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davidpsummers

Power for ISRU on Pol

Question

I put down a robotic miner onto Pol.  I hit one of the high areas show by the orbital survey scanner.  I put down my drills, and they pull in a bit less than .002 (whatever units).  A bit less than on Minmus but OK.  

But I'm so far out, I figured I would use fuel cells, since they seem to work OK as supplements to solar closer in.  But Drills + ISRU conversion is a net loss.  It takes more fuel to power them than they produce.  Am I just screwed?  What do people use for power in such a situation?  Just accept that you will need a bunch of heavy nuclear units?

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Out at Jool you kinda need a (preferably level 5) Engineer to increase the efficiency of the drilling. I think it's 25x better with a level 5 Engineer.

If you want to go fully robotic, I suggest stuffing the ship with RTGs, enough to power the drills and converter with enough left over to power the probe. It's probably a lot of those, though.

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5 hours ago, Geschosskopf said:

Well, you should NOT be able to power a mining and refining set-up on fuelcells burning the fuel produced by the system.  That would violate the 2nd Law of Thermodynamics.  At least in stock, where unrefined Ore doesn't burn.  Now, if you've got mods that have Ore-burning generators, then Ore can be regarded as analogous to a fossil fuel and you're just releasing stored energy, not chasing your tail, so no 2nd Law violation.  But OTOH, how did fossil fuels get on lifeless planetoids?

So, when I'm mining out where the sun don't shine, I use nuclear reactors, either fission or fusion depending on what mods I have in that game.

That doesn't follow.  Refined ore is fuel.  The fact that it takes two steps (mine, refine) to get the product that gives you the energy doesn't mean that violates the second law.
The two steps both take a lot of energy, but nothing compared to what is stored in the final product.

I suppose that high concentrations of hydrocarbons and oxygen bearing minerals on lifeless bodies probably requires some suspension of disbelief, but I don't think there's any suspension of the second law going on.

Happy landings!

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6 hours ago, Geschosskopf said:

Well, you should NOT be able to power a mining and refining set-up on fuelcells burning the fuel produced by the system.  That would violate the 2nd Law of Thermodynamics.

That is wrong. The second law of thermodynamics only applies to closed systems, and by definition a mining rig is not a closed system: ore is entering the system as you mine it.

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Just now, Geschosskopf said:

Refining Ore is the reverse of burning it

It seems to me that this is a misunderstanding.

Refining ore is only concentrating the useful bits.  The process of burning it is a specific chemical process releasing energy stored in the hydrocarbons.  Our refining process doesn't create the hydrocarbons from their constituent elements.  It just concentrates the hydrocarbons and throws away the junk.

Happy landings!

 

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Does the surface concentration change the ore per electricity rate or just use more electricity to mine more ore?

Engineer definitely increases efficiency as 5th says.

Edited by RizzoTheRat

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13 minutes ago, RizzoTheRat said:

Does the surface concentration change the ore per electricity rate or just use more electricity to mine more ore?

Engineer definitely increases efficiency as 5th says.

I'd have to test it, but I think concentration only affects how much ore you get per second. You always use the same amount of power per second.

I think also lower ore concentration generates more heat.

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In my main career save in the Jool system I have two miners and no fuel cells (embarrassing, but I forgot to add them). My ISRU/drill units are powered by one* RTG, one gigantor and four standard 6-segment panels. That means that in the dark (RTG alone) my drill works at a 14.5% load and the ISRU at 7.4% load (single process). In full daylight, that rises to 23% drill load and 12% ISRU load. They can't run together (and I can't use more than one ISRU process at a time) due to power shortage.

With that, and a one-star engineer (he was supposed to be on a training trip to the Jool system, but never went home...) I can get 1000 ore in about 6 days. However, in 1.0.5 (not sure about 1.1.x) there is a bug where warping at 1000x fills your batteries, not that I'd suggest exploiting this of course.

Long story short: yes, you can operate on solar power alone. I still consider an RTG a must-have for any important ship out in the Jool system.

 

*edit: @davidpsummers my apologies, I went back to check because the numbers were bugging me. In fact, that miner has three RTGs. That's how it gets a 14.5% load on the drill at night. The RTG output is 0.75 zaps per second. The drill requires 15 zaps per second, so the RTG provides 5% of that. 3*5= 15% if there is no other electricity usage. Likewise, one ISRU process requires 30 electricity per second, so one RTG gives 2.5% (max) load.

