Cold Fusion: The Cat is Finally out of the Box (French Article)

I missed this when it was published but today found a reference on the Journal of Nuclear Physics

Contrepoints.org, a popular French language self-described liberal website covering news and current affairs has published an article titled “Cold Fusion: the cat (E-Cat) is finally out of the box” by Jean-Pierre Cousty.

The article is largely positive in the coverage it gives of the Lugano test, and the author is enthusiastic regarding the future role that LENR will play in energy production.

Cousty writes:

We now have official proof that the future of global energy will not perhaps in the near future by wind or solar but the implementation of this cold fusion even if it takes decades. Several companies, such as Brillouin, are ready to market devices and futuristic dreams are good train as a Swiss company speaks to run cars continuously for months with a Stirling engine types. For example, problems of sea water processing freshwater have an off and inexpensive for Africa and countries in the developing solution …

Life of cold fusion will be able to resume its course with the 19th Conference Annual (ICCF-19) to be held this year in Padua in Italy April 13 to 17, 2015 with 500 participants from planned . . .

It is obvious that OPEC, EDF, AREVA etc. must adapt to the new conditions of cold fusion could quickly become commonplace. The coming years should be exciting and promising hopes in world energy.

Speaking of OPEC adapting, I have noticed Brent Oil futures prices dropped below $81 dollars a barrel today. I doubt that this is E-Cat related, but there are clearly macroeconomic pressures on OPEC, with non OPEC nations like the United States and Canada producing oil at levels not seen for 30 years. If LENR really does come online in the near future, it will surely be a further blow to the power of OPEC nations.

  • Obvious

    We are very complicatedly figuring out the resistance of the reactor resistor coils. This is needed to work out the correct current in the active run.

  • Obvious

    Ivan,
    Let me first apologize for the torture I have put you through.
    You are very patient, and for that I thank you.
    I have figured out what no textbook or website seems to be able to explain clearly.
    I hope that we can agree on this, and then move forward.
    The square root of three times line current is the vector sum of two 120° separated current flows for one phase.
    In other words, it already incorporates the vector math, and is the final result. (that is why it is so much smaller) than line current.
    The reason that sqrt(3) is the side of a triangle is due to: that for product of the vector of two equal length sides, the short cut to the same point is described by an a triangle with a sqrt(3)times the length sides.
    So, indeed, sqrt(3) is a vector, but is a completed vector operation for two of three equal vectors, (the third of which would return the vector solution to the origin).

    • Obvious

      And I’m not sure what this means, but I worked it out so I may as well save it for posterity.

      Using IL as the base of each equal 120° triangle formed inside a circle, and a using a line that exactly dissects one of these 120° triangles so that the line continues, superimposed on the common line for the other two triangles (same one as the circle radius), then the sum of the square of one complete base length plus the square one half of another (the one that is dissected) = Power
      (The three bases of the 120° triangles make an equilateral triangle within the circle, so the line dissects the equilateral triangle, and so we are using 1/2 an equilateral triangle).

      IE: 19.7^2 + 9.85^2 = 485.1125

      This means P = IL^2 + (1/2IL)^2

  • US_Citizen71

    On average yes 1/3 of the total power is dissipated in each resistor. But during any instant along a full cycle each of the the resistors will be dissipating a different amount of power. Also the circuit would need to be powered by a wye configuration with a floating neutral likely from a secondary on a transformer. This explains the increase of current and drop in voltage when compared with the line in from the main. So effectively only one phrase is being dissipated by each resistor. They simplified the math and the situation which does induce an error but this error reduces the calculated COP not increases it. So other than being the electrical equivalent of the grammar police you are pointing out that the reported COP in reality would be slightly higher.

  • Obvious

    Hold the correct constants.
    Wye has a higher V, Delta has a higher I. Holding P and R means that due to R in the denominator for P/1=V^2/R and as a factor in P/1=I^2R/1 that
    V and I are complimentary fractions of R.

  • US_Citizen71

    After a long conversation with a power systems engineer I will concede the current wouldn’t be half, but the power would be.

  • Obvious

    The sqrt(3) is the vector angle descriptor. When used to factor I, it describes the relative length of the base of an Isosceles triangle to its sides. The Isosceles triangle with a 120° angle point represents one phase of three phases of power. Connecting three of these triangles at the 120° point makes a set of three phases describing the inside of a circle so that all 360° are accounted for. This is the basis of all 3 phase-related mathematics.

    • Obvious

      You will like this. Bisect the triangle so that the base is divided in exactly 1/2, starting the line at the 120° corner (center of the circle).

      You will have two Right Angle triangles. Using Pythagoras’ Theorem:

      For one of these smaller triangles, the 90° angle’s side is sqrt(3) long, if the opposite side is 2 long and the base is 1 long. These ratios are forced since the angles of the triangle are 90, 60 and 30.
      Dividing 19.7 by the length of the sides by gives sides of 11.37, 9.85, and 19.7 .
      Very familiar numbers.

  • Obvious

    The equivalence of the Y to a delta allows the delta to be converted to a circuit that is intuitively easier to imagine as a DC circuit. It opens the delta up, eliminating the “cross-current” or “cancelling current” section across one side (from the viewpoint of one corner of the delta) that I described earlier, by turning it into a parallel branch.

