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A lake is fed by a polluted stream and a sewage outfall. The stream and sewage wastes have a decay rate coefficient (k) of 0.5/day (1st order units). Assuming complete mixing and no other water losses or gains, what is the steady-state pollutant concentration as mg/L in the lake? Incoming Stream: C = 10 mg/L, Q = 40 m^3/s Sewage Outfall: C = 100 ppm, Q = 0.5 m^3/s Lake: V= 200 m^3

Respuesta :

Solution :

Given :

k = 0.5 per day

[tex]$C_s = 10 \ mg/L \ ; \ \ Q_s= 40 \ m^3/s$[/tex]

[tex]$C_{sw} = 100 \ ppm \ ; \ \ Q_{sw}= 0.5 \ m^3/s$[/tex]

Volume, V [tex]$= 200 \ m^3$[/tex]

Now, input rate = output rate + KCV ------------- (1)

Input rate  [tex]$= Q_s C_s + Q_{sw}C_{sw}$[/tex]

                [tex]$=(40 \times 10) + (0.5\times 100)$[/tex]

                [tex]$= 2 \times 10^5 \ mg/s$[/tex]

The output rate [tex]$= Q_m C_{m}$[/tex]

                          = ( 40 + 0.5 ) x C x 1000

                          [tex]$=40.5 \times 10^3 \ C \ mg/s$[/tex]

Decay rate = KCV

∴[tex]$KCV =\frac{0.5/d \times C \ \times 200 \times 1000}{24 \times 3600}$[/tex]

            = 1.16 C mg/s

Substituting all values in (1)

[tex]$2 \times 10^5 = 40.5 \times 10^3 \ C+ 1.16 C$[/tex]

C = 4.93 mg/L

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