SWG - Amps Volts and bipolar cells

chem geek, no fuss, but I mast say that it is nice to see some one that "switched on" in electro-chemistry. I was working for the company for 10+ years and only come across 3-4 people who understand the process :)
Nice metting you :)
From your post above - I agree on all aspects. The only comment I have to make that you have described IDEAL conditions. For example, when electrolytic cell operates in real application it's efficiency is not 100% due to chemical reactions taking place some energy waste takes place. Some of it comes from the fact that in the reaction's happening NACL molecule get split and then recombined back as side reaction ( this is only account for minor % loss). Or other example, depending on catalyst used, say RuO is 90% energy efficient, so 90% energy will produce disinfectants and 10% will be side reactions. On the other hand Pt based anode material is only about 80% efficient. This is not to mention that different anode manufacturing technologies will produce anode material with different internal resistance ( that will waste some energy as you need extra voltage to drive it). And so on.... :-D
:-D

Mast add my head is pretty screw up as well. I can be in bed with my woman fouling in sleep and evaluating the design of new chlorinator and thermal load on the power components.....
As some one said once: people around you see car doing wheel spins burning rubber, you see heat generated combustion..." :roll:
 
mas985 said:
I for one am glad you took over this one.
Regarding your statement

Therefore, for a fixed voltage system, chlorine production rate should be proportional to salinity as well.

That should be true regardless of cell design. So even though these bipolar cells are like having multiple cells connected in series and therefore use a higher external driving voltage with lower external current, it would still be true that the chlorine production rate should be proportional to salinity (really, conductivity, but that's roughly proportional to salinity in the range you are talking about). If conductivity increases, the current increases (both internal and external), so production increases -- assuming constant externally applied voltage.

The statement Strannik made that it is the power that needs to be looked at to determine chlorine production output is a better way of looking at the problem, but isn't a precise relationship. Most electrolytic cells, regardless of monopolar or bipolar design, have a minimum voltage per cell element to start the electrolysis and then after that voltage and current work as if there is resistance (approximately). So, if we call this minimum voltage E0 and assume "n" cells connected in series, with all cells being equal, and if each cell has effective resistance to ion flow of R, then we have the following (where E and I are externally applied):

Ecell = E0 + IR
E = nEcell = nE0 + nIR
P = IE = nIE0 + nI2R
Chlorine Output (grams/hour) = n*1.32*I

So while it is true that looking at the externally measured current can throw you way off because it doesn't take into account "n", looking at the power also doesn't get you the right answer because the resistance is still a function of cell design (such as how far away the plates are or their area). It is a reasonable approximation if the resistance is low, which is generally a design goal for most cells since that minimizes energy loss to heat. Also, if one compares a monopolar cell that is the same overall size and distance between the plates as a bipolar cell that has its "outer" externally connected plates at roughly that same distance, then the "nR" factor on the resistive losses is close to the same (ignoring plate thicknesses, etc.). So that's probably where the "compare the power" rule comes from. It's a rough guide, but not an absolute rule -- it still has some dependency on cell design in terms of overall end-to-end resistance.

Richard
 
chem geek said:
mas985 said:
I for one am glad you took over this one.
Regarding your statement

Therefore, for a fixed voltage system, chlorine production rate should be proportional to salinity as well.

That should be true regardless of cell design. So even though these bipolar cells are like having multiple cells connected in series and therefore use a higher external driving voltage with lower external current, it would still be true that the chlorine production rate should be proportional to salinity (really, conductivity, but that's roughly proportional to salinity in the range you are talking about). If conductivity increases, the current increases (both internal and external), so production increases -- assuming constant externally applied voltage.

