Posts by mizunotadahiko

Daniel_Gはもはやミズノテクノロジーズ株式会社で働いていません。

Please tell me the schedule. I want to know the abstract submission deadline, hotel reservation deadline, payment deadline, etc.

After one test is completed, the calibration test is performed again with the heater. That is the 0 point of the overheat test. Therefore, there is no excess heat generated from the calibration test.

。

校正用ヒーターについても同じことが言えます。

では、なぜキャリブレーションヒーターで過剰な熱を確認できないのでしょうか。

これを説明してください。

The same discussion can be made about heaters for calibration.

So why can't we see excess heat with the calibration heater?

Please explain this.

The three years from 2019 were the worst of my life. In March 2021, my wife died, my cat died, the company was unjustly hijacked (in accussing), and in 2018, the company was hit by an earthquake and suffered great damage. I also thought about stopping the CF research as it is. But in the midst of misery, I have to say a lot about it. It seemed to be mentally useless just to deal with it. During that time, it was CF research that saved me. Fortunately, at the end of 2021, I got unexpected test results. Also this year, I was able to overcome four more walls. As a result of more than 30 years of trial and error, I was finally able to control the excess heat, so I think we were able to determine the conditions. I hope to be able to present at the conference.

I will attend the meeting.

There is no point in discussing the theory and mechanism of the CF phenomenon, which has not been completed yet. First of all, it is important to do many tests by yourself and accumulate data. In particular, it is meaningless to discuss the upper and lower limits of this phenomenon, such as the amount of excess heat generated, the duration, materials, and temperature.

I'm too busy right now to respond. There is a lot of new data, and we will announce it soon. I think it will please everyone.

Dear All

Mizuno Tec. Inc is the official page. I didn't even know MTI page before. Mr. Daniel made it ahead of time, causing confusion. I also contacted Mr. Daniel and gave him a lot of attention and suggestions. Mr. D is also a page I made for me. The pager was not well managed and caused distrust and confusion to many people. I will also be careful in the future.

Tadahiko Mizuno

私はこれ以上の入力データを持っていません。

[Google translate 'I don't have multiple input data' Alan.]

校正データーです。Control data.

図面は同じ試験炉で、熱損失補正前と補正後です。

Google translate

"The drawing is the same test reactor, with heat loss correction and after correction."

Quote from Ascoli65

A few more words from the cook would be very helpful to understand why.

水野：良いところに気付きましたね。出力計算は平均値を使っていたので、元ファイルを添付します。

いつでも校正後の値が大きいです。

Mizuno: You've noticed some good points. Since the output calculation used the average value, attach the original file.

The value after calibration is always large.

Quote from Bruce__H

I don't think so. Look at Figure 10 of Mizuno and Rothwell J Cond Matt Nucl Sci 29:1-12 (2019), then equation 2 of the same paper. The equation they fit to their calibration data is fractional power capture = O/I = 0.98 - [5.0811E-4 x T] where T is "the reactor temperature"

水野：

このときは炉温度で補正式を作りました。同じく校正データーも同じ形状、重さの炉で補正式を作りましたが、精度が悪く、正確ではなかった。そのために箱からの熱逃散を補正に使った。その結果、極めて精度が良くなった。

私が出すデーターは新しいものほど精度、正確性が良い。さらにどのような形の、重さの炉であっても補正を入れた一般式で表すことが出来るようになった。

Mizuno:

Before 2017, the correction formula was made with the reactor temperature. Similarly, for the calibration data, a correction formula was made with a reactor of the same shape and weight, but the accuracy was poor and it was not accurate. Therefore, the heat dissipation from the box was used for the correction. As a result, the accuracy became extremely good.

The newer the data I give, the better the accuracy and precision. Furthermore, it became possible to express any type and weight of furnace by a general formula with correction.

Quote from sam12

An update from replicator Desireless

and a question for DR Mizuno.

Desireless8 hours ago edited

New reactor module is giving COP 1.36 at 50W. When deuterium is introduced excess heat is coming from a different reactor part, not originating from the heater. If there is too much deuterium introduced then excess heat lasts only for hour or so and is decreasing slowly. It is being loaded by the mesh. It seems good pressure is needed in order to not load deuterium. Instead balance between absorption and degassing is the key. 2 
−

• Desirelessa day ago

  Can you ask Prof. Mizuno if he is using DC or AC power supply?

