Thank you mizunotadahiko for the graphs.
One proposal, it would be helpful for the not so well informed user if you could give a short comment to the graphs.
Many thanks!
Thank you mizunotadahiko for the graphs.
One proposal, it would be helpful for the not so well informed user if you could give a short comment to the graphs.
Many thanks!
the log versus 1/T(K) graph confirms earlier work that the reactor does not follow an Arrhenius dependency..
As the temperature rises the 'reaction'(s) appear to require more 'activation' energy
but thinking of in terms of one reaction, and it relying on Boltzmann-type kinetic collisions is probably erroneous..
Rough calculations of the energy input/output for the active R8
electrical heating input = ~ 9.3MJ
thermal output=~13.9MJ
Old Mizuno et al patent EP2597652A1 with interesting drawings
Much thanks to mizunotadahiko for sharing these results with us. As far as they are self explanatory they show active runs with a COP of 1,5 and control runs with COP close to 0,99, so it can be inferred that excess heat is being produced in the active cells. We really hope you can get the patents and then we will be able to know more about how you achieve this and see if others can obtain the same results as consistently as you. This is, IMHO, very good news and have greatly lightened my weekend.
Much thanks to mizunotadahiko for sharing these results with us. As far as they are self explanatory they show active runs with a COP of 1,5 and control runs with COP close to 0,99, so it can be inferred that excess heat is being produced in the active cells. We really hope you can get the patents and then we will be able to know more about how you achieve this and see if others can obtain the same results as consistently as you. This is, IMHO, very good news and have greatly lightened my weekend.
Actually, the raw COP is closer to 0.8, which is decent for his type of calorimeter.
The loss-corrected COP is very close to 1, (as it should be).
Thank you for sharing your work with us here.
Is the 393 C temperature the peak temperature for the 800 W active test?
How hot does the control reactor get at 500 W?
Can you please report the temperature delta for the control and active tests?
Actually, the raw COP is closer to 0.8,
there is no raw COP data presented ..
do not make assumptions.
Actually, the raw COP is closer to 0.8, which is decent for his type of calorimeter.
The loss-corrected COP is very close to 1, (as it should be).
I assume you are talking about the control run, I just said 0,99 for not saying 1 as it can’t be higher than that if all losses are accounted for an the only input is the heater power and no reactive cell.
The 13,94 output to 9,33 input from the numbers that robert bryant was gentle to calculate from the active run plots, gives 1,49.
Is the 393 C temperature the peak temperature for the 800 W active test?
How hot does the control reactor get at 500 W?
Does Paradigmnoia have a model for the R8 reactor..temperatures COPs and so on..
If so, could P . please report on the model..
perhaps this could be part of a new thread
there is no raw COP data presented ..
do not make assumptions.
Just under 500 W input, and a little less than 400 W measured output. Seems like a raw enough COP to me.
Maybe closer to 0.76 than 0.80 but close enough without scaling the diagram...
Or rather is the calorimeter heat recovery actually slightly less than 50% at 800 W ?
.
Does Paradigmnoia have a model for the R8 reactor..temperatures COPs and so on..
If so, could P . please report on the model..
perhaps this could be part of a new thread..
AFAIK P has done extensive work on replication of R20... but not of R8
and without the R20..
perhaps P could summarise the results so far..
Display MoreDoes Paradigmnoia have a model for the R8 reactor..temperatures COPs and so on..
If so, could P . please report on the model..
perhaps this could be part of a new thread..
AFAIK P has done extensive work on replication of R20... but not of R8
and without the R20..
perhaps P could summarise the results so far..
Does RBryant have an opinion on whether the graph above shows about 80 % or 50 % heat recovery?
I will note that the grey power trace extends out beyond the red trace at the beginning, and the red trace shows no magnification of measurement “noise“ effect typical of heat loss compensation factoring, so they appear to be two different measurements.
I assume you are talking about the control run, I just said 0,99 for not saying 1 as it can’t be higher than that if all losses are accounted for an the only input is the heater power and no reactive cell.
The 13,94 output to 9,33 input from the numbers that robert bryant was gentle to calculate from the active run plots, gives 1,49.
I see.
I personally don’t like to use the cumulative energy calculations until everything else seems tight. The reactor and control masses need to be the same, etc...
I see.
I personally don’t like to use the cumulative energy calculations until everything else seems tight. The reactor and control masses need to be the same, etc..
well, you are entitled to your different taste, you Can doubt the data all you want, but I think this is enough to maintain the interest in Mizuno’s work.
well, you are entitled to your different taste
The loss-corrected COP is very close to 1, (as it should be
Could P show ALL his calculations and ALL his assumptions for his assertion of a COP of 1 based on the R8 configuration and 800W input..
rather than based on his pipe contraption? and rather than just expressing 'taste'
My opinion of P's taste will be based on these calculations and assumptions.
The reactor and control masses need to be the same, etc...
No, they do not. You cannot detect any difference in the data from different reactors or control masses.
You seem to have a talent for finding pretend problems.
No, they do not. You cannot detect any difference in the data from different reactors or control masses.
You seem to have a talent for finding pretend problems.
If one compares all output energy measured from start to finish, comparing a control to an experiment device, then the masses, surface area, etc. of both devices must be the same. That is totally different from comparing control steady state to experiment steady state values.
Do you think a 100 g ceramic device will behave identically, from start to finish, to a 20 kg stainless steel device at the same constant input power level in a mass air flow calorimeter?
Could P show ALL his calculations and ALL his assumptions for his assertion of a COP of 1 based on the R8 configuration and 800W input..
rather than based on his pipe contraption? and rather than just expressing 'taste'
My opinion of P's taste will be based on these calculations and assumptions.
I did not assert a COP of 1.
The loss-corrected COP is very close to 1, (as it should be).
is an assertion
please calculations and assumptions for should be
and what exactly does close to 1 mean?
This indicates a certain amount of ignorant arrogance
Show Paradigmnoia data for calorimetry balance on R8 configuration..with 800 W input.
Display MoreThe loss-corrected COP is very close to 1, (as it should be).
is an assertion
please calculations and assumptions for should be
and what exactly does close to 1 mean?
This indicates a certain amount of ignorant arrogance
Show Paradigmnoia data for calorimetry balance on R8 configuration..with 800 W input.
So now you assert almost 100% recovery of heat in the calorimeter at steady state?
(That would be awesome, but it is not likely)
Edit: I was discussing the COP of the control with Curbina earlier. Not the the activated experiment. It is still very, very unlikely that this plot shows the control/calibration for measured heat power without a calorimeter heat loss correction factor included.