Constant head permeameter, seepage velocity

Given

$Q = 500$ ml in $t = 15$ min $= 900$ s, $D = 5$ cm, head loss $h = 40$ cm between two points $L = 15$ cm apart. Dry weight of sample $W_d = 4.86$ N, $G = 2.65$, $\gamma_w = 9.81$ kN/m³.

$$A = \frac{\pi}{4}(5)^2 = 19.63\ \text{cm}^2, \qquad q = \frac{500}{900} = 0.5556\ \text{cm}^3/\text{s}, \qquad i = \frac{h}{L} = \frac{40}{15} = 2.667$$

Coefficient of permeability

$$k = \frac{q}{A\,i} = \frac{0.5556}{19.63 \times 2.667} = 0.0106\ \text{cm/s} = 1.06\times10^{-4}\ \text{m/s}$$

Seepage velocity

Discharge velocity:

$$v = \frac{q}{A} = \frac{0.5556}{19.63} = 0.0283\ \text{cm/s}$$

Volume of sample $= 19.63 \times 15 = 294.5\ \text{cm}^3 = 2.945\times10^{-4}\ \text{m}^3$.

$$\gamma_d = \frac{4.86\times10^{-3}\ \text{kN}}{2.945\times10^{-4}\ \text{m}^3} = 16.50\ \text{kN/m}^3$$ $$e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.65 \times 9.81}{16.50} - 1 = 0.575, \qquad n = \frac{e}{1+e} = 0.365$$ $$v_s = \frac{v}{n} = \frac{0.0283}{0.365} = 0.0775\ \text{cm/s}$$

Answer: $k = 0.0106$ cm/s; $v_s = 0.0775$ cm/s ($n = 0.365$).