Criteria Values for Effective Flocculation
Type of Raw Water G, s-1 G to (dimensionless)
Low turbidity and colored 20-70 50,000-250,000
High turbidity 70-150 80,000-190,000
Chemical feed line
where the size cannot go on increasing. This limiting size is the critical size and depends upon the detention time and velocity gradient. Generally, the larger velocity gradients produce smaller critical sizes and the larger detention times produce larger critical sizes. Practice has it that this "mix" of detention time and velocity gradient can be combined into a product of G and to. Thus, the criteria values for effective flocculation are expressed in terms of two parameters: Gto and G. Table 6.3 shows some criteria values in terms of these parameters.
Figure 6.11 shows a longitudinal section of a flocculator. This flocculator belongs to the category of rotational mixers, only that it should be rotating at a very much slower speed. Notice that the compartments vary in size from the smallest one at the head end to the largest one at the exit end. This is an inverse relationship to the variation of the value of G; G decreases in value, instead. As the water or wastewater is introduced into the flocculation tank, the particles are smaller at the head end and then builds up until reaching the maximum size at the exit end of the tank. At the exit, the rotation of the paddle must be made much, much slower to avoid breaking up the flocs into pieces.
The paddle configuration in each compartment is called a paddle wheel. A paddle wheel is composed of paddle arms. In the figure, it is difficult to ascertain how many paddle arms there are in each paddle wheel, but what is obvious is that at least there are two arms per wheel. In each arm are then mounted the blades or slats. Assuming there are two paddle arms per paddle wheel, the total number of blades or slats per compartment is eight in the first compartment, six in the second compartment, and four in the third compartment.
The design of the flocculator of Figure 6.11 may be made by determining the power coefficients for laminar, transitional, and turbulent regime of flow field. We will, however, discuss its design in terms of the fundamental definition of power. Consider FD as the drag by the water on the blade; FD is also the push of the blade upon the water. This push causes the water to move at a velocity vp equal to the velocity of the blade.
The water is actually "touching" the blade, so the velocity that it attains on contact must be equal to that of the blade. Of course, as it departs, its velocity will be different, but this is not the critical point of power transfer. The blade transfers power to the water while still in contact. Upon detachment, the water parcel that got the power being transferred will then commence expending the power to overcome fluid friction imposed upon it by neighboring parcels; this process produces the velocity gradient required for flocculation to occur.
The paddle is rotational, therefore, vp is a tangential velocity referred to the axis of rotation at a radial distance rp as shown in the figure. Tangential velocity is equal to the radius times angular rotation o in radians per unit time. Thus, vp = rpo (6.36)
The drag stress in a fluid is proportional to the dynamic pressure ptvp 12, where Pi is the mass density of water. Thus, the force on a single blade, FD = CDApplvp2/2. CD is the coefficient of drag and Ap is the projected area of the blade in the direction of its motion. Power is force times velocity, so the power dissipation per blade is therefore p v2 p v3
The total power in the flocculator compartment is then the sum of the powers in each blade, thus,
Consider this equation as the flocculation equation. Apt is the sum of the projected areas of the blades. vp is the tangential velocity corresponding to rp; however, the location of the blades requires that there be several vp's. To use only one vp, the paddle tip velocity vpt is used multiplied by a factor a, as indicated in the equation. The product of vpt and a is the equivalent of the combined effects of the vp's of the several blades; it represents the conglomerate velocity for all the blades. The value of a is normally taken as equal to 0.75. With the expression for P found, the values of G and Gto may be checked to see if the flocculator performs at conditions of effective flocculation.
As mentioned before, higher values of G produce smaller flocs, while low values of G produce larger flocs. Although larger flocs are desirable, there is a time limit as measured by Gto to which they are allowed to form. If they are allowed to form much longer than some critical time, they will reach a critical size that, due to shearing forces, will simply break and crush to pieces.
