Indoor farming update

Antius · 2014-10-11 05:36:27

I ran the numbers for a 50m diamter dome, made from medium density polyethlene (yield strength 12MPa) and safety factor 5 (12.5cm thick). Inside temperature = 15C, outside temperature =-60C. This is the average temperature at a latitude 60N on Mars, the same latitude as Helsinki on Earth.

Initial results were a radiation and convection heat loss of 60W/m2 of dome, or 120W/m2 of internal floor. For conduction, I assumed a constant ground temperature of -20C at 1m depth and a thermal conductivity of 1W/mK for wet clay-like soil. Conduction heat losses are then 35W/m2 of floor.

Some of the results seem suspect to me, so i am going to double check. The blackbody heat loss at a temperature of 288K (15C) and background 213K (-60C) is 219W/m2 of dome surface, which would be 437W/m2 of floor. But we would expect some insulation from the dome itself.

louis · 2014-10-12 04:33:15

Antius wrote:

I ran the numbers for a 50m diamter dome, made from medium density polyethlene (yield strength 12MPa) and safety factor 5 (12.5cm thick). Inside temperature = 15C, outside temperature =-60C. This is the average temperature at a latitude 60N on Mars, the same latitude as Helsinki on Earth.
Initial results were a radiation and convection heat loss of 60W/m2 of dome, or 120W/m2 of internal floor. For conduction, I assumed a constant ground temperature of -20C at 1m depth and a thermal conductivity of 1W/mK for wet clay-like soil. Conduction heat losses are then 35W/m2 of floor.
Some of the results seem suspect to me, so i am going to double check. The blackbody heat loss at a temperature of 288K (15C) and background 213K (-60C) is 219W/m2 of dome surface, which would be 437W/m2 of floor. But we would expect some insulation from the dome itself.

Thanks for the help with that!

I went on a calculator site that gave me 1963 sq m for a 50m diameter circle. Multiplying 1963 by your 437 watts I get 857831 watts or about 858 Kws. Seems a lot but then a 50metre diameter dome is pretty big.

Although no doubt the figures don't translate, for a cubic hab of 4metres length, that would be a mere 7Kws approx., well within the capability of a PV panel and storage battery system.

And couldn't that be lowered through use of aerogel?

Last edited by louis (2014-10-12 04:34:05)

RobertDyck · 2014-10-12 06:32:34

Woo hoo! Someone who can do thermal calculations!

Antius, could you do some calculations for me? This is for a Mars Direct style science mission.

floor area: 4 metre x 10 metre
PCTFE film, tensile strength (Machine Direction/Transverse Direction) 49.8/49.5, bursting strength 19.4 psi
webbing straps every 2 metres, composed of Spectra uhmwpe (Ultra-High Molecular Weight PolyEthylene)
Straps held down with a "tent peg" at each end. For large tents on Earth, pegs are made of steel. However, titanium alloy is lighter, so we would use that on Mars just to reduce transport weight.
I have assumed 2 mil thick PCTFE film. As rip-stop, thermally bond fibreglass gauze to the inner surface. Is that sufficient?

Temperature control. There are several things we talked about to control heat loss. Two layers of polymer film, with argon gas between the two layers, the "gap" pressurized more than interior but less than Mars ambient, and spectrally selective coating. But as a first-cut, could you calculate for one layer of polymer film, and no coating?

For the day assume -8°C atmosphere temperature, and at night -77°C. Those were measured one solar day (sol) by Mars Pathfinder. Ground temperature is -55°C at 1 metre depth. Assume inside temperature +20°C.

Habitat air: let's keep it simple. Partial pressure of oxygen 2.7 psi, partial pressure of nitrogen 3.3 psi, which adds up to total pressure 6.0 psi. Apollo and Skylab had O2 pp 3.0, N2 pp 2.0, total 5.0; so this is slightly more total pressure. Maximum partial pressure nitrogen is 3.6 assuming suit pressure is 3.0 psi, and zero pre-breathe time for decompression. With pure oxygen in the suit, that's the same partial pressure O2 as Earth sea level, yet habitat partial pressure is more than many Earth cities, and with suit O2 at 10% more than habitat, the suit could endure 10% pressure loss with no danger to astronauts. Keeping habitat nitrogen a bit less than the maximum for zero-prebreathe time for decompression, it adds a safety margin to avoid the bends.

Last edited by RobertDyck (2014-10-12 06:53:37)

GW Johnson · 2014-10-12 14:02:29

Very interesting. Apollo/Skylab atmospheres were 3 psi O2 + 2 psi N2 for 5 psi total. That's 60% O2 at low pressure, and it was enough for adequate flash fire prevention? And the zero-decompression limit for N2 is 3.6 psi going into pure O2 in the suit? Those are handy criteria to have.

For longer-term habitations, are there any criteria governing exposure to more than 21% O2 and less than about 10 psi total pressure for successful pregnancies? I've heard there are problems with this, I'm guessing based on lab animal studies.

For the heat transfer calculations, what one has to remember here is that the clear-ish plastic domes or tents y'all are talking about are not the surfaces radiating IR away into the environment. It is the warm solid surfaces inside that are the source of the IR radiation emitted toward the environment, not the dome/tent itself. The inside warm-surface IR just has to transmit at less than 100% efficiency through your layers of plastic and interwall gas spaces.

