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A spherical interplanetary probe, with a diameter of 2 m, is sent out into the solar system. The probe surface is made of material having an emissivity of 0.9 and an absorptivity of 0.1. Signals from the sensors monitoring the probe surface temperatures are indicating an average value of 40C−40^\circ C for a space temperature of 0 K. If the electronics inside the probe is generating heat at a rate of 100W/m3,100 W/m^3, determine the incident radiation rate on the probe surface.

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The probe surface area is

A=πD2=π22=12.57 m2A=\pi D^2=\pi*2^2=12.57\ \text{m}^2

The probe volume is

V=π6D3=π623=4.19 m3\textbf{V}=\dfrac{\pi}{6}D^3=\dfrac{\pi}{6}*2^3=4.19\ \text{m}^3

Making a heat balance for the probe at steady state

E˙in+E˙gen=E˙outαE˙incident+Vq˙gen=εσA(Ts4Tsurr4)0.1E˙incident+4.19100=0.95.6710812.57(233404)\begin{align*} \dot{E}_\text{in}+\dot{E}_\text{gen}&=\dot{E}_\text{out}\\ \alpha \dot{E}_\text{incident}+\textbf{V}\dot{q}_\text{gen}&=\varepsilon \sigma A\left(T_\text{s}^4-T_\text{surr}^4\right)\\ 0.1*\dot{E}_\text{incident}+4.19*100&=0.9*5.67*10^{-8}*12.57*\left(233^4-0^4\right) \end{align*}

Hence, the incident radiation rate is

E˙incident=14.71 kW\color{#c34632}{\dot{E}_\text{incident}=14.71\ \text{kW}}

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