Stratified Aquifer: Effective Hydraulic Conductivity Calculation

This problem involves calculating the effective hydraulic conductivity ($K_{eff}$) for a stratified confined aquifer where flow is parallel to the layers. Darcy's Law governs the flow.

Problem Setup

The aquifer consists of three horizontal layers with the following properties:

• Layer 1: Thickness ($b_1$) = 10 m, Hydraulic Conductivity ($K_1$) = 15 m/day
• Layer 2: Thickness ($b_2$) = 5 m, Hydraulic Conductivity ($K_2$) = 20 m/day
• Layer 3: Thickness ($b_3$) = 5 m, Hydraulic Conductivity ($K_3$) = 30 m/day

The total thickness of the aquifer is $B = \sum_{j=1}^{3} b_j$. The flow is parallel to the strata, and the hydraulic gradient ($i$) is constant.

Calculation of Effective Hydraulic Conductivity

For flow parallel to the strata in a confined aquifer, the effective hydraulic conductivity ($K_{eff}$) is calculated as the thickness-weighted average of the individual layer conductivities:

$K_{eff} = \frac{\sum_{j=1}^{n} K_j b_j}{\sum_{j=1}^{n} b_j}$

Where $n$ is the number of layers.

Step 1: Calculate the total thickness ($B$).

$B = b_1 + b_2 + b_3$

$B = 10 \text{ m} + 5 \text{ m} + 5 \text{ m} = 20 \text{ m}$

Step 2: Calculate the sum of the products of conductivity and thickness for each layer ($\sum K_j b_j$).

$ \sum_{j=1}^{3} K_j b_j = (K_1 \times b_1) + (K_2 \times b_2) + (K_3 \times b_3) $

$ \sum K_j b_j = (15 \text{ m/day} \times 10 \text{ m}) + (20 \text{ m/day} \times 5 \text{ m}) + (30 \text{ m/day} \times 5 \text{ m}) $

$ \sum K_j b_j = 150 \text{ m}^2\text{/day} + 100 \text{ m}^2\text{/day} + 150 \text{ m}^2\text{/day} $

$ \sum K_j b_j = 400 \text{ m}^2\text{/day} $

Step 3: Calculate the effective hydraulic conductivity ($K_{eff}$).

$K_{eff} = \frac{\sum K_j b_j}{B}$

$K_{eff} = \frac{400 \text{ m}^2\text{/day}}{20 \text{ m}}$

$K_{eff} = 20 \text{ m/day}$

Conclusion

The effective hydraulic conductivity of the stratified aquifer, under the condition of parallel flow, is 20 m/day.