MIZUNO REPLICATION AND MATERIALS ONLY

Quote from Bruce__H

I wonder what the time course for cooling looks like. That could be pretty diagnostic. This group must have turned off the input power at some point and I would have expected then to see some interesting stuff.

Yes, that would be interesting.

Quote from Bruce__H

Overall, this isn't what I would expect to see from a system with a source of excess-heat that is, itself, activated by increased in temperature. There should be an inflection point as the excess heat production starts to create a positive temperature feedback to engage even more excess heat. I see no a hint of anything like that.

What about the big event at around 5000? And what about the fluctuations? They are not seen in the calibrations. Perhaps they are mini-inflection points. The heat is turning on and off.

Quote from JedRothwell

What about the big event at around 5000? And what about the fluctuations? They are not seen in the calibrations. Perhaps they are mini-inflection points. The heat is turning on and off.

It reminds me of an effect I sometimes see when the burner on my gas stove is not quite adjusted properly. The flame goes into quick (and noisy) oscillations. Changing the flow of gas either up or down eliminates the oscillations and once eliminated the flow can be adjusted back to the previous setting without reappearance of the oscillations.

Independent of this, there should still be an inflection point at relatively low temperatures and I don't see any such thing. It should be matched by and inflection point in the cooling curve ... this time at relatively high temperatures.

Quote from JedRothwell

What about the big event at around 5000? And what about the fluctuations? They are not seen in the calibrations.

My problem here is that the time courses look a lot like passive heating curves with fluctuations superimposed. I don't see how the fluctuations contribute to the substantial excess power that is obvious at longer times. In Figure 3 and 4 If one fitted an exponential or biexponential curve to the first 4000 seconds (or minutes or whatever those are) I think you would come up with a very fair extrapolation right out to the 16000 mark. That shouldn't be. The low temperature behaviour should be distinctively different from the higher temperature behaviour with the 2 regimes separated by an inflection point. Could the inflection points be the larger events around 5000 that you point out? For Figure 3 the answer is no. The initial part of that event is a decrease in temperature* which implies that the excess heating is already turned on ... so this cannot be the inflection point I am talking about. The large event in Figure 4 is more like it because the initial inflection is upwards, but then I don't understand why the event does not have a lasting effect and instead the curve goes right back to looking like an extrapolation of the loow-temerature behaviour.

For me, things are not making sense so far.

* Note: I am treating the adjusted power shown on the vertical axis as a proxy for temperature.

The peak power (un-corrected for losses) sits almost right on the calibration input power line, still, it seems.

Quote from Bruce__H

The low temperature behaviour should be distinctively different from the higher temperature behaviour with the 2 regimes separated by an inflection point.

Hold on. This is not low temperature behavior. The temperature shown in this graph is outlet minus inlet, not the temperature inside the reactor. The reactor inside temperature rises very quickly after the power is turned on. I don't recall how quickly, but it takes only minutes. it is like the inside of a toaster oven. The excess heat effect will begin immediately, if it begins at all. It takes a long time for the heat to leave the cell. That is, for stainless steel cell to heat up, and for the outside wall to begin radiating at the peak temperature.

Quote from Paradigmnoia

The peak power (un-corrected for losses) sits almost right on the calibration input power line, still, it seems.

Yes, but that proves there is a lot of excess heat, because it is far below the calibration input power line during all calibrations at these power levels. For it to be on the calibration input power line, the box would have to be perfectly insulated. Perfect insulation does not exist, and in any case, this is far from perfect. You can see the heat coming from the box with an IR camera.

My name is Jeff (username j9381) and I've been setting up a Mizuno replication experiment over the past 4 months. Coincidentally there is another "Jeff" here who has been posting about their Mizuno experiement under the username "Jeff".

Attached is a picture of my Mizuno replication experiment. I have nickel mesh (no palladium) that Jed sent me 5 months ago. I'll be starting calibration runs next week and hopefully start active runs with nickel mesh/palladium/deuterium within 4 weeks.

