stefan Member

(Quote from CWatters)


It turns out that we can make progress on the self-sustain issue without knowing the details of insulation etc. This is because Alan Smith has kindly supplied information regarding the average rate of energy input needed to hold…

Well, we have a very serious hard-nose Physicist here today, going through the data and looking at the hardware in operation. It will be interesting to see if we can make a dent in his skepticism about our work.

(Quote from Bruce__H)


The question is meaningful, as you intended, but the discussion is not yet at that stage. We haven't arrived at a decision yet. As for replication, that would not necessarily require publication, but might be achieved in other…

(Quote from Bruce__H)

if the reactor is free in the air, for air (convective) you have coefficients of the order 10-100 Power / Area / Kelvin probably 10.
for radiation you have W = sigma (T^4 - T0^4) = sigma (T^3+T^2T0+TT0^2+T0^3) (T - T0) => T=400K ->…

(Quote from CWatters)

Good point,

I think that the cooling dependence as a function on temperature has a knee. For small temperatures convection carries the heat away in a k(T-T1) fashion but at higher temperatures radiation may result and that's T^4…

(Quote from stefan)

If I've understood correctly... When switched off BOTH fueled and non-fueled reactors should cool assymtoticly towards room temperature. But the fueled reactor actually cools more slowly or starts cooling after a delay (this being the…

(Quote from Adrian Ashfield)

I bet you had chimneys above the furnaces, directly or further up, so that the heated air went directly to the outlet without mixing much with the rest of the room so much.
It all also depends on the size of the outlets and…

I'll answer for Russ on this, since he probably won't post any more. Right now we have something pretty amazing going on, but we don't yet know how useful it might be. Signs are that it will be very useful, but we need to do a lot more work to make it…

(Quote from Alan Smith)

This means that the controler kicks in at 200 or 230 which can be above both T1 and T2

(Quote from Bruce__H)

say T2 is maybe 275 degree C, and T1 is 220 degree C or whatever. No idea, we must wait for proper data to say anything. What I don't understand is that the curve seam to never go under
200 degrees.

OK. With Wreact = 0 for T less than some activation temperature (which for moderate temperatures qualitatively matches my model), and for k_cooling (k1) greater than k_heating (k2), I see that T should exponentially decline from some high temperature…

(Quote from Bruce__H)

I presented the formula and said T2 is the activation temperature meaning Wreact = 0 for T < T2

(Quote from Bruce__H)

your wrong, I modeled the T2 as the activation temperature and T2 > T0 so it will reach T2
and then become deactivated and no longer active e.g. no more heat generated and what remains is the
cooling only

(Quote from Bruce__H)


So let's be more rigorous, Wcool = k1(T - T1), Wreact=k2(T-T2), T2 activation temperature,
then Wreact-Wcool = (k2-k1)(T-(T2k2-T1k1)/(k2-k1)) = (k2-k1)(T - T0),
say that k2 = k and k1 = Ak (A > 1 if k1 > k2)) lead to the everall…

(Quote from Bruce__H)

Yes this is the case if Wreaction - Wcooling is positive, my point is that if Wreaction is k1T and Wcooling = k2T and k2 > k1 (assuming a = 0), then you will not get an increase as you suggest because it will cool just like with
…

(Quote from Bruce__H)

(Quote from Bruce__H)

The plot indicates this linear relationship in the region of interest, this condition would never be satisfied in this region else of cause for all possible temperatures you have a variation and your
analysis is…

(Quote from Bruce__H)

generated heat W = a + k T, k possitive

My point is that if the experiment is okey and there is an excess energy, then the data indicate that the reaction rate is indeed linear in the temperature band in question.

(Quote from Bruce__H)

This assumes a linear relationship between power and temperature, small intervalls are always linear, but the question is if the actual interval is small or not, we don't know.

Newtonion cooling means the the power is basically k1T, k1 constant
Assume the power of the LENR reaction is proprotional to T e.g. k2T, k2 constant

this means that the power added is (k2-k1)T

If we assume that k2 < k1 and take k = k1 - k2, we see…