10. A 15,000 ft" conventional activated sludge system has a recycle ratio of 0.38, an influent flow rate of 17,000 ft*/day, a mixed liquor volatile suspended solids (MLVSS) concentration of 280 mg/L, and an influent BOD, of 60 mg/L. What is most nearly the recycle rate of the system? (A) 2800 ft*/day (B) 3600 ft*/day (C) 5300 ft*/day (D) 6500 ft*/day A.

10. A 15,000 ft" conventional activated sludge system has a recycle ratio of 0.38, an influent flow rate of 17,000 ft/day, a mixed liquor volatile suspended solids (MLVSS) concentration of 280 mg/L, and an influent BOD, of 60 mg/L. What is most nearly the recycle rate of the system? (A) 2800 ft/day (B) 3600 ft/day (C) 5300 ft/day (D) 6500 ft*/day A.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

For an activated sludge basin with a flow rate of 500 L/s if the influent BOD5 = 100 mg/L, the MLVSS = 3000 mg/L and the volume of the aeration basin = 766100 gallons Find the F/M ratio If the WAS X= 10000 mg/L, find the WAS flowrate (m3/d)
The initial rate of infiltration of a watershed is estimated as 53 mm/hr, the final capacity is 5 mm/hr, andthe hydraulic conductivity, k as 0.4hr-1. Use Horton’s equation to find: • The infiltration capacity at t = 3 hr and 6 hr• The total volume of water infiltrated over the 6-hr period.
A rectangular sedimentation basin with a length twice the width and 3 m deep, receives a flow of 2400 m/day with a normal concentration of suspended solids of 120 mg/l Determine: 1. The surface over flow rate (SOR) when the detention time is 6 hours. 2 The water entering velocity to the basin (cm/min) 3. The efficiency of the basin when the total mass of sludge produced per day is 216 kg.
find the following : 1- Inflow discharge (cm/6-months) = ? 2- ΔS (cm/6-months)= ? 3- Evaporation (in/6-months) = ? 4- Evaporation (L/6-months) = ?
Q/ Find the BOD of trickling filter effluent if the BOD of raw sewage is 250 mg/l with 125% recirculation. The flow is 3.15 m/min and tank volume 860 m .Assume 35% BOD is removed by primary sedimentation tank using the NRC equation.
A wastewater treatment plant will receive a flow of 30,000 m3/day. The plant has a 3.6 m deep circular primary clarifier with a surface overflow rate of 60.0 m3/m2-day.a. Calculate the surface area (m2), diameter (m), volume (m3), and hydraulic retention time of the primary clarifier.b. How long will it take a 1.0 mm particle with a density of 1500 kg/m3to settle in the primary clarifier? show all work
What is the ultimate BOD 24 km downstream from the outfall? > Answer: 3.89 g/m3 What is the DO deficit 24 km downstream of the effluent discharge? > Answer: 3.192 g/m3 Question: At the location 15 km downstream of the effluent discharge, how much oxygen would have to be added to have the do at 7 mg/l?
Oil (SG = 0.89), with a kinematic viscosity of 0.007 ft2/s, flows in a 3-in.-diameter pipe at 0.049 ft3/s. Determine the head loss per unit length of this flow.
Assume that DLSUD lake is 100m2 and its average height is 5ft. Wastewater that flows into the lake are the following: Food square kitchen waste 7 m3/sec; Total Dissolved Solids (TDS) is 25,000 mg/L which operates 8hrs per day CIH 5m3 /hr kitchen waste; TDS is 20mg/L which operates 4hrs per day PCH 20m3 /day diluted laboratory waste ; TDS = 40mg/L (1)Determine the volumetric flow rate of the lake water can still accommodate prior to rainfall.
2.The annual evaporation from a lake is 150 cm. If the lakes surface area is 16 km2, what is the daily evaporation rate in cm and in.? 3. The storage existing in a river reach at a given time is 13 acre-ft, and at same time, the inflow is 450 cfs and the outflow is 500 cfs. One hours later, the inflow is 500 cfs and the outflow is 530 cfs. Calculate the change in storage at the end of the hour in cubic meters, and determine the new storage.
a) An activated sludge plant with recycle is evaluating its secondary clarification system. Settling column tests are conducted and analyzed for settling velocities at different initial solids concentrations. The data are provided below. The plant flow rate is 10,000 m3/d, with a recirculation ratio of 0.5. The MLSS in the reactor is 4000 mg/L. The desired underflow concentration is 17,000 mg/L. Solids concentration mg/L Settling velocity m/h 1500 6.5 3000 5.1 4500 4 6000 2.95 7500 2 9000 1.4 12,000 0.5 15,000 0.27 17,000 0.2 b) Design secondary clarifiers for the activated sludge plant of problem a), with a recycle ratio of 0.6. Calculate the required surface area if two clarifiers are used. Also calculate underflow rate, underflow velocity, and overflow rates. Clearly state your assumptions.
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…