 # Mizuno : Publication of kW/COP2 excess heat results

• THHuxley

"The real world is a bitch, and you learn not to make assumptions."

The assumption is yours. Jed was not talking about increased pressure due to the reaction but that due to the gram levels of reactants

Jed said : If enough O2 got in to cause significant heat, the pressure would rise"

He did not say "If the O2 reacted the pressure would rise"

The reactor pressure is monitored to be about 760pa, with only 3 milligrams of deuterium gas

If there were 3 grams of deuterium in the chamber the pressure would be 760,000 pascals or over 7 atmosphere's worth

To get significant heat such as 1- 5 MJ by combustion with O2 you will require to have a much higher mass of D2 /O2 the chamber.

.. gram levels.

As you have stated "Enthalpy of combustion of H2 is 280kJ/mol or 0.14MJ/g" and therefore that of D2 is 0.07MJ/g.

I'll leave it to you to work out the exact numbers ... but the pressure won't be 760 Pa.. or 1000 Pa due to the required gram levels of D2/O2 in the chamber

It will be much much greater than 7600 Pa.

However the pressure is around 760 Pa - 1000 Pa in the chamber throughout the 80 kilosecond test period.

And the minute quantities (3 milligrams of Deuterium) appear to produce megajoules of heat.

Mizuno designed the air calorimetry system to be able to detect down to about 0.2 MJ of heat

but even this smaller amount of heat would require 2.85 grams of D2 to combust with several times that mass of O2

• OK, I may have found the problem...please check me on this. I was able to download the spreadsheet "Mizuno 2017 120 W input excess heat" and look at it this morning.

I come up with several observations, but I will focus here on basically one. Namely, I used Mizuno's data to calculate an output power, and got massively different numbers from his. He computes Wout in col AR with the following equation (for the first row of data, row 12):

Col AQ is the air weight in kg

Col AD is the averaged calorimeter Tout-Tin (3 inlet T's and 2 outlet T's)

Air weight is calculated via this eqn:

=(AP12)*OutletArea*(3.391*EXP(-((E12)+273.2)/201.26)+0.41529)

OutletArea is 4.4x10e-3 (not 8.2x10e-3 as in paper)

Col AP is the "Blower m/s", i.e. the air velocity

Col E is the second Tout (note Mizuno uses the average in the prior equation, why not here?)

the expression following the OutletArea variable is the eqn for computing air density given in the paper to a few more sig. figs.

So we have air weight = velocity * cross-sectional area * density which is mass per unit time (kg/s)

The air velocity (col AP) is calculated via:

=-3.70072*EXP((AA12)/-1.37463)+3.98859

Col AA is the blower power in W. The above equation is the one given in the paper to compute velocity from power to a few more sig. figs.

Blower power is just blower current * blower voltage, the standard I*V expression.

But the mass calorimeter basic eqn. is Pout = mass flow rate * heat capacity * inlet vs outlet temperature difference

So Mizuno's eqn for Wout (= Pout) uses temp difference (AD) times mass of air per unit time (AQ) times some constants.

WHERE IS THE HEAT CAPACITY TERM???

My calcs (using degK) from Mizuno's data give Cp's of approximately 1.007, and I note you need to convert from kg to g, which adds a multiplier of 1000 in there, so the 1006 in Mizuno's equation *could be* that factor times a constant Cp of 1.006. But what are the 2.119 and 1.1429?

If I am doing this right, and please check me because I do make mistakes, Mizuno takes what is approximately the output power and multiplies it by 2.42! No wonder he sees excess heat!

Have I done this right?

What I did:

Computed average Tout

Computed density and Cp from the paper's equations and averaged Tout

Computed air velocity from the paper's equation using blower power (same as Mizuno)

Multiplied air velocity * OutletArea * density * Cp * dT to get my Pout values

Subtracted my Pout from Mizuno's Wout (col AR) - observed massive difference, follows shape of excess curve Mizuno presents, with a maximum difference of 238.5 W.

There are some other wrinkles and oddities, but I will hold off on them for now.

• Some notes on the ‘other wrinkles’ I mentioned. Again this drives off of the Mizuno spreadsheet uploaded by JedR.

First thing to do always is look at the data. So…

Looking at the temperatures, I see a lot of noise localized to the time frame the input power is non-zero. This is suspicious. Why the noise level changes is the question. This kind of behavior makes me suspect feedback noise, i.e. untrustworthy measurements.

