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- A wastewater with a BOD of 25mg/L is discharged through an outfall to a freshwater stream of mean velocity 0.1m/s. The DO downstream of the outfall is 8.5mg/L. Assume the deoxygenation rate is 0.25/d, the reaeration rate is 0.4 /d and the temperature of 20 degree C (DOs = 9.09mg/L). Compute for the ff: a) Critical distance b) minimum DO c) Oxygen deficit d) Critical timePoorly treated municipal wastewater is discharged to a stream. The river flow rate upstream of the discharge point is Qus = 8.7 cubic meters per second. The discharge occurs at a flow of Qd = 0.9 cubic meters per second and has a BOD concentration of 50 mg/L. Assuming that the upstream BOD concentration is negligible. (a) What is the BOD concentration just downstream of the discharge point, after mixing? (b) If the stream has a cross-sectional area of 10 square meters, what would the BOD concentration be 50 km downstream? BOD is removed with a first-order decay rate constant equal to 0.20/day.The BOD just downstream from a wastewater pipe is 12.9 mg/L and the DO (dissolved oxygen) is 8.6 mg/L. The saturated value of dissolved oxygen is 9.6 mg/L, the deoxygenation constant (kd) is 0.21/day, the reaeration constant (kr) is 0.42. Find the time where oxygen deficit is at a maximum and find the minimum value of DO.
- A municipal wastewater treatment plant processes an average of 14,000 m3 / day. The peak flow is 1.75 times the average. The wastewater contains 190mg/L BOD and 210 mg/L suspended solids at an average flow and 225 mg/L BOD and 365 mg/L suspended solids at peak flow. Determine the following for a primary clarifier with a 20-m diameter. Assume eff. BOD of 20 mg/L and effluent TSS of 30mg/L for average and peak flows. 1. Surface overflow rate and the approximate removal efficiency for BOD 5 and suspended solids @ average flow. 2. Surface overflow rate and the approximate removal efficiency for BOD 5 and suspended solids @ peak flow. 3. Mass of solids (kg/day) that is removed as sludge from average and peak flow conditions.A small town has been directed to upgrade its primary wastewater treatment plant to a secondary plant that can meet an effluent standard of 25.0 mg/L BOD5 and 30.0 mg/L suspended solids. They have selected a completely mixed activated sludge system. The following data are available from the existing primary plant. Existing plant effluent characteristics: Flow = 0.2000 m3 /s soluble BOD5 = 80 mg/L Assume that the BOD5 of the suspended solids may be estimated as equal to 50% of the suspended solids concentration, the microorganism concentration in the aeration tank (MLVSS) is 2000 mg/L, and the net activated sludge produced is 400.0 kg/d; also assume the following values for the growth constants: bacterial decay rate = 0.10 day-1 , yield coefficient = 0.60 mgVSS ∙ mg-1 BOD5 removed, maximum specific growth rate constant = 2.50 day-1 , and half saturation constant = 100 mg/L. (1) Estimate the volume of the aeration tank; (2) Compute the F/M ratio; (3) If the BOD5 is 60% of the ultimate…Chemistry A sewage treatment plant discharges 4 ML/d of effluent with 20.17 mg/L BOD5 into a stream. The stream discharge is 12 ML/d, and its initial BOD5 is 6.10 mg/L. The initial DO of the wastewater is 2.10 mg/L and in the stream, it is 7.91 mg/L. Assume the stream velocity is 0.31 m/s and the dissolved oxygen saturation concentration is 8.76 mg/L. The kd = 0.15 d-1 and kr = 0.31 d-1 values have been adjusted for temperature. Calculate the minimum DO in the stream and determine the distance (km) downstream that this minimum occurs. Is this a safe level for fish?
