As I'm seeing it the eV to BTU conversion is a killer. Richard, I'm sure you'll agree....
Assuming for simplicity you could fuse 1 whole mole of D2
That's 6.02E+23 Neutrons
multiplied by 2,400,000 for 1.44E+30 electron volts
multiplied by 1.519E-22 to give 219465120 BTUs
then to go from BTU to kWh you divide BTUs by 3412.
which gives 64325.22667 kWh
and since we're only at about $0.11 per kWh that's a whole
$7075.774934 worth of energy you've produced.
So as I see it that's 2 grams of D2 for only $7000 worth of energy.
(I haven't taken into account the actual facility and process losses and wholesale electricity costs. Conversions are from http://www.convert-me.com )
If you guys notice any significant errors please let me know.
Fusion Costs and Conversions (Beancounter Thread)
-
- Posts: 266
- Joined: Wed Mar 19, 2003 3:50 am
- Real name: Edward Miller
- Contact:
-
- Posts: 21
- Joined: Thu Dec 29, 2005 3:01 pm
- Real name:
- Contact:
Re: Fusion Costs and Conversions (Beancounter Thread)
haha. this is the best "if-only" post i've seen in awhile.
today i was talking to some guy who was POSITIVE fusion HAD been created as a viable energy source, but the costs of operating it were so high that they had to shut it down...... i ought to show him this formula
fusion + if only = $$
-brett
today i was talking to some guy who was POSITIVE fusion HAD been created as a viable energy source, but the costs of operating it were so high that they had to shut it down...... i ought to show him this formula
fusion + if only = $$
-brett
- Carl Willis
- Posts: 2841
- Joined: Thu Jul 26, 2001 7:33 pm
- Real name: Carl Willis
- Location: Albuquerque, New Mexico, USA
- Contact:
Re: Fusion Costs and Conversions (Beancounter Thread)
Fusion by Tokamak is close to being a "viable" energy source in the sense that it is considered physically practical even if the price point must be very high. ITER is under construction right now, and it will produce 0.5 GW(th) continuously by 2016 via DT fusion, with a power gain of 10 or so over the input requirements, if all goes according to plan. No efforts are planned to extract high-quality heat and make electricity with this machine; the heat will be dumped to the environment. Nonetheless, controlled-fusion power is right on the horizon. The "if only" is thought to be out of the equation though the "$$" could be a show-stopper.
-Carl
-Carl
Re: Fusion Costs and Conversions (Beancounter Thread)
Ed - I would do the calculation this way: (It gives a bigger result, too).
[*** I edited this to delete protons, and threw in a 33% thermal recovery percentage, for a real world $$ number. *** DC.]
AT 1 MOLE OF D2 FUSED: 6.02x10+23 molecules... which is 2 x 96500 coulombs of charge at 1.6x10-19 coulombs per particle.
So at 2.4 MEV per particle emitted, (2.4 million joules per coulomb) x 2x 96500 coulombs, this gives you joules...counting only the neutron. .
So results are pretty straightforward:
2 moles of neutrons at 2.4MEV = 4.632 x10+12 joules,
Counting just neutrons: 4.632 x10+12 j..
Now this converts directly to kwhr by dividing by 3.6 x10+6
So grand total is : 128,667 kwhr for neutrons only.
If you consider a thermal recovery only, then not more than 33% is a reasonable fraction. If you consider something exotic - yet to be developed, then pick a value, and might as well use 100%.
Over here, (USA) in expensive energy-land So. Calif. at upper tier prices of )$0.23/kwhr, that's a respectable $59.2 Kx 1/3 = $19.73K. And not shabby at even $0.11/per kwhr.
'Course... does anyone know how to do this???
Dave Cooper
[*** I edited this to delete protons, and threw in a 33% thermal recovery percentage, for a real world $$ number. *** DC.]
