Is there any reason the nose wouldn't be pointed at the sun? I'd always assumed that would be the orientation. The cross sectional area would be more like 70 sq m.If that were true, and the nose had MLI under a vacuum, wouldn't the heat transfer to the tanks be mainly through the skin and structure connecting the nose to the rest of the 2nd stage? In that case the conduction cross sectional area would be pretty small I'd think. Perhaps some kind of "chomper" opening on the lee side of the nose cone would radiate some of that in addition to decreasing the cross sectional area. Maybe the chomper door actually could be a boom extending forward to put a reflective shield in front of the nose?In that case, I'd think the Earth's radiation would be the main effect. That's something like a quarter of the Sun's and would vary depending on where the ship was in the orbit with the maximum heat flux being at local noon with the tail pointed at Earth. Obviously at night it would be less.
Quote from: Twark_Main on 01/16/2025 09:54 pmStarship is a cylinder and not a pancake. This only increases the back-side radiation by a factor of pi/2 (ie negligible), but on the illuminated side you probably want to numerically integrate over the curvature, accounting for cosine loss.When you integrate along the curve with the cosine losses to the irradiance, you find that the net incident radiation is proportional to the cross-section. So a pancake works fine.Emissivity is a different story, but to tell it you need to know a whole bunch of stuff about the orbit. Even the night side of Earth is pretty hot in IR, and the angular width of the Earth from VLEO is large enough that considerably less than πrL will be radiating into cold space.
Starship is a cylinder and not a pancake. This only increases the back-side radiation by a factor of pi/2 (ie negligible), but on the illuminated side you probably want to numerically integrate over the curvature, accounting for cosine loss.
Quote from: Overtone on 01/16/2025 11:39 amThis has been a great discussion about absorbance, emissions, and insulation for depots. Thank you everyone for sharing your insights.Good discussion, but I haven't seen the math.Let's take a Starship with heat shield of 6cm thick (including the tiles, the blankets, and the backup ablative), give it an R value of 30 in Imperial units (5.28 in MKS), for an Reffective of 0.32 m2K/W. We'll give the heat shield and absorption of 1.0 (probably pessimistic) and emissivity of 0.98.Now for the simplest scenario: The heat shield is facing the Sun, 1300 W/m2. What will be the equilibrium temperature and heat flux into the Starship at the boiling point of LOX, 90K?Now we have two simultaneous equations to solve:emitted heat = 5.67e-8 * 0.98 * Texternal4heat flux through insulation = ΔT/R = (Texternal - 90)/0.32That's a 4th order equation, but fortunately spreadsheets are really good at this kind of thing, and the equilibrium answer is:outer temp: 320Kinner temp: 90Kheat flux through insulation: 726W/m2So the heat shield is only dropping the heat flux from the sun by 44%. Very sad.However, the other side of the Starship is facing deep space in our simplified scenario. So we get to solve the equation again, but this time in the other direction:outer temp: 88Kinner temp: 90Kheat flux: -3 W/m2So the net heat flux is 723W/m2 into the Starship. Still very sad.How much LOX will that heat flux boil?Q=m⋅LQ = heat into the system = 723 J/sL = latent heat of vaporization for LOX = 213kJ/kg. m = mass of the LOX being boiledsolving for m, we get 0.00339 kg/secNow the fuel tank for LOX is about 20mx9m = 180m2 (this is a square cow, the real one is cylindrical), so every second we are boiling .6kg of LOX. That's 2.2t per hour.Is that right? Someone check my math please.Now this is the worst case scenario - deep space, or a highly elliptical orbit. In VLEO the sun is occluded 1/2 the time, and so now we have to start calculating rates. The system will heat up for half an orbit, and cool down for half an orbit. The tiles themselves absorb heat, which we would have to take into account. Will leave that for another post.https://docs.google.com/spreadsheets/d/1esN27mUh7s2gFvXzmvM6x1aU_AnVjfvRYwnGs5U2plw/edit?usp=sharing
This has been a great discussion about absorbance, emissions, and insulation for depots. Thank you everyone for sharing your insights.
