E 0.94 , 894.5 C ----> E 0.46 --> 1510 C ----> E 0.39 --> 1712 C ----> E 0.39 --> 1712 C
The coolest part of the device at the time of the above was 727.2 C ---- ... ----> ~1300 C , leading to a 'COP' of 4.0, ignoring that 3/5 of the device was greater than 850 C (to be conservative).
Rossi has $10M or so. You should ask him for money to further develop this interesting COP 4 device.
Rossi has $10M or so. You should ask him for money to further develop this interesting COP 4 device.
If I did my math correctly, the average temperature of the cylinder at the end of testing was 814 C, so about a 'COP" of 5 may have been accomplished.
That is using from left to right, 2.1 cm @ 727 C, 1.0 cm @ 850 C, 1.5 cm @ 894 C, 1.0 cm @ 880 C, 1.0 cm @ 767 C. (Measured carefully by pyrometer scanning and glow intensity).
The lengths of the cool outer ends are exaggerated somewhat, as they are the coolest temperatures recorded for each end, and as such are less than the actual averages over the ends.
If I did my math correctly, the average temperature of the cylinder at the end of testing was 814 C, so about a 'COP" of 5 may have been accomplished.
That is using from left to right, 2.1 cm @ 727 C, 1.0 cm @ 850 C, 1.5 cm @ 894 C, 1.0 cm @ 880 C, 1.0 cm @ 767 C. (Measured carefully by pyrometer scanning and glow intensity).
The lengths of the cool outer ends are exaggerated somewhat, as they are the coolest temperatures recorded for each end, and as such are less than the actual averages over the ends.
Nice work !
What power setting did you use for these results ?
Nice work !
What power setting did you use for these results ?
Maximum input was 358 W and 5.06 A (at ~ 120 V AC in) , including SSVR losses (according to the Kill-A-Watt), at 68.3 V true RMS measured right at the coil input.
I think (guess) the SSVR dissipates about 1 W per amp (it has a giant heat sink and is rated for 25A), and two indicator bulbs on the power supply use 0.3 W when the SSVR output is disconnected.
I also discovered that the cylinder convection-radiation calculator I was using assumes adiabatic conditions for the ends, so I need to add convection-radiation calculations to account for the ends of the cylinder so my watts in matches the calculated watts out better before working out accurate-ish 'COP's and guessing at the real total emissivity of the Durapot.
Edit: the Kill-A-Watt amps seem a little suspect... the watts should be good, but I will have to look into how the amps are dealt with when feeding a triac-like waveform. The power supply is plugged directly into the Kill-A-Watt, however.
This also seems to indicate that the difference in detector emissivity sensitivity between the Optris camera (7 to 13.5 micron) and the pyrometer (8 to 14 micron) is negligible.
I was curious if indeed the differences are minor.
So I recalculated my in band emissivities for the 8-14 um range. See below
The numbers are for the 8-14 um range somewhat lower, which is what is to be expected since near the 14 um the Alumina spectral emissivity curves are dropping of, resulting in more lower values integrated.
-----T C)----e Optris-- --e 8-14 um
------20.1------0.832------- 0.746
----277.9------0.878------- 0.820
----302.8------0.881------- 0.826
----330.0------0.886------- 0.832
----344.2------0.888------- 0.835
----402.0------0.896------- 0.848
----438.1------0.900------- 0.855
----502.9------0.908------- 0.868
----516.4------0.910------- 0.870
----560.8------0.914------- 0.878
----631.9------0.922------- 0.888
----702.9------0.928------- 0.898
----728.4------0.930------- 0.901
----801.2------0.935------- 0.908
----854.9------0.938------- 0.913
----914.0------0.941------- 0.917
----990.1------0.944------- 0.920
If I did my math correctly, the average temperature of the cylinder at the end of testing was 814 C, so about a 'COP" of 5 may have been accomplished.
That is using from left to right, 2.1 cm @ 727 C, 1.0 cm @ 850 C, 1.5 cm @ 894 C, 1.0 cm @ 880 C, 1.0 cm @ 767 C. (Measured carefully by pyrometer scanning and glow intensity).
The lengths of the cool outer ends are exaggerated somewhat, as they are the coolest temperatures recorded for each end, and as such are less than the actual averages over the ends.