Looking at the blower current, I note a dip in it. Was that supposed to be there? This tracks through to the blower power of course. Maybe into the blower voltage as well, but less noticeable. Maybe not as important as the temperature noise levels.

Next I computed the timestep per data row (i.e. time at row x minus time at row x-1). I found noise in this measure, which really surprised me. Again, it went away when the input power got shut off. Perfectly constant value after that, which is what I think it should have been throughout the whole run. This really bothers me, as it says that the data acquisition rate of the data logger is being impacted by something. I don’t think that should occur.

There is a column of data called “Wout/Avg” that has an associated note saying it is “500 second rolling average”. This is incorrect. If you look at the calcs, Mizuno is using a 100 row average, with an average of 24.47 sec per row, meaning his average covers 2447 seconds (40.78 minutes). Personally I think this is really weird. I can see doing a 5 to 9 point moving average to smooth noisy signals, but 100 points, covering 2/3rds of a hour???

Then as I noted in my prior post, I computed out and excess power. Looking at the excess power plot is informative. It starts at zero, since the first few rows are prior to turning the heater on. When the heater is turned on at ~120W input, the excess immediately drops to -120W. It then follows the curve up to about +25-30W average, but is very noisy. Then when Mizuno turns the input power off, the excess immediately takes a big jump up to ~125W and then falls slowly down to zero. This is about what one would expect if you had no excess at all, except that final 25-30W excess. During the 120W input period, the calorimeter never reaches steady-state, meaning it never balances out at a constant average power level. That means all we are dealing with are transients, which makes it harder to give any value to the apparent excess heat of ~25W. Once the input power was turned off, the fall in the excess is expected and must be compared to the normal cooling curve of a ‘no excess’ case to see if there is actually any excess keeping it hotter longer than it should be. We can’t do that unless we get cal runs data. Mizuno’s use of Wout,avg in his Figures obscures this behavior, as does his massively long averaging period.

So aside from the apparent computational error I think he made as noted in my prior post, the data seems to strongly suggest a lot of noise problems. This makes me wonder about how much we can trust the data. Those cal runs where no excess was supposedly observed are probably really important to study.

• What's the highest temperature for their calibration run vs the highest temperature for their claim of excess heat? If the highest temperature for their calibration run does not match the highest temperature for their claim of excess heat, I consider the result to be inconclusive at best.

• The real world is a bitch, and you learn not to make assumptions.

In this case we have:

2D2 [+ O2] -> 2D2O

There was no oxygen in the cell. The cell is connected to a mass spectrometer. You would see the oxygen. It was practically a vacuum in the cell. Even if there had been oxygen. there was not enough gas to produce a measurable level of heat.

This is described in the paper, so let me suggest you RTFM.

• I received some spreadsheets from Mizuno for the data in this paper:

Mizuno, T., Preprint, Observation of excess heat by activated metal and deuterium gas. J. Condensed Matter Nucl. Sci., 2017.

http://lenr-canr.org/acrobat/MizunoTpreprintob.pdf

The sheets included:

BCD60520.xls, 120 W calibration, date 2016/5/20

BND60519.xls, 120 W excess heat, date 2016/5/19

I uploaded a copy of the latter here:

I have not yet uploaded the first one.

The method of computing excess heat is still a little unclear to me. There was some confusion about the area of the outlet, which we now believe should be 0.0044 m^2. This affects the formula for the total weight of air that moves through the calorimeter per second. It will increase the estimate of excess heat in the paper. I will soon upload a revised pre-print reflecting this change. The figures and graphs in this report are based on the outlet area of 0.0044 m^2.

I thought I would make a simplified, preliminary analysis putting aside these issues. I do not need to know that absolute level of heat to compare the calibration to the (apparent) excess heat test. I need only compare temperatures. I made one assumption: Because the blower power is 4.7 W for both runs, I assume the air flow rate and calorimetry are the same in both runs.

COMPARISON OF CALIBRATION AND EXCESS HEAT RUNS

During the calibration run, the Delta T temperature (outlet minus inlet) rose to 4.7°C. That is the average at 10,000 s for about 30 minutes (Fig. 1).

During the excess heat run, the Delta T temperature rose to 7.3°C at 20,000 s (Fig. 2).