Quote from: Overtone on 01/16/2025 11:39 amThis has been a great discussion about absorbance, emissions, and insulation for depots. Thank you everyone for sharing your insights.The Skylab astronauts solved their heat problem with a jury rigged sunshade.Why is that solution not also appliucable to the depot?Even if there is need to develop retractable shades to allow refuelling, would it not be sufficient to simply shield the depot from sun and earthshine?
Quote from: etudiant on 01/16/2025 07:54 pmQuote from: Overtone on 01/16/2025 11:39 amThis has been a great discussion about absorbance, emissions, and insulation for depots. Thank you everyone for sharing your insights.The Skylab astronauts solved their heat problem with a jury rigged sunshade.Why is that solution not also appliucable to the depot?Even if there is need to develop retractable shades to allow refuelling, would it not be sufficient to simply shield the depot from sun and earthshine?The design I've seen for a depot sun shade was a conical structure a bit like a stiff cape that surrounds the tanks. The wider open end points away from heat sources. This doesn't work for two belly to belly ships. So it either needs to be able to reliably open up for transfers or go another route entirely. If a sun shade is in the works it will need to be as dynamic or more dynamic than the ISS solar panels and radiators. And it can't get in the way of transfers.It's one of those 'I wonder how SX will do it' types of things.
Possibly a dumb question, but can't there be some kind of solar powered active cooling system? That radiates the excess heat on the side not pointing towards the sun, to reliquidify the boil-off?
Quote from: OTV Booster on 01/20/2025 12:43 amQuote from: etudiant on 01/16/2025 07:54 pmQuote from: Overtone on 01/16/2025 11:39 amThis has been a great discussion about absorbance, emissions, and insulation for depots. Thank you everyone for sharing your insights.The Skylab astronauts solved their heat problem with a jury rigged sunshade.Why is that solution not also appliucable to the depot?Even if there is need to develop retractable shades to allow refuelling, would it not be sufficient to simply shield the depot from sun and earthshine?The design I've seen for a depot sun shade was a conical structure a bit like a stiff cape that surrounds the tanks. The wider open end points away from heat sources. This doesn't work for two belly to belly ships. So it either needs to be able to reliably open up for transfers or go another route entirely. If a sun shade is in the works it will need to be as dynamic or more dynamic than the ISS solar panels and radiators. And it can't get in the way of transfers.It's one of those 'I wonder how SX will do it' types of things.Such a collar would be a large structure and would probably need to be launched separately before unfolding, much like the JWST. Depot would then dock to it. But if Depot can dock there, it can also undock and emerge to perform transfer operations before re-docking. Depot would also need to be able to boost it to transfer to other orbits.
The JWST shade launched separately? Didn't know that.
Quote from: DanClemmensen on 01/20/2025 12:59 amQuote from: OTV Booster on 01/20/2025 12:43 amQuote from: etudiant on 01/16/2025 07:54 pmQuote from: Overtone on 01/16/2025 11:39 amThis has been a great discussion about absorbance, emissions, and insulation for depots. Thank you everyone for sharing your insights.The Skylab astronauts solved their heat problem with a jury rigged sunshade.Why is that solution not also appliucable to the depot?Even if there is need to develop retractable shades to allow refuelling, would it not be sufficient to simply shield the depot from sun and earthshine?The design I've seen for a depot sun shade was a conical structure a bit like a stiff cape that surrounds the tanks. The wider open end points away from heat sources. This doesn't work for two belly to belly ships. So it either needs to be able to reliably open up for transfers or go another route entirely. If a sun shade is in the works it will need to be as dynamic or more dynamic than the ISS solar panels and radiators. And it can't get in the way of transfers.It's one of those 'I wonder how SX will do it' types of things.Such a collar would be a large structure and would probably need to be launched separately before unfolding, much like the JWST. Depot would then dock to it. But if Depot can dock there, it can also undock and emerge to perform transfer operations before re-docking. Depot would also need to be able to boost it to transfer to other orbits.The JWST shade launched separately? Didn't know that. Or maybe the shade stays attached and opens up like Dracula spreading his cape. Ditching the shade when the depot needs ullage acceleration sounds like more of a problem than some reinforcement to handle the low g acceleration. The side where the heat shield would normally be is available for a ship length structure to house a shade and deployment mechanism during launch.
