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.
mizunotadahiko
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 Member since Mar 5th 2020
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Posts by mizunotadahiko


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."



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.

I don't think so. Look at Figure 10 of Mizuno and Rothwell J Cond Matt Nucl Sci 29:112 (2019), then equation 2 of the same paper. The equation they fit to their calibration data is fractional power capture = O/I = 0.98  [5.0811E4 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.

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?
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.

Desirelessa day ago

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 ( Temptime) 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 ( TEMPtime)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.
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

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 ( Temptime) 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 ( TEMPtime)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.
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..
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.

Robert
この計算は不正確だ。熱の箱からの逃げる量の補正をしていない。
This calculation is incorrect. The amount of escape from the heat box is not corrected.

古いデーターは正確では無い。測定系も日々改良している。その度に校正試験を行う。補正係数も日々正確になる。新しいデーターを見てほしい。総合的な報告書は特許申請後行う。
Old data is not accurate. The measurement system is improving daily. A calibration test is performed each time. The correction coefficient is also update every day. Look at the new data. Comprehensive report will be made after patent application.

DR Mizuno
Questions from replicator
Desireless posted on ECW..
Can you ask Prof. Mizuno if he developed a way how to run activated reactant with no excess heat?
Will it run with Protium only? Is there a comparison when running with Deuterium and Protium with the same reactant?活性化した炉を不活性化するのは簡単です。Heを入れたり、空気を入れると直ちに過剰熱は出なくなります。再活性の程度は入れたガスや温度、さらした時間によって変わります。軽水素でも熱は出ましたが、100％H2のガスは無いので、重水素の影響を無視できません。
Deactivating an activated furnace is easy. Excess heat does not come out immediately when He or air is added. The degree of reactivation depends on the gas and temperature used and the exposure time. Even with light hydrogen, heat was generated, but since there is no 100% H2 gas, the effect of deuterium cannot be ignored.

The graph showing Wex/g against the temperature (1/T) Is that power produced per gram of reactant (Ni mesh+Pa or D2)?
The graph showing Wex/g against the temperature (1/T) is that power produced only per gram of reactant Ni mesh.


I'm new to posting. I have posted the same one.
Jed is having a hard time answering, so let me talk. This is the only data I can publish now, but let's make a detailed announcement after applying for a patent. The CF phenomenon is absolutely true. It can be controlled very easy. However, the mechanism is still unknown. When the data comes out, an excellent researcher will give an answer. There is still much to study. Let's do our best. Don't lose to Corona. Believe in the future and advance.
