Question 4.2: Consider nonideal emission occurring from real surfaces. (a)...

Consider nonideal emission occurring from real surfaces.

(a) Determine the surface-radiation emission per unit area for a surface at T = 275^{\circ }C made of (i) a highly polished aluminum, (ii) an oxidized aluminum, and (iii) a Pyrex glass. The surface is depicted in Figure. Use the average, total emissivity listed in Table.

(b) Assuming that these surfaces are gray and opaque, determine their surface reflectivity.
(c) Comment on which one of these surfaces is closest to a blackbody surface.

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(a) The surface-radiation emitted per unit area is given by q_{r,\epsilon }\equiv q_{e,\epsilon }\equiv E_{r}(T)\equiv \epsilon _{r}E_{b}(T)\equiv \epsilon _{r}\sigma _{SB}T^{4} , i.e.,

q_{r,\epsilon }= E_{r}(T)= \epsilon _{r}E_{b}(T)\equiv \epsilon _{r}\sigma _{SB}T^{4}

where E_{r} is total emissive power

From Table, we have (using linear interpolation)
(i)  \epsilon _{r}(highly polished aluminum) = 0.04157 Table
(ii) \epsilon _{r}(aluminum oxide) = 0.63 Table
(iii) \epsilon _{r}(Pyrex glass) = 0.9446 Table
Then the surface-radiation emitted per unit area is

(i) highly polished aluminum:
q_{r,\epsilon} = 0.04157 × 5.67 × 10^{−8}(W/m^{2}-K^{4})(275 + 273.15)^{4}(K)^{4}= 2.128 × 10^{2} W/m^{2}

(ii) aluminum oxide:
q_{r,\epsilon} = 0.63 × 5.67 × 10^{−8} (W/m^{2} -K^{4} )(250 + 273.15)^{4} (K)^{4} = 3.225 × 10^{3} W/m^{2}
(iii) Pyrex glass:
q_{r,\epsilon} = 0.9446 × 5.67 × 10^{−8} (W/m^{2} -K^{4} )(250 + 273.15)^{4} (K)^{4} = 4.835 × 10^{3} W/m^{2}

(b) For gray, opaque surfaces, the total reflectivity is given by \alpha _{r}+\rho _{r}=\epsilon _{r}+\rho _{r}=1        or       \rho _{r}=1-\alpha _{r}=1-\epsilon _{r} as

\rho _{r}=1-\epsilon _{r}          gray, opaque surface.

Then
(i) highly polished aluminum:   \rho _{r} = 1 − 0.04157 = 0.9584
(ii) oxidized aluminum: \rho _{r} = 1 − 0.63 = 0.37
(iii) Pyrex glass:   \rho _{r} = 1 − 0.9446 = 0.05540.
(c) For a blackbody surface, \epsilon _{r} = 1  . Then from \alpha _{r}+\rho _{r}=\epsilon _{r}+\rho _{r}=1        or       \rho _{r}=1-\alpha _{r}=1-\epsilon _{r}

\alpha _{r} =\epsilon _{r} = 1      blackbody surface
\rho _{r}= 1 −\epsilon _{r} = 0      blackbody surface.
Thus the Pyrex glass is the closest to a blackbody surface, among the three surfaces.

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