FULL DESIGN OF COMPONETS
OF
SEWAGE TRETMENT PLANT (200 KLD)
RECEIVING CHAMBER
ABOUT: -
1. It is the structure to receive the raw sewage collected through Under
Ground Sewage System from the city.
2. It is a rectangular shape tank constructed at the entrance of the Sewage
Treatment Plant (STP).
3. The main sewer pipe is directly connected with this tank.
DESIGN: -
Average. Sewage per hour = 200/24 = 8.33 m3/Hr
Peak Factor = 3
Design Flow capacity = 8.33*3 =25.00 m3 /Hr. = 0.00695 m3/sec
Design flow = 0.00695 m3/sec
Retention Time = 60 Seconds
Volume Required = Design Flow*Detention Time
= 0.00695*60 = 0.4167 m3
Provide, Depth = 3.0 m
Area = 0.4167/3 = 0.1389 m2
Let, Length:Breadth = 2:1
Area = Length*Breadth (L*B) = 2B*B = 2B2 = 0.1389
B = 0.2635 m and L = 0.5270 m
Hence, Receiving chamber is designed for size of 527mm*263mm*3000
mm+FB
SCREENING
GENERAL: -
Screening is the very first operation carried out at a Sewage Treatment
Plant and consists of passing the raw sewage through different types of
screens so as to trap and remove the floating matter such as tree leaves,
paper, gravel, timber, pieces, rags, fibre, tampons, cans, and kitchen refuse,
etc.
PURPOSE OF SCREENING: -
Screening is essential in sewage treatment for removal of materials which
would otherwise damage the plant, interfere with the satisfactory
operation of treatment unit or equipment.
COARSE SCREENS: -
a. The coarse screens essentially consist of steel bars/flat placed 30o to
60o inclination to the horizontal.
b. The opening between bars are 50mm or above.
c. These racks are placed in the screen chamber provided in the way of
sewer line.
2. It is a rectangular shape tank constructed at the entrance of the Sewage
Treatment Plant (STP).
3. The main sewer pipe is directly connected with this tank.
DESIGN: -
Peak discharge of Sewage = 0.00695 m3/sec.
Assume velocity at average flow is not allowed to exceed 0.8 m/s.
The net area screen opening required = 0.00695/0.8
= 0.00869 m2
Clear opening between bars = 30 mm or 0.03 m.
Size of bars = 75 mm*10 mm.
Assume width of the channel = 1.0 m.
The screens are placed at 60o to the horizontal.
Velocity through screen at peak flow = 1.6 m/s.
Clear Area = 0.00869/(1.6*sin60) = 0.00628 m2.
No. of clear opening = 0.00628/0.03 = 0.21 Nos.
Width of Channel =