SOLUTIONS. F 0 λ1 T = (1) F 0 λ2 T = (2) ε = (6) F 0 λt = (7) F 0 λt = (11)

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1 Federal University of Rio Grande do Sul Mechanical Engineering Department MEC Thermal Radiation Professor - Francis Name: Márleson Rôndiner dos Santos Ferreira SOLUTIONS Question 2-5: The total hemispherical emissivity of this surface is at a temperature of 2000 K. Thus, the hemispherical absorptivity is: λ 1 = 1.0 µm = λ 1 T = 2000 µmk λ 2 = 2.0 µm = λ 2 T = 000 µmk F 0 λ1 T = (1) F 0 λ2 T = (2) α = 0.75 F 0 λ1 T (F 0 λ2 T F 0 λ1 T) (3) α = ( ) () α = (5) Therefore, by Kirchhoff s Law (α = ε), we have the total hemispherical emissivity: Question 2-6: (a) For a diffuse surface with a temperature of 6000 K. Therefore, the total absorptivity is: ε = (6) λ = 1.2 µm = λt = 7200 µmk (b) For a surface with a temperature of 600 K. F 0 λt = (7) α = 0.95 F 0 λt (1 F 0 λt ) (8) α = ( ) (9) α = (10) λ = 1.2 µm = λt = 720 µmk F 0 λt = (11) 1

2 Thus, the total absorptivity is: α = 0.95 F 0 λt (1 F 0 λt ) (12) α = ( ) (13) α = (1) Therefore, by Kirchhoff s Law (α = ε), we have the total emissivity: Question 2-12: The surface is diffuse. ε = (15) (a) We can calculate α for incident radiation from a gray source at 100 K in the following way: λ 1 = 2.0 µm = λ 1 T = 2800 µmk λ 2 =.0 µm = λ 2 T = 5600 µmk Thus, the total hemispherical emissivity is: F 0 λ1 T = (16) F 0 λ2 T = (17) ε = 0.6 F 0 λ1 T (F 0 λ2 T F 0 λ1 T) (1 F 0 λ2 T) (18) ε = ( ) ( ) (19) ε = (20) Therefore, by Kirchhoff s Law (α = ε), we have the total absorptivity: (b) We can calculate α, computing the following ratio: α = (21) α = ε ε source (22) α = (23) α = (2) Question 2-1: The sheet is placed in the orbit around the sun where the solar flux is 1350 W/m 2. The other face of the sheet is coated with a diffuse gray coating of hemispherical total emissivity ε = 0.6. (a) Note that for the solar radiating temperature T sun = 5780 K and λ = 2.1 µm, we have λt sun = µmk = F 0 λtsun =

3 Thus, as we know ε = α, we have α = 0.9 F 0 λtsun (1 F 0 λtsun ) (25) α = ( ) (26) α = (27) We can calculate the temperature T s by the equation (38). (ε a + ε b )σt s = αq i,sun, (28) where ε b = 0.6, q i,sun = 1350 W/m 2 and σ is given from the Table 1. Thus, we perform an iterative process by Algorithm 1 and in appendix, i.e., we give an initial value T s = 100 K, compute ε a, compute T s again and so on. Therefore, for a two iteration, we have: T s = K (29) (b) For a gray side α = ε b, we perform an iterative process by Algorithm 2 and in appendix. Thus, by equation (38) and an initial value T s = 100 K, we have the temperature for a gray side facing normal to the sun with two iteration: T s = K. (30) (c) We can calculate the hemispherical total reflectivity for diffuse surface noting that: From the item (a), we have α = Thus, ρ + α = 1 (31) ρ = 1 α (32) ρ = (33) ρ = 0, 13 (3) Question 2-16: The surface is in earth orbit around the sun and has the solar flux 1353 W/m 2 incident on it in the normal direction. Note that for the solar radiating temperature T sun = 5780 K and λ = 1.7 µm, we have Thus, as we know ε = α, we have λt sun = 9826 µmk = F 0 λtsun = α = 0.9 F 0 λtsun (1 F 0 λtsun ) (35) α = ( ) (36) α = (37) We can calculate the equilibrium temperature T by the equation (38). (ε a + ε b )σt = αq i,sun, (38) where ε b = 0 (because the perfect insulation on back side), q i,sun = 1350 W/m 2 and σ is given from the Table 1. Thus, we perform an iterative process by Algorithm 3 and in appendix, i.e., we give an initial value T s = 300 K, compute ε a, compute T s again and so on. Therefore, for a three iteration, we have: T s = K (39) 3

