For an industrial activated sludge treatment process, the biomass concentration (X) is 2,200 mg VSS/L and the substrate concentration (S) is 50 mg BOD5/L. Using the following values, find the net microbial growth rate (rg'). Max substrate utilization rate (k) = 5 mg BOD5/mg VSS-d Half saturation constant (Ks) = 60 mg BOD5/L Microbial yield coefficient (Y) = 0.6 mg VSS/mg BOD5 Endogenous decay (death) rate constant (kd) = 0.06 /d

Q: The moisture content of a sludge is 95% with a volume of 1000 m³. If the volume of dewatered sludge…

A: Moisture content of a sludge- P1 = 95 % Volume- V1 = 1000 m3 The volume of dewatered sludge is only…

Q: MEK, Methyl ethyl ketone (CH3C(O)CH2CH3) has waste code D035. Why does MEK have high solubility in…

A: Methyl Ethyl Ketone  is a colorless, liquid chemical substance that is used in many industrial…

Q: Consider a well-mixed aeration lagoon to remove BOD from a waste stream flowing at 5 MGD. The lagoon…

A: Data given: Discharge = 5 MGD Dimensions of lagoon Length (L) = 100 m Width (B) = 30 m Depth (D) = 3…

Q: Digester volume = 40,000 ft, Raw sludge volume 5,000 gpd, Sludge concentration %3D = 5% Volatile…

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Q: A wastewater treatment plant discharges 0.1 m3/s of treated wastewater to a river having a mean flow…

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Q: P.C. V, X, S S.C. Qs Xg, Sp Act. Sludge Basin RAS Q, S, Xp WAS Qw, Sun Xw

A: 1. V=5222 m3   2. Qw=121.8 m3/day   3. Qe=25798.2 m3/day   4. t=4.8 hours

Q: 2) BODs of a wastewater is determined to be 78 mg/L at 20°C. The temperature correction coefficient…

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Q: 1 The lower the water turbidity, the higher the allowable scour velocity in sedimentation tank.( ) 2…

A: 1 True - Lower water turbidity allows for higher allowable scour velocity as the particles in the…

Q: 3. The concentration of total solids in water can be considered as a conservative chemical property.…

A: Given, Q1 = 10 MGD  Q2 = 4 MGD Q3 = 0.5 MGD

Q: what are Observation methods used in groundwater pollution?

A: Groundwater pollution:- This pollution is occur when pollutants are released the ground and make…

Q: A container of Atrazine (500 g) is dumped into a river 2.3 km upstream of a municipal water intake…

A: Given, M = 500 gm or 0.5 kg L = 2.3 km or 2300 m  A = 5.7 m2  Q = 8.25 m3/s D = 0.1 m2/s

Q: sludge to meet their NPDES permit of 15 mg/L BOD5.  The mixed liquor concentration is 1,000 mg/L…

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Q: of a waste discharge. The wa . The stream has a reaeration dissolved-oxygen concentratio en deficit?…

A: Given: V=10 km/day x=0 Km

Q: After primary treatment, a municipal wastewater contains 150 mg/L BOD5. The wastewater flow of 0.44…

A: To determine- Limiting value of biotreatment safety factor.

Q: A completely mixed activated sludge plant treats wastewater at a rate of 2,565 m³/d. The influent…

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Q: If Grad=20.462m for point A What is the Grad of point B and C Which distance between two points…

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Q: You and your team have decided to build an aerated lagoon to treat the waste generated by a local…

A: Given:Q = 40 m^3 / daySo = 250 mg / L , BOD5S = 15 mg / L, BOD5Ks = 100 mg-BOD5 / Lkd = 0.1 d^-1mu =…

Q: The organic carbon content of an industrial wastewater sample is characterized by a COD level of…

A: COD is measure of amount of oxygen in water consumed for chemical oxidation of pollutants.  BOD is…

Q: (d) Biomass and substrate mass balance equations can be merged giving QwXR = () Y/Qo` X VX (SS) - ka…

A: Given:- SRT=20days HRT=20H So=350mg/l S=100mg/l Y=0.15 Kd=0.22per day Calculate :- 1) prove Given…

