We are dealing with a pollutant discharge into a completely mixed lagoon, and we are tasked with determining:

• (a) The steady-state concentration of the non-reactive pollutant in the lagoon effluent.
• (b) The concentration of the pollutant after a sudden increase in influent concentration after 10 days.

Given:

• Initial influent concentration, $C_0 = 15 \, \text{g/m}^3$
• Wastewater flow rate, $Q = 0.1 \, \text{ML/day} = 100 \, \text{m}^3/\text{day}$
• Lagoon detention time, $T_d = 10 \, \text{days}$
• Lagoon is completely mixed.

Assumptions:

1. The pollutant is non-reactive, so it does not decay or react in the lagoon.
2. At steady state, the concentration in the effluent equals the concentration in the lagoon due to complete mixing.
3. After the sudden increase in influent concentration, the system behaves according to the first-order differential equation for the completely mixed lagoon.

Solution:

Part (a) Steady-State Concentration

At steady state, the concentration of the pollutant in the lagoon effluent, $C_e$, is the same as the influent concentration $C_0$. This is because the lagoon is completely mixed and no reaction or decay occurs.

Thus, the steady-state concentration of the non-reactive pollutant in the effluent is: $C_e = C_0 = 15 \, \text{g/m}^3$

Part (b) Concentration 10 Days After the Influent Increase

Now, we need to calculate the concentration in the lagoon after 10 days, following a sudden increase in influent concentration from $C_0 = 15 \, \text{g/m}^3$ to $C_1 = 150 \, \text{g/m}^3$.

For a completely mixed lagoon, the concentration $C(t)$ at time $t$ after the change is governed by the first-order mixing equation: $C(t) = C_1 + (C_0 - C_1) e^{-t/T_d}$ Where:

• $C(t)$ is the concentration in the lagoon at time $t$,
• $C_0 = 15 \, \text{g/m}^3$ is the initial concentration,
• $C_1 = 150 \, \text{g/m}^3$ is the new influent concentration,
• $T_d = 10 \, \text{days}$ is the detention time,
• $t = 10 \, \text{days}$ is the time after the concentration change.

Substituting the known values: $C(10) = 150 + (15 - 150) e^{-10/10}$ $C(10) = 150 + (-135) e^{-1}$ Using $e^{-1} \approx 0.3679$: $C(10) = 150 - 135 \times 0.3679$ $C(10) \approx 150 - 49.67$ $C(10) \approx 100.33 \, \text{g/m}^3$

Thus, 10 days after the sudden increase in influent concentration, the concentration of the non-reactive pollutant in the effluent will be approximately $100.33 \, \text{g/m}^3$.

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Let me know if you would like further details or clarifications!

Related Questions:

1. How is the detention time of a lagoon calculated?
2. What factors affect the steady-state concentration in completely mixed lagoons?
3. How would the results change if the pollutant was reactive or decayed over time?
4. Can we use the same equations for partially mixed lagoons?
5. What would happen to the effluent concentration if the flow rate doubled?

Tip:

For systems like lagoons, the concept of "complete mixing" simplifies calculations because it assumes uniform concentration throughout, but real systems might exhibit partial mixing, which would require more complex models.