Assuming the conversion to power is correct, during calibration, it took ~4,500 s for output to reach the level of input power. Output lags because the thermal mass of the two cells in the insulated box is large, with 100 kg of stainless steel, which is approximately equal to the thermal mass of 10 kg of water. In the excess heat run, output reached input after ~3,550 s elapsed time, about 20% faster. This indicates there is more heat going into the system during the excess heat run.

In the excess heat run, power was turned off at 20,000 s. At ~40,000 s the cell stopped cooling for a while (Figs. 2 and 5). The temperature leveled off. This indicates a source of energy was present. The cell started cooling again at ~42,000 s.

Unfortunately, I cannot compute total energy output from the calibration run because the spreadsheet data stops at 19,000 s, before power is turned off and the cell cools, so I cannot estimate how much heat is left in the system. I cannot compare total output energy from the calibration to the excess heat run. However, the ratio of output to input power during the calibration indicates that the calorimetry is working correctly. Output is close to input. From the time the cell reaches a stable temperature at 5,000 s to the end of the data at 19,000 s, average output was 117 W and average input was 120 W (Fig. 3).

I can compare the energy from the calibration run to the excess heat run, both up to 19,000 s:

Calibration output 2.2 MJ, input 2.3 MJ, ratio 0.96

Excess heat output 4.5 MJ, input 2.2 MJ, ratio 2.03. A suspiciously round number.

Again, assuming the conversion to power is correct, during the entire excess heat run (Fig. 5), output was 7.4 MJ and input was 2.5 MJ. The ratio is 3.0, another suspiciously round number!

GRAPHS

Here are some graphs illustrating these comparisons.

Figure 1. 120 W calibration Delta T temperature (outlet minus inlet temperature). The average is ~4.7°C at 10,000 s, measured for 30 minutes.

Figure 2. 120 W excess heat run. The Delta T temperature average is 7.3°C at peak (20,000 s). Temperature stops falling at ~40,000 s, resumes falling at ~42,000 s.

Figure 3. Temperature measurements converted to power (electric input power and output heat). During the calibration, power is turned on at 80 s. Output power reaches the same level as input power after ~4,500 s elapsed. Data collection stopped at 19,000 s. From 5,000 s to 19,000 s, average input power was 120 W and average output was 117 W (98% recovery).

Figure 4. Excess heat run up to 20,000 s. Power is turned on at 835 s. Output power reaches the same level as input after 3,550 s elapsed, sooner than with the calibration. This indicates there is some excess heat.

Figure 5. The entire excess heat run of 25 hours.

• It is academic, because the IH replication failed.

"It is academic" seems to imply you know for a fact this experiment in Sapporo failed, you just don't know why. The IH replication did fail, but that does not mean this experiment also failed. Sometimes, an experiment will work in one place with one reactor, but not in another place with another reactor, for various reasons.

• WHERE IS THE HEAT CAPACITY TERM???

My calcs (using degK) from Mizuno's data give Cp's of approximately 1.007, and I note you need to convert from kg to g, which adds a multiplier of 1000 in there, so the 1006 in Mizuno's equation *could be* that factor times a constant Cp of 1.006. But what are the 2.119 and 1.1429?

1006 is the heat capacity, as you say kg to g.

2.119 and 1.1429 don't make any sense to me, but I haven't looked at the spreadsheet. The only reason I can think of for those to be there is as some kind of conversion factors, maybe to include some non-metric data inputs?

Have I done this right?

Yep, that's the right way to work out POut.

Is this subtracted graph you've made the exact negative of Mizuno's excess power curves, Kirk?

Seems a little odd, to say the least.

• JedRothwell said "Excess heat output 4.5 MJ, input 2.2 MJ, ratio 2.03. A suspiciously round number."

By spreadsheet analysis. I get 3.06 MJ thermal out, 2.48 MJ electrical in which gives excess of 0.58 MJ, ratio 0.19.

By chromatic analysis of your Figure 2 graph I get a time-temp area of 150,500 second.deg

http://www.geotests.net/couleurs/frequences_en.html (Merci, L.Jegou)

By calculation using average values this area leads to 3.04 MJ out. (see graph)

This contrasts with your thermal out of 7.7 MJ.

It is not about who is right but who is less wrong?

Sannin yoreba Monju no Chie

In the end we will be wise. • THHuxleynew wrote: "It is academic, because the IH replication failed. Still, I'd like to debug this"

It's Sunday morning here.. the kookaburras are noisy at 4.00 am in the morning

Noisy buggers.