It won't take the outer layer or two very long at all to chill very cold, and so be thermally-embrittled (vulnerable to shattering), even if everything starts out warm (which it won't). You'll have to heat this stuff as you lay it out on the cold, cold ground setting things up. Otherwise it'll break as you erect it.

GW

RobertDyck · 2014-10-12 16:30:25

For fire protection, the issue is partial pressure of O2. Earth at sea level is 14.7 psi total pressure, with 20.9% O2, so 3.07 psi partial pressure O2. Today the barometric pressure for Boulder, CO, is "29.88 in" according to www.weather.com; assuming that's inches of mercury at 0°C, that works out to 14.675 psi. With 20.9% O2, that works out to 3.067 psi partial pressure O2. You have to get high for significantly lower O2.

Combustion rate is directly proportional to partial pressure O2. We had this discussion before; many members thought other gasses such as nitrogen would "buffer" combustion. But nitrogen and argon don't, they're irrelevant. Water can, but you need very high humidity to significantly affect combustion rate. Fire fighters often flood a room with water to cool it down, and buffer combustion.

Apollo used 3.3 psi pure oxygen in their suits, so 10% pressure loss would still be 3.0% psi O2. They started with KSC ambient before launch, but decompressed as they ascended. No pumping necessary if you just let air vent into the stratosphere or vacuum of space.

Results of serious studies by divers and the Navy found a ratio to prevent the bends. For zero pre-breathe time, the maximum partial pressure of N2 is 1.2 times the total pressure of the lower pressure environment. Apollo stayed way below that. They used 2.0 psi partial pressure N2, and suit total pressure 3.3 psi. For that suit pressure the maximum partial pressure N2 would be 3.3 * 1.2 = 3.96 psi, but Apollo used 2.0 psi. Huge margin of safety. But lower cabin pressure also meant less stress on the hull. And less decompression time in the air lock. And the air lock never scavenges all air, so less air loss when using the air lock.

RobertDyck · 2014-10-12 16:48:14

GW Johnson wrote:

It won't take the outer layer or two very long at all to chill very cold, and so be thermally-embrittled (vulnerable to shattering), even if everything starts out warm (which it won't). You'll have to heat this stuff as you lay it out on the cold, cold ground setting things up. Otherwise it'll break as you erect it.

This is one reason I want to use PCTFE. It becomes embritted at -240°C; the coldest temperature I saw in data from Mars Global Surveyor the first Mars year was -140°C at the south pole. So PCTFE doesn't become embritted until 100°C colder than the coldest temperature on the planet. This stuff just loves cold! Actually, it becomes soft if heated above +132°C, and exterior of suits in Earth orbit can get to +150°C. However, the warmest temperature I saw on Mars was +24°C. Some people claim they saw a report of +30°C, but I haven't confirmed that. Never the less, that's way below +132°C. So PCTFE must be packed with no stress during transport, and just let it cool down on Mars before opening. Don't you just love a material that has to cool before opening on the cold, cold ground of Mars?

Another feature of PCTFE is that it's ridiculously transparent. The second most transparent polymer ever invented. Extremely low optical index and over 88% light transmittance deep into UV. Over 90% into near IR. Only Teflon AF is more transparent; that stands for "Amorphous Fluoropolymer". Its molecular structure is amorphous, like glass. Not as hard as glass, but harder than most plastics. But that means Teflon AF is not a flexible film.

By the way, what is the embrittlement temperature of Spectra?
::Edit:: Found it! Wikipedia says it becomes brittle below -150°C.

Viking 2 recorded surface temperature for more than one Martian year. The low it recorded was -111°C. I checked the entire dataset. Don't have data from Spirit, Opportunity, or Curiosity. Raw data from them is best described as a flood, so I'm sticking with published articles.

Last edited by RobertDyck (2014-10-12 17:06:11)

GW Johnson · 2014-10-12 18:36:54

The "official" elevation of Boulder, CO, is 5430 feet above mean sea level according to a quick internet search. That corresponds to an altitude pressure of 24.50 inches mercury in a standard atmosphere. Sea level standard is 29.92 inches mercury or 14.696 psia. Also 101.325 KPa.

The weather barometer as reported is corrected to sea level, so 29.88 inches of mercury is what a barometer would read at the bottom of a well 5430 feet deep (down to sea level). Standard 29.92 minus 29.88 says the atmosphere was 0.04 inches mercury lower than at-elevation standard that day. Subtracting .04 inches from the at-elevation standard value of 24.50 inches says that the local at-surface (uncorrected) barometer in Boulder was about 24.46 inches mercury that day. That's 12.014 psia.

Using 20.94% by volume (partial pressure) oxygen, the partial pressure of O2 that day in Boulder should have been 2.516 psi. But that figure would be for absolutely zero humidity. The partial pressure of the water vapor at any given humidity displaces the dry air pressure by that same amount. On a very humid day, that might be 3% of an atmosphere, leaving not 12.014 psi of dry air, but only 11.654 psi of dry air. 20.94% of that reduced value is O2, for a partial oxygen pressure of 2.440 psi.

Ordinary people function just fine at lower O2 partial pressures than that, all the time. Dry air partial pressure O2 at 10,000 feet (standard) is 2.116 psi, less if humid, less if a weather low on any given day. There are people who live at 15,000, even 20,000 feet.

