A stream carries a flow of 2000 m3/day with TDS concentration of 650 mg/l. At a point, 100 m3/day of industrial wastewater is discharged into the stream with a TDS concentration of 500 mg/l. Find out the resulting TDS concentration in the stream after industrial waste is completely mixed with the stream flow. No accumulation takes place
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- Question 2 ) A primary wastewater treatment uses centrifugation for dewatering of primary sedimentation sludge. The raw wastewater contains 300 mg/L TSS and 250 mg/L BOD5. The average daily flow to the treatment plant is 5000 m3/day. In primary sedimentation basin, TSS removal is 55%, and BOD5 removal is 30% of the incoming flow. The primary sludge contains 3.5 % solids. The TSS and BOD5 capture efficiency of centrifugation is 90 % each, and the moisture (water) content of the sludge cake is 70 %. Assume the specific gravities of the primary and sludge cake are 1.01 and 1.06 g/cm3. a) Calculate the primary sludge flowrate (underflow rate) (m3/day). b) Calculate sludge cake flowrate (m3/day).Poorly 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.a. The wastewater from a manufacturing treatment plant is discharged to a municipal `treatment plant for processing with domestic wastewater. The characteristics of the wastewater from the manufacturing treatment plant are 3500 mg/l BOD, 4500 mg/l SS, 70 mg/l nitrogen and 2 mg/l phosphorous. The characteristics of the domestic wastewater are 350 mg/l BOD, 285 mg/l SS, 50 mg/l N and 11 mg/l P. If the required BOD/N/P weight ratio is 100/5/1, what is the minimum quantity of domestic wastewater required per 1000 gallons of manufacturing wastewater to provide adequate nutrients for biological treatment and the concentration of SS in the combined wastewater? b. What additional of NH4OH and H3PO4 would be required in pounds per million gallons of wastewater to provide the necessary nutrient levels for the treatment of the manufacturing wastewater described above (a) if the domestic wastewater had not been combined with it.
- A waste with a 5-day BOD (BOD)=200mg/L and k = 0.1 day¹ is discharged to a river at a rate of 1 m/s. The river has flow rate of 9 m/s and a background ultimate BOD (Lo) of 2 mg/L upstream of the discharge. (a) Calculate the ultimate BOD (Le) of the waste. (b) Furthermore, assuming instantaneous mixing after discharge, calculate Lo of the river water after it has received the waste. Ans.: (a) Ultimate BOD of wastewater = 508.3 mg/L, (b) Ultimate BOD of river water after discharge of waste-52.63 mg/L.The wastewater from a manufacturing treatment plant is discharged to a municipal 'treatment plant for processing with domestic wastewater. The characteristics of the wastewater from the manufacturing treatment plant are 3500 mg/l BOD, 4500 mg/l SS, 70 mg/l nitrogen and 2 mg/l phosphorous. The characteristics of the domestic wastewater are 350 mg/I BOD, 285 mg/I SS, 50 mg/lN and 11 mg/I P. If the required BOD/N/P weight ratio is 100/5/1, what is the minimum quantity of domestic wastewater required per 1000 gallons of manufacturing wastewater to provide adequate nutrients for biological treatment and the concentration of SS in the combined wastewater?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.
- The characteristics of the synthetic textile wastewater are 1500 mg/L BOD, 2000 mg/L SS, 30 mg/L nitrogen, and no phosphorus. The characteristics of the domestic wastewater are 200 mg/L BOD, 240 mg/L SS, 35 mg/L nitrogen, and 7 mg/L phosphorus. Find the nitrogen concentrations in the mixture. The ratio of flow rate is domestic/textile = 4/1A municipal wastewater treatment plant is installing completely mixed activated sludge to meet their NPDES permit of 15 mg/L BOD5. The mixed liquor concentration is 1,000 mg/L MLVSS. The wastewater flowrate going to secondary treatment is 0.170 m3/s and the BOD5 in that stream is 110 mg/L. Laboratory measurements of the growth constants show: Ks = 60 mg/L BOD5, um = 2.8 d-1, kd = 0.05 d-1, Y = 0.73 mg MLVSS/mg BOD5. Determine the required volume of the aeration tank.The discharge from the Renenutet sugar beet plant causes the DO at the critical point to fall to 4.0 mg/L. The Meskhenet Stream has a negligible BOD and the initial deficit after the river and wastewater have mixed is zero. What DO will result if the concentration of the waste(L_w) is reduced by 50%? Assume that the flows remain the same and that the saturation value of DO is 10.83 mg/L in both cases.
- Question: A wastewater with a BOD5 of 326 mg/L and a Total Suspended Solids (TSS) of 392 mg/L is sent to primary treatment. The average flow rate is 0.050 m³/s. a. If the BOD5 removal efficiency is 33%, how many kilograms of BOD5 are removed in the primary settling tank each day? b. If the TSS removal efficiency is 57%, how many kilograms of TSS are removed in the primary settling tank each day?A cheese production facility in Winchester Ontario is having problems meeting the Provincial discharge limit of 25 mg/L BOD;. The facility produces and discharges 200 m³/d of wastewater with an influent concentration of 5,000 mg BODS /L. The current well mixed aerated lagoon has an operating volume of 10,000 m². What is the effluent concentration with the current system? What increase in volume is required to meet the discharge limit? The kinetic rate for BOD reduction is 0.5 d1. State any assumptions.The treated domestic sewage of a town is to be discharged in a natural stream with following data given: (i) Population = 50,000 (ii) BOD contribution per capita = 0.07 kg/day (iii) BOD of stream on u/s side = 3 mg/It. (iv) Permissible maximum BOD of stream on d/s side = 51 (v) Dry weather flow of sewage = 140 litres per capita per day 5 mg/lt. (vi) Minimum flow of stream = 0.13 m3/sec Choose the correct statement(s): BOD of treated sewage should be 8.21 gm/lt. Total BOD produced per day is 35000 kg/day C BOD of untreated sewage is 500 mg/It. Correct Option D Percentage treatment required is 98.358% Correct Option