# Mizuno reports increased excess heat

• This is true, unless someone else has already communicated the idea openly an made it public earlier.

Correct. Or, someone else may have filed similar invention earlier (that is not yet public).
As part of the patent application process a search for prior art will be performed.

If prior art is found the patent application process will be stopped or the claims will be modified or slimmed down.

Thanks for filling this in.

• And AFAIK we still don't know the precise positioning and shape of the 2 meter long heating element that ended up inside that much smaller reactor (or did I miss the answer?).

Yes, that what I was asking for.

• Some rough calculations on Mizuno's R20 reactor from atomic perspective to trigger further thoughts

(please, fill in any errors, assumptions and suggestions):

(1) Surplus heat @ 3KW: 2700W - Energy release 2700 W/s = 1.7 * 1022 eV

(2) Number of Deuterium atoms (n) in the reactor using ideal gas law PV = nRT : n = PV/RT (mol)

P = 0.0002 (@ 200 P)

V = 5.7 liters (given by R20 cylinder)

R = 0.08314 (given constant)

T = 673 K ( given by 400 degrees C as an R20 estimate)

n = 0.00020374 mol Deuterium = 2.4449 * 1020 free atoms in reactor space.

Assuming that absorbed Deuterium atoms do not take part in the energy generation, but serve as a 'gas reserve' that by means of an Deuterium equilibrium will be released in reactor space due to gas pressure and/or gas temperature changes.

Let's further assume that 1% of the free space atoms participate in energy release at the surface of the PD/Ni mesh.

That would result in 100 *(1)/(2) = 7 KeV per participating atom. Not a fusion result (would require several MeV/atom) if given assumptions would be correct.

• Ensure the control and active reactors are identical in size, shape and surface color (photographs)

The control and active reactors should be identical for any other detail, including the characteristics of the main components. But look at the spreadsheets of the active and control tests performed in May 2016, both run at a nominal input power of 120 W:

- active test, held on May 19, 2016 (link to the spreadsheet published in (1));

- control test, held on May 20, 2016 (link to the spreadsheet published in (2)).

The spreadsheets show a big difference in the DC parameters:

 V/DC I/DC Win Active test (-)49 V ca (-)2.4 A ca. 120 W ca. Control test (-)67 V ca. (-)1.8 A ca. 120 W ca.

So, either the resistances of the DC heaters are very different (about 20 and 40 ohm respectively) , or at least one DC measurement is wrong.

Assuming that the reactors are identical and the wrong DC parameter is a voltage, the real values of Win should have been proportional to I2. Now, the ratio between the I2 is about 1.8, a value compatible with the maximum Wout/Win reported in the active spreadsheet, which was about 1.5, but still on the rise when Win was zeroed.

I don't mean that this was the case, but - as it was already pointed out by you and others - a basic prerequisite to consider the validity of such extraordinary claims is the availability of much more detailed infos, such as the raw data recorded during the tests.

• Good point. For example, these different conditions mean that errors in measurement - e.g. average ac measurements of a non-sinusoidal waveform - could deliver different errors.

Not knowing of an error is different from knowing there are no errors.

The latter is what is needed here for results such as these to be convincing to all except an already convinced few.

• The control and active reactors should be identical for any other detail, including the characteristics of the main components. But look at the spreadsheets of the active and control tests performed in May 2016, both run at a nominal input power of 120 W:

Agreed.

Why is the finish not the same on the control vs the active reactor? (See figure 6, red box -- reactor is dull. Emmissivity of weathered stainless steel is 0.85 vs. polished stainless steel 0.075)

This throws off the calorimetry which is clearly a combination of conduction, radiation, and convection on different components of the rig. The calibration needs to be repeated. I would prefer if it was repeated using the heater inside the R20 unit and not this calibration reactor.

• I don't like to pick on details but it is part of a scientific process, as this discussion is, or should be. I hope my main objections will prove unfounded and that Mizuno's efforts will become rewarded, but the more I have looked into the data, the more questions are raised. The results seemed robust after my first reading.

The exponential excess heat/temperature curve in fig. 8 transforms into a line if the two(only) data points at 0 power input are omitted. And they should. With 0 power and the reactor wall temperature 23 C we should not expect any measurable output. The 2 W presented should be considered within the error limits.

In my experience (and according to the common energy dependencies), a straight line is not the expected behaviour for a temperature dependent reaction. It does not mean that the presented data must be erroneous, but it is a warning sign.

