Mizuno reports increased excess heat

Quote from JedRothwell

There is some understandable confusion about the outlet. Here is some information from an unpublished paper. This is translated by me:

"The blower exit is rectangular, 58 × 38 mm. The wind velocity in a square outlet is uneven. It is better to measure it some distance from the fan, so a 200-mm long cover pipe cylinder made of paper (Fig. 1-4) was attached in front of the blower outlet. One end is rectangular to fit the blower, and the other end is circular, 66 mm in diameter. This cylinder can be made of paper or plastic. 2-mm-thick urethane insulation is wrapped around the outside of the cylinder."

The traverse test covers 6 cm, so that leaves out only 6 mm to the edge, 3 mm on each side. The wind speed 3 mm from the wall of the pipe is the same as the wind speed in the middle, to the limits of precision with this anemometer. It is not possible the wind speed is zero 2 mm from the wall, or 1 mm from it. Somewhere at a fraction of a millimeter, the speed is down to zero. That's basic physics. I wouldn't know where, but you can see from the calibrations that the error is not significant.

The Reynolds number is 18,000, which is far about the textbook value of 2300 where turbulent flow begins.

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Writing a summary on another thread, and having there been accused of laziness, I realise I need to revisit this.

Are the anenometer air velocity measurements used to calibrate the blower made in the 58X38mm outlet, or in the 66mm tube? I believe the 66mm tube.

Are the traverse tests used determine average velocity made in the 58X38mm outlet or the 66mm tube?

I believe the 66mm tube.

Assuming all is in the 66mm tube we have at 3.5 m/s Re=12,000 (18,000 would be 5.25 m/s)

This is well into turbulent flow but that does not mean velocity profile is flat.

From: https://www.chegg.com/homework…ile-shown-figur-q13459661 or many other places.

we get v = (1 - r/R)^0.2 (power law, Re=10,000, answer is not highly sensitive to Re).

Set R=1 to do the integration of 2*pi*v(r)dr / pi*r*r

average velocity is then integral^{r=1}_{r=0} 2*pi*r(1-r)^0.5 / pi = int (2r*(1-r)^0.2) = 0.76

You can see that looks about right from the picture (noting that larger values, nearer the edge, comprise more of the total area).

This is a 24% difference between calculated and actual heat flow. Worth noting (basically subtract 20% or maybe 24% from the results). It is unclear why this was not seen in the calibration data, maybe this was done at a different time, or on a slightly different system? Or maybe there is some other exactly compensating error.

{ Using, RB please note, online Wolfram math Integrate[2x(1-x)^0.2,{x,0,1}] }

Confirming the ~ 20% difference between middle of tube velocity and average over area velocity.

It's interesting that so many people are eager to rush to replicate what Mizuno claims to have done. It would be quicker and more efficient for one or more of the more skilled of these folks to try obtaining an invitation to Mizuno's lab to go over his work in meticulous detail. And oldguy , nobody seems to get the concept of using a simple heat flow meter on the hotter and more powerful Mizuno reactor to do a "quick and dirty" measurement of surface heat flow and consequent rough output power calculation to see if the result is even in the ballpark. It's not rocket science. It's quick and inexpensive. And I love it when JedRothwell concludes that I know nothing about heat flow measurements, I offer to link him (privately) to my peer-reviewed published technical papers on the subject and he ignores it.

Suppose an amateur scientist in his garage replicates (or fails to replicate) Mizuno's result. Are we really going to know that much more? Especially since some advocate varying and deviating from the cookbook method.

Ah well... LENR quest is one messy and inefficient affair and it is easy to see why.

Quote from RobertBryant

The THHnew radiative forcing calculation is in error because of the omission of the area variable

and because the model is wrong. The reactor emits and receives radiation from box..

It does not receive and emit radiation from the ambient directly.

One CANNOT model the heat transfer this way.,,its just wrong.

One needs to take into account the area of the reactor and the area of the box.

The box area at 2m² is ten times the area of the reactor at 0,2 m²

The emissivities differ, reactor ~ o.2... box ~o,o4.. There are three variables involved,, not one.

Area ,Emissivity and T.

Heat radiated = constant x Area x Emissivity x T^{4 ..} The 67.5C calculation is meaningless.

The real situation is much more complicated as the convective exchange to the air exiting the box

does not increase linearly with temperature

because the convective currents within the box get stronger with temperature.

In addition we cannot assume emissivities stay constant with temperature

For example btw 240 and 380C the emissivities of stainless steel decrease by 10%.

In addition there is no guarantee that Stefan's T⁴ dependence holds strictly

when the reactor surface may be affected by strong magnetic fields.

It is better to do an experimental model with a cooktop as I have done.

A 488W input to the 0.027m2 cooktop results in a surface temperature of about 500C,

This compares with the 295W output of the Mizuno reactor giving 380 C.

