At 05:09 PM 6/23/2011, Joshua Cude wrote:
On Thu, Jun 23, 2011 at 3:37 PM, Abd ul-Rahman Lomax
<<mailto:[email protected]>[email protected]> wrote:
My guess is that it would still reach boiling point, at roughly the
time predicting by extrapolation of the rate of temperature rise,
because the device is insulated and most heat will not leave unless
the water starts boiling. It's a 600 watt steam kettle, I'd expect
such to boil water. Just not as quickly as seen.
Nooooooo. Not you, too. Now I find myself defending the Rossi crew
against a LENR advocate.
Hey, that's cool. I'm going to drop "pseudo-" from your "skeptic"
title, effective immediately. I'd ask, though, that you drop
"advocate" from my own title. I'm interested in LENR. Until Rossi
came along, I had little hopes of it becoming practical, I was almost
purely interested in the science of it. Now, there is Rossi, who is
frustrating indeed. I'm just taking it as it comes. In a way, you are
right. I do have a small commercial interest in LENR, and Rossi will
draw a lot of attention away from that, so it might have some
negative impact, but there is nothing I can do about that. For now,
it's just fine that Rossi is considered bogus, but the reality will
be revealed, I'm sure, and funding and interest and all that will
start to ramp up drastically. For that approach. Fortunately, I can
sell my materials.... I was never going to make more than a tiny
profit anyway, and I can do something else, perhaps related. I think
LR-115 is way cool, I want to use it to simply study background
neutron radiation, if nothing else.
It's flowing water, not a kettle. So the input power can only heat
it so much.
Depends on the flow rate, eh? Look, from the output temperature
readings, water is being raised to the boiling point, it's highly
likely that some of it is boiling, eh?
Power in = mass-flow-rate * specific heat * temperature difference
So,
temperature difference = 300W / (1.73 g/s * 4.2 J/K g) = 41 K
If the input temperature is 20C, then the maximum output is 61C. If
you accept the numbers as given.
So, are you convinced that the reactor is putting out energy?
Otherwise what is taking this to 100 C?
It's not like a kettle. The reason the graph shows a gradual
increase in the temperature when the power is first applied, is that
the reactor has to heat up first, and that absorbs some of the
power. When it reaches equilibrium temperature, all the power goes
in to heating the water.
That's strange. I'd think that it all starts at the same temperature,
and would rise in temperature more or less evently, except that the
reactor core itself, which is directly being heated, would get hotter
than boiling (apparently the heating element is being controlled
based on reactor temperature, about 450 C.).
If the only heating is from the element in the core, and that's
constant, all those temperatures would rise at some rate, linearly,
until they begam to approach equilibrium. The water will carry off
heat, if the cooling pump is on at this point, that will merely slow the rise.
In the chart of temperature, a sudden change in rate of temperature
rise appears, at 60 degrees C. I assume that this represents the time
when the core reached turn-on temperature.
There are lots-o-mysteries here. The original point was that the
thing shows no sign of settling at 60 degrees without excess heat, it
was rising linearly to that point. The only place where there would
be rapid alternation of the rate of change of temperature would be
where the water starts boiling, which would start to draw off far
more heat than the flowing water would remove.
