Integral charge 3 quark bound system with binding energy

by
U.V.S. Seshavatharam
DIP QA Engineer, Lanco Industries Ltd, Srikalahasti-517641, A.P, India
E-mail: seshavatharam.uvs@gmail.com

Prof. S. LAKSHMINARAYANA
Department Of Nuclear Physics, Andhra University, Vizag-530003, AP, India.
E-mail: lnsrirama@yahoo.com


Abstract
In the previous paper [1] it is suggested that there exists integral charge effective quark fermi-gluons and quark boso-gluons.
Effective quark fermi-gluons generates charged ground state baryons and quark boso-gluons generates ground state neutral mesons.
In this paper it is suggested that with a binding energy of 939 MeV any 3 (effective) quark fermi-gluons couples together to form a charged ground state baryon.
Square root of any 2 quark fermi-gluons or cubic root of any 3 quark fermi-gluons can be called as `hybrid’ quark fermi-gluons.
Hybrid quark fermi-gluons of up and down are 746 MeV, 779 MeV and 813 MeV. Out of 6 quark fermi-gluons, for a three quark bound system (with binding energy 939 MeV) different combinations of quark fermi-gluons and hybrid quark fermi-gluons can be possible and hence different ground state baryons can be generated with different quark flavors.
If n=1, 2, 3,.. excited energy levels follows

X sum of 3 quark fermi-gluons rest energy.
Another interesting thing is that light quark bosons like up boson mass=1.94 MeV and down boson mass=4.2 MeV couples with these ground or excited states to form doublets and triplets.
3 up quark fermi-gluons having rest energy 3×685 MeV and binding energy 939 MeV generates a ground state charged baryon of rest energy (3×685)-939≈1116MeV.
Up boson mass =1.94 MeV couples with this charged state and generates a neutral baryon at 1118 MeV.
Two up and one down quark fermi-gluons having binding energy 939 MeV generates charged (2×685+885)-939≈1316MeV .
One up and two down quark fermi-gluons having binding enegy 939 MeV generates charged (685+2×885)-939 1516MeV.
Thus 1177 MeV and 1377 MeV ground state charged baryons can be generated.
This idea can be applied to other heavy quark fermi-gluons.

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254 comments to Integral charge 3 quark bound system with binding energy

  • Andrea Rossi

    Dear Chris Johnson:
    Thank you for this interesting information.
    Warm Regards,
    A.R.

  • Chris Johnson

    Eng. Rossi,

    I thought this 10MW Supercritical CO2 turbine project (targeted for concentrated solar power) might be of interest.

    From http://www1.eere.energy.gov/solar/sunshot/csp_sunshotrnd_nrel_turbine.html

    “The proposed s-CO2 system uses no water, which is significant given that CSP plants are typically located in hot, dry climates where water is scarce. Researchers plan to demonstrate the inherent efficiencies of the s-CO2 power turbine and associated turbomachinery at a scale relevant to commercial CSP projects. Success in this endeavor will provide a foundation for solar applications that exceed the SunShot Initiative’s goal of 50% net thermal-to-electric conversion efficiency.”

  • Koen Vandewalle

    Joseph,
    I think you are right with the spheres. So there is not much to learn from errors, there.
    Then, let us learn from what is right.
    That is why it took probably hundreds or more different reactor designs before a stable and powerfull e-cat was created.
    Your sphere can be dropped in boiling water, and continue to produce 10kW @ 900C in its core.
    But if you drop it in sand, a few minutes later, a meltdown will occur in the core, because it does not produce temperature but power.

    So, my suggestion may be completely wrong, but that would mean that Andrea found a way to cut off the LENR power at a given temperature.

    Since there is no patent, Andrea has to keep it a secret what happens in the inside of the core. But this does not prevent us from inventing our own theories in our mind. And these theories live their own lives.
    If we look at the drawings on the patent documents, it is all tubes that we see, so i think that the suggestion even comes from an interpretation of that document.
    One of the mandatory conditions for having a patent granted is that a skilled independent person should be able to reproduce the invention. I do not know how skillful that person should be, and how much effort he is assumed to do in reproducing the device, but a little thinking and open minded creativity would help. It is a patent, not a book about DIY LENR for dummies.
    Kind Regards
    Koen

  • Joseph Fine

    Hank, Koen, Andrea:

    For a heat source inside a spherical shell, the equation for thermal propagation in Joules/second or Watts is:

    Qdot = 4k*pi*(Thot – Tcold) / ( 1/R1 – 1/R2)

    In the above eqn., R1 is the inner radius and R2 is the outer radius of a sphere.

    The inner radius, R1 is 1 cm (1E-2 m) and the outer radius, R2 is 2 cm (2E-2 m), so the

    shell thickness is 1.0 cm.

    These numbers were chosen to simplify the calculation.

    You may use other values for radii, temperature or heat conductivity.

    If Thot = 900 C and Tcold = 100 C, then (Thot – Tcold) = 800 C.
    (You don’t have to use degrees Kelvin here.)

    As an example, for a heat conductivity of k = 50.

    (Thermal conductivity for copper is 400.)

    Qdot (Watts) = 4*50*Pi*800/ [1/1.0E-2 – 1/2.0E-2] =

    4*50*Pi*800*(2/100) = 200*16*Pi = 10.053 kW (kJoules/sec.)

    Over how much surface area? 4*Pi*R^2 = 5E-3 sq meters.

    That’s quite a lot of heat coming from a small surface!

    Maybe I can learn from my errors.

    Joseph

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