Research Team in Japan Reports Excess Heat - (Nissan Motors among otheres)

Quote from bocijn

Are you suggesting that the HEX error is up to 0.26 +20%= 0.31W??? or =+20% of 5W = +1W??.

I am calculating the fractional standard deviation of the measured output power per the simplfied equation given in the prior post. To 'suggest' an actual error I need reliable error estimates on all the relevant experimental parameters, which are not available. But that is neither of which you suggest. In fact in my last post I explicitly told you what I was guessing were the noise bands based on a standard error calc. I suggest you re-read my last post.

Quote from bocijn

You wrote B-O-E . What values of ɳ , ΔT , C and ρ are you plugging into your B-O-E?

Doesn't matter. When you use the simplified fractional standard deviation approach, all you need to discuss is the fractional (easily converted to percentage) error of each experimental variable, which were not all given, so I am unable to give a definitive error estimate. I can only approximate from reasonable assumptions, which you may disagree with. But I won't argue about it because the paper's authors were supposed to give them, not make us guess.

Quote from bocijn

For the 80W input energy values Iwamura et al have written error estimates.

oil flowrate error = 0:012(ml/min);

oil temperature change error = 0:261(K);

input energy rate error = 0:031(W);

excess heat energy rate error (HEX) = 0:260 (W):

How did they determine these? Since they do NOT seem to know how to do standard error analysis, I have to suspect their given estimates are likewise determined in a non-standard fashion, and thus are effectively meaningless, since I have no idea how to apply them to calculate the standard errors. Usually, for some strange reason, when people determine errors in a non-standard way, they are underestimated. Go figure.

Quote from bocijn

What are your estimates

oil flowrate error = ? (ml/min);

oil temperature change error =? (K);

input energy rate error = ? W);

I did not estimate these numerically. I applied the usual 'sensitivity analysis' approach of examining the impact if the standard deviations were 1, 5 and 10% of the nominal or maximal variable values. That is all that is needed to estimate the fractional standard deviation of the output power. Then I took the 134W reported output power and calculated my estimated noise bands. My conclusion was they are working in the noise, always a bad thing to do if you want to draw definitive conclusions.

The excess power is another calculation, both for the value and for the standard deviation, but they are simple equations that you can figure out easily. Historically, input energy is usually reasonably well measured, as are temperatures (and thereby differences), but flowrate can be a problem in some cases.

Quote from bocijn

excess heat energy rate error (HEX) = ? (W):

Of course that error will be the square root of the sum of the variance in the input and output powers. (There, I told you one of the equations from the answer just above.)

Quote from bocijn

input energy rate error = ? W);

Pin = I * V or Pin = I^2*R . Use the POE equation to answer your question. You will probably have to guess at the errors since we are not informed how the supposed 'delta' is related to the 'sigma', and error calculations use 'sigma' (technically 's', the estimate of sigma).

Kirkshanahan

"I can only approximate from reasonable assumptions,which you may disagree with"

Iwamura et al estimate

excess heat(HEX) energy rate error (HEX) = 0:260 (W): for the 80W input, 5W excess .

The error in the HEX energy rate is crucial.

You said something like 20% error in variables. Does this mean that you estimate the error could be up to 20% of 5 W = 1W?

For HEX energy rate value a range 5W+/-1W is much less robust than one of 5W+/-0.26W

Plug in any ɳ , ΔT , C and ρ etc values, reasonable assumptions you think realistic.

What error in the HEX do you estimate for the 5W HEX case ?

1W, 5W,10W, 100W?

RobertBryant

Having trouble following things aren’t you? I already answered you above...

OK, last time…

As show earlier, our best estimate of the relative standard deviation (which is the standard deviation of a measured quantity divided by the nominal value of that quantity) (RSD) of the output power of a flow calorimeter, Pout (= k*F*r*Cp*(Tout-Tin), where k = 1/eta=1/n, and Tout-Tin=dT) is given by the Propagation of Uncertainty equation, i.e.

( s_Pout / Pout )^2 = (s_k/k)^2 + (s_F/F)^2 + (s_r/r)^2 + (s_Cp/Cp)^2 + (s_dT/dT)^2 [1]

where (s_dT/dT)^2 = (s_Tout/Tout)^2 + (s_Tin/Tin)^2 [2]

I’m doing a two-step process here where I first calculate the RSD of dT, and then use that in the calculation of the RSD of Pout. (Actually here I'm not ever really worrying about the dT error per se.)

