No one to my knowledge is showing data that the heat after pulling the plug
continues at the rate it had before power-off for a full 15 minutes. My
interpretation of Levu's comment in Part 3 of the Krivit video is that the rate
natually declines until after 15 minutes it was judged that steam production
had ceased. Either way thermal inertia plays a role. You're really stretch
credulity to ask me to believe you calculation shows only a few minutes is
possible. You haven't set it up carefully enough, i.e. it is flawed. For one
thing it would appear that more than 1 MJ would be stored in the case you
discuss. Your temperature seems low. I remember Rossi saying he was able to
heat a working fluid to 450C so the thermal mass would seem to get hotter than
that. Your calculation of time constant is clearly unacceptable as heat output
cannot remain constant. The analysis, if done properly, leaves no doubt about
the correct conclusion. To say its all off for a factor of 3 is laughable in my
judgment, esp. when you've underestimated values and overestimated outout and
failed to understand the decay of output.
Also it seems hydride formation probably explains any anomalous heat produced-
that is if it can be determined that its produced and how much. So far this has
been an enormous fiasco.
----- Original Message -----
From: Horace Heffner
To: vortex-l@eskimo.com
Sent: Sunday, August 28, 2011 8:58 AM
Subject: [Vo]:Corrections to "heat after death" calculations
On Aug 27, 2011, at 12:51 PM, Joe Catania wrote:
For the umpteenth time it is not an assertion. The thermal mass of the
reactor is about 1MJ (based on specific heat), the energy outflow is a mere
fraction (~1kW). OK?
There has been no demonstration that output is higher than inout. Steam
quality is not measured, therma; inertia not accounted for. and there is
Rizzi's determination that flow is over estimated. I hope I don't have to
repeat these facts again. The source of heat in the 15 minutes is thermal
inertia since it would account for all steam produced. Cold fusion is not
indicated by what Levi has said. I have not seen the graphs you speak of and
I'm not sure they are coincident with cutting the power but thermal inertia
needs to be accounted for. So show me the data. And all I can say is one does
not assume cold fusion to prove cold fusion. CF proof is totally elusive by the
means exploited. Its more likely a flaw in technique of measurement. But if
there is proof of anomalous heat it has eluded my detection so far. The
properway to do the calorimetry is not with flow as I've detailed before.
Levi said steam stopped after 15 minutes so it seems you need to get on the
same page.
My two cents on this is it is a typical one of a kind anecdote - with no
solid measurements to back it up. We don't really know if the device was
initially outputting 5000 W or just the input wattage, for example.
For the sake of discussion, let's just assume the story is correct and the
device was outputting 5 kW as advertised.
Let's also be generous with regard to mass, and assume it was equivalent to
20 kg of copper, and stored 1 MJ of energy as specified above.
Using a heat capacity of copper, 0.385 J/(gm K), a 20 kg mass requires
delta T = (10^6 J)/((0.385 J/(gm °C))*(2*10^4 gm)) = 130 °C
to store the 1 MJ thermal energy. If we assume inlet temperature of 23°C
then this is an absolute temperature of 153°C.
The thermal mass, Cth, is given by:
Cth = (0.385 J/(gm °C)*(2*10^4 gm) = 7700 J/°C
Assume the device transfers 5 kW of output heat when the internal temperature
is 153°C and inlet temperature is 23°C, i.e. delta T is 130°C. This gives a
thermal resistance of
R = (130°C)/(5^10^3 W) = 2.6x10^-2 °C/W.
The decay time constant, tau, for the 1 MJ thermal mass, C, is is given by:
tau = R*Cth = (2.6x10^-2 °C/W)*(7700 J/°C) = 200 s
We now have the thermal decline curve:
T(t) = T0 * e^-(t/tau) = (153 °C) * 1/e^(t/tau)
If we want steam to disappear at time t, then T(t) = 100°C. So:
(100°C) = (153 °C) * 1/e^(t/tau)
(t/tau) = ln((153°C) /(100 °C)
t = ln((153°C) /(100 °C)) * (200 s)
t = 85 s
So, if all is as assumed above (very unlikely!) the device should not be able
to output steam for 15 minutes, or even more than 2 minutes, unless a source of
heat was present after the power was cut off. The problem is we just do not
have enough data to make the above calculation credibly. This is not a new
kind of problem with regard to the E-Cat.
Hopefully in any case the above example is useful to others for theorizing.
We just have to wait until October to see what happens. I hope for the
best. I hope we don't see non-credible delays and moving target objectives as
we have seen before in similar situations. I wish Rossi great success. Even
the most minor technical success for Rossi would be one of the greatest
scientific breakthroughs ever, and have great importance for all mankind.
Rossi is not a young man. I hope he considers how limited his time on earth is
and makes the right decisions.
Horace Heffner
http://www.mtaonline.net/~hheffner/
