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.

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 230°C. This gives a thermal resistance of

   R = (230°C)/(5.10^3 W) = 0.046 °C/W.

The decay time constant, tau, for the 1 MJ thermal mass, C, is is given by:

   tau = R*Cth =  (0.046 °C/W)*(7700 J/°C) = 354 s

We now have the thermal decline curve:

   T(t)  = T0 * e^-(t/tau) = (230 °C) * 1/e^(t/tau)

If we want steam to disappear at time t, then T(t) = 100°C.  So:

   (100°C) = (230 °C) * 1/e^(t/tau)

    (t/tau) = ln((230°C) /(100 °C)

    t = ln((230°C) /(100 °C)) * (354 s)

   t = 294 s ~= 5 min

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 5 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.

BTW, most anyone in the LENR field knows well what "heat after death" means. It was used much in discussions on sci.physics.fusion over 15 years go, and is still in use in the present literature.

Horace Heffner
http://www.mtaonline.net/~hheffner/



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