Ans) Given,
Q = 20000 m^{3} /day or 0.231 m^{3} /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 m^{3}
Profile area = Volume/ width
= 624/14
= 44.57 m^{2}
Assuming square compartments, and length as thrice of its depth(d),
therefore, d x 14 x 3d = 624
d^{2} = 14.86
or d = 3.85 m
Hence, length(L) = 3 x 3.85
= 11.55 m
3) P = C_{d} x A x x (v^{3} /2)
where, C_{d} = Coefficient of drag
A = Paddle area (m^{2})
= density of liquid
v = velocity of paddle relative to water
P = 1.5 x (3 x 0.15) x 1000 x ( v^{3}/2)
Peripherial velocity(V) = RPM x d2 x / 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 1.57^{3} / 2
= 1306.08 W
4) Power to be imparted (P') to the water in each compartment
P' = G^{2} V'
where, V' = basin volume
= dynamic viscosity of water (0.00131 N -s/m^{2})
G = velocity gradient for each compartment
P' ( for first compartment) = 0.00131 x 60^{2} 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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