Chapter 14, Problem 14.3QPP

Q2: Design an activated sludge process to treat a waste flow of 15000 m³/day with a BODs of 180 mg/L following primary treatment. The effluent BOD, and SS are to be 20 m mg/L. Assume Xr = 15000 mg/L, X = 2500 mg/L, 0c = 10 day, Y = 0.60, kd=0.05. Determine 1- the reactor volume, 2- the Sludge production rate, 3- the circulation rate, 4- the hydraulic retention time, and 5- the oxygen required? C₁-Ce C 1 XV = 1 +0.532 QxC₁₂ VXF F= - 1+r (1+0.1r)² = Qr= dt Өс Qx Xr-X V R = [= Q YQ(So-S) Oc dx XV 1+Kd0c O₂demand =1.47 (So-S)Q-[1.14Xr(Qw)] Qw total sludge production/X, S = BOD, effluent - 0.63 * SS W

مارت حولة ملانول 60 Design Deceleration tank and screen for W.W.T.P of flow-144000 m² day Design Data: D.T=1-3 min 1-3b, d=1-1.5 m 60 Design Data: velocity= 0.6: 1.5 m/sec . A-Qdfv=b.d, b=2d 233 Design Data: bars used circular of p= 1.5:3 cm. rec. of (1:2 cm) x (2:6 cm) spacing between bars 2.5:5.0 cm fine screen, 2.5 7.5 cm coarse screen Sloping angle on hz (6)=45:60 = depth of water depth in approaching channel=d = ⚫nbars nspacings +1 ⚫b=nbars xo+nspacings I spacing net inclined area = 2 Aapproach channel=nspacing. spacing. L. 1=d/ sin Check: = 1.13 C 2.56 1134(6.05 P. 45.30 *velocity just befor the screen=v₁ = Qa(m³/sec) screen-b.d ≥ 0.6 m/sec. * velocity through the screen=v2 16mm use 2 Qa(m³/sec) nscreens (spacings Spacing).d ≤1.5 m/sec IsP. 22-65 12²² head losses = Ah = 1.4 x ≤0.1 m 2g bav usp = Ubar Q 23.65 k 148.72

Can you please explain the steps of this problem especially the variables for Q when solving for the shear force? Thanks.