[Spaceships] Reasonable numbers for Refinery Systems / 'Artificial' Fuel?

[vierasmarius]

Originally Posted by Anthony

This is called 'spaceships economics is broken'. Which isn't really news. Though I would divide by 10 years, not 50.

Yeah, I'm not worried so much about getting stats that provide an accurate economic model, as getting stats that don't blatantly break the laws of physics.

[vicky_molokh]

Originally Posted by Anthony

This is called 'spaceships economics is broken'. Which isn't really news. Though I would divide by 10 years, not 50.

This is why I'm asking about reasonable numbers. I think my 50% efficiency WMG is too crude.

[cccwebs]

Originally Posted by Ulzgoroth

Makes no sense when your refinery is, for instance, electrolyzing water. Not all 'refining' is extracting a trace component.

I was only using an example as given to illustrate that the discussion should be more on the efficiency of the refinery and the lack of information on the ratio of raw materials to processed goods for the various fuels. Even electrolyzing water will have some loss going from water to hydrogen and oxygen.

[Sunrunners_Fire]

Originally Posted by vicky_molokh

This is why I'm asking about reasonable numbers. I think my 50% efficiency WMG is too crude.

Ah. In that case, I would suggest utterly discarding the spaceships information and use real-world stuff.

If you insist on using Spaceships,

Given that a refinery module in a SM+5 spaceship is rated at 0.5 tons of fuel produced per hour and that this costs the spaceship 0.5 tons of water and 1 power point allocation for that hour and that a mhd turbine SM+5 module runs for six hours on internal fuel or twenty-four hours from a fuel tank and that it provides two power points when running and that a fuel tank sm+5 ship module holds 1.5 tons of fuel ...

So it produces 48 pp-hours from 3 pp-hours, given 1.5 tons water as an input. Erm.

3 pp-hours of input should produce between 1.5 pp-hours of output at 50% efficiency. So the new time-per-tank should change from 24 hours to 40 minutes assuming you keep everything else the same.

... this doesn't look very competitive with the batteries.

Change the efficiency? Even at 100% efficiency, you shouldn't get more energy out of a given amount of gas than it costs to create that gas mix from water especially given that the gas mix converts back to water as it reacts.

I dunno. I suspect I'm misunderstanding something.

[Sunrunners_Fire]

Originally Posted by vicky_molokh

I'm assuming that a there's some waste of energy in the process.

O2 + 2×H2 -> 2×H2O + energy
energy + 2×H2O -> 2×H2 + O2

Isn't it?

(Note that I am postulating the use of inert, ubiquitous components, not something that already stores a significant amount of chemical energy. A typical space-operatic example would be mining ice asteroids and turning them into rocket fuel.)

2 H2O -> 2 H2 + O2
Water (liquid) to Hydrogen (gas) + Oxygen (gas)
2 (-285.8 kj/mol) -> 2 (0 kj/mol) + (0 kj/mol)

Costs no less than 285.8 kj per 18 grams of liquid water reacted, producing 2 grams of hydrogen gas and 16 grams of oxygen gas.

Reversing the reaction gives you no more than 285.8 kj, though the usual efficiency is about 50% and so you'd only get 142.9 kj.

0.5 tons is 454,000 grams, and so should cost no less than 7,208 mj to react. Burning the hydrogen and oxygen to produce should give around 3,604 kj.

Assuming I'm not messing up the physics or my math, of course.

[Ulzgoroth]

Originally Posted by Sunrunners_Fire

2 H2O -> 2 H2 + O2
Water (liquid) to Hydrogen (gas) + Oxygen (gas)
2 (-285.8 kj/mol) -> 2 (0 kj/mol) + (0 kj/mol)

Costs no less than 285.8 kj per 18 grams of liquid water reacted, producing 2 grams of hydrogen gas and 16 grams of oxygen gas.

Reversing the reaction gives you no more than 285.8 kj, though the usual efficiency is about 50% and so you'd only get 142.9 kj.

0.5 tons is 454,000 grams, and so should cost no less than 7,208 mj to react. Burning the hydrogen and oxygen to produce should give around 3,604 kj.

Assuming I'm not messing up the physics or my math, of course.

I think you're ignoring the contribution of entropy there and temperature and pressure should factor in...

Definitely the right line of reasoning to get the energies involved. Unfortunately Spaceships tries hard to make the actual energies it uses inaccessible so it's hard to relate that to power points.

[vicky_molokh]

Originally Posted by Sunrunners_Fire

[ . . . ]
Change the efficiency? Even at 100% efficiency, you shouldn't get more energy out of a given amount of gas than it costs to create that gas mix from water especially given that the gas mix converts back to water as it reacts.

I dunno. I suspect I'm misunderstanding something.

It seems that indeed some sort of misunderstanding occurred.

When I say 50% efficiency, I mean that it would take a refinery X PP×hours of energy input in order to synthesise (or otherwise produce) as much fuel as will be required to get X×50% PP×hours when burned.