1. Not finding help here? Sign up for a free 30min 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!

Fission Reactor

  1. Nov 14, 2009 #1
    A fission reactor operates at 2700 MW level. Assume all this energy comes from the 200 MeV released by fission caused by thermal neutron absorption by 235U. At what daily rate is the mass of 235U used? (In practice, of course, the energy conversion is not 100% efficient, nor is all the 235U in a fuel cell used.)

    Answer is in kg/day

    I really don't know how to solve this question. Am I supposed to convert the 2700MW first into MeV/day and then get it into kg by using Avagadro's Number??? I have no idea. Please help!!!

    If you are supposed to do that then u would do:
    2700MW * 1e6 (convert into watts) * (1ev/(1.6e-19)) * (1MeV/1e6)*(86400s/day) ???
     
  2. jcsd
  3. Nov 14, 2009 #2

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Exactly, just convert 2700MW into J/s
    Then work out how many 200Mev this is per second, then per day
    Finally how many kg of U235 is this many atoms ( using Avagaro's number)

    hint - it might help to have an order of magnitude estimate before you put so many large numbers into your calculator
     
  4. Nov 14, 2009 #3
    what I don't understand is what the 200MeV energy that is created by fission is used for? is it suppose to equal something?
     
  5. Nov 14, 2009 #4

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Making electricity - that's not really part of the question
     
  6. Nov 14, 2009 #5

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Moved: this is homework.
     
  7. Nov 14, 2009 #6
    ok i am still not getting the answer...

    So this is what I am doing...

    http://img404.imageshack.us/img404/1023/webass.png [Broken]

    after that i dont know how to factor avagadro's number...so i just straight out multiply or do i need to use the molar mass. if i use the molar mass then the kgs cancel out...

    plus am i using the conversion factor of 1.6605e-27 kg / 931.5Mev/c2 correctly???


    PS: sorry about it being in the wrong section..
     
    Last edited by a moderator: May 4, 2017
  8. Nov 14, 2009 #7

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    2700MW = 2.7 E9 J/s
    200 Mev = 200E6 * 1.6E-19 J = 3.2E-11 J
    This gives you the number of U235 fissions/second and so per day - that you need to make to generate that amount of power.

    You are just using the 200Mev to count the number of U235 used up.
    Each U235 atom has a mass of 235 amu, 1 kg is 6.02e23 amu (this is avagadros number)
    Then it's just a case of being careful entering the numbers.

    hint - how big an answer do you expect?
     
  9. Nov 14, 2009 #8
    lol

    "2700MW = 2.7 E9 J/s" ---- Got that!

    "200 Mev = 200E6 * 1.6E-19 J = 3.2E-11 J" ------- Got that!

    "This gives you the number of U235 fissions/second and so per day - that you need to make to generate that amount of power." --- what does -- dividing 2.7e9J/s by 3.2e-11?? i am really sorry im being an idiot right now..i have lots of midterms next week and just cant focus properly

    You are just using the 200Mev to count the number of U235 used up.

    Each U235 atom has a mass of 235 amu, 1 kg is 6.02e23 amu (this is avagadros number) ---- isnt 1g = 6.02e23 amu so 1kg = 6.02e26? rite???

    Then it's just a case of being careful entering the numbers.

    hint - how big an answer do you expect? ---- I really don't know how big to expect -
     
  10. Nov 14, 2009 #9

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Yes, one 200Mev event is formed by a breakdown of a single U235 atom.
    So to get 2.7E9 J/s from 3.2E-11J events you need 2.7E9/3.2E-11 of them per second
    .
    Correct - a mole is the number of atoms in 235grams of U235 (that's what I meant about checking numbers)
    Finally you know that you don't need train loads of uranium arriving at a power station every day, so you can guess whether you expect an answer of a few grams or a few tons.
     
  11. Nov 15, 2009 #10
    hey thanks a lot for ur help it worked

    sorry for being an idiot

    lol
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fission Reactor
  1. Fission reactor (Replies: 0)

  2. Fission of nucleus (Replies: 17)

Loading...