معلوب H.W: An activated sludge process operates under the following data: Q=0.21 m³/s, So= 250 mg/l, S=6.2 mg/l, MLVSS (X) = 3500 mg/1, 0c= 10 days Return sludge concentration (X)= 10000 mg/l, Y=0.5, Ka =0.06 d¹, volatile super solid represent 80% of SS, Find:

Repeat Problem 2.14 with the following values: Gs = 2.75, temperature of water = 21C, t = 88 min, and L = 11.7 cm. 2.14 A hydrometer test has the following result: Gs = 2.65, temperature of water = 26 C, and L = 10.4 cm at 45 minutes after the start of sedimentation (see Figure 2.25). What is the diameter D of the smallest-size particles that have settled beyond the zone of measurement at that time (that is, t = 45 min)? Figure 2.25 ASTM 152H type of hydrometer placed inside the sedimentation cylinder (Courtesy of Khaled Sobhan, Florida Atlantic University, Boca Raton, Florida)

Two materials, A and B, are to be separated using two unit operations, 1 and 2. The feed has 10 tons/h A and 4 tons/h B. The split (fraction of material rejected by each operation) is given in Table P5-14. a. Which sequence of operations, (l?2) or (2?1), will yield the greatest recovery of material A? b. What will be the purity of material A using that sequence? c. What will be the efficiency of separation for the entire process train with regard to material A using the Worrell-Stesed equation?