4 Question 2-17: (a) We calculate the hemispherical emissivity for this surface by: ε = 1 π But the function are given by: 2π π/2 0 0 ε(θ) cos(θ) sin(θ) dθdϕ. (0) ε(θ) = 0.9, ε(θ) = 0.3, 0 θ π/6, π/6 θ π/2. Thus, ε = 1 π 2π π/ π π/2 0.9 cos(θ) sin(θ) dθdϕ (1) cos(θ) sin(θ) dθdϕ (2) π 0 π/6 ε = 1 [ 2π 2π ] dϕ dϕ (3) π 0 0 ε = () ε = 0.6 (5) (b) As we have a gray surface, the fraction emitted is the same that the fraction absorbed, i.e., (c) The rate of energy that must be added, is given by: where T = 800 K and σ is given from the Table 1. Thus, α = ε (6) α = 0.6 (7) Q = εσt, (8) Q = [ WK m 2 K ], (9) Q = 13935, 85 W m 2 (50) Question -1: Consider the temperature T source = 000 K, so for a wavelength λ = 1.5 we have λt source = 6000µmK = F 0 λtsource = (51) Thus, the flux absorved is given by: q a = [0.95F 0 λtsource (1 F 0 λtsource )]q i (52) q a = [ ( )]q i (53) q a = 0.712q i (5)

5 For other side, with T eq = 1000 K, we have Thus, the flux emitted is given by: λt eq = 1500µmK = F 0 λteq = (55) q e = ε(t eq )σt eq (56) q e = [0.95F 0 λteq (1 F 0 λteq )]σteq (57) q e = [ ( )] (58) q e = (59) Finally, equaling (5) and (59), since we have a gray-body source, follows q a = q e (60) 0.712q i = (61) q i = 888, 06 W m 2 (62) Porto Alegre - October 16,

6 Appendix Definição Símbolo e valor Constant in Planck s spectral energy (or intensity) distribuition C 1 = [W µm /(m 2 sr)] Constant in Planck s spectral energy (or intensity) distribuition C 2 = [µm K] Constant in Wien s displacement law C 3 = [µm K] Constant in equation for max. blackbody intensity C = [W/(m 2 µm K 5 sr)] Stefan-Boltzman Constant σ = [W/(m 2 K )] Speed of the Light c = m/s Table 1: Radiation Constants Wavelength Temperature Blackbody product λt Fraction µm K F 0 λt E Table 2: Blackbody Function from the Siegel s Book Algorithm 1: Calulate the temperature in MATLAB - Question 2-1a 1 clear ; clc ; 2 sigma = *10^( -8); % [W/(m^2 * K ^)] 3 q =1350; % [W/m ^2] 5 lambda =2. 1; % mu m 6 T =100; % K 7 EA1 =0.9; 8 EA2 =0.1; 9 EB =0.6; 10 alpha =0.8569; for j =1:2 13 EA=EA1 * F_lT ( lambda,t) + EA2 *(1 - F_lT ( lambda,t)) 1 T =(( alpha *q )/(( EA+EB )* sigma ))^(1/) 15 end 6

7 Algorithm 2: Calulate the temperature in MATLAB - Question 2-1b 1 clear ; clc ; 2 sigma = *10^( -8); % [W/(m^2 * K ^)] 3 q =1350; % [W/m ^2] 5 lambda =2.1; % [mu m] 6 T =100; % [K] Initial Guess for T_s 7 EA1 =0.9; 8 EA2 =0.1; 9 EB =0.6; 10 alpha =0. 6; for j =1:2 13 EA=EA1 * F_lT ( lambda,t) + EA2 *(1 - F_lT ( lambda,t )); 1 T =(( alpha *q )/(( EA+EB )* sigma ))^(1/) 15 end Algorithm 3: Calulate the temperature in MATLAB - Question clear ; clc ; 2 sigma = *10^( -8); % [W/(m^2 * K ^)] 3 q =1350; % [W/m ^2] 5 lambda =1.7; % [mu m] 6 T =300; % [K] Initial Guess for T_s 7 EA1 =0.9; 8 EA2 =0.1; 9 EB =0.6; 10 alpha =0.829; for j =1:3 13 EA=EA1 * F_lT ( lambda,t) + EA2 *(1 - F_lT ( lambda,t )); 1 T =(( alpha *q )/(( EA )* sigma ))^(1/) 15 end Algorithm : Function to calculate the blackbody fraction in MATLAB 1 function F= F_lT ( lambda,t) 2 3 C2 = ; %mu m*k csi =C2 /( lambda *T); 5 6 sum =0; 7 for n =1:30 8 sum = sum + exp (-n* csi )*( csi ^3+3* csi ^2/ n + 6* csi /n^2 + 6/n ^3)/ n; 9 end 10 F= sum *15/ pi ^; 11 end 7