Posts by anonymous

Quote from dartin

Full Range Appion Vacuum gauge:

www.amazon.com/Appion-AV760-Range-Digital-Vacuum/dp/B00VKGQ6OW/ref=asc_df_B00VKGQ6OW/

You can do better for the money with a Pirani type digital vacuum gauge which can go to 0.1 micron i.e. 10^-4 Torr from Lesker or Instrutech for around $400. Google for it and you will find the manufacturer and distributors.

You may be able to use a cheaper rougher (backing) pump on your Turbopump as $2700 seems high and this one only goes to 250 mTorr (i.e. 2.5x10^-1 Torr). See what the specs are for the backing pump on your turbopump. Without doing more research, I feel as if this pump has more capacity (i.e. liters per minute) than you need for a backing pump to your turbopump. The turbopump is the high cost item on your vacuum system.

Didn't look at the rest carefully. How much for your Turbopump and how much for the RGA. Good luck.

Quote from Longview

>>> 10e-4

0.001

If true, suggests the writers of Python may have been seriously short in the exponents of 10 business. For a simple example, 10 raised the the "minus one" power is by definition simply 1/10 = 0.1, likewise and by definition, 10 raised to the minus 2 power is simply 1 divided by 10 squared, that is 1/10^2, or 0.01 or "one hundredth".

As we all know, 1 raised to any positive power is simply 1, and by inference 1 raised to any negative power is simply 1/1, 1/1^2, 1/1^3 etc. And hence also simply "one". Such notation ostensibly from Python above seems at odds with basic exponential notation and its related algebra

We're going to see more of this, I fear. What concepts, and programmatic idiosyncrasies, rather an actual old fashioned engineering training, might have led to the failures of the Boeing 737 MAX MCAS/ACAS system?

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No Longview -- it's the standard and you are just not reading 10e-4 as described in the standard. But that's OK. It literally is DEFINED as 10 x 10^-4 in the standard. If you want to read it in a non-standard way, that is up to you. But we can be certain that "the writers of Python" have not "have been seriously short in the exponents of 10 business".

"If true". If you have a Mac, open a shell and type "python<enter>" and try it yourself and you will get the same answer. If you have a PC you will have to download and install Python. On Mac, Python is installed by default. If you don't want to install Python, do it on Excel on your PC. Try entering 10e-4 in a cell on an Excel spreadsheet. All the programming languages implemented the standard developed in Fortran II in 1958. A good standard last forever.

What more can I say -- nothing. But if you are communicating with the remaining people who know what 10e-4 means, you are confusing them because it means on any computer language 1/1000 and nothing else. It's a mistake.

Quote from oldguy

No typo- it is an the old "classic way" of writing Scientific notation see for example: http://albert-cordova.com/_science/exponent.html

10E-4 t is 0,0001 torr or 0.1 microns.

Notice many high vacuum gauges use the notation: https://mse.ndhu.edu.tw/ezfile…ST18-U8-pressuregauge.pdf

It seems you are living in the virtual computer world of notation and not in real life laboratory world.

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Whatever world I am living in, 10e-4 is not normalized. And this comes from the days of Fortran II (1958) and surely even if you are oldguy, your use of this convention doesn't predate 1958. The convention carried thru to today in all languages for their default display.

The mantissa is suppose to be between 1.000... and 9.999... and 10>9.999... I don't see any non-normalized numbers in Ming-Show Wong's 2018 paper that you cited, and regardless, it is non-standard and thus confusing. The Tracy Albert html page on "exponential notation" you cited is not any standard that exists and therefore is wrong (showing all non-normalized non-standard "scientific notation") and would needlessly confuse any high schooler learning scientific notation. What Mr. Albert means when he shows an "E" is the "^" which on ASCII is the way to write an exponent, thus 1.0*10e-1 should b 1.0*10^-1 which if he had a non-ascii character set would be 1.0*10^-1.

The correct way of displaying what it is I think you mean by 10E-4 is:

10^-4 = 0.0001 = 1/10000 = 1e-4.