• Lower emissivity on test reactor vs control makes calibration inaccurate:

If the emissivity on the test reactor is decreased to 0.075 vs. 0.75 on the control reactor, less heat leaves the test reactor by radiation at a given temperature. This means that:

1) the test reactor surface temperature will get higher in equilibrium;

2) more heat will leave the test reactor by convection into the air mass flow calorimeter.

This means that the air mass flow calorimeter efficiency (what is called in the paper the % of heat captured by the calorimeter) would be higher.

Worst case assume the efficiency of the calorimeter is 100% instead of the reported lower numbers. This lowers possible COP on the 50 watt R20 test reactor from 300/50 = 6x to 225/50=4.5x, i.e. the grey line in Figure 6 instead of the orange line. Still a strong result, but not as good.

It does however call into question the entire R19 reactor series data in table 1 (the June 18, 2019 paper) where it lowers many of the COPs to around 1.1 to 1.2 from 1.34 to 1.51. For example, look at the first table entry on 2/20 with the reactor at 238.90 C and Out/In at 1.39. The calorimeter calibration correction factor is show in equation 2 on page 11 of the JCMNS 29(2019)1-12 paper as O/i = 0.95 - 5.0811e-4 x t. Substitute 238.9 and get O/i = 82.86%. Multiply the COP of 1.39 by 82.86% and get COP = 1.15. Clearly a lot less signal in the noise.

(Note: the new correction factor equation was not published in the June 18th paper, but instead "The calorimeter performance has been stable. These critical parameters values have not changed significantly since 2017.)

Now compare the best run on 4/15 at T=381.00 and O/I = 1.51. Correction factor from the JCMNS equation 2 is 72.86%. The corrected COP now becomes 1.1x.

I believe we (and Mizuno/Rothwell) should prioritize the R20 COPs > 2 as these experiments are a lot more forgiving, if calibrated with published fan input power, then the COP 1.1 results.

• Adjustment for D2 gas at different pressure, vs control:

D2 gas, like it's H2 cousin and helium are very thermally conductive even at low pressures. Their thermal conductivity is reduced as pressures drop below 1 Torr (133 Pa) , falling to less than 1% of their 1 bar value around 4e-4 Torr (0.05 Pa).

For this experiment, the D2 is contained in the conflat except at the gas/vacuum supply line. Because all heat must pass out of the conflat thru either the walls, or the supply line, we need to estimate the variability of the heat being removed from the supply line vs the control at different temperatures and pressures.

So:

1) The control reactor should be mechanically (and emissivity) identical to the test reactor (if not the same) so that any different conduction, convection, or radiation paths thru the metal are the same.

2) The control reactor if possible should be measured with D2 in the test (but without Ni/Pd mesh) to eliminate the difference in thermal conductivity.

3) Absent (2) or (1), we need to estimate the change in the thermal conductivity from the reactor out thru the supply line (either because the control reactor doesn't have a supply line, or because the control reactor doesn't have D2 in the supply line). My suggestion would be to put a thermocouple on the conflat at the supply line exit, and then to put a thermocouple on the supply line at a the lesser distance of a few cm short of the first mechanical branching off (where conduction out would be expected), or 0.25 meters away from the conflat. This will allow (by having two temperature sensors) an estimate of the conduction losses thru the supply line and its gas.

In general introducing high conductivity gas in the supply line will remove heat from the test conflat, making heat rejected into the air mass flow calorimeter lower, and thus being conservative. However any relative increase in heat (temperature) noted at lower pressures significantly below 133 Pa may be due to lower gas conductivity out the supply line compared to pressures above 133 Pa.

Lastly, it might be interesting to note if there is a significant temperature difference between the longitudinal center of the cylinder near the mesh or the heater, and the end far away from the heater with and without D2 (i.e. under vacuum < 0.05 Pa). if the temperature is more evenly distributed by a significant amount (say more than 5 degrees) in the case with D2 loaded, the heat will have more surface area to conduct out the supports and to convect to the air mass flow. By measuring the temperature difference it can be established that the upper bound is not significant for the calorimetry. If it is significant, again, I prefer that the calibration be done in the test reactor with the mesh removed under same D2 pressure.

• Why is the finish not the same on the control vs the active reactor? (See figure 6, red box -- reactor is dull. Emmissivity of weathered stainless steel is 0.85 vs. polished stainless steel 0.075)

This throws off the calorimetry which is clearly a combination of conduction, radiation, and convection on different components of the rig. The calibration needs to be repeated. I would prefer if it was repeated using the heater inside the R20 unit and not this calibration reactor.