For some reason this Bible verse came to mind

"but those who hope in the Lord will renew their strength. They will soar on wings like eagles; they will run and not grow weary, they will walk and not be faint"

LENR research is not for the fainthearted.

I respect the efforts of many.. including those sponsored by IH.

In Japan, university retirement at around 65 is compulsory.

Mizuno started a small research company in Sapporo .

After seven years with limited financial resources (

(but not intellectual resources) he has produced good work.

Maybe this "debug"was an off the cuff remark by THHuxley but it really irritated me.

Miizuno has 'lived' this project for the last few years.. like so many LENR researchers

and the notion that it is a simple software program..to be debugged..is arrogant.

IH has sponsored similar work with plasmas, transition metals deuterium and reaction chambers

They have achieved much more modest excess heat...6%.. so far , than the up to 100% achieved by Mizuno

Perhaps if THHuxley is getting tired he could look to "debugging" this IH work first.. it is a smaller bug.

But perhaps he could do some ethical reading, read the researchers' papers first in detail

and show some respect for effort, rather than making uninformed effortless remarks.

https://www.lenr-forum.com/att…claytortsummaryoftri-pdf/ • By spreadsheet analysis. I get 3.06 MJ thermal out, 2.48 MJ electrical in which gives excess of 0.58 MJ, ratio 0.19.

I do not understand how you arrive at these numbers. Mizuno and I use the method shown cell AU2. The formula is:

=SUM(AR12:AR3484)*24.47

That is the sum of all instantaneous power readings (in Watts) in all rows, multiplied by the duration of each row, which is 24.47 s. For example, if there were 2 rows:

5 W

10 W

That's 15 watts * 24.47 s = 367 watt-seconds (Joules).

Is there a problem with this method?

The totals from the spreadsheet are:

 Input/J Output/J Out/In ratio 2,484,372 7,422,987 2.99
• If I am doing this right, and please check me because I do make mistakes, Mizuno takes what is approximately the output power and multiplies it by 2.42! No wonder he sees excess heat!

No, you are doing it wrong. If you were correct about this, Mizuno's calibration output power would not equal input power. The only differences between the calibration and the excess heat runs were the Delta T temperatures (outlet minus inlet), and the time it took to cool down (longer in the excess heat runs, including a segment with no cooling). The same equations were applied to both sheets, so obviously, with a larger temperature difference, the excess heat sheet had to show more heat. You have to explain that.

If, as you claim, there should have been a heat balance in the excess heat run, that would have to mean the calibration is 2.42 times too large. One or the other has to be wrong. It seems unlikely the caliorimeter is only recovering 50 W with 120 W input, so I expect the calibration is correct.

• Kia Ora Jed

I am not Japanese.

If a reviewer asked to see my raw data, I would find that most unusual.

I am of Maori Irish and Viking(way back) heritage.

I would probably physically take them out.. I am not Japanese.

I use the raw data and its processing up to column AX and no further.

I multiply by 24.47 to get mass per interval rather than mass per second.

This makes the AX column become Joules per interval

I sum the AX column to get total Joules =3.09 MJ

I think the 'rolling 500 is a nicety

I think the 1.5 hr system lag .. which is a consequence of its accuracy

... its a trade off ) is only significant after the reactor is hot

(and not before) because basically the reactor and the air are same temp then.

It is necessary to continue the calorimetry for this time lag to

pick up all the reactor heat.

I do not know what 2.119 1.1429 are..I don't use em. they may be relics of a modifed setup.

They are the reason why you get 7.4 MJ... I don't see evidence that Mizuno used them in his final

calculations .

If I look back through my own research spreadsheet data I have stuff that takes me

a long time to remember why I put it there. Luckily (for them)

the reviewers never asked me for it. (the review world is so small)

It is important to prepare a report.. but I am happy from my calculations

what is stated in Mizuno's work is strong evidence for 18% Xs heat at 120W

shorttime and ~43% xs heat 100W longtime.

Now I have a headache.. spreadsheets do that to me. esp ones with 3485 rows and up to AZ...

Otherwise some warlike person would use it as a weapon. • I do not know what 2.119 1.1429 are..I don't use em. they may be relics of a modifed setup.

They are the reason why you get 7.4 MJ... I don't see evidence that Mizuno used them in his final

calculations .

You have the calibration data, which I have not yet been able to upload yet. When you remove these constants from that spreadsheet, do you still see a balance of input and output?

• I don't have the calibration, inactive, data spreadsheet for 120W. I have not checked it.