See:

https://www.quora.com/What-does-1e-4-stand-for

Modern computer languages do the same:

R:

> .0001

[1] 1e-04

Python:

>>> 1e-4

0.0001

>>> 10e-4

0.001

Wikipedia:

https://en.wikipedia.org/wiki/Scientific_notation

Quoting the Wikipedia:

Normalized notation

Main article: Normalized number

Any given real number can be written in the form m×10n in many ways: for example, 350 can be written as 3.5×10² or 35×10¹ or 350×10⁰.

In normalized scientific notation (called "standard form" in the UK), the exponent n is chosen so that the absolute value of m remains at least one but less than ten (1 ≤ |m| < 10). Thus 350 is written as 3.5×10². This form allows easy comparison of numbers, as the exponent n gives the number's order of magnitude. In normalized notation, the exponent n is negative for a number with absolute value between 0 and 1 (e.g. 0.5 is written as 5×10⁻¹). The 10 and exponent are often omitted when the exponent is 0.

Normalized scientific form is the typical form of expression of large numbers in many fields, unless an unnormalized form, such as engineering notation, is desired. Normalized scientific notation is often called exponential notation—although the latter term is more general and also applies when m is not restricted to the range 1 to 10 (as in engineering notation for instance) and to bases other than 10 (for example, 3.15×2²⁰).

E-notation

A calculator display showing the Avogadro constant in E-notation

Most calculators and many computer programs present very large and very small results in scientific notation, typically invoked by a key labelled EXP (for exponent), EEX (for enter exponent), EE, EX, E, or ×10x depending on vendor and model. Because superscripted exponents like 10⁷ cannot always be conveniently displayed, the letter E (or e) is often used to represent "times ten raised to the power of" (which would be written as "× 10ⁿ") and is followed by the value of the exponent; in other words, for any two real numbers m and n, the usage of "mEn" would indicate a value of m × 10ⁿ. In this usage the character e is not related to the mathematical constant e or the exponential function e^x (a confusion that is unlikely if scientific notation is represented by a capital E). Although the E stands for exponent, the notation is usually referred to as (scientific) E-notation rather than (scientific) exponential notation. The use of E-notation facilitates data entry and readability in textual communication since it minimizes keystrokes, avoids reduced font sizes and provides a simpler and more concise display, but it is not encouraged in some publications.[3]

Examples and other notations

In most popular programming languages, 6.022E23 (or 6.022e23) is equivalent to 6.022×10²³, and 1.6×10⁻³⁵ would be written 1.6E-35 (e.g. Ada, Analytica, C/C++, FORTRAN (since FORTRAN II as of 1958), MATLAB, Scilab, Perl, Java,[4] Python, Lua, JavaScript, and others).

Quote from oldguy

I

You might use an oil pump for rough, and some can get to 10E-4 t but if you use them you need a foreline trap, inline molecular sieve filter or something similar.

Likely Typo: 10E-4 in computer scientific notation is 10 x 10^-4 = 10^-3. You probably meant 1E-4 = 1x10^-4 = 10^-4 = 1/10000

Quote from JedRothwell

I do not know whether capping is more likely to prevent leaks. However, if there are leaks, I am sure you will detect them with the recommended configuration.

You have to use a high performance turbo molecular pump and the pressure gauge that comes with it. I don't think you should do this experiment without them. Other kinds of pumps will contaminate the reactor, I think. The pressure gauge will show if there is any air leaking. There is no chance a significant amount of air, enough to affect the result, will leak in and you will not notice it.

Plus, you need a mass spectrometer. You need to sample the gas periodically. You will see if air and water have leaked in.

You cannot do this experiment blind.

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I agree with the turbopump, otherwise can't get to <10^-4 torr and also run risk of oil backflow contamination.

The mass spec will not make the experiment work, but it will definitely help the experimenter debug why it fails. If the experiment fails and there is not a mass spec, it is not the same configuration and the experimenter will not know what was different. I think the whole rig can be built for $10K with reasonably reconditioned used equipment. It could be built for less (i.e. with equipment of used but unknown condition), but the experimenter will spend more time trying to fix stuff.

