Answer :
Answer:
Explanation:
To estimate the concentration of 1,4-dioxane in the aquifer at a given distance from the canal after a certain period, we need to use the principles of solute transport in groundwater, specifically considering the effects of advection and dispersion. Since 1,4-dioxane is a non-degradable compound, we will assume it does not undergo any chemical reactions or degradation.
### Given Data:
- **Concentration of 1,4-dioxane in canal, \( C_0 \)**: 1.245 mg/L
- **Average linear pore velocity, \( v \)**: 0.544 m/d
- **Distance from canal, \( x \)**: 25 m
- **Time, \( t \)**: 45 d
### Assumptions:
- **Steady-state conditions**: Assuming that the system has reached a steady state where the concentration profile is stable.
- **One-dimensional transport**: The flow is assumed to be one-dimensional perpendicular to the canal.
- **Non-degradable and non-reactive compound**: 1,4-dioxane does not degrade or react in the aquifer.
### Solution:
To determine the concentration at a distance of 25 meters from the canal, we use the **advection-dispersion equation**. Since we're dealing with non-degradable and steady-state conditions, the simplified equation used here is:
\[ C(x, t) = C_0 \cdot \text{exp}\left(-\frac{v \cdot x}{D \cdot t}\right) \]
where:
- \( C(x, t) \) = Concentration of the solute at distance \( x \) and time \( t \)
- \( C_0 \) = Initial concentration of the solute
- \( v \) = Average linear pore velocity (m/d)
- \( D \) = Dispersion coefficient (m²/d)
- \( x \) = Distance from the canal (m)
- \( t \) = Time (d)
However, to use this equation, we need the dispersion coefficient, \( D \). Since the dispersion coefficient is not provided, and assuming that it is either not available or negligible, we can simplify the problem to consider the advection component.
### For a more exact calculation, you'd typically need the dispersion coefficient. Without it, let’s focus on a simplified conceptual approach where dispersion is minimal or not considered:
1. **Advection alone**:
If dispersion is negligible, the concentration at distance \( x \) and time \( t \) can be estimated as:
\[ C(x, t) = C_0 \]
2. **Including dispersion** (if we had \( D \)):
You would calculate the dispersion effect, which would typically lower the concentration compared to the initial concentration. This would require knowing the value of \( D \).
**Given the information provided and assuming negligible dispersion:**
At a distance of 25 meters from the canal after 45 days, and if dispersion is not considered, the concentration of 1,4-dioxane would be approximately:
\[ C(x, t) \approx C_0 = 1.245 \text{ mg/L} \]
In summary, without the dispersion coefficient, the best estimate of the concentration of 1,4-dioxane 25 meters from the canal at 45 days is approximately **1.245 mg/L**.
