Question

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 1

Answer: Input rate = 2.288 x 10⁶mg/s

Output rate =78 x 10³Cmg/s

Decay rate = 28.94 x 10 ⁵C mg/s

Step-by-step 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 =

Answer 2

24.78 mg/L

Step-by-step explanation:

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