Sorry about that. The other two RTGs were hidden away at the back of the service bay. So that does confirm what you were saying about needing a "bunch" of nuclear units...

Edited by Plusck

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Fuel cells work fine but I find you need a margin of extra fuel to power the cells while all the mining / refining equipment warms up to top efficiency.

Happy landings!

Edited by Starhawk

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6 hours ago, Starhawk said:

Fuel cells work fine but I find you need a margin of extra fuel to power the cells while all the mining / refining equipment warms up to top efficiency.

Happy landings!

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?

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17 minutes ago, 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?

Yeah, that's too low.  I'm not sure exactly, but the lowest I've worked with is something like 0.01 or so.  About 5 times what you're seeing.

The initial orbital resource scan with just the M700 is very low-res.  It sounds like you just ended up on a very low concentration spot.  Using a robotic miner means you have no Engineer bonus to your ore rate so you really need to be on a good location.  I would recommend using both the M700 Survey Scanner and the M4435 Narrow Band Scanner.  Hopefully that will enable you to find a better spot to drill.

Happy landings!

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20 hours ago, davidpsummers said:

I put down a robotic miner onto Pol.  I hit one of the high areas show by the orbital survey scanner.  I put down my drills, and they pull in a bit less than .002 (whatever units).  A bit less than on Minmus but OK.  

But I'm so far out, I figured I would use fuel cells, since they seem to work OK as supplements to solar closer in.  But Drills + ISRU conversion is a net loss.  It takes more fuel to power them than they produce.  Am I just screwed?  What do people use for power in such a situation?  Just accept that you will need a bunch of heavy nuclear units?

Well, you should NOT be able to power a mining and refining set-up on fuelcells burning the fuel produced by the system.  That would violate the 2nd Law of Thermodynamics.  At least in stock, where unrefined Ore doesn't burn.  Now, if you've got mods that have Ore-burning generators, then Ore can be regarded as analogous to a fossil fuel and you're just releasing stored energy, not chasing your tail, so no 2nd Law violation.  But OTOH, how did fossil fuels get on lifeless planetoids?

So, when I'm mining out where the sun don't shine, I use nuclear reactors, either fission or fusion depending on what mods I have in that game.

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4 hours ago, Geschosskopf said:

Well, you should NOT be able to power a mining and refining set-up on fuelcells burning the fuel produced by the system.  That would violate the 2nd Law of Thermodynamics.  At least in stock, where unrefined Ore doesn't burn.  Now, if you've got mods that have Ore-burning generators, then Ore can be regarded as analogous to a fossil fuel and you're just releasing stored energy, not chasing your tail, so no 2nd Law violation.  But OTOH, how did fossil fuels get on lifeless planetoids?

So, when I'm mining out where the sun don't shine, I use nuclear reactors, either fission or fusion depending on what mods I have in that game.

I consider "ore" to be a mix of compounds in which oxygen and hydrogen are rather loosely bound, and can be extracted for less than the energy that can be gained recombining them to form water. Otherwise it would just be a question of "ore" being "water".

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@Plusck, @Starhawk, and @Grumman.....

Because Ore doesn't burn at all in stock KSP but its components do, and there's zero mass loss in refining, using a fuelcell that burns the fuel produced by a mining-refining base that produces that fuel is a closed loop.  Refining Ore is the reverse of burning it.  There's no getting around that.  Therefore the 2nd Law of Thermo is broken because in such systems, you can't ever capture as much energy by burning the products as you expend separating those products from their low-energy state.  It doesn't matter that new Ore is coming into the system, it's the reactions in the fuelcells and ISRU that limit it.  A mining-refining-fuelcell set-up is exactly the same as trying to run a water electrolysization plant on the energy released by burning the H2/O2 cracked by the electrodes.  This is a clear violation of the 2nd Law of Thermo and the KSP system also violates conservation of energy by not only keeping itself running, but also still manages to radiate away hundreds of kW PLUS produce enough surplus to refuel other ships.  It's totally ludicrous.

Now, the toy-sized stock KSP universe clearly runs on radically different physical laws than our own.,  It routinely violates every conservation law we know, and the 4 fundamental forces of nature are obviously extremely different than what we're used to.,  Because this means the nuclear forces are different, KSP has no elements in common with our own periodic table, and KSP chemistry must therefore be totally different as well.  So, if you can accept that NOTHING in the KSP universe should be "realistic" in human terms, then by all means power your mining-refining bases with fuel cells.  Just toss conservation of energy and the 2nd Law of Thermo on the pile with all our other physical laws that you don't care about.  But if you hold the 2nd Law of Thermo to be sacred, then don't do it.

