The plant operators at the Acme WWTP maintains the 10,000 gal aeration basin shown below with a constant 1500mg/L mixed liquor suspended solids (MLSS). The suspended solids are separated in a clarifier with recycle of the separated sludge. The recycle flow rate is 5,000 gal per day and 500 gal of recycle are wasted daily. Given that the plant treats 25,000 gals of liquid waste per day and the effluent contains 30 mg/L MLSS, find the solids retention time assuming the system is at steady-state

The plant operators at the Acme WWTP maintains the 10,000 gal aeration basin shown below with a constant 1500mg/L mixed liquor suspended solids (MLSS). The suspended solids are separated in a clarifier with recycle of the separated sludge. The recycle flow rate is 5,000 gal per day and 500 gal of recycle are wasted daily. Given that the plant treats 25,000 gals of liquid waste per day and the effluent contains 30 mg/L MLSS, find the solids retention time assuming the system is at steady-state

Solid Waste Engineering

3rd Edition

ISBN:9781305635203

Author:Worrell, William A.

Publisher:Worrell, William A.

Chapter4: Mechanical Processes

Section: Chapter Questions

Problem 4.2P

See similar textbooks

Related questions

Q: A sewage treatment plant discharges 4 ML/d of effluent with 20.17 mg/L BOD5 into a stream. The…

Q: oluble BOD5 = 80 mg/L Assume that the BOD5 of the suspended solids may be estimated as equal to 50%…

A: According to given data A small town has been directed to upgrade its primary wastewater treatment…

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

A: Given: Influent BOD, S0=150mg/L Discharge through the aeration tank=0.44cumec HRT=6hr…

Q: The 550-bed Atlanta Hospital has a small, activated sludge plant to treat wastewater from the…

A: Given, V = 10 m x 10 m x 5 m N = 550 q = 1.50 m3 per day per bed To find : Aeration time

Q: 2. An activated sludge system is to treat wastewater from a population of 10,000 PE. The wastewater…

A: Given,Q=10000 PEBOD5 =300 mg/LSS=400 mg/LY=0.5 kg/kgkd=0.05 per day MLSS=3000 mg/LX=3000 mg/LFor…

Q: A complete mix activated sludge secondary treatment plant has an effluent standard of 30 mg/L SS and…

A: Solution: Given Data: Effluent standard of 30 mg/L SS BOD5=20 mg/L Plant produce an effluent with…

Q: 2. A scrubber separator is used to remove sulfur dioxide (SO2) from a gas stream. A 0.5 m/sec inlet…

Q: A Granulated Activated Carbon (GAC) adsorption laboratory column 10cm in diameter and 3m deep, is…

A: Given Area=7.854×10−3m3 Volume= 6.3660m/hr t=0.47123 Q=2.122 bed volume /hr

Q: The wastewater contains 7.5 x 105 fecal CFU per liter. The permissible fecal coliform concentration…

A: Given: Initial coliform concentration, Co = 7.5 x 105 CFU/L Maximum permissible coliform…

Q: A municipal wastewater treatment plant processes an average of 14,000 m3 / day. The peak flow is…

A: Surface overflow rate is defined as the ratio of the volume of the water applied in unit time per…

Q: A coagulation treatment plant with a flow of 0.5 m3/s is dosing alum at 20 mg/l. Concentration of…

A: According to given data A coagulation treatment plant with a flow of 0.5 m3/s is dosing alum at 20…

Q: A wastewater with a BOD of 25mg/L is discharged through an outfall to a freshwater stream of mean…

A: BOD = 25 mg/LMean velocity of freshwater stream = Vmean-fresh = .1m/sDOdownstream =…

Q: The 550-bed Atlanta Hospital has a small, activated sludge plant to treat wastewater from the…

A: Given, BOD5 after primary settling, So = 450 mg/L MLVSS = 2,600 mg/L To find : F/M ratio

Q: A Reverse Osmosis treatment plant is treating salty water on a farm and operates 24 hours a day. It…

A: Data given: Capacity of treatment plant = 10000L/day Treatment capacity = 50% Rate of infiltration =…

