problem 4:A wastewater treatment plant with an average daily flow of 0.25 m³/s must meet 30 mg/L of totalBODs and 25 mg/L of suspended solids standards. The suspended solids is 65% BODs. Thewastewater is treated in an activated sludge plant. Assume that the soluble BODs after theprimary clarifier is 115 mg/L and the concentration of the microorganism in the aeration tank is2500 mg/L. Use K-60 mg/L; ka=0.06 d¹; m = 5 and Y = 0.6. Determinea) The concentration of the soluble BODs in the effluent of the aeration tankb) The mean cell residence timec) The hydraulic retention timed) The required volume of the aeration basin

Part e) is not in your question but if you want it then only write otherwise leave that.

Answered: problem 4: A wastewater treatment plant…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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After primary treatment, a municipal wastewater contains 150 mg/L BOD5. The wastewater flow of 0.44 m³/s (10 MGD) is then treated in an activated sludge system with an SRT of 12 d and an HRT of 6 h. Given the biokinetic parameters tabulated below, determine the aeration volume (m³). Parameter Value 3d¹ 60 mg/L BODs 0.10 d¹ 0.6 mg VSS/mg BOD5 Hmax Ks ka Y
An activated sludge tank is 30 m long and 10 m wide and has an SWD of 4 m. The wastewater flow is 4.0 ML/day and Compute: a) the BOD concentration which will be applied to the aeration tank b) the food-to-microorganism ratio for the system. O a) 105 mg/L b) F/M=0.35 O a) 195 mg/L b) F/M=0.65 O a) 105 mg/L b) F/M=0.22 O a) 195 mg/L b) F/M=0.35 the raw 5-day BOD is 300 mg/L. Assuming that 35 percent of the raw BOD is removed in the primary clarifier, and the MLSS concentration is 1000 mg/L.
- An aerobic digester with a total capacity of 50,000 ft³ stabilize waste-activated sludge from an extended-aeration treatment plant without primary clarification. The average daily sludge flow pumped to the digestion tanks is 32,000 gpd with 1.5% solids, of which 65% are volatile. Calculate a) sludge detention time in the digesters b) the amount of sludge produced daily and c) the anaerobic digester dimensions. The population being served is 75,000. SS and BOD = 350 mg/l. %3D
An industrial wastewater discharged at a flowrate of 1000 m³/d having 220 mg/l initial concentration of total suspended solids (TSS). If this wastewater is to be treated individually by one of the following processes: 1. Primary settling tank which should remove 60% of TSS. 2. Coagulation unit using 40 g ferric chloride (FeCl3) as a coagulant /m²³ of wastewater. The coagulation unit will remove 85% of TSS. Estimate the mass and volume of the sludge generated from each suggested unit. Assume the following sludge specification: Source of sludge Primary settling tank Coagulation tank Specific gravity 1.03 1.05 Moisture content 94% 92%
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 = 4/1
C A trickling filter plant treats 1500 cum per day of sewage with a BOD, of 220 mg/L and a suspended solid concentration of 250 mg/L. Estimate the total solid production assuming that primary clarification removes 30% of BOD and 60% of influent solids. Take the solid production in the trickling filter as a 0.5 kg/kg of the applied BOD.
In a BOD test, 5 ml of wastewater (with no dissolved oxygen) is diluted with 295 ml of water containing 8.5 mg/1 of dissolved oxygen. After 5 days incubation at 25°C, the dissolved oxygen content of the mixture is 5 mg/1. Temperature coefficient can be taken as 1.047 and BOD rate constant is 0.23 per day (base e) at 20°C. The ultimate biochemical oxygen demand of the wastewater is
A complete mix activated sludge process treats 5 MGD of wastewater containing a total BOD5 concentration of 225 mg/L BOD5 to 20 mg/L BOD5, which is 75% soluble in both the influent and the effluent. The volume of the aeration basin is 111,400 ft3 and the MLSS concentration is 2500 mg/L. Determine the following: a) Detention time in the aeration basin (hr) b) F/M ratio c) Specific substrate utilization rate d) Substrate removal efficiency
A system is designed to treat contaminant X with two PFRs in series. The concentration of contaminant X entering the first PFR is 120 mg/L and the inflow is 420 L/s. The first PFR is 5000 m³ in size, and the contaminant reacts via first order kinetics with a k value of 0.50 (1/hr). The second tank is being built so that the treatment facility can meet the permit level for contaminant X of 9.4 mg/L. What size do they need to make the second tank in m³?

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