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I am thinking how you can estimate the mass loss of the fusion energy for 1 kWh.

  • I think you cannot use Einstein's $E=mc^2$ to calculate the mass loss in the fusion reaction of the Sun. How can you estimate the uncertainty of using $E=mc^2$ directly here in calculating the mass loss?

The corresponding reactions are p-p, d-p and He-He, and their corresponding energies per reaction:

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My proposal for the differential:

  • assume there are two materials interacting with concentrations $c_{A}$ and $c_{B}$

\begin{align} \Delta E &= \Bigg| \frac{ \partial E }{ m } \Bigg| \Delta m + \Bigg| \frac{ \partial E }{ \eta } \Bigg| \Delta \eta + \Bigg| \frac{ \partial E }{ P } \Bigg| \Delta P + \Bigg| \frac{ \partial E }{ c_{A} } \Bigg| \Delta c_{A} + \Bigg| \frac{ \partial E }{ c_{B} } \Bigg| \Delta c_{B} \\ &\approx \frac{0.1 keV}{ 25 000 keV} \\ &= 4.01 \times 10^{-6} \%, \end{align}

where $\Delta P = 0.1 \, keV \cdot s$, $\Delta m = 4.01 \cdot 10^{-11}$ kg (rob's answer), and $\Delta \eta = \text{random}$. So

  • each experiment has some device which has power so the differential of the power included in the definition of the energy
  • $\Delta m$ is unknown. Probably this one is in the power of ten to $-14$. What do you think about the uncertainty of the mass?
  • there must be some other terms so $\Delta \eta$ needs definition
  • $\Delta c_{A}, \Delta c_{B}$ are unknown
  • \begin{equation} \frac{ \Delta E }{ \Delta x } = c_{A} \left[ \frac{ \Delta E }{ \Delta x } \right]_{A} + c_{B} \left[ \frac{ \Delta E }{ \Delta x } \right]_{B} \end{equation} (Sauli) which gives me $\Delta E = 2.6 \, \text{keV/cm}$, $N_{P} = 27$ ion pairs/cm, and $N_{T} = 93$ ion pairs/cm in one example of GEM foil -experiment. I think this equation can be used to estimate the uncertainty of the hole energy by differentials. Actually, I do not like this subscript and square brackets way of writing this equation; if you can understand it better, feel free to edit.

Sources

  • Original Sauli, F. Title number 386531. Nucl. Instr. Methd A. 1997. Also in his book, Gaseous Radiation Decetors: Fundamentals and Applications.

How can you estimate the mass loss of the 1 kWh fusion of the Sun?

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closed as unclear what you're asking by Rob Jeffries, Bill N, user36790, ACuriousMind, John Duffield Sep 30 '15 at 12:09

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  • $\begingroup$ It is the analogous thinking that one uses to get the nuclear binding energy curve. en.wikipedia.org/wiki/Nuclear_binding_energy . The nucleus has less mass than the sum of the masses of the nucleons that composed it. $\endgroup$ – anna v Sep 28 '15 at 18:53
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    $\begingroup$ I do not understand how you think that the energy from mass changes from fusion should be different from the energy from mass changes from annihilation. Mass is being turned into (or taking the form of) energy. So, for a given energy value (kWh) you get a certain amount of mass transformed. Done. $\endgroup$ – Jon Custer Sep 28 '15 at 19:28
  • $\begingroup$ I don't understand your edited question and your reference is not valid. $\endgroup$ – rob Sep 29 '15 at 18:25
  • $\begingroup$ Its his original manuscript send to the journal. I added also his book where he explains the formula. $\endgroup$ – Léo Léopold Hertz 준영 Sep 29 '15 at 19:52
  • $\begingroup$ I'm confused. $1\ \mathrm{kWh} = 1000\ \mathrm{W} \times 3600 $ s $= 3.6\times 10^6$ J. Divide that by $c^2$ and you get $4\times 10^{-11}$ kg, not your number. Where did you get your mass number? $\endgroup$ – Bill N Sep 29 '15 at 21:30
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You compute the mass loss exactly by doing $E=mc^2$. One kilowatt-hour corresponds to a mass conversion of \begin{align} m &= \frac E{c^2} \\ & = \frac{ \rm 1\,kW\,h }{ \left( \rm 3\times10^8\, m\,s^{-1}\right)^2 } \\ & = \frac{ \rm 1\,kW\,h }{ \left( \rm 3\times10^8\, m\,s^{-1}\right)^2 } \times \left( \frac{\rm 3600\,s}{\rm 1\,h} \right) \times \left( \frac{\rm 1000\,J}{\rm 1\,kW\,s} \right) \times \left( \frac{\rm 1\,kg\,m^2\,s^{-2}}{\rm 1\,J} \right) \\ \\m&= {4.01\times10^{-11}\,\mathrm{kg} } \end{align} This is quite different from your value of 0.012 kg because I remembered to square the $c$ and you didn't. (Don't feel bad: I forgot too until I was pretty deep into making the unit conversions come out right.)

The remarkable thing about relativity is that the conversion between energy and matter is the same no matter the mechanism: matter-antimatter annihilation (which is included in solar fusion following the emission of positrons), changes in nuclear binding energy, changes in chemical binding energy, and even addition of thermal vibrations to a solid. It's usually not necessary to think about chemical binding energies relativistically because the mass of a typical atom is several $\rm GeV/c^2$, while electronic transitions are only a few $\rm eV$; detecting the inertial difference between a neutral and an excited atom would be a part-per-billion experiment that I don't think has been done.

The uncertainty on this calculation comes mostly from the measured $Q$-values for the different reactions, or equivalently from the nuclear masses. My favorite reference reports "mass excesses," which make it easy to compute $Q$-values using only addition and subtraction, for hydrogen and helium to a precision of $0.1\rm\,keV$. Since the energy liberated in the p-p chain is roughly $\rm25\,000\,keV$, the uncertainty in the liberated energy is something like $$ \delta E = \frac{\Delta E}{E} \approx \frac{\rm0.1\,keV}{\rm25\,000\,keV} = 4\times10^{-6} = 0.00\,000\,4\% $$ To describe the behavior of the actual Sun it's much more important that the solar luminosity $L_\text{sun}$ is known only to about four significant figures.

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    $\begingroup$ The uncertainty comes from the nuclear masses and the reaction $Q$-values. Those uncertainties are quite small. $\endgroup$ – rob Sep 29 '15 at 14:12
  • $\begingroup$ Can you estimate how much from each factor? I think the differential uncertainty computation gives them. $\endgroup$ – Léo Léopold Hertz 준영 Sep 29 '15 at 14:19
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    $\begingroup$ My favorite reference reports mass excesses to 0.1 keV, for a fractional precision 4 parts per million on a 25 MeV reaction. $\endgroup$ – rob Sep 29 '15 at 14:23
  • $\begingroup$ Excellent! I did some work during the summer with particle physics where I noticed that there is a systematic error of 4 events per 100 000 events on reactions of more than 25 MeV. How is this 4 event error coming? Is it still 4 events with 10 million events? I noticed that the error of 4 events remain also with smaller selection sizes of events. What is the optimum size of events when such a 4 event error occur? $\endgroup$ – Léo Léopold Hertz 준영 Sep 29 '15 at 14:41
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    $\begingroup$ I've added a calculation of the uncertainty. Note that the "Uncertainty Principle" is not relevant here. $\endgroup$ – rob Sep 29 '15 at 16:31

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