I checked the thermal output from Fig 25, 100W calibration by 'chromatic' analysis.

The agreement between thermal/ electrical was good enough for me.

If someone makes a nice figure 2 version of the 120W calibration I will

check that by 'chromatic'. That takes 2 minutes.

My chromatics agree with my spreadsheet for the 120W active

I am busy with my report for my reviewers

that's why I'm cranky.

• I am inclined to agree with you - that v low pressure in active tests mitigate against such a mechanism. But - how is the pressure maintained? If not a sealed envt that would not apply. I'm thinking that H outgassing from Ni and Fe is possible (Ni it would be small total amt, Fe could be large amt). If we have a sealed envt throughout the active phase then I agree.

I'm not pushing this as likely problem - I'd expect something silly to do with blatant uncaught errors. This is only a preprint and has various known errors.

It is of some interest how these errors can creep in. The reason for being pretty sure in this case that there is some error is that IH replicated twice (with M's help) and found nothing. Had there been any ambiguity in that negative result I'm sure they would have continued. For example, if M had sai, well, only 50% of these electrodes actually work. IH will chase down 1% chances so they would not let this go unless convinced hand-on that the M results were erroneous.

Which, however, does not help us to understand how that comes about. Such understanding would be good to have, and help others.

• Hi Jed

I was more wrong..and I am happy about it

I learned some.

the calibration factors are necessary otherwise the

excess heat is underestimated. I was measuring only the capacitive heat, mc delta T

And not including the environmental heat, E. see explanation below.

For accuracy both E and C are needed.

In reality mcdelta T only could pragmatically be used as a parameter for improving the reactor conditions.

The goal now is to increase the Xs heat output from 100% to 200% of electrical input to make it economic.

Explanation:

Heat loss from the reactor is via

1.the flowing air heat capacity increase- C=m x cp x delta T

2. radiation loss directly from the reactor ,R

3. radiation/convection/conduction loss from the box, B

4. conductive losses from the floor. F

The flow air capacitive heat ,C, is measured via delta T  ,inlet /outlet ,and the flowrate.

The other three terms , grouped as environmental heat, E ,and are difficult to measure.

Total heat loss, T=E+C

The environmental heat ,E, increases with temperature and hence with power input into the reactor

For example it may increase from 5% of the total heat, T, at 100W to 24 % at 120W

Measuring the capacitive heat, C= m cp delta T only will underestimate the reactor heat loss , T..

To estimate E a calibration reactor is used with inactive fuel.

For a constant power input e.g 100W, P: for a time period,t

A thermal/electrical energy balance allows the

environmental heat to be calculated by.

E =Pt-C… because T=Pt

A calibration factor can then be calculated which relates the total thermal energy to the measured capacitive heat

This is f= T/C = Pt/C.

Runs with the active reactor have electrical energy ,Pt ,plus excess energy X.

Energy balance: Pt+X = E+C

C is measured and E is estimated so that X=Pt- C-E= Pt-fC

The calibration factor is larger at the higher energy inputs such as 250W,

because the higher resultant temperatures, and temperature differences

increase the proportion of environmental energy.

Radiation loss is proportional to the difference of the fourth power of the temperatures(K)

Stefan Boltzmann LAW

and convective/conductive loss is proportional to the first power of the temperature difference.

NewtonsLaw of Cooling.

For each active reactor run a calibration factor run was performed with an inactive reactor

at the same power.

• THHuxley wrote "how is the pressure maintained"

The pressure changes because of two things

1. The temperature change e.g 20 to 100C

2. Possible fusion to reduce two atoms to one… perhaps formation of deuterinos

The start pressure is 760 Pa

The text says " keeping the pressure intact" .

The meaning needs to be clarified. I think that it means the voume was constant and there were no leaks.

This would be the easiest thing to do. I do not think the pressure is maintained during the reaction

As the reactor heats the pressure increases according to PV =nRT. 760 Pa would go up to 968 kPA

When the reactor cools the pressure would return to 760 pa ( maybe 10 pascals less.)

Minus some deuterium because of"fusion" but not much in one day

Howeever after 30 days the pressure would down significantly in the ratio8/20

The final cool volume would be 304Pa.

However this is in reverse. They observed 304 Pa and inferred usage of 12 mls which was attributed to

fusion loss.. perhaps Mills dark matter??

I am sure Mizuno is intensely interested in whats happening.. the Tritium counts may be revealing in the next report