At 07:24 AM 7/22/2011, Jouni Valkonen wrote:
Craig, indeed that is true, liquid water does
not contribute to the pressure at all, because
water does not gently flow out of the E-Cat, but
is spilled due to rather violent boiling at kW range in closed container.
No, that's an error. The E-Cat operation, in the
demos, begins with water flowing through, due to
the pumping. That would be gentle flow. What
happens later is unclear, and that depends on
internal conditions that we cannot observe directly.
Only thing that contributes for the pressure is
steam flow pressure out of the E-Cat and in the
hose. Steam flow resistance is roughly the same
in all E-Cat setups, therefore steam temperature
is depended directly and comparably on total water heating power.
An analysis of temperature and pressure, I just
wrote for the CMNS list. I'll add it below.
It was well established that wetness of the
steam was something in order of 1-2% that is
typical for normal boiling in closed container
where there is lots of spilling and water droplet density is high.
It is very much not well-established. However,
there are two problems, both somewhat semantic in
nature. The real question is how much water is
vaporized. Failure to vaporise a known flow can
come from two sources: literal overflow of liquid
water and wetness of steam. Both would be
expected to some degree. What degree is actually
found? We don't know, we have inadequate data,
and that inadequacy has been maintained by Rossi
unwillingness to allow definitive demonstrations.
from my post to CMNS:
Assume Temperature of chimney: 100.5 degrees.
Assume Boiling at ambient pressure of 99.6 degrees.
Interpolate pressure in chimney from
http://www.engineeringtoolbox.com/saturated-steam-properties-d_457.html
pressure: 1 bar, temperature 99.63 C.
pressure: 1.1 bar, temperature, 102.32 C.
Interpolated pressure at 100.5 C.: 1.036 bar
Interpolated steam density: 0.610 kg/m^3
Raw steam flow if no hose. Assume orifice from chamber, 1/2 inch.
Estimate from http://www.engineeringtoolbox.com/steam-flow-orifices-d_1158.html
Overpressure, 0.036 bar, 1 psig = 68.948×10^3 bar, 0.522 psig
This is off the chart. However, assuming
linearity, I come up with 40 lb./hour, which is
18 kilograms, and the claimed flow rate is 5
g/sec, oir 18.5 kg/hr. That is an amazing
coincidence, and is not a confirmation of that
exact value, considering how rough the chart is.
However, there is a hose attached. If we assume
18.5 kg/hr flow, 3 meters of 15 mm ID hose, steam density of 0.590 kg/m^3,
http://www.engineeringtoolbox.com/steam-pressure-drop-calculator-d_1093.html
gives a drop of 10978 Pa. 1 Pa = 10^-5 bar. That
would be a drop of about 0.1 bar, as has been
stated. However, we don't, in this marginal
calculation, have any pressure left, the flow
through the orifice was estimated based on the
pressure difference with atmospheric.
In fact, the steam flow will be reduced because
of back pressure from the hose, so that the
figures match, with the sum of pressures
equalling the total elevation of chimney pressure over ambient.
The expansion of the steam into the hose is a
factor of 1.40 by area ratio. Steam cannot cool
until the wetness approaches 100%, but the steam
will become wetter, reducing the steam volume, so
that "steam flow" is reduced. If I take the Mats
Lewan report as indicating that half the steam
condenses in the hose, this will reduce the
"steam flow" to 9.25 kg/hr, reducing back pressure to 2744 Pa.
I'm not going further with this. If I take the
data straight, as it is, and assume accuracy
(which is unreasonable, but it does allow us to
see what ball-park estimates could be), I come up
with an indication that 75% vaporization, very
roughly and without doing more exact math than is
found above, seems quite reasonable. Given the
roughness of the data, it is not impossible that
there is full vaporization, and it is possible
that vaporization is below 50%, I have not done an exact analysis.
And the reason for that is not only lack of time,
but that this is not going to nail anything down,
there is so little data. I return to my basic
conclusion, we don't have enough data to be sure
about the Rossi E-Cat either way.
However, claims that the data is contradictory,
on the basis of steam pressure calculations, seem to fail.
