Question

A water treatment plant is designed to process 100 ML/d (mega liter per day). The flocculator is 30 m long, 15 m wide, and 5 m deep. Revolving paddles are attached to four horizontal shafts that rotate at 1.5 rpm. Each shaft supports four paddles that are 200 mm wide, 15 m long and centered 2 m from the shaft. Assume the mean water velocity to be 70% less than paddle velocity and CD = 1.8. all paddles remain submerged all the time.

Calculate the following:

a. Difference in velocity between paddles and water
b. Value of G
c. Retention time

d. Camp number.

Answer 1

Answer:

A) 0.22 m/sec

B) 10.717 sec^-1

C) 32.4 min

D) 20833.848

Explanation:

A) calculate the difference in velocity between paddles and water

Vr = Vp - Vw

Vp = paddle velocity

Vw = water velocity

Vw = 0.3 Vp therefore Vr = 0.7vp

also ; Vp = ωr  =  $\frac{2\pi N}{60} r$ =  $\frac{2*3.14*1.5 *2}{60}$   =  0.314 m/sec

therefore

Vr = 0.7 * 0.314 = 0.22 m/sec

B)  Value of G

attached below is the detailed solution

C) Retention time

Td = V / Q  = Volume / Discharge = $\frac{30* 15*5*24}{100*10^6*10^-3} * 60 min$

    =  32.4 min

D) Camp number

camp number = G * t  

                       = 10.717 sec^-1 * 32.4 * 60

                       = 20833.848