After primary treatment, a municipal wastewater contains 150 mg/L BOD5. The wastewater flow of 0.44 m³/s (10 MGD) is then treated in an activated sludge system with an SRT of 12 d and an HRT of 6 h. Given the biokinetic parameters tabulated below, determine the limiting value of the minimum SRT (d). Parameter Value 3 d¹ 60 mg/L BOD, 0.10 d ¹¹ 0.6 mg VSS/mg BODS Ks ka Y

After primary treatment, a municipal wastewater contains 150 mg/L BOD5. The wastewater flow of 0.44 m³/s (10 MGD) is then treated in an activated sludge system with an SRT of 12 d and an HRT of 6 h. Given the biokinetic parameters tabulated below, determine the limiting value of the minimum SRT (d). Parameter Value 3 d¹ 60 mg/L BOD, 0.10 d ¹¹ 0.6 mg VSS/mg BODS Ks ka Y

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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After primary treatment, a municipal wastewater contains 150 mg/L BOD5. The wastewater flow of 0.44 m³/s (10 MGD) is then treated in an activated sludge system with an SRT of 12 d and an HRT of 6 h. Given the biokinetic parameters tabulated below, determine the effluent BOD5 (mg/L). Parameter Himax Ks ka Y Value 3 d ¹ 60 mg/L BOD, 0.10 d ¹ 0.6 mg VSS/mg BOD, K
a. Find the time and distance downstream at whoch the oxyfen deficit is a macimum b. Find the minimum value of DO c. Sketch the DO concentration profile
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
After primary treatment, a municipal wastewater contains 150 mg/L BOD5. The 3 wastewater flow of 0.44 m³/s (10 MGD) is then treated in an activated sludge system with an SRT of 12 d and an HRT of 6 h. Given the biokinetic parameters tabulated below, determine the limiting value of the biotreatment safety factor (dimensionless). Parameter Value 3d¹ 60 mg/L BOD, 0.10 d¹¹ 0.6 mg VSS/mg BOD5 Ks ka Y 2
Two wastewater streams A and B, having an identical ultimate BOD are getting mixed to form the stream C. The temperature of the stream A is 20 °C and the temperature of the stream C is 10 °C. It is given that • The 5-day BOD of the stream A measured at 20 °C=50 mg/l • BOD rate constant (base 10) at 20 °C-0.115 per day Temperature coefficient = 1.135 The 5-day BOD (in mg/l, up to one decimal place) of the stream C, calculated at 10 °C, is ●
Two wastewater streams A and B, having an identical ultimate BOD are getting mixed to form the stream C. The temperature of the stream A is 20 °C and the temperature of the stream C is 10 °C. It is given that The 5-day BOD of the stream A measured at 20 °C-50 mg/l BOD rate constant (base 10) at 20 °C=0.115 per day Temperature coefficient = 1.135 The 5-day BOD (in mg/l, up to one decimal place) of the stream C, calculated at 10 °C, is X EE
A municipal wastewater treatment plant is installing completely mixed activated sludge to meet their NPDES permit of 15 mg/L BOD5. The mixed liquor concentration is 1,000 mg/L MLVSS. The wastewater flowrate going to secondary treatment is 0.170 m3/s and the BOD5 in that stream is 110 mg/L. Laboratory measurements of the growth constants show: Ks = 60 mg/L BOD5, um = 2.8 d-1, kd = 0.05 d-1, Y = 0.73 mg MLVSS/mg BOD5. Determine the required volume of the aeration tank.
A wastewater sample was diluted 10% with tap water for conducting BOD Test. The DO content in wastewater sample was 1.2 mg/l and water was 6 mg/l. After 5 days incubation at 20°C, the DO of the sample was 0.5 mg/l. Calculate (i) 5 day BOD of wastewater and (ii) ultimate BOD. Assume K₂0 = 0.12/day. ion:
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
After primary treatment, a municipal wastewater contains 150 mg/L BOD5. The wastewater flow of 0.44 m³/s (10 MGD) is then treated in an activated sludge system with an SRT of 12 d and an HRT of 6 h. Given the biokinetic parameters tabulated below, determine the aeration volume (m³). Parameter Value 3d¹ 60 mg/L BODs 0.10 d¹ 0.6 mg VSS/mg BOD5 Hmax Ks ka Y
C. What is the concentration UI SUINUS 5. Determine the size (diameter and depth) of the secondary clarifier (s) at a municipal activated sludge treatment facility: The wastewater flow is 25,000 m³/d, the recycle ratio (recycle sludge flow/wastewater flow) is 0.25. The suspended solids concentration is assumed to be 2500 mg/L in the activated sludge tank. The over flow rate is 50 m3/d- m2 and solids loading is 125 kg/d-m2. Estimate detention time if the depth is 4.5 m. What would be the weir loading? Department of Civil & Environmental Engineering P.O. Box 519 Mail Stop 2510, Prairie View, Texas 77446 w.pvamu.edu Phone (936) 261-1660/1665(dept) Fax (936) 261 1662 Ensuring o Using Zoom o