For the same reason that an accountant will dig into the data if there is a 1 cent discrepancy
Perhaps the reason for this obvious nitpicking is different from an accountant's
An accountant is accountable for what they write.. and there is some small remuneration involved.
However neither THHnew or Ascoli are accountable... although there may be some peanuts
Yes, it would be wonderful if everyone involved would voluntarily step forward, identify themselves, and tell us exactly what they are doing, who is paying them, and what they have seen
bob2 thh2 ascoli65 .. perhaps they can put the first foot forward?
The anenometer is specified +/- 8%.
where is +/- 8%. specified?
is the accountant fudging the figures? or is there other data?
I had an interesting idea regarding testing the quality of the output air flow rate measurements while sitting in a plane today.
where is +/- 8%. specified?
is the accountant fudging the figures? or is there other data?
Re: 5%
If I haven mucked up the math, that works out to be about +/- 1 W inside a 65 mm diameter tube at 20 C at a reading of 4 m/s. A positive 2.5 mm diameter measurement (67.5 mm) error makes about 1 W error also at 20 C and 4 m/s air velocity.
(There is a good chance that I did muck something up since it is late, I’m tired, and I didn’t make a very thorough job of it by using a manual calculator and several unit conversions to get the answer. But I think that it is close...)
Re: 5%
If I haven mucked
This does not relate to the anemometer specifications... but to your calculations
http://www.trinergy.co.th/prod…?gid=1-001-008-009&id=665
Here , the CW60 anemometer , used by Mizuno has 3% accuracy
Display MoreSorry, THH, maybe I'm not able to explain correctly what I mean, but no, this is not my point. Forget the ripples, please. Assume there is none.
As you correctly said, V*I and wattmeter results should always be ROUGHLY the same, as shown in the second graph of my jpeg (*), but NOT EXACTLY the same as shown in the first graph of the same jpeg.
As for the control test, the discrepancy P-VxI - which ranges from -0.4 to +0.4 W - and the difference in the P and VxI shapes assure us that the "Input power" signal came from the wattmeter. Pay attention, please, not from the cheap portable wattmeter (Sanwa WattChecker, 75$) brought with him by JR, but from the expensive wattmeter (Yokogawa PZ 4000, 16000 $) which was connected to the HP data logger.
On the contrary, the substantial identity between the "Input power" values and the "V/DC*I/DC" product shown in the active test spreadsheet strongly suggests that the original data coming from the Yokogawa wattmeter were substituted, in this only case, by the values obtained by simply multiplying the V/DC and I/DC columns.
No, as said before, forget any AC component. A COP of 2 or 6 can only come from a big mess in the input circuitry. IMO the most suspicious candidate to be the cause of these huge errors is the shunt resistor used to measure the DC current.
No need to manually enter the wattmeter values. As said before, the Yokogawa PZ 4000 wattmeter is directly connected to the HP data logger. You are making confusion with the portable Sanwa WattChecker wattmeter plugged in by JR.
Yes, I assume that the Yokogawa PZ 4000 wattmeter is placed upwards of all the circuitry, hence upwards of the PSU too.
Anyway, I repeat the main question: why these "Input power" values directly measured by the Yokogawa wattmeter and recorded by the HP data logger appear only on the control test spreadsheet and not in the active test one?
Hi Ascoli,
I'm not quite sure what you think the difference between us is. To try to capture this:
(1) V * I calculated power vs Wattmeter measured power - we are agreed about the evidence here, and what was done.
(2) Ripples. Logically, using average V*I leaves this open. Small ripples would not matter in this context. But an unregulated PSU would matter - and given the two different measurement methods noted in (1) it would generate COP=2 without much difficulty and without this being obvious. I don't expect this to happen, but can't rule it out.
(3) Yokogawa Wattmeter connected directly to automatic logging somehow - agreed that looks likely, in which case why not use it all the time? A question worth asking.
(4) Yogokawa Wattmeter on input side. See below.
