MIZUNO REPLICATION AND MATERIALS ONLY

Quote from magicsound

Why is this level of vacuum stability so important although not mentioned by Jed? The Mizuno protocol requires maintaining 300-1000 Pa of D2 in the reactor for days or weeks, without any contamination of oxygen (air) getting in.

I do not think it stability per se is important. I think Mizuno said this to demonstrate that the reactor is leak-tight. That's what matters. No air leaks in to contaminate the material. Holding it at one particular pressure for a long time is not the point. So, for example, when you hold it a low pressure for a week, and a little gas leaks out the reactant over time and the pressure gradually rises, that would be a good thing. If it is leaking in from outside, that's a bad thing. I guess you need a mass spec to tell these situations apart.

If a lot of contaminants leak from the cell walls or from welding, that would be a bad thing.

If a lot of hydrogen appears to leak out of the reactant over several days or a week, I suggest you wait for it to stop coming out, fill the cell with deuterium at 5000 Pa for a day or two, and then pump it down again. Maybe repeat that a few times. That's what Mizuno suggested the other day. Cycle the gas through the reactant until you get more deuterium than hydrogen.

Quote from magicsound

This result suggests that the temperatures quoted by Jed in the report are not possible with the 500 watt heater he specified.

What temperatures? 110 deg C? Surely that depends on how think the steel is?

Quote from JedRothwell

What temperatures? 110 deg C? Surely that depends on how think the steel is?

No, 200 °C as specified in step 13. The thickness of the steel will have a relatively small effect on the external thermcouple positioned directly over the core heater. This is because the thermal conductivity of 304 stainless steel is fairly low, about 1/4 that of mild steel and just 7% of Aluminum. For my attempted bake out of the cell, I covered it with 1 inch thick tubular fiberglass pipe insulation. The heat loss through the insulation is pretty well controlled, as shown by the thermal image below. Much of the loss is through the left end cap where the thermowell is mounted. Even with this insulation, my 300 watt heater is not capable of bringing the center of reactor shell to 100 °C.

Quote from magicsound

OK now for some results. My reactor is 1/2 scale of Mizuno's (1/4 volume). The heater is a 300 watt cartridge in an axial thermowell. At full power with vacuum it only heats the reactor outside wall to ~80C, even with 1 inch of insulation over the entire tube. The thermal rise time is about 30 minutes. This result suggests that the temperatures quoted by Jed in the report are not possible with the 500 watt heater he specified. The Chinese replication reported in August used a 1500 watt heater element, and the cell was in a well-insulated calorimeter box. So my next improvement will be a 600 watt heater element installed naked through the 3/8" swagelok fitting in place of the thermowell. Then more leak checking and hopefully a proper bake-out of the cell.

AlanG

Magicsound Lab

FWIW, the outside of my sheet metal cylinder, 120 mm diameter & 500 mm long, gets to around 80 C closest to (beside) the internal heater at 200 W input. (It is a bit tricky to get an overall idea of the temperature peak and distribution while inside the calorimeter box.)

Edit: That is with the fan on. It would get much hotter without forced air cooling.

Quote from magicsound

Even with this insulation, my 300 watt heater is not capable of bringing the center of reactor shell to 100 °C.

I suggest you discuss this with Mizuno. He may be hard to reach for some time.

Quote from JedRothwell

I do not think it stability per se is important. I think Mizuno said this to demonstrate that the reactor is leak-tight. That's what matters. No air leaks in to contaminate the material.

He made this claim about stability when discussing a leak test with no mesh in the cell. There was no mesh, and the pressure as about low as he could make it. The pressure hardly changed for a week, indicating there were no leaks and no gas coming out of the cell walls. It was a test for leaks. As I said, I do not think he meant that very stable gas pressure is needed for the experiment itself.

Quote from magicsound

Even with this insulation, my 300 watt heater is not capable of bringing the center of reactor shell to 100 °C.

Look at the last page here. You can see the temperature is uneven. With 300 W input, it varies from 70 to 170 deg C.

https://www.lenr-canr.org/acrobat/MizunoTsupplement.pdf

Quote from JedRothwell

I suggest you discuss this with Mizuno. He may be hard to reach for some time.

One thing I noticed is that H₂ (or D₂) gas has a high thermal conductivity, particularly at relatively high pressures. This phenomenon was observed in a Celani-type of experiment, where the temperature of the internally heated element was monitored with an IR thermometer. Interestingly, the H₂ thermal conductivity (as measured by the difference in heater element temperature for a constant ambient temperature) did not change appreciably until the H₂ pressure was reduced to a few tens of milli Torr. This effect may not impact the temperature of the external containment vessel, but it would have an effect on the temperature of the Ni mesh if the mesh were heated via an internal heating element.

yes jeff you are right !

We have seen that only 1 mbar changes things to lowering IRs influence VS direct conduction.

Quote from jeff

One thing I noticed is that H₂ (or D₂) gas has a high thermal conductivity, particularly at relatively high pressures. This phenomenon was observed in a Celani-type of experiment, where the temperature of the internally heated element was monitored with an IR thermometer. Interestingly, the H₂ thermal conductivity (as measured by the difference in heater element temperature for a constant ambient temperature) did not change appreciably until the H₂ pressure was reduced to a few tens of milli Torr. This effect may not impact the temperature of the external containment vessel, but it would have an effect on the temperature of the Ni mesh if the mesh were heated via an internal heating element.

