Question
What is the flow rate in a 2.0-m long sample of clean saturated sand with a cross-sectional area of 0.05 m2 and hydraulic conductivity of 10^-3 m/s?
Answers
GPT-4o mini
To calculate the flow rate through a saturated sandy soil sample, we can use Darcy's law, which describes the flow of fluid through porous media. The equation is given by:
\[
Q = k \cdot A \cdot \frac{\Delta h}{L}
\]
where:
- \( Q \) = flow rate (m³/s),
- \( k \) = hydraulic conductivity (m/s),
- \( A \) = cross-sectional area (m²),
- \( \Delta h \) = hydraulic gradient (change in head, dimensionless),
- \( L \) = length of the sample (m).
In this case, we need to define the hydraulic gradient, which is typically the change in hydraulic head (\( h \)) over the length of the sample (\( L \)):
\[
\frac{\Delta h}{L} = \frac{h_1 - h_2}{L}
\]
Assuming the hydraulic head difference (\( \Delta h \)) across the length of the sample (\( L \)) is 1 meter for the calculation, we proceed with:
- \( L = 2.0 \, \text{m} \)
- \( A = 0.05 \, \text{m}^2 \)
- \( k = 10^{-3} \, \text{m/s} \)
- Assuming \( \Delta h = 1 \, \text{m} \)
Now, substituting these values into the equation:
1. Calculate the hydraulic gradient:
\[
\frac{\Delta h}{L} = \frac{1 \, \text{m}}{2.0 \, \text{m}} = 0.5
\]
2. Substitute into Darcy’s law:
\[
Q = k \cdot A \cdot \frac{\Delta h}{L} = (10^{-3} \, \text{m/s}) \cdot (0.05 \, \text{m}^2) \cdot 0.5
\]
3. Perform the calculation:
\[
Q = 10^{-3} \cdot 0.05 \cdot 0.5 = 10^{-3} \cdot 0.025 = 2.5 \times 10^{-5} \, \text{m}^3/\text{s}
\]
Thus, the flow rate through the saturated sand sample is
\[
\boxed{2.5 \times 10^{-5} \, \text{m}^3/\text{s}}
\]
\[
Q = k \cdot A \cdot \frac{\Delta h}{L}
\]
where:
- \( Q \) = flow rate (m³/s),
- \( k \) = hydraulic conductivity (m/s),
- \( A \) = cross-sectional area (m²),
- \( \Delta h \) = hydraulic gradient (change in head, dimensionless),
- \( L \) = length of the sample (m).
In this case, we need to define the hydraulic gradient, which is typically the change in hydraulic head (\( h \)) over the length of the sample (\( L \)):
\[
\frac{\Delta h}{L} = \frac{h_1 - h_2}{L}
\]
Assuming the hydraulic head difference (\( \Delta h \)) across the length of the sample (\( L \)) is 1 meter for the calculation, we proceed with:
- \( L = 2.0 \, \text{m} \)
- \( A = 0.05 \, \text{m}^2 \)
- \( k = 10^{-3} \, \text{m/s} \)
- Assuming \( \Delta h = 1 \, \text{m} \)
Now, substituting these values into the equation:
1. Calculate the hydraulic gradient:
\[
\frac{\Delta h}{L} = \frac{1 \, \text{m}}{2.0 \, \text{m}} = 0.5
\]
2. Substitute into Darcy’s law:
\[
Q = k \cdot A \cdot \frac{\Delta h}{L} = (10^{-3} \, \text{m/s}) \cdot (0.05 \, \text{m}^2) \cdot 0.5
\]
3. Perform the calculation:
\[
Q = 10^{-3} \cdot 0.05 \cdot 0.5 = 10^{-3} \cdot 0.025 = 2.5 \times 10^{-5} \, \text{m}^3/\text{s}
\]
Thus, the flow rate through the saturated sand sample is
\[
\boxed{2.5 \times 10^{-5} \, \text{m}^3/\text{s}}
\]