Now, this particular breach of human reality has a simple fix.  Just make it so raw Ore burns and loses mass in the refining process,  Then instead of being analogous to water, Ore would be analogous to fossil fuels and you wouldn't have a closed loop system with your fuelcell-powered mining base.  And then explain away the presence of fossil fuel analogs on lifeless bodies by invoking the alien geological processes and chemical make-up of the KSP planets.

But anyway, back to the OP.....  Use nuclear reactors to power your bases where there sun don't shine.

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In fairness, the stock ISRU doesn't throw away the junk, so it has to be a different kind of process.  Otherwise, there would be mass loss.  Perhaps you could throw in the idea that the drills operate as some kind of first-pass classifiers and the junk gets thrown out before what we call Ore enters the ship, but @Geschosskopf still has a point.  If the ISRU is anything but thermoequilibrated (which it cannot be, because it does work), the energy that is moving around has to be coming from somewhere and be going into somewhere else.  And if Ore is pre-concentrated by the drills, then it should be subject to some kind of direct consumption.  Though, in fairness again, there may be practical engineering reasons why stock KSP engines don't allow for direct consumption of Ore:  finding hydrocarbons in space doesn't mean we need to build rockets that run on crude oil.  The idea of refining is a good one, and the process exists for good reasons.

On the other hand, this presupposes that Ore is in a low-energy state and the refining process puts it into a high-energy state.  We know, given that liquid fuel and oxidiser burn, that they are high-energy.  What if raw Ore were higher-energy yet?  I believe the Karbonite mod uses something like this line of thinking (the drawback being that Karbonite is dangerous, unstable, and inefficient--though only the inefficiency is modelled), but of course the question becomes one of how such high-energy Ore came to be so evenly distributed across the solar system without violently exploding or something.

Also, Titan is covered in methane lakes.  Philae found acetone and propanal.  Hydrocarbons and other organic molecules don't require life to form; life is simply good at forming them.

 

Edited by Zhetaan

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36 minutes ago, Starhawk said:

It seems to me that this is a misunderstanding.

Refining ore is only concentrating the useful bits.  The process of burning it is a specific chemical process releasing energy stored in the hydrocarbons.  Our refining process doesn't create the hydrocarbons from their constituent elements.  It just concentrates the hydrocarbons and throws away the junk.

Happy landings!

 

Yes, this is similar to how I understand "ore".

Although I readily admit I have no idea where the oxidizer would come from. I tried looking for more info on this just now (I haven't really kept up-to-date on chemistry) but only found one example which suggested using a total of 39GWh to get a tonne of oxygen out of Moon rock... which doesn't really cut it since sticking that back into a fuel cell would (if my numbers are right) give about 8MWh.

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11 minutes ago, Zhetaan said:

In fairness, the stock ISRU doesn't throw away the junk, so it has to be a different kind of process.  Otherwise, there would be mass loss.

Ahh yes.  I see @Geschosskopf already pointed that out and I missed it.  Sorry about that.

It seems to me that the lack of mass loss is a design flaw.  Or perhaps we can hand-wave our way out of it.  Perhaps the hydrocarbons are tainted with small amounts of impurities that would impede the oxidization process and the refinery removes those.  If they have negligible mass compared to the fuel created this would approximate the way it works in-game.

Happy landings!

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Just now, Starhawk said:

It seems to me that the lack of mass loss is a design flaw.  Or perhaps we can hand-wave our way out of it.  Perhaps the hydrocarbons are tainted with small amounts of impurities that would impede the oxidization process and the refinery removes those.  If they have negligible mass compared to the fuel created this would approximate the way it works in-game.

The retention of mass was a deliberate thing by @RoverDude, as indicated by his comments in the various Ore-related config files.  He was concerned about the conservation of mass but I believe he was misapplying that.  The word "refining" means separating the good stuff from the bad stuff, which necessarily means you keep less than all of the mass you started with.  I think RoverDude kept all the mass to avoid having to create a new resource for the by-products, and tanks to put it in, and then a mechanism for jettisoning it.  But I'd be fine with the by-product mass just disappearing, on the assumption that it's being vented from the ship automatically.  If you want to get fancy with that, add a stream of dust blowing out of the ISRU and say that's the byproduct mass leaving the ship.