Quote from oldguy

OH yes, when you get a mass spec or RGA (my choice for such things) be sure that your pump will get low enough. You usually will need to get to 10E-4 torr or better to run most of them (to high and you burn out your bulb if not automatically controlled). The turbovac can be fitted with sensors for RGA/s.

If $$ is not critical, you may want to check out bake out systems - if you go with Ideal vacuum co, they can set you up on such things (no, I have no $$ interest with them). a reasonable US supplier of Leybold pumps, other distrib. for EU.

If you don't bake out, try an electric blanket over the system and pump down for a few days on the assembly put in some H2 and pump again to react any O2 on the walls.

Bake out with the heater inside the unit keeping the cylinder insulated in a high temperature blanket (i.e. reflective fiberglass). You just need to get the walls somewhat above 150C while under vacuum and the water should evaporate off the stainless steel and nickel fairly quickly. Monitor the RGA or vacuum gauge and when the vacuum stops going down after a few days, you got all the water and volatiles out. You eventually want to go to the maximum calibration temperature that you expect to run with while producing excess heat so you know there is nothing else to bake off.

Quote from RobertBryant

I agree,

especially when instant experts cannot do heat transfer calculations

even halfway right.

Even if they could, it is a complex system with multiple components and multiple heat transfer paths, i.e. a real time waster, i.e.

P12 = k12c*(T1 - T2) + k12r*(T1^4 - T2^4) + f_naturalConvection(T1,T2)

P23 = k23c*(T2 - T3) + k23r*(T2^4 - T3^4) + f_naturalConvection(T2,T3)

P13 = k13c*(T1 - T3) + k13r*(T1^4 - T3^4) + f_naturalConvection(T1,T3)

P14 = ...

P = P12 + P13 + P14 ...

As many elements as parts in the reactor can be modelled. Big time to setup and solve. It is possible that Jed is correct and emissivity doesn't matter, but a definitive solution would take time.

R20 is the immediate goal. My preference is to change as little as possible between experiment and control, and that includes emissivity.

Quote from IH Fanboy

Jed, while I agree with your statement, I think what anon is contending (he can confirm my interpretation of his position) is that radiative heat causes the air circulating within the calorimeter to be heated more slowly than convective heat, and therefore the difference in the kind of heat transfer matters for air flow calorimetry. Can you comment to this specific critique?

I suppose my response would be that the inside of the calorimeter soon reaches a steady-state temperature, and therefore any difference in the rate of heating of air between radiative/convective would be insignificant to the calorimeter measurements. And of course for R20, the COP is so large, such a difference wouldn't matter anyway.

No IH FB -- I am looking at steady state so that the speed or rate of transfer is not significant once it finally reaches equilibrium.

I am concerned about a different amount of radiation leaving the reactor and striking the walls of the calorimeter (even with emissivity claimed of 0.04) and thereby more effectively heating up the walls of the calorimeter which in turn would radiate, conduct, or convect the heat to the outside without being measured by the air flow Delta-T.

None of this matters for R20 with its "uncorrected" power output of 225 watts for 50 watts in, and at least some of our friends around the world will make their own replications using identical emissivity so that this will not be a factor.

If I was in high school and needed a science fair project it could be "Does Emissivity of a heat source effect an Air Mass Flow calorimeter's Heat Capture". It's easy enough for a high schooler to build -- fans, acrylic boxes, home depo insulation. It also lends itself to enough high school algebra to model the results. I hear all the well thought out arguments here on both sides but in the end, it just doesn't matter for R20 so we should move on. If we can replicate an R20, this debate here is a time wasting footnote.

In response to Jed:

"anonymous wrote: It doesn't say the procedure was to valve off the tank after the reactor cylinder was pressurized at the appropriate pressure (i.e. 300 Pa) ."

Jed's response: "It does not have to say that. I suppose anyone would know that from the many things the paper says."

Well -- I didn't know that (and I am an anyone) but you clarified that the tank was valved off here in your previous comments. If you want to clarify it in the paper that is up to you.

