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.
Bubblefoil on the inside? How does it stay in place? If you have any photos with the bubble foil installed I would like to see some.
Anyway, at R 1.1, it won’t make much difference except the heat won’t radiate out.
I'm liking this thread -- we have at least two active experimenters who are doing the real work feeding back to us in the audience with questions, conclusions, and requests for ideas.
Keep up the good work all and above all else -- remain polite to each other. We all share a common interest here if not a common goal. Great community. Good luck in the New Year all!!!
Fired up the box again.
Returned to the 6 x 9.5 cm opening, but now the opening cover is sealed on and the auxiliary 5 cm hole capped with tape. Also impaled the inlet TC into a strip of bubble foil to block IR from the calibration cylinder. Ran for 4 hours without input heat to test fan heat leakage, which is delta 0.25 C .
Now back to 200 W input and see what else bumps the delta T, if anything, once the calibration settles.
Stood in front of the calorimeter 20 cm from the inlet. Raised inlet temperature 0.7 C, but only 0.2 at the outlet. Maybe it might catch up at the outlet but I didn’t want to stand there for an hour to find out.
Reset the blower voltage to 10.45 from 8.55 to more closely match the Saito values.
Delta T dropped to 13.5 from 15.
Stood in front of the calorimeter 20 cm from the inlet. Raised inlet temperature 0.7 C, but only 0.2 at the outlet. Maybe it might catch up at the outlet but I didn’t want to stand there for an hour to find out.
Reset the blower voltage to 10.45 from 8.55 to more closely match the Saito values.
Delta T dropped to 13.5 from 15.
In addition to your body heat it could be you are interrupting the airflow from a more distant source (i.e. the door or a vent) to the inlet. This is to be expected and is noise. Because you are going to run for 1000's of seconds this noise will be filtered away. Or you could run a regression on it if you have a mechanical way to measure your position so you can subtract out the expected temperature increase noise of your walking around. If you don't want to see the noise you might need to move the unit so that when you are working on it you are not between the airflow source and the unit. I think because most of your heat removal is delta-T between input and output, this 0.7 degree rise is already not significant, but some of it is the conduction, natural convection and radiation to the environment from the rig and you can build that also into the equation. I think it is so second order compared to the signal from the heater control or the active unit that you don't need to worry about it
In addition to your body heat it could be you are interrupting the airflow from a more distant source (i.e. the door or a vent) to the inlet. This is to be expected and is noise.
There is no chance you could detect either of these things with this equipment. It measures plus/minus 2 W or so. The effects you describe would be a fraction of a milliwatt. (Not actually a milliwatt of heat, but an error on that scale.)
Known sources of noise such as ambient temperature changes and fluctuations from thermocouples are orders of magnitude larger than the kinds of things you speculate about here.
Stood in front of the calorimeter 20 cm from the inlet. Raised inlet temperature 0.7 C, but only 0.2 at the outlet. Maybe it might catch up at the outlet but I didn’t want to stand there for an hour
Well, you are good for around 200W body heat,. which should be detectable.
a person can be compared to a 100W bulb, a big fraction of a 200W test, must act carefully 
Like I said before, the inlet thermocouple knows I am present. Normally the inlet temperature climbs 0.1 to 0.2 C if I am in the vicinity of calorimeter. I normally keep clear while things are running, but take a note when I am adjusting something.
I am looking forward to looking at the last batch of data. After the new fan speed settled, I set up a board to reflect the outlet air back over the calorimeter for several hours.
Reflecting outlet air seems to have done nothing to the Delta T.
Body heat barely a blip (It was only 4 minutes).
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Next test I will use willpower to raise the delta T. (There is no try, only do or do not).
However, I am uncertain which way is best to do this. Shall I concentrate on heating the outlet thermocouples, the air inside the box, the reactor, or lowering the inlet thermocouple temperature?
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
the specific heat of the stainless steel is one third that of the acrylic
in the case of Mizuno
The thermal mass of the reactor= massxCp=20,000x0.5 =10000 J/C
The thermal mass of acrylic = 15000x1.5 = 22500 J/C ..
The acrylic thermal mass is over twice that of the reactor..
If insulation on the acrylic reduces the effect of the acrylic thermal mass dynamically
then the response time of the calorimeter should be shortened by this insulation
Well, you are good for around 200W body heat,. which should be detectable.
I see how body heat raising the temperature of the inlet air can be detected. This is the same as putting a resistance heater next to the inlet. However, the rising inlet temperature will be readily apparent. Heat from the reactor is outlet minus inlet, so this will not change the heat balance. It will make it less accurate; it is noise.
It is not possible to have the heat from a 200 W body outside an air-flow calorimeter actually go into the calorimeter and increase overall heat inside. Not with an instrument that can only measure to the nearest 1 or 2 W. The inside of the calorimeter is much warmer than the surroundings, including the area around your body, so heat cannot pass into it. At most, the heat from your body might retard heat radiated from the walls, but that would be a tiny effect, far less than 1 W. Even if you held your hand on the outside wall, there is no way it could measure the effect of this.
