1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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
  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


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook