5. A pilot column breakthrough test has been performed with a wastewater that has a trichloroethylene (TCE) concentration of 400 mg/L as TOC (total organic carbon). Pertinent design data are ID = 0.095 m, length = 1.04 m, mass of carbon = 2.98 kg, liquid flow rate = 17.42 L/hr, unit liquid flow rate = 0.679 L/s-m², and packed carbon density = 401 g/L. The breakthrough data are given in Table 1. Determine: a. The liquid flow rate in bed volumes per hour and the volume of liquid treated per unit mass of carbon using an allowable breakthrough concentration of 35 mg/L of TOC. in kg/kg. b. The kinetic constants ki in L/s-kg and qo Table 1. Throughput Volume Effluent TOC (mg/L) (L) 15 69 159 273 379 681 965 1105 1215 1287 1408 1548 12 16 24 16 16 20 28 32 103 211 350 400

5. A pilot column breakthrough test has been performed with a wastewater that has a trichloroethylene (TCE) concentration of 400 mg/L as TOC (total organic carbon). Pertinent design data are ID = 0.095 m, length = 1.04 m, mass of carbon = 2.98 kg, liquid flow rate = 17.42 L/hr, unit liquid flow rate = 0.679 L/s-m², and packed carbon density = 401 g/L. The breakthrough data are given in Table 1. Determine: a. The liquid flow rate in bed volumes per hour and the volume of liquid treated per unit mass of carbon using an allowable breakthrough concentration of 35 mg/L of TOC. in kg/kg. b. The kinetic constants ki in L/s-kg and qo Table 1. Throughput Volume Effluent TOC (mg/L) (L) 15 69 159 273 379 681 965 1105 1215 1287 1408 1548 12 16 24 16 16 20 28 32 103 211 350 400

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

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.
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?
Assume that all solutions are prepared at 25 degrees celsius and the solutes arr completely dissolved. In the lead content analysis of a wastewater sample, 75.00 mL of the original sample was taken and diluted with 125.00 mL deionized water. The diluted sample was found to have a lead content of 19.2 ppb. What is the lead content (in pbb) of the original wastewater sample?
What is the standard level of potassium concentration that can be found on wastewater ?
The PEF, Inc. firm desires to utilize a complete-mix activated-sludge system with recycle to treat their municipal wastewater after primary sedimentation. The design parameters are: Flow rate=0.92m³/s Influent Soluble BOD5 = 292mg/L MLVSS=2,500mg/L Mean cell residence time = 7.24 d For any missing parameters, use typical values of growth constants for domestic wastewater (provided in your text, page 494). Assume that the solids lost in the secondary clarifier effluent are minor and can be neglected. Assuming a BODs (S) concentration of 24 mg/L, the detention time (residence time or theta) in the activated sludge system is most nearly: O 28 hours O 6.5 hours O 168 hours O 53 hours
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
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.
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.
Asap
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
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 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).