折板絮凝池计算例题(国外英文资料).docVIP

折板絮凝池计算例题(国外英文资料) Example 1-2 design water is Q = 100,000 m3 / d, self-water coefficient is 1.08. Solution: (1) set up a set of two flocculates Single pool design flow rate (2) the volume of the flocculation pool and the total volume of the flocculation pool are determined 1) the flocculation time T = 13min 2) the flocculate requires a net volume V = 2QT = 2 times 0.625 x 13 times 60 = 975m3 3) flocculating pool partition, and water room, the folding plate occupies 30% of the volume, and the actual volume of the flocculation pool is 1.3 volts 4) the net volume of a single flocculation pool V = QT = 487.5 m3 5) reference design of advection sedimentation tank size, pool L = 12.50 m wide, effective depth H = 3.5 + H1, H2, the H1 for flocculation pool water head loss, H2 for flocculation tank to tank head loss The effective water depth H = 3.5 + 0.4 + 0.1 = 4.0 m High 0.3 m, 0.6m The pool width of a single flocculation pool is taken from B = 9.75 m (3) the intake pipe calculation 1) set up a feed pipe, and its design traffic is equal to 1.25 m3 / s = 1250L/s Take the velocity v = 1.11 m/s The tube diameter is DN1200, and a feed pipe takes two flocculates (4) design with water The net length is 5.7 m, and the net width is 2.5 m It enters a flocculate velocity v = 0.7 m/s, and D = 1.06 m, relative to the depth of 2m The size of the water distribution is 2.5 times 5.7 x 2.0 m3 (5) the division of the division is calculated 1) the flocculation pool is a multi-channel flocculation pool, which is fitted with a foldboard box, which is a parallel folding plate Fourth, the flow rate is respectively V1 = 0.3 m/s; V2 = 0.25 m/s; V3 = 0.20 m/s; V4 = 0.15 m/s 2) the first calculation The first tranche is divided into eight squares, each width of 1.3 m The lattice each net long m L = 1.60 m long The actual flow rate In one of the folded box, there are five folded plates, divided into six squares The spacing of the folded plates in the folding box B = 0.25 m The fold Angle is 90 deg