Question 4.7: In an electrically heated, open radiation oven, shown in Fig...

(a) This is a three-surface enclosure with surface 3 being the two ends with $\epsilon _{r,3} = 1$ (i.e., an imaginary blackbody surface is used to complete the enclosure). The heat generation $\dot{S} _{e,J}$ leaves surface 2 by radiation, i.e., the outer surface of the ceramic is ideally insulated, $Q_{2} = 0$, and there is no energy stored in the ceramic (due to the steady-state assumption). The thermal circuit diagram is shown in Figure (b).

(b) There are seven view factors (because $F_{2-2} \neq 0$). Using the reciprocity rule $A_{r,i}F_{i-j}=A_{r,j}F_{j-i}$, these are reduced to four. Using the summation rule $\sum\limits_{j=1}^{n}{F_{i-j}} =1$, these are yet reduced to two. These two are $F_{2-2}$ and $F_{2-1}$. The relations for these are given in Figures (d) and (e) and are

$-\frac{l^{\ast }}{2\pi R^{\ast }}\left\{\frac{(4 R^{\ast 2}+l^{\ast 2})^{1/2}}{l^{\ast }}sin^{-1}\frac{4( R^{\ast 2}-1)+(l^{\ast 2}/ R^{\ast 2})( R^{\ast 2}-2)}{l^{\ast 2}+4( R^{\ast }-1)}-sin^{-1}\frac{R^{\ast 2}-2}{R^{\ast 2}} +\frac{\pi }{2}\left[\frac{(4R^{\ast 2}+l^{\ast })^{1/2}}{l^{\ast }} -1\right] \right\}$    Figure (d)

$F_{2-1}=F_{2-1}(R^{\ast },l^{\ast })=\frac{1}{R^{\ast }} -\frac{1}{\pi R^{\ast }}\left\{cos^{-1}\frac{b}{a}-\frac{1}{2l^{\ast }}[(a+2)^{2}-(2R^{\ast })^{2}]^{1/2}cos^{-1}\frac{b}{aR^{\ast }}+b \sin^{-1}\frac{1}{R^{\ast }}-\frac{\pi a}{2} \right\}$      Figure (e)

where

$F_{2-3} = 1 − F_{2-2} − F_{2-1}          F_{1-3} = 1 − F_{1-2}$           summation rule

$F_{1-2}=\frac{A_{r,2}}{A_{r,1}} F_{2-1} ,                     F_{3-2}=\frac{A_{r,2}}{A_{r,3}} F_{2-3} ,                  F_{3-1}=\frac{A_{r,1}}{A_{r,3}} F_{1-3}$        reciprocity rule

Using the numerical values, we have

a = 1, 624,     b = 1,576

= 0.2 − 0.06366(0.2437 − 0.0125 × 4.152) = 0.1867

(c) Now, since $Q_{2} = 0$ and $\dot{S} _{1} = \dot{S}_{3} = 0, -Q_{1}+\dot{S}_{1}=Q_{r,1}=\frac{(q_{r,o})_{1}-(q_{r,o})_{2}}{\frac{1}{A_{r,1}F_{1-2}} } +\frac{(q_{r,o})_{1}-(q_{r,o})_{3}}{\frac{1}{A_{r,1}F_{1-3}} }$ through $-Q_{3}+\dot{S}_{3}=Q_{r,3}=\frac{E_{b,3}(T_{3})-(q_{r,o})_{3}}{\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}}\right) _{3}}       ,E_{b,3}(T_{3})=\sigma _{SB}T_{3}^{4}$ become

Note that surface 3 is blackbody and from $-Q_{3}+\dot{S}_{3}=Q_{r,3}=\frac{E_{b,3}(T_{3})-(q_{r,o})_{3}}{\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}}\right) _{3}}       ,E_{b,3}(T_{3})=\sigma _{SB}T_{3}^{4}$ with $(R_{r,\epsilon })_{3} = 0$, we have

and have used this above for $(q_{r,o})_{3}$ .

We cannot simply solve for one unknown at a time by successive substitutions. The simultaneous solution requires iterations or use of a solver, for example MATLAB.
We now solve for the six unknowns $(q_{r,o})_{1}, (q_{r,o})_{2}, (q_{r,o})_{3}, T_{1}, Q_{2}$, and $Q_{3}$ and the numerical results are

$(q_{r,o})_{1} = 1.789 × 10^{4} W/m^{2}$
$(q_{r,o})_{2} = 1.327 × 10^{5} W/m^{2}$
$(q_{r,o})_{3} = 4.179 × 10^{2} W/m^{2}$
$T_{2} = 983.9 K$
$Q_{1} = 5.994 × 10^{3} W$
$Q_{3} = 4.006 × 10^{3} W$.