Actual SWCG chlorine rate compared to theory

[Knetsel]

Hello,
If I'm not wrong, an SWCG should produce about 1.3 g/h of chlorine per Amp. through the cell.
When reading specifications in user's manuals, they claim it is about three times more...
I surely missed something. Can you tell me what ?
Thanks in advance

 

[Texas Splash]

Hello! While amps are an important factor, the amount of chlorine produced can also vary by manufacture, type cell, and other variables. Did you have a specific make/model SWG in mind? We have had several discussion on the forum this season about chlorine "advertised" versus the actual amount we've seen produced. In many cases, the amount of chlorine produced seems to fall short of what a manufacture is advertising, but if you take the advice of purchasing an SWG rated at 2X the size of your pool it generally does a good job meetings your FC demands.

Below is one such discussion.

Circupool RJ-45 Production Test

Fellow RJ-45 owners, I am curious to know what type of FC production you are seeing with your SWG - specifically an overnight GAIN test. For my pool, just under 18K, the APP shows that my cell should provide ~3.9 ppm of chlorine in 7 hours at 100% output. I have conducted four overnight gain...

www.troublefreepool.com

 

[tim5055]

Salt water chlorine generators come in multiple sizes. They do not all produce the same amount of chlorine.

 

[SoDel]

Hello,
If I'm not wrong, an SWCG should produce about 1.3 g/h of chlorine per Amp. through the cell.
When reading specifications in user's manuals, they claim it is about three times more...
I surely missed something. Can you tell me what ?
Thanks in advance

If you can link to any papers laying out the theoretical SWG Cl production per amp per hour in pool water, I’d be interested and also would be happy to dive in to try to determine why the variance between theoretical and actual production. The only paper tangentially related that I’ve located to date just casually poking around over the past couple months is this one: Optimization of free chlorine, electric and current efficiency in an electrochemical reactor for water disinfection purposes by RSM

I also believe I’ve seen @JamesW make the theoretical production calculation (if I remember right) with ease and can likely shed more light on the question.

 

[JamesW]

The Hayward T-Cell uses 13 plates (blades).

The 2 white wires go to the center plate and one black wire goes to each outer plate.

This makes the box and cell work like a battery charger where the water between the plates is the batteries.

Assuming 24 volts DC and 6 amps, it’s like there are 2 sets of (6) 4volt batteries in series being charged with the sets in parallel.

The total amps are 6 amps x 6 cells or 3 amps x 12 cells (36 amps either way).

Amps are the measure of the flow of electrons.

The amperage is directly proportional to the chlorine production.

On one side of a plate, a chloride ion loses an electron to become a chlorine radical and then combines with another chlorine radical to create chlorine gas.

So, one electron, one chlorine radical produced.

One amp is defined as 6.28 x 10^18 electrons per second

36 amps x 60 x 60 x24 = 3,110,400

3,110,400 x 6.28 x 10^18 = 1.95 x 10^25 electrons per day

1 mole of atoms = 6.022 × 10^23 atoms

1.95 x 10^25 electrons can produce 32.4 moles of chlorine atoms.

Chlorine is 35 grams per mole.

That’s 1,134 grams per day = 2.5 lbs per day maximum theoretical production at 6 amps.

The units shut down at 8 amps, so 7.9 amps is the maximum possible production.

At 7.9 amps, the maximum theoretical production is 3.29 lbs per day.

So, maybe that is where they are getting the number.

In any case, I would like to see a report from a qualified lab that shows the actual measured production of chlorine gas.

 

[JamesW]

If you have 6 identical 4 volt, 12 watt light bulbs, each will require 3 amps and a total of 72 watts.

If you power them with a 4 volt power supply, you will put the bulbs in parallel and the power supply needs to be able to deliver 4 volts and 18 amps.

18 amps x 4 volts is 72 watts.

However, if you had a 24 volt power supply, you could not connect the bulbs directly to the power supply because the voltage is too high.

So, you would put the 6 bulbs in series and then the 24 volts is divided evenly and each bulb gets 4 volts.

The power supply sees 24 volts and 3 amps, which is 72 watts, but each bulb is getting 4 volts and 3 amps.

If you wanted to add 6 more bulbs to the 24 volt supply, you would create another series of 6 bulbs and add it in parallel.

So, the power supply sees 24 volts and 6 amps, but each bulb is getting the same 4 volts and 3 amps.

If you used a 4 volt power supply, it would need to deliver 36 amps.

For a 13 plate cell, the water between the plates becomes the individual load that requires 4 volts and 3 amps.