Also, excessive velocity gradients can simply break the flocs to pieces. To prevent excessive velocity gradients between paddle tips, a minimum distance of 0.3 m should be provided between them. Also, a minimum clearance of 0.3 m should be provided between paddles and any structure inside the flocculator. Paddle tip velocity should be less than 1.0 m/s (Peavy et al., 1985).
Important parameters to be determined in design of flocculators are the dimensions of the blade. Thus, the flocculation equation, Equation (6.38), may be solved for Apt; once this is known, the dimensions of the blade may be determined. Apt is, of course, the sum of all the projected blade areas. The dimensions of each blade may be arbitrarily chosen based on this Apt. As shown in the figure, blades are normally rectangular in shape with length b and width D. Solving for Apt,
The value of the coefficient of drag is a function of the Reynolds number Re = Dvp/v; v is the kinematic viscosity. Assuming a blade of D = 0.25 m, the corresponding Re at vp = 1.0 m/s is 0.25(1)/10-6 = 2.5 x 105 at 20°C. The kinematic viscosity of water at 20°C is 10 m/s (Peavy et al., 1985). At an Re equals 105, the formula for CD has been determined empirically as (Munson et al., 1994).
The cD predicted by this equation applies only to a single blade. Work needs to be done to determine the value of cD for multiple blades.
Determination of electrical power input. To determine the electrical power input to the flocculator tank, P is divided by n, the brake efficiency, to obtain the brake or shaft power. The brake or shaft power is then divided by the M, the motor efficiency, to obtain the input electrical power. The electrical power input, of course, determines how much money is spent to operate the flocculator.
Example 6.7 A flocculator tank has three compartments, with each compartment having one paddle wheel. The ratio of the length of the paddle blades of the longest compartment to that of the length of the paddle blades of the middle compartment is 2.6:2, and the ratio of the length of the paddle blades of the middle compartment to that of the length of the paddle blades of the shortest compartment is 2:1. The flocculator is to flocculate an alum treated raw water of 50,000 m /d at an average temperature of 20°C. Design for (a) the dimensions of the flocculator and flocculator compartments, (b) the power requirements assuming motor efficiency M of 90% and brake efficiency n of 75%, (c) the appropriate dimensions of the paddle slats, and (d) the rpm of the paddle wheels. Assume two flocculators in parallel and four paddle blades per paddle wheel, attached as two per arm.
Solution: (a) Assume an average G of 30s- and a Gto of 80,000. Therefore, to = 44.44 min; vol. of flocculator = (50,000/2X44.44)^^^] = 771.53 m3. To produce uniform velocity gradient, depth must be equal to width. Thus, assuming a depth of 5 m and letting L represent the length of tank,
5(5)L = 771.53; L = 30.86 m Ans length of compartment no.1, the shortest = 1+2 + 2 6 (30 86)
Width and depth of compartment no.1 = width and depth of compartment no. 2 = width and depth of compartment no. 3 = 5 m, as assumed Ans Length of compartment no. 2, the middle = 2/5.6(30.86) = 11.02 m Ans Length of compartment no. 3, the longest = 30.86 - 5.51 - 11.02 = 14.33 m Ans Width of flocculator = depth of flocculator = 5 m, as assumed Ans
Assume the following distribution of G:
Compartment no. 1, G = 40 s-Compartment no. 2, G = 30 s-Compartment no. 3, G = 20 si = 10( 10-4) kg/m - s; V1 = (5)(5)(5.51) = 137.75 m3
Therefore, P¡ for compartment no. 1 =
0.75 (0.9) = 326.52 N m/s = 0.44 hp Ans ¥2 = 25 (11.02) = 275.50 m3
Therefore, Pt for compartment no. 2 = ^^0 752(059o0|)(30 ) = 0.44 hp Ans V3 = 25 (14.33) = 358.25 m3
-n, f m f , , Q 10(10-4)(358.25)(202) noc, . Therefore, Pt for compartment no. 3 = —-—0 775 ^^ 90)- = 0.28 hp Ans
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