industrial effluent andwaste water residues

water treatment

1. Tutorial CDB 4223 Industrial Effluents & Waste residue

2. L2-1 Example 2-4: The following test results were obtained for a wastewater sample taken at the headworks to a wastewater-treatment plant. All of the tests were performed using a sample size of 50 ml. Determine the concentration of total solids, total volatile solids, suspended solids, volatile suspended solids, total dissolved solids and total volatile dissolved solids. The samples used in the solids analyses were all either evaporated, dried, or ignited to constant weight. Data: Tare mass of evaporating dish = 53.5433 g Mass of evaporating dish plus residue after evaporation at 105ºC= 53.5794 g Mass of evaporating dish plus residue after ignition at 550ºC = 53.5625 g Tare mass of Whatman GF/C filter after drying at 105ºC= 1.5433 g Mass of Whatman GF/C filter and residue after drying at 105ºC= 1.5554 g Mass of Whatman GF/C filter and residue after ignition at 550ºC = 1.5476 g

3. Solution : Determine total solids: TS = TS = = 722 mg/L 2. Determine total volatile solids TVS = TVS = = 338 mg/L  

4. 3. Determine the total suspended solids TSS = TSS = = 242 mg/L 4. Determine the volatile suspended solids VSS = VSS = = 156 mg/L  

5. 5. Determine the total dissolved solids TDS = TS-TSS = 722-242 = 480 mg/L 6. Determine the volatile dissolved solids VDS = TVS –VSS = 338-156 = 182 mg/L

6. Exercise Problem 2-7: The following test results were obtained for four different waste water samples. The size of the samples was 100 ml . Determine concentration of total and volatile solids, expressed as mg/L, for one of the samples. Item Unit Weight , gm A B C D Sample size mL 90 100 120 200 Tare mass of evaporating dish g 22.6435 22.6445 22.6550 22.6445 Mass of evaporating dish plus residue after evaporation at 105ºC g 22.6783 22.6832 22.6995 22.6667 Mass of evaporating dish plus residue after ignition at 550ºC g 22.6768 22.6795 22.6832 22.6433

7. Determine total solids: TS = TS = = 348 mg/L 2. Determine total volatile solids TVS = TVS = = 15 mg/L  

8. Exercise Problem 2-9: The following test results were obtained for a waste water sample taken at an industrial facility, All of the tests were performed using sample size of 100 ml . Determine concentration of total solids, total suspended solids and dissolved solids for one of the samples. Item Weight, gm A B C D Tare mass of evaporating dish 54.6422 54.6423 54.6424 54.6423 Mass of evaporating dish plus residue after evaporation at 105ºC 54.7022 54.7173 54.7224 54.7148 Mass of evaporating dish plus residue after ignition at 550ºC 54.6722 54.6893 54.6801 54.6818 Tare mass of Whatman GF/C filter 1.5348 1.5347 1.5347 1.5346 Mass of Whatman GF/C filter plus residue after drying at 105 ºC 1.5553 1.5586 1.5622 1.5571 Mass of Whatman GF/C filter plus residue after ignition at 550ºC 1.5453 1.5454 1.5471 1.5418

9. Solution : Determine total solids: TS = TS = = 750 mg/L 2. Determine total volatile solids TVS = TVS = =280 mg /L  

10. 3. Determine the total suspended solids TSS = TSS = = 239 mg/L 4. Determine the volatile suspended solids VSS = VSS = = 132 mg/L  

11. 5. Determine the total dissolved solids TDS = TS-TSS = 750-239 = 511 mg/L 6. Determine the volatile dissolved solids VDS = TVS –VSS = 280-132 = 148 mg/L

12. Calculation of BOD Determine the 1-day BOD and ultimate first stage BOD for a wastewater whose 5-day BOD is 200 mg/L. The reaction constant k (base e) = 0.23 d -1 . What would have been the 5-day BOD if the test had been conducted at 25ºC? Solution : Determine the ultimate carbonaceous BOD BOD 5 = UBOD - BOD r = UBOD(1- ) 200 = UBOD(1- ) = UBOD (1- ) = UBOD(0.6833) 200 = UBOD( 0.6833) UBOD = 293 mg/L 2. Determine 1 day BOD BODt = UBOD (1- ) BODt = 293(1- = 293(1-0.7945)= 60.1 mg/L 3. Determine 5-day BOD at 25ºC k 1 25 = k 1 20 (1.056) T-20  

13. k 1 25 = 0.23(1.056) 25-20 = 0.30 day -1 BOD 5 = UBOD(1- ) = 293( 1- ) = 293(1- ) BOD 5 = 293 (0.765) = 227.6 mg/L Exercise Problem 2-28: The 5-d 20ºC BOD of a wastewater is 185 mg/L . What will be the ultimate BOD (UBOD)? What will be the 10-d demand? If the bottle had been incubated at 33ºC, what would be the 5-d BOD have been? K= 0,23 d -1 Determine the ultimate carbonaceous BOD BOD 5 = UBOD - BOD r = UBOD(1- ) 185 = UBOD(1- ) = UBOD (1- ) = UBOD(0.684) 185 = UBOD( 0.684) UBOD = 270 mg/L 2. Determine 10-d BOD BOD 10 = UBOD (1- ) BOD 10 = 270(1- = 270(1-0.10025)= 243 mg/L  

14. k 1 33 = 0.23(1.056) 33-20 = 0.467 day -1 BOD 5 = UBOD(1- ) = 270( 1- ) = 268.2 mg/L  

15. Exercise Problem 2-29: The 5-d BOD at 20ºC is equal to 350 mg/L for three different samples, but the 20ºC k values are equal to 0.25 d -1 , 0.35 d -1 and 0.46 d -1 . Determine the value of BOD(UBOD) of each sample . Determine the ultimate carbonaceous BOD when k= 0.25 d -1 BOD 5 = UBOD - BOD r = UBOD(1- ) 350 = UBOD(1- ) = UBOD (1- ) = UBOD(0.713) 350 = UBOD( 0.713) ; UBOD = 490 mg/L (ii) When k= 0.35 d -1 BOD 5 = UBOD - BOD r = UBOD(1- ) 350 = UBOD(1- ) = UBOD (1- ) = UBOD(0.826) 350 = UBOD( 0.826) ; UBOD = 423mg/L ( iii) When k= 0.46 d -1 BOD 5 = UBOD - BOD r = UBOD(1- ) 350 = UBOD(1- ) = UBOD (1- ) = UBOD(0.899) 350 = UBOD( 0.899) ; UBOD = 389 mg/L