The spectral emissive power E1 for a diffusely emitting surface is E1 = 0 for λ < 3 μm; E1 = 150 W/m2-μm for 3 < λ < 12 μm; E1 = 300 W/m2-μm for 12 < λ < 25 μm; E1 = 0 for λ > 25 μm. The total emissive power of the surface over the entire spectrum is
5250 W/m2
Understanding Total Emissive Power Calculation
This problem requires calculating the total emissive power of a surface, given its spectral emissive power ($E_1$) across different wavelength ranges. The total emissive power is the sum of emissive power over all wavelengths.
Spectral Emissive Power Breakdown
The spectral emissive power ($E_1$) is provided in distinct intervals:
- For wavelengths less than 3 $\mu$m ($\lambda < 3 \, \mu m$): $E_1 = 0 \, W/m^2-\mu m$
- For wavelengths between 3 $\mu$m and 12 $\mu$m ($3 \, \mu m < \lambda < 12 \, \mu m$): $E_1 = 150 \, W/m^2-\mu m$
- For wavelengths between 12 $\mu$m and 25 $\mu$m ($12 \, \mu m < \lambda < 25 \, \mu m$): $E_1 = 300 \, W/m^2-\mu m$
- For wavelengths greater than 25 $\mu$m ($\lambda > 25 \, \mu m$): $E_1 = 0 \, W/m^2-\mu m$
Calculating Total Emissive Power
The total emissive power ($E$) is found by integrating the spectral emissive power ($E_1$) over the entire range of wavelengths. Since the spectral emissive power is zero outside the ranges 3 $\mu$m to 12 $\mu$m and 12 $\mu$m to 25 $\mu$m, we only need to consider these intervals.
The formula for total emissive power is:
$$ E = \int_{0}^{\infty} E_1(\lambda) d\lambda $$Breaking this down into the given intervals:
$$ E = \int_{0}^{3} E_1(\lambda) d\lambda + \int_{3}^{12} E_1(\lambda) d\lambda + \int_{12}^{25} E_1(\lambda) d\lambda + \int_{25}^{\infty} E_1(\lambda) d\lambda $$Substituting the given values:
$$ E = \int_{0}^{3} (0) d\lambda + \int_{3}^{12} (150 \, W/m^2-\mu m) d\lambda + \int_{12}^{25} (300 \, W/m^2-\mu m) d\lambda + \int_{25}^{\infty} (0) d\lambda $$The integrals over the ranges where $E_1 = 0$ are zero. We calculate the remaining integrals:
Contribution from 3 $\mu$m to 12 $\mu$m:
$$ \int_{3}^{12} 150 \, d\lambda = 150 \, W/m^2-\mu m \times (12 \, \mu m - 3 \, \mu m) $$ $$ = 150 \, W/m^2-\mu m \times 9 \, \mu m $$ $$ = 1350 \, W/m^2 $$Contribution from 12 $\mu$m to 25 $\mu$m:
$$ \int_{12}^{25} 300 \, d\lambda = 300 \, W/m^2-\mu m \times (25 \, \mu m - 12 \, \mu m) $$ $$ = 300 \, W/m^2-\mu m \times 13 \, \mu m $$ $$ = 3900 \, W/m^2 $$Adding the contributions from both intervals gives the total emissive power:
$$ E = 1350 \, W/m^2 + 3900 \, W/m^2 $$ $$ E = 5250 \, W/m^2 $$Final Result
The total emissive power of the surface over the entire spectrum is 5250 W/m2.
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