Question

A well-mixed sample (volume = 30 mL) of mixed liquor from an aeration tank of an activated sludge wastewater treatment plant was divided into two halves. The first half was filled in a 22.285 g crucible and placed in a drying oven at 105°C for 24 hours. The weight of the crucible and the residue after drying was 22.344 g. The crucible was then placed in a furnace at 550 °C for 30 minutes. After that the crucible with the ash weighed 22.310 g. The second half was filtered through a 0.151 g glass-fiber filter. After that the filter was placed in a drying oven at 105°C. The weight of dried filter with the residue was 0.196 g. The filter was then placed in a furnace at 550 °C for 30 minutes. After that the filter with the ash weighed 0.167 g.

a) Calculate (in mg/L): Total Solid (TS), Total Suspended Solid (TSS), Total Dissolved Solid (TDS), Total Volatile Solid (TVS), Volatile Suspended Solid (VSS), and Volatile Dissolved Solid (VDS).

b) Which value could be assumed the microorganism concentration in the aeration tank (X)

c) If the influent of the aeration tank of this activated sludge plant has a flow rate of 0.1 m3 /s, find the volume of the tank, if the required detention time is 2 hours.

d) In the influent BOD5 is 95 mg/L, find the food to microorganism ratio (F/M) for this activated sludge plant in ?? ??∙? . e) If a 2- Liter graduated cylinder was filled by an activated sludge sample from the same plant and the settled volume after 30 minutes was 800 mL, find the SVI.

Answer 1

(a) Total solids TS = The residue remaining after sample being evaporated and dried at 105⁰C = 22.344 - 22.285 = 0.059 g/15mL = 3933.33 mg/L

Total volatile solids TVS = Solids burned off when TS is ignited at 500⁰C = 22.344-22.310 = 0.034/15mL = 2266.66 mg/L

Total suspended solids TSS = Portion of TS retained on 0.151 g glass filter after being dried at 105⁰C = 0.196-0.151= 0.045 g/15mL = 3000mg/L

Volatile suspended solids VSS = Solids burned off when TSS is ignited at 500⁰C = 0.167-0.151 = 0.016g/15 mL = 1066.66 mg/L

Total dissolved solids TDS = TS - TSS = 3933.33 -3000 = 933.33 mg/L

Volatile dissolved solids VDS = TVS - VSS = 2266.66 - 1066.66 = 1200 mg/L

(b) Microorganism concentration = X = Total Suspended solids = 3000 mg/L

(c) influent flow rate Q = 0.1m³/s

Detention time t = 2 hours

Volume of tank V = Q x t = 0.1 x 2 x 60 x 60 = 720 m³

(d) Influent BOD Yo= 95 mg/L = 95000 mg/m³

F/M = Q X Yo = 0.1 x 95000 = 9500 mg/s = 9500x60x60x24 = 820.8x10⁶ mg/d

(e) Sludge volume index = V_ob / X_ob

V_ob = Quantity of sludge in the liquor in ml/l = 800/2 = 400 ml/l

X_ob= MLSS = 3000 mg/l

SVI = 400/3000 = 0.133 ml/mg = 133.33 ml/g