MIZUNO REPLICATION AND MATERIALS ONLY

Quote from RobertBryant

Thankyou Paradigmnoia

So you were inferring... based on data that is not Mizuno's data

Please learn to read context properly

This series of graphs has been posted too many times on this thread

They have been a distraction

If Paradigmnoia has any 500W input data from his own experiments

that are relevant to the discussion of the recent Mizuno 500W/800W results

I would be interested to see them.

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If you have any relevant experimental data of your own to add to this discussion, this would be a great time to share it with us.

I can tell you that my acrylic box with no insulation (except on the bottom), at 200W input, managed a recovery of just under 63%

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Here are all the measurements for the experiment:

The only reason I was using 200 W is that my other power supply was too unstable, with +/- 15W excursions commonplace. So I changed to a dedicated 200 W DC supply to get the measurements under much better control. The heater currently installed can withstand 1378 W continuously. I will upgrade with other DC power supplies in due time, now that the rest of the system is working properly (excellently). However, I am almost 1000 km from the calorimeter and will stay that far away for a while, so don't expect much fresh data in the near future.

So what about this recent one brought to us by Dr. Mizuno himself.

Does it show about 48% recovery?

Am I reading something wrong?

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Quote from sam12

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 
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• 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?

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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.

Quote from Paradigmnoia

So what about this recent one brought to us by Dr. Mizuno himself.

Does it show about 48% recovery?

Am I reading something wrong?

.

Like Paradigmnoia, it isn't clear to me why, if one is just a corrected version of the other, the red and grey traces should cross near the beginning. The efficiency of the recovery should always be less than 100% and so the correction factor should always be greater than 1. Apply a correction like that to the grey trace and you should get a corrected version that is everywhere as big as, or larger than, the original.

Possible explanations for this are 1) that at low temperatures the proportion of recovered heat appears greater than 100% because of heating by the fan or other equipment and 2) that the red trace not a corrected version the grey one but of some other run.

So perhaps you can't calculate % recovery at all from this figure.

To DR. Mizuno:

Quote

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.

It would be also interesting to know the temperature of the heater and its active area.

Thanks

The heat recovery graph that shows the percentage of thermal wattage lost through the walls of the box versus input power (for the flow calorimetry that Mizuno does) is made from steady state conditions. Fast changes in temperature of the atmospheric air inside the box at start up leads to errors in the output watts. So there is going to be errors in the output power graph that can be acceptable if they are small portions of the overall trend. It would be time consuming to determine the appropriate equation to reduce that start up error - but it could be reduced with enough effort. I don't think it is necessarily worth the time if the runs are long enough.

Q andA replicator Desireless

Alain Samoun  Desirelessa day ago

Can you give us the temperature of reactor and Ni mesh as well as the pressure of D2 when getting excess heat?

• Desireless  Alain Samoun5 hours ago

  Reactor core 450°C, Reactor shell 120°C, pressure 10 mbar.

Quote from j9381

The heat recovery graph that shows the percentage of thermal wattage lost through the walls of the box versus input power (for the flow calorimetry that Mizuno does) is made from steady state conditions. Fast changes in temperature of the atmospheric air inside the box at start up leads to errors in the output watts. So there is going to be errors in the output power graph that can be acceptable if they are small portions of the overall trend. It would be time consuming to determine the appropriate equation to reduce that start up error - but it could be reduced with enough effort. I don't think it is necessarily worth the time if the runs are long enough.

Perhaps this is in answer to my observations on the grey and red traces crossing in Mizuno's 800W output plot? If so, then I understand and agree that the heat recovery graph most likely made from steady state observations. This is good for exactly the reasons you outline. But I don't see what this has to do with the grey and red traces crossing. One trace (the red one) is supposed to be just the other trace multiplied by a temperature-appropriate correction factor that is always greater than 1. So even if the grey trace contains errors of the type you outline, the red trace should just be an enlarged copy of these errors -- and its not.

No, there must be some other explanation.

the equation for start up could be complicated and involve the factors of time, heat capacity of the vessel, heat capacity of the calorimeter - in other words as complicated as you want to make it (though best is just to keep it simple and reduce the start up error to an acceptable level). The evaluated equation can be above or below the value of 1.

One way to do the mathematics of flowing calorimetry (ignoring the startup issue above) is to make a calibration chart of thermal power captured in the flowing air at steady state with the control reactor (no nickel mesh, no D2, no Pd) and this chart would always show a fractional value (i.e. less than 1) because some heat escapes through the wall of the box. This fractional value is a function of the power into the control reactor.

So for example: 130 watts of electrical resistor input, 100 watts measured in flowing air and 30 watts emitted by box at steady state. The fraction captured in the flowing air is 100/130 = 0.769.

When running active reactor (with nickel mesh, D2, palladium etc.) , if 100 watts is seen in the flowing air at steady state then the total emitted power is 100/.769 = 130 watts. The key is to note that the fractional capture value changes as a function of power emitted by the reactor and is mostly (perfectly?) valid at steady state. If the temperature is fluctuating/increasing/decreasing then it is a judgement call when using the steady state fractional capture value. In most cases it is adequate and certainly adequate for long runs.

On top of all that, when the input energy is turned off and the temperature returns to room temperature, the thermal power in the flowing air can be integrated over time to give the total amount of energy created by the reactor.

j9381

I understand what you are saying. I just don't think it is related to the crossing of the grey and red traces in the Mizuno 800W output figure.

