MIZUNO REPLICATION AND MATERIALS ONLY

Quote from Cipolla

frankly, or there is a measurement error or a cheating,

Please explain what benefit might accrue from cheating. What motive would there be? It isn't as if people get funded or otherwise rewarded for reporting a positive cold fusion result. On the contrary, that usually leads to attacks. University administrators threaten to fire the researcher; funding is withdrawn. People like you accuse the researcher of cheating. If anyone is going to cheat, it will be in the other direction. That is, the researcher will see excess heat and then cover it up with and say there is no heat, the way they did at MIT and NEDO.

Quote from RobertBryant

The expectation that the mesh is a homogenous device is contrary to my hypothesis..

My hypothesis is that there are thousands of hotspots which switch on at different temperatures..

but at higher temperatures a lot more switch on

and encourage others cooperatively to switch on

All cold fusion metals fit this profile, as far as I know. They are never uniform. Every method that has been used to identify active areas -- such as IR cameras or looking for melted or transmuted spots -- has shown that the reaction only occurs in scattered areas, and these areas turn on and off. If anything is wrong with these results it would be that they are too uniform. The heat is too stable. Except in Fig. 6.

Quote from JedRothwell

Not with an air flow calorimeter after a while. With different gases (or a vacuum) the heat come out of the reactor more quickly or more slowly. This changes the shape of the initial curve. But after everything stabilizes and the curve flattens out, the power level will be the same. It is the same no matter what gas is in a cell, or with a bare resistance heater and no cell.

On the face of it, your statement should be correct.

I have some concerns about the exterior heating method, since it is directly heating the air as much as, if not more than, the reactor or dummy.
Anything that increases the outlet temperature will appear as increased power, all else being the same.

Quote from RobertBryant

Calorimetry in LENR requires long time periods... if only you had a gamma spectrometer..10K$ worth

you might have more than dribbling to do..

The expectation that the mesh is a homogenous device is contrary to my hypothesis..

My hypothesis is that there are thousands of hotspots which switch on at different temperatures..

but at higher temperatures a lot more switch on

and encourage others cooperatively to switch on

there may even be a few spots that operate for significant periods at room temperature

however testing this by looking at bulk air temperatures .. won't show anything

gamma spectrometers.. I believe will show more... especially if there are several mounted over different spots in the reactor

Display More

I can borrow a gamma spectrometer, but see no need to at present.

If more electrical heat increases the reaction, then it looks like electrons are the fuel.

Perhaps heating by another method (piped in hot fluid in tubing wrapped around the reactor?) could be attempted to erase those doubts.

Quote from RobertBryant

External electrical heating was used by Mizuno..less effective than internal heating.. but effective..

its hard to see how electrons get thru the stainless steel wall..

And yet external heating is what we have an image of for the Hokkaido replication, with stellar results.

Better electrons sneaking in than the gammas that would be strong enough to exit through the thick stainless walls.

Quote from Paradigmnoia

On the face of it, your statement should be correct.

Extensive testing with a variety of different reactors and resistance heaters, including bare resistance heaters, shows that my statement is correct. I am not making this stuff up. I am not speculating. I am telling you what the data shows from three air flow calorimeters (two from Mizuno and one from Saito), and from many water flow calorimeters.

Quote from Paradigmnoia

I have some concerns about the exterior heating method, since it is directly heating the air as much as, if not more than, the reactor or dummy.

Why would this matter? It is not clear to me what "directly heating" means in this context, but whatever it means, how can it transfer more energy over time? I can see why it might heat up the air more quickly, if it emerges from the reactor sooner, or if there is less thermal mass. But over the total course of the reaction the heat balance will be exactly the same no matter with any heater. The heater does not measurably affect the inlet or outlet thermocouples. They are not directly exposed to the calibration heater or the active reactor.

Quote from Paradigmnoia

Anything that increases the outlet temperature will appear as increased power, all else being the same.

How can any configuration of the heater increase the outlet temperature? This would not "appear" as increased power; it would be increased power. It would be a perpetual motion machine. The terminal outlet temperature must be the same at a given power level for any heater, with any geometry. If it is not the same, I suppose the outlet thermocouple is directly exposed to the heat source, or there is some other problem. The whole point of calibrating with different kinds of heaters is to ensure there is no problem of this nature. I have never seen a problem of this nature, but anyway, if there were one, calibrations with different heaters should reveal it. If you don't think so, what test do you propose to show there is "direct heating"? How can we detect this?