This is an important detail that I have considered carefully and commented on in other contexts. Below is a chart[1] of the correction factor to be applied to a conduction-type (Pirani) pressure gauge. These gauges depend on the thermal conductivity of the gas, so the chart directly shows the variation at low pressure and gas composition. You can see that Hydrogen has the highest thermal conductivity of the gases, and that the curves converge as the pressure declines below 1 mbar. However, a nominal 1kPa (10 mbar) H2 pressure measured with a Pirani type gauge would represent only ~1 mbar of actual gas pressure. Therefore a capacitance gauge or physical manometer should be used for accuracy.

Neither of the two reports from Mizuno gives specific details of the vacuum system components used. Figure 5 of the JCMNS article [2] shows what may be a capacitance gauge but it would be good to have this confirmed.

Regarding the chart of thermal gradient measurements Jed posted, the reactor in the image appears to be the earlier one with the heater sheath wound externally around the body. The measurements shown seem consistent with the expected thermal behavior of such a metal construction. The relatively massive end caps act as heat sinks for distributed thermal input, resulting in a gradient with the hottest areas furthest from the ends. Therefore the conclusion that the hotter spots result from excess heat should be reconsidered, absent a null calibration data set from the same reactor and measurement procedure.

[1] https://www.pfeiffer-vacuum.co…ent-pressure-measurement/

[2] https://www.lenr-canr.org/acrobat/MizunoTexcessheata.pdf

We work by comparison to look for XH.

We put 2 mesh one next to the other surrounding the same central heater in the same tube , so same IRs and same pressure.

there is a TC at each mesh but these can have identical or different definitions.

Conduction problems are pushed on backside like that we only are focused on Ni/Pd definition.

RobertBryant Nicely done!

I suspect the discrepancy can be somewhat reduced by adjustment of the convective heat transfer coefficient. You used h = 4 which seems a bit low. In the example here 4.3 is used, and I've seen other examples using 5. The actual number would have depended on the ambient air temperature, pressure and humidity when the cell measurements were done.

The emissivity of stainless steel is another difficult parameter to specify. It will surely be lower than 0.83 for the polished reactor surface, for which 0.5 would be more typical. But there are what appear to be sections of film or coating on the tube, with unknown properties. Some exotic high-emissivity surface treatment might make the reactor Ɛ = 0.83 without it appearing black in visible light. Unfortunately, we just don't know enough to make the model really accurate.

Quote from RobertBryant

Looking at the temperature profile

of reactor shown in the report and modelling heat losses(output) as

radiative emission with an emissivity of 0.83 plus passive convection with a HTC of 4

the temperatures obtained seem a bit low, not too high

,, by my rough calculations assuming ambient T= 20C..

I can only get 216 watts output...whereas it was stated that the output should be 300W.

What would be nice is a temperature profile of the control reactor..and some accurate dimensions..

	Left end	1	2	3	4	5	6	7	8	9	10	RE	TOTAL
Temperature	51	91	115	132	130	122	131	127	116	120	85	47
Area sq.m	0.02	0.018	0.018	0.018	0.018	0.018	0.018	0.018	0.018	0.018	0.018	0.02	0.22
Radiant W emissivity =0.83 ambT=20 :Watts	3.4	8.63	13	16.5	16.1	14.4	16.3	15.4	13.2	14	7.7	2.9	141.53
Passive convection 4x delta Tx Area :Watts	2.48	5.112	6.84	8.064	7.92	7.344	7.992	7.704	6.912	7.2	4.68	2.16	74.408
							215.938

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Robert,

Did you in your calculation include the dissipated heat of the vertical outher walls ?

The convective heat transfer coefficient for a vertical wall is more effective and has a value twice that of the round area which you quoted for the left and right ends.

Including this will increase the total calculated power

Hello Robert, robert bryant

In my opinion your assumed average convective heat transfer coefficient of 4 is too low.

Calculating the convective heat trannsfer coefficient with the method used in the Lugano report I get different values.

Note that I verified the Lugano method by Computional fluid dynamic simulation which showed that the method used in the Lugano report was accurate.

For your sections I get, using the Lugano calculation method, the following results for the convective heat transfer coefficients :

-LE------1-------2--------3-------4--------5-------6-------7-------8-------9------10------RE

4.84---6.47---6.90---7.14---7.11---7.00---7.13---7.07---6.91---6.97--6.34--4.68

Correcting the average powers you calculated with an average value of 4 now gives the following convective power per section

-LE------1--------2--------3--------4--------5---------6-------7---------8--------9------10-----RE

3.00---8.27--11.80--14.39--14.08--12.85--14.25--13.62--11.84--12.55--7.42--2.53

The total convective power now becomes 126.60 Watt

Total power now becomes 141.53 + 126.60 = 268.13 Watt

However as I said in my previous post, also the radiated and convective power on the lefthandside and righthandside should be inclused.

A quick calculation shows that the total radiated and convective power from left and righthandside will be about 20 Watts total

This brings the total up to 268 + 20 = 288 Watt.

That is close to the 300 Watt input power

I don't know what the story is here. You people need to talk with Mizuno directly if you want to find out what's what.