But I need a low cost palladium ingot. Does anyone have one to sell? I assume I need 10 grams or more because a 1 gram sample is too difficult to rub onto nickel mesh. I realize it needs to be soft so I'll use a torch with slow cooling to soften it.

I'm not using air flow for calorimetry as Mizuno did.

In my set up, I determine the cell temperature constant (the thermal heat loss for a given surface temperature) from calibration and use that to determine the heat output based on 8 thermocouples on the outside surface of the vessel (under the aluminum foil in the photo). I also determine the specific heat capacity of the vessel based on the 8 external thermocouples by inserting a square wave heat pulse into the heater during calibration. I then use this specific heat capacity and the cell thermal power constant from calibration to determine the power being created at any instant of time during the active experiment.

I used this calorimetry technique 2 years ago.

I plan to use Argon, helium and deuterium first for calibration without any nickel mesh in the vessel. After that I will use deuterium for the active cell experiments containing nickel mesh.

I have a 1" diameter x 17" long ceramic tube with tungsten wire (.25 mm) wrapped around it to create a 400 watt heater at 120 VAC. This experiment uses conflat seals in most places and ISO quick flanges elsewhere. I have a RGA (residual gas analyzer) but I don't have it running yet and will do the initial experiments without it. The turbo pump works well and gets the pressure below 10E-6 Torr.

The vessel is approximately 4" diameter and 20" tall.

Jeff (aka j9381)

Quote from j9381

But I need a low cost palladium ingot. Does anyone have one to sell?

I suggest you ask Mizuno to send you a chunk of his Pd.

Quote from JedRothwell

Yes, but that proves there is a lot of excess heat, because it is far below the calibration input power line during all calibrations at these power levels. For it to be on the calibration input power line, the box would have to be perfectly insulated. Perfect insulation does not exist, and in any case, this is far from perfect. You can see the heat coming from the box with an IR camera.

Yes, but not much more insulation would put the power trace somewhere else, and then no one would mention it again.

Quote from Paradigmnoia

Yes, but not much more insulation would put the power trace somewhere else, and then no one would mention it again.

I do not recommend changing the insulation, the blower, or any other component. Do that, and you have to toss out weeks of calibration data. If it isn't broken, don't fix it.

No one should mention this in the first place. It has no scientific significance. It is a coincidence that the output including excess heat happens to equal input in most cases. In a few cases it is above. Would the results be much less convincing if the output happened to fall below the input but above the calibration? I don't see why. The only things that matter are the signal to noise ratios, and the consistency of the calibrations. Input is always 1.4 times higher than output at these power levels, because of losses from the box. Always. Consistently. These losses have been measured and confirmed by other methods. So, there is no question at all why the unchanging ratio is 1.4. So I do not see how there can be any question the excess heat is above the calibration.

Saying this is a problem resembles Seven_of_twenty's repeated assertion that input power is noise. No, it isn't. You can subtract out nearly all the input power, so it does not reduce confidence.

Google (UBC/MIT/LBNL) post Nature updates.

Quote from JedRothwell

I do not recommend changing the insulation, the blower, or any other component. Do that, and you have to toss out weeks of calibration data. If it isn't broken, don't fix it.

No one should mention this in the first place. It has no scientific significance. It is a coincidence that the output including excess heat happens to equal input in most cases. In a few cases it is above. Would the results be much less convincing if the output happened to fall below the input but above the calibration? I don't see why. The only things that matter are the signal to noise ratios, and the consistency of the calibrations. Input is always 1.4 times higher than output at these power levels, because of losses from the box. Always. Consistently. These losses have been measured and confirmed by other methods. So, there is no question at all why the unchanging ratio is 1.4. So I do not see how there can be any question the excess heat is above the calibration.

Saying this is a problem resembles Seven_of_twenty's repeated assertion that input power is noise. No, it isn't. You can subtract out nearly all the input power, so it does not reduce confidence.

Google (UBC/MIT/LBNL) post Nature updates.

Display More

Maybe I’m wrong, but printing graphs with estimated output watts instead of the actual measured output watts because the actual output watts land on the input watts trace is a promotional problem.

