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#21 |
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Join Date: Aug 2004
Location: Hamilton, Ont. CANADA
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In a perfectly efficient fuel cell or MHD turbine setup, one ton of water should be produced (BTW how do we store that? Do empty tanks have a launch mass?) for each ton of LOH fuel used. Using a "chemical refinery" (SS1 p.19) to crack it back, an SM+x system should process one SM+x tank in 3 hours (if x is odd) or 3h20m (if x is even) for one power point. This is VERY surprising because according to SS1 p.20 a fuel cell produces 12 PPh (Power Point hours) per fuel tank at TL7, 24 PPh/tank at TL8, 48 PPh/tank at TL9 and 96 PPh/tank at TL10+ (an MHD turbine has the same efficiencys at TL9 and TL10+). That's an energy output that is 4 to 32 times the energy input! What ever happened to TANSTAAFL?
Dalton "how do I submit this to Murphy's Laws?" Spence
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#22 |
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Untagged
Join Date: Oct 2004
Location: Forest Grove, Beaverton, Oregon
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The issue is that nearly everyone considers the production facilities and refineries/mining to be grossly over powered. At least one or two Spaceship issues suggest dividing their outputs by 24, if I remember correctly.
__________________
Beware, poor communication skills. No offense intended. If offended, it just means that I failed my writing skill check. |
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#23 |
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Join Date: Jul 2008
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Unfortunately even with that you're getting an efficiency out of the closed loop of 32/24 > 1 at TL10+.
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I don't know any 3e, so there is no chance that I am talking about 3e rules by accident. |
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#24 | |
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Untagged
Join Date: Oct 2004
Location: Forest Grove, Beaverton, Oregon
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Quote:
Batteries or at least capacitors are nice for the enormous but very short lived burst of power needed for exotic modules.
__________________
Beware, poor communication skills. No offense intended. If offended, it just means that I failed my writing skill check. |
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#25 | |
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Join Date: Aug 2004
Location: Hamilton, Ont. CANADA
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Quote:
Dalton "I still think it makes a good Murphy's Laws submission" Spence
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#26 | |
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Join Date: Feb 2005
Location: Berkeley, CA
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Quote:
It is possible to estimate the value of an energy point from performance stats. Looking at a 100 ton (SM +6) ship: Ion Drive: thrust = 100 lbf or 450N, vEx = 60 mps or 96 kps, power = 21.6 MW. Mass Driver: thrust = 2000 lbf or 9000N, vEx = 96 kps, power = 43.2 MW. Weapon, Major Battery: 30 MJ beam output, rate of fire 1/20 (1/10 if 'improved'), power output 1.5 or 3 MW. Now, a fuel cell is 1 ep. The energy content of LOx/LHd fuel is about 15 MJ/kg, so with the entire volume filled with fuel (4,500 kg) total energy available is 18.75 MWH. It is likely part of the problem here is the drives, since that's a power density of ~5 kW/kg for the ion drive, ~10 for the mass driver, and thus requires ~10 kW/kg for a TL 8 nuclear reactor, which is at least an order of magnitude too high. If we reduce power to a merely optimistic 1 kW/kg, reduce MD thrust by a factor of 10, and ion drive thrust by a factor of 5, we can justify an hour or two at TL 8, except that fuel cells have nothing like that power density (they don't even have a tenth that). Last edited by Anthony; 06-20-2012 at 06:57 PM. |
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#27 | ||
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Join Date: Jul 2008
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Quote:
Quote:
__________________
I don't know any 3e, so there is no chance that I am talking about 3e rules by accident. |
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#29 | |
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Join Date: Feb 2006
Location: Not in your time zone:D
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Quote:
Put simply, instead of output per hour, output per 48hrs at TL7, to per 36hrs at TL10. Although, if the numbers in mining/refining are right, that'd make all the power systems wrong (brain melting)
__________________
"Sanity is a bourgeois meme." Exegeek PS sorry I'm a Parthian shootist: shiftwork + out of country = not here when you are:/ It's all in the reflexes |
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#30 |
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Join Date: Feb 2005
Location: Berkeley, CA
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