Ion Exchange Backwash in a WWTP 1

[esass13]

Is there a limit to the amount of brine backwash from an IX filter that can be put in a small WWTP? The filter backwash has high chlorides, TSS, TDS, etc. The WWTP is small and treats about 4,000 gpd. I could dose the backwash water into the WWTP, but at what concenctration levels will the brine not kill the bugs in the plant?

thanks

 

[bimr]

The salt concentration would have to exceed several thousand mg/l before you would have an effect on the activated sludge process.

http://www.jstor.org/pss/25035591

The salt wastewater actually originates in the brining process and the first part of the rinse, not in the backwash water. Therefore, the volume is much less than if it was occuring with the backwash.

Dosing into the wastewater stream is preferred, but only if dosing can be accomplished economically.

 

[arguellesmanuel]

You should check the volumes of waste vs. the final volume to download, and also review the chloride discharge limits in your area.
Usually biological systems are limited in the removal of high amounts of chlorides.

M.Argüelles

 

[jartgo]

I've been told by several folks that the threshold for salt in an activated sludge process is 8 parts per thousand. Actually there were a range of numbers presented, but 8 seemed to be on the lower end of the threshold. That said, I've performed the subsequent calculation like this:

There are essentially three stages of regeneration:
1) Backwash: 4gpm/s.f. for 10 mins....in my case 480 gallons

2) Recharge: 5# of salt per c.f. of resin, in my case I have 40 c.f. of resin in each vessel, therefore requiring 200# of salt per vessel. Brine is assumed to be 2.5# of salt per gallon of brine, so it will pull 80 gallons of brine into a flow of 1/2 gpm per c.f. This gives 900 gallons of water containing 80 gallons of brine.

3) Rinse: 10 min at 4 gpm/sf...in my case 480 gallons per vessel.

The total flow to the waste stream is then 1860 gallons containing 200# of salt, for each vessel. In my case I had two vessels so I used a SAY value of 4,000 gallons and 500# of salt.

So, we have:
500/(4,000 * 8.34) = 0.015 (15 parts per thousand) as a slug flow to the waste stream. In reality one might argue that this flow mixed with the other wastewater already in the wastestream might be under the 8 parts threshold. However, in my case, that wasn't an agreeable approach so we dosed it into the sewer over a day.

The receiving WWTP has an average flow of 60,000gpd. As a general check, 500/(60,000 * 8.34) = 0.001 (1 part per thousand).

What's frustrating is that when you look at the big picture, the WWTP is not going to remove the salt, so it doesn't really make any sense to me to put the salt through the plant...the salt is ultimately going to end up in the river....but I just do what the regulators tell me.

Hope this helps.

 

[bimr]

jartgo, it appears that you have an error in your calculations.

If one uses the salt volume of 200 lbs per regeneration, that would equate to around 6,000 mg/l of salt concentration in a 4,000 gpd treatment plant, or;

if the salt volume of 200 lbs per regeneration, that would equate to around 400 mg/l of salt concentration in a 60,000 gpd treatment plant, or;

if the salt volume of 500 lbs per regeneration, that would equate to around 15,000 mg/l of salt concentration in a 4,000 gpd treatment plant, or;

if the salt volume of 500 lbs per regeneration, that would equate to around 1,000 mg/l of salt concentration in a 60,000 gpd treatment plant.

The calculatins assume a 24 hour retention time aeration tank in the treatment plant. The actual salt concentration would be slightly less as the salt will be discharging with the effluent at the same time.

This is the correct calculation:
Salt load (in lbs/day) = X mgd * X mg/l * 8.34

 

[jartgo]

bimr,
I think we're using the same calculation with the same result. That is, 1 part per thousand is equal to 1,000 parts per million (or 1,000 mg/L). I'm just converting gallons of water to pounds of water, then dividing pounds of water into pounds of salt. The way I suggested is the same as yours with the equation already rearranged and using the units as described in the problem.

 

[bimr]

Ok. your unusual terminology of parts per thousand threw me off.