It's even worse inside the wet lungs, where the water vapor pressure is the equilibrium value (straight out of the standard steam tables) for body temperature, independent of total air pressure. That's about 6% of a std atmosphere displaced dry air.

BTW, I like the cold service temperature for the PCTFE plastic. That's better than anything I ever heard of before. Should work fine. Does it outgas volatiles in any way, and is that outgassing a function of external atmospheric pressure? If so, it will age (and embrittle) quite fast in the laboratory vacuum that is the Martian atmosphere. Just something else to be worried about. Myself, I dunno.

And I really like that 1.2 ratio for preventing bends. That's a really good, useful criterion.

GW

Last edited by GW Johnson (2014-10-12 18:47:26)

Antius · 2014-10-13 08:11:53

louis wrote:

Antius wrote:
I ran the numbers for a 50m diamter dome, made from medium density polyethlene (yield strength 12MPa) and safety factor 5 (12.5cm thick). Inside temperature = 15C, outside temperature =-60C. This is the average temperature at a latitude 60N on Mars, the same latitude as Helsinki on Earth.
Initial results were a radiation and convection heat loss of 60W/m2 of dome, or 120W/m2 of internal floor. For conduction, I assumed a constant ground temperature of -20C at 1m depth and a thermal conductivity of 1W/mK for wet clay-like soil. Conduction heat losses are then 35W/m2 of floor.
Some of the results seem suspect to me, so i am going to double check. The blackbody heat loss at a temperature of 288K (15C) and background 213K (-60C) is 219W/m2 of dome surface, which would be 437W/m2 of floor. But we would expect some insulation from the dome itself.
Thanks for the help with that!
I went on a calculator site that gave me 1963 sq m for a 50m diameter circle. Multiplying 1963 by your 437 watts I get 857831 watts or about 858 Kws. Seems a lot but then a 50metre diameter dome is pretty big.
Although no doubt the figures don't translate, for a cubic hab of 4metres length, that would be a mere 7Kws approx., well within the capability of a PV panel and storage battery system.
And couldn't that be lowered through use of aerogel?

I have corrected my spreadsheet, it contained an error before.

Louis: The heat loss from your 4m cube is 40W/m2 of surface. I assumed a 1cm thick layer of aerogel, with thermal conductivity of 0.005W/mK. External temperature is taken to be -60°C and internal temperature is 15°C. For a perfect cube, this gives a total heat loss of 3200W from the upper 5 surfaces by radiation and convection, and 560W through the floor by conduction. That equates to 235W/m2 of internal floor area.

Some important points: Most materials transparent in visible light are completely non-transparent to infrared light. Any heat lost by radiation from the dome internals must first be transmitted to the dome interior surface, conducted through the dome and then lost by radiation and convection to the Martian surroundings.

(1) The heat losses are proportional to the surface area of the structure. In order to minimise heat losses, the surface area of the structure should be minimised. So a dome shaped structure approximating a sphere is much better than a cube if you are interested in conserving heat. A dome with high radius of curvature is better still and best of all would be a continuous roof, parallel to the floor.

(2) Heat loss from the exterior is absolutely dominated by thermal radiation. Convection in the thin Martian atmosphere accounts for only ~10% of the heat loss and this is even further reduced is a Plexiglas shield dome is used, as convection is no longer forced by wind.

(3) Most of the heat loss takes place at night. If average daytime temperatures (12 hours) are -40°C and average night time temperatures (12 hours) are -80°C, then some 60% of total heat loss takes place at night. Heat loss can therefore be cut by at least half by covering the dome with reflective covers during night-time. One arrangement might be to have aluminium foil coated plastic ‘petals’ that are hinged at the bottom and attached by cable to a winch-house at the apex of the dome. Sometime after dawn the petals are winched down and are pulled up again at sun-set. These could also be raised during dust storms to protect the dome from abrasion.

(4) The effect of aerogel insulation is very significant. If you change your greenhouse from a 4m cube to a hemisphere with the same floor area, then you total heat loss is 80W/m2 of internal floor from convection and radiation, plus 35W/m2 by conduction. However, when I performed the same calculation for the 12.5cm thick polyethylene dome (which has thermal conductivity of 0.4W/mK) total heat losses due to radiation and convection were 230W/m2 of floor, plus 35W/m2 conduction. Adding just 1cm of aerogel cuts total heat loss by 57%.

For a hemispherical 50m diameter greenhouse, equipped with 1cm of aerogel insulation and including petals, the total heat losses (including conduction) would be ~70W/m2 of floor area. This can be compared to time-averaged insolation levels at different latitudes on Mars (ignoring atmospheric effects):

• Equator: 190W/m2.
• 30°N (during equatorial summer): 165W/m2.
• 30°N (local summer solstice): 181W/m2.
• 30°N (local winter solstice): 99.2 W/m2
• 60°N (during equatorial summer): 95W/m2.
• 60°N (local summer equinox): 210 W/m2
• 60°N (local winter equinox): 10.62W/m2

These figures suggest that hemispherical domes, with aerogel insulation and night-time reflectors, can remain warm without any additional heating at all latitudes during summer. In fact, keeping them cool may become a problem at the height of summer.