Jed wrote:

"I believe that is the average. But you are missing the point. The paper says, in several places, the pressure should be low. Around 1% of atmospheric pressure. You and THH were asking whether the heat might come from burning deuterium and air getting into the cell. If that were the case, the supply of D2 and air would have to come in all day long, 24 hours a day, in large amounts. So the pressure would be 1 atm all the time. Neither the intraday nor the average would be ~1000 Pa. Also, when Mizuno says "keep the pressure low" and "be sure to exclude water and contamination" surely you understand that opening up the cell to a large flow of air and D2 gas ... flames would destroy the heater"

1) Now that I know the unit is sealed (rather than running under continuous regulated low pressure D2 supply) it is clear that the excess heat cannot be coming from the D2 that was already in the reactor. The possibility of making a low pressure D2 regulator that maintained the system at set pressure (i.e. 300 Pa) that included excess vacuum regulation capacity to remove any leaked air or heavy water vapor product was a possibility before your clarification. But you have clearly stated that this is not the case.

2) Burning and flames ... would destroy the heater -- disagree. Stainless steel or platinum when heated act as a catalyst allowing chemical recombination of hydrogen (or deuterium) and oxygen to occur without flame and at low pressures and hydrogen concentrations. These are used "autocatalytically" (generating their own heat once started) as a safety mechanism in nuclear power plants to prevent hydrogen explosions, to combine the hydrogen released from hot zirconium fuel rods reacting with steam to be safely converted into water vapor without flame. These failed at Fukushima, resulting in the top of the reactor being blown off in a hydrogen/air explosion. This paper shows that the temperature of the catalytic plates can be as low as 450K or 175C. No flames, just heat. See https://www.degruyter.com/down…-0042/aoter-2013-0042.pdf I will not get into the technical calculations to prove or disprove whether catalytic recombination could have occurred at these pressures (i.e. 300 Pa) in the unit, but the possibility did exist that this could have occurred with no flames, just excess heat from the recombination. When you ruled out the resupply of D2 to the reactor in your comments, this ruled out this effect being significant in any way on the small and thus limited (finite) amount of D2 in the reactor when the test is run.

Jed: "I think you are completely wrong. No flow calorimeter can detect a difference in emissivity from the reactor inside it. It would never collect less heat for these reasons. The heat will not escape from the calorimeter chamber no matter what color or type of steel you use or whether you substitute ceramic or some other material."

We disagree technically with each other. My suggestion remains to use a reactor calibration cylinder of the same thermal (conduction, radiation, convection) characteristics. I respectfully agree to disagree and suggest we both move on.

Thank you Jed,

Anonymous

Quote from THHuxleynew

\ Another thing - variation in air flow between the two reactors, if it existed, would have a similar affect, by allowing one to be cooled more effectively than the other and therefore have a lower case temperature. As would variation in the thermal resistance of the block on which the reactor is placed (possibly we are told those blocks are identical).

I skimmed your discussion on airflow. I did some of my own research and I decided that air mass flow is proportional to fan power input. If he is running at 6.5 watts the fan, then I believe that the mass flow will be about the same regardless of the accoutrements in the calorimeter. I also decided mass flow is independent of temperature. I have less concern about this part of the experiment. I see no need to measure the airflow at different spots, i.e. this is "good enough" for his proof.

Quote from JedRothwell

Question 1) Do you suppose cold fusion heat has some special quality that allows it to escape from the reactor in a way that resistance heating does not? That seems kind of far-fetched.

Question 2) Why would the ratios of conduction to radiation to convection be different with cold fusion, after the heat passes through a steel wall?

Answer 1): [Cold fusion heat has a special quality] No not at all. That is NOT my point. But perhaps that is a rhetorical device on your part. Of course it is far fetched -- I agree!

Answer 2): Because you are doing the cold fusion in a reactor that is of different emissivity. The reactor steel cylinder which is shiny has a different emissivity (about 0.075) than the calibration steel cylinder (about 0.85), which is flat grey. It's not the "cold fusion" heat that is different, it's the cylinder emissivity that is radiating the heat to the walls of of the calorimeter.