Such a collar would be a large structure and would probably need to be launched separately before unfolding, much like the JWST. Depot would then dock to it. But if Depot can dock there, it can also undock and emerge to perform transfer operations before re-docking. Depot would also need to be able to boost it to transfer to other orbits.
This would be for a single layer shade? Wouldn't a multi layer shade do much better?I couldn't follow the math past fourth order equations but I would expect the .98 emissivity would be for each layer in a multi layer, each layer emitting from both sides and each layer inward having less to emit. And wouldn't reflectance play into this?Or is my model wrong? IAN...
Quote from: DanClemmensen on 01/20/2025 12:59 amSuch a collar would be a large structure and would probably need to be launched separately before unfolding, much like the JWST. Depot would then dock to it. But if Depot can dock there, it can also undock and emerge to perform transfer operations before re-docking. Depot would also need to be able to boost it to transfer to other orbits.We are not trying to cool a telescope to near CMB levels, 90K is sufficient, and no the JWST went in the same package.It would be pretty trivial to deploy a rollable sun shade out of a cargo bay. The best thing to do would be to dual purpose it as a flexible solar panel, that way batteries can stay charged while you are in the shade.You only need to shade the side pointing to the Earth. As shown in the math above, facing the nose at the sun by itself is sufficient if you insulate the header tanks well.
Quote from: ChrML on 01/19/2025 10:10 pmPossibly a dumb question, but can't there be some kind of solar powered active cooling system? That radiates the excess heat on the side not pointing towards the sun, to reliquidify the boil-off?Not dumb, but there are a lot of challenges to overcome to make this work. Since SpaceX already announced they aren't planning to do this, it hasn't attracted a lot of interest.
Quote from: RDoc on 01/19/2025 05:08 pmIs there any reason the nose wouldn't be pointed at the sun? I'd always assumed that would be the orientation. The cross sectional area would be more like 70 sq m.If that were true, and the nose had MLI under a vacuum, wouldn't the heat transfer to the tanks be mainly through the skin and structure connecting the nose to the rest of the 2nd stage? In that case the conduction cross sectional area would be pretty small I'd think. Perhaps some kind of "chomper" opening on the lee side of the nose cone would radiate some of that in addition to decreasing the cross sectional area. Maybe the chomper door actually could be a boom extending forward to put a reflective shield in front of the nose?In that case, I'd think the Earth's radiation would be the main effect. That's something like a quarter of the Sun's and would vary depending on where the ship was in the orbit with the maximum heat flux being at local noon with the tail pointed at Earth. Obviously at night it would be less.Earth Flux averages over a 300km orbit about 350W/m2, or 350/1361 = .26 so 1/4 is reasonable.
Quote from: TheRadicalModerate on 01/17/2025 05:28 amQuote from: Twark_Main on 01/16/2025 09:54 pmStarship is a cylinder and not a pancake. This only increases the back-side radiation by a factor of pi/2 (ie negligible), but on the illuminated side you probably want to numerically integrate over the curvature, accounting for cosine loss.When you integrate along the curve with the cosine losses to the irradiance, you find that the net incident radiation is proportional to the cross-section. So a pancake works fine.Emissivity is a different story, but to tell it you need to know a whole bunch of stuff about the orbit. Even the night side of Earth is pretty hot in IR, and the angular width of the Earth from VLEO is large enough that considerably less than πrL will be radiating into cold space.Assuming 460W/m2 earth shine in the sun, the equilibrium temperature of the tiles is 193K, emissions, 77W/m2, and net 325W/m2 transmitted into the tank through insulation (remainder is reflected).
22*9 = 200m2 * 325W/m2 = 65kW net influx. I note heat on the nosecone is rounding error.
