MIZUNO REPLICATION AND MATERIALS ONLY

Quote from benbarrowes

Magicsound. I'm curious as to your next steps and tests with your setup.

I've finished the fabrication and assembly of the MR3 cell, and successfully baked it out to 8E-6 Torr (very good!). I discovered a leaking valve in the gas manifold during the first D2 soak, and will be dealing with that in the next few days. So calibration should be possible this coming week, then disassembly to install two already prepared meshes, and bake out again. If all goes well, MR3.1 with Deuterium should begin in a week.

Desireless is claiming he has made

a successful Mizuno replication

on ECW.

https://e-catworld.com/2020/06…nr-experiment-desireless/

Quote from Shane D.

We do have a new report of 2 more successes in addition to Desireless.

Can you share those, please?

Quote from Shane D.

That's all we have. Just 2 Japanese companies. No data, although i believe Jed may have that info?

No, I have never heard of any of them.

I wonder how Saito et al obtained such clean outlet temperature traces.

Longer averages?

MR3 Cal1 is now streaming if anyone is bored enough to watch. A longer heater cartridge is installed so the temperature profile is a bit different from MR2

https://www.youtube.com/watch?v=iUD9YVerJ1M

After fully cooling, the cell pressure had increased from 305 Pa to 343 Pa. So there is still some leakage after cleaning the sampling valve. I've ordered a diaphragm-sealed valve that is better suited to retaining Hydrogen, and that will be installed before the next calibration. Please stand by....

Unfortunately it appears Desireless is another replicator that has discovered the high thermal conductivity of Hydrogen/Deuterium and confused it as excess heat. Desireless is measuring external reactor temperature with and without Deuterium for a constant input power and is interpreting increased external reactor temperature when Deuterium is introduced as excess heat. What it appears they are observing is just the high thermal conductivity of the Deuterium increasing the heat conduction to their measurement location. You need a calorimeter for these experiments or at the very least you have to be as meticulous as magicsound.

Quote from Brian Albiston

Desireless is measuring external reactor temperature with and without Deuterium for a constant input power and is interpreting increased external reactor temperature when Deuterium is introduced as excess heat. What it appears they are observing is just the high thermal conductivity of the Deuterium increasing the heat conduction to their measurement location.

If the temperature is measured on the outside wall, and if you wait for it to stabilize, then the internal conductivity make no difference. The outer wall will come to the same temperature no matter what. With highly conductive gas, it will reach this temperature quickly. With less conductive gas, it will take longer. The final temperature of the outside wall is a function of the power level and the total area of the outside surface only. No other parameters affect it.

The temperature inside the reactor, at the center, will vary depending on gas conductivity. Less conductive gas will make the temperature higher, after the terminal temperature is reached.

Quote from JedRothwell

If the temperature is measured on the outside wall, and if you wait for it to stabilize, then the internal conductivity make no difference. The outer wall will come to the same temperature no matter what. With highly conductive gas, it will reach this temperature quickly. With less conductive gas, it will take longer.

Let me add that if that were not the case, you could make a perpetual motion machine. Suppose you can bring the surface temperature of two identical cylinders up to two different temperatures with same input power to both. You can then have heat flow from the hotter vessel to cooler one, to run a heat engine, that powers both of them.

In other words, if the temperatures are different, with the same surface area, more heat has to be radiating from the hotter vessel.

A hotter surface with the same surface area MUST mean there is more energy. A hotter inside temperature may well be because the insulation is better, where the insulation in this case is the conductivity of gas. Less conductive gas is like putting a heavier blanket over yourself to stay warm in winter, with the same body heat (~100 W while sleeping; basal metabolism). The outside of the blanket is the same temperature no matter how thick or thin the blanket is. YOU are warmer; the blanket surface is not.

Quote from Brian Albiston

Unfortunately it appears Desireless is another replicator that has discovered the high thermal conductivity of Hydrogen/Deuterium and confused it as excess heat. Desireless is measuring external reactor temperature with and without Deuterium for a constant input power and is interpreting increased external reactor temperature when Deuterium is introduced as excess heat. What it appears they are observing is just the high thermal conductivity of the Deuterium increasing the heat conduction to their measurement location. You need a calorimeter for these experiments or at the very least you have to be as meticulous as magicsound.