Yet with the hotplate at 500C , the blower body only situated 20cm above the hotplate only gets to 33C

I would suspect from these results that in the R20 expt ..the blower body never gets higher than 40C

which replicators can verify. The convective air currents cool the blower

to less than the adjacent box regions because of the higher air velocities in the blower

The THH indirect - heat- radiation- to -the- RTD hypothesis is groundless.

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We agree that it is difficult to model. You are the one claiming you know the value is < 40C.

Could be. Or not. All I can say is that your experiment proves nothing if your cooker top is 221C. If you can tell me you have measured 500C then I will agree your experiment has merit.

However, if you are comparing heater top temperature vs power with Mizuno reactor the same, this is clearly wrong. The cooker is not surrounded on all sides by foil-coated insulation, so for given power may be much cooler!

I'm not that interested in this because it is OT for this thread, but if we do try to estimate things and make definite statements we should do so correctly?

Quote from THHuxleynew

You are the one claiming you know the value is < 40C

You claim /calculate 67.5 C.. then you say too many variables to calculate. why even bother??

I put 488W and 500C 20 cm below blower outlet and get 33C..

this is definitely below 40C

Mizuno's reactor is cooler than 500C...with less output W

I expect the temperature to be less than 40C.

Quote from THHuxleynew

I'm not that interested in this because it is OT for this thread

Since when is it OT for this thread!!!!... because you finally decided its a foolish argument on your part?

You spent 9 hours of posts on this very issue.

Quote from RobertBryant

Since when is it OT for this thread!!!!... because you finally decided its a foolish argument on your part?

You spent 9 hours of posts on this very issue.

Calm down?

It is OT because you think I am doing something nefarious, and argue, when I am just casually interested in working things out, with no particular point (for this thread) since it does not seem to go anywhere. But, when you say things that are not substantiated, I feel it is my duty to call you out, as above.

Quote from THHuxleynew

because you think I am doing something nefarious

You are not nefarious .. just unrealistic... your claims are just not supported by evidence

Your calculations are unrealistic... dreamed up with no practical experience

but they are definitely not OT.. I commend your on your sacred duty.

Note about 500C: area of hot plate = 0.028m2.. rt=25C, power output =488W

Use emissivity =1.0(perfect black body ) check that

you get a temperature of 476.5 C or so for 488W

then check the hot temp for a realistic emissivity of 0.85.

Quote from THHuxleynew

RB - I'm in awe of this experimental effort.

If you want to check your rangehood flowrate with a velocity traverse.

I suggest an S-type pitot and a water manometer .. <$2

You can put a small hole in the outlet pipe tape over afterwards.

Mine was 12 cm ID but I managed to get seven readings across it

with a fairly flat velocity profile.

I got pressure readings of average 0.05 cm and 0.22 cm water on low and high settings

You know, I even managed to do pressure readings 0.5 cm outside the duct.. either side... the pressure reading was 0.0cm!!

This was just make sure I did not get a 20% or 64% error

as you have suggested for a standard velocity traverse .. which engineers have been doing for yonks... even when I was young.

I am in awe of Mizuno's experimental effort over 30 years. He is far better than me as an experimentalist.

Quote from THHuxleynew

Difficult to understand how you get X60 - this might be seconds to minutes etc, but that does not make sense here. Voltage measured over R as proxy for current would mean a resistance of 0.0167 - which is not an E24 value. I guess it could be a homebrew 1/60 potential divider on the voltage measurement, but that is a weird division factor to choose.

To measure high currents, you need to put a low resistance shunt in series with the load, otherwise the shunt will burn up. Shunts may be specified in amps-per-volt or max current plus measurement voltage instead of milliohms. It looks like he used a 60A per V shunt (maybe 6 A, 100mv). That is not a weird value for a shunt. They come in many values and current ratings, not just the standard resistance values for 1% resistors.

High precision shunts are much better than homebrew shunts which would be subject to resistance changes when they heat up. Some test equipment has high-current shunts built in for measuring currents.

Quote from Robert Horst

homebrew shunts

You can make a reasonable homebrew high current shunt by taking a copper clad carbon welding rod and removing some of the copper near the middle of length.

Quote from Robert Horst

To measure high currents, you need to put a low resistance shunt in series with the load, otherwise the shunt will burn up. Shunts may be specified in amps-per-volt or max current plus measurement voltage instead of milliohms. It looks like he used a 60A per V shunt (maybe 6 A, 100mv). That is not a weird value for a shunt. They come in many values and current ratings, not just the standard resistance values for 1% resistors.

High precision shunts are much better than homebrew shunts which would be subject to resistance changes when they heat up. Some test equipment has high-current shunts built in for measuring currents.

