Question

4000 m3/d of wastewater with a total suspended solids (TSS) concentration of 200 mg/L is subjected to treatment by coagulation and flocculation followed by sedimentation. The ferric chloride dose is 25 mg/L and the TSS removal is 90%. Estimate the daily volume of sludge produced if the sludge has a specific gravity of 1.05 and moisture content of 92.5%. The following reaction is applicable: 2FeCl3 + 3Ca(OH)2 ? 2Fe(OH)3 + 3CaCl2

Answer 1

volume of waste water = 4000 m³/day = 4*10⁶ L ( 1 m³ = 1000 L)

TSS concentration = 200 mg/L = 200*10^-6 Kg/L ( 1 mg = 10^-6Kg)

Total weight of TSS = volume of waste water*TSS concentration = 4*10⁶ L*200*10^-6 Kg/L = 800Kg

Total weight of TSS removed = 90% of Total weight of TSS = 0.90 *800 Kg = 720 Kg

feric chloride dose = 25 mg /L =25*10^-6 Kg/L

weight of ferric chloride = volume of waste water*feric chloride dose = 4*10⁶ L*25*10^-6 Kg/L = 100 Kg

Total Solid weight = Total weight of TSS removed+ weight of ferric chloride = 720Kg+100 Kg = 820Kg

% of solid in sludge = (100 - moisture content) = 100-92.5 = 7.5 %

Assume total weight of sludge = X Kg

Total solid weight = 7.5% * X = 0.075 X

0.075 X = 820 Kg

X = 10933.33Kg

Specific gravity of sludge = 1.05, density of sludge = 1.05*10³ Kg/ m³

Volume of sludge = Total weight of sludge /density of sludge = 10933.33Kg/ 1.05*10³ Kg/ m³ =

10.413 m³

Volume of sludge=10.413 m³

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