I picked up one of these voltage regulators, which will allow me to set the San Ace fan voltage exactly, when connected to to my 12.1 V supply. The instructions claim that the adjustment pot is not designed to be frequently adjusted, but I think that it should be OK to run it through a few voltage steps without wearing it out.
MIZUNO REPLICATION AND MATERIALS ONLY


I did the 1 W to 2.3 W fan input steps last night, attempting to replicate the Figure 9 (JCMNS 2019 Mizuno report) plot. It is interesting, but not difficult, matching the power for each step, because the blower fan does not have a constant resistance across the voltage range (because it is a motor).

Filled in the steps up to (including) 3.6 W now.
The results are fascinating.


Despite what looks like a huge range of values when Down and the Left Side traverse measurements are added to the plot, in general, the average of the Right Side and Up, plus Center anemometer traverse measurements is only about 0.25 m/s higher than the average of all of the traverse points.
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Regarding the above plots of averages, note that the velocity is almost unchanged from 4.8 W to 6.0 W. The RPM, however, increases at about the usual rate relative to W. I suspect that the start of this flat spot is the peak mechanical efficiency of the fan.
The fan will blow harder if I increase the voltage further, but the air restriction of the fan housing itself will increase the cost of average velocity in terms of power. (There might still be a higher voltage/RPM sweet spot for the motor/fan).

Coming together...

Using solely a 5 cm air inlet hole for the calorimeter makes the blower fan protest very noticeably. Major drop in RPM and significantly more motor noise. I doubt Mizuno used only the 5 cm air inlet for very long. The larger inlet is required to keep the calorimeter from developing a vacuum inside (and possibly burning out the fan motor) as well as helping with plumbing clearances.
Edit: Blocking the 10 x 20 cm inlet, so that all air entering the calorimeter box must enter only through the 5 cm diameter inlet hole causes the fan to increase RPM from 3000 to 3450 RPM (at 8.60 V). Presumably this is due to fan having lower air resistance due to a partial vacuum inside the box.

Can we briefly discuss the figure on page 19 of the ICCF 21 Mizuno paper?
You show 13.4 L/s air in your calculations, however I get 13.54 L/s air using the values supplied.
This leads to 202.6 W rather than the 198 W you calculated, using 1.293 kg/m^{3}as suggested.
However, consulting an online calculator https://www.engineeringtoolbox…essuredensityd_771.html , the suggested mass of air for 29.32 C is 1.167 kg/m^{3}, which lowers the power to 182.9 W.
Is there something obvious that I have missed?

Sorry if im being dumb but neither of the two papers in #1 (ICCF 21 and 22) have 19 pages.
1.293 kg/m3 seems a pretty high density to use. Think you would need dry air, temperatures around 510C and pressures around 10401060mbar.


Is there something obvious that I have missed?
At sea level and at 15 degrees C, the density of air is 1.225 kg/m^{3}. 1.29 at 0 degree. But this is for dry air!

Can we briefly discuss the figure on page 19 of the ICCF 21 Mizuno paper?
No such page. ???

No such page. ???
Excess heat generation by simple treatment of reaction metal in hydrogen gas, Tadahiko Mizuno, Jed Rothwell, ICCF21 38 June 2018, Fort Collins, CO, International Conference for Condensed Matter Nuclear Science, PDF page 21, numbered page 19.
My version of this has 29 numbered pages, including bonus slides, and ends with a poster consisting of 6 more numbered pages (16).
The first two sentences below the figure are: “Let me talk about the 99% recovery rate. As I said, it seems too good to be true.”
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At sea level and at 15 degrees C, the density of air is 1.225 kg/m^{3}. 1.29 at 0 degree. But this is for dry air!
OK, so what do you propose?

Excess heat generation by simple treatment of reaction metal in hydrogen gas, Tadahiko Mizuno, Jed Rothwell, ICCF21 38 June 2018
The PowerPoint slides:
https://www.lenrcanr.org/acrobat/MizunoTexcessheat.pdf
You show 13.4 L/s air in your calculations, however I get 13.54 L/s air using the values supplied.
This leads to 202.6 W rather than the 198 W you calculated, using 1.293 kg/m3as suggested.
I am not going to quibble about a 4 W difference. The error margin is larger than that.
However, consulting an online calculator https://www.engineeringtoolbox…essuredensityd_771.html , the suggested mass of air for 29.32 C is 1.167 kg/m3, which lowers the power to 182.9 W.
I don't see where on that page you found those numbers. Anyway, I used an online calculator (cheat sheet) and that's what you see in the slide.


The PowerPoint slides:
https://www.lenrcanr.org/acrobat/MizunoTexcessheat.pdf
I am not going to quibble about a 4 W difference. The error margin is larger than that.
I don't see where on that page you found those numbers. Anyway, I used an online calculator (cheat sheet) and that's what you see in the slide.
I am not quibbling about a 4 W difference, (other than that gives a very slight positive COP for the calibration), I am quibbling about the 17 W difference, which means a recovery of 91.5 %, rather than 99%.
There was a calculator on the Engineering Toolbox that gave the density of air at a usersupplied temperature, with a popup answer box. (Online Air Density Calculator) That is where I got the 1.167 kg/m^{3} value from. https://www.engineeringtoolbox…pecificweightd_600.html
Really, I am not even quibbling about the numbers per se, I am trying to get my calculations to jive with Mizuno’s calculations. That means first I should be able to get the same answer as Mizuno does with his numbers. Now that I do not have that conformity, I need to know if there are other parts missing in my calculations. For example, does the calculation used for the L/s include a correction for standard conditions which must be applied so that all values are normalized to standard conditions?

Will look into this next week.

I just noticed that there are two pages numbered ‘3’ and ‘19’ in that version, (which is the version I was referring to above).
Maybe the original Power Point handled the pages a bit differently.

OK, so what do you propose?
Just extrapolate it linearly to the temperature of the air. Mizuno is close to sea level.

Just extrapolate it linearly to the temperature of the air. Mizuno is close to sea level.
I am within 100 m of sea level elevation also.
I did account for temperature and normal sea level air pressure. That is where the discrepancy seems to come from.
Even humidity has little effect on air density below 30 C (increased humidity makes air density less).
Here is an air density calculator that includes humidity.
https://www.omnicalculator.com/physics/airdensity
Note that for dry air it gives the same value as the Engineering Toolbox. So at 80% humidity, the density for 29.32 C air is 1.1556 kg/m^{3} instead of 1.167 kg/m^{3} for dry air. So at 13.54 L/s, that is now only 181 W for 80% humid air, which represents 90.5 % heat recovery (using these values) instead of 99% recovery.
