MIZUNO REPLICATION AND MATERIALS ONLY

Quote from RobertBryant

as long as we are nitpicking

Adiabatic temperature increase

The outlet pressure is 300 pascals more than the inlet pressure..from fanspec.

Pin 101300, Pout 101600 pascals

To/Tin =(Po/Pin) exp(0.286) … the exponent is related to Cp, Cv via theory.

For Tin = 25C, 288K

To = 288x(101600/101300) exp(0.286)

=288.2432 K…

25.2432C

Adiabatic temperature rise= 0.243 C

which is a large proportion of 0.35 C.(actually measured)

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I am struggling with this. Maybe I am "missing something" too.

How is this in any way relevant to anything?

The fan is there to cause movement of air through the calorimeter. It does this by behaving like a compressor, causing a pressure gradient which generates the flow.

Whatever heat it inputs to the system is needed due to the need for air flow. This will come out in the calorimetry. I can't be bothered to look for it, but presumably TM has allowed for this. Even if he hasn't, it doen't make any difference.

I like rough approximations to start with, to make sure one is not calculating to 5 decimal places whilst being an order of magnitude out of ball park.

I have done a rough approximation based on the consistency of the calorimeter with R19 and R20. In one paper, reporting on R19, the cal run is 100W in, and it results in delta T of ~5C. With the excess heat run, the cop is 1.8 and the delta air T is 9C. So, 100/5 = 20W/C. If we then apply this to the excess heat run, we have 80W of excess, which calcs as 80W/(20W/C) = 4C. Then, 5C + 4C = 9C, which is the reported delta T. With the high cop R20 run, it is a bit more difficult because the 50W input for cal is difficult to discern on the scale, but it looks like about 60% towards half way, which would mean 2.6C. Thus, 2.6 * 20 is 52C, so within the likely variations, this is consistent enough, for me with this type of analysis.

So, the 0.2432C rise that you have calculated as adiabatic increase would give us .024(32) * 20 = 4.8 W. This, as expected, is the power of the compressor. It is a) allowed for, and b) totally to be expected.

As an aside, 25C ambient is actually 298.15 K. I have also calculated the more accurate answer as 298.15 *(101600/101300) ^0.286 = 298.402. I took the liberty of rounding that to + 0.25C. Which gives us 0.25*20 = 5W ie, no difference in the context.

So what is the 0.35C measured that you refer to as this being a large proportion of?

Also, what is the point please? The data in the reports is such that it renders the seemingly never ending arguments over minutiae completley irrelevant when considered in the context.

Quote from THHuxleynew

Adiabatic temperature rise= 0.243 C

which is a large proportion of 0.35 C.(actually measured)

Oh, wow. I've not been reading here carefully. Is it really true that the results here are based on such a low temperature difference (out - in)? Surely not! It would be unsafe in lots of ways. And unnecessary because in any air calorimeter flow can be reduced to increase out - in temperature difference.

They think there is a 150 to 250 watt excess heat. These adiabatic+fan power temperature rises are around 1 watt. So the temp chart (not provided) is showing a 150/1*.25 to 250/1*.25 = 37 to 62 temperature rise. We would like to see the temperature chart, but it must be something of the order I just estimated, not 1/4 degree.

If one simply multiplies the delta T at (example) 40 C from 20 C (a delta of 20 degrees or 20 K) and use the air density and air specific heat for 40 C, instead of calculating the energy in 20 C air and subtracting it from 40 C air energy per cubic m, then a 10% excess can be incorrectly calculated per cubic m.

Quote from Bruce__H

It depends on the details of the LENR mechanism, but broadly speaking you can think of fuel/reactor system as being multistable.

Usually there are optimal T points for LENR reactions. You can have multiple stable points. But currently you need a stimulus to uphold the proliferation of H*/D*. Once we are able to produce better fuels we could do it without stimulation. But then the control will be a severe challenge...

Quote from RobertBryant

One would not simply do that..

Are you sure?

.

.

Note that the Cp shown is for 20 C air, and the Density shown is for 0 C air.

Quote from RobertBryant

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.

That looks like the volumetric thermal expansion rate for air from 20 to 40 C, for which I have 6.82 %.

So... 0.932 m³ at 20 C goes in, and comes out at 1 m³ at 40 C....

...redo some math...

Neato! 0.1% difference between methods. That seems essentially irrelevant, which makes me feel better about the vanilla delta T method.

That also means that the hot wire anemometer use will lead to about the same error (6.8%) if it is not corrected from NTP velocity to Actual velocity for the Actual conditions at 40 C.

Quote from RobertBryant

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?

Quote from RobertBryant

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?

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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A huge part of the thermal drag on obtaining equilibrium is heating the calorimeter box, because it resists being heated.

The inlet TC can sense my presence.

15.3