Seepage through anisotropic soil — equivalent isotropic permeability (2D flow) A soil mass has principal permeabilities kx = 9 × 10^-7 cm/s (horizontal) and ky = 4 × 10^-7 cm/s (vertical). Determine the transformed coefficient of permeability k_eq for an equivalent isotropic soil in two-dimensional flow analysis.

Introduction / Context:
Natural soils often exhibit anisotropic permeability because of stratification and fabric. For plane seepage problems, it is convenient to transform the anisotropic medium into an equivalent isotropic one so that Laplace's equation applies in a simple rectangular coordinate system. The equivalent permeability summarizes the directional conductivities into a single representative value for the transformed plane.

Given Data / Assumptions:

• Principal permeabilities: kx = 9 × 10^-7 cm/s, ky = 4 × 10^-7 cm/s.
• Two-dimensional flow in the x–y plane with axes aligned to principal directions.
• Steady-state Darcy flow; homogeneous within each principal direction.

Concept / Approach:
For two-dimensional anisotropic flow aligned with principal directions, a standard transformation (coordinate stretching) maps the domain to one in which the governing equation is isotropic. The equivalent isotropic permeability for this mapping is the geometric mean of the principal permeabilities:
k_eq = sqrt(kx * ky). This preserves flow net orthogonality and flux-potential relationships in the transformed plane.

Step-by-Step Solution:
Compute product: kx * ky = (9 × 10^-7) * (4 × 10^-7) = 36 × 10^-14.Take square root: sqrt(36 × 10^-14) = 6 × 10^-7.Therefore, k_eq = 6 × 10^-7 cm/s.

Verification / Alternative check:
Sanity check: k_eq falls between kx and ky (4 × 10^-7 and 9 × 10^-7), as expected for a mean; the geometric mean is appropriate for multiplicative properties like permeability in orthogonal directions.

Why Other Options Are Wrong:
9 × 10^-7 and 4 × 10^-7: These are the individual directional values, not the equivalent isotropic one.
13 × 10^-7: Arithmetic sum; physically meaningless for k_eq.
3 × 10^-7: Too small; not consistent with geometric averaging.

Common Pitfalls:

• Using arithmetic or harmonic means instead of the geometric mean for 2D anisotropy transformation.
• Failing to align axes with principal directions before applying the transformation.

Final Answer:
6 × 10^-7 cm/s