A pollutant entered into a confined aquifer of 5 m thick and porosity 25%. If the seepage discharge through the aquifer is 0.25 m² / s, the travel time of pollutant for 100 m in the aquifer will be (in seconds)
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Q: A pollutant entered into a confined aquifer of 5 m thick and porosity 25%. If the seepage discharge…
A: Given: Seepage discharge = 0.25 m2/s Depth = 5 m Porosity = 25 % = 0.25 Length = 100 m
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- Consider a lake with 100 × 106 m2 of surface area for which the only source of phosphorus is the effluent from a wastewater treatment plant. The effluent flow rate is 0.4 m3 /s, and its phosphorous concentration is 10 mg/L. The lake is also fed by a stream having 20 m3 /s of flow with no phosphorous. If the phosphorous settling rate is estimated to be 10 m/yr, estimate the average phosphorous concentration in the lake. What level of phosphorous removal at the treatment plant would be required to keep the average concentration below 0.010 mg/L?A waste water stream (flow=2 m³/s, ultimate BOD=90 mg/l) is joining a small river (flow=12 m³/s, ultimate BOD=5 mg/l). Both water streams get mixed up instantaneously. Cross sectional area of the river is 50 m². Assuming the de-oxygenation rate constant, k'=0.25/day the BOD (in mg/l) of the river water, 10 km downstream of the mixing point isA large stream has a reoxygenation constant of 0.4 per day. At a velocity of 0.85 m/s; and at the point at which an organic pollutant is discharged, it is saturated with oxygen at 10 mg/L (D, = 0). Below the outfall, the ultimate demand for oxygen is found to be 20 mg/L and the deoxygenation constant is 0.2 per day. What is the dissolved oxygen 48.3 km downstream?
- A large stream has a reoxygenation constant of 0.4 per day (on base 10). At a velocity of 0.85 m/s; and at the point at which an organic pollutant is discharged, it is saturated with oxygen at 10 mg/L (D= 0). Below the outfall, the ultimate demand for oxygen is found to be 20 mg/L and the deoxygenation constant is 0.2 per day (on base 10). What is the dissolved oxygen at 48.3 km downstream?Dissolved chloride in a concentration of 500 mg/L is being advected with flowing groundwater at a seepage velocity of 0.3 m/d in an aquifer with a porosity of 0.25. Groundwater from the aquifer discharges into a river. a. What is the mass flux of chloride into the river (in g/(m2d))? b. How long does it take the chloride to travel 100 m through the aquifer (i.e., the travel time for a 100 m distance)?5. A stream has a cross-sectional area of A= 85 m2, length L = 1000 m, velocity U=185 m/hour, containing BOD with a concentration of 10 g/m3. At one point, the river receives a waste stream with a discharge of 100 m3/hour and a BOD concentration of 100 g/m3 where the reaction rate of BOD k = 2/hour. If the rate of diffusion is 2085 m^2/hour, determine: General solution of the BOD concentration function in a steady state stream.
- (e) A large stream has a reoxygenation constant of 0.4 per day (on base 10). At a velocity of 0.85 m/s; and at the point at which an organic pollutant is discharged, it is saturated with oxygen at 10 mg/L (D = 0). Below the outfall, the ultimate demand for oxygen is found to be 20 mg/L and the deoxygenation constant is 0.2 per day (on base 10). What is the dissolved oxygen 48.3 km downstream?Fifty kilograms of toxic material is spilled uniformly across a canal. The canal is trapezoidal with a bottom width of 3 m, side slopes of 2.5:1 (H : V ), and a depth of flow of 1.7 m. The discharge in the canal is 16 m3/s, and the longitudinal diffusion coefficient is 7 m2/s. Cross sectional area = A = by+my^2 = (3m)*(1.7m)+(2.5)*(2.89 m^2) = 12.325 m^2 What length of canal is contaminated 2 hours after the spill?The discharge from a food processing plant is the only source of BOD in a river. The oxygensag curve shows a minimum DO concentration of 3.0 mg/L. The DO concentrationimmediately downstream of the discharge is at saturation (10.0 mg/L). Other characteristics ofthe system are as follows: u = 60 miles/day; kr = 0.80 days-1; kd = 0.20 days-1.a. Currently the food processing plant does not treat the waste. What percent removal ofthe BOD would be required to ensure that the DO of the river at any locationdownstream of the plant is always greater than 5.0 mg/L?b. How far downstream would the lowest DO levels occur?c. What ultimate BOD concentration of the mixture of the river and waste immediatelydownstream of the discharge would result in a minimum DO concentration of 5.0mg/L?
- . A filtration system in a water treatment plant consists of four (4) filter boxes of the size = 45m^2 design flow= 40000 m^3/d, also obtained from Table 2.1. The filter bed is uniform with particles of size 0.50 mm, with specific gravity of 2.6, a porosity of 0.4. and a bed thickness of 0.75 m. Normally, the Reynolds number for flow through a rapid sand filter media is less than 5. ii. Check that the conditions satisfy the low Reynolds number condition. iii. Determine the expected head if clean water is passed through the filter at the design flow.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.In a homogeneous unconfined aquifer of area 3.00 km², the water table was at an elevation of 102.00 m. After a natural recharge of volume 0.90 million cubic meter (Mm²), the water table rose to 103.20 m. After this recharge, ground water pumping took place and the water table dropped down to 101.020 m. The volume of ground water pumped after the natural recharge, expressed (in Mm² and round off to two decimal places), is
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