125 m³/s of a sewage is discharged in a river which is fully saturated with O, and flow at a minimum rate = 1600 m³/s with a minimum velocity of 0.12 m/s. If the 5 day BOD of the sewage is 300 mg/l. Find out where the critical DO will occur in the river. Assume coefficient of purification of river = 4.0 k₁ = 0.11/day BOD = 1.25 BOD Assume sat DO at 20° 9.2 mg/l

125 m³/s of a sewage is discharged in a river which is fully saturated with O, and flow at a minimum rate = 1600 m³/s with a minimum velocity of 0.12 m/s. If the 5 day BOD of the sewage is 300 mg/l. Find out where the critical DO will occur in the river. Assume coefficient of purification of river = 4.0 k₁ = 0.11/day BOD = 1.25 BOD Assume sat DO at 20° 9.2 mg/l

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

2 A groundwater (Q1 = 5.0 L/s) reaches a point of convergence (mixing) in a river (Q = 1.0 m3/s) where an industrial wastewater discharge (Q2 = 10 L/s) also converges. /s). If the groundwater has ferrous iron with concentration [Fe2+] = 10.0 mg/L, the industrial wastewater has organic matter with concentration TOC = 50 mg/L, determine the COD of the river downstream of the mixing point.
It takes 22 hours for water to flow from point A to Point B in a river. The flow rate and BOD5 in the river are 20m/h and 2mg/L, respectively. A wastewater of flow rate 2m/h containing BOD5 of 150 mg/L is discharged into the river at point A. Calculate BOD5 concentration at downstream Point B by treating the river as 11 CMFRS in series with equal hydraulic detention times. The first order decay coefficient k of the BOD; is assumed to be 0.05/h.
A stream with a flow rate of 5 m²/s is having an ultimate BOD of 30 mg/litre. A wastewater discharge of 0.20 m3/s having BOD5 of 500 mg/litre joins the stream at a location and instantaneously gets mixed up completely. The cross-sectional area of the stream is 40 m² which remains constant. BOD exertion rate constant is 0.3 per day (logarithm base to e). The BOD (in mg/litre, round off to two decimal places) remaining at 3 km downstream from the mixing location, is
4. The stream shown below has a flow of 28,000 m/d, and a cross-sectional area of 35 m2. The stream is saturated with dissolved oxygen (C* = 10 mg/L) and its BOD, concentration is 0.5 mg/L. At point (A)7 km upstream from the stream's confluence with another much larger river, a food processing industry is continuously discharging 1,500 m/day of wastewater with an average organic composition of CH,0 (which is 90% degradable). The wastewater has a dissolved oxygen concentration of 2 mg/L. A stream sample taken just before the confluence of stream and river has a BOD, 20°C of 3.0 mg/L ( k = 0.24/day for BOD assimilation in the lab). a) Calculate the concentration of the wastewater (in units of mg CH,0/L) before it is mixed in the river. Relevant constants are given below. b) Determine whether the D.O. level in the stream remains above 5 mg/L between point A and confluence of the stream with the larger river. You may assume the following: i) plug flow ii) instantaneous mixing of stream and…
Q:A city of 200,000 people disposes of 1.05 m. s' of treated sewage that still has a BODU of 28.0 mg. L' and 1.8 mg. L' of DO into a river. Upstream from the outfall, the river has a flow rate of 7.08 m. s' and velocity of 0.37 m.s'. At this point, the BODU and DO in the river are 3.6 and 7.6 mg. L', respectively. The saturation value of DO ( at the temperature of the river ) is 8.5 mg. L'. The deoxygenation coefficient, ka is 0.61 day, and the reaction coefficient, k is 0.76 day". Assume complete mixing and that the velocity in the river is the same upstream and downstream of the outfall. 1- What is the oxygen deficit and the BODU just downstream from the outfall (just after mixing, before any reaction can occur)? 2- What is the DO 16 km downstream? 3- Calculate the critical time and distance. 4- What is the minimum DO?
A wastewater treatment plant will receive a flow of 30,000 m3/day. The plant has a 3.6 m deep circular primary clarifier with a surface overflow rate of 60.0 m3/m2-day.a. Calculate the surface area (m2), diameter (m), volume (m3), and hydraulic retention time of the primary clarifier.b. How long will it take a 1.0 mm particle with a density of 1500 kg/m3to settle in the primary clarifier? show all work
The treated domestic sewage of a town is to be discharged in the natural stream in which minimum flow is equal to 0.14 m³/s. The BOD in the stream on the upstream side is 3 mg/L. The treatment plant receives the waste water from a town with a population of 60000 persons with a dry weather flow of ( 153 liters/capita/day. Assume per capita BOD contribution as 0.07 kg/day. a) Determine the percentage of the purification required in the treatment plant if the permissible BOD on the downstream side of the stream is 5 mg/L. b) Draw a labeled diagram to show the process of self-purification of the natural stream if it receives the treated effluent.
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 Hmax Ks ka Y Value 3 d¹ 60 mg/L BOD, 0.10 d ¹¹ 0.6 mg VSS/mg BODS
The treated domestic sewage of a town is to be discharged in a natural stream with following data given: (i) Population = 50,000 (ii) BOD contribution per capita = 0.07 kg/day (iii) BOD of stream on u/s side = 3 mg/It. (iv) Permissible maximum BOD of stream on d/s side = 51 (v) Dry weather flow of sewage = 140 litres per capita per day 5 mg/lt. (vi) Minimum flow of stream = 0.13 m3/sec Choose the correct statement(s): BOD of treated sewage should be 8.21 gm/lt. Total BOD produced per day is 35000 kg/day C BOD of untreated sewage is 500 mg/It. Correct Option D Percentage treatment required is 98.358% Correct Option
A stream flows through a small town where an industry dischargesits effluent into the stream. Upstream characteristics prior to industrial discharge are as follows: fow rate 1000 m3/d, BOD5 2.5 mg/L,nitrates 2.0 mg/L, and temperature 19°C. The industry dischargesat 50 m3/d. According to regulatory requirements, the maximumallowable values in the stream following any discharge are BOD5 30mg/L, nitrates 4.0 mg/L, and temperature differential of 4°C fromupstream conditions. Calculate the maximum allowable values forthe industrial discharge.
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
Derive the Streeter-Phelps sag curve equations. What is the minimum oxygen level in the river and the oxygen level 20 km downstream of the sewage outfall for the following case: ka=0.2 day', k=0.3 day', BOD in the stream before a sewage outfall is 2 mg/l, the river flow rate is 10 m³/s. the sewage outfall rate is 0.1 m/s with a BOD of 300 and the river velocity is 0.2 m/s.