Page 882 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological
P. 882

Appendix F:  Alum Data and Conversions                                                           837


              Example F.3 Determine Monthly Cost of Alum       et al., 1993). For a ‘‘standard’’ tank=mixer design (i.e., H(tank)=
                                                               D(tank)   1, D(impeller)=D(tank) ¼ 0.33, and for a six-blade
              Given                                            flat-surface impeller), this amount of circulation occurs if the
                                      _
              Result from Example F.2, i.e., M(alum monthly) ¼ 51,848  factor, nt T ¼ 39 at R   2000, in which n ¼ rotational velocity of
              kg Al 2 (SO 4 ) 3   14H 2 O =month               impeller in rev=s, t r ¼ time for five passes through the impeller,
                                                                                             2
              Required                                         and R ¼ Reynolds number, or R ¼ nD a r=m.
              Monthly cost of alum                                Regarding alum dissolution, the 99% completion of mix-
              Solution                                         ing is merely for the alum to be dispersed uniformly through-
              Monthly cost of alum:                            out the solution which occurs by five circulation passes
                                                               (McCabe et al., 1993); it does not ensure that the alum will
              Cost per 100 lb bag Al (SO 4 ) 3   14H 2 O ¼ $11 15(pick $15)  have dissolved. If we wish that the fluid have 10 passes
                                2
                                                               through the impeller (as a further hedge of our bets that the
                         2
                                                  ð
              Cost per kg Al (SO 4 ) 3  14H 2 O ¼ $15=100 lbð  Þ  2:205 lb=kgÞ  alum will dissolve, then we would have 99.99% completion
                                      ¼ $0:33=kg Al (SO 4 ) 3  14H 2 O
                                                2              of the mixing, and nt T ¼ 78.
                                                                  As an index of the propensity of alum to dissolve, from
                    Cost(monthly) ¼ M(monthly)   Cost
                                                               Table F.1, 10 g Al 2 (SO 4 ) 3   14H 2 Oat208C (688F) will dis-
                                ¼ 51,840 kg=m   $0:33=kg
                                                               solve 100% within 45 s with high-speed mixing in 1000 mL
                                ¼ $17,146=mo
                                                               water. Therefore, it would seem that if we choose nt T ¼ 78 (to
                                                               give about 10 circulation passes), the probability would be
              (Cost is about $11–15 per 100 lb (45 kg) bag, if purchased  high of say  99% dissolution of the granular alum being
              in large quantities.)
                                                               fed to the reactor.
              Example F.4 Determine the Cost of Alum Based
              upon Price per Unit of Al 3þ                     F.3.2 DISPERSION
                                                               A portion of the incoming solution will pass immediately
              Given
              Cost of alum is $0.33=kg Al 2 (SO 4 ) 3   14H 2 O  through the reactor (see, for example, Figure 4.11) and will
                                                               not be around for the 10 passes through the impeller. From
              Required
                                                               Figure 4.11, about 0.37 fraction of the solution remains in the
              Cost per unit of Al 3þ
                                                               reactor when t=u ¼ 1.0, about 0.14 fraction when t=u ¼ 2, and
              Solution                                         about 0.05 fraction when t=u ¼ 3. This means that if the time
              To convert cost from $=kg Al 2 (SO 4 ) 3   14H 2 Oto$=kg Al 3þ
                                                               for 10 passes, t T , is such that t T =u   0.2, then 0.90 fraction of
              (solid granular form),
                                                               the contents will have passed through the impeller 10 times,
                                                               i.e., 0.90 fraction will have remained in the reactor.
                $=kgAl 3þ  ¼ $=kg Al (SO 4 ) 3   14H 2 O
                                2

                            MW Al 2 (SO 4 ) 3   14H 2 O=MW(Al )
                                                    3þ
                                                               F.3.3 CALCULATIONS
                                   2
                        ¼ $0:33=kg Al (SO 4 ) 3   14H 2 O   594=54ð  Þ
                                                               From the discussions above, two criteria must be satisfied: (1)
                        ¼ $3:63=kg Al 3þ
                                                               nt T   K, in which K is specified for the reactor and is for five
                                                               passes through the impeller, and (2) t T =u   0.2, for 90% of the
            F.3 DISSOLUTION                                    contents of the reactor to be retained. Example F.5 illustrates
                                                               how these criteria may be applied.
            Factors that affect the rate of dissolution of a substance in a
            stirred reactor include the propensity of the solid to dissolve,
            the target concentration after dissolution, the size of particles,  Example F.5 Illustration of Method
            water temperature, intensity of mixing, and the kind of mixer=  to Determine u, n
            basin configuration. The residual fraction (solids not dissolved)
            depends upon the product of the dissolution rate times the  Given
            detention time, u. Since the objective is not to have a residual,  Standard design reactor
            this product should be adjusted until the residual is near zero.  Required
            This can be done for a given u, by adjusting n, the mixing  a. Determine u for 10 passes through the impeller
            speed. Actually, the mass rate of dissolution decreases as the  b. Determine n
            mass of substance in the reactor declines.
                                                                  Solution
                                                                    1. Apply criterion
            F.3.1 CRITERIA FOR MIXING
            Complete mixing (defined as 99% blending) occurs if the con-             t T
                                                                                        0:2,
            tents of the volume mixed are circulated five times (McCabe              u
   877   878   879   880   881   882   883   884   885   886   887