Question

4. (20 pts) A rapid mixing basin is designed to process 3.0 MGD of water. The basins will be square, and the depth is assumed to be 1.times of the width. Use a detention time of 30 s and velocity gradient of 750 s'. Two basins will be constructed each having a capacity to process exactly half the entire water flow. Determine: (a) Dimensions of the basin (L, W and H) (b) Power input (hp) (c) Rotational speed (rpm) of the impeller, assume Np = 0.2 for a radial impeller of 0.9 m diameter

Answer 1

Given = os Discharge=3MGD = 13638 m3/day Detention time Velocity gradient=75051 Depth (H)=loswiath (W) We know that Assume the velocity of flow through the channels as 0.3 m/sec volume of water required to be treated - (Discharge x detention time) =(13638 m3/ day X 0.000347222203) 4.73 m3 Volume of one basin = (4.73)= =2136 m3 Length of flow = (velocity & detention time) = (6.3 X 30) = gm The required cross-section area of each channel = (236) = 0.26m² side of channel = 0.26 =0.5m i L=W=0.5m H105W=105x0.5 K W=0.5m =0.75m X H=0.75m L=0.5m

ی We know that, Р 6² = u 700 51 Dynamic viscsity u = o G = velocity gradient to 3 Ns/m2 four water rolume of water = 2.36 m² Power in walt = ? Putting the above values egr ①, we get P. cn 2 Р (750) To 3 x 2.36 p = 1327.5 watt 1 hp = 746 watt where, p = P = 1327-5 1.779 hp 74 € Power input 10779 hp = 1.779 hourse power Answer. We know that, P = Np (n) 3 (D; JS, Power = 1327.5 watt Np = 0.2= impeller Constant Di = diameter of impeller 8 = density of water, Kg/m3 = 1000kg/m3 the values 0-2(1)(0.9)+1000 n = 2.24 revolution per sec n=2.24x60= 134.4ž 134 revolution per = 0.gm Putting 1327.5 - minute, Answer 2 .