Exercise 4A treatment plant has the following process diagram. Determine the flow rate expression for each unit. The inflow isQi. The sludge waste flowrate is Qw. The return flowrate rate is Qr. The mixed liquor return flowrate is Qm.Mixed-liquor returnSecondaryclarifierEffluentInfluent-AnoxicAerobicAnoxicAerobicReturn activated sludge(e) Bardenpho™ (4-stage)SludgeExercise 5Derive the ODE system for above process. There are two species involved in the reaction, as shown below. The processrate for each tank can be simply represented by r1, 12,...15. The gravitational sedimentation is only applied to speciesX in the clarifier and can be represented by a reaction rate of r5 = 3.5X5. The influent are Xi and S₁.S=X.

Step 1:The plant uses a Bardenpho ™ process, which is a type of activated sludge process.The process…

Answered: Exercise 4 A treatment plant has the…

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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3. A 100 mL of wastewater sample was drying on a 0.5 g empty dish weight at 104 °C to weight 0.5625 g. Another 100 mL of the same wastewater was first filtered and the filtrate was dried on a 0.5 g dish at 104 °C to weight 0.5325 g. Determine in mg/L the:a) TSb) TSSc) TDSd) VSS assuming it represents 70% of TSS
Chemistry A sewage treatment plant discharges 4 ML/d of effluent with 20.17 mg/L BOD5 into a stream. The stream discharge is 12 ML/d, and its initial BOD5 is 6.10 mg/L. The initial DO of the wastewater is 2.10 mg/L and in the stream, it is 7.91 mg/L. Assume the stream velocity is 0.31 m/s and the dissolved oxygen saturation concentration is 8.76 mg/L. The kd = 0.15 d-1 and kr = 0.31 d-1 values have been adjusted for temperature. Calculate the minimum DO in the stream and determine the distance (km) downstream that this minimum occurs. Is this a safe level for fish?
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
Problem 1. A wastewater sample is examined for its solid contents. The ionic strength (μ) of wastewater is 0.01. It is known that 30% of total dissolved solids (TDS) is organic. The volumetric settleable solid concentration is determined as 20 mL/L using Imhoff Cone. It is assumed that 40% of total suspended solids (TSS) is equal to dry-settleable solids. The volumetric settleable solids composes of 2% of dried solids and 98% of water. The density of volumetric settleable solid concentration is given as 1020 g/L. Use information given in problem, calculate: a) Total Dissolved Solids (TDS) concentration as mg/L. b) Total Suspended Solids (TSS) concentration as mg/L. c) Total Solids (TS) concentration as mg/L.
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 47
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
• An average operating data for conventional activated sludge treatment plant is as follow:- • Wastewater flow =35000m³/d, volume of aeration tank =10900m2, influent BOD to aeration tank=162.5mg/l, effluent BOD=20mg/I, mixed liquor suspended solids32500mg/l. Based on the information above determine: - Aeration period, F/M, BOD load ,Percentage efficiency of BOD removal.
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?
The BOD just downstream from a wastewater pipe is 12.9 mg/L and the DO (dissolved oxygen) is 8.6 mg/L. The saturated value of dissolved oxygen is 9.6 mg/L, the deoxygenation constant (kd) is 0.21/day, the reaeration constant (kr) is 0.42. Find the time where oxygen deficit is at a maximum and find the minimum value of DO.
Q2: A coagulation treatment plant with a flow of 0.5 m/sec is dosing alum at 23 mg/l. No other chemical is being added. The raw water suspended solids concentration is 37 mg/l. The effluent suspended solids concentration is measured as 12 mg/l. The sludge content is 1% and the specific gravity of sludge solids is 3.01. What volume of sludge must be disposed of each day? Based on your result, remark with tolerance to water treatment process.
A 14 mgd wastewater treatment plant has a primary clarifier that treats an influent with 800 mg/L of solids and a removal efficiency of 55%. Sludge at the bottom of the tank is pumped at a rate of 0.1 mgd. 1. What is the effluent solids concentration from the clarifier (in mg/L)? What is the solids concentration in the sludge at the bottom of the tank (in mg/L and %)? The primary sludge is further processed in a digester following treatment. What weight of solids (in lbs) need to be handled each week? а. b. с.
5 ml wastewater is added to a 333 ml BOD flask. The initial dissolved O2 concentration was 13 mg/l and final dissolved O2 concentration was 4 mg calculate the BOD?

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