Instead of a light bulb creating light, the water produces hydrogen gas and chlorine gas.

So, for the calculations, for a 13 plate cell, you need to multiply the amps seen at the power supply by 6.

 

[JamesW]

This is from a T-15.

As you can see, the center plate and both outer plates get power.

The plates in between the powered plates are not directly powered.

The plates not shown go in the slots between the powered plates.

The center plate gets one polarity (either positive or negative) and the two outer plates get the opposite polarity.

This creates two sets of six cells where the six cells are in series and each set is in parallel.

 

[JamesW]

Note that the efficiency will not be 100%, so you have to account for that, but I don't know exactly what the efficiency should be for this process with this water chemistry.

There will be some oxygen produced, but I don't know how much compared to the production of chlorine.

The oxygen is still an oxidizer, so it might do some work that chlorine would do anyway.

For a 13 blade cell, below is the maximum theoretical chlorine production (Assuming that I am doing the calculations correctly).

The Hayward T-15 is a 13 blade cell and it has amperage in the 6 to 7 range for typical operation.

That puts the theoretical production at 2.5 to 2.91 lbs per day but Hayward says that the cell is rated for 1.47 lbs per day.

So, the question is, why does Hayward say 1.47 lbs per day?

Is the process 100% efficient?

Probably not.

1.47 lbs per day is about 1/2 of the possible output at 7 amps.

Does that suggest that the process is only 50% efficient?

Maybe Hayward has made an error in their calculations or testing or maybe a typo?

Maybe Hayward is being extra conservative to avoid liability?

Amps..........lbs per day

5.0..............2.08

5.2..............2.17

5.4..............2.25

5.6..............2.33

5.8..............2.42

6.0..............2.50

6.2..............2.58

6.4..............2.67

6.6..............2.75

6.8..............2.83

7.0..............2.91

7.2..............3.00

7.4..............3.08

7.6..............3.17

7.8..............3.25

T-15 has 13 Titanium Plates, 150 x 63mm. Produces 1.47 lbs/day (Per Hayward rating).

T-9 has 13 Titanium Plates, 101 x 63mm. Produces 0.98 lbs/day.

T-5 has 7 Titanium plate, 150 x 63mm. Produces 0.735 lbs/day.

T-3 has 7 Titanium Plates, 101 x 63mm. Produces 0.53 lbs/day.

For the 7 plate cells, you have one set of six cells in series.

The below reference shows 30 grams per hour at 5 amps, which is 720 grams per day or 1.59 lbs. vs the theoretical amount of 2.08 lbs. per day, which implies an efficiency of about 76% assuming they are using a 13 plate cell.

For an 11 plate cell, 5 amps should produce about 1.73 lbs per day.

If they are using an 11 plate cell, then it is about 1.59/1.73 = 92% efficient.

The IC40 uses 11 plates, which means that you multiply the amps at the power supply by 5 vs 6.

So, 6 amps at the power supply is 30 amps total for the IC40.

6 amps at an Intellichlor power supply makes as much chlorine as 5 amps at an Aquarite T-15 power supply.

 

[SoDel]

There’s even another way to calculate it using aggregate power and Coulombs.
24 volts at 6 Amps is 144 Watts.
144 Watts = -144 Joules = ∆PE
total charge moved q = ∆PE/∆V = -144/24 = -6 C
Number of electrons per second is total charge divided by charge per electron is
-6 / -1.6 x 10^-19 = 3.75 x 10^19
x 60 x 60 x 24 = 3.24x10^24 electrons per day
1 mole = 6.022 × 10^23 atoms
3.24x10^24 electrons can produce 5.38 moles (assuming one electron per atom) =
Cl @ 35 g / mole = 35 x 5.38 = 188.3 g Cl = 0.42 lb / day

Very different result. I’m not sure where I went wrong but it’ seems so far off, I must have went wrong..

edit — I‘m thinking the ∆V should be the voltage drop across the cell and not the potential across the plates — might be where I went wrong?

 

[JamesW]

Watts = -144 Joules

Watts is power and joules are energy, so you are mixing units.

 

[JamesW]

Looking at the power used compared to how much it should take to make hydrogen, the efficiency seems to be pretty low.

Chlorine is produced at the positive side of the plate and hydrogen is produced at the negative side of the plate.

The design is not optimized for production like it would be in a production factory.

There is probably a lot of current that just goes through the water without producing any chlorine or hydrogen.

There is some oxygen produced and some heat, but I don't know how much of either is produced.

 

[JamesW]

This is the conductivity of the water at 3,200 ppm.

