A sand sample of \(35 \text{cm}^2\) cross-sectional area and 20 cm long was tested in a constant head permeameter. Under a head of 60 cm , the discharge was 120 ml in 6 min . The dry weight of sand used for the test was 1120 g , and \(\text{G}=2.68.\) Determine the seepage velocity.

Concept:

Seepage velocity is related to Darcy’s velocity (discharge velocity, v) and porosity (n):

\(v_s=\frac{v}{n}\)

Calculation:

Given Data:

• Cross-sectional area (A) = 35 cm² = 35 × 10⁻⁴ m²

• Length of specimen (L) = 20 cm = 0.2 m

• Head (h) = 60 cm = 0.6 m

• Discharge (Q) = 120 ml in 6 min
  → Q = 120 ml = 120 × 10⁻⁶ m³ in 6 min
  → Time = 6 min = 360 sec
  → Q/sec = (120 × 10⁻⁶) / 360 = 3.33 × 10⁻⁷ m³/sec

• Dry weight of sand (Wd) = 1120 g = 1.12 kg

• G (specific gravity) = 2.68

Discharge Velocity: \(v=\frac{Q}{A}=\frac{3.33\times10^{-7}}{35\times10^{-4}}=9.51\times10^{-4}m/s\)

Volume of specimen = A × L = \(35\times10^{-4}\times0.2=7\times10^{-4}m^3\)

Volume of voids: \(\frac{Mass}{G\times\gamma_w}=\frac{1.12}{2.68\times1000}=4.179\times10^{-4}m^3\)

Void volume = Total volume − Volume of solids = 2.821 x 10^-4 m³

Porosity: \(n=\frac{V_v}{V}=\frac{2.821\times10^{-4}}{7\times10^{-4}}=0.403\)

Seepage Velocity: \(v_s=\frac{v}{n}=\frac{9.51\times10^{-4}}{0.403}=2.36\times10^{-3}m/s\)