Of course, if you lose mass during refining, that makes non-combustible Ore even more of a violation of the 2nd Law of Thermo than it already is.  Now the ISRU reaction most definitely is irreversible because not only are you losing energy as waste heat, but you're also losing mass.  Which means that the stock game must have a way to burn raw Ore.  Only when Ore itself burns in stock can we reasonably analogize Ore to hydrocarbons.

So now let's talk about hydrocarbons.   Crude oil is a mix of many different compounds with varying properties.  The ones we use as fuels are amongst the lighter varieties.  The heavier compounds get used a low-energy fuel oil, turned into plastics, or used to pave highways.  So if you're looking to make rocket fuel, which the Rocket Equation dictates must be high-energy and low-mass, the mass loss would be quite significant.

This leads to the question of finding industrial quantities of hydrocarbons on lifeless worlds.  Sure, hydrocarbons are scattered in relatively tiny amounts all over space, but are only highly concentrated in a few places (as far as we know), like Earth and Titan, and (as far as we know) something like crude oil only exists on Earth.  But that's our universe.  In the totally alien KSP universe, where chemistry is totally different from ours, we can assume that magic Ore is magically everywhere.

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Actually, the original issue was that in order to have asteroids 'worth it' (they were to be the primary ore source, planetary ore was secondary), we either had to increase asteroid mass (which broke contracts), break conservation of mass with the drills (i.e. get more ore out of an asteroid than it had mass), or compromise with a lossless conversion (which is the option we chose).

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Hmm.  Once I looked at the configuration file myself, I noticed that nowhere is the word 'refining', or any variant thereof, used.  The description does mention hydrogen and oxygen, which would imply water (no ore in the oceans can be explained away by saying that liquid water corrodes the converter, and <insert hand-wave here>), but I'm equally willing to accept that ore is a hydrocarbon soaked into oxygen-bearing rock, or possibly a collection of oxides, or even comes from all the rocket exhaust that was sprayed everywhere before version 1.0--that one explains why it so readily converts into both rocket fuel and monoprop, too.  The word ore only signifies what is worthwhile to dig up; it's nothing more specific than that.

Lossless electrolysis, hydrocarbon cracking, or conversion of organic oxides into some cousin of methalox or kerolox is possible, of course, but the inputs must be pure for obvious reasons.  That implies that the drills really can discriminate the wanted ore from the unwanted cuttings--a notion supported indirectly by the fact that engineers bump the drill efficiency--but that's beside the point.

The point is that in a lossless conversion, there ought to be a veritable river of energy running into the process, because such a process must move around a lot of heat.  Really, the question--and if I'm putting words into anyone's mouth, you have my sincerest apologies--is only one of energy input and output, not one of the particular makeup of ore.  There would be no problem with the notion that ore can be made of anything at all that can be converted into some kind of propellant, so long as the total energy put into that process is thermodynamically equal to the energy coming out of it, in light of the heat and chemistry involved.  Energy is lost as heat energy and as molecular bond energy, and it must come from somewhere.  That also constitutes work and therefore there is a thermal change, as well.  Either the energy to do this must be present in the ore harvested (which you see in the uranium-harvesting mods and is why nuclear reactors can justifiably power their own refineries), or it must be supplied by something outside the system.  If ore needs to be cracked by electrolysis, it cannot have that energy:  electrolysis supplies that energy.  Solar panels, RTGs, and mod-supplied nuke reactors pose no problem because they get their energy from outside the system.  The alternator on a Rhino engine poses no problem; it gets energy from the system but gives back less, resulting in a net loss.  The real issue is the fuel cells, because they are the only things that can supply more energy to the system than the system supplies to them.  They shouldn't produce enough energy to make their own fuel indefinitely, if the fuel-making is a lossless process.

The answer:  I think I'll play with the configs a bit and see whether I can find some settings for consumption or production I like and that also obey thermodynamics.  This may not be possible within the game environment.  I'll find out.