My understanding (which may be wrong!) is that the red trace is just a mathematical transformation of the grey one, i.e., it is the result of multiplying the grey one by a correction factor that depends on temperature. This should be true whether the entire grey trace is near steady state (which it is probably isn't) or whether the entire thing is far from steady state.

Now, I do think that the correction factor is derived from steady-state observations -- Mizuno and Rothwell aren't totally clear on this point but I suspect it is true -- so I think that the correction factor is determined from a calibration chart generated in pretty much just the way you outline ...

Quote from j9381

One way to do the mathematics of flowing calorimetry (ignoring the startup issue above) is to make a calibration chart of thermal power captured in the flowing air at steady state with the control reactor (no nickel mesh, no D2, no Pd) and this chart would always show a fractional value (i.e. less than 1) because some heat escapes through the wall of the box. This fractional value is a function of the power into the control reactor.

As you point out, the fractional value should always be less than 1. But that means that the corresponding correction factor needed to transform the measured power into corrected power (i.e., measured power plus estimated losses) should be greater than 1.

I further take your point that since the correction factor is steady state, it may not properly apply to the initial transient phases of the plot. But this is a difficulty for later interpretation and is not my immediate concern. My immediate concern is that the mathematical procedure here is to take the grey trace and multiply by a correction factor that is everywhere greater than 1. That seems to be what is done. So why is the resulting red trace sometimes smaller than the grey one? Wouldn't that reflect a situation where, with an inactive reactor (i.e., the one used for calibration), there is more heat being captured in the calorimeter than is being inputted into the reactor?

I don't know how they did the math in that particular graph you are referring to.

But whether the factor is above or below 1 on start up is different for every set up. The factor will be below 1 after some amount of time after starting for all set ups. There are lots of factors ... power emitted, surface area, reactor thermal heat capacity, calorimeter box thermal heat capacity, air flow rate, shape of the reactor vessel, shape of the calorimeter, insulation thickness etc.

There's no real need to be concerned with such a small portion of the full run. Someone could spend weeks running different control (i.e. null) experiments trying to make an equation to decrease the error on start up and that would be wasted time. It's better to do a long run and then the start up portion with its errors will be a tiny percentage.

Quote from j9381

I don't know how they did the math in that particular graph you are referring to.

But whether the factor is above or below 1 on start up is different for every set up. The factor will be below 1 after some amount of time after starting for all set ups. There are lots of factors ... power emitted, surface area, reactor thermal heat capacity, calorimeter box thermal heat capacity, air flow rate, shape of the reactor vessel, shape of the calorimeter, insulation thickness etc.

There's no real need to be concerned with such a small portion of the full run. Someone could spend weeks running different control (i.e. null) experiments trying to make an equation to decrease the error on start up and that would be wasted time. It's better to do a long run and then the start up portion with its errors will be a tiny percentage.

As far as I know, the calibration plot used to determine the correction factor is based on steady state observations. So I don't think it is a case of the correction factor being below 1 initially and then greater than one once initial transients become small. And I particularly don't understand why the correction factor used in the 800W plot should ever be is less than 1 when Mizuno and Rothwell's procedure seems to be to find fractional efficiency by fitting a straight line to their steady state calibration data and (apparently) constraining the line to pass through 1 at reactor temperature = 0 Celsius. This produces a fractional heat recovery that is strictly less than 1 for temperatures above 0C and hence a correction factor that is always greater than 1 at all temperature they were actually working at.

I don't think that the crossing of the red and grey traces is associated with transients. It's something else.

For what it is worth, I don’t really care about the start up part. (Although the lines crossing is unusual, and as such has been noted).

I base most of my observations on the steady state portion of the traces. Once at steady state, whatever heats the calorimeter is irrelevant to the power measurement. The calorimeter should react the same at the input power level no matter what heats it. Therefore a calibration run showing the same steady state output power (or reasonably close) as an anomalous heat test at steady state is the best cross check for the anomalous result.

I don’t see how this is controversial.

Quote from Bruce__H

As far as I know, the calibration plot used to determine the correction factor is based on steady state observations.

Definitely. Many hours at the terminal temperature for that power setting.

Quote from Bruce__H

And I particularly don't understand why the correction factor used in the 800W plot should ever be is less than 1 when Mizuno and Rothwell's procedure seems to be to find fractional efficiency by fitting a straight line to their steady state calibration data and (apparently) constraining the line to pass through 1 at reactor temperature = 0 Celsius.

I'm guessing that it crosses at a delta T of 0 Celsius (in other words, a temperature difference of 0 Celsius).

It's not crossing at 0 Celsius (freezing water temperature). The delta T is between the room air and the tube air flow temperature.

Quote from j9381

I'm guessing that it crosses at a delta T of 0 Celsius (in other words, a temperature difference of 0 Celsius).

It's not crossing at 0 Celsius (freezing water temperature). The delta T is between the room air and the tube air flow temperature.

I don't think so. Look at Figure 10 of Mizuno and Rothwell J Cond Matt Nucl Sci 29:1-12 (2019), then equation 2 of the same paper. The equation they fit to their calibration data is fractional power capture = O/I = 0.98 - [5.0811E-4 x T] where T is "the reactor temperature"