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  • Analyze this(not a movie review).......

    Hi guys. I spent some time yesterday learning how to calculate and check for PAR. Today, I ran the same test piece that I have been, only this time "watching" for PAR. The good part is, the end result was MUCH better then it has yet been since I switched to the LCD method. The bad part is....im not sure why I got the results I did. Id like your thoughts on.......well, everything.

    Part size- 1.5x2.75x.375

    I calculated 11.4375 sq.in.(.07943 sq.ft.). I then calculated that I would need .357A(at 4.5/sq.ft.). I am using a .1175ohm(4 .47ohn resistors in parallel) "current sensing" resistor......so I calculated that I need .042V across my resistors.

    Additional things that calculated this time.....
    PAR= 31.48ohms
    PAR voltage= 11.24V

    Other needed to know things...
    ano tank temp= 68degrees
    dye temp= 140degrees

    I recorded the voltage from cathode to anode at random times throughout the process today. Im not using a CC power supply, and the current didnt stabilize before the 2 minute mark, so thats where I started.

    2 minutes- 11.6V
    6 minutes- 11.81V
    11 minutes- 12.22V
    20 minutes- 12.30V
    35 minutes- 12.47V
    47 minutes- 12.70V
    52 minutes- 12.90V
    1:23 minutes- 13.16V
    1:40 minutes- 13.40V
    1:53 minutes- 13.50V
    2:00 minutes- 13.50V
    2:30 minutes- 13.50V

    After it stayed at 13.50V for over a half hour, I figured that it was at PAR and would EVENTUALLY drop in voltage. I dyed the part red and it turned out a nice deep red. Red has been the color that has been giving me more problems then anything else, and the Ph of my red dye is supposedly too high to work properly. I was doubting it was the dye due to the fact that it was only used once before, and I always rinse really well between tanks. I was about to replace it actually...and still may just to see what new red dye actually measures.

    Anyhow, did I do my calculations correctly? Why was my ending voltage 13.50V when it should have been 11.24V? Why did PAR take an extra half hour to reach(2 hours)? Thanks for any input guys!

  • #2
    Nice work Sid03.

    I would have said that your value for PAR is:

    I = V/R = 13.50V / .357A = 37.82 Ohms.

    It's convenient to normalize PAR to 1 sq.ft. This makes it easy to compare your results with others:

    37.82 Ohms x .07943 sq.ft. = 3.00 Ohms per sq.ft.

    Three Ohms per square foot means that your anodic coating is thicker than mine, which from a dyeing standpoint is always welcome. The very good dyeing results illustrates this. Recall that the higher the value of PAR, the thicker the coating.

    It appears that you have mastered electrical connections; a high value for PAR, AND very good dyeing. As you have read; if PAR is high, but the dyeing sucks, connections have degraded.

    The long elapsed time with no change in PAR indicates that in this case you have achieved equilibrium. This occurs when the rate of anodic growth is equal to the rate of dissolution; i.e., the coating is dissolving as fast as it is forming. If you changed something (current density, temperature, acid concentration, surface area, etc.) the two chemical reactions would again become unbalanced, and PAR will go up or down. Its easier to obtain equilibrium at low current densities than at high.

    In normal Type II coatings, the anodize forms about 1/3 of its thickness above the aluminum surface, and about 2/3 below it. This is 1/2 : 1/2 in Type III layers.

    I have not been able to confirm this in the available technical literature; but I believe that equilibrium forces the anodize to burrow deeper into the aluminum, which seems favorable for dyeing. Since PAR is constant, the layer isn't getting thicker or thinner. I THINK I see the overall thickness of the work getting a little smaller, measured with a 5 1/2 digit electronic digital micrometer. The dimensional changes here are on the order of 0.0001 - 0.0005 inches or so.

    Can you post a picture of the finished piece?

    Comment


    • #3
      The PAR values I calculated were done before I started anodizing. I thought I was reading where you guys used 2.5ohms and pre calculated what the ending voltage should be....?? Here's what I did....2.5 x (1 / .079427) = 31.48 Ohms. Then I used that value to find the predicted ending voltage: 31.48 x .357A= 11.24V. Thats how I got those values. Is that incorrect?

