MIZUNO REPLICATION AND MATERIALS ONLY

Quote from JedRothwell

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.

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.

Quote from Paradigmnoia

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.

If the total power is the same, the calorimeter will measure the same power level. A slight difference in convection versus conduction will not make a measurable difference in the heat that escapes through the walls. The R-value for the insulation is high for both convection and conduction.

Quote from Desireless

In other words: Higher pressure will cool down the core much more than a lower pressure. So internal temperature with higher pressure is always smaller but external temperature is always higher.

If you can make the surface temperature higher for a given power level, you can make a perpetual motion machine. You have violated the laws of thermodynamics.

Quote from Paradigmnoia

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.

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.

Quote from Ascoli65

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.

Quote from Paradigmnoia

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.

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.

(*) The NEDO Initiative - Japan's Cold Fusion Programme

Quote from JedRothwell

Do you mean the outer temperature of the reactor? If so:

1. No, the hydrogen cannot change the terminal outer temperature of the reactor. It can only make it reach that temperature slower or faster. The terminal temperature if a function of the power inside, only. It cannot be affected by the gas. If it could be, you could set up two cells with different gases, have one hotter than the other, and you would have a perpetual motion machine in violation of thermodynamics.

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.

This is not complicated. I suggest you learn more about flow calorimeters before speculating about them.

Display More

While that is true to first order, it will not be exactly true. how inexact would depend on reactor topology.

The key thing is that the heater will be hotter than the reactor, and its temperature. If there is any thermal transmission from heater through the reactor body to metal outside this varying temperature alters external temperature equilibrium, losses, etc.

Even if not, the reactor may not appear is an isotherm (in fact it has been posted here that in many cases the reactor case temperature is non-uniform). In that case the interior gas can alter the temperature distribution on the outside of the reactor.

In an air flow calorimeter with low losses and no other issues (e.g. accurate air temperature measurement in and out without artifacts related to temperature of other objects) none of this matters (as you state).

In the case that there are other issues it could matter.

THH

Quote from Ascoli65

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.

(*) The NEDO Initiative - Japan's Cold Fusion Programme

Display More

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.

Photos (old)

Uninsulated with SS calibration cylinder

Insulated box

Quote from Desireless

Another aspect is - Mizuno recommend to fill reactor with Deuterium when it is OFF. But pressure is changing very noticeably during heating.

In my case from 200 Pa at room temperature to 2000 Pascals at the max. heater power.

Is it indended that pressure is still at 200 Pascals during whole run? Because this factor is extremely dependent on the reactor construction. Shall we try to hold the constant pressure?

What if pressure is changed by the reaction or by fuel during run? Calibration will then not match likely.

1) Thermal conductivity of gas is not effected too much by pressures above 1 Torr = 133 Pascal at for example 600K -- less than 1%. It is a second order effect if you are running these pressures.

But even at lower pressures below 0.01 Torr where it makes a significant difference...

2) Because there is no other path for the heat to escape other than conduction (which is limited to the input pipes and the supporting structure), radiation (which is constant for a given physical shape, emissivity, and temperatures for the rig vs the surrounding environment (because some radiation is reabsorbed back in the rig), and convection (which is being measured by the airflow), I don't think different interior gas conduction, or interior convective gas heat transfer makes _any_ significant difference here. The proof is to run some calibrations evacuated to where there is no conduction (i.e. 0.0001 Torr) and compare to where there is gas (1 Torr). I think this would be proven in calibration by a careful experimenter.

Here is last nights experimenting.

Restricting the inlet definitely has an effect, but still only raised the outlet temperature (Delta T) 0.6 to 1.0 C . It may be a little higher, because the inlet temperature climbed a bit also, even with fairly high velocity air passing over it due to the restriction. However, the overall effect is obvious.

.

(The heater is inside the sealed 20 x 50 SS cylinder.)

Quote from RobertBryant

From your photos the thermal mass of the calorimeter box may be significant in increasing the response time of the calorimeter..

In the case of Mizuno's 20 kg reactor the calorimeter box mass (assuming 1/4 inch thickness with rough calc.. ) is ~15kg

The thermal mass of the reactor= massxCp=20,000x0.5 =10000 J/C

The thermal mass of acrylic = 15000x1.5 = 22500 J/C ..

I found that it was necessary to include the thermal mass of the acrylic to get a reasonable calorimeter heat balance at start up.

Display More

I mentioned before that the calorimeter is responsible for a significant part of the delay in steady state operation.

I recommend that the acrylic not be used at all for calorimeter design. I am building a new box with the same interior dimensions, but with 1“ isocyanate foam (R6) with aluminum facing, covered by R10, 2“ pink Corning foam board. With an access panel on one side.

Quote from Paradigmnoia

I mentioned before that the calorimeter is responsible for a significant part of the delay in steady state operation.

Do you mean because of the thermal mass of the calorimeter components?

Quote from JedRothwell

Do you mean because of the thermal mass of the calorimeter components?

Yes, exactly. Mostly due to the acrylic.

The acrylic looks great, and is easy to verify is sealed well where the panels come together.
But probably is a waste of money, and certainly, cumulatively, is a waste of experiment time.

It probably smooths out the outlet heat a bit, so it isn’t all bad.

Quote from Paradigmnoia

The acrylic looks great, and is easy to verify is sealed well where the panels come together.
But probably is a waste of money,

It is very cheap.

The specific heat is 0.35 to 0.50 Btu/lb/°F, which is 1.5 to 2.0 J/gK. I do not think that is a high thermal mass. Also, the bubble insulation is inside the acrylic, so very little heat reaches the acrylic.