Consider two gray, diffuse, and opaque surfaces exchanging radiation heat and being surrounded by a far-away surface, as depicted in Figure. The two surfaces are at T_{1} and T_{2} , while the surrounding surface is at T_{∞} .
(a) Describe an imaginary surface 3 that can be used to complete the enclosure.
(b) Write the energy equations -Q_{i}+\dot{S}_{i}=Q_{r,i}=\frac{E_{b,i}(T_{i})-(q_{r,o})_{i}}{\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}}\right)_{i} } ,E_{b,i}(T_{i})=\sigma _{SB}T_{i}^{4} and \frac{E_{b,i}(T_{i})-(q_{r,o})_{i}}{\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}} \right)_{i} } -\sum\limits_{i=1}^{n}{\frac{(q_{r,o})_{i}-(q_{r,o})_{j}}{\frac{1}{A_{r,i}F_{i-j}} } } =Q_{r,i} for the resulting three-surface enclosure.
There are no surface energy conversions (S˙i = 0) .
Question 4.4: Consider two gray, diffuse, and opaque surfaces exchanging r...
(a) The far-away surface or unbounded volume can be represented by an imaginary surface. This surface is shown by the dotted lines in Figure. These imaginary surfaces complete the enclosure, i.e., a three-surface enclosure is formed. These surfaces are considered as blackbody surfaces, because they
absorb all the radiation that comes from the enclosure surfaces. Therefore, the emissivity of the surface, i.e., surface 3 in Figure , is \epsilon _{r,3}=1 . The temperature of this surface is that of the surrounding, i.e., T_{3} = T_{\infty}. Note that surface 3 may be made of several segments (here two segments are shown). However, it is shown as one collective surface.
(b) Since \epsilon _{r,3}=1 , then -Q_{i}+\dot{S}_{i}=Q_{r,i}=\frac{E_{b,i}(T_{i})-(q_{r,o})_{i}}{\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}}\right)_{i} } ,E_{b,i}(T_{i})=\sigma _{SB}T_{i}^{4} gives
E_{b,3}-(q_{r,o})_{3}=-Q_{3} \left(\frac{1-\epsilon }{A_{r}\epsilon _{r}}\right) _{3}=0or
E_{b,3}=(q_{r,o})_{3} for a blackbody surface with equal emissive power and radiosity.
Here we assume planar surfaces such that F_{i-i} = 0 .
Using this, we will replace (q_{r,o})_{3} by E_{b,3} in the energy equations -Q_{i}+\dot{S}_{i}=Q_{r,i}=\sum\limits_{j=1}^{n}{\frac{(q_{r,o})_{i}-(q_{r,o})_{j}}{\frac{1}{A_{r,i}F_{i-j}} } } =Q_{r,i}=\sum\limits_{j=1}^{n}{\frac{(q_{r,o})_{i}-(q_{r,o})_{j}}{(R_{r,F})_{i-j}} } and -Q_{i}+\dot{S}_{i}=Q_{r,i}=\frac{E_{b,i}(T_{i})-(q_{r,o})_{i}}{\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}}\right)_{i} } ,E_{b,i}(T_{i})=\sigma _{SB}T_{i}^{4} and we have
-Q_{1}=Q_{r,1}=\frac{(q_{r,o})_{1}-(q_{r,o})_{2}}{\frac{1}{A_{1}F_{1-2}} } +\frac{(q_{r,o})_{1}-E_{b,3}}{\frac{1}{A_{1}F_{1-3}} }-Q_{2}=Q_{r,2}=\frac{(q_{r,o})_{2}-(q_{r,o})_{1}}{\frac{1}{A_{2}F_{2-1}} } +\frac{(q_{r,o})_{2}-E_{b,3}}{\frac{1}{A_{2}F_{2-3}} }
-Q_{3}=Q_{r,3}=\frac{E_{b,3}-(q_{r,o})_{1}}{\frac{1}{A_{3}F_{3-1}} } +\frac{E_{b.3}-(q_{r,o})_{2}}{\frac{1}{A_{3}F_{3-2}} }
-Q_{1}=\frac{E_{b,1}-(q_{r,o})_{1}}{\left(\frac{1-\epsilon _{r}}{A\epsilon _{r}}\right)_{1} }
-Q_{2}=\frac{E_{b,2}-(q_{r,o})_{2}}{\left(\frac{1-\epsilon _{r}}{A\epsilon _{r}}\right)_{2} }
where A_{r}’s, \epsilon _{r}’s, and F_{i-j} ’s are known and the geometry and materials are specified.