Question

A cylindrical soil sample of 10 cm diameter is tested in a constant-head permeameter. A volume of 250 cm$^3$ of water is collected in 5 minutes when the constant-head difference between tapping points 15 cm apart is 5 cm. Considering Darcy flow, the absolute value of coefficient of permeability in cm/s is .............. ($\pi = 3.14$) [round off to 3 decimal places]

Hint:

For permeability tests, always remember to include sample length $L$, head difference $h$, and discharge per unit time. Use $k = QL/(Aht)$.

Answer 1

Step 1: Formula for coefficient of permeability (Darcy's law).
\[ k = \frac{QL}{A \, h \, t} \] where $Q$ = discharge volume, $L$ = length between tapping points, $A$ = cross-sectional area, $h$ = head difference, $t$ = time.

Step 2: Given values.
- Diameter = 10 cm $\Rightarrow$ radius = 5 cm.
- Cross-sectional area: \[ A = \pi r^2 = 3.14 \times 5^2 = 78.5 \, \text{cm}^2 \] - $Q$ = 250 cm$^3$
- $t$ = 5 minutes = 300 s
- $L$ = 15 cm
- $h$ = 5 cm

Step 3: Calculate discharge rate.
\[ Q/t = \frac{250}{300} = 0.833 \, \text{cm}^3/\text{s} \]

Step 4: Substitute into Darcy's law.
\[ k = \frac{(0.833)(15)}{78.5 \times 5} = \frac{12.495}{392.5} \approx 0.0318 \] Oops — recheck with correct form: \[ k = \frac{Q \cdot L}{A \cdot h \cdot t} \] \[ k = \frac{250 \times 15}{78.5 \times 5 \times 300} \] \[ k = \frac{3750}{117750} = 0.0318 \, \text{cm/s} \] On correcting rounding and significant figures: $\approx 0.032$ cm/s.

Final Answer: \[ \boxed{0.032 \, \text{cm/s}} \]