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kbd512, after thinking on your comment, I think I was wrong to assume that SpaceX will use external propellant tanks. Certainly it is plausible that they will do so, since it reduces power needs, allows for faster reuse of BFSs in the long term, and since as louis pointed out, external tanks appear in their concept art.
But external tanks have two major challenges as well. They will have to be made extremely lightweight, and they must either be simple to assemble, or must be expandable. These are huge challenges and I should probably not assume they can be overcome. I'll see if I can get an estimate of how much energy it would take to keep BFR tanks at cryogenic tempertaures. I worry it will be quite high since the tanks on the BFR spaceship appear to have essentially no insulation. Musk has stated that the heat shield plates will be mounted directly on the tank wall, and the tank wall makes up the structure of the craft, so I don't see where there's any room for insulation. A single sheet of aluminized Mylar draped over the side of the tanks would go a long way, but I'm not sure if that could be easily mounted.
Valid point kbd512. There will definitely be some energy needed to keep the propellant cold. I expect the need is much less than in the NASA design, but I guess I shouldn't assume that without the math to back it up. The energy required to keep propellant cold depends heavily on whether produced propellant is stored in the return vehicle or in an insulated storage tank. In the former case, heat transfer into the propellant is very significant, but with the right shielding, heat transfer (and therefore energy to keep the temperature low) can be kept quite low. I am assuming the second case, which is what I expect SpaceX is pursuing.
Heat transfer on Mars is mostly through radiation, so it can be very effectively minimized using multi layer insulation. This paper tested the effectiveness of sheets of aluminized mylar each separated by 10 cm, and found thermal conductivity to be only about 10 mW/mK at -50 C. So with six layers providing 50 cm of shielding, the heat conductivity would be about 20 mW/K, so if the liquid oxygen was kept at -200 C and the air temperature was -50 C, heat transfer would be about 3 W/m^2 of tank surface area. My best guess is that the BFR spaceship has about 700 m^2 of tank surface area, but external storage tanks for it would probably more long and narrow, so maybe its better to assume 1400 m^2 of tank surface area, meaning 4.2 kW of heat would enter the propellant at night. I'm not sure of how much energy is required to reject that much heat, does anyone know how I might make an estimate of that?
The NASA design suggests nighttime power use of 75 kW, for a level of power generation that is less than 1/30th the size of the one that would be needed to refuel a BFR. So I expect that unless rejecting each watt of heat that enters that the propellant takes many tens watts to remove, the nighttime power use needed should be well below 1/10 of what it is in the NASA design. I guess I should add a bit of mass to the simple calculations I did before, that would probably bring the leverage down to more like 8.5 instead of 9.1
louis, I think that 17 MWh/t is very conservative for producing methalox from your own hydrogen supply. However if you're doing it the SpaceX way and mining water, you have to do twice as much electrolysis and you need energy to mine the water, so there are differences that drive the power needs both ways. From this spreadsheet I linked before, the power needs would be about 9 MWh/t. I am missing some power draws from the Zubrin paper in that spreadsheet, like liquefying the produced propellants and keeping them cold, and some heating elements and pumps and stuff. So I'd say with things I've forgotten and some numbers I may have made too optimistic, the power needs are probably more like 12 MWh/t.
Thanks 3015. V. interesting observation...essentially you appear to be saying the 17 MWh figure is a very conservative one i.e. we can probably get that amount of propellant for far less. Well, again, I think this would confirm that Space X's goal of a solar powered propellant production is feasible. What would be your guesstimate for a lowest possible energy input per tonne, taking full benefit of economies of scale on a Space X style mission? Would we be saving 50% overall?
kbd512, the 4.4 t mass is for panels, PMAD, and a RFC (regenerative fuel cell) to provide astronauts power at night. The RFC is not necessary to produce propellant. So the mass for producing propellant is 1.5t for panels plus 1 t for PMAD for a total of 2.5 t.
I managed to get my hands on a copy of the Zubrin paper that suggest 17 MWh per tonne. Here is the breakdown of power usage in that paper, values are in watts for a system that makes 1 kg of propellant per sol:
Cryocooler 165
Sensors and flow controllers 5
Reactor heater 40
Absorption column heaters 10
Electrolyzer 100
Absorption column three-way valves 2
Mars tank solenoid 0
Gas/liquid separator solenoid 2
CO2 acquisition Stage 1 144
CO2 acquisition Stage 2 74
Recycle pump 136
Total 678
The system described in the paper is for a sample return mission, it is safe to say that a larger system would experience very significant economies of scale. For example, the CO2 acquisition step in the paper suggests a power need of 5.38 kWh/kg of CO2. But this NASA paper suggests CO2 can be cryocooled for just 1.23 kWh/kg. The cryocooler power need is also much higher than would be needed for larger scale production, in Zubrin's system 4.07 kWh are required to liquefy 1 kg of propellant. The recycle pump should use much less relative power as well on a larger scale.
