On Aug 29, 2011, at 6:48 AM, Joe Catania wrote:
> 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.
That is correct - it is called an exponential decay function. I am
an old guy with some memory problems. I forget simple words and
sometimes it takes me a long time to write because I have to find the
words. As anyone knows who has read my posts here for the last 15
years, I am also prone to clerical errors, arithmetic errors, and
seldom get things right the first pass. However I wrote my first
thermal transfer finite element code in the 1970's, so I am no
stranger to these issues. You should note the function used/provided
is correct, namely:
T(t) = T0 * e^-(t/tau)
There is no need to hash this out.
There is no need to discuss Rossi at all, since there is insufficient
data provided by him to determine anything. All such discussion is
at best hypothetical, if not even based on obvious untruths.
However, this has not stopped either you or Jed from repeatedly
taking positions without foundation in fact. I think it is clear an
analysis can rule out some assumption sets as self-inconsistent.
That is more useful than repeated useless argument based on differing
assumptions.
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.
Again I say, 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.
If you think I have arithmetic or analytical errors, calculation
errors, please point them out specifically.
----- 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/
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