Technically, I need to do the same thing for r and Cp, since they are given as functions of temperature, and I showed that they varied significantly over the temperature range specified. Using an average temperature to calculate them as Takahashi, et al do makes this and makes it difficult to estimate the real variation in those quantities. I would also note a) I suggested they stop using Tavg and put the equations in directly and b) the equations themselves are fits to data and they have an error too, which should be folded in to the computation as well. But these are fine points.

The big problem is that Takahashi, et al, state the error as being (‘d’ is delta)

d(Hex) ~= d(F) * r*Cp*dT/n + d(dT) * F*r*Cp/n + d(W) (W=Pin, Pex = Hex = Pout-Pin) (I’ve left off the absolute value symbols.)

So, by their logic, d(Pin) + d(Pout) = d(Pex) and thus

d(Pout) ~= d(F) * r*Cp*dT/n + d(dT) * F*r*Cp/n

Presumably they want us to believe their delta is the standard deviation. Dividing by the calorimetric expression for Pout we then can supposedly calculate the RSD by their logic.

d(Pout)/Pout = d(F)/F + d(dT)/dT [3]

So now compare equations [1] and [3] and you will see that their method ignores 3 terms, and incorrectly adds the standard deviations instead of the variances. If they were correct in only including two terms, their approach will overestimate the error. To explain, assume for the moment the RSD of the two variables is the same. Their method then gives the RSD of Pout as twice the RSD of the variables. The correct method squares them, add them, and then takes the square root, which means in this case multiplying one of them by the square root of 2 (~1.4) instead of 2 as they use.

Getting back to eq. [1], the reason one converts to RSD is so that one can talk of fractional or percentage errors (as %RSD). One does that because there are common values for %RSD that reflect the ease of conducting an analysis. A 10% RSD technique is easy to do. 5% requires a bit more work to get and maintain, and 1% techniques require considerable work to achieve and lots of work to maintain. Getting better that 1% RSD is very, very hard and usually is limited to certain classes of measurements such as voltages and currents. In any case, if one hasn’t explicitly evaluated the RSDs, use of 1, 5 and 10% RSDs is typical to evaluate the level of effort that would be required to ensure you are working ‘above the noise’.

By eqn. [1], if only one term was important to the Pout’s RSD, then whatever %error it had would be the %error in Pout. Since I have no information on the variation of n, I did the standard B-o-E thing and calculated %RSD_Pout for a 1%, 5% and 10% RSD in n, while ignoring the other terms. Likewise for B-o-E purposes, I temporarily assume the error in Pin is unimportant, thus the RSD of Pout = the RSD of Pex.

In one case Pout was 134W for example. 1, 5, and 10% of that is 1.34, 6.7 , and 13.4W (we have now converted from RSD to SD as these Watt numbers are the actual estimated standard deviations). To decide if you are working out of the noise you must decide how wide the error band is. The error band width is usually calculated as some integer multiple of the SD (SD is sometimes referred to as ‘sigma’ which is usually technically incorrect, statistically speaking). In routine work, ‘3’ is the usual choice, but as I noted before, for higher importance work, that is sometimes increased to 5. That gives us:

3-sigma band = +/- 4.0, 20.1, 40.2W for 1, 5, and 10%RSD

5-sigma band = +/- 6.7, 33.5, 67.0W for 1, 5, and 10%RSD

The numbers above are excess heats that would be considered ‘in the noise’ for the 6 cases we have considered here.

Reported excess heats are on the order of 10W or less. That would imply they would need to be working with 1%RSD techniques. That is very difficult to do. As such my tendency is to believe they are actually ‘in the noise’. Without more replication, they cannot expect people to accept their results blindly.

You should observe at this point that your repeated requests for ‘what value of x did I use’ are not relevant. I used ‘standard’ values for %RSD to evaluate the work, primarily because no actual statistical info was given and it seemed they were ignoring potentially very important parameters in their attempt to guess at their error. And as I noted before, their guesses are grossly too small.

You should also note that when you are attempting to compute the standard error of a calculated quantity as we do above, you shouldn’t be minimizing things. You do that when you design your experiment and choose you instruments. If you instead play games with the POE calc, you just set yourself up for exactly what we see here, namely ‘convincing’ yourself that you are out of the noise band when you’re not. If anything, you should slightly increase the numbers in the POE calc just to increase the confidence your results aren’t in the noise.

Quote from kirkshanahan

OK, last time…

As show earlier, our best estimate of the relative standard deviation (which is the standard deviation of a measured quantity divided by the nominal value of that quantity) (RSD)

One last time: Shanahan's theories cannot explain:

Why the calibrations produce a zero balance.

