- Member since Oct 23rd 2015
Posts by Paradigmnoia
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OK, now I better understand your situation. Did you ever published a photo of your experimental setup?
Anyway, the base room temperature is not a problem and not even the cooling or heating of the box. Heating and cooling of a warmer than air component by a fan heater (not an IR one) or an AC unit cause the same response, the component cool down! See what happened during the INFN test in 2012 (*). The only important thing to replicate is the increase of the air speed around the experimental setup, in your case the box, therefore a simple fan is more than sufficient.
As for the inlet temperature to the box, it has no or just a minimal influence on the estimate of the output power. The main variations are on the outlet temperature and, IMO, its value is strongly influenced by the air speed around the box.
There are photos of my arrangement in either in this thread or the sealed Mizuno air calorimetry thread.
The inlet temperature should be the room ambient temperature, and therefore affects the exterior temperature of the calorimeter box, and the rate at which the box loses heat to the environment, (or the rate at which the outer environment can affect the box). I can confirm that without the air gap, the exterior bubble foil was quite warm at just 200 W input, especially the top. Since the bubble foil has very low emissivity, convection carries most of the heat away to the room from the outside of the box, so air speed around the box would have an especially strong effect.
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It's obvious that the overall time constant is dominated by the reactor mass, which is 10x greater than the box mass, but the down spike + recovery transients only involve the box and the air inside it, so no wonder that their time constant is 1/10 of the global one.
Please consider that the air inside the box, whose exit temperature provides an estimation of the heat produced inside the reactor, is heated by the reactor and cooled by the internal walls of the box. The first delta T is 10x the second one, but the surface of the box walls are 10x to 100x greater than the reactor external surface, so that the temperature of the internal air is very sensitive to the temperature of the box walls, which in turn are very sensitive to the velocity of the external air. Therefore during the down spikes, only the box mass is involved in the transients, not the reactor mass. The temperature of the reactor is not affected, because the internal air always flows at the same speed and the delta T between the reactor and the internal air is much greater than the temperature fluctuations of the box walls.
You have the opportunity to easily verify this behavior by testing it. If you wish.
At the moment it is a bit tricky for me to test the effect of cooling the box exterior, even with the air gap removed between the bubble wrap and acrylic, because my shop is currently about 9 C, and the heater (not used for these experiments) is an IR unit.
Conversely, heating the exterior could act similarly but oppositely to cooling the exterior.
However, I don’t see how this escapes notice of the inlet thermocouple. At 10 C, my inlet thermocouple increases by 0.1 to 0.2 C just from me being within 4 m of the calorimeter. -
The problem is that the MFMP actually measured the emissivity of their device.
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Another 200 W run going now with the inlet restricted to 5 cm diameter hole. Changed the inlet hole size after steady state was again reached with the previous inlet size. I think this should be a bit restrictive and drive the delta T upwards a bit by decreasing the total volume of air through the calorimeter.
Unplugged the fan for 60 seconds and got a nice long heat burst. So fan drawing current all the time is important. Eventually it reverts to the normal steady state after the fan is plugged back in.
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2. Even if the temperature was different, and it was not a magic violation of thermodynamics (meaning it did not mean more energy was coming out of the hotter cell), this magic would have no effect on the calorimetry. Because the calorimetry is not based on the reactor surface temperature. It is not based on the internal temperature either.
I ran the numbers back when this thread was new. A high emissivity reactor could be cooler than a low emissivity reactor in the calorimeter. The temperature of a low emissivity reactor climbs to where convection and conduction rates improve enough to match the loss in radiant power compared to a high emissivity reactor. In this case total power remains the same, even with a hotter or cooler reactor surface due to emissivity.
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And a fan underneath!
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So, you are confirming that the calorimeter at Hokkaido University of Science is totally inadequate to isolate the interior space from the external air.
Is it possible, in your opinion, that the 3 down spikes at the end of the "excess heat" test at 72 W, held on October 23, were simply caused by the late activation (during the evening) of a fan heater, operated in an on-off cycle?
Couldn't be possible that the regular trend of the corresponding calibration test at 72 W, held on December 2, does converge to a lower value due to the permanent activation from the beginning of the same fan heater?
(Here you can find the temperatures in Sapporo: https://www.timeanddate.com/we…storic?month=12&year=2019 )
The 72 W power drop/recovery part is something other than a reactor cooling heating effect, in my opinion, because it took 12000 seconds to raise the power from 60 W to 70 W the first time, and around 1200 seconds the next few times. Without overshooting the previous maximum of 70 W. The largest thermal mass in the calorimeter is the reactor, which as seen from the calibration and excess periods, requires a long time period to change temperature and therefore the outlet air temperature from which power is calculated.
