Question

You are hired to design a rectangular sedimentation tank (length to width ratio = 3:1 and detention time ~ 2 h) for a new water treatment plant (flow rate: 0.04 m/s; temperature: 20 ). Suspended particles can be removed when the overflow rate is higher than the settling velocities of particles. You have finished characterizing the influent particles (assuming all of them are spherical) and obtained the following results: Particle diameter (m) Particle density (kg/m) Percentile 50 1.4 x 10 90 60 1.5 x 100 80 Assume Type I sedimentation. For practical construction purpose, each dimension (W, L or H) will be rounded to an accuracy of 0.1 m. 1) What will be the dimensions (W x L x H, in m) of the sedimentation tank for removing 80% of suspended solids? 2) What will be the dimensions (W x L x H, in m) of the sedimentation tank for removing 90% of suspended solids?

Answer 1

Given data- L = 3 B 능 detention time td=2h =3600x2 = 72000 d=60mm flow rate Temp of water = 206 8 = 0.04 m/s f = 998 kg/m3 7-80% density of particle 8 = 1500 g/m² particle us loix 10"mals Assume G= 2.65 (not inm) we know that percentage removal e efficiency 7 = V x 100 where is'-settling velocity Us - surface overflow rate 80 = Vs' X100 Vs Us Us'= 0.8 V g (G-1) d² =0.8& 18 ♡ 9.81 (2-65-1) (0,06)2 =018 Vs 18 1.01815-6 Vs = 4006156 m/s =4006.56 (LXB) 0.04 =4006.56 LXB LXB = 9,868106 3B229.868106 1.824x10 3 m L = 51478x10m (but L-36] B = Q=V LXBXH=288 td V = 0.04X7200 288 m3 {H = 28.85x10m

>> LXB= 678462 1-617X10 f=1400kg/m² y = 90% particle dia d = 50mm Again use formula percentage removal efficiency n = Vi x100 Us toogus Ve x100 Us 90 = 0.9ų f. (69) 2 0.3 9.8) (2.65-1)(0,05)2 = 0.9 Us =Us = 2473'18 m/s 18 1601810-6 = 2473.18 (LXB) 0.04 =2473118 LXB 3B² = 1.6178105 B = 2.39x103m L=6.96X103m Q=V Ed V = 0.048 7200 L x B x H 288m3 H = 288 2.32x10 3 x 6.96810 {Hz Hz 17835909. 63m