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6-42   WATER AND WASTEWATER ENGINEERING











                            Around-the-end baffled channel type (plan)  Over-and-under baffled channel type (section)
                            FIGURE 6-22
                            Baffled channel flocculation system.


                              Example 6-9.   Design a flocculation basin by determining the basin volume, tank dimensions,
                            required input power, impeller diameter, and its rotational speed using the following parameters
                            and the manufacturer’s data:
                                                                   3
                                                                      3
                                             Design flow rate     11.5     10   m  /d
                                             Flocculation  t      30 min
                                                                                         1
                                            Three flocculator compartments with  G      70, 50, 30 s
                                            Water temperature     5 	 C
                                            Place impeller at one-third the water depth

                            From manufacturer’s data the following impellers are available:


                            Impeller type  Impeller diameters (m)  Power number (N p )
                            Radial       0.3     0.4     0.6        5.7
                            Axial        0.8     1.4     2.0        0.31


                                  Solution.  Two flocculation basins are provided for redundancy at the design flow rate. The
                            maximum day flow rate is assumed to be twice the average day flow rate.
                                                   3
                                                             3
                                                      3
                                  a.  Convert 11.5     10   m  /d to m  /min.
                                                                 3
                                                              3
                                                      11 5  10 m /d                3
                                                        .
                                                                       7 986 or  8 0 m /min
                                                                               .
                                                                       .
                                                                                    m
                                                        , (
                                                        1 440 min/d)
                                 b.  With two flocculation basins, the maximum day flow through each will be
                                                                   3
                                                              80 . m /min       3
                                                          Q                40 . m /min
                                                                   2

                                 c.  Using  Equation 6-13, determine the volume of the flocculation basin.
                                                                   3
                                                     V     Qt   (40 m /min )(30 min )   120 m 3
                                                               .
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