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Calculate the following for an industrial furnance in the form of a black body and emitting radiation of 2500°C : 

(i) Monochromatic emissive power at 1.2 µm length, 

(ii) Wavelength at which the emission is maximum, 

(iii) Maximum emissive power, 

(iv) Total emissive power, and 

(v) Total emissive power of the furnance if it is assumed as a real surface with emissivity equal to 0

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Given : T = 2500 + 273 = 2773 K ; λ = 12 µm, ε = 0.9 

(i) Monochromatic emissive power at 1.2 µm length, (Eλ)b :

According to Planck’s law,

where, C1 = 3.742 × 108 W. µm4/m2 = 0.3742 × 10–15 W.m4/m2 and 

C2 = 1.4388 × 10–2 mK 

Substituting the values, we get

(ii) Wavelength at which the emission is maximum, λmax

According to Wien’s displacement law,

(iii) Maximum emissive power, (Eλb)max

(Eλb)max = 1.285 × 10–5 T5 W/m2 per metre length 

= 1.285 × 10–5 × (2773)5 = 2.1 × 1012 W/m2 per metre length. 

[Note. At high temperature the difference between (Eλ)b and (Eλb)max is very small]. 

(iv) Total emissive power, Eb :

Eb = σT4 = 5.67 × 10–8 (2273)4 = 5.67 \(\left(\cfrac{2773}{100}\right)^4\) = 3.352 × 106 W/m2

(v) Total emissive power, E with emisivity (ε) = 0.9

E = ε σT4 = 0.9 × 5.67 \(\left(\cfrac{2773}{100}\right)^4\) = 3.017 × 106 W/m2.

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