(Some people here -- not you -- have a bad habit of suggesting there might be thus-and-such a problem, but when I ask: "Okay, how would you detect this problem? What test will reveal it?" they do not respond. That makes their assertion not falsifiable. Meaning it is not true or false. If you cannot describe a test to demonstrate a problem, that problem does not exist. It is imaginary.)

Quote from Paradigmnoia

And yet external heating is what we have an image of for the Hokkaido replication, with stellar results.

That image is not part of the calorimetry. As far as I know, the image does not show any excess heat. It is not a stellar result. It just shows how the heat is distributed with this reactor geometry. The heat distribution is not uniform, meaning you cannot use the surface temperature or an IR reading of the reactor surface temperature to do calorimetry. You have to measure the heat after it leaves the reactor. You have to do this some distance away from the reactor, where the reactor cannot affect the temperature sensors.

I do not think that a direct, temperature-based measurement would work with this reactor. It can be done with a small sample totally immersed in well-stirred water, in a properly shaped container, for example in an electrochemical cold fusion experiment. You might say this reactor is totally immersed in well-stirred air. You have to confirm the air is well stirred, with a traverse test.

Quote from JedRothwell

That image is not part of the calorimetry. As far as I know, the image does not show any excess heat. It is not a stellar result. It just shows how the heat is distributed with this reactor geometry. The heat distribution is not uniform, meaning you cannot use the surface temperature or an IR reading of the reactor surface temperature to do calorimetry. You have to measure the heat after it leaves the reactor. You have to do this some distance away from the reactor, where the reactor cannot affect the temperature sensors.

I do not think that a direct, temperature-based measurement would work with this reactor. It can be done with a small sample totally immersed in well-stirred water, in a properly shaped container, for example in an electrochemical cold fusion experiment. You might say this reactor is totally immersed in well-stirred air. You have to confirm the air is well stirred, with a traverse test.

I mean the image in Figure 1. The open side calorimeter box with the cruciform reactor.

Quote from JedRothwell

Why would this matter? It is not clear to me what "directly heating" means in this context, but whatever it means, how can it transfer more energy over time? I can see why it might heat up the air more quickly, if it emerges from the reactor sooner, or if there is less thermal mass. But over the total course of the reaction the heat balance will be exactly the same no matter with any heater. The heater does not measurably affect the inlet or outlet thermocouples. They are not directly exposed to the calibration heater or the active reactor.

How can any configuration of the heater increase the outlet temperature? This would not "appear" as increased power; it would be increased power. It would be a perpetual motion machine. The terminal outlet temperature must be the same at a given power level for any heater, with any geometry. If it is not the same, I suppose the outlet thermocouple is directly exposed to the heat source, or there is some other problem. The whole point of calibrating with different kinds of heaters is to ensure there is no problem of this nature. I have never seen a problem of this nature, but anyway, if there were one, calibrations with different heaters should reveal it. If you don't think so, what test do you propose to show there is "direct heating"? How can we detect this?

A poor comparison, but recall Rossi’s early pipe kludge that pre-heated the cooling water.

Anything that improves heat transfer to the calorimeter air instead of something else will increase the outlet temperature.

Edit: What I suggest as a reality check is a heater that heats the air as effectively as reasonably possible, with as low a mass as possible, operated at the same calibration input power levels inside the calorimeter to set a roof on the peak air heating combined with a minimal time to reach steady state. A very thin, long Kanthal wire with as little support structure as possible might work.

Quote from Paradigmnoia

I mean the image in Figure 1. The open side calorimeter box with the cruciform reactor.

Oh. Do you mean the cruciform reactor is heated on the outside? Yes, I think it is. The Saito results are not so stellar. It produces about 150 W. Mizuno's results seem to indicate that an internal heater works better, assuming all else is equal. But all else is never equal, so it is difficult to say if this is a real finding or a coincidence.

Granted, 150 W is pretty darn good. I am happy with it! I wish the others trying to replicate would see this much heat. Heck, I would be happy if they saw 10 W, as long they could measure it with confidence. 150 W is much better than Mizuno was getting a few years ago, and much better than most cold fusion experiments.