  What will happen if he will connect positive electrode from power supply to the reactor shell? Has it impact at excess heat?

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水野：　以前の試験では90％が直流電源です。交流電源も使用しましたが、加熱では試験結果に違いは有りません。直流の方が測定は楽です。

Mizuno: 90% of the previous tests were DC power supplies. An AC power supply was also used, but there is no difference in the test results with heating. DC is easier to measure.

Quote from RobertBryant

The amount of escape from radiation and convection

is not corrected in Figure 20 or Figure 26..

the corrected power will depend on the calibration factor..

which depends on assumptions

The corrected power for the active reactor

depends on the actual calibration factor

which is not stated in the 2017 paper

... this is a rough way of calculating it.

by using ( Temp-time) areas in yellow

the spreadsheet method used by MizunoSan is more accurate

but it needs the numerical data

the uncorrected heat can be calculated roughly by using

delta H(J) = delta T (K) x flowrate ( m3/ s) x time(s) x density(kg/m3) x Cp(J/kg.K)

=deltaTemp x time (ks,deg)  x flowrate x density x Cp

The power P can be calculated by dividing by TIME 85 ks/82 ks .... from the graph..

there is an averaging error..(time/TIME ) which underestimates both uncorrected powers.

there is also the use of average air densities.

in addition there is an error due to the fact that the active reactor

is hotter than the inactive reactor , despite equal 100W inputs,

which increases the heat escape.,, so the 1.27 factor is too small..

However the uncorrected power for the active reactor will

be much more than the input electrical power of 100W.

In addition the uncorrected output of the active reactor

is much more than the inactive reactor.

Comparing the ( TEMP-time)areas makes this obvious..

590 versus 335. is equivalent to 73% more

The Fig 29 .. derived by spreadsheet anlysis

gives a corrected output for the 100W

case of approximately 74% more (74W) for the active reactor

corrected power.

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この校正データーは今では使用しない。前に書いたように校正試験はいつもアップデートしてるので、注意して欲しい。水野忠彦

This calibration data is no longer used. As I said before, the calibration test is always updated, so please be careful. Mizuno Tadahiko

Quote from RobertBryant

The amount of escape from radiation and convection

is not corrected in Figure 20 or Figure 26..

the corrected power will depend on the calibration factor..

which depends on assumptions

The corrected power for the active reactor

depends on the actual calibration factor

which is not stated in the 2017 paper

... this is a rough way of calculating it.

by using ( Temp-time) areas in yellow

the spreadsheet method used by MizunoSan is more accurate

but it needs the numerical data

the uncorrected heat can be calculated roughly by using

delta H(J) = delta T (K) x flowrate ( m3/ s) x time(s) x density(kg/m3) x Cp(J/kg.K)

=deltaTemp x time (ks,deg)  x flowrate x density x Cp

The power P can be calculated by dividing by TIME 85 ks/82 ks .... from the graph..

there is an averaging error..(time/TIME ) which underestimates both uncorrected powers.

there is also the use of average air densities.

in addition there is an error due to the fact that the active reactor

is hotter than the inactive reactor , despite equal 100W inputs,

which increases the heat escape.,, so the 1.27 factor is too small..

However the uncorrected power for the active reactor will

be much more than the input electrical power of 100W.

In addition the uncorrected output of the active reactor

is much more than the inactive reactor.

Comparing the ( TEMP-time)areas makes this obvious..

590 versus 335. is equivalent to 73% more

The Fig 29 .. derived by spreadsheet anlysis

gives a corrected output for the 100W

case of approximately 74% more (74W) for the active reactor

corrected power.

Display More

Quote from Paradigmnoia

Here is the loss-corrected 120 W plus excess plot.

.

Quote from RobertBryant

警告

It is not advisable to feed fingers to Paradigmnoia.

Or natto

Btw where in P's world did this graph come from

It's not in my world?? did P manufacture it..

MIZUNO REPLICATION AND MATERIALS ONLY

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この120W入力で350W近くに達する過剰熱データーは私は知りません。風量測定を始めた2017年からのデーターを全て再確認しましたが、有りません。私のデーターでは無いと思う。

I don't know the excess heat data reaching 350W with this 120W input. I reconfirmed all the data from 2017, when I started measuring the air volume, but there are none. I don't think it's my data.