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How do i calculate this -- Using numerous parabolic mirrors to focus light

  1. Sep 27, 2014 #1
    So for my semester project, I was given something that most scientists think is "impossible". Using numerous parabolic mirrors, i need to focus the reflected light onto a mutual point, and achieve a temperature of about 3000*C. Why 3000? That is when thermal decomposition of H2O occurs. What i want to calculate first, is the amount of energy i will need to make this happen. Am i supposed to go with Q=MCdT? The specific heat of H2O changes a lot after so many degrees. My idea was to first calculate from room temp to 100*C, then calculate from 100*C to the degree where the specific heat starts to change again, and then im completely lost, because what specific heat would i use to calculate the rest? Please someone give me a magical equation to solve all my problems:S
  2. jcsd
  3. Sep 27, 2014 #2


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    I can't help with expertise nor magic, but your project caught my eyes with interest. I'm attaching a link to a tech document from Chromalox, an industrial heat corporation. It doesn't offer details up to 3000 C, and yet there seems to be valuable info regarding superheating steam (that's what I searched on). Have you seen this or similar? I smiled when I read a comment in it about "always allow 20% for unknown loss" oo)
    http://www.chromalox.com/...information/...Heater-Selection-Steam-Heating.pdf [Broken]

    What kind of vessel would you contain the water in? What's the part the scientists think is impossible?
    Last edited by a moderator: May 7, 2017
  4. Sep 28, 2014 #3
    The link is broken.
  5. Sep 28, 2014 #4


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    Every temperature range has its value for heat capacitance, and I'm sure there are databases or publications covering that.
    I wouldn't be worried about the water so much. Whatever you use as container, at 3000°C it will glow white and lose significant power to the environment. You will need a really powerful light source to counter that.
  6. Sep 28, 2014 #5


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  7. Sep 28, 2014 #6
    This is not true. Water dissociates at lower temperatures. 3000 is nothing but an arbitrary number.
  8. Sep 28, 2014 #7


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    I'm not sure of the validity of this but as a stab in the dark...

    Perhaps you could assume that the object being heated glows white hot and behaves like a black body. Then use Boltzmann's law to work out the power loss at 3000C.


    Suppose the object is a 1cm cube and has a surface area of 6cm^2. The power loss would be around..

    5.6*10^-8 * 3000^4 * 6 * 10^-4 = 2.7KW

    Solar power being around 1000W/m^2 at mid day on a cloudless day. So I reckon you will need around 3-4 square meters of mirrors at least?

    Is it possible to focus a home made mirror that large down to such a small area?

    This old article has some comments on that ...

  9. Sep 28, 2014 #8


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    It is possible. The maximal temperature you could reach via mirrors is the same as the temperature of the surface of the sun, and you reach it if the object "sees" the sun in every direction. That doesn't make it easy, however.
  10. Sep 30, 2014 #9
    That's too simple because,
    1. C changes with Temperature
    2. you also need energy for the thermal decomposition and
    3. you would neglect the loss of energy due radiation and thermal conduction

    Let's start with the energy collected by the mirrors:

    [tex]\dot Q_{in} = A_m \cdot E_0[/tex]

    where Am is the total area of the mirrors and Eo the local solar constant.

    The heat loss by radiation is

    [tex]\dot Q_{rad} = A_c \cdot \sigma \cdot T^4[/tex]

    where Ac is the area of the collector.

    The heat loss by thermal conduction is

    [tex]\dot Q_{cond} = k \cdot \left( {T - T_0 } \right)[/tex]

    Now it's going to be interesting. The heat required for heating and thermal decomposition of water is the difference of the formation enthalpies. If water with the Temperature To flows into the reactor and a mixture of water vapor, hydrogen and oxygen with the temperature T comes out the corresponding change of enthalpy is

    [tex]\dot Q_H = \dot n_{H_2 O} \cdot \Delta _f H_{H_2 O} \left( T \right) + \dot n_{H_2 } \cdot \Delta _f H_{H_2 } \left( T \right) + \dot n_{O_2 } \cdot \Delta _f H_{O_2 } \left( T \right) - \dot n \cdot \Delta _f H_{H_2 O} \left( {T_0 } \right) [/tex]

    The stoichiometry of the reaction

    [tex]H_2 O \Leftrightarrow H_2 + {\textstyle{1 \over 2}}O_2[/tex]

    results in the following amounts of substances

    [tex]\dot n_{H_2 O} = \left( {1 - x} \right) \cdot \dot n[/tex]
    [tex]\dot n_{H_2 } = x \cdot \dot n[/tex]
    [tex]\dot n_{O_2 } = {\textstyle{1 \over 2}}x \cdot \dot n[/tex]


    [tex]\dot n = \frac{{\dot m}}{{M_{H_2 O} }}[/tex]

    is the total flow into the reactor.

