Consider a wastewater flow of 10 MLD with a BOD 5 value of 600 mg/L. It is proposed to treat this waste using an activated sludge process (ASP). The size of the proposed aeration tank will such that the water is detained for 8 hours. Complete mixing is assumed in the aeration tank. The BSRT ( ) of the system is 10 days. In addition to the BOD5 , this water also contains nitrogen and phosphorus. The nitrogen concentration expressed as Total Kjeldahl Nitrogen (TKN) is 40 mg/L as N, and phosphorus concentration is not limiting. Assuming that nitrification occurs in this system, calculate the effluent TKN concentration, and the additional oxygen requirement for nitrification.
X r = 10000 mg/L; Y T = 0.5; Kd = 0.05 /d; K =0.1 L/mg/d.
(KS ) N = 2.0 mg/L; (Y T ) N = 0.2; (Kd ) N = 0.05/d (q m ) N = 2 /d
Biomass may be represented as C5 H 7 O 2 N
Nitrogen incorporation in heterotrophic biomass must be accounted for.
Nitrogen incorporation in autoprophic biomass may be neglected.
Oxygen Requirement (in Kg/d) = 1.5.Q.(S o - S e ) -1.42.( D X)
Formula weight of biomass, C5 H7 O2 N = 5(12)+7(1)+2(16)+1(14) = 113
Therefore, production of 1990 Kg/d biomass requires:
Influent Nitrogen concentration = 40 mg/L (as N), or
Therefore, Nitrogen available for nitrification = 400 - 247 = 153 Kg/d
Or, [TKN]o = 15.3 mg/L (as N)
0.25 MLD of sludge is generated from the activated sludge process of a wastewater treatment plant. The solids concentration in this sludge is 13,000 mg/L. This sludge is first processed in a sludge thickener, where the solids content is increased to 4 percent (weight basis). Next the sludge is treated in the anaerobic digester for reduction in sludge solids. The digested sludge is then applied to sludge-drying beds for reduction of water content to 55%. Determine the weight and density of dried sludge that will be produced per day.
Density of Water = 1000 Kg/m3 ; Dry Density of Sludge Solids = 2200 Kg/m3
COD reduction efficiency of sludge digester = 60 %; ;
For sludge digestion, YT = 0.06 mg/mg, and Kd = 0.03 /d on COD basis
Consider 1 m3 of sludge after thickening
Let the sludge density be
Weight of this sludge =
Therefore, Weight of solids in 1 m 3 of sludge = (0.04).(1).(1022.305) =40.89 kg.
Solids Concentration (X o ) = 40890 mg/L
Let the discharge of thickened sludge be Q sl
Therefore, assuming 100 percent solids capture,
Q sl . (40890) = Q w . (13000) or, Q sl =
Solids Loading ( D X) =
; Digester Volume (V d ) = ;
Assuming 60% treatment efficiency,
Also, , Also,
Anaerobic Sludge Production ( =
Therefore, Reduction in Solids =
Therefore, Solids Loading to Sludge Drying Beds = (3250-1891) = 1359 kg/d