Power for ISRU on Pol

[cantab]

Getting back on topic, having an engineer around will make an ISRU unit produce more fuel per unit of electricity consumed, and a sufficiently skilled engineer is a requirement for a fuel-cell powered miner to be self sustaining.

So for autonomous mining you need another source of power. In the outer solar system, whether you use solar power - and you can get usable power out of Gigantors at Jool - or RTGs, you will probably just have to accept a slow production rate. I'd argue that in most cases that's fine. A probe exploring the different Joolian moons is not in a hurry. If you're sending robots ahead of Kerbals, they've got a full year to work if you send the robots one launch window and the Kerbals the next, and that's plenty of time to make some useful fuel. Even longer if you let the robots arrive and check they work before the Kerbals even launch.

[Zhetaan]

I think the higher Ore concentration increases electrical consumption because the ISRU has more to work with.  I understand the mechanic as this:

A drill consumes a set amount of EC/second.  The EC consumed does not affect the Ore produced; it is only altered by the concentration (or an Engineer, but let's assume a robotic miner).  This is because the drill digs up x tonnes of 'rock', not x tonnes of Ore, so having the drill on a 1% deposit or a 100% deposit is going to show the same electrical draw.

The ISRU consumes Ore (and makes fuel) much faster than a typical drill can supply it.  Thus the ISRU constantly stalls; it never gets up to full speed, so it never gets up to full electrical consumption.  Once you supply it with more Ore (because you are in a better spot and are digging up more Ore), it works harder and takes up more power.  There is a limit at which you supply Ore faster than the ISRU can process it (easily seen as filling Ore Tanks), but it's pretty high up compared to typical drill output.  Once you hit that limit, though, your EC usage plateaus.

In terms of the practical use of a robotic miner, @cantab is correct; in most cases, the best thing to do with a mining/processing outpost is to let it work for a year before the Kerbals arrive to take the fuel, a la the MAV in The Martian.

[Geschosskopf]

On 6/10/2016 at 11:47 AM, Zhetaan said:

t is this second idea of taking reaction products, splitting them back into reactants, and combining them again for a net energy gain that is the problem.  First, the evidence:  the fact that the fuel cells in the game use LF+O is evidence that the process is chemical.  The fact that the parts are called 'fuel cells' is more evidence.  The descriptions of the ISRU parts go further and heavily imply that the process is, if not hydrolysis of water, then a direct analogue.

That's where the issue lies:  if Ore simply happened to contain something usable as fuel and something else usable as oxidiser, and the only preparatory steps needed were extraction and purification, then it would be as you describe in your deuterium analogy and I'd have no concerns about the process being energy positive (getting 100% LF or 100% Ox from the same Ore is another discussion).  But if Ore is convertible into LF+O via hydrolysis, then the theoretical minimum energy you need to accomplish that is, by dint of thermodynamics, exactly equal to the theoretical maximum energy you can extract from recombining those converted products in the same (reversed) process.

Bingo.  Because the ISRU process can only really be described as electrolysis, the very best the fuel cell could do is run itself, not provide any surplus, and this only in a 100% reversible, ideal system.  Which of course can't exist---energy is always lost along the way.  Which brings us back to the radiators....

Radiators are places where vast amounts of energy leaves the system---they're all rated in hundreds of KW.  A mining/refining base is spewing energy into space when operating.  Which of course should make the process irreversible and non-ideal.  But the game totally ignores all this waste heat in its energy calculations, only worrying about the tiny amount needed to run the coolant circulating pumps.  Where do these hundreds of KW of waste heat then come from?  They must ultimately come from the fuel cell, because the fuel cell is powering the machinery, and the friction and whatnot in the machinery is creating the waste heat.  This all makes the fuel cell even more thermodynamically ludicrious because it can produce a surplus of fuel above its own needs while still tossing away hundreds of KW of waste heat.

[bewing]

Well, I disagree. It has been stated that Eve's oceans are made of liquid fuel. Therefore, liquid fuel is not merely hydrogen. Therefore the ISRU process is most definitely not electrolysis. So that whole line of reasoning is shot.

[THX1138]

Did you only do an orbital survey? You need to use a surface scanner to find out the exact concentration in the particular area you are drilling.

[Jesrad]

On 10/05/2016 at 9:27 PM, davidpsummers said:

I did check how I did when warmed up.  I'm still not breaking even.  I guess my rate of 0.0019/second is too low.  anyway of calculating how much you need?

I did the maths back then.

Quote

<400 energy per half unit of ore mined

...

you can only "refuel at a profit", energy-wise, using a level 3+ (advised 4 or 5) engineer off a moon (assuming ~5% ore concentration) operating a single drill, or one unmanned drill off an asteroid (75% or better ore concentration).

 

Edited June 14, 2016 by Jesrad