AT 1 MOLE OF D2 FUSED: 6.02x10+23 molecules... which is 2 x 96500 coulombs of charge at 1.6x10-19 coulombs per particle.
So at 2.4 MEV per particle emitted, (2.4 million joules per coulomb) x 2x 96500 coulombs, this gives you joules...counting only the neutron. .
So results are pretty straightforward:
2 moles of neutrons at 2.4MEV = 4.632 x10+12 joules,
Counting just neutrons: 4.632 x10+12 j..
Now this converts directly to kwhr by dividing by 3.6 x10+6
So grand total is : 128,667 kwhr for neutrons only.
If you consider a thermal recovery only, then not more than 33% is a reasonable fraction. If you consider something exotic - yet to be developed, then pick a value, and might as well use 100%.
Over here, (USA) in expensive energy-land So. Calif. at upper tier prices of )$0.23/kwhr, that's a respectable $59.2 Kx 1/3 = $19.73K. And not shabby at even $0.11/per kwhr.
'Course... does anyone know how to do this???
Dave Cooper
Re: Fusion Costs and Conversions (Beancounter Thread)
Yes - just ain't got the Dollars - sorry, Beans!
Re: Fusion Costs and Conversions (Beancounter Thread)
- if I have - does that make me a 'has beans'?
- Richard Hull
- Moderator
- Posts: 15023
- Joined: Fri Jun 15, 2001 9:44 am
- Real name: Richard Hull
Re: Fusion Costs and Conversions (Beancounter Thread)
Of course, as we have discussed before, if ITER works and runs 24-7 for a year or two at 10X output, a 10GW fusion power plant would need a 1 GW crank up source assuming once it got going you could use part of its output power recycled after a day or two.
The big questions are, is material science up to GW fusion levels 24-7? Will ITER really work? If a lot of tweeks are needed to get it to work, another, bigger, highly modified ITER might be needed to prove further concepts. (add twenty years after 2016 for solid 24-7 proof of concept and maybe another 10-15 years for the first on-grid power reactor. ]
Of course, all this could shake out to fusion grid power in only 15 years from today if a world wide enegy crisis forces us to "manhattan project" fusion power.
OR
We could never see fusion power if world-wide infrastructure collapse due to war, anarchy, massive nuclear exchange, etc., leaves only 100 million inhabitants on the planet. (most of these would be in the more low tech non-bombed or fourth world areas. Places where the people were already scratching the earth for a living. Not many fusion techies here to do fusion.
Putting food in one's mouth and not getting killed in the next 24 hours by marauding bands of thugs is always a priority over electrical power to watch the Simpsons on TV.
So very much depends of the most delicate of societal balances and world conditions remaining fairly status quo. We already see this erroding like sand grains under our feet as the ocean water sweeps back out to sea.
Time will tell and is the final arbiter in such matters.
The overall thrust of the original post is correct "power is money", but it needs to be reversed to "money is power" if one is to really understand the world of people or of fusion. Only in the most desparate of times is power, money.
Richard Hull
The big questions are, is material science up to GW fusion levels 24-7? Will ITER really work? If a lot of tweeks are needed to get it to work, another, bigger, highly modified ITER might be needed to prove further concepts. (add twenty years after 2016 for solid 24-7 proof of concept and maybe another 10-15 years for the first on-grid power reactor. ]
Of course, all this could shake out to fusion grid power in only 15 years from today if a world wide enegy crisis forces us to "manhattan project" fusion power.
OR
We could never see fusion power if world-wide infrastructure collapse due to war, anarchy, massive nuclear exchange, etc., leaves only 100 million inhabitants on the planet. (most of these would be in the more low tech non-bombed or fourth world areas. Places where the people were already scratching the earth for a living. Not many fusion techies here to do fusion.
Putting food in one's mouth and not getting killed in the next 24 hours by marauding bands of thugs is always a priority over electrical power to watch the Simpsons on TV.