You have said here that some of the heat is lost thru the external ("Home Depo") reflecting insulation. We agree there. The HD insulation's inner surface emissivity is likely around 0.05% (i.e. it's very shiny), and it is likely backed with additional insulation, but heat radiation still gets through it. You know this because at 360 C only 77% of the heat is captured in the calorimeter by the mass flow, raising the exit air temperature. Those are quoted directly in the paper. 77% means that 23% of the heat is escaping, most likely mostly through this insulation. Now you lower the amount of thermal radiation that gets to the reflecting insulation by a factor or 7.5/85 = 1/11.3x at a given temperature. Making the reactor shiny is like putting more reflective insulation inside the calorimeter. So less heat now gets out of the calorimeter wall insulation, because less radiation reaches it at a given reactor surface temperature by factor 11.3x. That's a BIG difference. Where is the heat going to go? Out thru the airflow. The calorimeter now goes from 77% efficiency to what, 90%? We don't know how much it goes up because we didn't measure (calibrate) the calorimeter+reactor with an identical shiny reactor cylinder.

Am I not clear? Is there something I missed.

Respectfully,

Anonymous

P.S. This doesn't change the basic conclusion that the R20 reactor is making at least 225 watts out for 50 watts in, or COP of 4.5x. That is still a revolutionary improvement.

Quote from JedRothwell

It is right there in Table 1! It shows the pressure. Do you see anything like 101,000 Pa? How could he keep it at 1000 Pa if he did not valve off the D2?? Leave the needle valve open and it will fill to whatever the pressure in the tank is. Of course you have to shut the valves to keep it at 1% of atmospheric pressure.

What a weird comment.

Jed, I understand that you are frustrated with answering what you think are ridiculous questions, but table 1 only shows a set of pressures on each day. It doesn't say anything about the intraday history of the pressure after the day's test began. It doesn't say when in the day the pressure was measured. It doesn't say the procedure was to valve off the tank after the reactor cylinder was pressurized at the appropriate pressure (i.e. 300 Pa) . There is no schematic of the D2 and vacuum system. I am "skilled in the art". I did not mean to upset you with another ridiculous question. I am not a skeptic. I am trying to prove self-consistency of the data from your paper. My initial hypothesis was that the rig had some kind of D2 pressure+vacuum regulator that would allow you to maintain the stated low pressure. You then told me me that that was incorrect, and that the tank was simply valved off. I have used high vacuum systems, and their valves, and I know from experience that if the valve is functioning, it does indeed valve off the supply of gas (or vacuum). So your answer (that the reactor was valved off from the D2) corroborated the remainder of the paper and was good enough for me. But it was in no way obvious to me before you answered my question. I am a mature adult like you, and one skilled in the art of vacuum systems. It is also not in my opinion obvious to third party readers who will read your paper without the benefit of reading all 850 comments on this thread.

So my "ask" is simple: do other readers the favor of explaining this with a simple few sentences in the next revision of the paper:

"The reactor pressure was set to the desired pressure shown in Table 1 by letting in a small amount of D2 from the tank via a valve. If the pressure was to be lowered, by evacuating the system with the vacuum pump, the pressure was lowered via a different valve connected to the vacuum pump. Once the desired test pressure was obtained, the D2 tank and the vacuum were both valved off from the reactor. The pressure in the reactor was monitored with [type of vacuum measurement equipment, e.g. Pirani Gauge] and did not rise more than [xx%] for the 24 hour period of each test, thereby proving that there was no significant leaks or outgassing during each run that would effect the result."

Thank you Jed. I appreciate your contribution to LENR now and over the years. Please continue -- you're achieving success with your work.

(P.S. I am your anonymous supporter here, but when you make fun of me for asking a question, it makes it more difficult. Fortunately I can check my ego at the door.)

Quote from JedRothwell

Actually, I understood THH perfectly. So did Robert Bryant. THH repeatedly claimed the D2 in the cell might cause the heat. He wanted to know how much was absorbed in the reactant. Robert and I repeatedly told him there only enough for ~1 s, but he kept saying the same thing. If he had meant a leak from the D2 tank, he would have said that. He just now made that up, I suppose to save face. I pretended it was my misunderstanding, because that's the proper form in an academic discussion.

