yes, i think the best way to be well uniform should be chemical plating, here by a beam we should have small thickness variations mostly we stay at nanometers level.
Only a SEM will show something. Here we see nothing with a classical microscope.
Now, with today understandings, only the experiment will prove which method is best.
yes, i think the best way to be well uniform should be chemical plating, here by a beam we should have small thickness variations mostly we stay at nanometers level.
The nanometer thickness may well be fine.
Not sure about the uniform application though.
Mizuno scored with very heterogeneous Pd application. His success could be down to hitting something random, but highly dimensionally sensitive.
Best of luck though, it will be interesting to see your results. It's all data to work the problen with.
Just a couple of questions for my own interest.
I've noticed from previous posts that you are into sputtering. I was wondering if the type of sputtering process you use still allows the CaCO3 crystals to be present on the mesh?
If so, does the applied Pd build so that the CaCO3 crystals protrude through it, or would it cover them if they were present?
If I am poking my nose in too much, I apologise. 
yes, I think surface topology should be very important but not mainly for absorption.
Crystalline structure is something else. In positive experiments, an amorphous structure is often reported as for Japanese powders. Sputtering can help to make this kind of structure too.
I especially think that it helps for hydrogen flow speed. High adsorption or not, NAE or not, I don't think that is the most important. I will rather say stœchiometry, as in chemistry.
Like everyone else I'm trying things around Pd / Ni way, we'll see
It looks like you don't like any kind of oxide present if you have removed NiO too.
Do you reckon that the D2 dissociative adsorption is driven by surface topology/Ni crystal structure alone then? Assuming you are using D2 and agree with that as the first step of the process, of course.
Crystalline structure is something else.
Sure, in some respects. The reason I put it together with surface topology though, is that there seems to be a deal of evidence that the step and terrace structure of ordered crystals is dimensionally in the right ball park for assisting adsorption.
But then again, if you rate a stoichiometric aspect above adsorption anyway, then I can see how you would not put the two together.
Of course, that does raise the question : "stoichiometric to what?", but perhaps you regard that as your personal deductions.
However, at the end of the day, we will indeed see. Thank you for the discussion.
What voltage is Mizuno operating the San Ace blower fan at for 4.3 W (or whatever it is in recent R20 vintage tests)? It is reported that the V and I were recorded.
I am home for a few days and should be able to test the air velocity with traverses in better detail. A quick preliminary test a couple weeks ago showed that substantially lower voltages than 12.1 V (DC) does flatten the velocity profile (at least with a 60 cm tube). I would like to match as close as possible the voltage to the fan used by Mizuno, and assuming that I can get it close enough and also verify the flat velocity profile using the 25 cm tube (65 mm diameter), I can connect the fan and tube assembly to the calorimeter box and finally begin testing out the calorimeter.
The spreadsheet equation for blower power is: (V*A)*0.3333-(A*A)/3
Example:
V = 10.08
A = 1.28
Power = 3.76 W (with rounding)
Display MoreThe spreadsheet equation for blower power is: (V*A)*0.3333-(A*A)/3
Example:
V = 10.08
A = 1.28
Power = 3.76 (with rounding)
Thank you.
I last had my fan running at about 9.4 V and 0.5 A (at a resolution of 0.1 A so I have to fix that up). I had the fan running at as low as 7.89 V and 0.3 A
The specifications for the fan suggest a voltage range of 7.2 V to 13.8 V DC
(I think that at 13.8 V the fan body itself would be restrictive to the amount of air it is trying to push.)
Display MoreThe spreadsheet equation for blower power is: (V*A)*0.3333-(A*A)/3
Example:
V = 10.08
A = 1.28
Power = 3.76 W (with rounding)
Sorry, but I seem to be a bit thick today.
What is everything after (V*A) accomplishing exactly? Is it PWM?
Dunno. Ask Mizuno.
Hmmm. That reminds me that I can loop the fan power wire 10 times and possibly improve my current resolution.
The spreadsheet equation for blower power is: (V*A)*0.3333-(A*A)/3
Sorry, but I seem to be a bit thick today.
What is everything after (V*A) accomplishing exactly? Is it PWM?
It looks like he is measuring blower power as the total power from the supply minus the power lost in the current sense resistor.
Resistor power is I2R, and the (A*A)/3 corresponds to the power through a 1/3 ohm (.3333 ohm) resistor.
Not sure why V*A is multiplied by .3333 though. Seems like it should be (V*A)-(A*A)/3
It looks like he is measuring blower power as the total power from the supply minus the power lost in the current sense resistor.
Resistor power is I2R, and the (A*A)/3 corresponds to the power through a 1/3 ohm (.3333 ohm) resistor.
