Question

The consolidation process of a clay layer of 4 meter thickness is solved by a numerical procedure. The consolidation coefficient is \(c_{v}=10^{-6} \mathrm{~m}^{2} / \mathrm{s}\). The layer is subdivided into 20 small layers. What is the maximum allowable magnitude of the time step?

Short answer

To determine the maximum allowable time step for the numerical procedure of solving the consolidation process of a clay layer that is 4 meters thick, divided into 20 smaller layers, and with a consolidation coefficient of \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}\), we first calculated the smallest layer thickness, \(\Delta h\), as \(0.2 \; \mathrm{m}\). Then, using the von Neumann stability analysis, we found that the maximum allowable time step, \(\Delta t\), should satisfy the stability criterion \(\frac{c_{v} \Delta t}{\Delta h^2} \leq 0.5\). Plugging in the values, we determined that the maximum allowable time step is 20 seconds.

Step-by-step solution

Calculate the smallest layer thickness

First, we need to find the thickness of each smaller layer by dividing the total thickness by the number of layers. The total thickness is given as 4 meters, and there are 20 smaller layers; hence, we can find the smallest layer thickness, \(\Delta h\). $$ \Delta h = \frac{\text{Total Thickness}}{\text{Number of Layers}} = \frac{4 \text{m}}{20} = 0.2 \text{m} $$

Determine the time-step limit for numerical stability

To ensure numerical stability in the consolidation process, the governing equation of consolidation is discretized using the finite-difference method. The von Neumann stability analysis helps us find the stability criterion for a time-step, \(\Delta t\). For the consolidation process, the stability criterion for the time-step can be expressed as: $$ \frac{c_{v} \Delta t}{\Delta h^2} \leq 0.5 $$ In this scenario, the consolidation coefficient, \(c_{v}\), is \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}\), and the layer thickness, \(\Delta h\), is \(0.2 \text{m}\).

Calculate the maximum allowable time step

Using the stability criterion, we can calculate the maximum allowable time step, \(\Delta t\). $$ \Delta t \leq \frac{0.5 \Delta h^2}{c_{v}} $$ Plugging in the values, we get: $$ \Delta t \leq \frac{0.5 (0.2 \text{m})^2}{10^{-6} \mathrm{~m}^{2} / \mathrm{s}} \Rightarrow \Delta t \leq 20 \mathrm{s} $$ Hence, the maximum allowable magnitude of the time step is 20 seconds.