(a) The thermal circuit diagram is shown in Figure (b). From this diagram, for the solid ceramic layer (SiC) and for the radiating surface, we have the energy equation
Q_{k,H-1}=\frac{T_{H}-T_{1}}{R_{k,H-1}} =a_{1}\dot{S}_{r,c} ceramic-layer conduction resistance
Q_{r,1} = Q_{r,1-2} + Q_{r,1-\infty } = a_{1}\dot{S}_{r,c} radiating surface energy equation.
(b) To determine T_{1}, we need to write Q_{r,1-2} and Q_{r,1-\infty } in terms of the temperatures. Because we have all blackbody surfaces, all the surface grayness resistances vanish. So, all the surface radiosities and emissive powers are equal [this is determined from -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) } ,E_{b,i}(T_{i})=\sigma _{SB}T_{i}^{4}]. Then we use -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}}} for surface 1, and replace the (q_{r,o})’s with E_{b}’s. The result is
Q_{r,1} = Q_{r,1-2} + Q_{r,1-\infty } =\frac{E_{b,1}-E_{b,2}}{\frac{1}{A_{r,1}F_{1-2}} } +\frac{E_{b,1}-E_{b,\infty }}{\frac{1}{A_{r,1}F_{1-\infty }} }
Then the radiating surface energy equation becomes
A_{r,1}\sigma _{SB}[F_{1-2}(T^{ 4}_{ 1} − T^{ 4}_{ 2} ) + F_{1-\infty }(T^ {4}_{ 1} − T^ {4} _{\infty })] = a_{1}\dot{S}_{r,c}.
Solving for T_{1}, we have
T_{1}=\left[\frac{a_{1}\dot{S}_{r,c}+A_{r,1}\sigma _{SB}(F_{1-2}T_{2}^{4}+F_{1-\infty }T_{\infty }^{4})}{A_{r,1}\sigma _{SB}(F_{1-2}+F_{1-\infty })} \right]^{1⁄4}
The view factors F_{1-2} and F_{1-\infty} are related through the summation rule </p>
<p>(q_{r,o})_{i}=1\times (q_{r,o})_{i}=\sum\limits_{j=1}^{n}{F_{i-j}(q_{r,o})_{i}} , i.e.,
F_{1-2} + F_{1-\infty} = 1, or F_{1-\infty} = 1 − F_{1-2}.
The view factor F_{1-2} is determined from Figure (b). The input parameters are
a^{\ast }=\frac{a}{l} and w^{\ast }=\frac{w}{l}
The numerical values are
a^{\ast }=\frac{30(a)}{5(a)}=6
w^{\ast }=\frac{30(a)}{5(a)}=6
F_{1-2} \simeq 0.76 Figure (b)
F_{1-∞} = 1 − 0.76 = 0.24 .
Then
T_{1}=\left\{\frac{10^{4} (W) + 0.3(m) × 0.3(m) × 5.67 × 10^{−8} (W/m^{2}-K^{4})}{0.3(m) × 0.3(m) × 5.67 × 10^{−8} (W/m^{2}-K^{4})×(0.76+0.24)}×\frac{[0.76 × (773.15)^{4} (K^{4} ) + 0.24 × (637.15)^{2} (K^{4} )]}{1} \right\} ^{1/4}
=\left[\frac{10^{4} + 5.103 × 10^{-9} (2.716 × 10^{11} + 4.928 × 10^{10})}{5.103\times 10^{-9}} \right] ^{1/4}
= 1,229 K.
(c) From the energy equation, we have
Q_{r,1-2}=\frac{E_{b,1}-E_{b,2}}{\frac{1}{A_{r,1}F_{1-2}} } = A_{r,1}F_{1-2}(E_{b,1} − E_b,2)
= A_{r,1}F_{1-2}\sigma _{SB}(T^{ 4}_{ 1} − T^{ 4}_{2} )
= 0.3(m) × 0.3(m) × 0.76 × 5.67 × 10^{−8} (W/m^{2}-K^{4}) × [(1,229)^{4} (K^{4}) − (773.15)^{4} (K^{4})]
= 7.459 × 10^{3} W.
(d) From above
Q_{r,1-\infty} = a_{1}\dot{S}_{r,c} − Q_{r,1-2}
= 10^{4} (W) − 7.459 × 10^{3} (W) = 2.541 × 10^{3} W.
(e) From the conduction resistance in the ceramic layer, given above, we have
\frac{T_{H}-T_{1}}{R_{k,H-1}} =a_{1}\dot{S}_{r,c}
From Table , for the planar slab, we have
R_{k,H-1}=\frac{l_{1}}{A_{k}k_{1}} Table
Solving for T_{H} ,
T_{H} = T_{1} + \frac{(a_{1}\dot{S}_{r,c})l_{1}}{A_{k}k_{1}}
where A_{k} = A_{r,1} . Using the numerical values, we have
T_{H} = 1,229 K + \frac{10^{4} (W) × 3 × 10^{-3} (m)}{0.3(m) × 0.3(m) × 40(W/m-K) } = 1,237 K.
