D) Design the circular sedimentation with the same SOR of rectangular sedimentation tank with dimensions (L-15m W=8.8446m depth=3.5m), treating wastewater for a community of area 5km² and population density is 5000cap/km², the average waste water generated per day is 300L/cap, T=2hr. b)Determine the overflow weir loading rate at average flow and maximum flow. c)The suspended solids concentration in the primary sedimentation effluent flow, removal efficiency of the tank 30%, suspend solids influent flow=800ppm.

two sedimentation tanks operate in parallel. The combined flow to the two tanks is .1 m^3/s. the depth of each tank is 2 m and each has a detention time of 4 h. what is the surface area of each tank and what is the overflow rate of each tank in m^3/d*m^2?   Please don't provide handwritten solution

A rectangular primary sedimentation basin is to be designed for an activated sludge treatment plant serving a municipality. The regulatory agency criteria for primary sedimentation basins used for activated sludge system are as follows: peak overflow rate (S) = 80 m3/d*m2, average overflow rate (S) = 40 m3/d*m2, peak solids loading = 50 kg/d*m2, peak weir loading = 372 m3/m*d. The average flow to the basin is 12 000 m3/d. a) Calculate the tank surface area in m2. b) Calculate the peak flow rate in m3/d. c) Calculate the total suspended solids concentration in the incoming flow to the basin (mg/L) d) Calculate the total weir length (m)

A circular secondary clarifier will be used to treat an average daily flow (ADF) of 6,000 m3 /d from a high-purity oxygen activated sludge process with a MLSS concentration of 8,500 mg/L. The RAS flow is 2,500 m3 /d. Key design criteria include an overflow rate of 15 m3 /(d.m2 ) and a solids loading rate of 100 kg/(d.m2 ). Determine the minimum diameter (m) of the clarifier to achieve these criteria.

Secondary Clarifier  An activated sludge plant has a design flow (average daily flow) of 11 MGD. Return sludge ratio R = 0.35.  Determine the dimensions of final settling tanks.  Check weir overflow rate.  The design parameters: number of tank = 4.       Use circular tanks.   MLSS = 3000 mg/L   VSS/TSS = 0.80    overflow rate = 1000 gpd/ft2, detention time = 2  hr weir overflow rate = 10,000 – 15,000 gpd/ft        solid loading rate (lb/d/ft2) = 50 lb/d/ft2

find the following :  1- Inflow discharge (cm/6-months) =  ?  2- ΔS (cm/6-months)=  ?  3- Evaporation (in/6-months) =  ?  4- Evaporation (L/6-months) =   ?

A municipal wastewater treatment plant has an average flow of 0.5 m3/s, and two aerated grit chambers are to be used in parallel. Dual units are to be provided for flexibility in operation, since one could be used, should the other require temporarily maintenance. During normal operation, both chambers would be used. Each chamber is to be a rectangular tank with spiral-roll flow. The peak flow is 2.4 times the average flow. The tank width is 2 times the depth, and the length is 4 times the width. Determine tank width in meters, if the detention time is 3 min. Determine the total air flow in m3/min, if 0.35 m3/min per meter of tank length is to be provided.

Assuming the parameters for Philip’s equation are sorptivity, S = 6 cm/hr and K = 0.3 cm/h. Determine thecumulative infiltration after 0, 0.5, 1, 1.5 and 2 hr. Plot infiltration rate as a function of cumulativeinfiltration. Assume continuous ponded condition.

(a) The initial infiltration capacity fo of a watershed is estimated as 1. 5 in./hr and the timeconstant is taken to be 0.35 hr.-1. The equilibrium capacity fc (also known as field capacity orconstant infiltration capacity) is 0.2 in./hr. Use the Horton Equation to find (a) the value of fat t=10 min. 30 min., 1 hr., 2 hr., and 6 hr., and (b) the total volume of infiltration over the 6-hr period.

The 4 hourly rainfall data are given below. fo the initial value of the infiltration capacity is 6 mm/hr,and f0 the limit value of the infiltration capacity is 2 mm/hr and k is 0.4 1/hr. Hour: 1 Precipitation (mm): 4 Hour: 2 Precipitation (mm): 12 Hour: 3 Precipitation (mm): 17 Hour: 4 Precipitation (mm): 27 what is the runoff rate?