From: chuck@coplex.com (Chuck Sites)
Subject: Re: Comments on McKubre et al
Organization: Copper Electronics, Inc.
Date: Mon, 13 Apr 1992 07:02:32 GMT

   Hi all.  It's been a while since I've added anything to the discussion,
(business kept getting in the way) but here are a couple of points I thought
might be useful.  I haven't heard this discussed yet, and I wonder if this
is important to the Takashai type experiments.  (D. Morrison's non-belief
in excess heat kind of gets me.  It should be obvious from the argument 
below there should be an effect of some sort). 

   Since we are discussing the heat of formation in PdD, here are some
more numbers to confuse the situation.  In the AIP conference proceedings
#228, Dr. Y.E.Kim of Purdue University presented a paper under the title
"Surface-Reaction Theory of Cold and Warm Fusion" were he discusses a
hysteresis effect in PdD and PdH loading.  He references A. Iervert, W.Danz
Z. Physik. Chem (B) 34, (1936, & 1937) who showed the solubility of D & H
in Pd as a function of temperature (Figure 1).  Kim uses this to characterize
a hysteresis cycle in the electrolysis type CF experiments.  There are two
stages in the temp/PdD loading; the desorption phase at Temp-Pd from 103C
to 113C where D/Pd decreases from 0.55 to 0.06 by rapid desorption, and the
absorption phase at Temp-Pd from 73-86C where the D/Pd ratio increases from
0.05 to 0.56.  What he suggests is that the electrolysis of Pd-D2O starts
slowly in the reabsorption phase, which upon loading to D/Pd 0.56 leads a 
rapid desorption stage, which then cause a reabsorption phase. The result
causing a hysteresis cycle.

         | o* o                                          
         |     * o o o                                    
     0.6 +         *  o o o                      Isobaric solubilities
         |           *       o                   of deuterium in palladium
         |            .       .                  as a function of temp.
     0.5 +             .       .                 Reported in Y.E. Kim's
         |              .      .                 paper.   
         |              .      .                          
     0.4 +   D2 (*)     *      .                          
D/Pd     | Absorption   .      |    D2 (o)                
Loading  |              *      v  Desorption              
     0.3 +              .      .                          
         |              .      .                          
         |              ^      .                          
     0.2 +              |      .                          
         |              .      o                         
         |               .      .                         
     0.1 +               .       .                       
         |                *       o                       
         |                 * * * * oo*o*o*o*o*o*o*o*o*    
     0.0 +----|----|----|----|----|----|----|----|----|----|
         20       60        100       140       180       220
                         Temp Pd (degrees C)

   To describe the effect let me throw in the numbers.  This is extracted
from my notes on the Kim paper. It's pretty much verbatim except for my little
bit of speculation at the end.  To start,

D2O + Qa  -> D2 + 1/2 O2   (Qa = 70 kcal/mole of D2 or 4.87E-19 J/D2)   [1]

The power disapated by the reaction is given by infamous P=1.53IW, and the
rate of D2 production is given by R_a=(0.3125E19)I (D2's/sec).  The other
reaction Dieter mentioned several notes back is: 

2Pd + D2 -> 2(PdD) + Qb    (Qb = 9 kcal/mole of D2 or 6.36E-20 J/D2)    [2]

The enthalpy for H absorption into Pd in alpha phase is -19 kJ/mole
of H2, and for beta phase is -46 kJ/mole of H2.  For a common alpha+beta
mix with H/Pd <= 0.5 the enthalpy is aprox. -38 kJ/mole of H2.  The heat
of desorption (endothermic) is the same except for the change in sign. As
a common test case, at 25C, Pd can be loaded to 0.66 D/Pd.  Assuming a 
Pd density 6.767E22 Pd's/cm^3 the deuteron density is 1/2(0.66) n_Pd =
2.23E22 D's/cm^3. The energy released from 2Pd + D2 -> 2(PdD) + 38kJ/mole
gives Q_b n_D2 = 1.40kJ/cm^3 released by loading Pd from 0 to 0.66.

   Now during the initial stages of loading, the rate of absorption is
approximately the same as the rate of D2 production from reaction 1.
The power produce by [2] is then, P_b = Q_b R_a = 0.196I W. Finally one
can estimate the time to load a volume of Pd (V_Pd), by:

Delta T = V_Pd Q_b n_D2  / P_b = V_Pd n_D2 / R_a.

After this period of absorption, the temp will have increased from Joule
heating, and from reaction 2.  The desorption is expected to be slow, 
until the desorption stage of the hysteresis cycle is reached, at which
time almost all absorbed gases will be release.  For a Pd rod 2cm diam,
5 cm length, the desorption time has been measured to be aprox 80 sec.
with a total hysteresis cycle time of 10min. During desorption the number
of deuterons released is f(n_D2) where f <= 1. Because of the short time
period, the result is non-equilibrium condition where an amount of D2
gas is release in competition with the rate of D production from reaction 1.
This can create a double layer electric field, with a large potential of
10E9 Volts/meter across the surface. This and the fact that desorption 
would create a large D+ D2+ ion current at the surface.  Kim goes on
to argue that given all the correct conditions, one can expect an apparent 
11%+-3% excess heat from the electrolysis of Pd-D2O. 
            
   One reason why I find this interesting is that in the Takasashi type 
electrolysis experiments, they use a sawtooth voltage followed by
a long period square wave.  With in the concept of a hysteresis cycle,
the long gradual increase in power applied would charge the cell to the point
of a desorption, at which time (assuming the timing is correct) the power
into the cell drops to zero, allowing the discharge without creation of 
the intense D-layer. This may allow for an easier loading during the next
cycle than would be normally possible by a constant current.  In addition
there is the reaction:

 2D+ + 2e- => D2 + 31.7eV       (This is equal to 3056 kJ/mole D2).  [3]

  It's believed that D absorbed looses it's electron to Pd electronic band
state, and if rapid desorption occurs, the D released would likely favor
D+ ion state.  With the sawtooth current, this may allow for re-absorption
of the D+ ions, far easier than energy required by reaction 1. This may be
the reason why doping Pd with Ag is rumored to generate heat. The extra
electron donated to Pd by silver, either participates as in [3] or cause
reabsorption without applied current.  But this is just speculation on
my part.  The main focus of the Kim paper is his development of a theory
that suggests a very low energy resonance state in D+D fusion maybe 
responsible for the nuclear oddities in cold and warm fusion experiments.
Considering all of the evidence in cold & warm fusion, this may not be
a bad assumption.  It would certainly be hard to detect in normal colider
experiments. I suspect the only way to test this hypothesis with any
mechanism acceptable to the general science community would be via muon
catalyzed fusion, or by the creation of a very low energy collider
system with extremely fine controls (perhaps laser steered D ion 
collisions?)  It's a theory that should be considered if not for D+D
fusion, then perhaps for other easily nuclear reacting particles. Who
knows, it seems to be unexplored territory.

   Anyway, Kim's theory on the formation of an apparent excess heat by
means of hysteresis cycle is a very convincing argument upon scrutiny.
It's something that I think skeptic and believer can agree upon as a
base line for heat formation in PdD(x).  I still hold out on my
heavy heat hypothesis as a possible additional effect.  Even more so
now that hydrogen banding has been observed and measured on the surface
of Ni! (see Nature N&V article "Hydrogen Band Together").

Have fun and experiment. 


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