Well, more than half the nose of any EDL-capable Starship is TPS, which you almost certainly can't paint.
Quote from: Greg Hullender on 08/10/2022 06:41 pmQuote from: TheRadicalModerate on 08/10/2022 06:10 pmGiven that you want to recover that tanker (otherwise, you expend one for every lunar mission), you're probably restricted to bare metal tanks and TPS. But then that makes it essential minimize transit boiloff without solar white coatings, sunshields, etc. Table 1 in the article I linked to above said that applying 5 mils (about 1/8 mm) of Solar White paint on top of stainless steel reduced absorption of solar energy from 53% to 8.5%. That seems like a very fine improvement for very little cost. Especially if you only need to paint the nose.Well, more than half the nose of any EDL-capable Starship is TPS, which you almost certainly can't paint. The question would be whether to clean off the solar white after EDL would be worth it, vs. just soaking up the rays. Remember, the nose for a tanker is kinda like the outside of a dewar. Only a fraction of the heat that's radiated inward, into the payload bay, will warm the top of the LCH4 tank. I'll bet a hunk of MLI in the payload bay to cover the dome would be vastly easier to engineer than figuring out how to paint the nose. The paint won't survive entry. I have no clue what scorching it would do to the heat distribution during EDL. Probably nothing good.
Quote from: TheRadicalModerate on 08/10/2022 06:10 pmGiven that you want to recover that tanker (otherwise, you expend one for every lunar mission), you're probably restricted to bare metal tanks and TPS. But then that makes it essential minimize transit boiloff without solar white coatings, sunshields, etc. Table 1 in the article I linked to above said that applying 5 mils (about 1/8 mm) of Solar White paint on top of stainless steel reduced absorption of solar energy from 53% to 8.5%. That seems like a very fine improvement for very little cost. Especially if you only need to paint the nose.
Given that you want to recover that tanker (otherwise, you expend one for every lunar mission), you're probably restricted to bare metal tanks and TPS. But then that makes it essential minimize transit boiloff without solar white coatings, sunshields, etc.
The average albedo OTOH, is not so helpful. VLEO gives such a restricted view of the surface that worst case needs to be explored. It's not impossible that some areas will be so far from the average that mitigation aimed at the average will still allow significant boil off. This may not be, but it should be looked at.
The outer surface of the tiles is black vitrification. In other words, a glaze, and glazes are routinely pigmented. Is there a chance the black is choice, not chance? It would be very Elon like.
Quote from: OTV Booster on 08/11/2022 12:47 amThe outer surface of the tiles is black vitrification. In other words, a glaze, and glazes are routinely pigmented. Is there a chance the black is choice, not chance? It would be very Elon like.There's a reason they call it "Black Body Radiation"The black maximizes the Stefan-Boltzmann coefficient, which is the primary method of removing heat from a surface that can reach 1300degC. The equation has a term where the energy moved is proportional to temperature change to the 4th power, times that coefficient. Stainless steel is 0.35, the black borosilicate about 0.95 or better. The black paint they added behind the forward flaps by itself improves the coefficient to above 0.9 (but without all other benefits of the tile)Maybe this will help you if you your depot spends half time in the shade and half time in the sun in LEO. The half time in the shade allows the black surface to maximize the emission of the energy gained while being in the sun, while the time in the sun heats up the tiles (which by themselves have about 10MJ / square meter heat capacity, if I'm remembering the math correctly), with the tiles and the mineral fiber mat underneath acting as insulation.You'd have to do actual math to figure out if this is better or worse than reflecting the energy from the Sun before it is absorbed. There's a lot of that kind of math on the heat shield thread.The tiles themselves were designed to be bathed in intense heat for about 30 minutes and are allowed to pass a lot of that heat through to the interior as long as the skin doesn't exceed 800degC or so. Not exactly what you are looking for, but your scenario has vastly different things happening.If your depot needs to polar orbit the moon, there is no shade, so I doubt the black surface would be better than a sun shade or white surface.
One takeaway: Y2O3 exhibits infrared photoluminescence when laser pumped which explains its high emissivity in the IR.
I did a plot of black-body radiation distribution from 0 to 8 microns for the Starship tiles (1650 K) and the surface of the sun (5778 K). I picked 8μ because Solar White reflects light for shorter wavelengths (down to 0.25μ, below which neither the sun nor the tiles emit much of anything). It's pretty amazing because it reflects all but 0.2% of the sun's light.Unfortunately, this range is so broad that it's only going to have 5% emissivity for the tiles. That pretty much breaks them completely.You can't compromise and use something that's a bit less reflective because then it won't keep cryogenics cold. So I think this completely rules out the idea of painting the heat-resistant tiles with something like Solar White. It just won't work.Note: numbers on the x-axis are meters.
Quote from: OTV Booster on 08/11/2022 01:41 amOne takeaway: Y2O3 exhibits infrared photoluminescence when laser pumped which explains its high emissivity in the IR. I wouldn't get too hung up on the details, because I doubt that this is exactly what SpaceX will use on the LSS.My dumb explanation of solar white:1) It's really reflective in UV, visible light, and mid-infrared.2) It's really emissive in low infrared.It's the combination of these two properties that makes it effective.Edit: Hopelessly ninja'd. That's what happens when you don't pay attention to the "you might want to recheck the thread" warning after leaving something sitting overnight.
