Homework Solutions #7

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					                                  Homework Solutions #7
                                   CE260/Spring 2000
Chapter 13: Coagulation Flocculation
1. Colloids are charged entities and they derive their stability from their charge. More
   highly charged ions are more strongly attracted to colloids and neutralize them,
   allowing them to agglomerate. The Schulze-Hardy rule states that there is roughly an
   order of magnitude increase in coagulation ability with each unit increase in charge of
2. First look at the overall equation (Table 13.2) for the chemistry of each chemical.
   Alum for this question only has 14.2 waters attached to it unlike that in Table 13.2
   which has 18 waters. The only thing this changes is the numbers of waters on the
   right hand side. Just write it out and check to make sure it is balanced.
   Need to compare the number of equivalents of the species being consumed to CaCO3.
   Using that relationship, stoichiometry and molecular weight of the chemical being
   added you can calculate alkalinity for each chemical in the following way:
         40 mg / L alum          2 meq Ca ( HCO3 )  1 meq CaCO3  100.1 mg 
Alk                     3 mmole                   
                                                                        
                                                                                
        597.9 mg / mmole              mmole         1 meq Ca ( HCO3 )  2 meq 
                            Alkalum  20.1 mg / L as CaCO3
                         Alk ferrous sulfate  14.4 mg / L as CaCO3
                           AlkFeCl3  37.0 mg / L as CaCO3
                         AlkFerric sulfate  30.0 mg / L as CaCO3

7. From equation 5 in Table 13.2 you can calculate then the grams of
   Ca(OH)2/Fe2(SO4)3 needed (which is the lime requirement).
   a) Calculate the daily addition of ferric sulfate (D=QC) and from that you can get the
      daily lime dosage needed = 375 kg/d (assuming little lime present).
   b) Again using eq. 5 Table 13.2 calculate the amount of solids produced. From that
      calculate the daily load of Fe(OH)3 and then calculate the daily load of incoming
      solids add the two together to get the total amount of solids produced (1103
      kg/d).Using the 90% removal rate you get the answer to be 993 kg/d produced
      from the sedimentation basin.
8) From what is given determining the basin dimensions will be no problem. From
   detention time and flow rate you can calculate volume and then using the
   relationships given for the depth you find the dimension to be: L = w =1.41m, and
   D = 1.76 m.
   Use equation 13.6 along with table values for viscosity and density at 15C to
   determine the input power to be = 3156 w.
   Remember that power dissipation in an unbaffled square tank is 75% of power
   imparted in a baffled square or baffled circular tank. At the high G value, flow is
   assumed to be turbulent. From Table 13.4, for 6 flat blade vane disk turbine, K =
   6.30 for turbulent conditions. Define the imparted in a circular baffled tank as Pc.
   Use the following formulas (found on page 392) to calculate N and then be sure to
   check your assumption of turbulent conditions by calculating the Reynolds number.
                               ,   K Re p , for turbulent flow p  0
                          N D
                             3 5

                          Pc  KN 3 d 5     P  0.75 Pc  0.75 KN 3 d 5 
                                            N  92 .8 rpm
9) Rapid mixing is required for coagulation to rapidly disperse the coagulant to allow it
   to react with the particles in the water before it reacts with water itself. Mixing in the
   flocculation basin must be in an intermediate range where the velocity gradients cause
   fluid movement and particle contact. If mixing is too intense large particles will be
   sheared apart.
19) a)Calculate the Vc (volume of individual compartments) = 83.2 m3 = wD2.
    Remember the area of the paddles (Ap)should be 15-20% of the area of the
    compartment. Calculate the area of a single paddle (ap) and the paddle area per shaft
                                     a p  w p l p  0.375 m 2
                                   A ps  4 * 0.375  1.50 m 2

   The length of the paddle and associated clear space is in the vicinity of 2.50 + 0.60 =
   3.10m. Let w = n(3.10) where n is the number of paddle sets, therefore wD2 =
   n(3.10)D2 = 83m2. Based on the previous equation and keeping in mind the range of
   15-20% for the area of paddles n will be set at 3. The number of paddles per shaft is
   therefore = 12. With a wall spacing of 0.30m and the spacing between adjacent
   paddles set at 0.60 m the width of the compartment is 9.30 m and the length is 3.00
   b) The shaft will be located at mid-depth (1.50m). The outer edge of the outer
      blades will be at a distance of 1.50-.30 = 1.20m from the shaft. Placing the outer
      blade midway between the shaft and the outer blade, its midpoint is located at
      0.525 m from the shaft. Its inner edge will be a distance of 0.45 m from the shaft.
   c) Assume CD = 1.8, and k = 0.25. Using eq. 13.6 calculate the power to be 272 W.
      The power expenditure for paddle Pp has the following eq:
                                                                        
               Pp  1.44 x10 4 C D bN 1  k  ro4  ri 4  0.273N 3 ro4  ri 4   
                             For inner paddle Pp  0.0242N 3
                             For outer paddle Pp  0.234N 3
                          Total power exp enditure P  1.55N 3
                   First Chamber N  3 P            5.60rpm
                 Second Chamber N  3 P             3.52rpm
                   Third Chamber N  3 P            1.92rpm
d) The design rotational speed should be the middle of the motors. For the first
   compartment, the range is 2.80 to 11.20 rpm. For the second compartment 1.76 to
   7.04 rpm. For the third compartment 0.96 to 3.84 rpm.

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