Question

Problem 2 (20 points) 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 °C). 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: Percentile Particle diameter (um) > 50 > 60 Particle density (kg/m) > 1.4 x 10 > 1.5 x 10 90 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 (WxLxH, in m) of the sedimentation tank for removing 90% of suspended solids?

Answer 1

© solution :- (3) @) flow Given that, rate = 0.04 m/s @ temperature = 20°C ş, td > 2hx 36008th 느 IN = 3 72008 -3 1-002 x10 pas MIN Pw volume 998.2 kglma of tank axta < 0.04 x 7200 288m3 Particle diameter (num) particle perventile density 21.04 Xi03 >50 90 > 60 > 1.5x103 80 og jay removing 80% of suspended solids (VB OFR) g (pp - Puld² 18 -81 < (1500-998. 2) x86 x15 °) - 8 18 x 1.00 2x103 = 7.861x 10mis Nolu, plan area = Q = 0.04 "(taxL) OFR 7.861 x 10 = 50.88m2

frow, = 3W) → 3W2 = 50.88 W 50.88 - 4-12m, l = 3W>>12.36m 3 H VA =5.66M 50.88 .. IL 12:36 = 282 لدا 4. 12m @1 bori suspended solide H 5-66m removing X 90% (us work? 9.81 x (1400-998.2) x (50 xio Xooq 18 x 1.002 X103 = 4.917 x 104 mils Now, plan area (1x4) = 0/0FR = 0.04 4 4-917 X10 = 81-35m2 NOW (L=3 (11) -3W2 = 81.35 =) W = 5:21m, L= 3W = 15-63 m ☆ V 288 A = 3.54m 31:35 81.35 3 H- = 12 15.63m TW 5.21m H 3.54m