A waste water stream (flow=2 m³/s, ultimate BOD=90 mg/l) is joining a small river (flow=12 m³/s, ultimate BOD=5 mg/l). Both water streams get mixed up instantaneously. Cross sectional area of the river is 50 m². Assuming the de-oxygenation rate constant, k'=0.25/day the BOD (in mg/l) of the river water, 10 km downstream of the mixing point is

A waste water stream (flow=2 m³/s, ultimate BOD=90 mg/l) is joining a small river (flow=12 m³/s, ultimate BOD=5 mg/l). Both water streams get mixed up instantaneously. Cross sectional area of the river is 50 m². Assuming the de-oxygenation rate constant, k'=0.25/day the BOD (in mg/l) of the river water, 10 km downstream of the mixing point is

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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A waste water stream (flow=2 m³/s, ultimate BOD=90 mg/l) is joining a small river (flow=12 m³/s,
ultimate BOD=5 mg/l). Both water streams get mixed up instantaneously. Cross sectional area of the
river is 50 m². Assuming the de-oxygenation rate constant, k'=0.25/day the BOD (in mg/l) of the river
water, 10 km downstream of the mixing point is

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A waste water stream (flow = 2 m3/s), ultimate BOD = 90 mg/l is joining a small river (flow = 12 m³/s) ultimate BOD = 5 mg/I. Both water streams gets mixed up instantaneously. Cross-section area at the river is 50 m2. Assuming Deoxygenation constant = 0.251 day. Estimated downstream BOD of river flow.
Flow rate: 1.5 m3/s Pollution: pesticide permethrin Lake size: 100m x100m x 2.5m Steady state concentration of permethrin leaving the lake is 1.4 ppm Average cross-sectional area of the river is 2.7 m2. Half-life of permethrin in this environment is 19 hrs. What is the concentration of permethrin entering the lake and what distance downstream of the outlet would the concentration be less than 1 ppm?
Please ans a and b
A complete-mixed activated sludge process (CMAS) is to be designed to treat 6.0 MGD of primary effluent having a BOD5 of 190 mg/L. The requirement is that the effluent BOD5 and TSS concentrations be 20 mg/L or less on an annual average basis. The following biokinetic coefficient obtained at 15°C: Y = 0.6 mg VSS/mg BOD5, k = 2.85 d-1, Ks = 70 mg/L BOD5, and kd = 0.041 d-1. Assume that the MLVSS concentration in the aeration basin is maintained at 2,500 mg/L. Determine: a. The minimum MCRT (d) b. The MCRT necessary to meet the requirement (d) c. The volume of the aeration basin (ft3)