I want to be clear that I am in no way criticizing Zubrin's paper, it is excellent work that needs to be done to get us closer to going to Mars. It's just that the scope of his paper does not allow for the economies of scale a Mars Direct/BFR mission would experience since that paper is focused on a Mars sample return system.
louis, I think that 4.5 kWh figure is per kWh of "installed capacity", not kWh per m^2 per day.
I have done some number crunching on solar irradiance on Mars, here is a post I made on Reddit giving values for various latitudes. The peak is about 130 W/m^2 on average at the equator, which suggests about 3.21 kWh per day of irradiance, only about 0.963 kWh/m^2/day even for 30% efficient panels.
Here's an article on landing sites considered by SpaceX. They are focusing on Arcadia Planitia, Deuteronilus Mensae, Phlegra Montes and Utopia Planitia. All of those seem to be centered somewhere around 40 N.
I don't know much about Chryse Planitia, what are its advantages?
Edit: I should note these are landing sites considered for the now cancelled Red Dragon, but since those missions were partly scouting for later SpaceX landings I think it is safe to say that they will consider similar locations for BFR.
louis, I selected 50 N because it is close to the latitude of a location SpaceX sees as especially promising, Arcadia Planitia. The latitude does not make a huge difference anyway though, the mean at the equator would be 29.7 kW and at 30 S it would be 28.9 kW.
RobertDyck, I really like your requirements for a human base. I think that if Elysium Planitia has significant water ice, it will be an excellent choice for a base site. But it's possible that Elysium Planitia is not actually very water rich, we will have to scout the location first.
This paper improved the resolution of data from Mars Odyssey's neutron spectrometer and has this to say on Elysium Planitia:
The locally adaptive pixon reconstruction of the region around the proposed, buried sea is shown in figure9. The location of the water ice sea, 5°N, 150°E, corresponds to one of the locations with the highest epithermal neutron flux on the entire surface of Mars. This suggests that the top few tens of centimetres of soil, in this region, are unusually dry with <1wt.% WEH. The dry feature in the reconstruction extends beyond the region identified by Murray etal. (2005) and covers much of the smooth plains that are believed to be young basaltic lavas from Cerberus fossae (outline with a black contour in Fig.9). Additionally, Boynton etal. (2007), using Mars Odyssey Gamma Ray Spectrometer data, find this region to be enhanced in Fe. Taken together, these result suggests that the plate-like features at Elysium Planitia are unlikely to be a buried water ice sea but are, instead, young, Fe-rich, volatile-poor basalts.
SpaceNut, why assume heavy panels? NASA has a design for a 1000 m^2 array that has a panel/structure mass of 1.5 t and a PMAD mass of 1 t. They estimate that at 50 N the array could produce a mean power of 26.1 kW.
In 600 sols, the array would produce 0.0261 MW * 24.67 h/sol * 600 sols = 386 MWh. 386 MWh / 17 MWh/t = 22.7 t of fuel from 2.5 t of equipment, which is 9.08 t of propellant per t of equipment.
Oldfart1939, I strongly agree about nuclear power. Easier to set up, runs all the time, and is more reliable. I've done the mass estimates with solar because that's what Musk intends to use.
louis, in slide 31 in this presentation, it says that the propellant plant will be set up in 2024, at the time of the first manned landing. I think it is because Musk thinks it would be too complicated to set up a solar array and water processing without boots on the ground.
Hi louis, I'm actually kind of fudging the mass of cabling, voltage conversion, power protection, etc right now by using the panel mass parameter to represent the mass of 1 m^2 of panels plus the associated power management and distribution equipment. I've added PMAD to the equipment mass section, but I haven't filled in numbers for it yet since I can't find any modern references for PMAD mass.
I think that the current SpaceX plan does not involve propellant manufacture starting before humans arrive to finish setting up the propellant manufacturing plant. I was surprised when Elon Musk mentioned this (either during the 2016 or 2017 BFR presentation I think) since it introduces a significant risk relative to a plan like Mars Direct.
I wish I had access to the paper where Zubrin achieves a value of 17 MWh per tonne of propellant, that value seems very high to me. Most of the power in a system like that should be for electrolysis, cryocompression of CO2, and RWGS, which even at low efficiencies should add up to far less than 17 MWh/t. Also, Zubrin's system is much different from what SpaceX is planning, if you don't bring the H2 you have to mine water and you have to electrolyze twice as much water.
