H2S removal 1

[BRIS]

Groundwater has a pH value of 8 and hydrogen sulphide concentration of 8mg/L, (temperature 31 deg c).

At pH 8 it can be expected that only 10% of the hydrogen sulphide will exist in the unionized form as H2S. 90% will be in the ionized form HS-

we want to remove the hydrogen sulphide by simple aeration. (no chemicals)

Is trhis possible ?

 

[bimr]

No. Removal of hydrogen sulfide from sulfur waters by plain aeration is unsatisfactory unless the water is acidic.

 

[dalcazar]

could you diffuse CO2 into the water to lower pH and at the same time "carbon-dioxate" the water to pull H2S along? What volumes are we talking about here bris?

 

[semo]

Even if the pH were lower, Aeration would have a difficult time with that level of H2S. You will probably require some filtration (manganese greensand) and chlorination to effectively removal it.

Chlorination will require a feed of 2 parts CL2 to each part of H2S for oxidation. A greensand filter will also require permanganate feed to maintain the media.

If the source of the H2S in the well is sulfur reducing bacteria, you might contact a well driller to try shocking the well with CL2. That could lower the H2S levels until the bacteria build up again.

 

[bimr]

The chlorine dosage per each ppm of H2S ranges from 8.32 ppm at pH values of 6.4 or less to 5 ppm at pH values from 9 to 10.

 

[BRIS]

Thanks for the responses

1) The volume is 250,000 m3/day so we really have a problem with chemical oxidation. I had the rule of thumb 2ppm of Cl per 1 ppm H2S. This would be a problem. Cl dose at 8.33 ppm per 1 ppm of H2S would be a serious problem.

2) The aquifer is anaerobic fossil water. I suspect that sampling procedures have not been adequate to purge and flush the well and the high H2S may be from SRB contamination from the well. However the results from 15 different exploratory wells are consistent. we have recommended re sampling after running the artesian wells for several days to ensure that samples are representative of the formation.

3) The response confirms my concerns that we have a problem.

Brian

 

[bimr]

As a further explanation:

The amount of chlorine to oxidize 1 ppm of H2S to water and sulfur is 2.08 ppm. Depending on the pH of the water and the amount of sulfide present, it may take up to 8 atoms of chlorine to oxidize 1 molecule of hydrogen sulfide instead of the 2 atoms theoretically required to oxidize the hydrogen of the H2S to water and liberate the sulfur. This is because either all or a great part of the sulfides are oxidized to sulfates and this oxidation takes place to a great extent even when a deficiency of chlorine is added.

While the chlorination of raw waters containing high contents of sulfides would be rather expensive, chlorination is of great value in eliminating the small residuals from the effluent of other sulfide removal processes.

 

[BRIS]

bimr thanks for that pointer. Following up I find the chemical reactions are:

Cl2 + H2S == 2HCL + S
(2.1 mg/l of Cl to 1 mg/l H2S)

or

4Cl2 + H2S + 4H2O == 8HCl + H2SO4
(8.4 mg/l of Cl per 1 mg/l of H2S)

The first reaction predominates at pH values > 8.

 

[cameron54]

I have a question about an equation I have come across - conversion factor. Where does 7420 come from in the following equation?

(H2S ppm x air flow cfm)/7420 = lb/day h2s

 

[BRIS]

cameron54 you need to start a new thread.