  • US_Citizen71

    What a three phase transformer with three separate output windings? That would take the three phase power in and give three separate single phase line out.

  • Obvious

    Ivan,

    Lets quickly go back to step one: basic assumptions and first principles.

    The line current divided by square root of three for one phase is a vector sum of the currents coming from the other phases. The line current in a delta is composed of two
    currents that are out of phase by 120 degrees. This is where sqrt(3) comes from.
    The vector sum of the two phase current components is the square root of
    three times the current in any phase.
    So if you know one vector current in one phase, the other two phases must total to sqrt(3) times the current in the one phase. But since a third phase can’t also conduct through a wire at the same time as two others are conducting in one phase (it would short out), the current multiplied by the sqrt(3) is only in one other phase at a time. In this way power stays constant in the equation.

    • Obvious

      The math above is a bit whacked.
      19.7*sqrt(3) = 34.081 (two remaining delta sides, if one has 11.37 A)
      19.7/sqrt(3) = 11.37 (one delta side, C2 cable)
      total………….= 45.451
      All the above values are magnitudes, not respecting their sign (+/-). Their signs are assigned using the correct vector polarity, which is relative to an arbitrary choice of frame of reference.
      If you choose the start vector point at one corner of a delta (say x) leading to another (y), then the magnitudes of the values between xz and zy will resolve to relative vector quantities with appropriate sign compared to the reference point.
      If both C2 cables coming from a C1 have each 11.37 A, (total 22.74 A) then the last side has -22.74 to make the total zero.

      • Obvious

        The “short” version is this: If you use sqrt(3), you are using a vector quantity. The average RMS is integrated over time. Time cannot be ignored, but can be “halted” at an instant in time in order to simply things. But we must hold the time constant consistently in one location, using one description that describes what is happening at that one instant.
        So at the instant you freeze time at a point when one delta side sees 11.37 amps, then the rest of the equation must follow the consequences of that “measurement” at this instant. The average RMS values of the other sides are averages of time-integrated values. So they cannot be used when compared to a frozen instant of time in another side of the equation.
        Moving the frozen viewpoint of time to the C1 cable at RMS 19.7 A means that measured (or arbitrarily calculated RMS) values cannot be used for any other of the following parts of the circuit. You must use whatever math is appropriate based on only one RMS measurement per averaged factor (V and I). Resistance is not a time-sensitive RMS value, so it remains independent (time invariant). It is already “frozen in time” for the purposes of the relevant equations.
        We must also freeze P at some point. It must be tied to the same time frame as the other factors. Since P is a function of V and I, this follows that P is frozen in time with these two functions, and must be frozen from the same viewpoint.
        Since V is unknown, we must hold it constant, leaving only I as the variant to be held still from the viewpoint of any math applied to the circuit.

        • Obvious

          Just before I head out:
          The problem we keep tripping over in this power story is time. When measured in seconds, and t=1 in the denominator or as a factor, or both, and gets cancelled somewhere, then it is ignored for convenience sake and then we forget that it is still part of the original equations from which later calculations are made.

    • US_Citizen71

      What do you think about option 2 in the below PDF being inside of the control box? Wouldn’t that solve the problem as well?

      http://www.ft-transformers.co.uk/wp-content/uploads/2012/06/Supplying-Single-Phase-Loads-from-a-Three-Phase-Supply-Tra.pdf

      • Obvious

        Yep. I thought that switching to a wye and delta, or vice versa with the control might throw wrench in the works, and add a new dimension of electrical ugliness for discussion. But it seems that the delta uses the most power anyways. If the box ran it wye for dummy, then delta for the run, then the proportion of Joule heating between active and dummy might be non-linear or at least have a sudden “phase change”. lol

  • Obvious

    The power in a delta is higher than an equivalent wye. This is why there is delta-wye starters for some motors. You start the motor on wye so they don’t burn out at low starting speeds, but low speed torque is sacrificed. Current is 1/3 of a delta in a wye. (Electric motor torque is maximum at 0 rpm, but so is current. Dropping the current by a third keeps the windings cooler on start, as long it can still start turning.)

    The wye has less line voltage by V/sqrt(3), but 3 times the effective resistance (total impedance) comparable to a delta.
    The delta has line voltage, and 1/3 the effective resistance compared to a wye.
    If I can remember correctly, the total effective power difference (line V being the same) for a delta is about 1.4(?) times higher than a wye.

  • Obvious

    For (b) I had 3 (bold) because the equivalent resistor in a delta is 3 times one in an equivalent wye. This is just the equation for one r of the delta so far. Obviously the r doesn’t really change, but this is the larger impedance that the total circuit sees because the current increases due to increased V in the sides of the delta relative to a wye. Three times the r, times three again for each phase ends up with 9 times the power from which to derive the resistance value from.
    Where my brain is stalling for some reason is when I try to force the r to stay constant, what happens to power. It seems to end up as divided by 9 for the complete circuit, which should work out OK since the 9 times (3^2) cancels the 9 times power in the r calculations. Somehow.

  • Obvious

    Good day, Ivan. I’m back at this again. Lets see if we can put this to bed today.
    I had dreams of P R V etc floating around all night….
    I’m just going over your newer comments.
    For the sake of both our sanity, I won’t re-edit my earlier comments.