Also, a leak from the D2 tank is only marginally less preposterous than 3 mg producing 9 million times more heat than is possible. No one with experience doing experiments would seriously propose such a thing. Such a leak would be obvious in many ways, some of which I listed above. It would immediately bring the experiment to a halt. Such arguments are only intended to confuse the issue and make people think there may be problems where no problems exist. This is trolling, not a serious scientific discussion.

No one misunderstood anything here. It was perfectly clear, just as it was when THH proposed that drops of condensed water become invisible, move up against gravity, and that magically cancels out the energy needed to evaporate them. Okay, he didn't put it that way, but anyone can see that is what his hypothesis adds up to.

If Mizuno says he valved off the D2 as an ongoing source of chemical energy, I believe him. Can we move on. This will be proven by the replicators.

(Please note that for the future publication I would suggest making this fact explicit in the published paper. It tripped me up early on before Jed corrected me.)

Quote from JedRothwell

I believe you are confused. It makes no difference how the heat leaves the reactor. Conduction, convection or radiation all end up heating the air inside the calorimeter chamber. We measure the temperature of the air as it emerges. The air flows through, cool going in, warmer coming out. The way in which the reactor heated that air cannot be detected, and it makes no difference. Every joule of heat coming out of the reactor must end up heating the air. It is not possible for any heat to leave the reactor and not heat the air, because the chamber is well insulated with reflective insulation (see the ICCF21 paper).

Some of the heat from the air ends up escaping from the walls of the calorimeter chamber, rather than being captured by the emerging air. We know how much. See Fig. 2. That is the "heat capture efficiency" of this calorimeter. The heat capture efficiency from the reactor to the calorimeter chamber is 100% at all times, in all conditions, with any type of reactor, no matter how the heat emerges from the reactor into the air (conduction, convection or radiation).

Thank you for responding to my post. Respectfully Jed, I disagree.

Situation 1: (Emissivity 85% calibration unit) -- If the heat is radiated out of the surface of the stainless steel cylinder (reactor or control) more of it makes its way _without being absorbed_ through the air in the convection chamber to the walls, thereby bypassing the air. The insulation on the inside of the calorimeter chamber is not a perfect reflector and is likely more absorbent than the optically transparent air at IR wavelengths.

Situation 2: (Emissivity 7.5% test unit). Less heat is radiated out of the surface of the stainless steel cylinder so more heat is directly absorbed by conduction at the air/cylinder wall interface, thereby being transferred to the air in the air mass flow calorimetry (and raising its temperature). Less heat is absorbed thru the insulation to the calorimeter outer walls, so less heat escapes measurement by the mass flow method. The method the experiment used captures more heat when the cylinder has 7.5% emissivity then when it has 85% emissivity. This makes it more "efficient" than the provided calibration.

Quote from THHuxleynew

Worth noting that conservative errors are still problematic:

(1) They prevent calibration data from checking the "first principles" absolute measurements, unless the error can be proven identical in the control and active reactors.

(2) They make "first principles" measurement less easy to self-check. Some non-conservative error (normally leading to > 100% efficiency and therefore detected) might not be detected due to a conservative error, that was not present under different conditions.

The importance of cross-checks in this type of data cannot be over-emphasised, given there are quite a number of possible error sources.

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True and true. Right now we are going with what is published. Future experimenters replicating the experiment can fix this. That's scientific freedom. We can't hold Mizuno or Rothwell accountable as they are their own scientists. If we don't like their technique we can redo it ourselves. I only comment here to help other people figure this thing out. I think we will get proof positive shortly -- if we can replicate R20 225 watts out for 50 watts in (my correction of the 300 watts to a more conservative 225 watts).