6,000 microsiemens/cm.

So, you can calculate the amount of current passing through the water from plate to plate.

 

[Knetsel]

Hello,
The way I see it is the following :
Whatever the shape, the number and the connection system of the plates, there is (theoretical maximum) one mole of Cl- produced for each mole of e- passing through the cell.
One mole of e- corresponds to a charge of ~ 96500 Coulombs and one mole of Cl- has a mass of 35,5 grams.
One Ampere = 1 Coulomb /second thus 1 Amp

 

[JamesW]

there is (theoretical maximum) one mole of Cl- produced for each mole of e- passing through the cell.

You have to look at each volume of water between each set of plates as a separate load.

There are 12 loads that get 3 amps each which is equivalent to 36 amps.

If you used a 4 volt power supply with each load connected in parallel, it would require 36 amps.

 

[Knetsel]

Oups !
I inadvertendly sent part of a draft version of my post...Here-below the full post..

Hello,

The following is the the way I see it :

- electrolysis of NaCl occurs at concentrations high enough not to go into water electrolysis;

- whatever the shape, the number and the connecting system of the plates, there is one mole of Cl- produced for each mole of e- passing through the cell (theoretical maximum);

- one mole of e- corresponds to a charge of ~ 96500 Coulombs and one mole of Cl- has a mass of 35,5 grams.

- one Ampere = 1 Coulomb /second thus 1 Amp during one hour = 3600 C which corresponds to 3600/96500 moles of e- or Cl-.

- the mass of Cl- thus produced each hour is 35,5 x 3600 / 96500 = ~1.32 g.

- the power "consumed" at the cell is, of course, determined by the intensity and the voltage at the cell; these two parameters are, in turn, determined by the shape, the connections, the maximum intensity accepted by the plates per unit area, the conductivity, etc..

 

[Knetsel]

Hello,

Checking the specs of T15 cell, it is supposed to deliver 30 g/h of chlorine which means that the current flowing through the cell should be at least 30/1.3 =~ 23 A.
I didn't find any indication re nominal values for current and voltage of the cell but the values of 23.45 V and 6,47 A as an illustration of the display screen.
To me, any arrangement regarding the plates is a way to optimize the functioning of the cell.
The basis still remains : two cl- give one Cl2 and two electrons e-.

 

[Knetsel]

For SoDel, here below an extract of "White's handbook of chlorination..", page 9 :

======================================================================

The overall chemical reaction in a diaphragm cell is

NaCl + H 2 O + Electric current → NaOH + Cl 2 + H 2 . (1.3)

The principal anode reaction is
2 Cl − → Cl 2 + 2 e − (1.4)

Chlorine formed at the anode saturates the anolyte, and an equilibrium is established as follows:
Cl 2 + ( OH ) → Cl − + HOCl (1.5)
HOCl → H + + OCl − (1.6)

The principal cathode reaction is
2 H + + 2 OH − + 2 e − → H 2 + 2 OH − (1.7)
107 . 880/ 0 . 00111801 = 96,493 C (1.8)

Therefore, from Faraday ’ s law we know that 96,493 coulomb (1 f) will liber-
ate 1.0080 g of hydrogen and 35.457 g of chlorine in the electrolysis of salt.

Converting this to amperes per pound of chlorine per day, we get

A × 86,400 s / day / 96 , 493 C x 35 . 457 gCl 2 / 454 g /lb = 0 . 07 (1.9)

hence A × 0 . 07 = lb / day Cl 2

=================================================================
which is equivalent to A x 1,32 = g/h Cl2

 

[JamesW]

Checking the specs of T15 cell, it is supposed to deliver 30 g/h of chlorine which means that the current flowing through the cell should be at least 30/1.3 =~ 23 A.

You have to think about each individual cell between the plates.

If you powered each individually, each would require 3 amps.

It does not matter if you use a 48 volt 3 amp power supply and put all cells in series or if you use a 4 volt 36 amp power supply and put each in parallel.

The power supply does not see all electrons because the chloride ion that loses an electron passes it to the hydrogen ion on the other side of the plate.

So, you're getting electrons from the chloride ions and passing them to the hydrogen ions.

The power supply does not see those electrons.

The power supply only sees the electrons going in and coming out.

 

[Knetsel]

To JamesW :
I would very much like if you could link to any paper describing such a process as it is still difficult for me to understand how an anode can simultaneously be a cathode.
This being said, I do very much appreciate your teaching efforts. Thanks a lot.

 

[JamesW]

Google bipolar electrolysis cell.