Hi,

I was referring to the amazing results reported last year at ICCF21, in particular those shown on Fig.4 (1), not to the results allegedly obtained with R19 and R20 reactors and disclosed in (2). IMO, these last results can't be commented upon, due to the lack of detailed information. However, they are so sensational that they only deserve to be widely advertized in order to be replicated by any interested scientific or industrial party. I can only wish good luck to the authors.

On the contrary, the results reported at ICCF21, which were also described in 2017 at JCMNS (3), may worth a more in-depth analysis, thanks to the availability of much more information, in particular the spreadsheets with the raw data of the 120W active and control tests. In my experience, when a sufficient amount of experimental data is publicly available, it is always possible to find out those errors which explain in a simple and mundane way any seemingly extraordinary result.

• In general introducing high conductivity gas in the supply line will remove heat from the test conflat, making heat rejected into the air mass flow calorimeter lower, and thus being conservative. However any relative increase in heat (temperature) noted at lower pressures significantly below 133 Pa may be due to lower gas conductivity out the supply line compared to pressures above 133 Pa.

Worth noting that conservative errors are still problematic:

(1) They prevent calibration data from checking the "first principles" absolute measurements, unless the error can be proven identical in the control and active reactors.

(2) They make "first principles" measurement less easy to self-check. Some non-conservative error (normally leading to > 100% efficiency and therefore detected) might not be detected due to a conservative error, that was not present under different conditions.

The importance of cross-checks in this type of data cannot be over-emphasised, given there are quite a number of possible error sources.

• Official Post

JedRothwell is checking out the heater position, and when he has confirmed it, will let us know I'm sure.

• JedRothwellis checking out the heater position, and when he has confirmed it, will let us know I'm sure.

It is smack in the middle, as shown in the photo in Fig. 16. Here's the question. When it is shipped it is wound in a circle. See:

https://www.monotaro.com/p/7075/6017/

That will not fit into the reactor. So it has to be squished down, I suppose, or twisted, as shown in Fig. a and b below. The wire is flexible but it holds the shape, according to the specifications. Mizuno said it should be right at the center. So I suppose he squishes down the circle, or twists it. I asked him how he does it. He has not had an opportunity to respond. He has been busy, and out of touch.

I sent him this illustration of how I suppose it is shaped:

• 3) Absent (2) or (1), we need to estimate the change in the thermal conductivity from the reactor our thru the supply line (either because the control reactor doesn't have a supply line, or because the control reactor doesn't have D2 in the supply line).

No we do not. That's ridiculous. That could not change by 250 W! The supply line is a little bitty pipe. And if that were the problem, it would show up in calibration.

You people are suffering from acute speculation-itis.

• Why is the finish not the same on the control vs the active reactor? (See figure 6, red box -- reactor is dull. Emmissivity of weathered stainless steel is 0.85 vs. polished stainless steel 0.075)

This throws off the calorimetry which is clearly a combination of conduction, radiation, and convection on different components of the rig.

No, it does not. The calorimetry is done by measuring the air temperature at the outlet, and subtracting the inlet (and ambient) temperature. The conductivity, radiation and convection of the reactor has no effect on this. The size and shape of the reactor has no effect. The whole point of an air flow, liquid flow or Seebeck calorimeter is to eliminate such effects. The heat is measured after it leaves the reactor, not while it is leaving.

If you use the reactor surface temperature to do calorimetry, then the issues you raise might have an effect. Not much of an effect. There is no way they could cause a 250 W error.

• I think someone asked about the placement of the thermocouple outside the cell. It should be in contact with stainless steel reactor wall which is in contact with the mesh. So the temperature is close to that of the mesh. Do not put it far away from the part of the reactor that touches the mesh.

I should correct Fig. 13 to show this more clearly.

The temperature of the reactor wall is far higher with excess heat than with 50 W of resistance heating. As I recall, it is around 40 deg C during the 50 W calibration, and 380 deg C with 250 W of excess heat. This high temperature next to the mesh indicates the anomalous heat is coming from the mesh.