Q: It is a kind of wastewater treatment where chemical coagulation of sewer was done prior to…

A: Given:-  It is a kind of wastewater treatment where chemical coagulation of sewer was done prior to…

Q: If the BOD, of a waste water sample is 75 mg/ L and reaction rate constant K(base e) is 0.345 per…

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Q: sing alum at 23 mg/litre. No o ncentration is 37 mg/l. The efm and specific gravity of sludge is =

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Q: The moisture content of a sludge is 95% with a volume of 1000 m, If the volume of dewatered sludge…

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Q: Estimate sludge production (Px) based on the following conditions: microbial decay coefficient (kd)…

A: Given Influent fow= 1m3/s  Q=1000 liter/s  Q=10001×24×60×60 liter/day=86.4×106 liter/dayQ=86.4 MLD…

Q: After primary treatment, a municipal wastewater contains 150 mg/L BOD5. The wastewater flow of 0.44…

A: BOD5=150 mg/LQ=0.44 m3/s (10 MGD)SRT = 12dHRT=6h To determine- Limiting value of SRT

Q: Using Rihet equation, estimate aeration tank Volume (V) based on the following conditions:…

A: GIVEN- Saturation constant (Ks) = 80 mg/L microbial decay coefficient (kd) =0.15/d maximum specific…

Q: b. Find the DO at 12 10. A water treatment plant treats the water at the rate of 33 MGD. Average…

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Q: BOD is widely used for measuring the concentration of biodegradable organic matters in a wastewater.…

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Q: 1) Estimate the biomass concentration in a CSTR aeration tank with the following operating…

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Q: The waste flow 18000m³/day is treated by activated sludge process to reduce 97.5% of influent BOD5…

A: Step 1: Step 2: Step 3: Step 4:

Q: Q-100 m³ of sludge holds a moisture content of 95%. If its moisture content changes to 90%, the…

A: Given data: Sludge volume, V1 = 100 m3 At moisture content, P1 = 95 % For moisture content, P2 = 90…

Q: 2. The raw water to be coagulated has a pH of 7.5, determine how much alkalinity the water must have…

A: pH of coagulated water = 7.5 After adding 150mg/L of alum, we have to determine the change in…

Q: Microbial growth in an activated sludge system can be described by the Monod model. If the maximum…

A: µ=specific growth rate = .6 day-1   µmax=maximum value of μ = 1.2 day-1   KS=saturation constant =…

Q: Dissolved oxygen (D0) is influenced by which factors? Explain. How can you determine chloride…

A: Dissolved oxygen: Dissolved oxygen means the amount of non compound free oxygen present liquid or…

Q: QUESTION 4 phase of bacterial growth. Activated sludge runs at the O Decay Log Stationary O Lag

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Q: You have been treating a wastewater with bacteria that have the following kinetic coefficients: k=…

A: Data given k= maximum specific growth rate = 16 mg BODL/mg VSS-day Ks= half-saturation constant =10…

Q: Tosw 20 + Neg -245.20 0 EC. O -T. as 20 + Fa so -Te cas 20 + 0,2 Nog =o -o tG E

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Q: A groundwater (Q1 = 5.0 L/s) reaches a point of convergence (mixing) in a river (Q = 1.0 m3/s) where…

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Q: Poorly treated municipal wastewater is continuously discharged directly into a lake. The discharge…

A: The pollutant balance equation for the steady-state is: In-out-Decomposed=0  Also Accumulation=0

Q: A method known as multiple tube fermentation technique is used to check biological water quality.…

A: Given data,

Biological Processes

Biological solid waste includes bacterial and contaminated waste produced by living individuals. Its treatment is essential to prevent the spread of any air-borne or water-borne diseases or diseases transmitted from one organism to another in a food chain. There are many benefits of the biological solid waste treatment and some of them are: the minimized quantity of waste substance, waste control, eradicating any disease-causing pathogens, generation of biogas, composting, anaerobic digestion, bio drying, mechanical biological treatment, microbial fuel cell, phytoremediation, and biofiltration. The final products of the biological solid waste treatment may be recycled to obtain organic fertilizers or disposed of in SWDS (Solid Waste Disposal Site) after proper treatment.