I did a quick FEM simulation of your cylinder.
Material properties of Alumina where used.
Since I did not know the exact length and postion of the heater coil, I distributed it evenly over the full length. As a result my end temperatures are higher.
However the center temperature of the simulation was 892 degree C, very close to the 894 C measured.
LDM ,
What alumina spectral emissivity curve are you using for your in band calculations?
Are you accounting for the broadening of the high emissivity area with increased temperature?
(See Figure 5, Manara et al 2009, from my earlier reference.)
I don't remember where I got them from, but incorprated it in a computer program for calculating the in band emissivity. And I used the program to calculate for 8-14 um.
However i found that there are two types of Alumina curves seen in Literature, some like the one you referred to which become broader with higher temperatures, but also others which are shifting to the right with higher temperatures.
I don't know why for the same material (Alumina) there are two type of curves, but the ones shifted give closer results to the recomended setting of .95 for the in band emissivity.
But also for the curve you supplied, if you shift the band from 7.5-13 um to 8-14um then you incorporate a larger part (on the righthandside) with lower values and as a result your in band emissivity will drop.
I agree that incorporating more of the long wavelengths will reduce the integrated in band emissivity.
Perhaps since once a temperature is assigned to a particular emissivity using a particular bandwidth,
Are we talking about broad band or in band emissivities ?
And what do you consider bandwidth, the width of the band or the spectrum range of the band ?
translations to other emissivities at a new respective temperature using the same bandwidth is essentially independent of the bandwidth used.
If we are talking about the same spectrum range then indeed the radiant energy received in that band has to be constant and translating temperature with it's emissivity is for that spectrum range defined.
This is the basis of my in band radiant power matching scheme that I used in early calculations that ended up supporting Clarke's paper when I initially disagreed with it.
This assumes that the radiant power received by the detector is constant, no matter what the user selected emissivity setting and the respective temperature reported by the camera or pyrometer is.
Totally agree.
However the translation is also dependent on the ambient temperature and a factor "n"
See the formula for U on the righthand top of page 9 of the Optris IR basics document.
Since U is constant for a certain case (Constant radiant power received) , we can equate two times the same formula with each side having it's own emissivity and temperature. (The pyrometer temperature cancels out when equating).
For high temperatures we can discard the ambient temperature and the equation leads then to the conversion formula used by the MFMP in which they assume n to be 3.
For lower temperatures you can not discard the ambient temperature.
Also there is the question if n is indeed 3. Optris states that it is between 2 and 3.
Some calculations I did in the past gives me the impression that it is about 2.75, but the difference in calculations using 2.75 or 3 is only small.
I will attempt to email Mr. Rozenbaum and Mr. Manara and see if they are willing to share some spreadsheets of alumina spectral and total emissivity values.
Also ask about the transmittance of 9 %
Seems high to me, so I wonder if the Alumina they tested was partly doped or was processed to have a high density increasing the transmittance.
Again thanks for the work you put in your tests.
Combined with the short FEM simulation it shows that for calculations of Durapot the material data of Alumina can be used. (At least that is my conclusion)
I could not have come to that conclusion without your experimental data.
I will be going to Norway for two weeks and will start following the thread again after I return.
If we are talking about the same spectrum range then indeed the radiant energy received in that band has to be constant and translating temperature with it's emissivity is for that spectrum range defined.
Yes, that is what I meant.
Also there is the question if n is indeed 3. Optris states that it is between 2 and 3.
Some calculations I did in the past gives me the impression that it is about 2.75, but the difference in calculations using 2.75 or 3 is only small.
I have the definition for the derivation of n as used in the Optris equation somewhere. It also explains why n is 21 for the shortwave end of the IR spectrum, as in the image that Whyttenbach posted a long time ago. The number is not a fixed value, (as is represented in the image), but has to do with the relative proportion of the power that is contained in various wavelengths or spectral ranges within a true blackbody spectrum. I'll try and dig up the source so I don't muddy up the explanation further.
Also ask about the transmittance of 9 %
Seems high to me, so I wonder if the Alumina they tested was partly doped or was processed to have a high density increasing the transmittance.
Reviewing the Manara paper, it says it was a sample of "fused" alumina. It may or may not be equivalent to other types of alumina.
