Where's all the Methane coming from?

The atmospheres of Uranus, Neptune and Titan all contain significant amounts of methane (2.3%, 1.5% and 1.6%, respectively). That is why in On the Surface of Giants, the propellant depot is in such a close orbit - it is skimming methane from the atmosphere. And the methane is relatively easy to distill out of the atmospheric gases that are collected, as it boils quite a bit higher than nitrogen, hydrogen and helium, and quite a bit lower than ammonia.

And you are right in your terminology, the stuff you toss out the back is propellant or reaction mass (or remass for short), the fuel is what gives the propellant the energy. Propellant and fuel are the same only for combustion rockets.

Who needs cows when you have the pantries stocked with pork-n-beans.

Water is separated into hydrogen and oxygen via electrolysis of course. The reaction itself requires 285.83 kJ per mole of water. One mole of water produces one mole of hydrogen and a half mole of oxygen. A mole of hydrogen weighs 1.01g (see this calculator), which means that a you need at least 1000 * 285.83 / 1.01 = 283 MJ per kilogram of hydrogen thus produced. However, if your water is not 100% pure H2O there will be side reactions which will make your efficiency suffer. Additionally, you need to expend some energy to separate the gases, which depends on the method of separation, but it appears that this cost may be small compared to the energy for the reaction itself.

Sabatier process itself exothermic, i.e. it produces heat, so while you need to heat up the reactor to start the reaction (yes, apparently these devices are called chemical reactors), it will then support itself. In practice the reaction also needs a catalyst and it seems that different catalysts can change Sabatier reaction efficiency though I'm not clear on how that affects energy requirements.

The whole process in relation to methane production on Mars is described in this article. If you are wondering why is the delta H for electrolysis in this article is twice as large as what I cited previously, this is, as far as I understand, due to the fact they are talking about a chain of reactions, so that value is for two moles of water instead of one.

Yes, but you have to keep them in a vacuum.

FTFY.

So, under an ideal circumstances,

1 mole of water makes 1 mole of H2 for 285 kj. The sabatier process actually needs 4 moles of H2 per mole of CO2, so about a total of 1.14 MJ.

You then mix those 4 moles of H2 plus a mole of CO2 for 1 mole of Methane plus about 165 kj of energy, which might partially be able to offset energy costs, but probably not all that much.

So, 1 kg of Methane would be (1,000/16) 62.5 moles, which would require (62.5x4)=250 moles of H2, or close to 500 grams. But, as your link shows, that's inefficient since your actually making two water molecules for every methane molecule. By reusing the water molucules, you can reduce your need to two H2 per methane. Thus, you would only need 125 moles of H2 per Kg of Methane, about 250 grams of H2.

So, assuming you had a source of H2 available, you could make 1 kg of methane on mars using 500 g of H2 and require basically no energy, or make 1 kg out of 250g of H2 if you were willing to spend some 318 kj per mole of methane, or about 2 MJ of energy per kg. This is if you have H2 already available. If you need to produce H2 initially from water, that would seem to add some 3.6 MJ per kg if your reusing the water generated.

So, from scratch, it looks like you would need about 6 MJ of energy to make 1 kg of methane. Lets round up, somewhat arbitrarily, to 10 MJ in actual practice.

So, if you had a 4 GWatt powerplant dedicated solely to making Methane, it could theoretically make 400 kg a second. Over a day, that would be 35,000 tonnes a day at full output.

Now, the next question would be, how much production do you need?