Now, this all makes sense. But it is wrong. I do not claim to be expert on calorimetry etc, but I am on anything electronic. If the PSU is not regulated (for example a transformer feeding a full-wave bridge ) there will e substantial ripple. In any case that would not allow adjustment of the heater power. If the PSU is regulated, for example a DC lab PSU, as would be expected, then it will not be 100% efficient. Nor will it be 95% efficient. Linear (old fashioned) regulated PSUs are highly inefficient. For example, a linear 100V lab PSU, as Mizuno says he used for the R20 tests, would have efficiency < V / 100 where V is the output voltage,. At an output of 20V it would have efficiency of only 20% - less since the unregulated input to the regulator would have to have some headroom above 100V. In practice a PSU might have multiple switched voltage ranges, allowing higher efficiency, but the 20% figure is very possible and efficiency will vary between 80% and something small according to output voltage and range. A more modern switching PSU will be more efficient. Fixed voltage switching PSUs can be as high as 90%, although 75% is quite normal. Lab PSUs, which must have varying output voltage, have efficiency that depends on the output but ranges between 50% and 80% typically.
Regulated PSU lost power always has two components. (1) as Jed says quiescent power (usually small). (2) efficiency, as above. Depends on output voltage and is normally relatively independent of output current although it will also vary with current. A conversion from input-side Wattmeter measurement to output power would be approximately accurate if made as Pout = Pwatt*E - Q where E (eficiency0 and Q (quiesecent power) are TBC and E depends on the output voltage and the PSU range.
However, triac switched PSUs are as Jed says. there is no attempt at regulation, the average output voltage is adjusted by changing the duty cycle. In this case I can guarantee that the measurement methodology shown will lead to a large apparent COP, enough to explain R20. I would not think this likely here except that (unregulated) triac-switched supplies are very efficient as Jed says the PSU that Mizuno used was.
OK - so what does this wattgate issue mean?
I think we need more information about methodology. Best guess from spreadsheet data would be:
(1) the active run V*I calculated power is more or less correct (unless a spiky PSU is used, e.g. triac switched as per Rossi, in that case all bets are off).
(2) the calibration run calculated power from wattmeter measurement with some adjustment could be very inaccurate and would underestimate the real calibration heater power by a factor that could be anything.
Ironically, this actually would explain, in principle, the R20 room heating results. The R20 was driven from a lab PSU. If this is a relatively inefficient linear PSU it could dissipate large amounts of power, emitting more heat than the reactor and heating the room. I think 3kW would be highly unlikely. But 1-2kW, for an old high voltage high current PSU, possible. The difference between calibration and active results would come from the differing measurement methodologies used. However, there is one problem with this scenario. The V*I values for active run input would be quite accurate. The calibration input would be lower. That does not make sense given the calorimeter absolute response.
So I don't see this alone as explaining the R20 results, but it does, alone, explain the very different calibration and active run output, unless a triac switched PSU is used for the heater (unlikely). In that case the active V*i measurement is all wrong, the calibration measurement correct, and R20 is exactly explained.
This does not relate to the anemometer specifications... but to your calculations
http://www.trinergy.co.th/prod…?gid=1-001-008-009&id=665
Here , the CW60 anemometer , used by Mizuno has 3% accuracy
RB - no you've got that wrong. It is +-8% (roughly).
1% FS = 0.2. 1dg = 0.1 (for XX.Y - given 0.1 resolution) total 0.3 / 3 = 10%
3% + 1dg = 3*0.03 +0.1 = 0.19. 0.19 / 3 = 6.3%
The numbers vary according to which datasheet you use and what assumptions you make, we should take the largest bound when not sure. In addition, the (Japanese) datasheets I linked earlier had slightly different spec.
1% FS = 0.2. 1dg = 0.1 (for XX.Y - given 0.1 resolution) total 0.3 / 3 = 10%
3% + 1dg = 3*0.03 +0.1 = 0.19. 0.19 / 3 = 6.3%
why use 3m/s when the avg velocity is 4.
the velocity readings are 4m/s on average
giving btw 7.5% and 4.8% on your calculation
8% is an exaggeration.
Display More1% FS = 0.2. 1dg = 0.1 (for XX.Y - given 0.1 resolution) total 0.3 / 3 = 10%
3% + 1dg = 3*0.03 +0.1 = 0.19. 0.19 / 3 = 6.3%
why use 3m/s when the avg velocity is 4.
the velocity readings are 4m/s on average
giving btw 7.5% and 4.8% on your calculation
8% is an exaggeration.
Dear RB: I fully admit to a 0.5% exaggeration, based on the specific figures you use here. Mea culpa, please forgive me.