Edited by Zhetaan

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14 hours ago, Geschosskopf said:

Well, you should NOT be able to power a mining and refining set-up on fuelcells burning the fuel produced by the system.  That would violate the 2nd Law of Thermodynamics.  At least in stock, where unrefined Ore doesn't burn.  Now, if you've got mods that have Ore-burning generators, then Ore can be regarded as analogous to a fossil fuel and you're just releasing stored energy, not chasing your tail, so no 2nd Law violation.  But OTOH, how did fossil fuels get on lifeless planetoids?

So, when I'm mining out where the sun don't shine, I use nuclear reactors, either fission or fusion depending on what mods I have in that game.

Yes, it always bugged me, fossil fuel work as it can be burned with oxygen however this would not work outside of kerbin and laythe. 

In gameplay at least back in 1.04 the drills and ISRU did not use power then not loaded but it still produced ore. 
Has not tested this yet in 1.12 but its an high chance its still the same. Don't work the same way as timewarp, first the ore tanks fills up then the fuel tanks you have activated convention for. 
 

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Well, it took a month (real life strikes again!), but I've made some progress.  Warning:  long post ahead.

Summary (this is the on-topic part; I haven't forgotten you, @davidpsummers!):

There are two ISRU converters and two kinds of fuel cell.  I chose not to consider the drills in this first pass because drilling Ore is simply moving it around--it's supplemental, not integral, to the conversion process.  The conversion of electricity into fuel and then fuel into electricity is the loop that breaks thermodynamics:  because Ore doesn't also come out of the fuel cells for potential recycling, I cannot reliably gauge its involvement in the process.

The two kinds of fuel cell are interchangeable (with a scalar multiple) in terms of production of electricity and consumption of fuel:  the Fuel Cell Array produces and consumes exactly as much as twelve Fuel Cells (in spite of its visual model and the description in the editor).  The mass, EC capacity, impact and thermal tolerances, &c. are different, but again, they have nothing to do with power conversion, so I ignore them.

The ISRU converters are completely different from one another:  each requires the same amount of electrical charge (30 EC/s) for any of its conversion processes, but the smaller 125 takes fives times as much Ore to produce half the fuel in a given period of time compared to the larger 250.

In terms of mass efficiency per unit of Ore, the 125 is 10% mass efficient when producing LF, O, or LF+O, and 8% mass efficient when producing Monopropellant.  The 250 is 100% efficient when producing LF, O, and LF+O, and 80% mass efficient when producing Monopropellant.  This is, as I said above, not really relevant to the fuel-power-fuel feedback loop, but I do find it chemically interesting that the 250 is able to take the same piece of Ore and convert it into 100% LF or 100% Ox.  I'll be revisiting that balance later in my own playthrough.

In terms of electrical efficiency, the 125 costs 12 EC/kg for LF+O and 15 EC/kg for MP; the 250 costs 6 EC/kg for LF+O and 7.5 EC/kg for MP.  There are no MP-based fuel cells, but since it's less efficient anyway, the really important figure is the 250's LF+O conversion rate.  That's the cheapest one, meaning that you get the most fuel for the least power.

Compare the Fuel Cell (Array):  its electrical efficiency is 80 EC/kg:  more than thirteen times the electrical power produced as the 250 consumes.

5 Fuel Cell Arrays (90 ec produced) will exactly operate 3 Convert-O-Tron 250s (90 ec drained).

These will draw .10125 LF/s and .12375 Ox/s at a total mass loss of 1.25 kg/s.  The Convert-O-Trons will produce 1.35 LF/s and 1.65 Ox/s at a total mass addition of 3 kg/s.  This means that, so long as Ore is supplied and neglecting the radiator mechanic, such an assemblage will produce 1.75 kg/s of LF+O without any external energy input.  In other words, the quintessential effect is to pass Ore through it and get this much LF+O for absolutely nothing.  Providing the requisite Ore is a separate issue, a question of supply versus demand, and therefore more economic than thermodynamic.  The point is that if the Ore is available, then howsoever it may be supplied, this device causes it effectively spontaneously to turn into that much LF+O.  Because the conversion produces more energy (chemical) than is put into it, this device is a perpetual motion machine of the first kind.  Thermodynamically speaking, one may as well throw a tarpaulin over the assemblage, write PHILOSOPHER'S STONE on it in big, block letters, and go prospecting for lead.  Including radiators increases the electrical consumption by less than 1%, but even if the demand were 100% (i.e., so high as to require double the fuel cells), there would still be a net production of fuel.