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

A: To determine- Aeration volume (m3)

Q: The characteristics of the synthetic textile wastewater are 1500 mg/L BOD, 2000 mg/L SS, 30 mg/L…

A: Given: Textile plant: BOD = 1500 mg/l SS = 2000 mg/l Nitrogen = 30 mg/l Phosphorous = 0 Domestic:…

Q: A complete mix activated sludge process is to be used for biological treatment. This plant receives…

A: Q = 10000 m3/day BOD = 250 mg/L kd = 0.05 day-1 8 days

Q: What is the purpose of secondary wastewater treatment? a. To reduce toxic chemicals and heavy…

A: Purpose of secondry waste water treatment.

Q: It is desired to design an activated sludge process to treat 20,000 m3/d of settled wastewater with…

A: Solution-

Question

100%

4. The plant operators at the Acme WWTP maintains the 10,000 gal aeration basin
shown below with a constant 1500mg/L mixed liquor suspended solids (MLSS).
The suspended solids are separated in a clarifier with recycle of the separated
sludge. The recycle flow rate is 5,000 gal per day and 500 gal of recycle are
wasted daily. Given that the plant treats 25,000 gals of liquid waste per day and
the effluent contains 30 mg/L MLSS, find the solids retention time assuming the
system is at steady-state.
The variables for flow rates, Q, and MLSS concentra-
tions, X, can be placed on the schematic to help orga-
nize the solution to this problem.
VA = 10,000 gal
ХА 3 1500 mg/L
Xe = 30 mg/L
Qin = 25,000 gal/day
Qin
influent
Qu = 500 gal/day
Qr = 5000 gal/day
Xw = X, (unknown]
Q = [Qin + (a, - Qu}} – Q,
= Qin - Qw
%3D
aeration
basin
fo determine Xr, a solids balance must be taken at
he clarifier. Since the total solids entering the clarifier
nust be equal to the total solids exiting the clarifier for
teady-state flow conditions,
effluent
VA; XA
clarifier
X,
Q, - Qw
Qw
waste
Xw = X,
гecycle
(Qin + (Qr - Qu)]Xa = Q¢X¢ + Qr X,
%3!
=(Qin - Qu)Xe + QrX,

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Municipal solid waste characteristics and quantities