That is a creative idea, to look specifically for high current shunts. Could well be used, although the application here leads to relatively low currents (500W, 100V => 5A max).

However, the sample values we have recorded put some limit on that. From Jed's spreadsheet:

V = 1.1730

A = 0.5400

P = 38W

That would not work with unscaled V, because the heating element has a resistance of around 20 ohms.

In addition, I was not familiar with typical shunt resistor values. Looking at all current sense resistors from Farnell:

1, 1.5, 1.6, 2, 2.5, 3, 3.5, 4, 5, 7.5 (+ E24 values)

So you do get 1,2,2.5,3,4,5,7.5 as "round" numbers, but not 1.667

THH

Quote from JedRothwell

That does not work. It happens to be close to the numbers I posted with 4 digit accuracy, but it does not fit with more digits, or with another section of the spreadsheet chosen at random. Besides, why would anyone come up with those numbers? What physical constants do they represent? Here are the numbers I posted before, plus another section of the spreadsheet, and the computed value minus the measured value. Your formula comes up with different numbers. Mizuno's spreadsheet would give the same numbers shown here, if he had used that formula. Spreadsheets don't make rounding errors with this many digits. No formula with three terms will give the actual air speed numbers. You might come up with something if you keep piling on terms, but why would anyone do that? If Mizuno was trying to fool you, he would just use a random number generator. He would not go to the trouble to make a multi-term equation that hides the transformation. If he (or I) wanted to use the motor power instead of the anemometer reading, we would do that, and say that is what we did. We would not make an elaborate equation to do that. Multiplying by one constant (0.583) comes close enough for any practical purpose.

Blower power (W)	Air speed (m/s)	* 0.583436 + 2.010436	Actual - computed
3.713216	4.176865	4.176860	0.000005
...	...	...	...

This is the formula used to calculate the Air Speed (AS) values on the basis of the Blower Power (BP) values

AS = A exp(-BP/w) + B

where the values of the 3 parameters A, w, and B are very close to the following:

The following table shows the application of the above formula to the BP and AS values that you have published:

The last column demonstrates (one more time!) that the Air Speed values doesn't come from an anemometer reading, but they have been calculated starting from the Blower Power values by using a mathematical formula.

All 水野忠彦 JP applications and patents [1-20] and [21-38]:

https://astamuse.com/ja/search…9%87%8E%E5%BF%A0%E5%BD%A6

Use Google Translate, e.g. :

https://translate.google.com/t…ed%2FJP%2FNo%2F2018155708

[Electrode for heating device, heating device and heating method]

Quote from JedRothwell

Very close does not count.

It counts in the limit of the precision of the data that have been made available. You published a set of Blower Power (BP) and Air Speed (AS) pairs of data with six decimal digits and the relationship that I wrote above reproduces exactly all those data up to the sixth decimal digit!

If the BP and AS data in your spreadsheet has more than six decimal digits, please, let us know their values and I will try to improve the precision of the 3 parameters A, w and B.

Quote

The numbers you stuffed into the equation do not work, and even if they did, they do not represent any physics.

They work. The above formula and the 3 parameters allow to calculate AS values which - after their rounding off to the sixth decimal digits - are identical to the published values.

Quote

Why would Mizuno use such a weird equation when he has the numbers coming in from the anemometer? What you say makes no sense.

Good question. Ask him. In his 2017 article on JCMNS (1), he wrote:

BTW, if you have any doubt on the meaning of my equation, you might also ask him which physics is represented by his A, w and B parameters.

You recently replied to THH (2): "All measurements for all cells all done the same way, with the same instruments, in the same calorimeter. Why would he do it any differently for this reactor?"

May I ask you the same question? Why would he do differently this time?

(1) http://lenr-canr.org/acrobat/MizunoTpreprintob.pdf

(2) Mizuno reports increased excess heat

For the current shunt, you usually want to pick one rated just a little above the max current you need to measure in order to get a high enough voltage to minimize noise. Using a 100A 75 mv shunt to measure 1 amp gives you a reading of just .75 mv.

Here is a datasheet for a wide range of shunts with ratings from 1 to 20000 A

https://www.gmc-instruments.fr/src/download/dShunts_gb.pdf

They have one that is 6 A, 100 mv (60A per V) that would have the same divide by 60 as shown in the charts. With a 220V heater, that could measure heater power up to 1.32 KW.

But other values are easier to find, like this one that is 10A 100mv.

https://www.newark.com/empro/h…hunt-100mv-10a/dp/99B0936

Quote from Ascoli65

This is the formula used to calculate the Air Speed (AS) values on the basis of the Blower Power (BP) values

AS = A exp(-BP/w) + B

where the values of the 3 parameters A, w, and B are very close to the following:

A=	-6,2990950
w=	5,5507040
B=	7,4035103

T

The last column demonstrates (one more time!) that the Air Speed values doesn't come from an anemometer reading, but they have been calculated starting from the Blower Power values by using a mathematical formula.