At high latitudes (60°N), winter insolation would provide useful heating only for a few hours per day and would appear to be insufficient to keep internal temperature above 15°C. To remain warm here, the dome could rely upon stored summer heat. Some heat would naturally be stored in the structures and soil beneath the dome. At high latitudes in mid-winter, this may be insufficient to keep temperatures above freezing. In this case, hot water storage of summer heat may provide a good option. An average heat deficit of 50W/m2 for 90 days would be 108kWh per m2. This much heat could be stored in 2057 litres of water, with an initial temperature of 60°C. To store enough summer heat for a 50m diameter dome for a 90-day mid-winter, a 4million litre insulated water tank would be needed. If this were spherical, it would have a diameter of 20m. It could therefore be buried under the floor of the dome.

Additional heat could be provided by integrating living spaces and industrial areas into the domes. This would allow waste heat to compensate for environmental heat losses. Buildings within domes would be adobe construction. Agriculture could take place on their flat roof tops.

RobertDyck · 2014-10-13 18:20:42

GW Johnson wrote:

BTW, I like the cold service temperature for the PCTFE plastic. That's better than anything I ever heard of before. Should work fine. Does it outgas volatiles in any way, and is that outgassing a function of external atmospheric pressure? If so, it will age (and embrittle) quite fast in the laboratory vacuum that is the Martian atmosphere.

PCTFE technical specifications:
Aetna Plastics
Boedeker Plastics
Daikin Industries

Boedeker Plastics wrote:

PCTFE also has extremely low outgassing (0.01% TML, 0.00% CVCM, 0.00% WVR when tested per ASTM E-595-90), so it is suitable for use in aerospace and flight applications.

Daikin Industries wrote:

Gas Permeability: cm^3, cm/cm^2, sec, atm
       PCTFE         FEP
N2  0.18*10^-10    120*10^-10
O2   1.5*10^-10    370*10^-10
H2  56.4*10^-10  1,080*10^-10
CO2  2.9*10^-10    970*10^-10
Moisture permeability constant: 0.2 g/m, 24 hours
Water absorption 0.00 %, 24 hours
0.0 % by weight, 168 hours

RobertDyck · 2014-10-15 07:33:55

Antius, please, could you run the numbers for the greenhouse I described?
Light transmittance for PCTFE and the specrally selective coating are in post #15 on the first page of this discussion thread. Other criteria from post #28 above.
Final configuration is with the spectrally selective coating, and two layers with argon filling the gap.

For Mars temperatures, recent rovers have provided a flood of data that I find overwhelming. It's easier to use data from Mars Pathfinder. Weather data is available here. That page has tables of numbers, but I'll post graphs. Temperature for all days recorded:

And more high resolution data for a little over one solar day here:

This figure shows temperature measured by the Pathfinder ASI/MET instrument, 1.0 m above the level of the lander solar panels, for a period of continuous measurement during Sols 68 and 69. Measurements are obtained every 4 seconds.

Questions:

Can this greenhouse maintain temperature by sunlight alone?
Does it require an aluminized mylar curtain at night, or is the spectrally selective coating enough?
Do we need to insulate the floor?

One flooring option is "bubble wrap" filled with argon. That is, transport the polymer empty, but fill with argon extracted from Mars atmosphere and seal bubbles in the polymer sheets to create "bubble wrap". Lay that on the ground beneath soil trays. Perhaps plastic open grid floor tiles to talk on, with sides that raise the grid above the ground enough so we could put more "bubble wrap" under the floor tiles. That would also mean astronauts don't walk on the polymer film pressure envelope.
images?q=tbn:ANd9GcQFzfOuO_D2NL3uMD1aof9-r3yosPx4hxswvkvSKznkNGcK1oem

Antius · 2014-10-15 08:41:08

Input parameters

Inside temperature: 20C.
Outside temperature -8C (day), -77C (night).
Structure dimensions: hemispherical poly-tunnel, 4m in diameter, 10m long. Surface area = 75.4m2.
Thickness = 2mm.
Material thermal conductivity = 0.24W/mK.
Internal pressure = 0.4 bar
Internal heat transfer coefficient = 7.4W/m2K
External heat transfer coefficient = 1.4W/m2K (assumes a 10m/s wind speed). For such a low viscosity gas at low pressures, h can be assumed to be proportional to wind speed.
Pathfinder latitude = 19.5°.

Results

Conduction heat loss = 75W/m2 floor (3000W whole structure - both night and day).
Radiation and Convection (from dome surface) = ~95W/m2 (day)
Radiation and Convection (from dome surface) = ~275W/m2 (night).
Total heat loss rate (day) = 10,163W (254W/m2 floor)
Total heat loss rate (night) = 23,735W (593W/m2 floor)
All heat loss rates are accurate to within ±10%.
Heat loss proportions (night): Radiation 54.5%; Convection 32.7%; Conduction 12.76%.

Insolation at Pathfinder Site (~20° North)

Summer solstice: 4.735kWh/m2/day (average of 395W/m2 over 12 hour day)
Equinox: 4.2kWh/m2/day (average of 350W/m2 over 12 hour day)
Winter solstice: 3.07kWh/m2/day (average of 307W/m2 over a 10 hour day).

Conclusion

The poly-tunnel heat losses can be balanced by insolation at all times of the year during day. During summer months, keeping the tunnel cool during day may even become a problem.
Thermal protection will be required during night-time at all times of the year. Both radiation and convection heat losses can be cut to virtually zero by covering the poly-tunnel with a reflective cover at night. A double wall with xenon filling will reduce heat losses by 40% at most, as the gas-gap will be transparent to infrared radiation. Transparent solid filler (aerogel?) would be more effective, as it would block outgoing IR radiation, but would also attenuate incoming solar radiation.

