(a) For two concentric spheres radiation heat exchange is \[Q = \frac{A_{1}\sigma \left ( T_{1}^{4} – T_{2}^{4}\right )}{\frac{1}{\epsilon _{1}}+\left ( \frac{A_{1}}{A_{2}} \right )\left ( \frac{1}{\epsilon _{2}}-1 \right )}\]

Since sphere is black, therefore \[\epsilon _{1} = \epsilon _{2} = 1\]

So, \[Q = A_{1}\sigma \left ( T_{1}^{4}-T_{2}^{4} \right ) = \pi\left ( 0.5 \right )^{2}\left ( 5.67\times10^{-8} \right )\left [ 400^{4} – 300^{4} \right ]=779\,W\]

(b)(i) Net rate of radiation exchange between the surfaces if they are diffuse and gray with \(\epsilon _{1} = 0.5\) and\(\epsilon _{2} = 0.5\), \[Q = \frac{779}{\frac{1}{0.5}+\left ( \frac{0.5}{1} \right )^{2}\left ( \frac{1}{0.5}-1 \right )}=346\,W\]

(b)(ii) Error introduced by assuming blackbody behaviour for the outer surface , having all other conditions same, since \(\epsilon _{2} = 1\)

\[Q = A_{1}\sigma \epsilon _{1}\left ( T_{1}^{4} – T_{2}^{4}\right )=\frac{\pi}{4}\left ( 0.5 \right )^{2}\left ( 5.67\times10^{-8} \right )\left [ 400^{4} – 300^{4} \right ]= 389.5\,W\]

Error induced is,

\[\frac{389.5-346}{346}\times 100 = 12.57\, \%\]

Prandtl Number, \[Pr= \frac{\mu C_{p}}{k} = \frac{25 \times 10^{-6}\times 2 \times 10^{3}}{0.05}=1\]

\[\frac{\delta_{h} }{\delta _{t}} = \left ( Pr \right )^{\frac{1}{3}}=\left ( 1 \right )^{\frac{1}{3}}=1\]

\[\Rightarrow \frac{\delta_{h} }{\delta _{t}} = 1\]

\[\Rightarrow \frac{0.5 }{\delta _{t}} = 1\]

\[\Rightarrow \delta _{t} = 0.5\,mm\]

Time required to cool aluminium to 95°C.

\[T = 95^{\circ}C, T_{0} = 290^{\circ}C, T_{\infty} = 15^{\circ}C\]

\[\frac{T-T_{\infty }}{T_{0}-T_{\infty}}=\textrm{exp}\left [ -\left ( \frac{hA}{\rho cV} \right )t \right ]\]

\[V = \frac{m}{\rho}=\frac{5.5}{2700}=2.037\times10^{-3}m^{3}\]

\[V = \frac{4}{3}\pi R^{3}\Rightarrow R = 0.0786\,m\]

\[\Rightarrow L_{c} = \frac{R}{3} = 0.0262\,m\]

\[\frac{hA}{\rho cV} = \frac{3h}{\rho cR} = \frac{3\times58}{2700\times900\times0.0786}=9.1\times10^{-4}/s\]

\[\frac{95-15}{290-15} = \frac{80}{275}=\textrm{exp}\left ( -9.1\times10^{-4} t\right )\]

\[3.4375 = \textrm{exp}\left ( 9.1\times10t \right )\]

\[\Rightarrow t = 1357\,s\]

Efficiency of fin =

\[\eta =\frac{tan\,h\left ( mL \right )}{mL}\]

\[L=80\times10^{-3}\,m\,=0.080\,m\]

\[m=\sqrt{\frac{hP}{KA}}=\sqrt{\frac{h\times\pi d}{K\frac{\pi d^{2}}{4}}}=\sqrt{\frac{4h}{kd}}=\sqrt{\frac{4\times 55}{32\times 0.010}}=26.22\]

\[\eta = \frac{tan\,h\left ( 26.22\times0.080 \right )}{26.22\times0.080} = 0.4625=46.25\, \%\]

Efficiency of a plate fin,

Efficiency =\( \eta =\frac{tan\,h\left ( mL \right )}{mL}\)

\[L=1.5+\frac{0.2}{2}=1.6\,cm=1.6\times10^{-2}\,m\]

\[m=\sqrt{\frac{2h}{Kt}}=\sqrt{\frac{hP}{KA_{c}}}=\sqrt{\frac{2\times 285}{210\times2\times10^{-3}}}\]

\[mL=0.589\]

\[\eta = \frac{tan\,h\left ( 0.589 \right )}{0.589} = 0.899\,=\,89.9\, \%\]

Heat transfer coefficcient, h = \[\frac{T_{0}-T_{\infty}}{T-T_{\infty}}=e^{mx}\]

\[\Rightarrow e^{m\left ( 0.40 \right )} = \frac{500-27}{50-27}\]

\[\Rightarrow m = 7.55\]

\[\Rightarrow \sqrt{\frac{hP}{kA}}=7.55\]

\[\Rightarrow \frac{h \times \left [ \left ( 0.010 + 0.020 \right )\times2 \right ]}{205\times\left ( 0.010\times 0.020 \right )}=57.13\]
\[h = 39.03\,W/m^{2\circ}C\]

Diameter of fin is 5mm and length is 100 mm, \[m=\sqrt{\frac{hP}{KA}},P = \pi d,A = \frac{\pi}{4}d^{2}\]

\[\Rightarrow \frac{P}{A}=\frac{4}{d}\]

\[\Rightarrow m = \sqrt{\frac{4h}{Kd}}=\sqrt{\frac{4\times40}{400\times5\times10^{-3}}}=8.94\]

\[mL = 8.94\times0.1=0.894\]

\[mL<2.64\]

\[Q_{f_{in}} = mAK\left ( T_{0}-T_{\infty } \right )tan h\left ( mL \right )\]

\[\Rightarrow Q_{f_{in}} = 8.94\times\frac{\pi}{4}\left ( 5\times10^{-3} \right )^{2}\times400\times100\times tanh\left ( 8.94\times0.1 \right )=5.008\,W\]