      "It appears that you have mastered electrical connections"- Thats the first time ive heard you state that, im flattered! haha. Normally you state the opposite, which is what I expected since my ending results were not what I predicted.

      "if PAR is high, but the dyeing sucks, connections have degraded."- Could you explain how that happens, how PAR can end up being the same in two cases, but in one its good and the other its bad?

      The big thing im wondering is WHY it turned out to have a higher PAR, and why it took longer to achieve PAR then typical....??

      Thanks for your input Fibergeek, and if somebody can host the pic I can take one for you.

      Comment


      • #4
        The PAR of 2.5 ohm/sq ft was the approx. value that Fibergeek obtained in his testing. If you look at his document, you'll see his PAR value ranged from 2.2 - 2.8 ohms (or there about). The determination of PAR is dependent on a number of variables such as temperature and the thickness of the anodized layer, as well as to some extent the aluminum alloy being anodized.

        If the PAR value was high AND had you had poor dyeing it is most likely because of poor connections which increases the resistance in the electrical connection (only true if you have eliminated other reasons for poor dyeing..ie..contamination, spent dye, dye temps, etc). If you have a high PAR value with good dye qualities, it is because the layer is thicker...where the thicker layer also equals higher resistance.

        The PAR value and the voltage is variable, depent upon your setup and circumstances and everyone will have different readings. The point is to detect the slow rise in voltage to its peak, then the slow drop. As Fibergeek mentioned, you probably reached equilibrium in that the coating was being eaten away by the acid at the same rate it was being grown. I've run into the exact same situation where the voltage stayed relatively constant for a long period of time and never actually dropped and the dyeing was excellent.

        Comment


        • #5
          To say what Neilfj and I said yet another way:

          Setting PAR = 2.5 Ohms/sq.ft. is an average value that is useful in predicting how much voltage will be required for a given current density. You can adjust this value to suit your circumstances if you like.

          It has turned out (to my suprise) that explaining a good connection is actually harder than making one. Neilfj had one heck of a time with this until he figured it out. Now you know how to do it too. This means you get to help us explain it to others who may be having trouble with it. Everyone is the product of their own experiences, adding your way of explaining this can only help.

          To further explain "PAR is high but dyeing sucks"; like Neilfj said, some extra resistance caused by the connection degrading can fool you. You can't tell if all the resistance you see (calculate) is real or just the bad connection. There is a way to get around this; it's called in electronics a "Kelvin Connection". It involves adding another connection to the work to sense the resistance only, it carries no anodizing current. This is typically operated at very low current, like 1 microamp. Since V = I x R, if I is very small, V will also be very small for a given R. The degradation of the connection caused by the anodization will be vanishingly small, if there is even any at all. But this is another wire; and you will probably also need some amplification, we can avoid this with good connections.

          BTW, you don't have to reach equilibrium to grow a layer as thick as you did.

          Comment


          • #6
            Thanks for the replies. So whats your thoughts on my red dye? The Ph is MUCH higher then the dye manufacturer told Caswell it should be, yet it seems to work now. Also, im not sure that you have addressed the issue of it taking at least 30 minutes longer to achieve a decent ano layer. Im guessing that something im doing must be "wrong", or it would be done anodizing closer to 90 minutes. I realize there are variables, but if the effects of such variables can chage the timeframe so drastically im wondering how many others that purchase caswell's kits using the new LCD will have similiar experiences given the supplied instructions of 90 minutes ano time in the LCD instructions. IF it's working for everybody else closer to 90 minutes, why is it taking longer for me? Any thoughts? Thanks again

            Comment


            • #7
              It may be your red dye is working in spite of the too high pH. The pH values you were given are from the dye manufacturer, it may work better (or faster?) if you use the recommended pH.

              It probably wasn't necessary for you to continue the process for that long after equilibrium, but it didn't seem to hurt anything. Anodizing for say 10 minutes after PAR stopped changing could have been enough.

              I don't think you are doing anything "wrong". I was able to get an equilibrium condition several times at 3 A/sq.ft. current density. One time I let it run for an extra hour, another time I stopped shortly after it occurred, the dyeing was great both times. I have noticed that equilibrium occurs more easily if the work is very small for the volume of the tank. I have not seen it at all when the work was around 0.5 sq.ft. in 3 gallons of electrolyte, conditions don't stay uniform enough I guess.