Quote from Paradigmnoia

Maybe I’m wrong, but printing graphs with estimated output watts instead of the actual measured output watts because the actual output watts land on the input watts trace is a promotional problem.

Yes, you are wrong. No one is doing that, and your statement is a little obnoxious. Mizuno, I, and everyone else doing air flow calorimetry always shows graphs adjusted for losses, because losses tend to be high with this technique. Losses are high because the calorimeter chamber is much larger than with other methods. It is supposed to be larger. That's the whole point. It is made large to accommodate a large cell. If the cell were small, you use a smaller water flow or Seebeck calorimeter, which are much better in many ways. With a large cell, water flow does not work, and a Seebeck calorimeter would cost a fortune.

Without this adjustment, the results are confusing. It looks like there is no excess heat. You seem to have noted that. Surely you understand that is an optical illusion, and a coincidence with no deep meaning. As I said, if the results fell a little below the calibration line, would you dismiss them altogether? What sense does that make?

Mizuno and I made it quite clear in our papers that we are adjusting some (but not all) graphs to reflect losses. We spent weeks working on the losses. Saito et al. graphed the loss rate carefully, with 13 points, in the graph I posted above. That is at least 13 days of data collection, over several months, with some points tested multiple times. (And if they were to change the insulation, they would have to throw that out and start all over again.)

You can reduce losses with an air flow calorimeter, but other issues are more important. Especially ensuring a stable ambient temperature, and making sure the loss rate is linear and consistent, always the same at the same power level, with any shaped heat source.

Quote from JedRothwell

Hold on. This is not low temperature behavior. The temperature shown in this graph is outlet minus inlet, not the temperature inside the reactor. The reactor inside temperature rises very quickly after the power is turned on. I don't recall how quickly, but it takes only minutes. it is like the inside of a toaster oven. The excess heat effect will begin immediately, if it begins at all. It takes a long time for the heat to leave the cell. That is, for stainless steel cell to heat up, and for the outside wall to begin radiating at the peak temperature.

By low temperature behaviour I meant the behaviour of the entire reactor-calorimeter system when it starts out near ambient temperatures. I think what you are saying, however, is that local temperatures at the mesh can become high quite suddenly relative to the heating time constant of the rest f the system. In that case, by the time the calorimeter is showing an appreciable input-output temperature difference the excess heating phenomenon has already been long since engaged at a high mesh temperature .... the reactor-calorimeter system is acting like a lowpass filter.

In light of all this, I would hope that Mizuno or some of the replicators would consider using ramped inputs rather than step functions. This is the approach that Alan Smith initially took with his Ecalox reactors and it is capable of producing more persuasive results regarding the nature of whatever is producing excess heat (although his group has never cared to exhibit calorimetric data that are persuasive in this way).

Quote from Bruce__H

I think what you are saying, however, is that local temperatures at the mesh can become high quite suddenly relative to the heating time constant of the rest f the system.

Correct. As I said, it heats up about as quickly as a toaster oven. A toaster oven has more power, but it is less well insulated.

Quote from Bruce__H

In that case, by the time the calorimeter is showing an appreciable input-output temperature difference the excess heating phenomenon has already been long since engaged at a high mesh temperature .... the reactor-calorimeter system is acting like a lowpass filter.

Correct.

Quote from Bruce__H

In light of all this, I would hope that Mizuno or some of the replicators would consider using ramped inputs rather than step functions.

Why? What's the point? Is the purpose to discover the effect of heat on the reaction? I guess that would be interesting. I think this is the wrong kind of calorimeter to explore that. You want one with less latency and more direct detection of the reaction.

Quote from JedRothwell

Why? What's the point? Is the purpose to discover the effect of heat on the reaction? I guess that would be interesting. I think this is the wrong kind of calorimeter to explore that. You want one with less latency and more direct detection of the reaction.

Correct. One wants a calorimeter more suited to exploring the effect of heat on the reaction. This is supposed to be a heat-activated reaction isn't it?