The statement Strannik made that it is the power that needs to be looked at to determine chlorine production output is a better way of looking at the problem, but isn't a precise relationship. Most electrolytic cells, regardless of monopolar or bipolar design, have a minimum voltage per cell element to start the electrolysis and then after that voltage and current work as if there is resistance (approximately). So, if we call this minimum voltage E0 and assume "n" cells connected in series, with all cells being equal, and if each cell has effective resistance to ion flow of R, then we have the following (where E and I are externally applied):

Ecell = E0 + IR
E = nEcell = nE0 + nIR
P = IE = nIE0 + nI2R
Chlorine Output (grams/hour) = n*1.32*I

So while it is true that looking at the externally measured current can throw you way off because it doesn't take into account "n", looking at the power also doesn't get you the right answer because the resistance is still a function of cell design (such as how far away the plates are or their area). It is a reasonable approximation if the resistance is low, which is generally a design goal for most cells since that minimizes energy loss to heat. Also, if one compares a monopolar cell that is the same overall size and distance between the plates as a bipolar cell that has its "outer" externally connected plates at roughly that same distance, then the "nR" factor on the resistive losses is close to the same (ignoring plate thicknesses, etc.). So that's probably where the "compare the power" rule comes from. It's a rough guide, but not an absolute rule -- it still has some dependency on cell design in terms of overall end-to-end resistance.

Richard

According your calculation 25Amp 'should" generate approx. 33grams/hour ( 1 x 1.32 x 25 )
As per book you are correct. As per real life application, depending on number of factors, 25Amp generate about 25 grams/hour. This is an real major error that some "newly born" SWG manufacturers get wrong :)
 
Valera Orlingis said:
chem geek, no fuss, but I mast say that it is nice to see some one that "switched on" in electro-chemistry.
Thanks, though I would have preferred to have "switched on" to it a bit earlier so as not to gone down a conceptually wrong path. It would also help if y'all would just explain something instead of saying "I'm not going to disclose technical details as they are commercial secret". I understand you can get frustrated with our not knowing what you do, but we're not stupid and there isn't any secret sauce I disclosed in the basics of bipolar cells. Anyway, we've got that all resolved now.

I definitely got it about the losses. Even so, your SMC30T has an input of 300 Watts and an output of 25*6.9 = 172.5 Watts (maximum) so that's 58% efficiency just from the transformer and related electronics. The 33 grams per hour of chlorine production represents an "nI" of 33*0.76 = 25.1 Amps (which is "ideal" and doesn't account for losses I'll get into) at an "ideal" E0 of around 1.5V ignoring concentration effects (which are not insignificant) so that's around 38 Watts for making chlorine if there were no resistance so that leaves 135 Watts lost through all those inefficiencies you mentioned plus generated heat from the resistance from ion flow.

At least now I have a better idea of why some SWG systems are quoted at much higher output voltages than others. The higher voltage cells are more likely to be bipolar. I was concerned that the higher voltages, if in a monopolar cell, would lead to more side reactions, such as the production of oxygen. Now I know that these were probably just bipolar cells. That makes a whole lot more sense now.

Richard
 
Richard, Strannik just trying to be on the careful side as he bound by employment contract. So please understand.
In regards to SMC machine, the data published is first slightly out of date, second is estimation. In reality, SMC machine more efficient than what you calculated as we originally published (max power consumption) and average chlorine production (note not maximum). Powersupply uses Flex PCB China supplied for interconnection of the control and SMPS boards.
For example, due to the SMC chlorinator's ability to work in saline solution from 0.35 to 2.8% ( depending on Temperature) it will be producing about 38-40 grams/hour if seawater is used and at the same time will use much less power that 300 watts :) This is due to SMPS ability to become much more efficient when output voltage reduces.
 
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After reading these posts a couple times over I think I finally get it. I am well versed in electrical principles, 25 year career, but the chemistry is more challanging for me.

Let me read it back to see if I got it right.

The series distribution of cells acts as a voltage divider so the equivalent circuit is two parallel cells with half the volatage and same current thus the n factor chemgeek talks about.

The point I am a bit fuzzy on a few points:

Are there effectively three plate sections, two connected to the power supply (active) and one passive section in between the actives?

My cell is DC driven and it reverses polarity on each run cycle, does this mean it is bipolar?

Why is this configuration more efficient?

Are all the plates the same material?

Is there any reason that cell could not be driven by AC and thus no need to have polarity reversal to clean the cells. Why was DC original chosen?

