Federal University of Rio Grande do Sul Mechanical Engineering Department MEC0065 - 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 = 0.06673 (1) F 0 λ2 T = 0.8087 (2) α = 0.75 F 0 λ1 T + 0.5 (F 0 λ2 T F 0 λ1 T) (3) α = 0.75 0.06673 + 0.5 (0.8087 0.06673) () α = 0.25712 (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 = 0.25712. (6) λ = 1.2 µm = λt = 7200 µmk (b) For a surface with a temperature of 600 K. F 0 λt = 0.81919 (7) α = 0.95 F 0 λt + 0.15 (1 F 0 λt ) (8) α = 0.95 0.81919 + 0.15 (1 0.81919) (9) α = 0.8053 (10) λ = 1.2 µm = λt = 720 µmk F 0 λt = 1.8 10 6 (11) 1
Thus, the total absorptivity is: α = 0.95 F 0 λt + 0.15 (1 F 0 λt ) (12) α = 0.95 1.8 10 6 + 0.15 (1 1.8 10 6 ) (13) α = 0.15. (1) Therefore, by Kirchhoff s Law (α = ε), we have the total emissivity: Question 2-12: The surface is diffuse. ε = 0.15. (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 = 0.22789 (16) F 0 λ2 T = 0.70102 (17) ε = 0.6 F 0 λ1 T + 0.35 (F 0 λ2 T F 0 λ1 T) + 0.15 (1 F 0 λ2 T) (18) ε = 0.6 0.22789 + 0.35 (0.70102 0.22789) + 0.15 (1 0.70102) (19) ε = 0.372. (20) Therefore, by Kirchhoff s Law (α = ε), we have the total absorptivity: (b) We can calculate α, computing the following ratio: α = 0.372. (21) α = ε ε source (22) α = 0.372 0.822 (23) α = 0.22. (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 = 12138 µmk = F 0 λtsun = 0.9618. 2
Thus, as we know ε = α, we have α = 0.9 F 0 λtsun + 0.1 (1 F 0 λtsun ) (25) α = 0.9 0.9618 + 0.1 (1 0.9618) (26) α = 0.857. (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 = 13.1758 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 = 377.9560 K. (30) (c) We can calculate the hemispherical total reflectivity for diffuse surface noting that: From the item (a), we have α = 0.857. Thus, ρ + α = 1 (31) ρ = 1 α (32) ρ = 1 0.857 (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 = 0.91101. α = 0.9 F 0 λtsun + 0.1 (1 F 0 λtsun ) (35) α = 0.9 0.91101 + 0.1 (1 0.91101) (36) α = 0.829. (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 = 66.9338 K (39) 3
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/6 0 0 2π π/2 0.9 cos(θ) sin(θ) dθdϕ (1) + 1 0.5 cos(θ) sin(θ) dθdϕ (2) π 0 π/6 ε = 1 [ 2π 2π ] 0.9 0.125 dϕ + 0.5 0.375 dϕ (3) π 0 0 ε = 0.225 + 0.375 () 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 = 0.6 5.67051 10 8 800 [ 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 = 0.73779. (51) Thus, the flux absorved is given by: q a = [0.95F 0 λtsource + 0.05(1 F 0 λtsource )]q i (52) q a = [0.95 0.73779 + 0.05(1 0.73779)]q i (53) q a = 0.712q i (5)
For other side, with T eq = 1000 K, we have Thus, the flux emitted is given by: λt eq = 1500µmK = F 0 λteq = 0.01285. (55) q e = ε(t eq )σt eq (56) q e = [0.95F 0 λteq + 0.05(1 F 0 λteq )]σteq (57) q e = [0.95 0.01285 + 0.05(1 0.01285)] 5.67051 10 8 1000 (58) q e = 391.05 (59) Finally, equaling (5) and (59), since we have a gray-body source, follows q a = q e (60) 0.712q i = 391.05 (61) q i = 888, 06 W m 2 (62) Porto Alegre - October 16, 2013 5
Appendix Definição Símbolo e valor Constant in Planck s spectral energy (or intensity) distribuition C 1 = 0.59552197 10 8 [W µm /(m 2 sr)] Constant in Planck s spectral energy (or intensity) distribuition C 2 = 1387.69[µm K] Constant in Wien s displacement law C 3 = 2897.756[µm K] Constant in equation for max. blackbody intensity C =.095790 10 12 [W/(m 2 µm K 5 sr)] Stefan-Boltzman Constant σ = 5.67051 10 8 [W/(m 2 K )] Speed of the Light c = 2.9979258 10 8 m/s Table 1: Radiation Constants Wavelength Temperature Blackbody product λt Fraction µm K F 0 λt 700 1.8E-6 1500 0.01285 2000 0.06673 2800 0.22789 000 0.8087 5600 0.70102 6000 0.73779 7200 0.81919 9850 0.91101 12100 0.9618 Table 2: Blackbody Function from the Siegel s Book Algorithm 1: Calulate the temperature in MATLAB - Question 2-1a 1 clear ; clc ; 2 sigma =5.67051*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; 11 12 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
Algorithm 2: Calulate the temperature in MATLAB - Question 2-1b 1 clear ; clc ; 2 sigma =5.67051*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; 11 12 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 2-16 1 clear ; clc ; 2 sigma =5.67051*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; 11 12 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 =1387.69; %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