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Impervious dam seepage discharge

Short answer

Head difference H = 7 m. Seepage q = k H (Nf/Nd) = 3×10⁻⁶ m/s × 7 m × (9/12) = 1.575×10⁻⁵ m³/s per metre length (≈1.36 m³/day per m).

If the upstream and downstream heads of an impervious dam are 8 m and 1 m respectively, then find the seepage discharge when seepage of water takes place from upstream to downstream via the isotropic soil lying below the impervious dam. The total number of flow channels and equipotential drops are 9 and 12, respectively. Also, take the coefficient of permeability of the soil layer, k = 3×10−43 \times 10^{-4} cm/s.

Given

Upstream head 8 m, downstream head 1 m, so net head H=8−1=7H = 8 - 1 = 7 m. Nf=9N_f = 9, Nd=12N_d = 12. k=3×10−4k = 3\times10^{-4} cm/s =3×10−6= 3\times10^{-6} m/s.

Calculation

q=k H NfNd=3×10−6×7×912q = k\,H\,\frac{N_f}{N_d} = 3\times10^{-6} \times 7 \times \frac{9}{12} q=1.575×10−5 m3/s per metre length of damq = 1.575\times10^{-5}\ \text{m}^3/\text{s per metre length of dam} q=1.575×10−5×86400=1.36 m3/day per metreq = 1.575\times10^{-5} \times 86400 = 1.36\ \text{m}^3/\text{day per metre}

Answer: q=1.58×10−5q = 1.58\times10^{-5} m³/s per m (≈1.36\approx 1.36 m³/day per m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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