Question

For a liquid, the permeability of a sandy soil having a void ratio of 0.60 was determined as 0.14 cm/s. Considering the same liquid and by using Taylor's equation, the permeability (in cm/s) of this soil corresponding to the void ratio of 0.80 is (rounded off to two decimal places).

Hint:

Use Taylor's relation k is proportional to e^3/(1+e) to compare permeability at two void ratios.

Answer 1

Correct Answer: 0.29

Taylor's permeability-void ratio relation comes from treating the soil as a bundle of capillary channels: the flow through the pores depends on how much void space is available and how well the voids connect, which the function $e^3/(1+e)$ captures. This route finds the proportionality constant $C$ from the first test, then uses it directly to predict the second permeability, instead of working with a ratio.

The relation is

\[ k = C \cdot \frac{e^3}{1+e} \]

where $C$ depends only on the grain shape, size distribution and the liquid used, all of which stay fixed between the two tests.

From the first test, $e_1 = 0.60$ and $k_1 = 0.14$ cm/s. Compute the shape factor:

\[ \frac{e_1^3}{1+e_1} = \frac{0.216}{1.6} = 0.135 \]

so

\[ C = \frac{k_1}{0.135} = \frac{0.14}{0.135} = 1.037 \]

Now use this same $C$ at $e_2 = 0.80$:

\[ \frac{e_2^3}{1+e_2} = \frac{0.512}{1.8} = 0.2844 \] \[ k_2 = C \times 0.2844 = 1.037 \times 0.2844 = 0.295 \ \text{cm/s} \]

This matches the direct ratio method, since $C$ is just $k_1$ divided by its own shape factor. The result shows permeability grows fast with void ratio (a cube-power relation), so going from $e = 0.60$ to $e = 0.80$, only a 33% rise in void ratio, more than doubles the permeability.

The permeability at void ratio 0.80 works out to about 0.29 cm/s.

\[ \boxed{k_2 \approx 0.29 \ \text{cm/s}} \]