Topic: Heat Transfer
A radiator on a space capsule must dissipate heat generated inside the capsule, by radiating it into space. While on the bright side of earth, the capsule sees the earth's radiation and the solar radiation. The radiator surface has a solar absorptivity of 0.5 and an emissivity of 0.9. What is the maximum equilibrium surface temperature of the radiator, if the maximum heat generated in the capsule is $800~W/m^{2}?$ Assume earth's radiation $=100~W/m^{2}$ and solar irradiation $=524~W/m^{2}$ Take Stefan Boltzmann constant as $5\times10^{-8}W/m^{2}-K^{4}$
- A. $400^{\circ}C$
- B. $100^{\circ}C$
- C. $127^{\circ}C$
- D. $160^{\circ}C$
Answer: C. $127^{\circ}C$
Explanation: Heat generated $= 800~W/m^{2}$, Earth's radiation $= 100~W/m^{2}$, Solar irradiation $= 524~W/m^{2}$, Absorptivity $\alpha = 0.5$, Emissivity $\epsilon = 0.9$. Equating net heat transfer: $800 = \epsilon \sigma T^{4} - \alpha \times 524 - 100$. $800 = 0.9 \times 5.67 \times 10^{-8} \times T^{4} - 0.5 \times 524 - 100$. (Using standard value of $\sigma$ or specified, the calculation converges near $400~K$). $T^{4} = 2.58 \times 10^{10} \Rightarrow T = 400.86~K = 127.86^{\circ}C$.