Mizuno reports increased excess heat

OK, I did a quick sanity check with a hypothetical cylinder, using natural convection. Ambient at 20 C. (Just the trend is being examined.)

The cylinder temperature will increase while the heat output stays constant at that same input level, when the emissivity is strongly decreased.

The power required to reach a given temperature will decrease when the emissivity is strongly decreased, however the output power also decreases commensurately with the input power decrease, so that the total heat is conserved.

100.875 W in --------> E=0.85 -----------> 100.875 W out

................................... T=50.0 C .......................................

.

.

100.734 W in --------> E=0.08 ------------> 100.734 W out

................................... T=70.5 C .......................................

.

.

52.602 W in -----------> E=0.08 -------------> 52.602 W out

.................................... T=50 C ..........................................

.

.

190.728 W in ---------> E=0.85 ------------> 190.728 W out

.................................... T=70.5C ........................................

Quote from Cydonia

i never believed in magic powders

Neither do I.

for the magneto electric effect in Niobium 93

consult Cheng

https://www.google.com/url?sa=…Vaw0SSBJE93gjBcUGjbSBMJBd

Forsley , Cheng and Hagelstein

are talking of the mysterious other channel

it is not magic it is

A Dieu

Quote from RobertBryant

The main effect of putting low emissivity 0.04 foil on the calorimeter box is to

increase the share of passive convection heat loss and decrease the share of the radiation loss.

If one assumes 2 m2 SA for the box and an overall very crude HTC of 4 for passive laminar convection

the heat loss rate is 160W for a surface temp of 20+20 = 40C where 20 is ambient,

https://www.electronics-coolin…fficient-on-a-flat-plate/

The radiation loss is 10W for emissivity of 0.04

https://www.engineeringtoolbox…-heat-transfer-d_431.html

The reason for doing this is to more linearize the heat loss with respect to surface temp

of the box rather than having it vary more greatly

Convective loss is more linear(except for increases in HTC wrt to T)

than radiation loss, which varies according to Stefan's 4th power.

These calculations are indicative only.

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The key factor here is reactor temperature.

This is forced air cooling (not free convection) so convective transfer to air is roughly proportional to (case temperature - air temperature) (for M reactors input and output temperatures are roughly the same, case temp a lot higher e.g. 378C

From R19 5/1 290W, 378C. Given say 30C av air temp we have 1.2 W/C going out from reactor into exhaust.

From paper, at this case temperature, we have 77% efficiency. If we assume all losses are from the calorimeter box (there will be some also from the exhaust pipe, but not significant:

We have 290*0.23 = 60W lost.

An insulated acrylic box is used for airflow calorimetry. It is 400 mm × 750 mm, height 700 mm. During a test, the
inside of the plastic box is covered with 1.91 m2 of reflective padded aluminum insulation (shanetsu.com, Fig. 7).
This minimizes losses to radiation. These losses are low in any case, because the cooling air keeps the inside of the
box at ∼36◦C (16◦C above ambient). Similar insulation from a US vendor (US Energy Products) has an R-value of 11,
so the insulation radiates ∼3 W (16◦C/11 W/m2 × 1.9 m2
).

Obviously this is 20X larger than the example from the paper, with 320C across the insulation.

What is the thermal conductivity of the (radiative + convective) air gap, as opposed to the insulation?

Radiative transfer scales as (T0+dT)^4 - (T0^4) = 4*dT*T0^3 at 40C T0 = 300K. at 370C T0 = 640K. The ratio is thus (640/300)^3 = 9.7

Thus while convective transfer stays roughly proportional to dT, radiative transfer will be 10* larger at this high case temperature relative to convection.

From the above rough figures:

40C 10W vs 160W

380C 100W vs 160W (dT over the radiative/convective gap is not 10X larger with the higher temperature - obviously. The insulation inner foil will equilibrate to a temperature that balaces the radiative + convective heat flow over the gap with the heat flow through the insulation).

And for higher emissivity radiation becomes much more significant. 0.04 is very shiny.

Thus we can ignore radiation at low powers. For the type of power used in the R19 tests we cannot ignore radiation.

The whole system here is quite complex, and difficult to analyse. Both the inner foil and the reactor surface are cooled by the forced convective airflow. Probably the inner foil gets cooled a bit more at the same temperature because it will have a larger surface area. The actual inner foil temperature will be quite high - because the insulation and heat flow out forces this, but not as high as the reactor. At higher temperatures the larger radiative airgap component will make the foil temperature relatively higher as a fraction of the reactor - ambient temperature.

Key take home here is that most of the calculations here are linear, but the radiative power transfer is highly nonlinear and much larger (relatively) at higher powers. Even so the thermal resistance from calorimeter to ambient may be dominated by the (linear) insulation rather than the (nonlinear) radiation + convection. I'd not want to make assumptions about this complex system, especially because tests on it cannot be made instantaneously, it will have a significant time constant.

Quote from THHuxleynew

because tests on it cannot be made instantaneously, it will have a significant time constant.

This is HVAC rocket science ,, not electronics,

These heat measurements are made over hours..

do you think you can calculate a TC of hours length?

Show calculation pls.

So we agree

Loss from reactor is non linear.

Loss from box is linear wrt temperature

and calibratable.

Since the reactor loss= airflow loss(measured) + box loss(calibratable)

whether or not the reactor loss has a time constant of 2 minutes/10minutes or 2 seconds is irrelevant.

Quote from JedRothwell

It better be conserved! Otherwise, as Scotty would say to Captain Kirk, "She's gonna blow apart!"

I had to Monte Carlo by hand some of the temperatures and powers in order to match the example parameter, so I was manually conserving energy, based on historic precedent and some experience in other systems.

I haven't confirmed conservation of energy by testing these examples, however.

