Q3) A treatment plant treats 18000 m³/d wastewater. of SS concentration 300mg/l and BOD, 200 mg/l. The treatment plant consists of one primary circular sedimentation tank and two single stage trickling filters. The effluent SS concentration from primary sedimentation tank -120mg/l. Each trickling filter of depth 2m, r-120% and BOD removal - 80 %. Find: 1.Diameter of each trickling filters (m). 2.Total BOD removal in treatment plant (%).

Q3) A treatment plant treats 18000 m³/d wastewater. of SS concentration 300mg/l and BOD, 200 mg/l. The treatment plant consists of one primary circular sedimentation tank and two single stage trickling filters. The effluent SS concentration from primary sedimentation tank -120mg/l. Each trickling filter of depth 2m, r-120% and BOD removal - 80 %. Find: 1.Diameter of each trickling filters (m). 2.Total BOD removal in 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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Q4: A wastewater treatment plant as shown: The surface area of each unit 800 m². Depth of each unit 2 m. and %BOD removal in Primary and Secondary sedimentation tanks = 25%. Find: 1- Total % BOD removal in the plant. 2- BOD removal in the Trickling filter. 3- The flow rate (m³/d) in the plant. r-150% BOD, 180 mg/l Pri. sed. tank River BOD,-20 mg/1 Trickling filter Sec. sed. tank (25 marks)
A wastewater stream A with 200 m3 /h of flow rate and BOD of 300mg/I mixes with stream B having flow rate of 400m3/h and BOD 225mg/l. A completely mixed activated sludge process is used to treat the mixture. BOD of effluent stream is 10 mg/l. Given Y= 0.5, k=5 day", Kd=0.06 day ; Ks=100 mg/l and MLVSS= 2000 mg/l. Assume all other parameters if necessary Find (a) BOD and flow rate of mixture (b) Mean residence time (c) Oxygen requirement (d) F/M ratio (e) Recycle ratio (f) Sizing of the tank
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An aeration basin for an activated sludge facility has the following characteristics Length = 90 ft Width = 30 ft Liquid depth = 12 ft MLSS=4,000 mg/L The raw wastewater has the following characteristics: Flow = 12 MGD BOD, = 220 mg/L Suspended solids (SS) = 250 mg/L A primary clarifier is ovided that removes 25% of the BOD, an 60% of the suspended solids The return activated-sludge flow rate is 0.8 MGD If the primary sludge solids content is 4% solids and the specific gravity is 1.0, the primary sludge volume (ft³/day) is most nearly OA. 1,500 OB. 60,000 O C.600 O D.4,500
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
The design flow for a water treatment plant (WTP) is 4000 m3/d, the rapid mixing tank will have a mechanical mixer and the average alum dosage will be 50 mg/l. The theoretical mean hydraulic detention time of the tank will be 60 second and the depth is 1.4 m. Design a circular rapid mixing tank and determine the quantity of alum needed.
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/d. 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. 1. Calculate the primary sludge flowrate (underflow rate) (m3/d). 2. Calculate sludge cake flowrate (m3/d).
8 Determine the solid loading rate on a secondary circular clarifier (diameter = 20 m) if the plant influent load is 2 ML/d. The recycle flow from secondary clarifier to the activated sludge system (aeration basin) is set to 10% of the influent load. It is also known that the aeration basin produces an MLSS (mixed liquor suspended solids) concentration of 3000 mg/L.
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Exercise 4 A treatment plant has the following process diagram. Determine the flow rate expression for each unit. The inflow is Qi. The sludge waste flowrate is Qw. The return flowrate rate is Qr. The mixed liquor return flowrate is Qm. Mixed-liquor return Secondary clarifier Effluent Influent- Anoxic Aerobic Anoxic Aerobic Return activated sludge (e) Bardenpho™ (4-stage) Sludge Exercise 5 Derive the ODE system for above process. There are two species involved in the reaction, as shown below. The process rate for each tank can be simply represented by r1, 12,...15. The gravitational sedimentation is only applied to species X in the clarifier and can be represented by a reaction rate of r5 = 3.5X5. The influent are Xi and S₁. S=X.
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?
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.