Question

1) (20) Carbon adsorption is to be used to treat 4 ppm of TCE in 80,000L/hr of contaminated water. The density of the GAC is 460g/l and the Freundlich constants are K = 56.9 (mg/g)(L/mg)" and 1/n = 0.48. The treatment goal is 0.006 ppm and EBCT for the column is 12 minutes. Find the following: 1. The minimum carbon utilization Rate (CUR) (g GAC/L water treated) 2. The specific throughput = (L water treated/g GAC) 3. Total mass of GAC on the column 4. Total volume that can be treated with the column provided. 5. The bed life of the 10 min EBCT column

Answer 1

Solution : diven that: Inblaent concentralion > 4ppm = 4 ogle Frecendlich galensity paraméleos LYD=0.48 Treatment goal = 0.006 mgle EBCT - la miny Flow rale = 80 000 4 ho = 1333 33 L/min density of GAC = 460 gle So. 1. CVR = MGAC Cont A Quick De Cint k (Cint) in freundlich egn. 4 migl (56-9 (mg/g) (4 mg) Cungle jour Scanned with CamScanner

- 0.3393 g GAC/L water treated | 8. Specitic throughput: @tok MGAC Velna GBCT PE VE tok GB CT PE Wlhere, M "GAC = mag of GAC Kg tak = time to breakthrough at the trealment goal. to = Absorher density AYO, Carbon utilization våle cove) y the reciprocal of Specitic throughput so, Speci tic throughput sve 0.33939 GRC) I wälen treated] $.9472 (ewater treated lg GAC. Ics Scanned with CamScanner

3. Mass or Gec on the coların Volane Gert XaX PAC - (12 mjM) x (1333,33 /mfri) x (460 glx) = 7359981.69 = 7359.9816 kg y volume of water treated. mall OF GAC GAC utilization rate 7359981.69 (@.33937 G9/41 = 21691664 L Scanned with CamScanner