851 likes | 3.02k Views
Wastewater Treatment Processes. Solids in a Typical Municipal Wastewater. Solids in a Typical Municipal Wastewater. Wastewater Treatment Processes. Pre treatment Primary treatment Secondary treatment Advanced / tertiary treatment. Wastewater Treatment Processes. O 2. Secondary treatment
E N D
Wastewater Treatment Processes • Pre treatment • Primary treatment • Secondary treatment • Advanced / tertiary treatment
Wastewater Treatment Processes O2 • Secondary treatment • Aerobic, anaerobic lagoons • Trickling filter- activated sludge-oxidation ditch • Mostly BOD removal technology • Primary treatment • screening • grit removal • removal of oil • sedimentation • Tertiary treatment • Nitrate removal • Phosphorus removal • Disinfection
Statistical Example • Determine the average BOD5 concentration of a municipal wastewater if the results at the influent for 12 consecutive days are as listed below. Also, what are the 90 percentile and 50 percentile concentration?
Solution (X – 475)/(0.9-0.83)=(525-475)/(0.92-0.83) 90th percentile = X = 512
Design of the bar screen channel (Approach Channel) The cross section of the bar screen channel is determined from the continuity equation: Qd = AcVa Ac = Qd/ Va The head loss through the bar screen Qd = design flow, m3/s Ac = bar screen cross section, m2 Va = Velocity in the approach channel, m/s Hl = head loss Va = approach velocity, m/s Vb = Velocity through the openings, m/s g = acceleration due to gravity, m/s2 Usually, rectangular channels are used, and the ratio between depth and width is taken as 1.5 to give the most efficient section.
Example • A manual bar screen is to be used in an approach channel with a maximum velocity of 0.64 m/s, and a design flow of 300 L/s. the bars are 10 mm thick and openings are 3 cm wide. Determine • The cross section of the channel • The velocity between bars • The head loss in meters • The number of bars in the screen
Solution: 4) n* t bar + (n-1)Sc = W n x 1 + (n-1) x 3= 56 n= 14.75 = 15 1) Ac= Qd / Va= 0.3/0.64 = 0.47 m2 Ac= W x1.5W =1.5 W x W W = 0.56 m, Depth (d) = 1.5 W = 0.84 m = 0.84 x 0.56 (3/3+1) = 0.35 m2 2) From continuity equation: Va Ac= Vb Anet Vb= 0.64 x 0.56 x 0.84/0.35 = 0.86 m/s < 0.9 m/s ok 3. Head loss: = 0.024 m
Settling Theory Vh = scour velocity = Friction factor of particles = Darcy-weisbach friction factor Vs = settling velocity of particles = density of particles = liquid density d = particle diameter CD = drag coefficient
Example A suspension contains particles of grit with a diameter of 0.2 mm and specific gravity of 2.65. For particles of this size CD= 10, f= 0.03, and = 0.06. The suspension also contains organic solids of same size for which the specific gravity is 1.10 and and f are unchanged. Determine the settling velocity of the grit and the scour velocity of grit and organic material. Solution Settling velocity of particles = 2.1 cm/s Scour velocity of particles = 23 cm/s Scour velocity of organic solids = 5.6 cm/s
Grit Chamber Goals: Removal of inorganic matter which has high density > 2000 kg/m3 and particle size 0.1 to 0.2 mm in order to protect pumps from abrasion and to protect digesters from getting clogged. Vh A = W * H = Q/Vh Vs / Vh = H / L H A VS W L Example Design a grit chamber for Treatment plant having a daily flow of 11000 m3. Use the values of Vh and Vs from example 3. Solution A= W*H= 0.13 (m3/s)/0.23 (m/s)= 0.55 m2 Assume W= 1 m , then H = 0.55 m Vs/ Vh= H/L 2.1/23 = o.55/L L= 6.04 m
Using Stokes Law • Example: Design a horizontal flow grit chamber to remove grit of size greater than 0.2 nun if the through flow is 10000 m3/d. The specific gravity of the particles is 1.9.
Primary Treatment • Primary treatment is often called clarification, sedimentation or settling. • This is the unit process where the wastewater is allowed to settle for a period (~2h)in a settling tank to produce a clarified liquid effluent in one stream and a liquid-solid sludge (called primary sludge) in a second stream. • The benefits of primary treatment include: • Reduction in suspended solids • Reduction in BODs • Reduction in the amount of waste activated sludge (WAS) in the activated sludge plant . • Removal of floating material . • Partial equalization of flow rates and organic load