A diffuse surface is characterized by the spectral...

A diffuse surface is characterized by the spectral hemispherical emissivity distribution shown in the figure below, where $epsilon_1 = 0.75$, $epsilon_2 = 0.10$, and $epsilon_3 = 0.2$. For a surface temperature of $T_s = 300$ K, what is the emissive power for the surface?

Transcript

00:01 The given data we have our emissivity epsilon 1 equals 0 .75 epsilon 2 equals 0 .10 and epsilon 3 equals 0 .2 the temperature of the surface is given to be 300 kelvin and we are asked to calculate the emissive power of surface to calculate the emissive power we have a formula e equals emissivity of the formula equals emissivity times boltzmann -stefan constant and temperature raised to the power surface temperature raised to the power 4.

00:48 Now to calculate the total emissivity, we need to calculate the fraction of radiation corresponding to each range of wavelengths.

00:58 We have the wavelength range from lambda equals 0 micrometer to lambda equals 4 micrometer or micrometer then from lambda equals 4 micrometer to lambda equals 6 micrometer and from lambda equals 6 micrometer to lambda equals 10 micrometer to calculate the emissive power total emissive power we need to first calculate the fraction of radiation corresponding to each range of wavelengths so first we will we will calculate the product of wavelength and temperature so for lambda equals 4 we have lambda temperature product equals 4 micrometer multiplied by temperature surface that is 300 kelvin so it will be 1200 micrometer kelvin the corresponding fraction of radiation will be f 0 to 4 will be 0 point 0 0 to 1 3 g4 then for lambda equals 6 we have the temperature wavelength product equals 6 micrometer times the temperature of the surface that is rendered kelvin so we have 1800 micrometer kelvin the corresponding fraction will be f 0 to 6 we will have 0 .039341 so the fraction we have got the fraction range from f 0 to 4 equals 0 .002134 now to calculate the fraction for f 4 to 6 we will subtract f 0 to 6 minus f 0 to 4 so we will get 0 .039341 minus 0 .000002134 so we will get the value 0 .037207 and for the fraction range for f6 to f10 we will just subtract the value of f026 from 1 so we will get 1 minus 0 .039341...

Question

A diffuse surface is characterized by the spectral hemispherical emissivity distribution shown in the figure below, where $epsilon_1 = 0.75$, $epsilon_2 = 0.10$, and $epsilon_3 = 0.2$. For a surface temperature of $T_s = 300$ K, what is the emissive power for the surface?

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