On Oct 18, 2011, at 10:36 AM, David Roberson wrote:
Rossi has stated that the energy released by the LENR reaction is
in the form of moderate energy gamma rays(X-Rays?) These rays are
converted into heat within the lead shielding and coolant. If this
is true, heat to activate the core could be made to exit into the
coolant to slow down the reaction. The actual temperature within
the core section is perhaps 600(?) C degrees or more. You can
find his statement within his journal if it is important to you.
The 60 degree figure probably refers to the temperature of the
water bath when the core reaches its starting value.
Dave
Hi Dave,
Welcome to vortex!
I am happy to see your spreadsheet made it through the vortex
filter. Historically nothing made it through above 40KB without
special processing by Bill Beaty himself. Your post with spreadsheet
was 55.4 KB. Either a new limmit has been established or Bill Beaty
is closely watching (the latter seems to me unlikely.)
The implications that gammas heat the lead and coolant do not make
any sense. If they had the energy to make it out of the stainless
steel fuel compartment used in prior tests, then they would have been
readily detected by Celani's counter. This was discussed here in
relation to my "Review of Travel report by Hanno Essén and Sven
Kullander, 3 April 2011".
http://www.mail-archive.com/vortex-l@eskimo.com/msg51632.html
http://www.mail-archive.com/vortex-l@eskimo.com/msg51644.html
http://www.mail-archive.com/vortex-l@eskimo.com/msg51648.html
Excerpted below are the most relevant notes I made regarding the
April 3, 2011 test:
FIG. 3 NOTES
It appears the heating chamber goes from the 34 cm to the 40 cm mark
in length, not 35 cm to 40 cm as marked. Maybe the band heater
extends beyond the end of the copper. It appears 5 cm is the length
to be used for the heating chamber. Using the 50 mm diameter above,
and 5 cm length we have heating chamber volume V:
V = pi*(2.4 cm)^2*(5 cm) = 90 cm^3
If we use 46 mm for the internal diameter we obtain an internal
volume of:
V = pi*(2.4 cm)^2*(5 cm) = 83 cm^3
Judging from the scale of picture, determined by the ruler, the OD of
the heating chamber appears to actually be 6.1 cm. The ID thus might
be 5.7 cm. This gives:
V = pi*(2.85 cm)^2*(5 cm) = 128 cm^3
The nickel container is stated to be about 50 cm^3, leaving 78 cm^3
volume in the heating chamber through which the water is heated.
If the Ni containing chamber is 50 cm^3, and 4.5 cm long, then its
radius r is:
r = sqrt(V/(Pi L) = sqrt((50 cm^3)/(Pi*(4.5 cm)) = 3.5 cm
total surface are S is:
S = 2*Pi*r^2 + 2*Pi*r*L = 2*Pi*(r^2+r*L) = 2*Pi*((3.5 cm)^2 +
(3.5 cm)*(4.5 cm))
S = 180 cm^2
The surface material is stainless steel.
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.
COMMENT ON GAMMAS
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.
SUBSEQUENT COMMENTS
... can anything said about the inside of the E-cat be believed?
There are numerous self-inconsistencies in Rossi's statements, and
behaviors. These things may be justifiable in Rossi's mind to
protect his secrets. Whether justified or not, such things damage
credibility.
One thing is for sure: if the E-cat is operated at significant
pressure then 2 mm walls would be too thin at high temperatures.
Also, there are other limits to surface steam generation I have not
discussed, that take precedence at high power densities. One
limiting factor is the ability of the catalyst and hydrogen to
transfer heat to the walls of the stainless steel container, a
process which would likely be mostly very small convection cell
driven. Again, we know too little about the internals. Nothing much
new about that. A heat transfer limit is reached if a stable vapor
film is formed between the walls of the catalyst container and the
water. The top of the catalyst container may be exposed to vapor,
thereby increasing the thermal resistance, the effective surface
area. At high heat transfer rates bubbles can limit transfer rates.
It would be an interesting and challenging, though now probably
meaningless, experiment to put 4 kW into a small stainless steel
container under water and see what happens, see if the element burns
out, etc.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