The reactor temperature is important for the analysis, and it proves there is excess heat. There is no way 50 W could make it so hot. It is also much hotter than you would expect from the O/I ratio of 5:1 (250 W out, 50 W in). That ratio was measured with the air flow calorimetry. The air-flow is correct; the reactor temperature ratio of ~10:1 is not accurate. In other words, the reactor does not work well as an isoperibolic calorimeter. This is because the reactor surface temperature is not uniform. We don't know how non-uniform it is. Mizuno has not yet investigated this in detail.

• I think the way to go without air calorimetry is to switch off the power in a hot reactor with a dummy load and take cooling curves- many many cooling curves from all kinds of temperatures. They should be classic Newtonian curves of course. Then do it in a properly fuelled reactor and look for a difference.

And if the power is switched off, it eliminates many of the arguments about 'artifacts' .

This is data they probably already have. It would be good to run some checks on the existing data in that way.

• No, it does not. The calorimetry is done by measuring the air temperature at the outlet, and subtracting the inlet (and ambient) temperature. The conductivity, radiation and convection of the reactor has no effect on this. The size and shape of the reactor has no effect. The whole point of an air flow, liquid flow or Seebeck calorimeter is to eliminate such effects. The heat is measured after it leaves the reactor, not while it is leaving.

If you use the reactor surface temperature to do calorimetry, then the issues you raise might have an effect. Not much of an effect. There is no way they could cause a 250 W error.

Agree:

"If you use the reactor surface temperature to do calorimetry, then the issues you raise might have an effect. Not much of an effect. There is no way they could cause a 250 W error."

(By my calculation, it's at MOST a 60 watt error on the R20 300 watt out test.)

Disagree:

"The size and shape [and finish] of the reactor has no effect. The whole point of an air flow, liquid flow or Seebeck calorimeter is to eliminate such effects. The heat is measured after it leaves the reactor, not while it is leaving."

The reactor is an air cooled "radiator", in the sense of a car or airplane radiator. Radiators reject (transport out) heat by the 3 ways: conduction, convection, and radiation. If the radiator is polished silver, it rejects less heat by radiation than if it is painted black. Less heat by radiation means more heat by conduction and convection. Because most of the heat is being rejected by convection, but some of the heat is being rejected by other means (i.e. your 77% heat capture efficiency calibration at 360 degrees implies 23% is being rejected by conduction or radiation). By substituting a polished silver cylinder for a rough natural stainless finish, the balance of heat rejection by convection is changed. Exactly how is not a trivial task, so that the only way to eliminate this is by using a thermally identical polished silver cylinder for calibration. In my post I estimated the maximum error that this could have put into the COP (output/input) analyses.

• True and true. Right now we are going with what is published. Future experimenters replicating the experiment can fix this. That's scientific freedom. We can't hold Mizuno or Rothwell accountable as they are their own scientists. If we don't like their technique we can redo it ourselves. I only comment here to help other people figure this thing out. I think we will get proof positive shortly -- if we can replicate R20 225 watts out for 50 watts in (my correction of the 300 watts to a more conservative 225 watts).

• Alan Smith wrote:

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Or do what I suggested, take multiple 'no power' cooling curves for the same reactor with prepared fuel in and with plain mesh. Any deviations from the Newtonian can be significant assuming the external conditions remain closely aligned. A lot simpler and less prone to criticism than most other methods.

JedRothwell : This seems like something to request from Mizuno. I understand he's busy but if he has assembled reactors on hand, it would be relatively little work to perform and it should yield very valuable data.

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The reactor temperature is important for the analysis, and it proves there is excess heat. There is no way 50 W could make it so hot. It is also much hotter than you would expect from the O/I ratio of 5:1 (250 W out, 50 W in). That ratio was measured with the air flow calorimetry. The air-flow is correct; the reactor temperature ratio of ~10:1 is not accurate. In other words, the reactor does not work well as an isoperibolic calorimeter. This is because the reactor surface temperature is not uniform. We don't know how non-uniform it is. Mizuno has not yet investigated this in detail

So why not measure, or estimate from a cheap IR thermometer, the wall temperature of the reactor running in the fireplace at 3kW output? That would allow drawing a crude curve or comparison for zero out, 250W out and 3000W out vs temperature. Then, do the same for a control reactor with zero, 50W and 300W electrical power in. Just a trivial exercise but a valuable back of the envelope type of check. I suspect JedRothwell will offer some non-credible reason not to do either suggested tests which is typical of why I usually find looking at LENR frustrating and "slippery." Mizuno gave very interesting data so far but neglecting the more powerful reactor setup after showing the tease photo is, IMO, unconscionable if you want acceptance and research money.