Just curious.
 
My responses to your questions in bold as I understand it.

mas985 said:
The series distribution of cells acts as a voltage divider so the equivalent circuit is two parallel cells with half the volatage and same current thus the n factor chemgeek talks about.

Yup. If you are talking about a series of 2 cells, then this is equivalent (ignoring inefficiencies) to driving one cell with plates that have twice the area (remember that the bipolar plate has two sides so it's as if there are 4 half-sized plates in a bipolar 2 cell-like system) at half the voltage and twice the external current (though total current through all plates combined is the same in both cases). In practice, these systems have multiple numbers of bipolar plates so multiple "internal" cells. To have roughly the same output production of products, the single larger plate area cell has twice the distance between the plates as the 2-cell system distance between each of its pair of plates (so that total resistance from ion flow is about the same). The doubled distance in the single cell system is canceled out by the doubled area of the plates so that the net resistance to ion flow is the same as with the "two-cell" bipolar system.

The point I am a bit fuzzy on a few points:

Are there effectively three plate sections, two connected to the power supply (active) and one passive section in between the actives?

For two cells you are correct -- the outer two connected to a power supply while the middle plate is not connected to anything -- it just acts as a divider preventing ion flow between the cells (mostly) and as a conductor passing electrons from one side to the other. It's technically the same as having two separate cells with the anode of one connected to the cathode to the other. The main difference is that with a bipolar cell, there is inefficiency due to ion flow leakage between the cell compartments while in two separate cells connected together there is inefficiency in resistive losses in the wire connecting the anode of one cell to the cathode of the other cell.

My cell is DC driven and it reverses polarity on each run cycle, does this mean it is bipolar?

Depends on the definition of bipolar and I've seen it used both ways. Some say an electrode is bipolar if it can be used as either an anode or a cathode, so can be reversed in polarity (i.e. it's not optimized for only positive or negative). Others define bipolar in terms of the type of electrolytic cell and call the plates that act as an anode on one side and a cathode on the other and are not externally connected as "bipolar". It's confusing to have the same term used in different ways.

Why is this configuration more efficient?

Apparently the higher externally applied voltage can be maintained better, but there may be other reasons. Certainly, resistive losses are proportional to the square of the current so that may be part of the benefit of using higher external voltage with lower external current (similar to why transmission lines are stepped up in voltage).

Are all the plates the same material?

I believe so, but don't know for sure. That's probably proprietary :)

Is there any reason that cell could not be driven by AC and thus no need to have polarity reversal to clean the cells. Why was DC original chosen?

Unless the AC is VERY slow, electrochemistry won't work with AC. Essentially, the products of the reaction don't have time to build up and flow away so instead with AC you end up producing product and then converting it back again -- just moving back and forth rapidly in the equilibrium (chloride to chlorine gas to chloride to chlorine gas, etc. or hydrogen ion to hydrogen gas to hydrogen ion to hydrogen gas, etc.). The key to a good cell design is to get the products away from the plates as quickly as possible. The fluid flow dynamics inside these cells is where I'll bet a lot of the secret sauce is (as well as coatings on the plates to maximize one reaction over another and to enhance the life of the plates).
 
Thanks for the response and everything makes sense especially the AC thing. Didn't think about the chemical reaction time.

I suspect that my cell may be bipolar simply because of the voltage and amps that it draws. Specs indicate about 27.5 g/hr production rate (I've actually measured 30 g/hr in a spa test but volume & CL accuracy is not assured) and the cell currently runs at ~25.5v @ 6.3 amps. This would indicate a n factor of 4. At 6.3 amps and 4 sets of plates, 100% efficiency production should be about 33.14 g/hr so according to the specs, this makes my cell about 83% efficient. Does that sound about right?

If I go by the spa measurements then the efficiency is closer to 90% but that sounds a bit high. I might of had higher current for those tests but I unfortunately did not record it. I might try it again for kicks and see if it works out to the same efficiency.
 