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Agreed Ascoli. I ran this non-linear regression and got slightly different answers:

A = 6.38827916

w= 5.78290823

B= 7.53828388

My results also agree to Jed's to the 6th digit to the right of the decimal point, having a residual standard error of 2.1e-7.

Quote from Ascoli65

Good question. Ask him. In his 2017 article on JCMNS (1), he wrote:

The wind velocity at the flow meter was estimated by semi empirical Eq. (5). V = A exp(-Wb/w) + B; (5) where A is a constant, -3.7; B = 4; w = 1.375; Wb is the blower input (W);

BTW, if you have any doubt on the meaning of my equation, you might also ask him which physics is represented by his A, w and B parameters.

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Entirely agree Ascoli.

We know that the wind speeds and the power did not come from independent measurements in Jed's provided data, because they exactly match.

The equation V = A exp(-Wb/w) + B

is suspect, because the airspeed is non-zero at zero input power to the blower because the exponent goes to 1 as Wb goes to zero making V0 = A - B.

[Subsequent Edit: this equation fits empirically better]

At infinite power V goes to Vmax = B.

Are we suppose to believe that V0 is some kind of "convection" bias? Maybe. [Subsequent edit: probably but not certain that is why this equation fits better] But then it is surely based on the difference between reactor or control tube temperature and room temperature. See as it appears that both the blower and the output conduit are on the top of the calorimeter box, am having a problem believing that the V = 0.3 m/s for the 2017 experiment, or using your calibrations (or mine) 1.1 m/s. I suggest that B = A and by having a different B != A (!= does not equal) suggests an extra emperical degree of freedom in the power vs. airspeed formula to make it fit better which is undoubtedly contained somewhere but not shown to us.

I suggest that it makes more sense to fit real world

V = Vmax * (1 - exp(Wb/w)

where Vmax is a limit of wind speed due to turbulence effects and fan inefficiency at high back pressures.

[Subsequent edit: this equation doesn't work as well as the V = A exp(-Wb/w) + B -- see my subsequent numerical investigation in the following posts]

Of note it that the above equation when fit (you can use solver if you only have excel) gives Vmax = 5.9766382 and w = 3.0938366 for standard error of 1.43e-6 and a worst error of 0.000003. Therefore, I suggest at least for these data the use of the third factor B is unnecessary as we are overfitting.

To do this right, we need real calibration data for blower power and airspeed at input powers between 0 and 7 watts as the spreadsheet snippet is way too small. Mizuno et al has this in Figure 4 of the most recent pre-print data.

Quote from JedRothwell

The problem is that spreadsheets are all more precise than than 5 decimal places, and this one shows different answers than yours. So you got the wrong answer. If the function you wrote here gave the right answer, it would agree with Mizuno's numbers to more than 5 decimal places.

1) I fit the original 49 points (post 1076) to all the precision you gave us, (4 decimals to the right of the decimal point). You subsequently gave me 45 points (post 1111) or which points 16 to 45 are entirely new, but all to six decimals to the right of the decimal points. The new linear fit to the NEW DATA matches to 5 decimal points. This is because it appears (without seeing the formula in your spreadsheet and the parameters) that you generated airspeed from blower power using an exponential of the form

V = A * exp(Wb/w) + B.

The first derivative of V with resepct to Wb, i.e. dV/dWb is exactly the slope of the linear regression fit that I came up with. It fits, but it is not the formula that you used.

Could you please provide us with that formula and the used parameters for A, B and Wb?

2) Using Figure 4 from your Mizuno et al ICCF-22 preprint, I agree that for wider ranges, an exponential function would fit better. Is there any data at the zero point of the graph. Can you provide us with the table of Center V velocities vs. Blower power?

3) Do you have any airspeed (i.e. non-zero) measured at zero blower power from convection. What temperature was the control reactor cylinder at when that airspeed was measured, or was it zero? If the airspeed is zero then this equation would fit better: V = Vmax(1 - exp(Wb/w)) so that at zero blower power you get zero V.

4) I am almost certain that after calibration airspeed used for the calorimetry was derived from blower power rather than a live anemometer in the rig. Can you confirm that is the case?

[See my continuation of this analysis in post #1163 below]

Typo in the caption to Fig 8 p6 ?

Figure 8. Calibration with blower power only, and no power to the reactors. The temperature rise of ∼3.5◦C is probably caused by waste heat from the blower motor reaching the outlet RTDs.

But the graph shows about 0.35◦C : Vertical axis (Temp Difference) is 0.1 0.2 ... 0.5 C

Also : Is there a model number for the blower? Is it centrifugal, or a fan?