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If the inventory of the reactor of about 20,000kg

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data
    If the inventory of the reactor of about 20,000kg of Zr given the reaction:
    Zr+2H2O---> ZrO2+2H2, Q=616MJ/Kg-mole of Zr

    1- find the 10% of the heat generated from Zr oxidation by steam in MJ
    2- Find the 10% hydrogen generated from Zr oxidation by steam in kg and the volume fraction (given initial pressure 1 bar and volume 90,000m^3, treat hydrogen as an ideal gas.
    3- what is the energy released from the combustion of H generated and what will be the power if it is released within 10s, assumption is the amount of Oxygen is enough to burn the total H generated, heat of combustion is 240MJ/Kg-mol
    4- what will be the pressure increased by the H generation from 10% oxidation given initial pressure 1 bar and volume 90,000m^3, treat hydrogen as an ideal gas, ignorw the T rise.


    2. Relevant equations



    3. The attempt at a solution
    1- it is simple :
    molar mass for Zr is 91.22 kg/kmole

    10%x 20000kgx 616/91.22 =13.5x10^3 MJ

    2- it is by: 2kg/kmol of 2H, so 2x13.5x10^3/(616)=43.8kg
    I didnt understand the second part of volume fraction?

    3- energy release from the combustion of H:

    H2+1/2O2---->H2O+ 240MJ/Kgmol

    240x43.8=10.51x10^3MJ

    the power is going to be:

    10.51x10^3MJ/10=10.51x10^2MW

    4-I have no idea but I need a hint

    is my work true
     
  2. jcsd
  3. Oct 12, 2011 #2

    Borek

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    Re: chemistry

    I guess they mean volume fraction in the mixture after the reaction - it still contains remaining water AND produced hydrogen. However, it is not clear to me what is the temperature. Perhaps just 100°C, as that's enough for water to be a steam at 1 bar.

    Convert between mass/moles and volume using ideal gas equation, nothing more fancy.
     
  4. Oct 12, 2011 #3
    Re: chemistry

    so PV=nRT, so in this case 90,000m^3 going to be the total volume right?

    n=2kg/kmol for hydrogen once we find the voulme of hydrogen just divide it by the total volume of 90,000m^3 is it right?
     
  5. Oct 12, 2011 #4
    Re: chemistry

    the T=100c you are right
     
  6. Oct 12, 2011 #5

    Borek

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    Re: chemistry

    Yes.

    Note that this case is specific - total number of moles of gas doesn't change (one mole of H2O is replaced with one mole of H2 - this is in the stoichiometric coefficients). It doesn't have to be true, which may add an additional step to calculations.
     
  7. Oct 12, 2011 #6
    Re: chemistry

    so the is going to be V of hydrogen=8.32x10^3x373x2/1x10^5=31m^3

    which is very small!

    volume fraction=31/90,000=3.45x10^-4!!

    What do you think
     
  8. Oct 12, 2011 #7

    Borek

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    Re: chemistry

    There is in general something wrong with your numbers.

    Why do you mix 616 (which is enthalpy change) with masses, when calculating stoichiometry?
     
  9. Oct 12, 2011 #8
    Re: chemistry

    oops yes you are right its mistake it should be 13.5x10^3/616=21.9kg>>what about now?
     
  10. Oct 12, 2011 #9

    Borek

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    Re: chemistry

    No, it shouldn't be. You can't calculate mass produced using enthalpy change (which would be obvious if you were paying attention to the units). You start with masses, you end with masses. This is a simple stoichiometry.
     
  11. Oct 12, 2011 #10
    Re: chemistry

    I confused really because this is not my background field!
    can you refer me to a wiki link for understanding more because I really lost
     
  12. Oct 12, 2011 #11

    Borek

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  13. Oct 12, 2011 #12
    Re: chemistry

    Zr+2H2O---> ZrO2+2H2
    91.1kg/kmol for Zr.....2kg/kmol of 2H
    10%x20,000kg of Zr......Y

    in this case Y is the required mass for this problem! which is equal to 43.91kg!! almost the same as what Igot before!
     
  14. Oct 12, 2011 #13

    Borek

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    Re: chemistry

    No, it is still wrong. 2 kg is a correct mass of a kilomole of hydrogen, but you are not producing 1 kilomole of hydrogen (H2) per kilomole of Zr.
     
  15. Oct 12, 2011 #14
    Re: chemistry

    91.1kg/kmol for Zr.....4kg/kmol of 2H
    10%x20,000kg of Zr......Y

    87.82kg what about now? this is what i understand from the link? really almost will be crazy from chemistry lol
     
  16. Oct 12, 2011 #15
    Re: chemistry

    can you give me example to understand this problem plz
     
  17. Oct 12, 2011 #16

    Borek

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    Re: chemistry

    Not 4kg/kmol of 2H but 2 times 2kg/kmol of H2. Think what each 2 in 2H2 means.

    That's the correct mass of hydrogen which is a first part of the answer. Not you have to convert it to the volume.
     
  18. Oct 12, 2011 #17
    Re: chemistry

    so the is going to be V of hydrogen=8.32x10^3x373x4/1x10^5=124m^3

    volume fraction=124/90,000=1.38x10^-3!! what about this one?
     
  19. Oct 12, 2011 #18

    Borek

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    Re: chemistry

    Units please - I am not going to guess what you are doing.

    I think you are wrong, but without knowing what each number represents I can't be sure.
     
  20. Oct 12, 2011 #19
    Re: chemistry

    so the is going to be V of hydrogen=8.31 x10^3(pa m^3/kmolK)x373 (K) x4 (Kmol)/1x10^5(pa)=124m^3

    volume fraction=124/90,000=1.38x10^-3!! what about this one?

    this is after showing the units!
     
  21. Oct 12, 2011 #20

    Borek

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    Re: chemistry

    Why 4 kmols? Is that amount of hydrogen produced?
     
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