A well-mixed sewage lagoon is receiving 500 m3/d of sewage. The lagoon has a surface area of 10 hectares and a depth of 1 m. The pollutant concentration in the raw sewage is 200 mg/L. The organic matter in the sewage degrades biologically (decays) in the lagoon according to first-order kinetics. The reaction rate constant (decay coefficient) is 0.75 d-1. Assuming no other water losses or gains (evaporation, seepage, or rainfall) and that the lagoon is completely mixed, find the steady-state concentration of the pollutant in the effluent.

Answer:

Steady-state concentration of the pollutant in the effluent= 1.3245033 mg/L

Explanation:

Given Data:

Amount of sewage received=500 m^3/d

Surface Area= 10 hectares=10*10^4 m^2

Depth=1 m

Pollutant concentration=200 mg/L

Decay coefficient=0.75 d-1

Required:

Steady-state concentration of the pollutant in the effluent= ?

Solution:

Volume=Surface Area * Depth

[tex]Volume=10*10^4 *1\\Volume=10*10^4\ m^3[/tex]

Time to fill the lagoon=[tex]\frac{Volume}{Amount\ received\ per\ day}[/tex]

[tex]Time\ to\ fill\ the\ lagoon=\frac{10*10^4\ m^3}{500\ m^3}\\ Time\ to\ fill\ the\ lagoon= 200\ days[/tex]

Formula for steady State:

[tex]A_t=\frac{A_0}{1+kt}[/tex]

where:

A_t is the steady state concentration

A_0 is the initial concentration

k is the decay constant

t is the time

[tex]A_t=\frac{200\ mg/L}{1+0.75*200}\\ A_t=1.3245033\ mg/L[/tex]

Steady-state concentration of the pollutant in the effluent= 1.3245033 mg/L

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