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