To bring the top 1m of soil up to 20°C, some 50kWh/m2 of heat is needed. On this basis, conduction heat losses will be much greater than the steady state conditions indicated for several weeks after erecting the poly-tunnel. Conduction would probably be sufficient to hold internal air temperature beneath zero for the first week or two. So build your poly-tunnel and give it a week or two to warm up before planting anything.

RobertDyck · 2014-10-15 09:14:57

Wonderful! Thank you!

So we don't need bubble wrap on the floor. Just wait a couple weeks. That's a lot simpler.

A couple points. I proposed polymer film thickness of 2 mil. That converts to 0.0508 mm.

Second point: your response appears to imply that you did not include any spectrally selective coating. Notice the second graph in post #15; I propose using the coated for line 5. For IR in the 1.0 micrometre wavelength, transmittance is approximately 10%. For the 3.0 micrometre wavelength, transmittance is approximately 45%. That material is intended to cool buildings in the US, because it transmits more radiant heat from warm things like the floor and furniture, less from extremely hot things like the Sun. I don't have any other graph. That graph is for "low-e glazing", which I believe only uses silver oxide. NASA uses that plus gold and nickel to block more UV. I don't have a graph for NASA's coating. But one question was whether a reflective curtain is necessary if we use this spectrally selective coating. The coating does reflect some IR.

Impaler · 2014-11-10 01:03:20

I'm skeptical that transparent roofing for the use of natural will even be worth it, rather fully enclosed and artificially lit growing environments will be the way to go.

Optimal plant growth can be done with 22 hours per day of light, so relying on sun light alone your giving up 10 hours right off the bat. But Mars has axial tilt and a varrying day length through a year that is twice as long as Earths, unless your going to store food for nearly a year (which will take a whole set of freezers, cans and various equipment) you need to size your greenhouse for solar minimums which are going to be 8 or less. So your already 36% potential from duration effects alone.

Then account for Mars's weaker light, that 44% of Earth intensity, so your growth rate is now 15.8% of the fully enclosed alternative, if the transparent roof is 90% transparent (a very good rate) it's down to 14%, and that before we crank up the light intensity (per unit time) in the enclosure above Earth equivalent by using selective grow-light frequencies.

Their is also the nature of light at the Martian surface, it is being heavily scattered by dust and is attenuated in the blue end of the spectrum but if this is the the part of the spectrum plants need and if it is more or less severe then the attenuation of the same wavelengths that happen on Earth I don't know, I'm not able to find a detailed analysis of Martian sunlight that examines it from a photosynthesis perspective, only from a solar power perspective. Generally I think plants will care much less about diffuse light then a solar panel will. Then you also have to factor in the compensation point of photosynthesis, as plants are respiration continually they don't achieve a net fixation of carbon until light is above a certain level, this means growth decreases with reduced light are worse then linear, a a 70% reduction in light could mean an 80% reduction in growth rates.

Now also consider that in an artificially lit growing environment we can stack our plants at least 3 levels high, but with natural light were limited to single layer aka 1:1 or lower floorspace:growingspace ratio. So the stacked and artificially lit greenhouse could come out at only 5% of the size of the naturally lit one while growing the same food output.

Sure it will take considerable energy to light it but the compact size means were saving a LOT of building materials to make it (which could be allocated to power instead) and the materials can be far more durable STEEL and ALUMINUM not transparent plastics which have short lifespans in harsh conditions. And it's simpler to heat, though the math so far dose not seem to indicate that heating is going to be that demanding in energy and cooling might actually be needed. I think the degree of consistency and compactness that full enclosed and artificially light growing environment is going to win out over the energy costs.

RobertDyck · 2014-11-10 07:12:05

Build on the equator, where it's relatively warm. That means no winter; sunlight during the day all year. And a temperature controlled greenhouse means crops all year; again no winter.

Nature of light at Mars surface: not scattered at all. Mars atmosphere is rediculously thin. Clouds are extremely rare, the number of clouds on the entire planet at any given time can be counted on one hand. That means direct sunlight; less scattering than Earth on a hazy day. I didn't say overcast, I said hazy. But plants do not care about diffuse vs scattered light, that is only an issue for solar concentrators. Photovoltaic panels also collect whatever light is available, scattered or not. Plants have evolved for conditions on Earth, so they use whatever light is available. Plant leaves work on a hazy day, they work when it's overcast.

So you want artificial light. Ok. Now your power system fails so badly that you can't get it working for 2 weeks. How do you breathe?

Impaler · 2014-11-10 18:03:05

If my power system fails for two weeks were dead from 10 different causes, such as freezing, inability to circulate air, cook food, pressurize the EVA chamber to even conduct a repair, etc etc. Also we likely have ISRU going and Oxygen (even if it is intended as propellent) has certainly been stored which can accommodate a period in which a Greenhouse is not producing oxygen.

Equatorial locations on Mars are more moderated in terms of light levels that's true but they are less desirable in other ways, also it is false that their is 'no winter' in an equatorial location, the winter/summer day length variation is just minimized, at best this gets you up to 21% of Earth surface productivity. I said explicitly that diffuse vs direct light doesn't matter for plants, only panels need direct and I did not knock the transparent greenhouse in the math for that (and yes the light IS very diffuse on Mars because of Dusty conditions).

