Can I get a rough estimate of the power (in watts) needed to produce light to grow enough food for one person, using vertical farming of the kind proposed for space habitats?

Any reasonable outpost/colony concept that grows most of its food and does not use direct light, but has information on the base's energy source (kWe of a nuclear reactor, area of photovoltaic arrays) would be useful for finding conservative estimates.

  • $\begingroup$ Different types of food have radically different efficiencies in terms of producing a given number of calories of food. I think the variation amounts to at least a couple of orders of magnitude. For example, lettuce and beef are both extremely expensive to produce per unit of food energy -- lettuce simply because it's not a very calorie-dense food. Should we assume that your colonists are willing to be vegetarian and eat a diet restricted to the most efficient crops? $\endgroup$ – Ben Crowell Jun 18 '17 at 23:20
  • 2
    $\begingroup$ @BenCrowell Yes, those assumptions match the purpose. I want a very general idea, to help decide if various concepts are feasible. This paper: ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20070032685.pdf seems to suggest 3 kW/person for all purposes, which seems awfully optimistic to me. $\endgroup$ – Deimophobia Jun 19 '17 at 0:16
  • $\begingroup$ @DJohnM Correct, sorry, I meant power. $\endgroup$ – Deimophobia Jun 19 '17 at 0:33
  • 2
    $\begingroup$ Perhaps a worst-case could be calculated based on an optimistic .25 hectares of land to support 1 person and approx. 1 kW/sq m solar flux for about 2500kW per person. Factor in daylight hours, solar incidence angle in most farming regions and it maybe drops to 500kW or so. But one sedentary person generates about 100W of heat, so there may be some huge inefficiencies in conversion of sunlight to farmed food, which may or may not exist in "space habitat" food production. $\endgroup$ – Anthony X Jun 19 '17 at 1:18
  • $\begingroup$ related: space.stackexchange.com/questions/19924/solar-panels-on-mars $\endgroup$ – Ben Crowell Jun 19 '17 at 1:26

The 3kw/person figure in the original report may just be for general life support - it's caveated as "in the habitat" and the report goes on to explicitly say that the manufacturing unit will have extra power. The farm is shown as a third module.

The South Pole greenhouse (entirely indoors) is estimated to run at an efficiency of "...assuming that fuel costs about \$4.50 per litre, greens cost about \$50 per pound". A litre of diesel seems to produce approximately 10kwh of power, so you're looking at 100kwh to produce a pound of vegetables. That could be anything from 100 calories (spinach) upwards; rice and wheat have much higher energy densities (500-1500 cal/lb) but they also have a lot of additional inedible weight, eg in the stalk. Let's say you can get 500 cal/lb with really intensive farming approaches and clever reuse of heat, probably a generous estimate.

This means you need enough energy input for about four and a bit pounds of growth per person per day, or 450 kwh. Assuming 24-hour farming, you're then looking at ~20kw of solar power on a continual basis.

This seems challenging. The whole ISS (assuming the arrays work about 50% of the time) would produce just enough power to feed a three-person crew with nothing left over for life support...


Looking around for data on indoor agriculture, the most relevant stuff I could find was for marijuana growing in High Times magazine. It appears to be fantastically energy-intensive, possibly using up as much as 1% of the US's electrical energy output. Growing a kilogram of marijuana seems to require about $E=2\times10^{10}$ J of energy. Let's assume we can grow a kilogram of wheat with the same energy input. Wheat has an energy density of about $d=3400$ kcal/kg. A human needs about $c=0.02$ kcal/s of food.

The result is $cE/d\sim100$ kW of electrical power to support one human being with indoor farming. Given all the very rough order-of-magnitude assumptions involved, this seems to be roughly in line with Anthony X's estimate in a comment of 500 kW. No matter what, this is a fantastic amount of energy, dwarfing the approximately 2 kW per person used at Antarctic bases, which ship in their food.

It seems that a self-supporting space colony would do well to use sunlight, not electrical lighting, for their agriculture.

For comparison with $E\sim10^{10}$ J to grow a kg of wheat indoors, it only costs something like $5\times10^8$ J to accelerate a kg to the same velocity as the earth's orbital velocity around the sun. This suggests that if you're somewhere that doesn't have enough sunlight (surface of Titan?), you might be smart simply to import your food.

  • $\begingroup$ It's an interesting choice for a plant model. I made a slight edit just to let people know better the name of the source they might be clicking on. But the statement "...would do well to use sunlight, not electrical lighting, for their agriculture." would need to be better defended. Mars sunlight is dim, about half of Earth's so plant growth would be slower without concentrators. Windows would have to insulate against the infrared exposure to space. On Earth, the ambient air keeps greenhouse glass warm. I'd say stick to answering the question, or find solid technical support. $\endgroup$ – uhoh Jun 19 '17 at 2:20
  • $\begingroup$ So I've asked Which is easier to build on mars per square kilometer; greenhouse windows or photovoltaics/LEDs? Let's think it through. $\endgroup$ – uhoh Jun 19 '17 at 2:41
  • $\begingroup$ Surface of Titan, indeed! All these seem to be based off of sunlit crops. Maybe we can find some figures on vertical farming, most of which (today) is done via UV lights, not windows? $\endgroup$ – Deimophobia Jun 19 '17 at 3:10
  • 1
    $\begingroup$ @Deimophobia: I don't understand your comment. There is nothing in my estimate that has anything to do with windows. This estimate is about artificial lighting, and it uses data from commercial agriculture that uses artificial lighting. $\endgroup$ – Ben Crowell Jun 20 '17 at 0:02
  • $\begingroup$ @BenCrowell Oh, apologies! I should have checked your link. $\endgroup$ – Deimophobia Jun 20 '17 at 0:03

If you accept the numbers at http://www.theoildrum.com/node/6252 (which are sourced, but that doesn't make them necessarily accurate), it takes 0.430 kwh of energy to produce 390 food calories (390 kcal) worth of corn, for an efficiency rating of 102%.

I thought this was impossible at first, since it would seemingly violate the Seecond Law of Thermodynamics, but the calculation only includes the light energy required: the additional caloric energy comes from the physical matter that makes up corn, primarily carbon, hydrogen, and oxygen. In other words, creating corn requires both energy and raw materials, and the raw materials themselves contain stored energy.

Assuming you could live primarily (though not entirely) on corn, this number is close enough to 100% that we can assume roughly 1-to-1 transfer of light energy to caloric energy.

The average healthy person burns at least 2000 calories a day (right around 100 watts), though this assumes very little activity. With a reasonable amount of activity, this is closer to 3000 calories (about 150 watts).

So, somewhat uninterestingly, the amount of energy you need is, assuming you have sufficient raw materials, is pretty much the amount of energy an average person burns.

  • 3
    $\begingroup$ Those numbers are extremely suspect for this context: if you check the linked study for the corn it does not appear to include solar energy as an input, in other words the 0.430kWh is only the energy inputs from human sources. I should also add that CO2 and H2O do not contain energy for purposes of plants - in fact energy is required to break the chemical bonds to get the hydrogen and carbon for making carbohydrates. $\endgroup$ – Blake Walsh Jun 21 '17 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.