Here is what Desireless responded on ECW:

"Of course there was calibration performed. Actually there were multiple calibrations, including with very same deuterium pressure. Actually multiple pressures to make sure I can run with the mesh at a different pressures.

Later on after excess heat results were clear the mesh was removed to re-run calibration. And it is matching with the original calibration. And of course there were measurements from all sides of the reactor. This work was done in a half year or so.

Interestingly internal pressure is not playing that important role, since surface temperature is different only by not more than 5C in pressure range from 0.001 - 0.1 Bar. Moreover behavior you can see at the plot can't be reproduced without activated mesh. You will not see temperature bump followed by a slow decrease."

Quote from JedRothwell

If the temperature is measured on the outside wall, and if you wait for it to stabilize, then the internal conductivity make no difference. The outer wall will come to the same temperature no matter what. With highly conductive gas, it will reach this temperature quickly. With less conductive gas, it will take longer. The final temperature of the outside wall is a function of the power level and the total area of the outside surface only. No other parameters affect it.

The temperature inside the reactor, at the center, will vary depending on gas conductivity. Less conductive gas will make the temperature higher, after the terminal temperature is reached.

JedRothwell what you say is true, but Desireless didn't wait for it to stabilize. The plot he has shared covers a period of 20 minutes. If his reactor is anywhere near the mass of Mizuno's then it takes hours for it to stabilize. You can see it starting to head back toward equilibrium in the tail of his plot. Even in the steady state the temperature distribution will change when Deuterium is added. In vacuum much of the heat is conducted out through the heater support resulting in the heater supporting end of the reactor being relatively warm. Once Deuterium is admitted the heat distribution will change to be a more consistent temperature over the surface of the reactor. I think you agree that this makes it very difficult to use temperature only as a measurement of excess heat. You have to do calibrations with multiple temperature measurement points at each Deuterium pressure and wait hours at each power step. If Desireless would like to publish these calibrations and full pictures of his setup as others of us have then I might take his claims more seriously.

Quote from JedRothwell

The outer wall will come to the same temperature no matter what.

That would be true (in time) only if the outer wall was not losing heat to the outside.

But because it is not a closed system, and because it is losing heat to the outside, the same wall temperature can't be achieved. The result will be two different equilibrium temperatures for the outside wall, and two different internal temperatures for the core.

Quote from Brian Albiston

JedRothwellwhat you say is true, but Desireless didn't wait for it to stabilize. The plot he has shared covers a period of 20 minutes

In that case, you are right!

Quote from Shane D.

The storyline is the same with each: they claim success, and after stringing us along for a time, they are eventually challenged to show the raw data, and photos. Many promises and excuses later, they disappear.

I'd settle for cooked data.

Quote from Mark U

That would be true (in time) only if the outer wall was not losing heat to the outside.

But because it is not a closed system, and because it is losing heat to the outside, the same wall temperature can't be achieved.

Nope. If the two cylinders have the same surface area, the same internal power, and they are exposed to the same outside conditions (air or water), the surfaces have to come to the same temperature. Or you have a perpetual motion machine, as I said.

It may take a while for them to come to the same temperature.

In my Mizuno replication experiment, I calibrate the vessel using the average of eight thermocouples on the outside of the 4.5" OD vessel (secured with pipe clamps). I determine the heat transfer coefficient K for a given gas at different pressures and steady state wattage inputs. I'll then use this data to determine energy output when I use the nickel/Pd mesh.

So far I have only done the control experiment with a stainless steel mesh (large hole mesh) and argon. Next week I'll start the same control experiment using deuterium at different pressures.

So far I've noticed that increasing the argon pressure to 10 Torr increases convection inside the vessel so that the upper thermocouples indicate higher temperatures and the lower thermocouples indicate lower temperatures compared to a near vacuum calibration at .015 Torr. The 4.5 inch diameter, 20 inch tall vessel is oriented vertically so Argon gas convection inside the vessel brings the hot gases upwards. The end result is that at 500 W, the heat transfer coefficient K, which is based on the average of eight thermocouples, changes a maximum of 4% over the pressure range of .015 Torr to 10 Torr.

I'm not doing flow calorimetry. Mine is based on the average temperature of the outside surface for a given wattage input. The calculated heat output also uses the specific heat capacity of the vessel determined in a calibration with a square wave thermal input pulse (multiple input powers).

For those that are curious, the details of this method are shown in a different experiment I did 2 years ago:

http://zhydrogen.com/?page_id=2120