I'd also like to point out that dust settling on a transparent greenhouse roof is going to be a BIG problem, your roof is huge and you can't go up and walk on it if it is some transparent plastic bubble, ground mounted panels are going to be a piece of cake to clear by simply walking past them with a compressed air gun and back pack compressor compressing ambient Martian air.

RobertDyck · 2014-11-10 19:00:04

Light that reaches Mars orbit is 47% of light that reaches Earth orbit. And Mars doesn't have clouds, fog, or haze. Not even an ozone layer. More sunlight reaches the surface than Earth (as opposed to orbit of the same planet).

Dust on the greenhouse is a good point. In other threads (plural) I have argued for a long narrow greenhouse, oriented perfectly east-west. With a mirror on each side, as long as the greenhouse and height above ground of the mirror top matching the top of the greenhouse. Greenhouse would be twice as wide as it is high. So this provides as much sunlight from the mirror as direct from overhead. And the mirror does not have to track the Sun's daily movement. Mirror angle will have to be adjusted for season, but that's 1° change every second week. That change is easy enough to do by hand, with a notched rod holding up the mirror. And if you do it by hand, you can clean the mirror at the same time. So a bi-weekly chore.

To address your point, a long narrow greenhouse could have some sort of arch with air nozzles that you "sweep" over the greenhouse periodically.

Ps. Flat photovoltaic panels don't care if light is direct vs diffuse. Only brightness and spectrum. Solar concentrators don't work with diffuse light, only direct. That's one reason Mars rovers use flat photovoltaic panels, not concentrators.

Impaler · 2014-11-10 22:19:11

Total percentage of light at top of atmosphere that reaches the ground (direct or diffuse) alone is not sufficient. We need to know which frequencies have being absorbed or scattered. Plants only photosynthesis with 2 very narrow frequencies of red and blue light within the visible spectrum (400-450 and 650-700). Both Earth and Mars surface light IS spectral different from light at the top of the atmosphere, all you have to do is look at ANY photo taken from the Martian surface, it dose not look like a picture taken on the Moon, the light is clearly diffused and pinkish. You can not claim that this is better OR worse then Earth until you look at the actual spectrum intensity.

Flat solar panels DO care if they are getting direct light at the right ANGLE, because direct light is almost never perpendicular to a fixed panel. Any off axis light directly lowers output, the loss is linear rather then total as with a concentrator system though. My point earlier was that if the same lumins are coming in IS diffuse rather then direct then a panel will produce the same light regardless of it's orientation, and Plants are functionally fixed solar panels, so we would expect no loss in growth rate from diffusion. And everything I read said that Mars light is partly diffused because of continually suspended dust, an actual dust storm is even worse. Estimates are that surface light intensity is cut in half and virtually all light reaching the surface is diffuse during the global dust storms. Solar systems would operate at half power and need frequent cleaning, but would not be completely inoperable.

I any case the direct/diffuse is one that I said dose NOT count against against Mars in any way so I don't know why it is at issue.

The 'tube' greenhouse with an arched air-blowing device sounds doable, but it comes at the cost of higher material for the pressure vessel per unit of interior floor space.

44% is the light intensity number I've seen everywhere I've looked, but we know Mars orbit is eccentric so it's just an average. According to my math Mars will vary between 37% and 54% of Earth light intensity just from moving closer to or further from the Sun during it's orbit, with no way to alleviate this due to landing site. The eccentricity based light effects compound with day-length changes in the South, but are counter-cyclical with the Northern Hemisphere so you may get a better light profile going North rather then equatorial. As I'm assuming fast crop cycles are mandatory and food storage is prohibitive that means the lowest light level is the bottle-neck any greenhouse running on sunlight has to get through, and winter is nearly 180 days long on Mars.

Here is another good article on the kind of system I see being practical, it's all based on enclosed artificially lit volumes, the Germans are developing it and plan to try it at their antarctic research station. http://motherboard.vice.com/en_uk/read/ … an-cuisine

Last edited by Impaler (2014-11-10 23:20:15)

Impaler · 2014-11-11 02:39:58

Antius wrote:

I have always been fascinated by the idea of crowing crops using artificial light. But the economics are always weak unless electric power is cheap. The problem comes down to the poor efficiency of converting electrical energy to glucose.
In sunlight, crop plants are typically 1-2% efficient, but this can be roughly doubled if light wavelengths can be tuned to red and blue parts of the spectrum. LEDs can be 35% efficient in the red and blue parts of the spectrum. To produce 2500 calories per day of food (2.5MJ) some 240MJ of electric power are needed (66kWh).

I think this math can be improved. First my understanding was that LED efficiency was in the 90% range, as in only 10% of output in in heating the diode, but if 35% refers to the portion of energy coming out at the desired wavelength then that would indeed be what matters. But again my understanding was that LED's that are very narrow in light spectrum were easy, that's why 'white' LED's took so long to arrive commercially.