The efficiency rating that you calculated is going from total current (externally applied plus presumed internal current flow at the bipolar plates) to chlorine production. I would expect that to be on the high side, but Valera listed several factors that make such efficiency lower. It sounded like some coating materials (he referred to them as the catalyst that is used because they probably lower the overvoltage aka activation energy of the desired product reaction) were 90% efficient while other less sophisticated (and probably less expensive) materials were 80% efficient. In this context, the lack of efficiency comes from side reactions that are produced (such as producing oxygen instead of chlorine).

The losses due to resistance don't affect the current, but do require higher voltages to achieve the same output so are less energy efficient. So all in all an 83% number sounds about right for the current-to-chlorine calculation. Of course, there are significant assumptions you made about an integral number of bipolar plates since that implies they are of the same plate area as the externally connected plates. If they aren't, then that obviously would throw off the calculations.

Anyway, it very much sounds like your system is of bipolar design.

So now we've got two of us "switched on" to electrochemistry. :)
 
One important comment about driving cell with AC voltage is it's lifespan. Every time you revere the anode you damage it. Reversing cell every 1/50th of the second will kill it very quick. Anode material manufacturers usually state maximum reversal rate when the supply anode to SWG manufacturers. The rate fluctuates from 2 - 8 hours ( as maximum reversal infrequency). Reversing it more often will reduce the nominated by manufacturer lifespan.
The whole concept of manufacturing powersupply module to operate the cell is about making the cell last the distance. There is a number of specialized circuits and logic that operates the cell module the particular way to maximize it life. The reason why cell module so carefully looked after is the cost. The cell cost to SWG manufacturers as much as the rest of the machine. Not many people realize how hight tech the anode material is and how much it cost...
 

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Here are a couple of comments. First ChemGeek is exactly right with the calculations regarding current density, this is the only factor when calculating chlorine output. There are many ways to connect the plates of a cell but the total current determines the chlorine output. Since Sodium is a conductive metal it sets the conductance of the solution(water), thus determines the current flow at a certain voltage. Since the current only flows one way, to avoid plating unwanted particles on the plate(electrode) the current is reversed periodically with most power supplies. The frequency of the reversals is not critical only that it happens. This is the easy part.

Pool water is not pure water with salt added, it contains other minerals and metals which obviously are conductive. Even the metal that is used in the
cell plates themselves is conductive. Passing a current between 2 electrodes placed in a solution is the exact process of electroplating. So if the cell is plated with a metal there is a danger that this plating will plate itself off during operation of the cell. This is why both plates are usually made
from the same material. External leakage must be held to a minimum to prevent unwanted migration of material from the plates. The solution(pool water) does contain oxidizers and acids that will etch the plates or oxidize the plates, then add to this a small amount of oils that will coat the plate and lower the conductance. Plate material choice becomes quite a challenge.

Since the water is not conductive and the sodium is, the sodium chloride will migrate to the cell plate, here the sodium is liberated from the chlorine,
also is some of the hydrogen from the water. These are both gases so the water flow through the cell washes them away. This process takes
some time that is why the cell reversal is usually longer term(hours). The process still works at a faster rate of reversal, but it is not as efficient.
In cell operation some of the metals contained in the water can be plated onto the cell plates, in some cases these metals will disturb the conductivity
of the plate itself. If they result in oxidation of the plate the plate will loose its ability to conduct current to the solution. As long as the cell is conducting
current it will produce chlorine as it is the current that produces the chlorine. Plate size will effect the conductance. The larger the plate the smaller the current density between the plates. There is a trade off on plate size and efficiency. If the plate size is too small the current will boil the water
do to its resistance. If the plate size is too big you will reduce the current density and thus the efficiency of chlorine generation.

Cliffs
 
Interesting data Cliffs, I have different test results on efficiency VS current density. Unless you are talking far extremes, I have tested variations in size of about 100% retuning stable, linear results of no change.
Also, there is another variable which is speed of water flow thru the plates, as cell produces Sodium Hydroxide that also combines with the Hydrochloric Acid to make Salt, Sodium Chloride, which is reused by the Cell - therefore loss of efficiency.
 
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