Respuesta :
Answer:
a) 2000 W/m² ; b) 636.94 W/m².sr ; c) 0.5
Explanation:
a)
The formula for calculation of total emissive power is:
Total emissive power = E = [tex]\int\limits^\alpha_0[/tex] E'λdλ
= [tex]\int\limits^a_0[/tex](0)dλ + [tex]\int\limits^b_a[/tex](100)dλ + [tex]\int\limits^c_b[/tex](200)dλ + [tex]\int\limits^d_c[/tex](100)dλ [tex]\int\limits^e_d[/tex](0)dλ
where a = 5; b = 10; c = 15; d = 20; e = 25
= 0 +100(10-5) + 200(15-10) +100(20-15) + 0
= 2000 W/m²
b)
The formula for total intensity of radiation is:
I[tex]_{e}[/tex] = E/π = 200/3.14 = 636.94 W/m².sr
c)
Fo submissive power leaving the surface in range π/4 ≤θ≤π/2
[E(π/4 ≤θ≤π/2)]/E = [tex]\int\limits^f_0[/tex][tex]\int\limits^g_0[/tex][tex]\int\limits^i_h[/tex] Icosθsinθ dθdΦdλ
where f = infinity, g=2π, h=π/4, i=π/2
By simplifying, we get
= (-1/2)[cos(2π/2)-cos(2π/2)]
= -0.5(-1-0)
=0.5
The total emissive power is 2000 W/m². The total intensity of the radiation is 636.62 W/m². The fraction of the emissive power leaving the surface is 0.500
The emissive power is described as the entire amount of energy in a body from which a fraction of radiation is emitted as compared to a black body at the same temperature.
The total emissive power can be estimated by using the integral formula;
[tex]\mathbf{E = \int ^{\infty}_{o} \ E_{\lambda }(\lambda) \ \ d \lambda }[/tex]
where;
- [tex]\mathbf{\ E_{\lambda }(\lambda) }[/tex] the energy emitted by the surface for the wavelength
- dλ = change in wavelength
[tex]\mathbf{E = \int ^{\lambda_2}_{\lambda_1}E_{\lambda1}(\lambda) \ d\lambda+ \int ^{\lambda_3}_{\lambda_2}E_{\lambda2}(\lambda) \ d\lambda+ \int ^{\lambda_4}_{\lambda_3}E_{\lambda3}(\lambda) \ d\lambda}[/tex]
where;
- [tex]\mathbf{E_{\lambda 1,2,3}}[/tex] = emissive power for wavelength
- λ1,2,3,4 = wavelengths
∴
replacing the values from the diagram, we have:
[tex]\mathbf{E = \int ^{10}_{5}(100) \ d\lambda+ \int ^{15}_{10}(200) \ d\lambda+\int ^{20}_{15}(100) \ d\lambda}[/tex]
[tex]\mathbf{E =\Big [100 \ \lambda \Big ]^{10}_{5}+ \Big [200 \ \lambda \Big ]^{15}_{10}+\Big [100 \ \lambda \Big ]^{20}_{15}}[/tex]
[tex]\mathbf{E =\Big [100 \ \Big ]({10-5)}+ \Big [200 \ \Big ]({15-10)}+\Big [100 \ \Big ]({20-15)}}[/tex]
E = (500 + 1000 + 500) W/m²
E = 2000 W/m²
b.
The total intensity of the radiation emitted is determined by using the formula:
[tex]\mathbf{I_c = \dfrac{E}{\pi}}[/tex]
[tex]\mathbf{I_c = \dfrac{2000 \ W/m^2}{\pi}}[/tex]
[tex]\mathbf{I_c =636.62 \ W/m^2}[/tex]
c.
The fraction of the emissive power leaving the surface is computed by using the formula:
[tex]\mathbf{\dfrac{E(\pi/4-\pi/2)}{E} = \dfrac{\int ^{2\pi}_{0} \ \int^{\pi/2}_{\pi/4 } I_e cos \theta sin \theta d \theta d \phi}{\pi I_e}}[/tex]
[tex]\mathbf{\dfrac{E(\pi/4-\pi/2)}{E} = \dfrac{\int^{\pi/2}_{\pi/4 } I_e cos \theta sin \theta d \theta \int ^{2\pi}_{0} \ d \phi}{\pi I_e}}[/tex]
[tex]\mathbf{\dfrac{E(\pi/4-\pi/2)}{E} = \dfrac{1}{\pi } \Big[\dfrac{sin^2 \theta }{2} \Big]^{\pi/2}_{\pi/4} \phi \BIg|^{2 \pi}_{0}}[/tex]
[tex]\mathbf{\dfrac{E(\pi/4-\pi/2)}{E} = \dfrac{1}{\pi}\Big [ \dfrac{1}{2} (1^2 -0.707^2)(2 \pi -0)\Big ]}[/tex]
[tex]\mathbf{\dfrac{E(\pi/4-\pi/2)}{E} = \dfrac{1}{\pi}\Big [0.2500755\times 6.283185307\Big ]}[/tex]
[tex]\mathbf{\dfrac{E(\pi/4-\pi/2)}{E} = 0.500}[/tex]
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