- Drainage from a former pesticide manufacturing site enters Bear River at river kilometer 240. The drainage contains DDT, and the concentration of DDT after complete mixing is 5 µg/L. The town of Bear Valley, adult population 15,000, uses the river at river kilometer 40 for its water supply with only chlorination. The drainage has occurred for 30 years. The river velocity is 5 cm/s. Assume the half-life of DDT is 70 days (a) The concentration of DDT in the town's drinking water is mostly nearly (A) 1.2 µg/L (B) 2.3 µg/L (C) 3.2 µg/L (D) 3.9 μg/L (b) With 2 L/day water consumption, adult body mass of 70 kg, and life expectancy of 70 years, the lifetime chronic daily intake of DDT for an adult living in Bear Valley is most nearly (A) 4 x 10³ mg/kg-day (B) 8 x 10 mg/kg-day (C) 6 x 10³ mg/kg-day (D) 2 x 10² mg/kg-day (c) Assume a slope factor of 0.34 (mg/kg-day)-¹ for DDT. The number of excess lifetime cancers that can be expected in the adult population of Bear Valley is most nearly (A)…3. Consider the design of a wastewater treatment plant (WWTP) for a community with the following design data: TABLE 1: Design Data Design data Flow-0.150m³. s-¹ Influent suspended solids-280mg L-¹ Influent BODs = 500mg L-¹ -1 Sludge concentration= 6.0% Efficiency= 60% Effective length= 40.0 m Width = 10.0 m Liquid depth = 2.0 m Weir length= 75.0 m Table 2: Acceptability Criteria Parameter Detention time (hours) Overflow rate (m- day) Weir loading (m³-day m¹') Acceptable range 1.5-2.5 80-120 <250 (c) Assume that the primary sedimentation process in this treatment facility removes 60% of the suspended solids (SS) and 40% of the BODs of the raw sewage. Determine the SS and BOD5 concentrations in the primary sedimentation effluent flow.Wastewater with DO concentration of 2 mg/L is discharged to a river. The temperature of the river is 10°C and the river is saturated with oxygen (11.33 mg/L). The flow rate of the river is 3.0 m³/s. The temperature and flow rate of the wastewater are 20°C and 0.3 m³/s, respectively. The temperature of the river after mixing is 11°C. The oxygen saturation level at that temperature is 11.08 mg/L. The oxygen deficit imme diately after mixing is most nearly: Only attempt when you are sure .of unsure let other tutors do.
- a) A city discharges 1.5 m3/sec of wastewater into a river whose flow rate is 10m3/sec. The velocity of the river is 1.0 km/hr. The temperature of the wastewater is 20oC, while that of the river is 15 oC. The 20oC BOD5 of the i) wastewater is 200 mg/L, while that of the river is 1.0mg/L. The wastewater contains no dissolved oxygen (DO), but the river’s DO is 90% saturated upstream of the discharge. At 20oC, BOD decay coefficient is estimated to be 0.3/day, while re-aeration rate coefficient is 0.7/day. Using temperature coefficients of 1.135for BOD decay coefficient and 1.024 for re-aeration, answer the following questions i. Determine the DO of the river before discharging the wastewater (saturation concentration of DO at 15oC is 10.1 mg/L)ii. Determine the temperature, DO and BOD of the mixture.iii. Determine the ultimate BOD5 of the mixture b) In a water treatment plant the responsible sanitary engineer designed a cone aerator for aeration of…The town of Pittsburgh discharges 0.126 m3/s of treated wastewater intoCherry Creek. The BOD5 of the wastewater is 34 mg/L. Cherry Creek hasa 10-year, 7-day low flow of 0.126 m3/s. Upstream of the wastewater outfallfrom Pittsburgh, the BOD5 is 1.2 mg/L. The BOD rate constants k are0.222 d21 and 0.090 d21 for the wastewater and creek respectively. Thetemperature of both the creek and the municipal wastewater is 208C. Calculatethe initial ultimate BOD after mixing.A municipal wastewater flow of 1,000 m³/d contains 400 mg/L BOD5 and is treated to be in a trickling filter. The filter will use a plastic filter medium with a specific surface area of 200 m³/m² and will operate with a recirculation ratio of 100% and a hydraulic loading of 25 m³/m² -d. Based on a typical volumetric loading of 0.6 kg BOD5/m³-d, what volume (m³) of plastic media would be required?