Net radiation heat exchange rate,

$$\sigma = 5.67 \times 10^{-8}\,W/m^{2}K^{4}$$

$$\frac{Q}{A}=\frac{\sigma \left ( T_{1}^{4} – T_{2}^{4} \right )}{\frac{1}{\varepsilon _{1}}+\frac{1}{\varepsilon _{2}}-1}= \frac{ 5.67 \times 10^{-8}\left ( 400^{4} – 300^{4} \right )}{\frac{1}{0.8}+\frac{1}{0.8}-1} = 661.5\,W/m^{2} = 0.66\, kW/m^{2}$$

The gravitational pull on the spacecraft is less than on the ground.  Natural convection is based on buoyancy force and this force is due to gravity. Since in the spacecraft there is low or no gravity effect there will be very low or no byoyancy force. This will lead to slower cooling of egg on the spacecraft than on the ground.

Using the properties of air 48.5 °C

(a) Assuming unit Width in Z direction

$$Re_{x}=\frac{u_{\infty}x}{v}=\frac{3\times0.25}{17.36\times10^{-6}} = 43200$$
$$Nu_{x} = 0.332\left ( Re_{x} \right )^{0.5}\left ( Pr \right )^{1/3}=0.332\left ( 43200 \right )^{0.5}\left ( 0.7 \right )^{1/3}=61.4$$
$$h_{x}=\frac{Nu_{x}\, k}{x}  = \frac{61.4\times0.02749}{0.25}=6.75\,W/m^{2}K$$
$$\bar{h} = 2h_{x}=2\times6.75=13.5\,W/m^{2}K$$
$$z = 1\,m, Q_{25\,cm} = \bar{h}A\left ( T_{w} – T_{\infty} \right )=13.5\left ( 0.25\times 1 \right )\left ( 70-27 \right )=145.1\,W$$

(b) Assuming unit width in z direction

$$Re_{x}=\frac{u_{\infty}x}{v}=\frac{3\times0.45}{17.36\times10^{-6}} = 77765$$
$$Nu_{x} = 0.332\left ( Re_{x} \right )^{0.5}\left ( Pr \right )^{1/3}=0.332\left ( 77765 \right )^{0.5}\left ( 0.7 \right )^{1/3}=82.4$$
$$h_{x}=\frac{Nu_{x}\, k}{x}  = \frac{82.4\times0.02749}{0.45}=5.034\,W/m^{2}K$$
$$\bar{h} = 2h_{x}=2\times 5.034 = 10.07 \,W/m^{2}K$$
$$z = 1\,m, Q_{45\,cm} = \bar{h}A\left ( T_{w} – T_{\infty} \right )=10.07\left ( 0.45\times 1 \right )\left ( 70-27 \right )=194.85\,W$$