Determine (a) the band fraction of emitted radiation energy, and (b) the blackbody radiation energy emitted, which are in the visible range (0.39 ≤ λ ≤ 0.77μm), when a thin tungsten wire (i.e., a filament) is heated to (i) T = 500 K (ii) T = 1,500 K, (iii) T = 2,500 K, and (iv) T = 3,500 K. (Thea

Question

Determine (a) the band fraction of emitted radiation energy, and (b) the blackbody radiation energy emitted, which are in the visible range [latex](0.39 ≤ \lambda ≤ 0.77 \mu m)[/latex], when a thin tungsten wire (i.e., a filament) is heated to (i) T = 500 K, (ii) T = 1,500 K, (iii) T = 2,500 K, and (iv) T = 3,500 K. (The melting temperature of tungsten is [latex]T_{sl} = 3,698 K [/latex].)

Accepted Answer

(a) The fraction of emitted radiation energy in a band [latex]\lambda_{1}T[/latex] to [latex]\lambda_{2}T[/latex] is found from the fraction in band [latex]0-\lambda_{1}T[/latex] and [latex]0-\lambda_{2}T[/latex] from [latex]F_{\lambda _{1}T-\lambda _{2}T}=\frac{\int_{\lambda _{1}}^{\lambda _{2}}{E_{b,\lambda }(T,\lambda )d\lambda } }{\sigma _{SB}T^{4}}= F_{0-\lambda _{2}T}-F_{0-\lambda _{1}T} [/latex], i.e.,

[latex]F_{\lambda _{1}T-\lambda _{2}T}= F_{0-\lambda _{2}T}-F_{0-\lambda _{1}T}[/latex]

(b) The energy emitted is then found from [latex]F_{\lambda _{1}T-\lambda _{2}T}=\frac{\int_{\lambda _{1}}^{\lambda _{2}}{E_{b,\lambda }(T,\lambda )d\lambda } }{\sigma _{SB}T^{4}}= F_{0-\lambda _{2}T}-F_{0-\lambda _{1}T} [/latex], i.e.,

[latex]\int_{\lambda _{1}T}^{\lambda _{2}T}{E_{b,\lambda }(T,\lambda )d\lambda } =F_{\lambda _{1}T-\lambda _{2}T}\sigma _{SB}T^{4}[/latex]

(i) For T = 500 K, we have

[latex] \lambda _{1}T=0.39(\mu m) imes 500(k)=195\mu m-K [/latex]

[latex]\lambda_{2}T = 0.77(\mu m) × 500(K) = 385 \mu m-K[/latex]

[latex]F_{0-\lambda_{1}T} \simeq 0, F_{0-\lambda_{2}T} \simeq 0[/latex] Table
[latex]F_{\lambda_{1}T -\lambda_{2}T} = F_{0-\lambda_{2}T} − F_{0-\lambda_{1}T} = 0 − 0 = 0,[/latex]

[latex] \int_{\lambda _{1}T}^{\lambda _{2}T}{E_{b,\lambda }d\lambda } =(F_{\lambda _{1}T-\lambda _{2}T})\sigma _{SB}T^{4}=0 imes 5.67 imes 10^{-8}(W/m^{2}-K^{4}) imes 500^{4}(K^{4})=0W/m^{2} [/latex]

[latex]F_{0-\lambda T}(\lambda T)=\frac{15}{\pi ^{4}}\sum\limits_{i=1}^{4}{\frac{e^{-ix}}{i} }\left(x^{3}+\frac{3x^{2}}{i}+\frac{6x}{i^{2}}+\frac{6}{i^{3}} ight)[/latex] curve fit for band fraction of total blackbody emissive power,

where [latex]x = 14,338 (\mu m-K)/\lambda T [/latex]

(ii) For T = 1,500 K, we have

[latex] \lambda _{1}T=0.39(\mu m) imes 1,500(k)=585\mu m-K [/latex]

[latex]\lambda_{2}T = 0.77(\mu m) × 1,500(K) = 1,155 \mu m-K[/latex]

[latex]F_{0-\lambda_{1}T} \simeq 0 , F_{0-\lambda_{2}T} = 0.00321 [/latex] Table
[latex]F_{\lambda_{1}T -\lambda_{2}T} = F_{0-\lambda_{2}T} − F_{0-\lambda_{1}T} = 0.00321 − 0 = 0.00321,[/latex]

[latex]\int_{\lambda _{1}T}^{\lambda _{2}T}{E_{b,\lambda }d\lambda } =0.00321 imes 5.67 imes 10^{-8}(W/m^{2}-K^{4}) imes 1,500^{4}(K^{4})=9.214 imes 10^{2}W/m^{2}[/latex]

(iii) For T = 2,500 K, we have

[latex] \lambda _{1}T=0.39(\mu m) imes 2,500(k)=975\mu m-K [/latex]

[latex]\lambda_{2}T = 0.77(\mu m) × 2,500(K) = 1,925 \mu m-K[/latex]

[latex]F_{0-\lambda_{1}T} = 0.000312 , F_{0-\lambda_{2}T} = 0.0565 [/latex] Table
[latex]F_{\lambda_{1}T -\lambda_{2}T} = F_{0-\lambda_{2}T} − F_{0-\lambda_{1}T} = 0.0562,[/latex]

[latex]\int_{\lambda _{1}T}^{\lambda _{2}T}{E_{b,\lambda }d\lambda } =0.0561 imes 5.67 imes 10^{-8}(W/m^{2}-K^{4}) imes 2,500^{4}(K^{4})=1.251 imes 10^{5}W/m^{2}[/latex]

(iv) For T = 3,500 K, we have

[latex] \lambda _{1}T=0.39(\mu m) imes 3,500(k)=1,365\mu m-K [/latex]

[latex]\lambda_{2}T = 0.77(\mu m) × 3,500(K) = 2,695 \mu m-K[/latex]

[latex]F_{0-\lambda_{1}T} = 0.00714 , F_{0-\lambda_{2}T} = 0.204 [/latex] Table
[latex]F_{\lambda_{1}T -\lambda_{2}T} = F_{0-\lambda_{2}T} − F_{0-\lambda_{1}T} = 0.197,[/latex]

[latex]\int_{\lambda _{1}T}^{\lambda _{2}T}{E_{b,\lambda }d\lambda } =0.197 imes 5.67 imes 10^{-8}(W/m^{2}-K^{4}) imes 3,500^{4}(K^{4})=1.676 imes 10^{8}W/m^{2}[/latex]

Question 4.1: Determine (a) the band fraction of emitted radiation energy,...

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