Quote from: InterestedEngineer on 01/16/2025 08:02 pmQuote from: Overtone on 01/16/2025 11:39 amThis has been a great discussion about absorbance, emissions, and insulation for depots. Thank you everyone for sharing your insights.Good discussion, but I haven't seen the math.Let's take a Starship with heat shield of 6cm thick (including the tiles, the blankets, and the backup ablative), give it an R value of 30 in Imperial units (5.28 in MKS), for an Reffective of 0.32 m2K/W. We'll give the heat shield and absorption of 1.0 (probably pessimistic) and emissivity of 0.98.Now for the simplest scenario: The heat shield is facing the Sun, 1300 W/m2. What will be the equilibrium temperature and heat flux into the Starship at the boiling point of LOX, 90K?Now we have two simultaneous equations to solve:emitted heat = 5.67e-8 * 0.98 * Texternal4heat flux through insulation = ΔT/R = (Texternal - 90)/0.32That's a 4th order equation, but fortunately spreadsheets are really good at this kind of thing, and the equilibrium answer is:outer temp: 320Kinner temp: 90Kheat flux through insulation: 726W/m2So the heat shield is only dropping the heat flux from the sun by 44%. Very sad.However, the other side of the Starship is facing deep space in our simplified scenario. So we get to solve the equation again, but this time in the other direction:outer temp: 88Kinner temp: 90Kheat flux: -3 W/m2So the net heat flux is 723W/m2 into the Starship. Still very sad.How much LOX will that heat flux boil?Q=m⋅LQ = heat into the system = 723 J/sL = latent heat of vaporization for LOX = 213kJ/kg. m = mass of the LOX being boiledsolving for m, we get 0.00339 kg/secNow the fuel tank for LOX is about 20mx9m = 180m2 (this is a square cow, the real one is cylindrical), so every second we are boiling .6kg of LOX. That's 2.2t per hour.Is that right? Someone check my math please.Now this is the worst case scenario - deep space, or a highly elliptical orbit. In VLEO the sun is occluded 1/2 the time, and so now we have to start calculating rates. The system will heat up for half an orbit, and cool down for half an orbit. The tiles themselves absorb heat, which we would have to take into account. Will leave that for another post.https://docs.google.com/spreadsheets/d/1esN27mUh7s2gFvXzmvM6x1aU_AnVjfvRYwnGs5U2plw/edit?usp=sharingThis would be for a single layer shade? Wouldn't a multi layer shade do much better?I couldn't follow the math past fourth order equations but I would expect the .98 emissivity would be for each layer in a multi layer, each layer emitting from both sides and each layer inward having less to emit. And wouldn't reflectance play into this?Or is my model wrong? IAN...
Personally I doubt all this talk about complex deployables (and especially re-deployables). This would add a bunch of complexity and failure modes. "The best deployment process is no deployment process."Hybrid SOFI-MLI can achieve great thermal performance without the complexity.
Quote from: Twark_Main on 01/20/2025 03:30 amPersonally I doubt all this talk about complex deployables (and especially re-deployables). This would add a bunch of complexity and failure modes. "The best deployment process is no deployment process."Hybrid SOFI-MLI can achieve great thermal performance without the complexity.Personally, I doubt Musk & Co are going to throw away 3.2km/sec of relatively free (20t dry) of deltaV by having a reusable Depot depart LEO without a heat shield and flaps. It's an order of magnitude more of fuel to do that by propulsion (mass ratio of 2.5, so fuel is 1.5x dry mass or ~225t). That's 225t that cannot be transferred to another ship.Of course, if it's NASA, they just throw away things, so maybe they'll just throw away the Depot if it has to leave LEO for Artemis.Then again, if it's impossible to one-pass aerobrake saving 205t of mass and at the same time not lose 205t of mass to evaporation, then I'm wrong about the heat shield. I wonder if it is possible to do both.
Quote from: OTV Booster on 01/20/2025 12:33 amThis would be for a single layer shade? Wouldn't a multi layer shade do much better?I couldn't follow the math past fourth order equations but I would expect the .98 emissivity would be for each layer in a multi layer, each layer emitting from both sides and each layer inward having less to emit. And wouldn't reflectance play into this?Or is my model wrong? IAN...The math was for no shade at all, going with "best part is no part". TL;DR - nose at the sun or hottest source in the sky works extremely well, no shade needed for deep space, for exampleIn LEO, however, there are two hot sources - the Sun, and 1/4 of that the Earth. Pointing nose isn't enough.I'm pretty sure now for long term LEO depots shading from earth-shine will be the most important thing to do, and any lightweight deployable shade would do the job, it only being 460W/m2 on the sun-facing side of the Earth. Just keep the nose always pointed at the Sun.