Environment Engg. Civil A surface water treatment plant operates round the clock with a flaw rate of 35 m³/min. The water temperature is 15°C and Jar testing indicated and alum dosage of 25 mg/l with flocculation at a Gt value of 4 x 104 producing optimal results. The alum quantity required for 30 days (in kg) of operation of the plant is
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- A surface water treatment plant operates round the clock with a flow rate of 35 m³/min. The water temperature is 15 °C and jar testing indicated an alum dosage of 25 mg/1 with flocculation at a Gt value of 4x104 producing optimal result. The alum quantity required for 30 days (in kg) of operation of the plant isA surface water treatment plant operates round the clock with a flow rate of 35 m³/min. The water temperature is 15 °C and jar testing indicated an alum dosage of 25 mg/l with flocculation at a Gt value of 4 × 10ª producing optimal results. The alum quantity required for 30 days (in kg) of operation of the plant isA surface water treatment plant operates round the clock with a flaw rate of 35 m3/min. The water temperature is 15°C and Jar testing indicated and alum dosage of 25 mg/l with flocculation at a Gt value of 4 x 104 producing optimal results. The alum quantity required for 30 days (in kg) of operation of the plant is
- The results of a jar test on a surface water supply sample revealed that the optimum dose of alum (Al2(SO4)3.14H2O) as coagulant is 17 mg/L. Furthermore, the water has total alkalinity = 50 mg/L as CaCO3; pH = 7.8; and residual turbidity at optimum alum dose = 4 NTU. The water treatment plant has a mean flow of 10m3/hour. Determine: The kg of alum per year needed in the operation of the plant. The residual alkalinity in the sample, following the addition of alum at optimum dose. Solve parts (a) and (b) if instead of alum, ferric chloride (FeCl3) were used as coagulant. Assume the numerical jar test results are the same as those with alum.Example (Activated Sludge Process) - Completely Mixed with Recycle A completely mixed activated sludge process, treating a municipal wastewater, has a primary clarifier effluent BOD, of 135 mg/L. The design MLSS is 3590 mg/L, and the MLVSS is 80% of the MLSS. The plant permit is for an effluent BOD, of 20 mg/L and 20 mg/L suspended solids. The effluent suspended solids have a BOD, of 0.65 mg BOD, / mg suspended solids. The underflow solid concentration (MLVSS) is 10,000 mg/L. If umax = 3.0 day, Ks 60 mg/L, Y= 0.50 mg MLVSS / mg BOD, removed, ka= 0.1 day, and the influent flow is 0.25 m/sec, determine the following: 1. The required reactor detention time, 0. 2. The reactor basin volume, V, in m³. 3. The net waste activated sludge produced each day (kg/day). 4. The mass of solids (both volatile and inert) to be wasted each day (kg/day). The return sludge flow rate the recycle ratio. 6. The F/M ratio. 7 June 2021 Dr. Dawood Eisa Sachit Biological Wastewater Treatment 47A wastewater treatment plant has been operating with the following specifications:flow rate = 7,630 m3/d, aeration tank volume = 1,270 m3, mixed liquor volatile suspendedsolids = 2,600 mg/l. The soluble BOD in the plant influent used to be 133 mg/l, but hasrecently changed to 222 mg/l. What is the new MLVSS this plant should use to maintain thesame F/M ratio as before? (Ans. Xnew = 4,340 mg MLVSS/l)
- 6. A tertiary treatment plant treats a municipal secondary effluent and is to employ fixed-bed carbon adsorption columns. Parallel treatment using two rows of two columns in series (in other words, a total of four columns) will be used. From batch-type slurry tests it has been found that 0.42 g COD/g carbon is adsorbed when Ce = Co. Pertinent data are as follows: flow = 5700 m³/d, COD in feed to carbon columns = 20 mg/L, contact time function based on an empty carbon bed = 30 min, and unit liquid flow rate= = 4.4 L/s-m². Each column is kept on-line until the entire mass of carbon in the column is completely exhausted. Determine: The volume of each column, m³. a. b. Page 2 of 3 C. d. e. The diameter and height of each column, m. The mass of carbon in each column if the packed density is 400 kg/m³. The kilograms of carbon exhausted per day if the COD removed is assumed to be 98%. The on-line time for each of the columns.The 550-bed Atlanta Hospital has a small, activated sludge plant to treat wastewater from the hospital. The average daily hospital discharge is 1.50 m3 per day per bed, and the average soluble BOD5 after primary settling is 450 mg/L. The aeration tank effective liguid dimensions of 10 m x 10 m x 5 m. The plant operating parameters are as follows: MLVSS - 2,600 mg/L; MLSS - 1.20 (MLVSS); Settling sludge volume after 30 min. = 200 mL/L. The aeration time is nearly?The 550-bed Atlanta Hospital has a small, activated sludge plant to treat wastewater from the hospital. The average daily hospital discharge is 1.50 m3 per day per bed, and the average soluble BOD5 after primary settling is 450 mg/L. The aeration tank effective liguid dimensions of 10 m x 10 m x 5 m. The plant operating parameters are as follows: MLVSS - 2,600 mg/L; MLSS - 1.20 (MLVSS); Settling sludge volume after 30 min. = 200 mL/L. The F/M ratio is nearly?
- A jar test reveals that the optimum alum coagulant dose is 50 mg/L at a pH of 8.1 and a temperature of 73 F. How many pounds per day of 49% alum are required for the whole plant? Total flow is 36 MGD with four parallel treatment plants, each designed for 12 MGD. Could use some guidance on setting up the equations.A coagulation treatment plant with a flow of 0.5 m/s is dosing alum at 23 mg/litre. No other chemicals are being added. The raw water suspended solid concentration is 37 mg/l. The effluent suspended solid concentration is 12 mg/l. Sludge content is 1% and specific gravity of sludge is 3.01. What volume of sludge must be disposed of every day.Homework - A sample of wastewater was incubated for 7-days at 20°C and showed a BOD of 208 mg/L.. Assuming k=0.15 day', calculate: (i) Its 5-days BOD( ii) Its 10-days BOD (iii) Its ultimate BOD