Question

A liquid droplet radiator (LDR) is used to liberate heat from the UT Space StationnamedDOLLYPARTON(Droplets OfLiquid LeaveYielding Particles Alleviating Radiation To Overwhelming Nothingness). The LDR works in the following way. Waste heat generated from daily operation of the space stationistransferred to an oil bath. The bath is ejected into space in droplet form effectively radiating energy into deep space.The droplets are recollected after travelling a distance (L)through spaceand reintroduced to the oil bath.

The average ejected droplet temperature is Ti = 550 [K] and the droplet velocity is v = 0.05 [m/s]. The oil has a density of

p =1000 kg/m^3 ,a specific heat capacity of cp= 2000 J/kg/K and an emissivity of E= 0.92.The average droplet diameter is 500 [\um]. Deep space can be thought of as surroundings with zero temperature (Tsurr= 0 [K]). The LDR is shielded from solar radiation.

The distance between the ejector and collector can be adjusted to optimize the temperature of the droplet when it is recollected. Ideally, the temperature should be Tf = 300 [K] because this is the mean temperature of the station interior. A lower temperature would liberate more energy making the system more efficient but complications with freezing could occur operating below 300 [K]. What ejector–collector separation (L) will yield a droplet recollection temperature of Tf = 300 [K]?

??P? ?? /?? = ???(? ^4 − ?^4????) [ ? /? ]

What is the total internal energy lost per droplet to space?

Answer 1