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June 23, 2008

Arata Cold Fusion follow up : how much excess heat

This is a follow up on the Arata cold fusion experiment which was showing excess heat


Physics world has a review of the Arata cold fusion research This information is a review of the Arata cold fusion work at the LENR-CANR site. [Low Energy Nuclear Reactions, also known as Cold Fusion. (CANR, Chemically Assisted Nuclear Reaction)]

Based on 0.5 to 1 degree celsius of excess heat for one liter of water for 5000 minutes then it appears to be about 200-400 kwh of excess heat.

A deuterium (cold fusion) versus hydrogen (ordinary chemical) experiment is performed by Arata. There is always a significant temperature difference between T (inside, Tin) and T(surface of the energy cell, Ts), indicating that the sample and cell are not reaching equilibrium. After 300 minutes the Tin of the deuterium experiment is about 28 °C (4 °C warmer than ambient), while Tin/Ts of the hydrogen experiment is at about 25 °C (1 °C warmer than ambient).

Arata claims that, given the large amount of power involved, this must be some form of fusion — what he prefers to call "solid fusion". This can be described, he says, by the following equation:

D + D = 4He + heat

The deuterium experiments remain 1 °C or more than ambient for at least 3000 minutes while still exhibiting the temperature difference between the sample and the cell, Tin and Ts.



20 °C calorie: the amount of energy required to warm 1 g of air-free water from 19.5 °C to 20.5 °C at a constant pressure of 101.325 kPa (1 atm). This is about 4.182 J. The experiment appears to have been dealing with one liter of water. So one extra degree for 5000 minutes would be 300,000 seconds times 4172 joules which is 1251.6 megajoules. This is 347.6 kwh. Being able to convert that thermal energy into electricity would not be efficient without boosting the temperature.

1 comments:

Bob said...

Oh Lord ...

PLEASE don't perform thermodynamics calculations without understanding thermodynamics.

In a system of an inner container surrounded by an external vessel, only radiative, conductive and convective heat transfer mechanisms exist for shunting the heat of the sample to the environment. That a sample that starts at 70ºC drops to ambient (23ºC) over several hours does not in any way indicate that some reaction is keeping it at an elevated temperature for the duration.

To me, from a the graduate thermo course I took at UCBerkeley, it looks like a classic binding-energy release curve. For whatever reason, D likes to bind to nanometer scale palladium, and upon so binding, releases a fair amount of caloric energy to the suspension water. Enough to take a liter and raise it to about 70ºC from about 23ºC (delta = 47ºC, at 73.3 J/K/mol = 191 kJ) or 0.05 kWh.

The part that isn't spoken of (seriously) is that activation 'bake time' ... which easily could require as much or more energy to activate the nanoparticles suspension.

If the experimenters had really wanted to show that there was a continuous evolution of energy after the obvious initial-phase activation energy release, then they would have compared a vessel heated to 70ºC by the admixture of 300 ml 23ºC and 700 ml 100ºC water into the inner vessel (no palladium), then measuring the temperature decrease over time as a reference system standard. I'd bet good money that the curves are VERY similar.

[PS: there may be an additional scavanging heat-of-formation release between the remenant partially unreacted palladium and the free D2 in solution ... again, it is chemical energy, not nuclear driving this, if measured.]