After thinking a little bit about the calculations I did (see below) and considering what I have learned from this year reading vortex, I came to the conclusion that the engineering approach proposed by Aussie Guy (and also Rossi) is the best approach forward in the LERN field... If one manages to show a LERN device that runs for days on its own and that is able to light a single small light (a LED would suffice) during its operation, we will see proper resources employed to do good research on LERN all over the World. This will result in the production of devices that will really change the way we live.
Aussie Guy: please build your kits. They will sell a lot. Cheers, Alberto. On Tue, Dec 27, 2011 at 3:52 PM, Alberto De Souza < alberto.investi...@gmail.com> wrote: > After some calculations, I think it is better to use the MPG-D751. See > below. > > On Tue, Dec 27, 2011 at 3:17 AM, Aussie Guy E-Cat < > aussieguy.e...@gmail.com> wrote: > >> The 2.5 x 2.5 mm device has a max power output of approx 0.8 mW at 10 deg >> K differential. Assuming 1 Watt excess with a COP 5 yields 200 mW input. >> Would need around 300 of the MPG-D615 devices with fitted finned heat sinks >> to each device's COLD side to get good thermal transfer into the air. >> Could be doable with 75 devices per finned heat sink assembly per side of >> a square container. > > > This is exactly the set up I had in mind. > > >> Optimal load resistance could be a issue. Something to look at in the >> future. >> > > Using the data in http://www.lenr-canr.org/acrobat/DashJcoldfusion.pdf(slide > 20), one can estimate that the average voltage required in a PdD > cell is about 5,7W / 1.5A = 3.8V (see also slide 11). One MPG-D751 can > provide 1.2V at about 1mA with 10 deg K differential (see > http://www.micropelt.com/down/datasheet_mpg_d651_d751.pdf voltage x > current graph), i.e., 1.2mW. Using the circuit shown in > http://www.national.com/pf/LM/LM2621.html#Overview we can elevate that to > 3.8V. Assuming an efficience of 50% (it should be better than that, see > http://www.ti.com/lit/ds/symlink/lm2621.pdf), we have 0.6mW per MPG-D751. > > To achieve 5.7W, we have to put 5.7W / 0.6mW = 9,500 MPG-D751 in parallel > (and use at least 2 LM2621 circuits). These will occupy about 9,500 x > 4.248mm * 3.364mm = 135,758mm2. This is a square of ~368mm on each side. > Using a rectangular recipient and putting these 9,500 units on its 4 > lateral sides, we have a minimum lateral side size of 184mm x 184mm, or > ~18cm x ~18cm. > > If I did the math correctly, it is doable. But we need a COP of 20 or more > (not considering peak power eventualy needed during reaction startup and/or > control). > > Cheers, > > Alberto. >
