2. (a) Alum is added in the aeration tank of activated sludge process for phosph
ID: 702928 • Letter: 2
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2. (a) Alum is added in the aeration tank of activated sludge process for phosphorus removal The effluent limit for total phosphorus is 0.5 mg/L. The alum is dosed at 140 mg/L. The average total phosphorus concentration is 9.0 mg/L effluent of the primary clarifier. Assume organic P removed in biological solid is 2% by weight (i) Determine the molar ratio of aluminum to phosphorus. (ii) Determine the weight ratio of alum dosed to the phosphorus content in the wastewater. (ii) Estimate the amount of sludge generated, assuming 0.5 mg/L of biological solids are produced by 1 mg/L of BOD reduction and the operational data are provided: influent BOD, 148 mg/L; effluent BOD, 10 mg/L; influent TSS 140 mg/L; effluent TSS, 12 mg/L (b) A sedimentation tank with 65% solids removal efficiency treats a wastewater from a population of 15,000. The per capita flow rate is 400 Id. The wastewater contains total suspended solids of 250 mg/L The sludge contains 4.5% total solids, and 60% of the solids are volatile. The specific gravity for volatile solids is 0.9, and that for fixed (i) Calculate the required surface area of the sedimentation tank if (ii) Determine the specific gravity of the sludge and the volume of (iii) The volume of sludge is reduced by 40% in a sludge thickening solids is 2.5. the surface overflow rate is 20 m*/m2d sludge produced daily. process. Determine the solids content in mg/L in the thickened sludge. State all assumptions made in your calculationExplanation / Answer
The principal source of aluminum for use in phosphorus precipitation is "alum", a hydrated aluminum sulfate, having the approximate formula A12(S04)3' 14H20 (molecular weight of 594). The chemical, which is also known as "filter alum" or "papermaker's alum", averages about 17% soluble aluminum expressed as Al20 3 or about 9.1% expressed as AI. Its reaction with Pol- may be written as follows:
AI2(S04)3' 14H20 + 2P043-~2AIP04 + 3S042- + 14H20
The sulfate remains in solution as S042-. The above reaction indicates that lIb-mole of alum (594 lb) will react with 2lb-moles (190lb) of P043- containing 62 lb phosphorus to form 2 lb-moles (244lb) of AlP04. The weight ratio of alum to phosphorus is, therefore, 9.6: 1. The alum requirement per pound of phosphorus may also be derived from the Al:P mole ratio as follows:
Mole ratio AI:P = I: 1
Therefore weight ratio Al:P = 27:31 = 0.87:1
Alum contains 9.1% Al
Therefore, alum required per lb of P =0.87:1=9.6 lb
Alum to P wt ratio 140/9= 15.5:1
Al to P wt ratio 15.55×0.091= 1.415
AI to P Mole Ratio 1 0.87
AI to P Mole Ratio 1.621:1
Sludge from Primary Treatment - Alum Addedd
Al Dose: = 140×-.091=12.74 mg/l
9 - 0.5= 8.5 mg/l P removal
8.5/31 = 0.27meg (milliequivalent)/ Al P04
12.74/27 = 0.47 meg/ I Al
0.47 - 0.27 = 0.2meg/l Al excess
0.27 x 122 = 32.94 mg/l Al P04
0.2 x 78 = 15.6 mg/l Al (OH)3
32.94+ 15.6 = 48.54 mg/l sludge produced
Sludge from BOD =( 148-10 )*0.5=69 mg/l
Tss = 140-12 = 128 mgl
Toton sludge = 245.54 mg/ l
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