> I can not see how the above remark is relevant in any way. Did you not see
> that I am providing the standard logarithmic decay function? The cutoff
> time for the logarithmic decay is intended to set a *boundary time* on when
> there is no more steam. If the temperature of the mass is below boiling
> point then it can not generate steam.
Your exponential (not log) function is not correct as it does not allow for
exponential decay in energy. There is no need to hash this out. My original
presentation and conclusions are correct. Your calculations are not correct,
nevertheless they do show 15 minutes is possible. There's nothing hypothetical
about thermal inertia. Perhaps you are referring to your errant guesswork. We
don't need to second guess mother nature.
I mentioned ~ 1MJ not exactly 1 MJ and of course the amount will vary with
mass, specific heat and temperature. The problem is that you used it without
understanding it. This is not a plug-in.
----- Original Message -----
From: Horace Heffner
To: vortex-l@eskimo.com
Sent: Monday, August 29, 2011 2:53 AM
Subject: Re: [Vo]:Corrections to "heat after death" calculations
On Aug 28, 2011, at 5:37 PM, Joe Catania wrote:
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.
I can not see how the above remark is relevant in any way. Did you not see
that I am providing the standard logarithmic decay function? The cutoff time
for the logarithmic decay is intended to set a *boundary time* on when there is
no more steam. If the temperature of the mass is below boiling point then it
can not generate steam.
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.
My calculations do not show that "only a few minutes is possible." My
calculations show the decay time *for some assumed conditions*. *Under those
conditions* only a very short time is possible. I explicitly stated this. I
am providing a model to assist others in make such calculations based upon
their own assumptions. It is important to keep in mind such an approach is
purely hypothetical, but might shed some light on boundaries to the
experimental conditions.
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.
I did not supply the 1 MJ number. It was suggested on this list. Further,
I don't believe 1 MJ number is correct and never did. It is merely a *sample*
assumption based on the postings of others, your postings perhaps.
Your temperature seems low.
The temperature is a result of the *assumptions*, specifically the assumption
of a 1 MJ energy storage, and the assumed power input. I set up the the spread
sheet so that temperature is specified and stored energy is computed, instead
of vice versa as was done in the manual calc. below.
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.
Have you not heard of logarithmic decay curves? Their shape is based on a
decay constant lambda, or half life, etc. Did you even read my post?
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.
I don't think we have communicated. Please explain what the following
comments, taken from my post quoted below by you, meant to you:
"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 ... "
"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."
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.
The fiasco is due to the lack of data and credible measurement of run
enthalpy.
I should note that I posted a corrected version of this, August 28, 2011
1:06:32 PM AKDT, including reference to a spread sheet with multiple
assumptions, but that post has not come back. I don't see it in the archive
either. I will resubmit it. I made reference to it in the following post:
http://www.mail-archive.com/vortex-l%40eskimo.com/msg50819.html
which includes an additional 15 sets of assumptions. The first spread sheet
also had 15 assumptions:
http://www.mtaonline.net/~hheffner/RossiThermal.pdf
http://www.mtaonline.net/~hheffner/RossiThermal2.pdf
I also added calculations to take into account flow rate and water heating
energy.
BTW, if you don't like my assumed numbers feel free to tell me what your
assumed numbers are for Mass, Thermal Power (before shutoff), Inlet Temp., Mass
Temp., and Inlet Flow. However, it should be self evident that the results
can be made to look as you please by choice of assumptions. The required data
is simply not available to make a determination. However, discussion based on
quantitative analysis should be more meaningful than discussion based on
personal feelings and arbitrary assumptions. At least some of the inconsistent
assumptions might be ruled out.
----- 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/
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
