Preliminary and Primary Wastewater Treatment

SOLUTION

 

  1. Design a grit chamber of rectangular cross-section.  Following information is provided: Design Flow (Q): 50 MLD; Settling velocity of the smallest particle to be removed completely is 0.0236 m/s; Specific Gravity of particles (Ss): 2.65; Horizontal mean flow velocity (V): 0.30 m/s; Theoretical depth (D): 1.4 m; Calculate the dimensions and slope (mm/m) of the grit chamber. While designing the actual grit chamber, add 25 percent to the depth for grit collection, and 0.25m freeboard.  Also add 50 percent to the theoretically calculated length.  Value of Manning’s ‘n’ is 0.020.  Assuming that the volume of grit in the wastewater is 0.15 m3/ML, and 100 percent grit removal in the grit chamber, calculate the total mass and mass of dry grit produced per day. The collected grit has a porosity of 0.5.                                                                                                             

 

Solution:

Theoretical depth of flow (D) = 1.4 m

Required detention time (T) =  = 1400 / 23.6 = 59.32 s (say 60 s)

Velocity of Flow (V) = 0.30 m/s

Theoretical length of channel (L) = V.T = 0.30 x 60 = 18.0 m

Flow (Q) = 50 MLD = 50000 / 86400 = 0.5787 m3/s 

Area of flow (A) = = 0.5787 / 0.30 =1.929 m2

Width of the Channel =  = 1.929 / 1.4 = 1.377 m (say 1.4 m)

Whetted Perimeter (P) = (1.4 + 1.4 + 1.4) = 4.2 m

 

Using Manning’s equation:

                  ,           Where, S = Channel slope, m/m

Using this equation, S was calculated to be 1.02 x 10-4

 

Thus,          Design Length (L’) = 1.5.(18.0) = 27.0 m

                  Design Depth (D’) = (0.35 + 1.4 + 0.25) = 2.0 m

                  Design Width (W) = 1.4 m

                        Slope = 2.75 mm / 27 m

 

Daily grit production = 0.15.(50) = 7.5 m3

Assuming porosity = 50 percent, volume of grit solids = 0.5.(7.5) = 3.75 m3
Mass of grit solids = (3.75).(2650) = 9937.5 Kg (say 9950 Kg)
Mass of Water in the grit (assuming saturated conditions) = 3.75.(1000) = 3750 Kg
Therefore total of grit produced per day = 9950 + 3750 = 13,700 Kg/d.                     

 

  1. Give three reasons of why even though, theoretically speaking, depth has no effect on the particle removal efficiency in a primary settling tank used in wastewater treatment, a depth of 3-5 m is provided in practice.                       

 

Solution:

 

Reason 1:         Provision of lower depth, without increasing the cross-sectional area would increase horizontal flow velocity through the tank, which may lead to scouring and consequent re-suspension of settled particles.

 

Reason 2:         Wind currents affect shallow settling tanks more.  The turbulence resulting from such wind currents may hinder settling efficiency.

 

Reason 3:         The settling theory, which suggests that settling efficiency in a settling basin is not affected by depth, assumes that particles settle discretely.  However, during sedimentation, may organic particles in wastewater flocculate, forming bigger particles.  The settling efficiency for flocculent suspensions in a sedimentation tank improves with depth.