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. What is the effluent solids concentration from the clarifier (in mg/L)? b. What is the solids concentration in the sludge at the bottom of the tank (in mg/L and%)? c. The primary sludge is further processed in a digester following treatment. What weight of solids (in lbs) need to be handled each week? а.

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. What is the effluent solids concentration from the clarifier (in mg/L)? b. What is the solids concentration in the sludge at the bottom of the tank (in mg/L and%)? c. The primary sludge is further processed in a digester following treatment. What weight of solids (in lbs) need to be handled each week? а.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

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?
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.
An aeration tank has an MLSS concentration of 2000 mg/L. After settling for 30 min. in a 1-L graduated cylinder, the sludge volume is measured to be 150 mL. Compute the SVI of the sludge.
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
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
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)?
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
In a municipal wastewater treatment plant, the influent flow rate into a complete mixing activated sludge system is 50 L/sec, its solid concentration MLSS is 30 mg/L, and the BODs in the influent is 245 mg/L. The MLSS concentration at the bottom of the secondary clarifier is 15,000 mg/L and 10 mg/L at the top of the |clarifier. The flow rate of returned activated sludge is 20 L/sec. a. What is the concentration of the MLSS in the activated sludge reactor (assuming that it is fully mixed and has constant concentration)? b. What is the activated sludge basin volume if the hydraulic retention time in the basin is 10 hours and the system is operated at the R/Q ratio given in the problem? |c. If the cross section of the basin is 20 m² (perpendicular to the flow), how long the basin will be? d. The plant has 96% BOD5 removal. What is the effluent BOD5? e. The flowrate of the waste activated sludge (WAS) is 0.2 L/sec. How many solids (in kg) are wasted every day through the secondary…
D Q1: Design an activated sludge process to treat a wastewater of 15000 m³/day with a BOD5 of 180 mg/l following primary treatment. The effluent BODs and SS are to be 20 mg/l. Assume Xr-15000 mg/l, X=2500 mg/l, ec=10 days, y=0.60, Kd=0.05. Determine the reactor volume, the sludge production rate, the circulation rate, the hydraulic retention time and oxygen required?
A wastewater with a BOD of 25mg/L is discharged through an outfall to a freshwater stream of mean velocity 0.1m/s. The DO downstream of the outfall is 8.5mg/L. Assume the deoxygenation rate is 0.25/d, the reaeration rate is 0.4 /d and the temperature of 20 degree C (DOs = 9.09mg/L). Compute for the ff: a) Critical distance b) minimum DO c) Oxygen deficit d) Critical time
Note:hand written solution should be avoided.
The aeration tank for a completely mixed aeration process is being sized for a designwastewater flow of 4500 m3/d. The influent COD is 150 mg/L. The design effluent COD is7 mg/L. Recommended design parameters are a sludge age of 10 days and MLVSS of 1400 mg/L. The expected Sludge Volume Index is 100 ml/g. The selection of these values takes into account the anticipated variations in wastewater flows and strengths. The kinetic constants from a bench-scale treatability study are Y = 0.60 mg VSS/mg COD and ke = 0.06 per day. Calculate:a. the hydraulic residence time and volume of the aeration tankb. the food/microorganism ratioc. the sludge production rated. the sludge recycle ratioe. the oxygen requirement