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Would the droplet cooling system from Mass Effect be actually possible in our universe? I'm interested in expanding the space industry and promote manufacturing in space. One of the biggest problems that seem to persist is heat dissipation, radiators are ok but there has to be a better alternative. I was thinking of the droplet cooling system from Mass Effect, here's the codex entry:

In a droplet system, tanks of liquid sodium or lithium absorb heat within the ship. The liquid is vented from spray nozzles near the bow as a thin sheet of millions of micrometer-scale droplets. The droplets are caught at the stern and recycled into the system. A droplet system can sink 10-100 times as much heat as DRA strips.

It seems like it could work but also seems like its just science fantasy. I guess it would depend on how long the droplets would be out in the vacuum of space.

So to make it simple, I'll give out some numbers, let's say 1,000,000,000,000 droplets (about 1 Liter worth I think) (each droplet with a radius of 5 microns, the normal size of mist droplets) of liquid sodium at 850°C (a bit below its normal earth atmospheric boiling temperature) out in space for 30 minutes. What Temperature would the returning sodium be? Keep in mind this isn't an accurate scenario, just some numbers thrown out. If you want to provide a more realistic scenario I would be more than happy. Thanks!

(No, it's not off-topic, do your bloody jobs right, don't look for excuses to not answer it. it's not engineering, I'm not asking someone to build it, I'm simply asking how much the temperature would go down.)

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closed as off-topic by Sebastian Riese, ZeroTheHero, Kyle Kanos, Jon Custer, Chris May 28 '18 at 1:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question appears to be about engineering, which is the application of scientific knowledge to construct a solution to solve a specific problem. As such, it is off topic for this site, which deals with the science, whether theoretical or experimental, of how the natural world works. For more information, see this meta post." – Sebastian Riese, ZeroTheHero, Kyle Kanos, Jon Custer, Chris
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  • $\begingroup$ 1,000,000 droplets each a 10 micron diameter sphere is only half a microliter of total volume, not 1 Liter. $\endgroup$ – pentane May 19 '18 at 7:18
  • $\begingroup$ Can you explain why such a complex, expensive and flammable solution is considered? Why not use a water based heat pipe? $\endgroup$ – my2cts May 19 '18 at 8:14
  • $\begingroup$ @my2cts we are talking about a system in space. Parts of the water would simply evaporate (or sublimate) and you would lose your coolant quickly. Sodium is neither expensive nor flammable when you are in space, since there is (almost) no oxygen in space. $\endgroup$ – Sebastian Riese May 19 '18 at 14:28
  • $\begingroup$ @pentane yeah I may have forgotten a few zeros, will fix (math is not my strong suit) $\endgroup$ – SpaceManTyler May 20 '18 at 8:03
  • $\begingroup$ I think this particular question would get a better reception on Worldbuilding. $\endgroup$ – knzhou May 20 '18 at 11:12
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In the vacuum of space, there would be no heat transfer by conduction nor convection. That leaves only radiation:

$$P = \epsilon \cdot \sigma \cdot A \cdot T^4 $$

$P$ = Power radiated in $Watts$
$\epsilon$ = emissivity, a unitless number between 0 and 1
$\sigma$ = Stefan–Boltzmann constant, in $Watts/m^2K^4$
$A$ = surface area in $m^2$
$T$ = temperature of the object in degrees $K$

The emissivity of the sodium (as other unoxidized metals) is probably between 0.05-0.2.

The most the surface area could be is 1,000,000,000,000 times the surface area of a single droplet, which is $4 \pi r^2$. That totals to 314 $m^2$.

$$P = 0.2 \cdot 5.67 \times 10^{-8}\ Watts/m^2K^4 \cdot 314\ m^2 \cdot (1123\ K)^4 = 5.6 \times 10^6\ Watts$$

If the droplets are outside the spaceship for 30 minutes = 1800 seconds. 1800 seconds at $5.6 \times 10^6\ Watts$ = $1.01 \times 10^{10}\ \text{Joules}$. The heat capacity of sodium at 850 °C is around 1.278 Joules/gram °C and the density is $0.968\ g/cm^3$. The total volume of sodium you suggested is 0.52 L = 520 mL. The mass of sodium is then:

$$520\ cm^3 \cdot\ 0.968\ g/cm^3 = 504\ g$$

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  • $\begingroup$ That's not the answer I'm looking for. I'm looking for the change in temperature, not the mass. $\endgroup$ – SpaceManTyler Jun 4 '18 at 3:36

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