. At most, the heat from your body might retard heat radiated from the walls, but that would be a tiny effect, far less than 1 W.
Rough calculation? ... Paradigmnoia front torso = 0.1 m2 T=37 emissivity =0.95(naked =0.97) Temperature of acrylic = 25 ..emissivity =0.95
Anyone?... I think we should be able to get a watt or more for the net radiative heat transfer..no willpower necessary
The inlet thermocouple is currently protected from emissivity effects by a strip of bubble foil bubbles.
The air going around my body is being sucked into the box (when I stand right in front of the inlet), and the room is currently quite cool (circa 10 C), so the air being drawn in is being heated by body. Certainly not very many watts worth, but some. This raises the inlet temperature 0.7 C in my experiment. It took about 3 minutes for the outlet temperature to report a 0.2 C increase. It would probably take a long while for this inlet 0.7 C temperature change to propagate fully to the outlet, but most assuredly it will, as would any constant air temperature change at the inlet.
Also in my case, the bubble foil is on the outside, with a 1 cm gap between the bubble foil and the acrylic box (for the most part), so emissivity effects in general are prevented from the outside of the calorimeter.
Similarly, the whole-body natural convection coefficient for the manikin fell within the mid-range of previously published values at 3.4 and 3.3 W/m2 per K when standing and seated respectively. In the forced convective regime, heat transfer coefficients were higher for hands, feet and peripheral limbs compared to the central torso region. Wind direction had little effect on convective heat transfers from individual body segments..
Certainly not very many watts worth, but some
True..
Put in.. effective area 0.1 m2 and stick in 37C and 10.C
3.4x 0.1x(37-10) =9W..(0.9C delta on the 2017 Mizuno calorimeter)
There is no chance you could detect either of these things with this equipment. It measures plus/minus 2 W or so. The effects you describe would be a fraction of a milliwatt. (Not actually a milliwatt of heat, but an error on that scale.)
Known sources of noise such as ambient temperature changes and fluctuations from thermocouples are orders of magnitude larger than the kinds of things you speculate about here.
Disagree Jed. If the cooling vent is on the rig, its temperature goes down. Has nothing to do with milliwatts, has to do with removal of heat by flow of air. We have all experienced same when standing outside on a still wind day at say 40 Fahrenheit and it seems warm, but add in a breeze and you are freezing. Convective air movement removes heat. I know this to be true from past experience.
Rigid foil covered PIR house insulation is quite cheap and readily available. If you used 100mm thick sheets glued and with aluminium tape on internal and external seams you could get away without an aluminium frame to support it. A plug door from two sheets would allow access.
PIR has a thermal conductivity of 0.022W/mK. If we estimate the surface area at 2sqm heat loss could be limited to less than 0.5W/K.
Mizuno in here..
http://lenr-canr.org/acrobat/MizunoTincreasede.pdf
..seems to run his calorimeter with a delta T of perhaps 10-20C?
So losses through the walls could be reduced to perhaps 5-10W.
Perhaps less as I've probably overestimated the surface area.
Display MoreRigid foil covered PIR house insulation is quite cheap and readily available. If you used 100mm thick sheets glued and with aluminium tape on internal and external seams you could get away without an aluminium frame to support it. A plug door from two sheets would allow access.
PIR has a thermal conductivity of 0.022W/mK. If we estimate the surface area at 2sqm heat loss could be limited to less than 0.5W/K.
Mizuno in here..
http://lenr-canr.org/acrobat/MizunoTincreasede.pdf
..seems to run his calorimeter with a delta T of perhaps 10-20C?
So losses through the walls could be reduced to perhaps 5-10W.
Perhaps less as I've probably overestimated the surface area.
That is pretty much my plan. I already have the polyiso R6 foil-faced foam board, and a sheet of R10 pink foam board.
I have the ogre of the hot wire vs vane anemometer tests to deal with first. That should be a barrel of fun...
Due to increasing the fan voltage, I calibrated the fan delta T contribution again. Now it is 0.45 C instead of 0.25 C. So the power into the fan seems be heating the air ultimately. Of course this extra bit of delta T is included in all output measurements, calibration and any excess events. It should probably be tested at several different, higher ambient temperatures, to be certain that the fan adds heat to hot air as much as to cold air.
Then I started to mess with heating up the fan... and after a few short experiments I worked out that any heating up of the fan requires ultimately heating the air going through it, and the same input power, minimum, as the delta T would normally require is required to raise the delta T at the downstream TC (if the air is mixed). So short of directly heating the outlet thermocouples there is no free lunch of delta T increasing due to heating the fan by some mysterious way.
I did shine a very bright incandescent flashlight at the outlet thermocouples with a possible 0.1 C increase.
And then I did a full 8-crossing (every 45 degrees) hot wire anemometer traverse, and also the vane anemometer capturing 100% of the airflow. More on that later.
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