Radiation

From the figure we can say that at 6000 \mathring{A} wavelength total hemispherical emissivity is maximum
\lambda _m=6000 \mathring{A} =0.6\mu m
Wein's law
\lambda _m T=C
\lambda _m T=2898 \mu mK
T=\frac{2898}{0.6}=4830K
Note: In this question, we need to find temperature at which total hemispherical emissivity will be highest.
But to calculate total hemispherical emissivity we require additional data like function chart and emissivity value at particular wavelength.
With the use of Wein's law we can find out temperature at which spectral emissivity is highest but in question it is asked temperature for total hemispherical emissivity.

Equilibrium Temperature =T_S=?
In this it is mentioned that entire flux is absorbed by the plate it means for solar flux absorptivity is 1. Kirchoff's law \alpha =\varepsilon =0.21
Surface of cast iron and surrounding fluid temperature difference is small due to this we can use Kirchoff's law
At equilibrium condition
Energy absorbed = Energy leaving
\begin{aligned} \alpha GA&=\varepsilon A\sigma (T_S^4-T_\infty ^4)\\ 1 \times 600 &=0.21 \times 5.669 \times 10^{-8}(T_S^4-298^4)\\ T_S&=491.34K\\ T_S&=218.34^{\circ}C\\ T_S&=218^{\circ}C \end{aligned}

\begin{aligned} r_{1} &=10 \mathrm{~mm} \\ A_{1} &=4 \pi r_{1}^{2} \\ A_{2} &=6 \times\left(30^{2}\right) \\ F_{12} &=1 \\ A_{1} F_{12} &=A_{2} F_{21} \quad \Rightarrow \quad F_{21}=\frac{A_{1}}{A_{2}}=\frac{4 \pi \times(10)^{2}}{6 \times(30)^{2}}=0.2327 \\ F_{22} &=1-F_{21}=0.7672 \simeq 0.77 \end{aligned}

From Wien's Displacement law,
\begin{aligned} \lambda_{m} T &=\text { constant } \Rightarrow \lambda_{m 1} T_{1}=\lambda_{m 2} T_{2} \\ \lambda_{m 1} \times 3000 &=1.2 \lambda_{m 1} \times T_{2} \\ T_{2} &=\left(\frac{3000}{1.2}\right) K=2500 K \end{aligned}

\begin{array}{l} \mathrm{F}_{\mathrm{LM}}=\mathrm{F}_{\mathrm{L}-\mathrm{a}}+\mathrm{F}_{\mathrm{L}-\mathrm{b}}+\mathrm{F}_{\mathrm{L}-\mathrm{c}}=0.8-\mathrm{c}\ldots(1) \\ \mathrm{F}_{\mathrm{N}-\mathrm{R}}=0.4 \end{array}
From the figure,
\begin{array}{l} \mathrm{F}_{\mathrm{L}-\mathrm{a}}=\mathrm{F}_{\mathrm{L}-\mathrm{c}}=\mathrm{F}_{\mathrm{P}-\mathrm{e}}(\text { similar }) \\ \mathrm{F}_{\mathrm{L}-\mathrm{b}}=\mathrm{F}_{\mathrm{N}-\mathrm{R}}(\text { similar }) \end{array}
From equation (1)
\begin{array}{l} \mathrm{F}_{\mathrm{L}-\mathrm{a}}+\mathrm{F}_{\mathrm{L}-\mathrm{b}}+\mathrm{F}_{\mathrm{L}-\mathrm{c}}=0.8 \\ 2 \times \mathrm{F}_{\mathrm{L}-\mathrm{a}}+\mathrm{F}_{\mathrm{N}-\mathrm{R}}=0.8 \\ \mathrm{F}_{\mathrm{L}-\mathrm{a}}=\frac{0.8-0.4}{2}=0.2 \end{array}
From the fig -3
\begin{aligned} \mathrm{F}_{\mathrm{P}-\mathrm{Q}}&=\mathrm{F}_{\mathrm{P}-\mathrm{d}}+\mathrm{F}_{\mathrm{P}-\mathrm{e}} =\mathrm{F}_{\mathrm{N}-\mathrm{R}}+\mathrm{F}_{\mathrm{L}-\mathrm{a}} \\ (\mathrm{F}_{\mathrm{P}-\mathrm{d}}&=\mathrm{F}_{\mathrm{N}-\mathrm{R}} \text{ and } \mathrm{F}_{\mathrm{P}-\mathrm{e}}=\mathrm{F}_{\mathrm{L}-\mathrm{a}}(\text { due to similarity })) \\ &=0.4+0.2\\ F_{P-Q}&=0.6 \end{aligned}

Given data:
\mathrm{d}_{1}=0.1 \mathrm{m}
\mathrm{d}_{2}=0.4 \mathrm{m}
\mathrm{F}_{1-1}=0 (from the geometry)
From the additive rule
\mathrm{F}_{1-1}+\mathrm{F}_{1-2}=1
\mathrm{F}_{1-2}=1

From Wein's displacement law
For black body,
\begin{aligned} \lambda_{M} T &=\text { constant } \\ \lambda_{M 1} T_{1} &=\lambda_{M 2} T_{2} \\ 1.45 \times 2000 &=\lambda_{M 2} T_{2}=2.90 \times T_{2} \\ \therefore \quad T_{2} &=\left(\frac{1.45}{2.90} \times 2000\right)=1000 \mathrm{K} \end{aligned}

E_{b} \propto T^{4} (Stefan Boltzmann Law of Radiation)
\frac{E_{b_{2}}}{E_{b_{1}}}=\left(\frac{2 T}{T}\right)^{4} \Rightarrow E_{b_{2}}=16 P

F^{12} =0.5

\begin{aligned} T_{1} &=800 \mathrm{K} \\ T_{2} &=600 \mathrm{K} \\ T_{\text {surounding }} &=300 \mathrm{K} \end{aligned}
Let (3) denote the surrounding environment
(Assume unit width perpendicular to plane of Fig.)
From summation rule:
\begin{aligned} F_{21}+F_{\beta 2}^{0}+F_{23} &=1 \\ A_{1} F_{12} &=A_{2} F_{21}=6 \times 1 \times 0.5 \\ &=(5 \times 1) \times F_{21} \\ F_{21} &=\frac{6 \times 0.5}{5}=0.6\\ \therefore \quad F_{23}&=1-F_{21}=0.4 \end{aligned}
since surrounding environment being large,
No surface has a surface resistance.

Heat transfer rate from (2) and (3)=\frac{E_{b_{2}}-E_{b_{3}}}{\frac{1}{A_{2} F_{23}}}
\begin{aligned} &=\frac{\sigma\left(T_{2}^{4}-T_{3}^{4}\right)}{1} \\ &=\frac{5.67 \times 10^{-8}\left(600^{4}-300^{4}\right)}{A_{2} F_{23}} \text { watt/metre }\\ &=13.778 \mathrm{kW} / \mathrm{metre} \\ \end{aligned}
Assume that heat exchange is per unit width perpendicular to plane of figure

\begin{aligned} Q &=\frac{\sigma\left[T_{1}^{4}-T_{2}^{4}\right]}{\frac{1}{\epsilon_{1}}+\frac{1}{\epsilon_{2}}-1}=\frac{5.67 \times 10^{-8} \times\left(1000^{4}-400^{4}\right)}{\frac{1}{0.5}+\frac{1}{0.25}-1} \\ &=11049.7 \mathrm{W} / \mathrm{m}^{2}=11.05 \mathrm{kW} / \mathrm{m}^{2} \end{aligned}