Now lets look at the inefficiencies of Photosynthesis and break them down

Wikipedia wrote:

100% sunlight → non-bioavailable photons waste is 47%, leaving
53% (in the 400–700 nm range) → 30% of photons are lost due to incomplete absorption, leaving
37% (absorbed photon energy) → 24% is lost due to wavelength-mismatch degradation to 700 nm energy, leaving
28.2% (sunlight energy collected by chlorophyl) → 32% efficient conversion of ATP and NADPH to d-glucose, leaving
9% (collected as sugar) → 35–40% of sugar is recycled/consumed by the leaf in dark and photo-respiration, leaving
5.4% net leaf efficiency.
Many plants lose much of the remaining energy on growing roots. Most crop plants store ~0.25% to 0.5% of the sunlight in the product (corn kernels, potato starch, etc.). Sugar cane is exceptional in several ways, yielding peak storage efficiencies of ~8%.

Right off the bat were not gonna waste 47% of incoming energy in the wrong wavelengths, the next inefficiency is from chemicals other then chlorophyll absorbing light, nothing we can do their other then maybe use a plant that is uber rich in chlorophyll like spinach, and while our Mars colonists might be quite healthy eating nothing but spinach I think calorie density would be too low, not to mention the massive forearm problem.

Wavelength mismatch would theoretically be eliminated by providing only 700nm (red) light but it seems plants need some blue light for proper growth, I think whats happening here is that the two separate 'systems' in the chloroplast actually need different energy levels to do two different activities, one pumps hydrogen to create an acidic environment and make ATP, the other breaks Water and creates NADPH. The system that only 'needs' low 700nm energy photons though is going to waste some of the light that falls on it in comparison to that light falling on the other system which needs that higher energy 400nm blue light. Likewise the heavy 32% efficiency of glucose production isn't going to be changing any time soon.

Respiration losses and photo-respiration though are quite amenable to reduction. Not turning off the light means plants never stop making glucose, but they still respire at a constant rate so we can expect about half as much respiration 'tax' on the plant. Photo-respiration is an inefficiency in the carbon fixation of C3 plants (C4's evolved a way to greatly minimize it) and it generally runs around 25%. Elevating CO2 levels or dropping O2 level may reduce this, but would probably come at the cost of making the growing space air unbreathable for humans. I'll estimate that under constant light and a just breathable these inefficiencies could come down to 20%

Lastly we are looking at 10% of sugar fixed into edible form, note that the Sugar canes great 8% (total) efficiency comes from it being a C4 plant (so low photo-respiration losses) and the fact that nearly all of its produced sugar stays as sugar in the cane juice and our processing of it extracts nearly all of that, it's not really that the plant is more efficient, it's cause were 'eating' nearly the whole plant that it looks so efficient.

So our irremovable inefficiencies are 30% (non chlorophyll absorption), 24% (wavelength mismatch), 68% (glucose synthesis), 20% (photo-respiration). That all comes out to 13.6% efficiency for Biomass production, but were still going to need to take a hit on the non-edible parts of the plant and that can vary a lot. Digging into the references on Wikipedia seems to indicate 5-6% is the theoretical maximum efficiency, but observed (outside on farms) is around 2-3% and both are for total Biomass, so I think the light optimization and CO2 concentrating looks to be consistent with that. Obviously avoiding pests, drought and other stresses will put the performance close to that new 13.6% theoretical maximum. In the end the single most important factor seems to be how much of the plant we eat and how much we throw away, so maybe we just need to have people eat an entire hydroponically grown lettuce plant, roots and all or process it into some kind of tofu. Feeding the inedible to some animals for protein production might work too.

Caloric math is a bit off too, a 'dietary calorie' is actually equal to 4.2 kJ meaning daily energy needs are 10.5MJ. If plants can hit 10% and our LED's are 90% efficient that means 116MJ electric which is only 32 kwh, this is right at about the average US homes daily energy consumption, not an unreasonable expenditure for one person to be feed on Mars.

Last edited by Impaler (2014-11-11 02:49:56)

Antius · 2014-11-11 06:10:37

Good work Impaler! At 32kWh per person per day, a 1400MWe fast breeder reactor can produce enough food for 1million people per day. A result that suggests that human beings equipped with a nuclear power supply should be able to live just about anywhere where there is a sufficient source of energy.

RobertDyck · 2014-11-11 08:15:50

Actually, no. Photovoltaic cells don't care about the direction or focus of light. They only care about how many photons strike the photovoltaic cell per second. Light could be scattered from haze or dust direct above, so it doesn't strike the cell. But other light strikes a scattering medium somewhere else in the atmosphere, resulting in it striking the photovoltaic cell. So scattering changes the light source from a point source to a plane, but doesn't change the total flux that strikes the cell. So again, a flat photovoltaic cell doesn't care whether light is scattered or not. That's why Mars landers and rovers have used flat photovoltaic panels.

Plant leaves also don't care if light is scattered or direct. However, yes, plants only utilize certain spectra. Halobacteria utilize green light, reflecting red and blue. That's why they look purple: they reflect red and blue. Green plants utilize red and blue, reflecting green. Exact absorption spectra has been characterized. Here is the spectra for one particular plant.

Note it isn't just "2 very narrow frequencies of red and blue light". The absorption spectrum is actually fairly broad. And plants use multiple photodyes to absorb as much light as possible. Chlorophyll has fairly narrow absorption bands, so they use other chemicals to absorb more light. It's not just chlorphyll A. And it's not chlorophyll A plus chlorophyll B. It also includes beta-Carotene, Phycoerythrin, and Phycocyanin.

louis · 2014-11-11 13:49:11

Antius wrote:

Good work Impaler! At 32kWh per person per day, a 1400MWe fast breeder reactor can produce enough food for 1million people per day. A result that suggests that human beings equipped with a nuclear power supply should be able to live just about anywhere where there is a sufficient source of energy.