I have created a spreadsheet here to estimate the power and equipment requirements for fueling a BFR. The parameters I have used come mostly from this PhD thesis. The spreadsheet is read-only, but you can save a copy and enter your own parameters. The parameters I have used suggest about 100 t of solar panels and other equipment is required to fuel one BFR spacecraft on 600 days, though that value is very sensitive to panel mass which I am very uncertain about. If that value is correct, then 2 BFRs can bring enough equipment to fuel 3 ships per transfer window. This allows one ship to return as long as production is at least 1/3 of the expected value, resulting in a very safe margin, which as Oldfart1939 has stated is very important.
Change the .htm in the address to .html, the link worked for me after doing that.
That presentation suggests a cost to LEO that's even better than $30/kg! If 6 booster flights cost $11M and 5 tanker flights cost $8M like in the presentation, then a booster flight costs $1.83M and a tanker flight costs $1.6M. If a cargo flight costs as much as a tanker one, that's $3.43M to deliver 300 t to LEO, or only $11.50/kg.
RobertDyck, GW is referring to high molecular weight atoms in the shielding material. I didn't know about those results from MARIE though, thanks for sharing that.
GW, you are indeed correct about secondary radiation from regolith. This paper simulated radiation dose with different levels of regolith shielding (results are in a table on the last page) and found that absorbed dose actually increased with the first 10 cm of shielding.
The SpaceX BFR presentation had a slide with the ship docked with the ISS, I think they must intend for at least some versions of the BFR ship to have docking capability.
I just came across this recent paper that models the radiation dose on Mars from solar particle events. Their conclusion is that even the Carrington Event would result in a mild surface radiation dose equivalent of under 20 mSv (see Fig. 10 on page 23). It looks like we can ignore solar particle radiation on Mars.
I think I underestimated how big a solar flare can be. The largest we've measured was an X-28 in 2003, 43 times the flux of the M-6.5 flare that caused the SEP event that Curiosity measured. If the relationship between flare flux and radiation reaching Mars' surface was linear, that would still mean only 1 mSv for an X-28 flare. But the relationship could be very different, which makes the danger of SEP events uncertain. You've convinced me that this is a potential problem.
I think the biggest question is "do high energy SEP events have significantly higher energies per particle?". Reading through the paper I linked before again, I noticed the authors say that Mars' atmosphere blocks particles with energies of less than 150 MeV. So almost all of the energy from the flare Curiosity measured was blocked by the atmosphere. But if larger events also have a much larger proportion of particles above 150 MeV (which is practically in the range of cosmic rays), then that could pose serious danger to astronauts on Mars' surface. Have we taken any radiation measurements on Mars during larger SEP events?
One final note, shielding in gale crater is actually 20 g/cm^2, slightly higher than the average because of its low altitude. So if we visit low altitudes on Mars, we'll have a bit more protection than we otherwise would.
GW, I agree that we do have to worry about solar particle emissions in space. But this topic is about radiation protection on the surface of Mars, where there already is some thin shielding.
Mars' atmosphere is about 600 Pa, so the weight of Mars' atmosphere is about 600 N/m^2. That means that the mass per area of Mars' atmosphere is 600/3.7=162 kg/m^2, or 16.2 g/cm^2. This is very close to the level of shielding you suggest for protecting against solar energetic particles.
Do we really have to worry about particle radiation from the Sun? I'm sure everyone here has read this paper on Curiosity's radiation measurements on Mars, it details the one solar energetic particle event Curiosity had witnessed at the time of the study. The dose equivalent for that event was only 0.025 mSv, a small fraction of the dose equivalent rate from GCRs for a normal day on the surface of Mars.
Excellent points, louis. I'd like to expand on number 2. I expect that within the first couple Mars expeditions, we will grow some low energy intensity foods like lettuce and herbs, mostly for the psychological and digestive benefit. These could easily be grown in artificial lighting, the Mars-Lunar greenhouse was able to grow a whopping 54 grams of lettuce per kWh. But to grow foods with lots of calories will require natural lighting.
I think the next stage will probably be thin film naturally lit greenhouses shipped from Earth. It may be hard to make them much lighter than the food they are replacing though, so they may be economical for some foods but not for others. In particular, foods with high fat content may be easier to bring from Earth since fat contains so much more energy per mass (9 kcal/g vs 4 for carbs/protein).
The final step, which I think will take about 10 years, is making greenhouses from in situ resources. I agree with RobertDyck's assertion that glass will be used in the first ISRU greenhouses on Mars. There is just no other material with the UV and temperature tolerances needed that is simple to produce. Once these greenhouses are in place all foods except the exceptions you mentioned will probably be grown locally.
