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Homework Help: Hydrogen to Helium Reactor, how much hydrogen to power earth?

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose we succeed in building a H-->He fusion reactor. How much hydrogen would have to be converted per second to supply the world's electricity consumption of 10^13kwh.

    2. Relevant equations

    This is what I am looking for, I think i can figure it out once I know the appropriate equation

    3. The attempt at a solution

    My attempt is to first figure out the energy output of when hydrogen turns into helium. So i take 4 hydrogen at 1.00794 amu a piece, then 1 helium at 4.002602 amu and subtract the totals. Thus:

    (1.00794 x4) - 4.002602 = .029158 amu I assume this number is the mass converted to energy?

    From my book 1g of H--> HE is 650,000,000 kw of energy output (converted from 6.5x10^18 erg) So from there couldnt I figure this problem out by just finding how many seconds are in a year. Then dividing 10^13kwh by the total amount of seconds, then with knowing the output of 1g convert my Kw per second into amount of material?

    I would like to know if im on the right track or if I am disregarding something important (efficiency isn't involved).
     
  2. jcsd
  3. Feb 9, 2012 #2

    Delphi51

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    Homework Helper

    That all sounds like it will work, but it is tricky.
    My instinct is to work with energy, converting 10^13kwh into Joules.
    "1g of H--> HE is 650,000,000 kw of energy" is not clear - the units kw are for power, not energy. Anyway, kind of awkward so better to use E = mc² to convert the .029158 amu (change to kg first) into Joules. Then you can divide to see how many 4-hydrogens must be fused.

    Are you sure of your reaction and the mass converted to energy?
    It might be worth reading
    http://en.wikipedia.org/wiki/Proton–proton_chain_reaction
     
  4. Feb 9, 2012 #3
    haha just so happen to be reading that as you posted..
     
  5. Feb 9, 2012 #4
    So I took E=mc^2 and i plugged in:

    (0.2916 amu x 1.66053 x10^-27 ) to convert amu to Kg

    then that by c^2 which is 9x10^16 m/s

    and got 4.358x10^-12

    is that joules released by combining 4H to 1 he?
     
  6. Feb 9, 2012 #5
    With my previous method before your help, I came to about 1.7g per second needed. That seem reasonable?
     
  7. Feb 10, 2012 #6

    Delphi51

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    Homework Helper

    Yes to the 4.358 E-12 for converting 4H.
    I end up with a 1000 times less mass per second than you got, but I'm not so great at calculating these days. When I converted the 1E13 KWH to Joules I got 4.35 E -12. Dividing by the energy of 4H gave me 8.28 E27 4H's that must be converted per year.
     
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