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Time for which RTG will supply the required energy output?

  1. Oct 29, 2016 #1
    1. The problem statement, all variables and given/known data
    A newspaper article stated that the NASA Galileo space probe to Jupiter 'contained 49 lb of plutonium to provide 285 watts of electricity through its radioactive thermonuclear generator (RTG)'.

    (Note: An RTG is a device for converting thermal energy produced by fission into electrical energy.)

    Assuming that the plutonium is 239Pu, which is built into a small nuclear reactor and that the efficiency of the RTG is 10 %, what is the maximum time for which the RTG will supply the required energy output?

    (Take the energy emitted for each nuclear disintegration of the 239Pu to be 32 pJ, NA = 6.0 * 1023 mol-1, 1 lb = 0.45 kg.)

    Answer: 6.2 * 1011 s.

    2. The attempt at a solution
    We have the mass m = 22.05 kg, power P = 285 W, element 239Pu, energy emitted per disintegration E = 32 * 10-12 J and the Avogadro number with the fact that the RTG is 10 % efficient.

    Power P = Work done W / Time t where, as I understand, t is what we need to find.

    We know the energy emitted, but how many disintegrations are there? It says "for each disintegration", but how to find how many of them are there are what kind of disintegration a nuclear disintegration is?

    So, I think it is required to find the number of disintegrations, then calculate the total work done and then calculate time. Though I'm not sure about the 10 % efficiency. Efficiency = Power output / Power input. So for the final calculation it should be used 285 * 10 % = 28.5 W (like 28.5 W = 32 * 10-12 J * X number disintegrations / Y maximum time for which the RTG will supply the required energy output).

    Is this logic correct? How to find the number of disintegrations?
     
    Last edited: Oct 29, 2016
  2. jcsd
  3. Oct 29, 2016 #2

    Jonathan Scott

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    Gold Member

    To find the disintegration rate, you need the half life of 239Pu. If it's not supplied in the question, you need to look it up.

    10% efficiency means that only 10% of the energy output is converted to electricity. If the requirement is for 285W of electricity, then the total energy output is obviously much higher.
     
  4. Oct 29, 2016 #3
    OK, so we have T1 / 2 = 2.4 * 104 years or * 365 days * 24 hours * 3600 minutes and seconds = 7.56864 * 1011 seconds.

    You mean 10 % of the energy INput is converted into electricity (OUTput)?

    So the requirement is to have 285 W as output, so input should be 2850 W.
     
  5. Oct 29, 2016 #4

    Jonathan Scott

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    That's close enough. By my way of thinking, the plutonium is producing a total thermal energy output of 2850W and the RTG converts 285W of that to electric power.

    Now you need to use the usual equation to determine the disintegration rate from the half life.
     
  6. Oct 29, 2016 #5

    gneill

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    Staff: Mentor

    Something tells be that at some point the original problem was "updated", changing the isotope from PU 238 to Pu 239. Both the Cassini and Galileo missions used Pu 238 RTGs.

    That said, I have my doubts that the original spec of 49 lb of plutonium would produce enough energy to harvest 285 W if it were Pu 239 and the RTG only 10% efficient. Even initially, never mind over time.

    Specific activity of Pu 239 is only 0.06133 Ci/gm (that's curies* per gram) compared to Pu 238's 17.13 Ci/gm. See:
    LLNL-TR-490356: Plutonium 239 Equivalency Calculations, J. Wen, July 2011.


    * A curie is a unit of activity. 1 Ci = 3.7 x 1010 decays per second.
     
  7. Oct 29, 2016 #6
    dN / dt = - ln 2 / T1 / 2 * N

    But how to find N?

    We have 239 gram of plutonium which have 6 * 1023 atoms. But we also have like 22.05 kg, which cannot contain so much atoms, right?

    Update
    I used N = m NA / Ar, where m is in grams = 22 050 * 6 * 1023 / 239 = 5.5 * 1025 atoms.

    dN / dt = - ln 2 / (7.6 * 1011) * 5.5 * 1025 = 5.1 * 1013 Bq.

    t = 32 * 10-12 * 5.1 * 1013 / 2850 = 0.563 s.

    In the original post I only changed "plutiniom" (or something close to it) to "plutonium". The numbers are all correct, checked them with the book. I think it's just a made-up problem, the only quote from the article they are giving is in
     
    Last edited: Oct 29, 2016
  8. Oct 29, 2016 #7

    gneill

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    Calculate the initial power output of your 49 pounds of Pu 239. You've determined that it has an activity of 5.1 x 1013 Bq, and you're told that each decay delivers 32 pJ of energy. So, what does that give you?
     
  9. Oct 29, 2016 #8
    Yes, and I calculated 32 pJ * 5.1 * 1013 Bq.
    Since Bq is disintegration per second, so if we multiply these two numbers we'll get total Work done W.
     
  10. Oct 29, 2016 #9

    gneill

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    No, I asked for the power output. You've gone and divided by 2850 for some reason. If that 2850 is a power in watts, then your calculation would yield a unitless number, not a time.

    In fact, if 2850 W is the necessary minimum power output of the Plutonium, what you've shown is that the RTG will only produce about 56 % of the required minimum amount.
     
  11. Oct 29, 2016 #10
    What shall I do then? I still don't understand what do you suggest, sorry.
     
  12. Oct 29, 2016 #11

    gneill

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    I'm saying that the problem is not solvable as it is given. The author of the problem may have changed some values and not properly checked the result to see that the problem scenario was valid.

    If you can prove that the problem is not workable then that is a valid answer.

    The problem author may have misinterpreted a result when he made checked it. I think he ignored a sign when he solved for the time and interpreted a negative time value as a positive one. Allow me to demonstrate.

    Suppose that when brand new the RTG produces a power output ##P_o##. The RTG's isotope has a decay constant ##λ##, so that the power output will decay over time according to:

    ##P(t) = P_o e^{-λt}##

    If the minimum usable power output is ##P_{min}## then solve for t in:

    ##P_{min} = P_o e^{-λt}##

    ##t = \frac{-ln\left( \frac{P_{min}}{P_o} \right)}{λ}##

    Now, if ##P_{min}## is less than ##P_o## then all is well: the ln function will return a negative value, cancelling the negative sign in front of it. But if ##P_{min}## happens to be greater than ##P_o##, the result of the ln is a positive value and the overall result will be negative.

    I plugged in the values associated with this problem and obtained a time of ##t = -6.24 \times 10^{11} s##.

    I think that this is pretty good evidence that the author mucked up :smile:
     
  13. Oct 29, 2016 #12
    t = - ln (285 / 2850) / (ln 2 / 7.6 * 1011) = 2.52 * 1012 s. What am I missing?
     
  14. Oct 29, 2016 #13

    gneill

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    You've used incorrect power values. The minimum required RTG output power ##P_{min}## is 2850 W (so that the 10% recoverable is 285 W). The initial power output of the RTG, ##P_o##, is the power output you calculated for the given initial mass of isotope. I asked you to calculate it in post #7.
     
  15. Oct 29, 2016 #14
    But power has no connection to Bq. At least I can't find any formula on that.

    I thought since dN / dt = disintegrations per second I could use that number to multiply on 32 pJ and find total work done W in Joules.
     
  16. Oct 29, 2016 #15

    gneill

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    Bq is decays per second. Each decay produces a bit of energy. Energy per second is power.
    Yes, that's right.
     
  17. Oct 29, 2016 #16
    Yes, the second P = 1622 W. Now I also get - 6.15 * 1011 s.

    Thank you!
     
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