The same applies to PV power for Moon and Mars. But with PV you can split it up and use it where you need it. If one part goes wrong, there are plenty of other parts to go right. A single nuclear reactor means you are putting all your eggs in one unwieldy basket.

Antius · 2014-11-11 15:32:27

If you had enough natural light for PV, why would you need PV at all for plant growth? Unless we are saying that it's easier/cheaper to build a solar power plant to feed plants with artificial light, than to put them in a greenhouse.

The main advantage I can see with artificial crop lighting is that it allows you to grow food in locations or in an abundance that would not be possible with the ambient energy available, or allows you to boost the productivity beyond what would be achievable with sunlight alone.

SpaceNut · 2014-11-11 18:07:21

Actually a tracking photovaic system will have a 30% greater output versus a none tracking unit during the summer and only 15% in the winter.

Good stuff on the wavelengths that plants want.

Last edited by SpaceNut (2014-11-11 19:20:37)

Impaler · 2014-11-11 21:41:13

RobertDyck wrote:

Actually, no. Photovoltaic cells don't care about the direction or focus of light. They only care about how many photons strike the photovoltaic cell per second. Light could be scattered from haze or dust direct above, so it doesn't strike the cell. But other light strikes a scattering medium somewhere else in the atmosphere, resulting in it striking the photovoltaic cell. So scattering changes the light source from a point source to a plane, but doesn't change the total flux that strikes the cell. So again, a flat photovoltaic cell doesn't care whether light is scattered or not. That's why Mars landers and rovers have used flat photovoltaic panels.

How many times do I have to repeat myself, a flat panel CARES about direct light when the ANGLE OF LIGHT STRIKING THE PANEL IS OFF AXIS, because this directly results in less photons striking the panel, the panel only absorbs the light from the cross-sectional 'shadow' it casts in the directional light stream.

When you say A (direction or focus of light) dose not matter and only B (how many photons strike the cell per second) is TECHNICALLY correct but a asinine thing to say when A directly determines B. It's like saying a bridge doesn't care how much weight is put on it, it only 'cares' when the force on the cables exceeds the tensile yield strength of the material, the former determines the latter.

Comparing PV area to solar greenhouse might be a good exercise, we should expect PV per unit Area to be a lot less massive if we use thin-film solar which are considered to be the best power/mass ratio, so raw area may not be the real factor of import but it may be illuminating. Thin films are 20% efficient right now, and we can expect further improvement. Panels can't completely carpet the ground, but neither would plants in a greenhouse so I'll say the gaps cancel out. Our Solar system suffers all the same power reductions from winter and day-length changes that a greenhouse dose so no different their, also the 90% efficiency of LED is comparable to the transparency of our greenhouse glass so thouse cancel each other. The only real factor left is the 47% inefficiency of natural light that is outside of plant desired spectrum vs the 20% efficiency of the solar cell. That means we just need just 2.6x more area for the panels then we would need for the greenhouse. So for example a 1000 m^2 greenhouse can be replaced with at 2600 m^2 solar field and an enclosed 50 m^2 growing area with 3 levels of growing trays. This effectively concentrates the Martian sunlight by a factor of 17 and when we look at photosynthetic frequencies of light it's nearly 35 times concentrated. That's why we can get the yield from that tiny growing area to rival the huge 1000 m^2 greenhouse.

A thin-film system of that size should still be very light, though it would be counter balanced by the need for batteries (or better yet Regenerative Fuel Cells which are NASA's preferred solution), it would likely still come to less then the material needed to create a safe transparent pressure tight greenhouse of such size.

RobertDyck · 2014-11-11 22:09:40

Impaler wrote:

How many times do I have to repeat myself

Please stop. It doesn't matter how many times you repeat yourself. That doesn't make it so.

I'll try this one last time. With directional light, you only get the light from the surface of the Sun on a direct line to the solar panel. With diffuse light, some of that is directed away from your panel. You may think that would reduce the light your panel gets. However, light on a path that would normally miss your panel will also get diffused. Some of that light will strike your panel. The Sun is not a point source, but with a distance of over a hundred million kilometres, it may as well be so. Angle from one side of the Sun to the edge of your solar panel is almost exactly the same as from the other side of the Sun to the same edge of your solar panel. The shadow of your solar panel with have a fuzzy edge because of that difference in angle, basically its a gradient, but the width of that gradient is so small that the shadow may as well be a hard edge.

But with diffuse light, the entire sky is a two dimensional plane source. Under overcast conditions on Earth, light is entirely diffuse. The result is smooth lighting from the entire sky. The only shadow is a fuzzy patch directly beneath your feet. You won't notice any shadow at all. Mars is not perfectly diffuse. In fact, shadows from images of Mars landers and rovers show it's almost entirely direct sunlight.

But again, under diffuse conditions, light your panel doesn't get because direct light is diffused away, is compensated for by light from kilometres off to the side that is diffused into your panel. Total light per unit area remains the same. The only way light could be reduced is if it isn't diffusion, rather simply occlusion. That means something opaque in the way.

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