Why palladium and heavy water work in platinum and light water do not. These differences cannot affect the calorimetry.

Why helium is correlated with the anomalous excess heat.

How a cell could remain probably hot for days, producing ~100 W, with no input power. Shanahan's explanation for this is a gross violation of common sense and the Second Law of Thermodynamics. He claims that the hot object is not a heater and an object that remains hot for days is not being heated. If he believes this he is a crackpot, and if he does not believe it he is trying to deceive stupid people in the audience.

Here we are, talking about powder samples in a gas cell and how to compute random error and up pops Jed talking about F&P systems that only have the use of flow calorimetry in common. Does Jed hate me or what?

Quote from JedRothwell

One last time: Shanahan's theories cannot explain:

Why the calibrations produce a zero balance.

Why palladium and heavy water work in platinum and light water do not. These differences cannot affect the calorimetry.

Already done many, many times, but for the newbies...

'Calibrations' are always assumed to be Pout = Pin. Your comment is actually nonsensical, i.e it doesn't make sense.

And I think you meant "Why palladium and heavy water work *AND* platinum and light water do not". First off, platinum does, just look at Ed Storms excellent work. Light water also works but with Ni electrodes and as I recall a different electrolyte. Second, anyone who thinks about the proposed chemical mechanism for the Fleischmann and Pons Effect will understand why this is so. And it is the getting of the Effect that affects the calorimetry, not the constituents of the cell. The constituents affect the getting of the Effect.

Quote from JedRothwell

Why helium is correlated with the anomalous excess heat.

Technically, it is known as 'small sample statistics'. IOW, it is an accident of the data set. This is easy to recognize when you realize there is little evidence there has ever been any real excess heat in F&P-type experiments (and probably elsewhere as well).

Quote from JedRothwell

How a cell could remain probably hot for days, producing ~100 W, with no input power. Shanahan's explanation for this is a gross violation of common sense and the Second Law of Thermodynamics. He claims that the hot object is not a heater and an object that remains hot four days is not being heated. If he believes this he is a crackpot, and if he does not believe it he is trying to deceive stupid people in the audience.

Ah, the Mizuno bucket anecdote again...fourth or fifth time you've repeated the same old lies isn't it?

Keep beating that dead horse Jed...

There are no liars here, Kirk, only people with different beliefs to you. Please remember that. Alan.

Kirkshanahan

"I already answered you above"

No you didn't

You wrote. "The excess power is another calculation, both for the value and for the standard deviation, but they are simple equations that you can figure out easily"

What error in the excess power (HEX )do you estimate for the 5W HEX case ?

1W, 5W,10W, 100W?

btw "CCS" = "clear communication society"? "Capsanthin/capsorubin synthase"?

Quote from kirkshanahan

Ah, the Mizuno bucket anecdote again...fourth or fifth time you've repeated the same old lies isn't it?

I am curious. What do you mean?

1. Are you saying the anecdote is a lie? That it never happened to Mizuno?

OR

2. Are you saying that you never claimed a hot object is not a heater, and an object that remains hot for days is not being heated?

#2 is a violation of the Second Law. If you deny saying that, I can look up your original quotes. Or, perhaps, you could explain what you meant. OR, you could take it back and agree this is not true.

Quote from snoopy

Giving JedRothwella chance: what do you mean by that? And how is it evidence for the hypothesis?

It is unclear to me who this message is addressed to, or what it means.

Anyway, I meant that Shanahan's theory cannot explain why calibrations work. That is, why calorimeters show a balance of input and output energy where there is no excess (or endothermic deficit). If his theories were correct, you would expect to see Pt-H produce excess heat as often as Pd-D, and both would produce an endothermic deficit as often as anomalous heat.

Shanahan claims his theories cover this, but I disagree, as do the authors he thinks are mistaken. See:

http://lenr-canr.org/acrobat/MarwanJanewlookat.pdf

KIrkshanahan

"CCS problem" From your first post on this thread.

Perhaps you expect CCS to be common knowledge.. but it isn't. not on Google at least

Eureka!!http://lenr-canr.org/acrobat/MarwanJanewlookat.pdf It does not mean "Clear Communication Society" after all

CCS = Calibration Constant Shift.

Did you employ your CCS hypothesis in your intermmetalic tritium research in the last few decades? such as

Tritium aging effects in LaNi_4.25Al_0.75 https://doi.org/10.1016/S0925-8388(03)00139-7

Quote from JedRothwell

Why helium is correlated with the anomalous excess heat.

