A city wishes to discharge wastewater primary effluent to a river which supports a trout. The state requires the city to show that the trout population is protected.  Trout require a minimum of 4 mg/l of dissolved oxygen to survive.  Determine the minimum downstream dissolved oxygen concentration from the following information about the wastewater discharge and river.  Does the minimum dissolved oxygen concentration meet the requirements?Wastewater discharge=0.45m3/s, river flow rate= 8.5 m3/s, river cross-sectional area=19m2, dissolved Oxygen concentration in the wastewater= 0.2mg/l, dissolved oxygen content of the river upstream = 8.1mg/l, saturation dissolved oxygen concentration at 20 C= 9.0 mg/l, BOD concentration in wastewater = 85 mg/l, BOD concentration upstream in the river= 0.2 mg/l, BOD decay constant = 0.2 d-1, and the reaeration constant for the river= 0.4 d-1.

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Asked Dec 12, 2019
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  1. A city wishes to discharge wastewater primary effluent to a river which supports a trout. The state requires the city to show that the trout population is protected.  Trout require a minimum of 4 mg/l of dissolved oxygen to survive.  Determine the minimum downstream dissolved oxygen concentration from the following information about the wastewater discharge and river.  Does the minimum dissolved oxygen concentration meet the requirements?

Wastewater discharge=0.45m3/s, river flow rate= 8.5 m3/s, river cross-sectional area=19m2, dissolved Oxygen concentration in the wastewater= 0.2mg/l, dissolved oxygen content of the river upstream = 8.1mg/l, saturation dissolved oxygen concentration at 20 C= 9.0 mg/l, BOD concentration in wastewater = 85 mg/l, BOD concentration upstream in the river= 0.2 mg/l, BOD decay constant = 0.2 d-1, and the reaeration constant for the river= 0.4 d-1.

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Expert Answer

Step 1

the following data can be calculated with given data:

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wastewater (given) stream (given) Wastewater and stream Parameter water mixture 0.45 m³ /s flow (m³ /s) 8.5 m³ /s Qmixture= 0.45+8.5=8.95 m³ /s Dissolved oxygen, 0.2 mg/L 8.1 mg/L DOmixture = (0.45*0.2+8.1*8.5)/(8.5+0.45) = 0.69 mg/L %3! BOD5 at 20°C, mg/L 0.2 mg/L 85 mg/L BODmixture= (0.45*85+0.2*8.5)/(8.5+0.45) = 4.46 0.2(No temp. correction required) (assumed for stream water) 0.4(No temp. correction required) %3D Oxygen consumption rate (K1 at °C) (1/day) 0.2 Oxygen reaeration rate (K2 at °C) (1/day) 0.4

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Step 2

Required minimum DO = 4.0 mg/L in stream water. As DO at critical location is 5.89 mg/L, greater than the recommended DO level, no modification in wastewater effluent characteristics is required.

Step 3

 [Note that 5 mg/L DO deficit is allowable and ...

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