Design of an activated sludge plant with recycleDefinitions Q: flow C: concentration X: particulates S: soluble M: mass Sludge recycle rate: R = Q4/Q1 Fractionation of COD . Fractionation of nitrogen . Treatment efficiency E = (C1-C3)/C1 or better: E = (M1-M3)/M1 . Volumetric loading BV = Q1*C1/V2 or better BV = M1/V2 . VSS.Sludge concentration Measured as SS (same as TSS). TS. VS . Sludge Mass MX = ΣV*X . Sludge loading BX = Q1*C1/(V2*X2) or better BX = M1/(V2*X2) . Sludge production FSP = Q3*X3 + Q5*X5 or better FSP = M3 + M5 . Excess (surplus) sludge production FOSP = Q5*X5 or better FOSP = M5 . Observed yield FSP = Yobs (C1-C3)*Q1 or better FSP = Yobs (M1-M3) . Sludge age / sludge retention time θX = MX/FSP . Typical sludge retention times . . aerobic/FSP MX.Aerobic sludge age θX. aerobic = ΣVaerobic*X . aerobic = MX. The ratios are used as an estimation of if enough organic matter is present.C/N ratio and C/P ratio For both denitrification and biological phosphorous removal enough organic matter is needed. . SVI • Take 1000 ml MLSS from an activated sludge tank and let it settle in a cylindrical glass for 30 minutes. • SVI = (ml sluge after settling) / (g SS/l before settling) .Sludge Volume Index. DSVI • Take 1000 ml MLSS from an activated sludge tank. 1:2. 1:4 and let it settle in a cylindrical glass for 30 minutes.g.Diluted Sludge Volume Index. • DSVI = (ml sludge after settling) / (g SS/l before settling but after dilution) . 1:1. dillute it e. 1:3. SVI determines the MLSS concentration in the tanks when the tank size is given . Treatment plants can be designed by: • Volumetric loading: BV = M1/V2 • Sludge loading: BX = M1/(V2*X2) • Sludge age (nitrification) and denitrification rate (organic matter quality) • Computer aided design – Should only be used for plant optimization not for basic design. Here “rule of thumbs” are safer! . Design of activated sludge treatment plants for nitrogen removal and phosphorous removal A manual . Estimate the degradability of the wastewater. BOD. COD. N. and P. BOD. Note that this is not the same as the outlet standards. N. Find the flow and the loadings of.Key data for design • • Chose the lowest temperature at which nitrogen removal must take place. Is it easily or not so easily degradable? • • . and P that the plant must be design to treat. COD. Find the expected discharges of SS. aerobic) .Design of nitrification and denitrification 1 4 5 2 3 • Estimate the necessary aerobic sludge age ( X. In general it is better to use the later unit as the yield chosen hence is independent on chemical additions to the plant.• Estimate the observed yield (Yobs) without chemical sludge production in either the unit of kgSS/kgBOD or kgCOD/kgCOD. . The same graph from another book . • Estimate the chemical sludge production CP.2-1. – In the first case.6 (mole Me) / (mole Pinlet). the needed molar ratio is 0.chem. In the second case. – The chemical sludge production can be calculated based on the assumption that MePO4 and Me(OH)3 are the end products of the process.e.4 (mole Me) / (mole Pinlet). The molar ratio needed to obtain good P-removal depends on whether chemical P-removal is the only process applied or if there is also biological P-removal.2-0. I. the needed molar ratio is 1. First all PO4 is precipitated by Me and then the rest of the metal goes into the side reaction producing metal hydroxide. but good enough for estimating the chemical sludge mass produced. This is not entirely true. These values are based on experiences in full-scale treatment plants. . SS: 5-15 g m-3. P: 0. Typical values are BOD: 2-3 g m-3.0 g m-3. .3-1.• Estimate the outlet concentrations from the treatment plant (C3). N: 4-6 g m-3. 7 gVSS/gSS. – NOTE that the ratio of kgVSS/kgSS decreases when chemicals are added for P-precipitation! For systems without biological P-removal the ratio is roughly 0.• Determine the sludge production (FSP) and the excess sludge production (FESP): – FSP = Yobs (MBOD1 – MBOD3) when the unit of Yobs is kgSS/kgBOD – If the unit of Yobs is kgCOD/kgCOD then FSP = Yobs (MCOD1 . – If there is chemical sludge production: – FSP = Yobs (MCOD1 .chem – FESP = M5 = FSP – X3 .MCOD3) + MP. But now the unit of FSP becomes kgCOD/d and one must convert from kgCOD/d to kgSS/d.5 kgCOD/kgVSS and then multiplying with the ratio of VSS to SS in the sludge. This is done by multiplying with a conversion factor between COD and VSS in the sludge. The later is around 0.6-0.MCOD3). typically 1.8 kgVSS/kgSS when no chemicals are added to the plant.4-1. The exact value can be found from the analysis of the actual excess sludge of the plant in question. – MX.• Determine the necessary aerobic sludge mass.aerobic FSP .aerobic = X. if these values are not available. . This is done either based on the actual analysis of the sludge or.denit). Estimate the nitrogen concentration in the excess sludge.• Find the nitrogen that is to be denitrified (MN. on a theoretic estimate of the N-content of sludge. • Estimate the denitrification rate (rX.S(NO3) . • Determine anoxic sludge mass. MX.anoxic = MN.denit / rX.S(NO3)) based on the biodegradability of the wastewater. .• Estimate the suspended solids (SS) concentration in the process tanks (X2). Typical values are 4-6 kg SS/m3. anoxic + MX.aerobic) – taerobic = MX. – tanoxic = MX.anoxic + MX.• For plants with recycle determine the aerobic and anoxic reactor volume.aerobic) . For alternating plants determine the total reactor volume and the fractions of time where the plant must operate in each phase.anoxic / (MX.aerobic / (MX. denit at the inlet to the denitrification reactor is > 4 kgBOD/kgN .• Check that ratio between CBOD and CN. Design of biological phosphorous removal • Estimate the volume of the anaerobic tank (1 – 3 hours retention time) • Check that the ratio between soluble COD and soluble P is > 10 gCODsol/gPsol. . If nitrate is present at the inlet to the anaerobic tank then this ratio must be increased.