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

12.60 The aeration tank for a completely mixed aeration process is being sized for a design waste- water flow of 7500 m³/d. The influent BOD is 130 mg/l with a soluble BOD of 90 mg/l. The design effluent BOD is 20 mg/l with a soluble BOD of 7.0 mg/l. Recommended design para- meters are a sludge age of 10 d and volatile MLSS of 1400 mg/l. Selection of these values takes stants from a bench-scale treatability study are Y into account the anticipated variations in wastewater flows and strengths. The kinetic con- 0.60 mg VSS/mg soluble BOD and k₁ = 0.06 per day. Calculate the volume of the aeration tank, aeration period, food ganism ratio, and excess biomass th
The town of Gatesville has been directed to upgrade its primary WWTP to a secondary plant that can meet an effluent standard of 30.0 mg/L BOD5 and 30.0 mg/L suspended solids (SS). They have selected a completely mixed activated sludge system. Assuming that BOD5 of the SS may be estimated as equal to 57 percent of the SS concentration, estimate the required volume of the aeration tank. The following data are available from the existing primary plant. Existing plant effluent characteristics Flow =0.150 m3/s BOD5 =82.0 mg/L Assume the following values for the growth constants: Ks= 100 mg/L BOD5; µm =2.5 d-1; kd =0.050 d-1; Y = 0.50 mg VSS/ mg BOD5 removed. Assume the concentration of MLVSS is 2000 mg/L.
The 550-bed Atlanta Hospital has a small, activated sludge plant to treat wastewater from the hospital. The average daily hospital discharge is 1.50 m3 per day per bed, and the average soluble BOD5 after primary settling is 450 mg/L. The aeration tank effective liguid dimensions of 10 m x 10 m x 5 m. The plant operating parameters are as follows: MLVSS - 2,600 mg/L; MLSS - 1.20 (MLVSS); Settling sludge volume after 30 min. = 200 mL/L. The F/M ratio is nearly?
The 550-bed Atlanta Hospital has a small, activated sludge plant to treat wastewater from the hospital. The average daily hospital discharge is 1.50 m3 per day per bed, and the average soluble BOD5 after primary settling is 450 mg/L. The aeration tank effective liguid dimensions of 10 m x 10 m x 5 m. The plant operating parameters are as follows: MLVSS - 2,600 mg/L; MLSS - 1.20 (MLVSS); Settling sludge volume after 30 min. = 200 mL/L. The aeration time is nearly?
Note:hand written solution should be avoided.
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) A wastewater treatment plant consists of primary treatment followed by an activated-sludge secondary system. The plant processes 7000 m3/d of wastewater with 180 mg/L of BOD5 and 150 mg/L of suspended solids. The primary sludge contains 500 kg of dry solids per day with 4.5% solids content. The plant produces 760 kg of total sludge (primary and secondary) per day. Assume 30% removal efficiency of the primary clarifier and F/M ratio of 0.2 for the aeration tank. b) The thickened sludge from problem a) is to be digested in a standard-rate mesophilic anaerobic digester (MAD). The sludge has 68% organic content, and approximately 60% of the organic fraction is digested after a 30-d period. The digested sludge has a solids content of 6% and is stored for a period of 90 d. Calculate the volume of the standard-rate digester
a) A wastewater treatment plant consists of primary treatment followed by an activated-sludge secondary system. The plant processes 7000 m3/d of wastewater with 180 mg/L of BOD5 and 150 mg/L of suspended solids. The primary sludge contains 500 kg of dry solids per day with 4.5% solids content. The plant produces 760 kg of total sludge (primary and secondary) per day. Assume 30% removal efficiency of the primary clarifier and F/M ratio of 0.2 for the aeration tank. Determine the following: a. Primary sludge volume and removal efficiency of primary clarifier b. Influent and effluent BOD5 of the secondary activated-sludge system c. Volume of secondary sludge with 1% solids content. b) In problem a), the primary and secondary sludges are mixed in a gravity thickener and sent for further treatment. The thickened sludge contains 4% solids. Calculate the percent volume reduction in the gravity thickener. c) The thickened sludge from problem b) is to be digested in a standard-rate mesophilic…
An activated sludge wastewater treatment system was designed and operated in a steadystate under the following conditions:• Reactor volume of 10,000 m3in total;• Sludge concentration (MLSS) of 3200 mg/L in the reactor;• Return sludge concentration at 8000 mg/L (the underflow of the secondary clarifier)• Each day, 400 m3 of underflow sludge (return sludge) was wasted as excesssludge;What is the average Sludge Residence Time (SRT)?
Name:.. 11. An activated sludge system has a single aeration basin with a volume of 300,000 gal and 2 - 25 ft. diameter secondary clarifiers. The influent flow is 1.1 MGD, and the influent BODs and TSS are 170 mg/L and 190 mg/L, respectively. The mixed liquor suspended solids concentration is 2500 mg/L, and the recycle suspended solids concentration is 6800 mg/L. The sludge recycle flow rate is 0.60 MGD and the solids residence time is 9 days. 10a. Calculate the volumetric BODs loading rate on the aeration basin in units of lbs/1000 ft³-d. ( 10b. Calculate the F:M ratio in units of g-BODs/g-TSS-d 9
A completely mixed activated sludge process is being designed for a wastewater flow of 3 MGD using kinetic equations. The influent BOD of 180 mg/L is essentially all soluble and the design effluent soluble BOD is 10 mg/L. The mean cell residence time is selected to be 8.0 days for sizing the aeration volume tank, and the MLVSS is 2500 mg/L. The kinetic constants from a bench-scale treatability study are Y = 0.60 Ib VSS/Ib BOD and Kd = 0.06 day-1. ---Determine the efficiency of soluble BOD removal. For this, you will have to recognize substrate concentration in the coming flow (So) and the concentration of substrate in the exiting flow (Se).
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.