May be Kirk should know that one source of Helium is extensive Soba-(noodles) consume....(They cook with the excess heat...)

Wyttenbach " one source of Helium is Soba'

Cold soba is nice with the temp. 35C like today..but helium is unlikely unless its a cold soba birthday party.

Tritium has been associated with LENR by number of researchers

Benyo et al 's 2017 Investigation of Deuterium Loaded Materials Subject to X-Ray Exposure

concludes "tritium by an unexpected nuclear effect".. https://arxiv.org/abs/1704.01183

Claytor, T. N., Tuggle, D. G., Taylor, S. F.; Evolution of Tritium from Deuterided Palladium Subject to High Electrical Currents, Frontiers Science Series No. 4, Proceedings of the Third International Conference on Cold Fusion., October 21-25 Nagoya Japan., Ed. H. Ikegami, Universal Academy Press Tokyo Japan., 1993, p217.

Tadahiko Mizuno, Tadashi Akimoto, Kazuhisa Azumi and Norio Sato, “Tritium evolution during cathodic polarization of palladium electrode in D2O solution”, The Electrochemical Society of Japan Vol.59, No.9, (1991) 798-799. in Japanese.

Radhakrishnan, T.P., et al., Tritium Generation during Electrolysis Experiment, in BARC Studies in Cold Fusion,

P.K. Iyengar and M. Srinivasan, Editors. 1989, Atomic Energy Commission: Bombay. p. A 6.

N. J. C. Packham, K. L. Wolf, J. C. Wass, R. C. Kainthla and J.O’M. Bockris Production of tritium from D_{2 O} J. Electroanal. Chem., 270 (1989) 451-458, Elsevier Sequoia S.A., Lausanne.

Quote from Alan

There are no liars here, Kirk, only people with different beliefs to you. Please remember that. Alan.

Oh really? Well, a) I don't think it's possible to be human and not have told a lie at one point or another (Honey, does this dress make me look fat?) , and what is one who tells a lie called? _______ (fill in the blank). So, we are all in that boat Alan.

But more to the point, b) what do you call a person who continually writes:

Quote from JedRothwell

Shanahan's explanation for this is a gross violation of common sense and the Second Law of Thermodynamics. He claims that the hot object is not a heater and an object that remains hot for days is not being heated. If he believes this he is a crackpot, and if he does not believe it he is trying to deceive stupid people in the audience.

when he knows this is false? A falsifier? Delusional? What?

Jed can't support this contention without grossly distorting what I've written multiple times Alan. What do you call a person who deliberately does that then?

Quote from bocijn

Did you employ your CCS hypothesis in your intermmetalic tritium research in the last few decades? such as

Tritium aging effects in LaNi4.25Al0.75 https://doi.org/10.1016/S0925-8388(03)00139-7

Absolutely.

Quote from JedRothwell

Shanahan claims his theories cover this, but I disagree, as do the authors he thinks are mistaken. See:

http://lenr-canr.org/acrobat/MarwanJanewlookat.pdf

Another example of Jed repeating something that he knows is incorrect, having been corrected multiple times, just because it contradicts what I say.

Quote from JedRothwell

Anyway, I meant that Shanahan's theory cannot explain why calibrations work. That is, why calorimeters show a balance of input and output energy where there is no excess (or endothermic deficit). If his theories were correct, you would expect to see Pt-H produce excess heat as often as Pd-D, and both would produce an endothermic deficit as often as anomalous heat.

Also repeating something false after having been corrected multiple times.

Quote

 JedRothwell wrote: 

Quote

Quote

Shanahan's explanation for this is a gross violation of common sense and the Second Law of Thermodynamics. He claims that the hot object is not a heater and an object that remains hot for days is not being heated. If he believes this he is a crackpot, and if he does not believe it he is trying to deceive stupid people in the audience.

Quote from kirkshanahan

when he knows this is false? A falsifier? Delusional? What?

Jed can't support this contention without grossly distorting what I've written multiple times Alan. What do you call a person who deliberately does that then?

I think you might have a point here Kirk... It was the first law that you horrifically cocked-up previously.

Quote from Zeus46

It was the first law that you horrifically cocked-up previously.

Another unsupportable without gross distortion contention.

Quote from kirkshanahan

Another unsupportable without gross distortion contention.

Been reading too much William Burroughs?!

Ta-da: The Playground

zeus

Your confused rendition of what I wrote does not prove I did what you say. You need to review the whole issue again, and actually follow the discussion and activities, instead of partially grasping them.