Good grief. Of course there is control or there would be no point in
having all those pumps.
On 8/21/2016 9:12 AM, a.ashfield wrote:
"a pump by itself does not regulate the level of the water. "
what is the point of having all those digitally controlled pumps if
there is no control?
On 8/21/2016 12:55 AM, David Roberson wrote:
Thanksfor the information. But AA, a pump by itself does not
regulate the level of the water. There must be some form of active
level feedback applied in order for this to occur. If no level
sensing and feedback is used then either the water totally fills up
the device or it gets boiled off when subject to a constant input flow.
The question can be clarified by finding a direct reference to level
sensing in Rossi's documentation in which case the pumps would need
to be controlled by software that cycles each of them on and off, or
possibly adjusts the flow dynamically. Can you point to such a
document. A picture of a pump is not adequate proof unless it is
specified to have a limited maximum pressure that would prevent the
boiling point of the water from reaching significantly above 102 C if
the ECAT device fills up.
I am sitting on the fence with respect to assuming that the test
results are measured accurately. Jed and others have presented a
fairly convincing arguement that all is not well. You and others
appear to be 100% convinced that Rossi is correct.
If Jed and allies are right then science must be able to explain what
is erroneous regarding the calibrated meter measurements. I am
seeking that explanation to fill in the blanks in the event that they
are found correct. To me that is the scientific method.
Let us put my latest hypothesis to rest which will enhance the proof
that Rossi is on the proper track. Is there any evidence of water
level control feedback that we can locate? I will look carefully at
Engineer48 photos on Ecatworld per your suggestion.
Dave
-----Original Message-----
From: a.ashfield <a.ashfi...@verizon.net>
To: vortex-l <vortex-l@eskimo.com>
Sent: Sat, Aug 20, 2016 8:10 pm
Subject: Re: [Vo]:Interesting Steam Calculation
If you look at the my original reference showing a link to photos of
Engineer48 on Ecatworld, it shows the many precision pumps for each
Tiger that maintain the correct water level in the reactors.
AA
On 8/20/2016 3:40 PM, David Roberson wrote:
Could youshow me a reference to level gauges in each of the
devices? I do not recall seeing one so far.
Dave
-----Original Message-----
From: a.ashfield <a.ashfi...@verizon.net>
To: vortex-l <vortex-l@eskimo.com>
Sent: Sat, Aug 20, 2016 3:00 pm
Subject: Re: [Vo]:Interesting Steam Calculation
That would mean the Tiger E-Cats would have to be completely
flooded. But the level gauges don't show that.
Why not suggest pixie dust?
On 8/20/2016 1:51 PM, David Roberson wrote:
Today I made an interesting calculation that some may find
relevant to the ongoing discussions.
According to steam tables, the following could be possible,
assuming that I did not make a mistake in my calculations.
Assume you have 1kg of water inside a solid container at 130
C and 39.2 psi absolute. Then you place a restriction device
that allows all of the liquid to eventually escape. Some of
the liquid will immediatly flash into vapor while most of the
1 kg remains in the liquid form as it exits the restriction.
If you assume that the resulting mixture ends up at 102 C and
15.75 psi absolute then it is possible to calculate the
amount of vapor and liquid that is present at that location.
The internal energy of the initial liquid at 130 C is 546.388
kj/kg which in this case yields 546.388 thousand joules. I
am assuming that this same amount of energy remains within
the liquid and vapor combintation of the lower temperature
and pressure stream.
When I solved the equation relating the quality of the
mixture to the various heat contents I determined that there
would be .053 kg or vapor and .947 kg of liquid water at the
output. On first glance, this result suggests that it should
be easy to separate the water from the steam, but actually
calculating the two volumes makes that not so evident.
The volume of the vapor would be .053 kg * 1.565 cubic meters
per kg = .0826 cubic meters. The volume of the liquid water
would be .947 kg* .001045 cubic meters per kg = .000989 cubic
meters.
Using the above numbers it appears that you would have 83.488
times as much vapor by volume as liquid. This is quite a
large ratio which suggests that it might well be possible to
mistake a stream of mass with this consistency as consisting
of only vapor. Especially if a visual technique were used.
I am not saying that this calculation reveals the source of
the Rossi test confusion, but that perhaps it might open
discussions that have not been considered so far. I do
recall that on earlier demonstrations that the temperature
within the ECATs was reported to be in the range of 130 C.
Perhaps some of our mathematically inclined vortex residents
can take a few moments to verify that my assumptions and
calculations make sense.
Dave