Quote from JedRothwell

No, it does not. The calorimetry is done by measuring the air temperature at the outlet, and subtracting the inlet (and ambient) temperature. The conductivity, radiation and convection of the reactor has no effect on this. The size and shape of the reactor has no effect. The whole point of an air flow, liquid flow or Seebeck calorimeter is to eliminate such effects. The heat is measured after it leaves the reactor, not while it is leaving.

If you use the reactor surface temperature to do calorimetry, then the issues you raise might have an effect. Not much of an effect. There is no way they could cause a 250 W error.

Agree:

"If you use the reactor surface temperature to do calorimetry, then the issues you raise might have an effect. Not much of an effect. There is no way they could cause a 250 W error."

(By my calculation, it's at MOST a 60 watt error on the R20 300 watt out test.)

Disagree:

"The size and shape [and finish] of the reactor has no effect. The whole point of an air flow, liquid flow or Seebeck calorimeter is to eliminate such effects. The heat is measured after it leaves the reactor, not while it is leaving."

The reactor is an air cooled "radiator", in the sense of a car or airplane radiator. Radiators reject (transport out) heat by the 3 ways: conduction, convection, and radiation. If the radiator is polished silver, it rejects less heat by radiation than if it is painted black. Less heat by radiation means more heat by conduction and convection. Because most of the heat is being rejected by convection, but some of the heat is being rejected by other means (i.e. your 77% heat capture efficiency calibration at 360 degrees implies 23% is being rejected by conduction or radiation). By substituting a polished silver cylinder for a rough natural stainless finish, the balance of heat rejection by convection is changed. Exactly how is not a trivial task, so that the only way to eliminate this is by using a thermally identical polished silver cylinder for calibration. In my post I estimated the maximum error that this could have put into the COP (output/input) analyses.

Adjustment for D2 gas at different pressure, vs control:

D2 gas, like it's H2 cousin and helium are very thermally conductive even at low pressures. Their thermal conductivity is reduced as pressures drop below 1 Torr (133 Pa) , falling to less than 1% of their 1 bar value around 4e-4 Torr (0.05 Pa).

For this experiment, the D2 is contained in the conflat except at the gas/vacuum supply line. Because all heat must pass out of the conflat thru either the walls, or the supply line, we need to estimate the variability of the heat being removed from the supply line vs the control at different temperatures and pressures.

So:

1) The control reactor should be mechanically (and emissivity) identical to the test reactor (if not the same) so that any different conduction, convection, or radiation paths thru the metal are the same.

2) The control reactor if possible should be measured with D2 in the test (but without Ni/Pd mesh) to eliminate the difference in thermal conductivity.

3) Absent (2) or (1), we need to estimate the change in the thermal conductivity from the reactor out thru the supply line (either because the control reactor doesn't have a supply line, or because the control reactor doesn't have D2 in the supply line). My suggestion would be to put a thermocouple on the conflat at the supply line exit, and then to put a thermocouple on the supply line at a the lesser distance of a few cm short of the first mechanical branching off (where conduction out would be expected), or 0.25 meters away from the conflat. This will allow (by having two temperature sensors) an estimate of the conduction losses thru the supply line and its gas.

In general introducing high conductivity gas in the supply line will remove heat from the test conflat, making heat rejected into the air mass flow calorimeter lower, and thus being conservative. However any relative increase in heat (temperature) noted at lower pressures significantly below 133 Pa may be due to lower gas conductivity out the supply line compared to pressures above 133 Pa.

Lastly, it might be interesting to note if there is a significant temperature difference between the longitudinal center of the cylinder near the mesh or the heater, and the end far away from the heater with and without D2 (i.e. under vacuum < 0.05 Pa). if the temperature is more evenly distributed by a significant amount (say more than 5 degrees) in the case with D2 loaded, the heat will have more surface area to conduct out the supports and to convect to the air mass flow. By measuring the temperature difference it can be established that the upper bound is not significant for the calorimetry. If it is significant, again, I prefer that the calibration be done in the test reactor with the mesh removed under same D2 pressure.