Let's look at it in terms of electrical energy:

One Convert-O-Tron 250 will supply enough LF+O to power 22.222222... Fuel Cell Arrays.  In integers, 9 Convert-O-Trons will power 200 Arrays.  The Convert-O-Trons will produce fuel at a cost of 270 EC/s.  The Arrays will produce 3600 EC/s from that fuel.  Thus, this device will produce a net of 3330 EC/s.  Turning it back into a more intuitive comparison, one Convert-O-Tron can ultimately drive the production of 400 EC/s--of which it consumes 30, for a net of 370 EC/s.

Put simply, the only restriction to powering anything with this arrangement is the availability of Ore.  Otherwise, you can just keep adding converters and fuel cells to satisfy any power need.  Therefore, the fuel refinery on Pol needs either a better location or better drill efficiency; those are the only two reasons that you should be running out of power there.

 

Now, for the thermodynamically inclined:

In my own game, I've experimented with both increasing the power requirement for ISRU and reducing the power production of the fuel cell.  My current settings (I haven't had the time to thoroughly test them) are to increase the conversion cost to 40 EC/s for all processes and to increase the fuel consumption of the fuel cells by a factor of 10.  This makes fuel and power production break even, but the need for radiators to have good efficiency forces the ISRU to cost more to run than the fuel cells can supply, in any configuration.  The best efficiency, insofar as I've calculated it, is 99.38%.  That's still unrealistically good, requiring a single solar panel or a single RTG to put it over the top, but it's not 100%, and in a world of magical momentum-dumping reaction wheels and infinite EVA fuel (just remember to come in out of the cold every few hours), I think that fits the overall style.  Realism is for the RO crowd.

Additionally, I really didn't like the '100% LF or 100% Ox' conversion potential of Ore, so I reduced the efficiency of the 250 by a factor of five, which means I pretty much need to send an engineer now.  It also reduces the value of asteroids a lot, but on the other hand, I was visiting asteroids back when they weren't worth anything at all, so it's not really a net loss from my point of view--especially when the blasted things keep spawning on me.

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I completely disagree with the "thermodynamic" argument by Geschosskopf. Obviously the drill is selective and throws away mass. That's the whole point of the .002 number. For every tonne of input material the drill inputs, you get 2kg of "ore". So if I translate the argument into another format: I have a "drill" that uses some energy, takes in water, throws away most of it, and outputs deuterium. I have a "converter" that uses energy, inputs deuterium, and encapsulates it in little glass beads with no mass loss. I have a "fuel cell" that does inertial confinement fusion on the glass beads and converts them into electricity. To claim that this process can never be energy positive is utter BS, and it's a direct analogy to the KSP system.

Just because ore doesn't burn directly doesn't mean it has no energy in it.

Edited by bewing

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@bewing, that's absolutely true, and there are other good examples of this, as well:  you can't put pitchblende in a reactor and expect results, but the uranium in pitchblende is obviously energetic.  Geschosskopf's contention about burning crude fuels is valid only for those fuels that release their stored energy via combustion (e.g. petroleum).  However, there is still a valid thermodynamic concern here.

To borrow your analogy, the idea that you can extract deuterium, fuse it, and get more energy than you put into extracting the deuterium is correct and fully supported thermodynamically.  What is not supported is the idea of taking the fusion products, fissioning them back into deuterium at low energy cost, and fusing them again to get even more energy.

It 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 electrolysis 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 electrolysis, 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.

That being said, a lot of it is moot for this game.  The fact Ore that can be converted 100% into LF or 100% into Ox simply breaks chemistry.  The devs acknowledged it and gave good gameplay reasons for why they chose to do this:  I simply don't agree with those reasons.  Fortunately for me (thanks again to the devs), five minutes and a configuration file later, there's a mod for that.

Edited by Zhetaan

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On 5/10/2016 at 7:01 AM, 5thHorseman said:

I'd have to test it, but I think concentration only affects how much ore you get per second. You always use the same amount of power per second.

I think also lower ore concentration generates more heat.

I'm 90% sure ore concentration increases both mining rate AND electrical consumption. 

Not long ago I made a ship with 4 RTGs and 6 fuel cells to complement them. No solar panels. Depending on ore concentration, I had to turn on the fuel cells. Small ISRU and small drill.

 

By the way, I tend to travel to jool alot, and after the update which nerfed solar panels, I don't ever bother with them anymore. RTGs, batteries and fuel cells for me.

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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.

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