    The degree of conversion x can be obtained from the equilibrium constant

    [tex]K = \frac{{x_{H_2 } \cdot \sqrt {x_{O_2 } } }}{{x_{H_2 O} }}[/tex]

    The amounts of substances give the mole fractions

    [tex]x_{H_2 O} = \frac{{2 - 2 \cdot x}}{{2 + x}}[/tex]
    [tex]x_{H_2 } = \frac{{2 \cdot x}}{{2 + x}} [/tex]
    [tex]x_{O_2 } = \frac{x}{{2 + x}}[/tex]

    resulting in

    [tex]K^2 = \frac{{x^3 }}{{\left( {2 + x} \right) \cdot \left( {1 - x} \right)^2 }}[/tex]

    The last step is the calculation of the equilibrium constant

    [tex]\ln K = - \frac{{\Delta _r G}}{{R \cdot T}}[/tex]

    from the change of Gibbs free energy:

    [tex]\Delta _r G\left( T \right) = \Delta _r H\left( T \right) - T \cdot \Delta _r S\left( T \right) [/tex]

    Now T, To, Am, Ac and dm/dt need to be adjusted in order to balance the sum of all heat flows out:

    [tex]\dot Q = \dot Q_{in} \left( {A_m } \right) - \dot Q_{rad} \left( {A_c ,T} \right) - \dot Q_{cond} \left( {k,T,T_0 } \right) - \dot Q_H \left( {\dot m,T,T_0 } \right) = 0[/tex]

    The required thermodynamic data can be obtained here:

    PS: How can I limit the formulas to a reasonable size?
  11. Sep 30, 2014 #10
    Its not the number which water dissociates, different percentages of H2O dissociate at various temperatures. 3000 is my goal, as i can get a decent 40% decomposition.
  12. Sep 30, 2014 #11
    We haven't decided on what type of vessel to contain the H2O in yet, but we are counting for an 80% heat loss. With all the losses at such high temperatures, rather then working out every loss, we concluded on a 20% efficiency.
  13. Oct 1, 2014 #12
    How are you going to separate the products of the reaction?
  14. Oct 1, 2014 #13
    Probably il make the separation through electrolysis. I dont know what anode and cathode i will use yet, but i will def let you know when i figure that out.
  15. Oct 1, 2014 #14
    Also..I will be giving the current using solar energy as well.
  16. Oct 1, 2014 #15
    Electrolysis can split water into hydrogen and oxygen (you already planned the thermal decomposition for this step) but not for separation of a gas mixture. Or do you mean using ion conductors?
  17. Oct 1, 2014 #16


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    3000 C, huh? That's pretty hot. How do you intend to hold the H2O in place while you heat it up? Are you going to hold it in a pressurized tungsten crucible? The pressure will get pretty high
  18. Oct 1, 2014 #17


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    Certainly easier to do that than build a solar furnace :-)
  19. Oct 1, 2014 #18
    yes electrolysis can split water into hydrogen and oxygen but once the split occurs the oxygen and hydrogen atoms go to the anode and cathode individually..Now, after the split is done using thermal decomposition, even with small current, the atoms will go in there directions to the anode and cathode. Think of it as using thermal decomposition to give a rapid boost to the electrolysis. How do i intend to hold it in place? no clue on that yet. Im currently working on the mathematical parts. All i know now is that i will make my focal point very tiny and will be using 100grams of H2O. My intention isnt to get 100% decomposition. My goal at the end is to strictly use solar resources to get hydrogen. At the end of the day, no matter how slow, this is 100% effeciency. Whats the loss?
  20. Oct 1, 2014 #18


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    I'll go with, "one-hundred grams of water." (Is this a trick question?) :confused:

    It's interesting to follow your design progress, Ali. "Good luck is the residue of preparation."
  21. Oct 2, 2014 #19
    That is a fine goal, but there is no need to re-invent the wheel.This is already being done on an industrial scale in Spain. Perhaps you should study their approach and other research in this field.

    CWatters earlier calculated the power loss of 2.7 kW for 1 cubic cm of water. You are planning to use 100 cubic cm, so your power loss is going to be roughly 60 kW, which means you need to collect solar energy from at least 120 square meters. Do you have budget for this sort of project?
  22. Oct 2, 2014 #20
    Hydrogen and oxygen do not go to cathode and anode. They are produced there.

    You are confusing atoms with ions. Atoms are not charged and will therefore not be affected by the electrodes. If you want to separate hydrogen and oxygen by electric or magnetic fields you would need to heat them up to plasma state. At normal pressure 3000 K wouldn't be sufficient this purpose.
  23. Oct 4, 2014 #21
  24. Oct 14, 2014 #22

    Would you be able to explain this equation a bit more, im kind of lost with the variables..

  25. Oct 14, 2014 #23
    Why are you using deltaF? isnt that the enthalpy for saturated liquid? shouldnt deltaG be used instead? where that would stand for saturated vapor?
  26. Oct 14, 2014 #24
    DeltaF is the enthalpy of formation. The equation describes the heat required for increase of temperature and thermal decomposition of water according to the (simplified) chemical reaction given above.

    DeltaG is the change of Gibbs free enthalpy and it is used for the equilibrium constant.
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