So very much depends of the most delicate of societal balances and world conditions remaining fairly status quo. We already see this erroding like sand grains under our feet as the ocean water sweeps back out to sea.
Time will tell and is the final arbiter in such matters.
The overall thrust of the original post is correct "power is money", but it needs to be reversed to "money is power" if one is to really understand the world of people or of fusion. Only in the most desparate of times is power, money.
Richard Hull
Progress may have been a good thing once, but it just went on too long. - Yogi Berra
Fusion is the energy of the future....and it always will be
The more complex the idea put forward by the poor amateur, the more likely it will never see embodiment
Fusion is the energy of the future....and it always will be
The more complex the idea put forward by the poor amateur, the more likely it will never see embodiment
Re: Fusion Costs and Conversions (Beancounter Thread)
You are almost right aka within an order of magnitude.
To get 6.02E+23 reactions you need 6.02E+23 D2 molecules, which
would weigh 4 grams. But, only 1/2 those reactions would give you a
Neutron so you only get 1/2 as many Neutrons as you thought.
(Which is not a bad thing but most fusers take into account the
energy released as those Neutrons destabilize other atoms in the
system…)
From an energy perspective 6.02E+23 Reactions would release 4.03
MeV (you get energy on both particles not just the Neutrons) * 6.02
E+23 > 2.43E+23 * 1.619E-22 > 3.93E+08 BTU.
Now for BTU > kWh > $$$ I would assume your operating at 40%
efficiency and selling at 6c/kWh wholesale, but I will also run it your
way for comparison.
My way:
(3.93E+08 BTU *.4) /3412 > 46046.79kWh *.06 $/kWh > 2762.8$
Your way:
3.93E+08 BTU /3412 > 115117kWh *.11 $/kWh > 12662.87$
To summarize:
When working though these numbers it best to think in terms of # of
reactions. That way you can see that 1 mole of D2 > 6.02 * 10 ^ 23
D-D reactions which would produce .5 mole of T (at 1.01 MeV), .5
mole of p (at 3.02 MeV), .5 mole of 3He (at .82 MeV) and .5 mole of N
(at 2.45 MeV). As to the weight thing 1 mole of D weighs about twice
what 1 mole of Hydrogen weighs and you have 2 moles of D so it’s
about 4 grams.
PS Check out http://en.wikipedia.org/wiki/Nuclear_fusion for a good
and fast look at fusion.
To get 6.02E+23 reactions you need 6.02E+23 D2 molecules, which
would weigh 4 grams. But, only 1/2 those reactions would give you a
Neutron so you only get 1/2 as many Neutrons as you thought.
(Which is not a bad thing but most fusers take into account the
energy released as those Neutrons destabilize other atoms in the
system…)
From an energy perspective 6.02E+23 Reactions would release 4.03
MeV (you get energy on both particles not just the Neutrons) * 6.02
E+23 > 2.43E+23 * 1.619E-22 > 3.93E+08 BTU.
Now for BTU > kWh > $$$ I would assume your operating at 40%
efficiency and selling at 6c/kWh wholesale, but I will also run it your
way for comparison.
My way:
(3.93E+08 BTU *.4) /3412 > 46046.79kWh *.06 $/kWh > 2762.8$
Your way:
3.93E+08 BTU /3412 > 115117kWh *.11 $/kWh > 12662.87$
To summarize:
When working though these numbers it best to think in terms of # of
reactions. That way you can see that 1 mole of D2 > 6.02 * 10 ^ 23
D-D reactions which would produce .5 mole of T (at 1.01 MeV), .5
mole of p (at 3.02 MeV), .5 mole of 3He (at .82 MeV) and .5 mole of N
(at 2.45 MeV). As to the weight thing 1 mole of D weighs about twice
what 1 mole of Hydrogen weighs and you have 2 moles of D so it’s
about 4 grams.
PS Check out http://en.wikipedia.org/wiki/Nuclear_fusion for a good
and fast look at fusion.