According to a Rozenbaum paper: (2009, Texture and Porosity Effects on the Thermal Radiative Behavior of Alumina Ceramics":
"As emissivity in the transparent zone depends on extrinsic parameters, it is not possible to predict the spectral value of emissivity only by the knowledge of intrinsic parameters (refractive index and extinction coefficient) and the thickness of a material. Indeed, as explained before, the modifications in the semitransparent and transparent zones were essentially due to the bulk texture and porosity. In the opaque zone, the increase and the spectral modification of the emissivity were mainly due to the structure of the ceramic surface. These results show for instance the necessity to have the exact characteristics of a ceramic to measure correctly its temperature with an optical pyrometer."
You should not try to use a cheap meter like the KILL-A-WATT to measure power delivered through a triac. The output of a triac is a chopped sinusoid with a sharp falling edge partway through the cycle. Even a good RMS meter can have trouble accurately measuring waveforms with sharp edges. The input current to the triac dimmer has the same sharp edges.
As I suggested earlier, if you instead deliver power through a Variac (autotransformer), the sinusoidal waveform is preserved and any decent meter will do the RMS measurements with high accuracy.
You should not try to use a cheap meter like the KILL-A-WATT to measure power delivered through a triac. The output of a triac is a chopped sinusoid with a sharp falling edge partway through the cycle. Even a good RMS meter can have trouble accurately measuring waveforms with sharp edges. The input current to the triac dimmer has the same sharp edges.
As I suggested earlier, if you instead deliver power through a Variac (autotransformer), the sinusoidal waveform is preserved and any decent meter will do the RMS measurements with high accuracy.
The Kill-a-Watt measures consumed power surprisingly effectively, even when a triac is supplied. The amps cannot be trusted with a triac, however, but I do record it anyways. As it is, I measured the true RMS voltage at the coil with a good meter, and measured resistance of the coil over the entire temperature range. The V2/R reports almost exactly the same number as the watts on the Kill-a-Watt, with some error due to resistance rounding on the meter to one decimal place.
Kill-a-watt devices are numeric wattmeters, but with modest sampling frequency, thus not measuring perfectly steep ramp of triacs... question is if that error is important or not.
Simply filtering the current would help the wattmeter be more precise, but a filter can interact with the dimmer and kill it...
I would not trust my opinion without testing.
By the way I have very bad experience with cheap wattmeters for computers (bronze corsair PSU, that should be filtered actively), with displaying 1000W, but integrating 900Wh per hour, which show a 10% error for integrating (while another wattmeter is few % near).
I have the definition for the derivation of n as used in the Optris equation somewhere. It also explains why n is 21 for the shortwave end of the IR spectrum, as in the image that Whyttenbach posted a long time ago. The number is not a fixed value, (as is represented in the image), but has to do with the relative proportion of the power that is contained in various wavelengths or spectral ranges within a true blackbody spectrum. I'll try and dig up the source so I don't muddy up the explanation further.
I am still interested in the explanation because doing some verifications it seems that the factor n could also be dependent on the measured temperature.
So I would appriciate if you can dig up the article for me.
LDM
Some progress with the CFD simulation of convection of a single and three stacked tubes.
However still trying to find out how to extract the amount of convected heat.
Stacked tubes convective heat flux distribution
The CFD simulation picture below shows for three stacked tubes of 3cm diameter and equilateral spaced with 1 mm distance at a temperature of 400K how the convective heat flux is distributed at the boundery between the tubes and the surrounding air. (Air temperature set at 300K)
It can be seen that in between the tubes the convective heat transfer is minimal. (Dark blue color)
Also the convective heat transfer of the part of the lower rods facing each other (about 50%) has a low value.
The upper rod contribution to the convective heattransfer is less then each of the lower rods which is to be expected since the heated convective air of the lower rods will surround the upper rod. Thus the upper rod will have a higher ambient background temperature resulting in a lower convective heat transfer.
The major part of the convective heat transfer of the stacked tubes comes from the outward facing halves of the lower rods.
While still not having found out how te extract quantative data from the CFD simulation in a correct way, I expect, based on the above picture, that the convective heat transfer correction factor for the stacked tubes will be lower then the factor 2/3 used by the Lugano team.
