The problem is that there’s not enough surface area. Assuming that the surface is completely black (the best case), the amount of heat radiated away is proportional to the surface area times the temperature to the fourth power. Well, the amount of heat we need to radiate is fixed: That’s the heat output of all of those people. And the surface area is fixed. Given those, we can calculate the temperature we’d need. I’ven’t done this calculation, but it seems quite plausible that this temperature is, as ftg claims, too high for human occupancy. This is not insurmountable, since only the surface needs to be at this temperature. One could, in principle, air condition the interior, and dump all the heat onto the surface, with good insulation in between. But it’d be one heck of an engineering problem.
Are you assuming the DS is a perfect sphere for this exercise? It seems that in reality the SA of a spacecraft is fixed only in the way that the SA of the lung or gut is fixed, ie it’s fixed, but fixed at an infinitely high level, or so close to infinite surface area as makes no difference. To give you some idea of what I’m talking about, the SA of the human lung is about 100 m^2, yet takes up less than 10l of volume.
I asume I’m missing somehting here. SA in this application will be entirely dependent on the scale of measurement won’t it? And with sufficent folds, crennelations etc. SA will continue to increase until we get to the micron scale at the very least? If we assume some simple fractal design on the skin of the death star, and the same scale on the terminal protrusions as found on the lung it seems like there would be plenty of SA for heat loss.
Something as simple as building large chunks of the skin with ‘waves’ of 100m tall parabolas will increase the SA to an astonishing degree. And those parabolas themselves could have parabolas built onto them. And the ‘top’ part of those parabolas that can see the sky can then have parabolas built in, and so ad infinitum.
As an exercise imagine a tree with a roughly spherical crown. What is the SA of that sphere? What is the leaf area of the actual tree? How much SA does that tree absorb radiation from? (BTW, most of the leaves of a tree are located on the outside of the canopy where they can catch the light, so the interior volume isn’t greatly affcted by this huge SA)
I don’t see why the temperature would be a problem.
All you’d need to do is build a small (say 2-3 meters wide) thermal exhaust port, right below the main port, leading to the core…
But the Death Star in Episode VI had an interior infrastructure and, presumably, more surface area than the mere “outside surface of a sphere” we’re talking about, yes?
Maybe that’s what all those trenches are for: to increase surface area for heat dispersion. How much surface area would they need, anyway?
Couldn’t you just channel the excess heat to the station’s power core? Then you’ll have a partially human-powered space station!
That pesky second law of thermodynamics again. You can easily capture waste heat as long as the place it’s going to is colder than the heat source. That’s why the heater in a car is a perfect use of waste heat. Considerably more difficult to capture that waste heat and use it as energy for the engine.
Diffcult, but not impossible. Mankind has been converting heat to electrical current for over a century. The details here are a bit different, but the principle’s the same.
Look at it this way - the station’s reactor will have to contain a huge amount of heat anyway. What’s a bit more?
No they haven’t. what they have been converting is the movement of heat from a hot place to a cold place. You can’t radiate heat into an engine that is hotter than your source. Technically you can but more will radiate back. Yes, I know about air conditioners. They work by adding yet more energy to make the heat transfer medium hotter than the place they are dumping it. We’ve already stated that there isn’t enough surface area for black body radiation to dump all the heat into space so adding more won’t help.
A though experiment for you next time you’re standing in front of a camp fire in a cold wood and you’re all nice and toasty. Drink several toddies until your bladder is full of 98.6 degree “heat transfer fluid.” The heated fluid will be hotter than the surroundings away from the fire. I want you to use the fluid to make the fire hotter.
That would be just peachy if you were sending the heat out to a fluid in contact with the surface. This is why the convolutions of lungs is useful, more surface area for molecultes to permeate. The problem with a radiator in space is that due to the geometry much of the energy will radiate right back into adjacent surfaces. Someone correct me if I’m wrong but I don’t think there will be any net increase in radiated heat.
There aren’t any useul loopholes in the second law of thermodynamics guys.
Excepting always hyperspatial rifts into regions of negative time and their like. 
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- I don’t think “excess heat” would be a problem… I would guess that you would have to use insulation to prevent heat from escaping off into space too fast:
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- first of all we note that the moon’s temperature rises to about 100C in the daytime, and falls to about -200C at night, but just a few centimeters below the surface the temperature stays around -40C all the time.
- when a skyscraper is built (on Earth), the basement is about 10% of the building’s volume, and the dirt removed from the basement usually weighs more than the loaded and occupied building would. Now it is true that the Earth and moon are not the same densities–but then, the building materials for each wouldn’t have to be the same densities either. The moon is 3/5 dense, but only has 1/6 gravity.
- the Death Star was supposed to be roughly as big as a moon.
- in space and concerning fairly-low temperatures, the main factor in heat retention is basically simple mass.
- So suppose that we built a “spherical building” (full of hallways and rooms, ect), named it the Dope Star. If you built the Dope Star the same diameter as the moon, the Dope Star would only have a rather small percentage of the density of the real moon, depending on what materials the Dope Star was made of exactly. If you built the Dope Star to the same mass as the real moon, then the Dope Star would have the same mass, but also have a lot more surface area. Either way it’s going to shed heat even faster than the moon does–and the moon averages -40C.