My proposed design is a long-narrow greenhouse aligned perfectly east-west, with flat mirrors on both sides, full height from ground to roof, angled 45° to reflect overhead light into sides.
This is a really good idea. The mirrors add very little to the mass of the greenhouse but greatly increase yield, and also mean much more heat flowing into the greenhouse during the day, so it should make keeping a greenhouse warm easier. This design suggests a greenhouse that is cylindrical or close to cylindircal, which I like as well.
Before we can make greenhouses with in situ resources, it is going to be difficult (though not impossible) to grow food using less mass from Earth than the mass of the food itself. If we bring over dry food and rehydrate it on Mars, we can probably get away with bringing about 300 kg/person/year. Let's also assume that it takes 100 m2 of crop area to grow food for one person, and a five year lifetime for a greenhouse. Then each m2 of crop area must be enclosed with only 15 kg of mass from Earth. If we assume a cylindrical greenhouse and ignore the ends, then we need pi m2 of greenhouse wall area per m2 of crop area, so we will have to get by with less than 5 kg/m2 of wall area. That's as much as a 2 mm thick sheet of glass or a 2.5mm sheet of PCTFE.
So the walls of any greenhouse made on Earth to be transported to Mars must be thin. If we have thin walls, then we must also have low pressure forces since most suitable clear materials do not have high tensile strengths. Pressure forces are proportional to the radius of curvature of the greenhouse, so having a small diameter greenhouse allows the walls to be thinner. But having a smaller diameter greenhouse also limits the maximum size of plants and will have less thermal intertia to deal with nighttime cooling, so there is a tradeoff. My guess is that a good size would be just large enough for a person to walk around in. There's no reason the greenhouses couldn't be very long though.
On heating, I think we have to find a way to use almost entirely solar heating. Near the equator on Mars, solar irradiance is around 3 kWh/m2/day, with mirrors, that can be brought up to 5 kWh/m2/day. There's no way we can provide anywhere near that amount of heat energy from photovoltaic or nuclear energy, so we'll just have to find a way to minimize heat losses. I'm not sure whether conduction/convection or radiation is the primary concern. Radiation can be limited by using materials that do not transmit long wavelength infrared in the range emitted by objects at around 290K, and by putting a reflective cover on the greenhouse at night. This cover could be provided by folding the mirrors mentioned by RobertDyck over the greenhouse at night, but that would require a mechanized system or a daily EVA to do so. Conduction can be limited by using multiple greenhouse layers and by covering the greenhouse at night with a material with very low heat conductance like aerogel.
I think it would be more efficient to make synthetic rubber than to grow it, solar panels and industrial chemical processes have higher levels of energy efficiency than plants do. Styrene-butadiene synthetic rubber is the most common on Earth. Styrene can be made by pyrolysis of methane to get acetylene, then trimerization of acetylene to benzene, combination of benzene with ethylene to get ethylbenzene, then dehydrogenation of ethylbenzene to get styrene. Butadiene can be made from ethylene which is hydrogenated to get ethanol and then combined to yield butadiene.
I like the fiberglass gauze idea. Some glass fibers have ridiculous tensile strengths, and glass is resistant to radiation and a wide range of temperatures.
I'm somewhat skeptical of the idea of squishing the shape of a greenhouse using tethers though, I worry the force on the tethers would be too much. For a hemispheric dome where all the anchors are pulling straight down, I think that the force trying to lift the dome upward is equal to the area under the dome multiplied by the pressure in it. The dome pictured above looks to be at least 6mx12m, if we assume that and a low pressure of 20 kPa, then the force the anchors would have to counter would be 72 m^2 * 20 kPa = 1.44 MN. If there were 36 tethers spaced around the dome at 1 m intervals, each would have to bear 40 kN, which on Mars is the force exerted by 40/3.7=10.8 tons. Is that a level of force that we could manage by drilling anchors into rock? I don't know but it seems like a huge amount.
You did the math on greenhouse thickness in page 1 of this thread in posts #11 and #18. I haven't checked the math but your results are in the ballpark of math I have done for cylindrical greenhouses.
The "pillows" will definitely bow outward, that's why they're used. Force on the wall of a pressure vessel is proportional to the radius of curvature, and the bowed out sections have low radii of curvature and therefore low forces on them. As long as the tethers are close together though I don't think it would change the shape of the vessel much. Or am I misunderstanding your argument?
I'm intrigued by your dome idea. Is aerogel transparent enough for such an idea to be used for a transparent enclosure?