Mea culpa, please forgive me
Now THHnew has to show how this impacts on the COP of the R20 of 300% or so.
THHnew of course maintains by waving hands that somewhere somewhere there is an error in the R20
but I guess the conclusive debugging will have to wait.
Perhaps a joint AscolianTHH debugging effort will yield something by 2021
The estimate COP is 8-10 at best.
Why do all these "children" discuss about 5% issues that are only relevant for one single measurement ? If you repeat a measurement with a calibration source then the error will be reduced if the result is always the same! E.g. 5 % on the data sheet is based on a random distribution...
But we have 800-1000% excess what's 2- 5%of out??
But we have 800-1000% excess what's 2- 5%of out??
...and they are actually heating a room! 
This does not relate to the anemometer specifications... but to your calculations
http://www.trinergy.co.th/prod…?gid=1-001-008-009&id=665
Here , the CW60 anemometer , used by Mizuno has 3% accuracy
Of course that is for the device in a laboratory calibration setting. Not as used by human in turbulent flow and perhaps not perfectly aligned towards the source, totally free of vibration.
Now THHnew has to show how this impacts on the COP of the R20 of 300% or so.
THHnew of course maintains by waving hands that somewhere somewhere there is an error in the R20
but I guess the conclusive debugging will have to wait.
Perhaps a joint AscolianTHH debugging effort will yield something by 2021
No - I have to do no such thing! If you read what I've consistently written throughout this thread you'd know that.
My initial guess was that the R20 results might be due to a mistake in input side measurement (wrong electrical power in).
Since then, check out wattgate above, there is some complexity discovered and not resolved. But this guess looks more likely now than it did initially.
Again - your comments here assume that understanding Mizuno's paper is a zero-sum game in which one side or the other wins, and you have a pre-declared side. That can't be true. Certainly from my POV I know vastly more about all the issues here than i did originally, and I'm still learning. It is fun.
The estimate COP is 8-10 at best.
Why do all these "children" discuss about 5% issues that are only relevant for one single measurement ? If you repeat a measurement with a calibration source then the error will be reduced if the result is always the same! E.g. 5 % on the data sheet is based on a random distribution...
But we have 800-1000% excess what's 2- 5%of out??
There are two sets of results here: R19 and R20.
R19 is large - but still possibly within what could be got from a collection of systematic errors
R20 is v large and I agree outside the range of calorimetry errors, though not outside the range of measurement mistakes
Why bother? it comes from a more realistic view of results where nothing is certain and it therefore matters how confident we are in any specific result. In this case, if you are convinced by R20 results, you should not be reading this thread - it does not matter to you. Otherwise the validity of R19 (smaller, but much more extensively tested) and R20 both matter.
...and they are actually heating a room!
Must be Rossi, but for me such anecdotal and unsupported comments do not convince...
your comments here assume that understanding Mizuno's paper is a zero-sum game
THHnew's debugging game hasn't changed since 2017
THH2017
"It is academic, because the IH replication failed. Still, I'd like to debug this. But not sure I have the motivation to spend long amts of time on it given the IH work."
The latest debug attempt = Wattgate
what oh!
Perhaps the reason for this obvious nitpicking is different from an accountant's
An accountant is accountable for what they write.. and there is some small remuneration involved.
However neither THHnew or Ascoli are accountable... although there may be some peanuts
I won't get into motivations or psychology. Ascoli just seems to have a skill that involves being able to pay attention to detail and digging into things that most would overlook.
It takes a lot of wasted energy to get upset at Ascoli's "nitpicking".
It would be much more productive for Jed to take a quick look at the most recent data and see if input power is still being calculated in a different manner for control and actual run. If yes, I would really wonder why.
I won't get into motivations or psychology. Ascoli just seems to have a skill that involves being able to pay attention to detail and digging into things that most would overlook.
People overlook these problems because they are imaginary. Only Acoli can see them.
It would be much more productive for Jed to take a quick look at the most recent data and see if input power is still being calculated in a different manner for control and actual run. If yes, I would really wonder why.
Nope. Same instruments and methods. Note also that the power supply overhead is modest. I don't recall how much, but when it supplies 50 W to a resistance heater, the power supply does not draw much more than that. Around 70 W, I think. There is no digital watt meter that would mistake 300 W for 70 W. There is no likelihood three meters would all make that mistake and come out with the same answer to within a fraction of 1%.