              If what you saw really is prevalent among other users I haven't heard about it.

              Comment


              • #8
                well, it looks like I didnt hit that equilibrium until 15-20 minutes after the "normal" 90 minutes were up. I was just wondering why it took longer for me?

                Comment


                • #9
                  The difference in measured voltage between 13.40V (1hr. 40min.) and 13.50V (1 hr. 53 min.) is less than 1% (13.40 / 13.50 = .9926).

                  The difference in voltage between 13.16V (1 hr. 23 min. = 83 min.) and 13.50V is about 2.5%. Your are actually very close to my times, but your anodize is thicker. PAR at this point is 2.93 Ohms per sq.ft. For any practical purpose, this should yield results just as good as what you had at 3 Ohms per sq.ft.. You could have stopped here.

                  Differences as small as these look like normal experimental error.

                  Try it again. If you can do this consistently you're doing real well.

                  Comment


                  • #10
                    It may look like a small difference on paper, but the difference is an easy to scratch ano that barely takes the dye compared to a good hard to scratch ano layer that takes the dye as it should. I have been doing the 90 minutes thing since I started LCD with unsatisfactory results, and this is the first time I have got results even close to the old method previously used. I have done that same sample piece about 4-5 times at 90 minutes, and this is the only time it took the red dye and didnt scratch VERY easily. Maybe I NEED to let it run at the peak voltage for awhile....maybe thats the key. Im simply trying to figure out the problems I was having, and if you say the time difference wasnt it.....im left wondering why it worked this time and not before. There was one difference, I used air agitation this time and havnt thus far with LCD since the bubbles during ano are hardly noticeable. Maybe that was the thing that made everything work this time? It would simply amaze me if that was the case.

                    Comment


                    • #11
                      I have noticed that running past PAR usually makes dying easier. I have done a few parts lately that were run past PAR at equilibrium for 10-20 minutes and dyed solid black in 2-3 minutes. Blue hasn't been a problem, I have found if the blue doesn't take then it is a bad anodize job. Red seems to dye best if run past PAR. I have anodized to PAR, and the red wouldn't get dark, so I have pulled parts back out of the dye, rinsed and anodized more. Sometimes, just a few more minutes in the anodizing tank makes a big difference, and the red dye takes.

                      My red PH is high too, 7-8. I noticed when I mix a new batch of red, it will not work as well untill it ages a day or so. I have read that distilled water will drop in PH as it is exposed to air, because it absorbs CO2 which is acidic. The red dye concentrate seems to be fairly high on the PH scale.

                      Comment


                      • #12
                        I think caswell should mix up some of their red dye and see what it measures new. Id be surprised if it is in the range that their dye supplier states it should be.


                        Today I ano'd another part to test things out again.....see if it would still work as it did yesterday. I ran a larger piece, and ran into something strange. After anodizing the part was very dark. Instead of a yellowish tinit, it was almost grey/black. I actually had to go get a camera to take a couple pics! Any ideas why something would get so dark? IT's suppose to be the same series of aluminum as I have been using also. It's dark enough that just sealing and having it look silver would be totally out of the question. Any ideas?

                        It's late, but if I did the math right PAR turned out to be 4.29ohms.

                        Comment


                        • #13
                          Sid, I believe you calculated PAR correctly. The high value of PAR indicates a very thick anodized layer. This sounds like a higher current density than 4.5 A/ft^2 was achieved. I have achieved finishes like that when I was going for higher current density anodizes. I usually don't see a dark gray layer formed until my current density is over 15 A/ft^2 or so... I'm not sure exactly why your part reacted as it did.

                          By chance, was your part bead blasted? I notice that my bead blasted finishes always appear more gray than a gloss finish.

                          Please post your pictures, I'd be interested to see them. I'll post some of my pictures later on today.

                          Comment


                          • #14
                            no, it was machine finished. It was either 6061 or MIGHT have been cast then machined. Would cast aluminum react like that? How can I post pics? Thanks

                            Comment


                            • #15
                              Most cast alloys that I have anodized have turned very dark gray. The only color that takes well is black.

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