…Also if you consider–there’s no reason to build the Dope Star “solid” all the way through. The local gravitational pull of a spherical object in space decreases as you get closer to its center–so as you get nearer to the center, the loads decrease until there are no loads to carry anyway. It would be a better utilisation of materials to build a larger-diameter hollow sphere. And if you do that, then (for any quantity of equvalent mass) you’re getting even more surface area–which would mean even faster heat loss.
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I figure, the outside surface of the ‘Death Star’ would radiate away energy (blackbody, at the very least) at a regular pace that would have to be determined. If the outer surface area is large enough, when coupled with the emissivity of the materials & their shape, you should be able to reach a comfortable equilibrium.
And let’s build a “Maxwell’s Demon”!, Moohoohoohaaahaaahaaaaaaa!
I think I remember it being proposed by IBM in the early 80’s that there might be some nano-tech answer to acheiving macro-scale mechanical motion from random molecular vibrations (associated with temp). I suppose it must not have worked - another victory for that annoying 2nd Law of Thermo.
Be careful. The pressure inside is not directly related to the gravitational force at that given point, but the integral of grav force/A on all the mass all the way up to the surface - that’s why fusion happens in the sun, even though the grav force is ~ 0 at the center.
Sorry, I should have read all responses - Agreed, reposted.
Is that assuming people inhabit the whole sphere? Maybe they only occupy a few levels below the surface, and the center is mostly empty (or filled with liquid helium)
Brian
For the purposes of radiating energy, surface irregularities and internal cavities contribute very little to the effective radiated area. (They do help for cooling by convection, but the Death Star isn’t air-cooled.) Putting a trench or tunnel in the surface doesn’t help you radiate much heat as most of the heat radiated by the sides of the trench will simply be reabsorbed by the other side. If it doesn’t have a direct line-of-sight to space, it won’t help you cool. To add effective radiating area you need to increase the amount of surface which can “see” space, which means either making the object larger, or adding fins or panels which protrude far enough from the surface to not have their “view” of space obstructed by other surface features. As the Death Star doesn’t have any large (compared to its overall size) radiator fins or towers projecting, its surface area is effectively very close to that of a perfect sphere. The small towers and trenches it does have aren’t going to make a significant difference to its ability to radiate heat. So the rate at which it looses heat will be roughly the same as an equivalent sized moon. (The density of the Death Star is irrelevent for determining its rate of heat loss, although it does matter for calculating thermal inertia.)
Our Moon maintains a fairly cold internal temperature because it is generating essentially zero heat internally. Its only source of heat input is sunlight, so its internal temperature is determined by the point at which the heat radiating from its entire surface is equal to the heat being absorbed by the sun. The Death Star will have significant sources of heat. In addition to all the people inside it, you’ve also got waste heat generated by every machine or electrical device inside it. Even the heat generated by keeping the interoir lit adds up over time. The Death Star may well have all manner of refrigeration and air conditioning systems to keep the interior at a liveable temperature, but if the Laws of Thermodynamics apply, every bit of that heat can only be gotten rid of by radiating out from the surface of the Death Star.
This isn’t just a problem for a spaceship the size of a moon. The Space Shuttle needs to keep its cargo bay open while in orbit in order to use the open doors as radiators. If the cargo doors won’t open, they have to do an emergency landing before the shuttle overheats. The International Space Station has radiator arrays that are used with a liquid cooling loop to keep the interior and electronics cool. The problem just gets worse as you get bigger - since the area available for radiator surfaces scales with the square of size, while the mass (and presumably total heat load) scales with the cube of size. The larger a ship becomes, the more problematic heat management becomes.
Now, the Death Star as portrayed in the movies is already breaking several fundamental laws of physics, so we shouldn’t be too worried about how it stays cool. If you’re already ignoring relativity and conservation of energy, ignoring the laws of thermodynamics isn’t much more to ask. But if you’re asking if we could build such a thing with today’s technology - or for that matter any plausible future technology - the answer is no.
we ought to build the weapon and point it toward our planet,
i think it would be a neat art peice 
pure genius
Excess heat is a problem even in the puny little space stations we have now. If you look at this picture of the ISS, all the vertical panels near the middle are radiator panels for dumping excess heat into space. The problem will get worse as you scale up, because you have less surface area per volume that you can use as radiator surfaces.
The Death Star is plausible if most of the interior is hollow, and the entire surface is used as a radiator. Is there any reason to believe that isn’t the case? A radiator can be white - it just needs to be “black” (high emissivity) in the infrared, it can still be white in the visible light. In fact that’s the best way to do it, because you want to maximize infrared radiation while minimizing absorption of visible light.
Although if you have the technology to build the thing, there might be other ways to dump heat. Like getting a big chunk of ice from a comet and dumping it into the middle of the Death Star periodically.
Maybe the death start IS black, it jsut appears white because it is so hot…
Brian
Well, let’s do the calculation. Assume that the only source of heat is the people on board the station. According to the site Squink linked to, the power consumption of the Death Star is on the order of 10[sup]30[/sup] watts. Again according to starwars.com, the Death Star is 120 km in diameter, and the Stefan-Boltzmann Law says that the thermal power radiated is given by
P = sigma A T[sup]4[/sup]
where sigma is the Stefan-Boltzmann constan, A is the surface area, and T is the temperature. Working through the numbers we get…
Oh. Oh dear.
A balmy 9 million Kelvins.
Obviously there are some kinks in the system that still need working out.