Lower emissivity on test reactor vs control makes calibration inaccurate:

If the emissivity on the test reactor is decreased to 0.075 vs. 0.75 on the control reactor, less heat leaves the test reactor by radiation at a given temperature. This means that:

1) the test reactor surface temperature will get higher in equilibrium;

2) more heat will leave the test reactor by convection into the air mass flow calorimeter.

This means that the air mass flow calorimeter efficiency (what is called in the paper the % of heat captured by the calorimeter) would be higher.

Worst case assume the efficiency of the calorimeter is 100% instead of the reported lower numbers. This lowers possible COP on the 50 watt R20 test reactor from 300/50 = 6x to 225/50=4.5x, i.e. the grey line in Figure 6 instead of the orange line. Still a strong result, but not as good.

It does however call into question the entire R19 reactor series data in table 1 (the June 18, 2019 paper) where it lowers many of the COPs to around 1.1 to 1.2 from 1.34 to 1.51. For example, look at the first table entry on 2/20 with the reactor at 238.90 C and Out/In at 1.39. The calorimeter calibration correction factor is show in equation 2 on page 11 of the JCMNS 29(2019)1-12 paper as O/i = 0.95 - 5.0811e-4 x t. Substitute 238.9 and get O/i = 82.86%. Multiply the COP of 1.39 by 82.86% and get COP = 1.15. Clearly a lot less signal in the noise.

(Note: the new correction factor equation was not published in the June 18th paper, but instead "The calorimeter performance has been stable. These critical parameters values have not changed significantly since 2017.)

Now compare the best run on 4/15 at T=381.00 and O/I = 1.51. Correction factor from the JCMNS equation 2 is 72.86%. The corrected COP now becomes 1.1x.

I believe we (and Mizuno/Rothwell) should prioritize the R20 COPs > 2 as these experiments are a lot more forgiving, if calibrated with published fan input power, then the COP 1.1 results.

Quote from Ascoli65

The control and active reactors should be identical for any other detail, including the characteristics of the main components. But look at the spreadsheets of the active and control tests performed in May 2016, both run at a nominal input power of 120 W:

Agreed.

Why is the finish not the same on the control vs the active reactor? (See figure 6, red box -- reactor is dull. Emmissivity of weathered stainless steel is 0.85 vs. polished stainless steel 0.075)

This throws off the calorimetry which is clearly a combination of conduction, radiation, and convection on different components of the rig. The calibration needs to be repeated. I would prefer if it was repeated using the heater inside the R20 unit and not this calibration reactor.

Quote from JedRothwell

Nope. It is around 380 deg C.

Hi Jed,

I've done some work and thus far all my calculations on your calorimetry tie out within a few percent. (Note your above 380 degree answer on the R20 reactor is close to what I expected for a 300 watt output.)

I am checking my work and have a few questions.

1) Can we assume that the R19 and the R20 reactors are the same length and outer diameter as whatever was used for calibration in the calorimeter.

2) I am convinced that for the air convective cooling is more or less proportional to mass flow which is constant at whatever motor power you are running (assuming equal impedance to the cooling air flow which is reasonable if the R19 and R20 reactors are of the same dimensions as the calibration unit). Is the motor power constant at 6.5 watts. Is the air flow impedance the same (i.e. wires, heaters around the units).

2a) Did you calibrate with an empty R19 or R20 unit before you loaded with D2? (This is not necessary if the calibration unit is thermodynamically identical to the test unit).

3) It would be helpful to me and others if you would release on your table 1 the following additional data: Tinput (air), Toutput (air), and motor power. This would help corroborate my internal calculations and the reasonableness of the calibration.

4) It would be helpful if you would release the following additional data as a table for the R20 test cited in the most recent paper: Tinput, Toutput, motor power, Treactor.

5) It would be helpful if for the calibration unit you would release the following additional data as a table: Tinput, Touput, motor power, Treactor [EDIT: and reactor electrical input power for each data point]. [Edit: ideal would be stepping the reactor heater in 50 watt increments from 0 to 300 watts (and giving it time to settle).]

I want to emphasize that this work looks good and that there is likely not a problem with the calibration. I am extremely hopeful that members of the community will be able to replicate.

Thank you,

Anonymous Helper