Continuing from the previous post, to first approximation the power lost from the calorimeter is proportional to the inner insulation foil temperature. Not quite because the outer box temperature is determined by free convection which will scale as (T-Tamb)^2 (or something).

The inner foil will equilibrate to some fraction F of the reactor temp (all referenced to Tamb):

Tfoil - Tamb = (Treactor - Tamb)*F

F will get closer to one at higher Treactor because of the more powerful radiative component of the air gap heat transfer.

If F was constant we would have a nice linear system with constant efficiency (at low temperatures the heat losses are less, but so is the output power. These two effects cancel).

Therefore losses will get higher (as a fraction of output power) at higher temperatures. That explains the change from 95% to 77% calorimeter efficiency.

We ignored the nonlinearity of free convection outside the calorimeter, this will also make efficiency lower at higher powers.

Quote from THHuxleynew

That explains the change from 95% to 77% calorimeter efficiency

I calculated that two years ago.. thanks for the revelation

Anything new? an hour long TC?

Loss from box is linear wrt temperature

and calibratable.

Since the reactor loss= airflow loss(measured) + box loss(calibratble)

whether or not the reactor loss has a time constant of 2 minutes/10minutes or 2 seconds is irrelevant.

As you can see from above efficiency is not linear with temperature, though that might be a decent approximation.

• Calibratable - only if Tamb is controlled constant.
• As GSVIT noted the time constants are significant if the ambient temperature is changing. As appears to have been the case in these tests. If that happens the errors scale with the reactor time constant.
• In any case because of the time constants calibration must be over a long enough period.
• Anyone replicating would want a thermostatically controlled constant room temperature so that the only time constant to worry about is the reactor (+ calorimeter box which will be shorter) and errors due to Tamb changing are reduced. There still will be some errors due to temperature gradients from calorimeter heat.
• You could also reduce these errors by reducing flow and therefore increasing output temperature. However that would reduce efficiency and maybe increase other errors.

For those not liking this complexity:

You need to consider all of it if wanting to evaluate the R19 results

You can ignore it if evaluating the much larger R20 results.

Dear all, I do enjoy these calcs. But not the company!

THHnew is a babe in the woods as far as heat transfer and fluid mechanics

(see the previous laminar/turbulent apocalypse)

Why are you revealing these basic calculations as a novo mundo?

Spare us

pls use the engineers toolbox

https://www.engineeringtoolbox.com

Quote from THHuxleynew

. But not the company

Its not about THHnew, Getover yrself.
(selves)

USe engineers toolbox,,, it saves time,

Quote from JedRothwell

You are confused. Understandably so, because I used different standards in the two examples. The 320°C is the reactor body temperature. The "∼36°C (16°C above ambient)" is the difference between the air temperature in the box and the ambient air outside. The insulation NEVER sees a difference as large as 320°C. I suppose the insulation would crinkle up and the bubbles would pop. It is basically aluminium foil with air bubbles. I do not think the air temperature is ever more than 50°C hotter than ambient (at ~1000 W * 0.05522 W/K, which is the limit of this instrument). That is the biggest difference across the insulation. The air is moving through the box at 13.4 L/s so it does not have time to get very warm.

Mizuno used the reactor temperature in Figs. 2 and 3 because it is measured more accurately and it is more sensitive. A graph of air temperature would be more scattered. However, you can do this with air temperature. I have done it. This allows you to use the manufacturer's spec sheet for the insulation and double check the result. These R-ratings are so high, and the heat losses so small at most temperatures, the numbers are only approximate. It is a good sanity check, but for precision I would stick to the numbers shown in Figs. 2 and 3.

That is interesting Jed. I wondered about it. Where am I wrong:

At Treactor = 380C efficiency = 77% => heat lost to exhaust is 23%.

At Treactor = 380C, P = 290W (5/1 test from R19 table) heat loss from calorimeter box is therefore 290W*0.23 = 60W approx.

At this heat loss (from box) the temp gradient across your insulation is 16C (your figure at 3W) X 20 = 320C

Otherwise you need a lot of this 60W lost from the output pipe? Surely not when the exhaust air temperature is so low.

Do you think most of the calorimeter heat loss is conductive? Maybe you don't have as good insulation on the base? In that case the time constant of whatever the calorimeter rests on heating up will be maybe very significant.

PS - the foil temperature can be much higher than the air temperature due to radiative transfer (see above for how this gets very significant - 10X more so - at Treactor = 380C compared with Treactor = 40C). This is slightly, but not very, counterintuitive.

Quote from THHuxleynew

In any case because of the time constants calibration must be over a long enough period.

Calculate an hour long TC .. if yourselves enjoy calc.

I have the whole morning to co-enjoy

Quote from THHuxleynew

Maybe you don't have as good insulation on the base

Dang.. Mizuno messed up there..

Quote from RobertBryant

Dang.. Mizuno messed up there..

So, are you suggesting he did? Or did not?

Anyway this is for Jed to answer - he probably knows.

Perhaps you could say whether you agree with the calculations showing this large temperature gradient over the insulation at high reactor temperatures? There are other options, but they all seem a bit weird. Unless Mizuno's calorimeter efficiency graph is all wrong.

Quote from THHuxleynew

So, are you suggesting he did? Or did not?

I am suggesting that this question is

like your one in 2017

"Do you maintain the low pressure with a vacuum pump?"

which I told you, was a gauche question

especially since in the 2017 paper you read

why Mizuno did not use a vacuum pump.

Please read 2017 again,,.

Another question ... do you think that in three -four years of work

that Mizuno would tolerate an hour long response time for his calorimeter system??

Do you think that he has babeinthe woods knowledge?

Or do you think that he might have designed the system to respond in minutes?