Sewage is being treated in two aerated lagoons in series. Each lagoon has a surface area of 5 hectares (ha) and depth of 1 m, and each operates as a CSTR at steadystate. The influent to the first lagoon contains 100 mg/L of biodegradable organic material in a flow of 8640 m3/d. At steady-state, the effluent from the second lagoon must contain no more than 20 mg/L of the organics. What first-order biodegradation rate constant must be reached to achieve the treatment goal?

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19-02-2020
Engineering

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Sewage is being treated in two aerated lagoons in series. Each lagoon has a surface area of 5 hectares (ha) and depth of 1 m, and each operates as a CSTR at steadystate. The influent to the first lagoon contains 100 mg/L of biodegradable organic material in a flow of 8640 m3/d. At steady-state, the effluent from the second lagoon must contain no more than 20 mg/L of the organics. What first-order biodegradation rate constant must be reached to achieve the treatment goal?

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AyBaba7 AyBaba7 · Answer 1 · 2020-03-03T09:16:50+01:00

Answer:

The first-order biodegradation rate constant that must be reached to achieve the treatment goal is

k = 0.003456 /day

Explanation:

Each of the lagoon can be treated as a continuously stirred tank reactor.

Let C₀ represent the initial concentration of the biodegradable organic material = 100 g/L = 100 g/m³

Let C₁ be the concentration of biodegradable organic material that leaves the first lagoon and entering the second lagoon.

Let C₂ be the concentration of biodegradable organic material leaving the second lagoon = 20 mg/L = 20 g/m³

Let the rate constant be k

Let V be the volume of each of the lagoons = 5 hectares × 1 m = 50000 × 1 = 50000 m³

Let F₀ be the flowrate of influent into the first lagoon = 8640 m³/day

Let F₁ be the flowrate of outlet from lagoon 1 and influent into lagoon 2.

Let F₂ be the flowrate of outlet from lagoon 2.

Since the volumes of the lagoon (reactors) are constant, F₀ = F₁ = F₂ = 8640 m³/day

The performance equation for a CSTR with a first order reaction going on for lagoon 1 is

(kC₀V/F₀) = (C₀ - C₁)/C₁

Make C₁ the subject of formula

C₁ = (F₀C₀)/(kC₀V + F₀) = (8640)(100)/[(k×100×50000) + 8640]

C₁ = 864000/(5000000k + 8640)

For the 2nd lagoon, performance equation is

(kC₁V/F₁) = (C₁ - C₂)/C₂

Make C₁ the subject of formula once more

C₁ = - F₁ C₂/(kC₂V - F₁)

F₁ = 8640 m³/day, V = 50000 m³, C₂ = 20 mg/L = 20 g/m³

C₁ = (-8640 × 20)/(k×20×50000 - 8640)

C₁ = - 172800/(1000000k - 8640)

Equating the C1 in both equations to each others

864000/(5000000k + 8640) = -172800/(1000000k - 8640)

864000 (1000000k - 8640) = - 172800(5000000k + 8640)

864 (1000k - 8.64) = - 172.8 (5000k + 8.64)

864000k - 7464.96 = - 864000k - 1492.992

864000k + 864000k = 7464.96 - 1492.992

1728000k = 5971.968

k = 0.003456 /day.