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Consider a 50 x 106 m3 lake fed by a polluted stream with a flow rate of 75 m3 /s and a pollutant concentration of 25.5 mg/L. There is also a sewage outfall that discharges 3.0 m3 /s of wastewater with a pollutant concentration of 125 mg/L. Stream and sewage wastes have a reaction rate coefficient of 5% per day. Find the steady-state (i.e., effluent) pollutant concentration and flow rate

Answer :

ajeigbeibraheem

Answer:

24.78 mg/L

Explanation:

The step-to-step explanation is written legibly with clear explanation in the diagram attached below.

${teks-lihat-gambar} ajeigbeibraheem
Darase

Answer:

Answer: Input rate = 2.288 x 10⁶mg/s

             Output rate =78 x 10³Cmg/s

            Decay rate = 28.94 x 10 ⁵C mg/s

Explanation:

Assuming that complete and instantaneous mixing occurs in the lake, this implies that the concentration in the lake C is the same as the concentration of the mix leaving the lake Cm

   Input rate = Output rate  + KCV

Input rate = Q₁C₁ + QwCw

                = (75.0m³/s x 25.5mg/L + 3.0m³/s x 125.0mg/L) x 10³L/m³

                = 2.288 x 10⁶mg/s

Output rate = QmCm = (Q₁ +Qw)C

                   =(75 + 3.0)m³/s x Cmg/L x 10³L/m³ = 78 x 10³Cmg/s

Decay rate = KCV = 5/d x Cmg/L x 50 x 10⁶ x 10³L/m³  

                                           24 hr/d x 3600s/hr

                     = 28.94 x 10 ⁵C mg/s

So,

                      =2.288 x 10⁶ = 78 x 10³C + 28.94 x 10 ⁵C  = 29.72 X 10⁵C

                      = 2.288 x 10⁶          

                        29.72 X 10⁵         = 0.77 mg/l

                   C =

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