Not sure why V*A is multiplied by .3333 though. Seems like it should be (V*A)-(A*A)/3
Good call -
But I think in that case 'A' could be the voltage across the resistor, with I = A/Rsens. . Then P = V*A*0.3333 => Rsens = 3. In which case Prsens = (A*0.333)^2 * 3 = A*A/3.
The equation is then correct, for a 3 ohm current sense resistor
Good call -
But I think in that case 'A' could be the voltage across the resistor, with I = A/Rsens. . Then P = V*A*0.3333 => Rsens = 3. In which case Prsens = (A*0.333)^2 * 3 = A*A/3.
The equation is then correct, for a 3 ohm current sense resistor
Yes, you are right that the equation works for a 3 ohm sense resistor. It is easier to see this way:
If A is the voltage across a 3 ohm resistor, then I=A/3 or A=3I. Substituting, into:
P= (V*A)*0.3333-(A*A)/3
you get
P = VI -3I2
and the power in the resistor is I2R = I2*3
But 3 ohms is large for a sense resistor. It will get very hot, even at 1 A (3W). At 2A (12W), you need a high wattage resistor to keep it from burning up. It also adds error to the calculations because the resistance changes when it heats up.
A typical wirewound resistor has a temperature coefficient of 400 ppm/deg C. If it heats up 100 C, that introduces a 4% error.
The error and wasted power is reduced if you use a smaller resistance value. DVMs and data acquisition systems can measure small voltages at high accuracy and it would be better to use something like 0.1 or 0.5 ohms.
Mass spec, reactor and a lot of messy-looking plumbing. Nearly ready to rumble, just waiting for cold trap parts (between reactor and mass spec) and also babying this through the next round of vacuum testing.
A few questions/comments:
I notice that some of the system appears to have a heater blanket, but some of the small diameter tubing does not. Assuming that you plan to bake out the system, does this unheated tubing make a difference?
The small diameter tube will really slow down the evacuation rate in the molecular flow region. The conductance varies as the cube of the tube diameter, so attaining a high vacuum at the far end may take some time.
How good a vacuum do you intend on achieving after bakeout? I ask this question in part because I'm constructing a similar setup.
Once D2 is introduced it will be difficult to remove. T/M pumps have a hard time pumping light gasses due to the high velocity of the molecules compared to the rotor tangential velocity. Do you intend to achieve a high vacuum only to guarantee that the chamber is contaminant free and then assume that D2 in some quantity will always remain?
Jeff
So the San Ace fan is operated at 8.8 V, which means I can insert another IN5404 in series with my 9.4 V that I currently have to get it, or alternatively use a 3 ohm 10W resistor in series with 10 V supplied to the fan (which is a better match to the Mizuno set up).
1. I notice that some of the system appears to have a heater blanket, but some of the small diameter tubing does not. Assuming that you plan to bake out the system, does this unheated tubing make a difference?
2. The small diameter tube will really slow down the evacuation rate in the molecular flow region. The conductance varies as the cube of the tube diameter, so attaining a high vacuum at the far end may take some time.
3. How good a vacuum do you intend on achieving after bakeout? I ask this question in part because I'm constructing a similar setup.
4. Once D2 is introduced it will be difficult to remove. T/M pumps have a hard time pumping light gasses due to the high velocity of the molecules compared to the rotor tangential velocity. Do you intend to achieve a high vacuum only to guarantee that the chamber is contaminant free and then assume that D2 in some quantity will always remain?
1. The small tubes can be heated for bakeout purposes with a blowtorch or hot-air gun so not really a problem
2. You are quite right - pumping this system right down can take 24 hours. But that's ok -we can wait for it. Also there are three vacuum pumps- 2 stage pump on the reactor end (on the right) and a 2 stage roughing pump and turbo on the Mass Spec on the left.
3. We have been as low as 10-8 mB. That's plenty low enough.
4. We hope to keep D2 out of the Mass Spec by using a cryo-trap- not shown here. This is designed to stop D2 getting into the MS because it's hard to discriminate between D2 and He - and we are on the hunt for helium.
OK, after some fiddling I now have 9.75 V total, 1.38 V across a 3 ohm 5 watt resistor in series with the fan, and 8.37 V across the San Ace fan. Close enough for now. Time for some anemometer traverses...and let it run a bit to see if the values stay stable.
Edit:
After several hours of continuous operation the total voltage settled at 9.80 V, 1.35 V across the 3 ohm current sense resistor, 8.45 V across the fan ..... and a kinda-sorta flat velocity profile with a 48 sample average of anemometer points of 3.69 m/s. Fan speed of 2847 RPM.
