Homework SourseA spacecraft which can be considered to be a hollow sphere w
A spacecraft which can be considered to be a hollow sphere with a diameter of 2 m, made of a material
with very high thermal conductivity, essentially being isothermal experiences a radiation heat flux from
the sun of 8,000 W/m2. The only heat transfer mechanism is due to radiation.
Its surface has an emissivity of
a) How much heat is absorbed by the spacecraft if the temperature of sun is Tsun=5,750 K?
b) What is the steady state temperature of the spacecraft?
(Hint: Assume that the spacecraft behaves as a grey body for thermal emission and justify that
assumption at the end of this part.)
At some point, a circular hatch (diameter 0.5 m) pointed directly at the sun opens in the sphere and stays
open. The resulting cavity may be considered a black body.
c) How much heat is now absorbed by the spacecraft?
d) What is now the temperature of the sphere after the spacecraft reaches steady state conditions?
(Hint: In the given geometry, the surface of a sphere cap such as the hatch above with a diameter D
may be approximated by the surface area of a circle without introducing large errors.)
Solution
a) The hear absorbed by the spacecraft is a portion of the heat flux from the sun, and is equal to the emmisivie power of the spacecraft.
q = epsilon x sigma x Ts^4 where e = 0.5
q = 0.5 x Q where Q = sigma x Ts^4
q = 0.5 x 8000 = 4000 W / m2
b) The steady state temperature is the heat absorbed due to difference in radiative surfaces.
q = sigma x Tabs^4 where Tabs is the spacraft\'s absolute temperature.
Tabs^4 = (q / sigma)
log Tabs = log (q / sigma ) / 4
log Tabs = log (4000 / 5.67 x 10^-8) / 4 = 2.71
Tabs = 515.37 K
The steady state temperature is Ts - Tabs = 5750 - 515.37 = 5234 K
c) The emissivity of the surface reduces due to the cavity of 0.5 m in diameter which simply leads to the value of emissivity (epsilon) to be 0.2.
The heat absorbed by the spacecraft with cavity is ,
q = 0.2 x sigmax Ts^4
q = 0.2 x 8000 = 1600 W/m2
d) The steady state temperature of the spacecraft with cavity is Ts - Tabs, cavity
q = sigma x Tabs, cavity ^4
log Tabs = log (q / sigma ) / 4
log Tabs = log (1600/5.67x10^-8) / 4
Tabs = 409.85 K
Steasy state temperature Ts - Tabs = 5750 - 409.85 = 5340.15 K