Nice plot, Can you include an estimate for the solar white paint?
Does anybody have an estimate for how much solar wing and radiator area it would take just to have a depot with bare metal tanks and a brute-force cryocooler that could keep up with VLEO levels of boiloff?
So it's 11-14kW to get rid of. What's the expected efficiency of practical cryo heat pumps? I suspect it's awful, AFAIR 10-20%. So 55 to 140kW range for the electric power.
You will have some radiation hitting some heathsield because the earth takes like 1/3 of the sky so it will illuminate the sides, but at a slant angle.
So BOTE result is 15kW to get rid of if you can't average and you have to deal with the momentary peak flux on the day side. If you could average things by for example allowing the pressure to rise on the day side you could get the power to get rid of down to 11-14kW depending on inclination.
So, stainless towards the earth it is.
Quote from: sebk on 08/12/2022 01:22 pmYou will have some radiation hitting some heathsield because the earth takes like 1/3 of the sky so it will illuminate the sides, but at a slant angle.Radiative flux from a sphere to a cylinder has a computable view factor. It's horrid, as you can see below, but you can use the pretty graph and see that it's going to be about 45% in VLEO. So that helps a little.
Re: Above.Of course this was only the Earth-side heating. It assumes pointing the nose at the sun effectively eliminates heat-xfer to the tanks. IMO, that's not a given, since the nose isn't reflective and isn't designed for the job of being a sun-shield.Quote from: TheRadicalModerate on 08/12/2022 05:49 pmQuote from: sebk on 08/12/2022 01:22 pmYou will have some radiation hitting some heathsield because the earth takes like 1/3 of the sky so it will illuminate the sides, but at a slant angle.Radiative flux from a sphere to a cylinder has a computable view factor. It's horrid, as you can see below, but you can use the pretty graph and see that it's going to be about 45% in VLEO. So that helps a little.It's going to be complicated because the angle of the ship to Earth isn't constant, but rotates 360 degrees each orbit if the nose points to the sun.
Quote from: sebk on 08/12/2022 01:22 pmSo BOTE result is 15kW to get rid of if you can't average and you have to deal with the momentary peak flux on the day side. If you could average things by for example allowing the pressure to rise on the day side you could get the power to get rid of down to 11-14kW depending on inclination.Using 15kW as the worst case, but at 24hr/day. (Ie, 15kJ/second.) And a heat-of-vaporisation figure for methane of 511 kJ/kg, that means boil-off-only cooling would cost 2.5 tonnes of methane per day.If your refuelling mission cycle takes 30 days from empty, through accumulation, to final target-vehicle transfer and empty again, you lose 75 tonnes of methane. I don't think it's reasonable to assume a faster cycle for a first generation depot.(Oxygen has worse heat-of-vap (213kJ/kg), so you're only hurting yourself if you let it evaporate. Likewise, proper heat capacity of methane is around 2 kJ/kg per K so ignored at this level of chicken scratchings.)Quote from: sebk on 08/12/2022 01:22 pmSo, stainless towards the earth it is.There's no reason to have bare stainless steel. Even if you intend to re-enter the depot-ships, you can paint the dorsal side with something reflective. Just going from your 60% best-case reflectance to a hand-waving 90% for a white coating, and assuming 2/3rds of your 15kW is coming from the bare SS and 1/3rd through the heat-tiles on the sides, that coating alone reduces heat-input by half. And much better if you can also give at least the side-tiles a non-black coating. Between them, coatings alone could get the heat-input down below 4kW. Boil-off-only cooling is thus reduced to below 2/3 tonnes of methane per day, maybe down to half a tonne/day. 30 day mission ops for fuel accumulation costs you 15 tonnes of methane.And 15 tonnes feels viable for a first generation depot.
Quote from: sebk on 08/12/2022 01:22 pmSo it's 11-14kW to get rid of. What's the expected efficiency of practical cryo heat pumps? I suspect it's awful, AFAIR 10-20%. So 55 to 140kW range for the electric power.So 150kW would be about 370mē of panels. That's 15% of an ISS, all to deploy and stow robotically--reliably--if you want to get the Starship back. Plus, you need the same amount of heat rejection. Presumably, you deploy that as the back of your solar array.That's not an easy engineering task. Not impossible, but not the simplest thing that SpaceX could do (which is what they'll do). That makes a non-recoverable depot a lot more likely.PS:Quote from: sebk on 08/12/2022 01:22 pmYou will have some radiation hitting some heathsield because the earth takes like 1/3 of the sky so it will illuminate the sides, but at a slant angle.Radiative flux from a sphere to a cylinder has a computable view factor. It's horrid, as you can see below, but you can use the pretty graph and see that it's going to be about 45% in VLEO. So that helps a little.The other computation that we need is how much heat power a particular boiloff rate gobbles up and expels. This obviously strays away from full ZBO, but if you have some max amount of electrical power in your budget, it should let you compute the boiloff rate.
That's not an easy engineering task. Not impossible, but not the simplest thing that SpaceX could do (which is what they'll do). That makes a non-recoverable depot a lot more likely.