Design of sedimentation tank

Given Data:-

Maximum Daily demand of water : 8 MLD

Detention Period :- 4 Hours

Flow through Velocity 25 cm/min

Assume depth of water 4 meter

Free board = 0.5 meterSolution:-

Quantity of water for detention period of 4 Hours = Demand of waterXdetention period/24

Quantity of water for detention period of 4 Hours = 1333.33 m3

Length of Tank= Velocity X Detention Period

= 60 Meter

Cross sectional area = Quantity of water for detention period / Length of tank

= 22.22 m2

Width of Tank = Cross sectional area/Depth

Width of Tank = 5.56 meter

Overall Depth of tank= 4.5 meter

Henc dimensions of tank are, 60 X 6 X 4.5

Result:-

Daily demand Detention time Velocity Capacity of tank Length of tank(MLD) (hr) (m/min) (lit) (m)

8 4 0.25 1333330 60

Demand of waterXdetention period/24

Length of tank Width of tank Overflow rate(m) (m) (lit/hr/m2)

60 6 925.93

Design of Filtration Unit

Given :- Population to be served= 300000 Persons

Average rate of demand= 150 lit/day/capita

Rate of filtration= 3000 lit/hr/m2

L/B ratio= 2

Maximum daily demand= 1.8 times Average Daily demand

Number of units= 4 Units(1 stand By)

Solution:- Average Daily demand= population X Rate of water supply

Average Daily demand= 45000000 lit/day

Maximum Demand= 1.8 X Average daily demand

Maximum Demand= 81000000 lit/day

Discharge per hour = 3375000 lit/hr

Area required for filtration= Discharge per hour / Rate of filtration

Area required for filtration= 1125 m2

Area of each unit= Area required for filtration / no. Of units

Area of each unit= 375 m2

As L/B=2 ,

2B2= 375 m2

B= 13.69 m

L= 27.39 m

Hence provide 4 units of size ,

28 m X 14 m

Result:-

Population Average demand Maximum demand Rate of filtrationPer day Per day (lit/hr)

300000 45000000 81000000 3000

Rate of filtration Rate of filtration Total Area No. Of Length of (lit/hr) (lit/day) (m2) Filter filter

units (m)

3000 72000 375 4 28