I suppose the power is much higher than most experiments simply because there is so much more reactant. Most cold fusion experiments have only a few grams of palladium, and who knows how much of that is active. If the high power is mainly a function of the total mass of material, I suppose that fact does not contribute much to theory, or to the understanding of the reaction. But it makes it easier to measure. And more convincing. Those are important advantages. I guess it must be an improved reaction in some ways, for who-knows-what reason. Because similar large masses of mesh years ago barely produced ~10 W, and most did not produce any excess heat at all.

Quote from Paradigmnoia

Anything that improves heat transfer to the calorimeter air instead of something else will increase the outlet temperature.

No, I do not think so. Anything that improves heat transfer to the calorimeter air will reduce the time it takes to reach the peak power output. But it will not increase the total energy measured over the course of a calibration test from beginning to end. Or the total energy from the beginning of the test until it reaches a stable power level, with a flat output curve. The only thing that can increase the outlet temperature over time is more energy. Otherwise, you could use the calorimeter as a perpetual motion machine. That is to say, over the course of a test, you put in X Joules of power, and you measure Y Joules of output, which is X minus losses from the walls (and a few other small factors). If you can find a way to get out more than Y Joules without changing the wall configuration or losing less heat, you are creating energy out of nowhere. You cannot change Y by changing the geometry, shape or anything else about the heater. If you could, the calorimeter would not be working right. The calibration would show there is an error, such as unstirred air. (Unstirred water in a water flow calorimeter is a common problem. So is measuring the flow rate wrong. The point is, you find these problems when you look for them.)

Quote from Paradigmnoia

What I suggest as a reality check is a heater that heats the air as effectively as reasonably possible, with as low a mass as possible, operated at the same calibration input power levels

There is no possible difference in the efficiency of heating the air. All heaters in an enclosed calorimeter impart all heat to the air. There is no place else for the heat to go. That's the beauty of envelope calorimeters, as opposed to isoperibolic ones.

Low mass heaters have been tested. That would be a bare resistance heater. You can't make the mass any lower. All kinds of heaters have been tested: low mass, high mass, big, little, cruciform, tubes . . . you name it, Mizuno has tried it. (Partly because I asked him to. We thought of this.) They all produce exactly the same total energy output over the full course of the test. Not the same power level at the beginning, as I said. You can tell what kind of heater is in the calorimeter by the initial shape of the curve. Compute total energy output and they all look the same. There is no way to sneak the heat directly from the heater to the inlet or outlet thermocouple. I suppose you could design a calorimeter with the outlet TC placed close to the reactor, and unshielded. I guess that would give the wrong answer. I can't say for sure because I don't know anyone who has done this. People do not deliberately make instruments they think will not work. Except for Steve Jones. He wanted to show the F&P's calorimetry did not work, and that it produced recombination. So he made an isoperibolic calorimeter the wrong shape. F&P warned against doing that, and said it would not work. Then he ran it at a power level about a thousand times too low, which will definitely produce recombination. Mel Miles said, 'why not throw some palladium powder into to the electrolyte while you are at it, to ensure 100% recombination.'

I will just have to do it myself.

Quote from Alan Smith

Agreed- a watt is a watt wherever. I was unclear- I meant that it would make a difference to the appearance of the calibration curves.

Yes! The initial shape of the curve varies with the type of heater, and the thermal mass of the heater.

I did a 200 W calibration run, and now am repeating it with two fire bricks raising the SS tube assy about 6.5 cm to see what happens. This is preliminary to removing the bare resistor cartridge from the 20 x 50 cm long SS tube and running it open at 200 W.

Should be fun...

Two steps forward, one step back... The first attempt to operate the T/M pump yielded an acceptable vacuum on the range of 1e-8, as measured with an ionization gauge. At this time I decided to shut the system down for the day. Somehow during the shutdown process the forepump must have backstreamed some oil into the T/M pump, despite the presence of a trap. The next time the system was evacuated it was impossible to achieve better than 1e-4 vacuum. So I had to tear down everything and wash it out with acetone (including the T/M pump).

The next step is to install a solenoid-operated isolation valve (Edwards type PV25EK) that is slaved to the forepump power. That should take care of backstreaming. Another thing learned: install viton CF gaskets at first to make sure everything works. Then, and only then, replace them with Cu gaskets. Besides being expensive, the latter are a royal pain to install and remove.