More similar to someone opening the side panel and putting it back on.
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Thank you.
Your box is much better insulated than the Saito calorimeter shown in Figure 1 (1). The latter has a metal frame, probably in aluminum, and only one layer of bubble foil which separates the interior space from the external air. Needless to say, I would be very curious to see what would happen if you remove the interior acrylic box and leave only the exterior bubble box. Maybe it could be the right sort of artifact inducement to attempt.
As was discussed on this thread earlier, bubble foil is terrible as insulation if it can conduct heat anywhere. The reflective properties of bubble foil are only effective when used with an air gap on both sides, in still air. (The R value of bubble foil is 1.)
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Have you tried to direct the air flow from a fan heater to the exterior walls of your box?
Maybe it's possible to reproduce the strange behavior shown in Figure 6. "Calibration and excess heat at 72 W."
See also The NEDO Initiative - Japan's Cold Fusion Programme
I am attempting several sorts of artifact inducement. So far the calorimeter is remarkably robust.
My box currently has an air gap between the exterior bubble foil and the acrylic box, so outside influences are minimal. The interior acrylic can heat up, but transfers little to the outer surface (or vice-versa, presumably). Previous tests showed about a 10% improvement in heat recovery from adding the air gap.
Previously, before the air gap was introduced, active cooling of the outside would probably have lowered the maximum interior temperature, and therefore decrease the outlet air temperature. -
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With a 4 MOhm load, the MOT provides the expected ~20x voltage multiplication.
When I reduced to 10 kOhm, there is a buzzing noise and it becomes a useless 0.90x multiplier. I guess the transformers in the variac and MOT are fighting each other?
Next I will try wall voltage attached directly to the MOT once I have a suitable, safe arrangement for that.
Disclaimer: MOTs are deadly, use proper high voltage safety.
The MOT saturates very easily, so it needs a ballast or it will probably burn. Often additional MOTs with shorted secondaries installed in series are used for that purpose.
Edit: A few years ago while scrounging around for MOTs, two appliance repairpersons from different shops told me the same thing:
"If you are messing around with MOTs to get sparks, you need two more things on hand at all times: Someone to give you CPR and someone to call the ambulance while CPR is going on."
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Data from last test plotted up and examined.
Ended up with a best Delta T of 15.9, which is 0.1 C less than the previous maximum. With input power bouncing around a little bit, that is basically the same.
However, steady-state operation began an hour earlier than previous tests, so reducing the calorimeter air inlet opening (but not actually causing an air restriction), was a good improvement.
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The inlet TC can sense my presence.
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A huge part of the thermal drag on obtaining equilibrium is heating the calorimeter box, because it resists being heated.
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Restricted the calorimeter inlet to 9.5 x 6 cm with a chunk of bubble foil and about an hour has elapsed from start at 210 W. Delta T now 13 C, two more hours to settle, and better than about 16.5 C delta is the benchmark.
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In other news I just tested feeding a microwave oven transformer with a variac. The test load was four, 1 megaohm resistors (total of 4 megaohm). I had a voltmeter across one of the 1 MOhm resistors. It seemed to be fine at 400V (total 1600V) where the load current should have been 0.4 mA, and 0.16W per resistor. The 10A fuse popped in the variac as I dialed up around 500 V per resistor.
I'm not sure what the problem was. Maybe there was some temporary short in the voltmeter (which is rated to 500V) or some issue with the variac transformer interfering with the MOT transformer. The voltmeter seems to working still. Perhaps the "absolute voltage" (rather than the voltage difference) inside the voltmeter which could be much higher than 500V was to blame.Edit: I think the voltmeter screen went blank or at the moment of failure so perhaps it is to blame.
Also, there is a full bridge rectifier between the MOT and resistors.
I am probably content to stay below 1000V with this setup...
A CRT or oil heater igniter flyback transformer is much safer than a MOT
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Calculations please..
0.25 C = 10 W, 150/10 = 15, 15/0.25 = 3.75
Therefore 3.75 C = 150 W
Unless you meant something different originally? -
Are you sure? One=I .... wouldn't
The cp/density difference btw 20/40 C air is 6.76 %
I would also make sure to check my calorimeter RTD's
to check that they are not underreading or overreading by 0.25C
if 10W is a significant quantity.
Now I can deal with this comment.
So, is it then about a 4 C difference from calibration that reports as 150 W extra?
