On Sep 21, 2011, at 7:51 PM, Alan Fletcher wrote:
HEAT FLOW THROUGH THE NICKEL CONTAINING STAINLESS STEEL COMPARTMENT
If the stainless steel compartment has a surface area of
approximately S = 180 cm^2, as approximated above, and 4.39 kW heat
flow through it occurred, as specified in the report, then the heat
flow was (4390 W)/(180 cm^2) = 24.3 W/cm^2 = 2.4x10^5 W/m^2.
The thermal conductivity of stainless steel is 16 W/(m K). The
compartment area is 180 cm^2 or 1.8x10^-2 m^2. If the wall thickness
is 2 mm = 0.002 m, then the thermal resistance R of the
compartment is:
R = (0.002 m)/(16 W/(m K)*(1.8x10^-2 m^2) = 1.78 °C/W
Producing a heat flow of 4.39 kW, or 4390 W then requires a delta T
given as:
delta T = (1.78 °C/W) * (4390 W) = 7800 °C
The melting point of Ni is 1453°C. Even if the internal temperature
of the chamber were 1000°C above water temperature then power out
would be at best (1000°C)/(1.78 °C/W) = 561 W.
Most of the input water mass flow necessarily must have continued on
out the exit port without being converted to steam.
That presumes that the heat exchange takes place on the surface of
the core.
But the heat is (supposedly) produced by thermalization of gamma
rays, which could be anywhere nearby. Rossi has said that it is
partly in the copper tubing and partly in the lead shielding. The
total available area is easily 10 times that of the core, so the
delta T could be 780C, not 7800C.
This is not likely, because no gammas were detected. As I have shown,
if the gamma energies are large, on the order of an MeV, a large
portion of the gammas, on the order of 25%, will pass right through 2
cm of lead.
The lower the energy of the gammas, the more that make up a kW of
gamma flux. Consider the following:
Energy Activity (in gammas per second) for 1 kW
-------- ----------
1.00 MeV 6.24x10^15
100 keV 6.24x10^16
10.0 keV 6.24x10^17
The absorption for low energy gammas is mostly photoelectic. The
photoelectric mass attenuation coefficient (expressed in cm^2/gm)
increases with decreasing gamma wavelength. Here are some
approximations:
Energy mu (cm^2/gm)
-------- ----------
1.00 MeV 0.02
100 keV 1.0
10.0 keV 80
We can approximate the gamma absorption qualities of the subject E-
cat as 2.3 cm of lead.
Given a source gamma intensity I0, surrounded by 2.3 cm of lead we
have an activity:
I = I0 * exp(-mu * rho * L)
where rho is the mass density, and L is the thickness. For lead rho
= 11.34 gm/cm^3.
For 1 kW of MeV gammas we have:
I = (6.24x10^15 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) *
(2.3 cm))
I = 3.7x10^15 s^-1
For 1 kW of 100 keV gammas we have:
I = (6.24x10^16 s^-1) * exp(-(1.0 cm^2/gm) * (11.34 gm/cm^3) *
(2.3 cm))
I = 2.9x10^5 s^-1
For 1 kW of 10 keV gammas we have:
I = (6.24x10^17 s^-1) * exp(-(0.80 cm^2/gm) * (11.34 gm/cm^3) *
(2.3 cm))
I = ~0 s^-1
So, we can see that gammas at 100 keV will be readily detectible, but
much below that not so. However, it is also true that 0.2 cm of
stainless will absorb the majority of the low energy gamma energy, so
we are back essentially where we started, all the heat absorbed by
the stainless, and even the catalyst itself, in the low energy range.
If the 2 mm of stainless is equivalent to 1 mm of lead, for 1 kW of
100 keV gammas we have:
I = (6.24x10^16 s^-1) * exp(-(1.0 cm^2/gm) * (11.34 gm/cm^3) *
(0.1 cm))
I = 2x10^16 s^-1
and an attenuation factor of (2x10^16 s^-1)/(6.24x10^16 s^-1) = 32%.
Down near 10 keV all the gamma energy is captured in the stainless
steel or in the nickel itself.
To support this hypothesis a p+Ni reaction set including all
possibilities for all the Ni isotopes in the catalyst would have to
be found that emitted gammas only in the approximately 50 kEV range
or below, but well above 10 keV, and yet emitted these at a kW level.
This seems very unlikely. If such were found, however, it would be a
monumental discovery. And, it would be easily detectible at close
range by NaI detectors, easily demonstrated scientifically.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
