w rate of 260 md and a detention time of 20 h size sedimenta

w rate of 26.0 m/d and a detention time of 2.0 h, size sedimentation tank for the average flow at Cynusoidal City a primary g-1). What would the overflow rate be for the unequalized Proimum flow? Assume 15 sedimentation tanks with length-to-width ratio of 4.7. rs: Tank dimensions 15 tanks at 2.17 m deep by 4.34 m by 20.4 m. Maximum overflow rate 39.3 m/d

Solution

To Determine

SOLUTION:-

H=hight L= length B= width , A = Area ,V= volume

Given information:-

Concept used and formula

CALCULATION:-

flow rate =26mld

detention time =2 hr

volume of tank = flow rate *detention time

( 26*2)*^10^3 /24 = 2166.67 m^3 --------------------------------------------------------(1) euation 1

for maximum flow take volume = 1.33*2166.67 =2881.67

now ,

volume of one sedemention tank = 2881.671/15 =192.11 m^3

(VOLUME = AREA *HIGHT)

Assume Hight =2.17m

Area = volume/hight = 192.11 /2.17 = 88.531 m^2

Given L/B ratio= 4.7 , L = B*4.7 ,

L*B = 88.531 m^2

4.7*B^2 = 88.531 ,, B = 4.340m and length L=4.7*4.340 = 20.40m

so , from here hight = 2.17 , length =20.40 , width =4.340m  

Now ,

from equation 1

Orignel volume of tank = 2166.67

so orignal volume of single tank =2166.67 /15 = 144.44m^3

(AREA = VOLUME / HIGHT )

Hight = 2.17m

so,

Area = Volume/hight 144.445/2.17=66.560m^2

Then Maximum overflow rate = given flow rate / Area = (26.0/66.560)*100 =39.060m^3/d/m^2

(this maximum flow is less tha 40m^3/d/m^2 then result is Ok )