MIZUNO REPLICATION AND MATERIALS ONLY

Quote from Curbina

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).

mizunotadahiko ,

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?

Quote from RobertBryant

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 ?

.

Quote from RobertBryant

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..

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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.

Quote from Curbina

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...

Quote from Paradigmnoia

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.

Quote from JedRothwell

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?

Quote from RobertBryant

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.

Quote from RobertBryant

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.

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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.

Dr. Mizuno, ( mizunotadahiko )

Is it possible to make a plot of the not-corrected heat power (W) of both the active and calibration at 500 W input? I think this would be easier to understand and reduce petty arguments about what is being presented.

Optionally, an additional calibration power trace at a power level close to the activated reactor heat level may be extra helpful, if the final plot is not too crowded.

Thank you.