε + ε λ ( E λ /Eb) dλ 1F( 0 0.4μm) F( μm) 3F( 0.7μm ) ( ) ( ) ε = ε1f( 0 0.4μm) + ε F( 0 0.7μm) F( 0 0.4μm) + ε3( 1 F0 0.7μ m ) From Table 1.1, with T = 3000 K λ μ μ ( 0 0.4μm) λ μ μ ( 0 0.7μm) T = 0.4 m m K : F = T = 0.7 m 3000 K = 100 m K : F = ε = ( ) + 0. ( ) = 0.49 < (b) For the tungsten filament, ε = ε1f( 0 μm) + ε( 1 F( 0 μm) ) With λt = 6000μm K, F(0 μm) = ε = ( ) = < Assuming, no reflection of radiation from the bulb back to the filament and with no losses due to natural convection, the power consumption per unit surface area of filament is P 4 elec = εσ T. Continued..
7 PROBLEM 1.30 (Cont.) Pelec = W / m K K = W / m P elec = W / m K 3000 K = W / m Zirconia: ( ) 4 Tungsten: ( ) 4 Hence, for an equivalent surface area and temperature, the tungsten filament has the largest power consumption. < (c) Efficiency with respect to the production of visible radiation may be defined as ( ) ε 0.4 λ E λ,b d λ ε 0.4 λ E λ,b /E b ε η vis vis = = = F E ε ε m With F ( μm) = for T = 3000 K, Zirconia: η vis = ( 0.8 / 0.49) = 0.63 Tungsten: η vis = ( 0.45 / 0.358) = ( μ ) Hence, the zirconia filament is the more efficient. < COMMENTS: The production of visible radiation per unit filament surface area is E vis = η vis P elec. Hence, Zirconia: Tungsten: E 6 5 vis = W / m = W / m E 6 5 vis = W / m = W / m Hence, not only is the zirconia filament more efficient, but it also produces more visible radiation with less power consumption. This problem illustrates the benefits associated with carefully considering spectral surface characteristics in radiative applications.
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PROBLEM L. (3) Noting that since the aperture emits diffusely, I e = E/π (see Eq ), and hence
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