If the void ratio and discharge velocity for soil is 0.5 and 6 × 10-7 m/s respectively, what is the value of seepage velocity (m/s)?
18 × 10-7
This question asks us to determine the seepage velocity of soil given its void ratio and discharge velocity. To do this, we need to understand the relationship between these parameters and the porosity of the soil.
The discharge velocity ($\text{v}$) and the seepage velocity ($\text{v}_\text{s}$) are related through the porosity ($\text{n}$) by the following equation:
\(\text{v} = \text{n} \times \text{v}_\text{s}\)
Therefore, the seepage velocity can be calculated as:
\(\text{v}_\text{s} = \frac{\text{v}}{\text{n}}\)
Porosity ($\text{n}$) is related to void ratio ($\text{e}$) by the formula:
\(\text{n} = \frac{\text{e}}{1+\text{e}}\)
We are given the following values:
First, we calculate the porosity (\(\text{n}\)) using the given void ratio:
\(\text{n} = \frac{\text{e}}{1+\text{e}} = \frac{0.5}{1+0.5} = \frac{0.5}{1.5}\)
To simplify the fraction:
\(\text{n} = \frac{0.5 \times 2}{1.5 \times 2} = \frac{1}{3}\)
So, the porosity is \(\frac{1}{3}\).
Next, we calculate the seepage velocity (\(\text{v}_\text{s}\)) using the discharge velocity and the calculated porosity:
\(\text{v}_\text{s} = \frac{\text{v}}{\text{n}}\)
Substitute the given values:
\(\text{v}_\text{s} = \frac{6 \times 10^{-7} \text{ m/s}}{\frac{1}{3}}\)
Dividing by a fraction is the same as multiplying by its reciprocal:
\(\text{v}_\text{s} = (6 \times 10^{-7}) \times 3 \text{ m/s}\)
\(\text{v}_\text{s} = 18 \times 10^{-7} \text{ m/s}\)
The calculated seepage velocity is \(18 \times 10^{-7}\) m/s.
| Parameter | Value | Unit |
|---|---|---|
| Void Ratio (\(\text{e}\)) | 0.5 | Dimensionless |
| Discharge Velocity (\(\text{v}\)) | \(6 \times 10^{-7}\) | m/s |
| Calculated Porosity (\(\text{n}\)) | \(1/3\) | Dimensionless |
| Calculated Seepage Velocity (\(\text{v}_\text{s}\)) | \(18 \times 10^{-7}\) | m/s |
Based on the given void ratio of 0.5 and a discharge velocity of \(6 \times 10^{-7}\) m/s, the calculated seepage velocity is \(18 \times 10^{-7}\) m/s.
| Property/Velocity | Symbol | Definition | Common Relationship |
|---|---|---|---|
| Void Ratio | \(\text{e}\) | Volume of voids / Volume of solids | \(\text{n} = \text{e} / (1+\text{e})\) |
| Porosity | \(\text{n}\) | Volume of voids / Total volume | \(\text{e} = \text{n} / (1-\text{n})\) |
| Discharge Velocity | \(\text{v}\) | Flow rate / Total area | \(\text{v} = \text{n} \times \text{v}_\text{s}\) |
| Seepage Velocity | \(\text{v}_\text{s}\) | Flow rate / Area of voids | \(\text{v}_\text{s} = \text{v} / \text{n}\) |
The flow of water through soil is often described by Darcy's Law, especially for laminar flow conditions. Darcy's Law relates the discharge velocity to the hydraulic gradient and the permeability of the soil.
\(\text{v} = \text{k} \times \text{i}\)
Where:
While Darcy's Law gives the discharge velocity, the seepage velocity represents the actual speed of water particles moving through the pore channels. The tortuosity of the flow path (the winding nature of the pores) is one reason why the seepage velocity is greater than the discharge velocity.
Understanding the difference between discharge velocity and seepage velocity is crucial in geotechnical engineering and hydrogeology for analyzing groundwater flow, pollutant transport, and consolidation settlement.
Which one of the following equations correctly gives the relationship between the specific gravity of soil grains (G) and the hydraulic gradient (i) to initiate 'quick' condition in sand having a void ratio of 0.5?
In the graphical method of obtaining flow nets, if the lowest flow line confirms to the bottom boundary conditions, the flow net:
The hydraulic gradient between two adjacent equipotential lines is given by:
A soil has a discharge velocity of 6 × 10 -7 m/s and a void ratio of 0.5. What is its seepage velocity?