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1. A cross- flow, horizontal shaft, paddle-wheel flocculation basin is to be designed for a flow...


1. A cross- flow, horizontal shaft, paddle-wheel flocculation basin is to be designed for a flow of 20.000 m ld a mean veloci
1. A cross- flow, horizontal shaft, paddle-wheel flocculation basin is to be designed for a flow of 20.000 m ld a mean velocity gradient of 33.33 s (at 10 C), and a detention time of 45 min. The GT value should be from 50,000 to 100,000. Tapered flocculation is to be provided, and three compartments of equal depth in series are to be used, as shown in Figure 1. The G values determined from laboratory tests for the three compartments are Gl= 60 s-1, G2 = 30 s. G3-10 s". These give an average G value of 33.33s-The compartments are to be separated by baffle fences, redwood baffle fences, and the floor of the basin is level. The basin should be 14.0 m in width. The velocity of a paddle blade relative to the water is approximately 0.75 of the peripheral blade velocity Determine: I. The GT value. 2. The basin dimensions 3. The paddle- wheel design. 4. The power to be imparted to the water in each compartment 5. The rotational speed of each horizontal shaft in rpm. 6. The peripheral speed of the outside paddle blades in m/s. Assume, side wall clearance is 30 cm, consider 4 paddle wheel shafts with three blades on each side. Paddle length : 3 m, width 0.15 m. Area of blade should be 15 to 20% of tank profile area. Round th numbers up. Make any necessary assumption. Please sketch a longitudinal section of the tanks(s and a sectional view of one paddle-wheel-all with proper dimensions Given: Di-2.81 m; D2=2.50m; D3= 1.54m; μ at l 0°C 1.307*10-3 Ns/m2, C,-1.5
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Ans) Given,

Q = 20000 m3 /day or 0.231 m3 /sec

Mean velocity gradient (G) = 33.33 /sec

Detention time(t) =45 min or 2700 sec

G1 = 60/sec , G2 = 30/sec , G3 = 10/sec

Width of basin = 14 m

Velocity of paddle = 0.75 times peripheral blade velocity

1) GT value

GT = G x t

= 33.33 x 2700

= 89991

Since the GT value is between 50000 to 100000, therefore the detention time is satisfactory

2) Basin dimensions

Basin volume(V') = flow x detention time

= 0.231 x 2700

= 624 m3

Profile area = Volume/ width

= 624/14

= 44.57 m2

Assuming square compartments, and length as thrice of its depth(d),

therefore, d x 14 x 3d = 624

d2 = 14.86

or d = 3.85 m

Hence, length(L) = 3 x 3.85

= 11.55 m

3) P = Cd x A x \rho x (v3 /2)

where, Cd = Coefficient of drag

A = Paddle area (m2)

\rho = density of liquid

v = velocity of paddle relative to water

P = 1.5 x (3 x 0.15) x 1000 x ( v3/2)

Peripherial velocity(V) = RPM x d2 x \pi / 60

Assuming Rotation of paddle wheel as 4 RPM,

V= 4 x (2.50 x 4) x 3.141 /60

V = 2.09 mps

Therefore, v = 0.75 x 2.09

= 1.57 mps

Therefore, P = 1.5 x 3 x 0.15 x 1000 x 1.573 / 2

= 1306.08 W

4) Power to be imparted (P') to the water in each compartment

P' = \mu G2 V'

where, V' = basin volume

\mu = dynamic viscosity of water (0.00131 N -s/m2)

G = velocity gradient for each compartment

P' ( for first compartment) = 0.00131 x 602 x 624

= 2920 W

P'(for second compartment) = 0.00131 x 30 x 30 x 624

= 736 W

P' (for third compartment) = 0.00131 x 10 x 10 x 624

= 82 W

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