---
title: "Seepage in inclined permeable layer"
subject: "Soil Mechanics"
chapter: 6
key: "semester_4/soil_mechanics:inclined-permeable-layer-seepage"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/inclined-permeable-layer-seepage
site: BCT Engineering
---

# Seepage in inclined permeable layer

Soil Mechanics · Chapter 6: Principle of Effective Stress, Capillarity and Permeability

- Asked in: 2074 Bhadra, 4 marks
- Video: [Soil Mechanics | Permeability: Seepage Through an Inclined Permeable Soil Layer](https://www.youtube.com/watch?v=7aOzFKxd6Ps) — Engr. Marc Laurina - Engineering Lectures
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/inclined-permeable-layer-seepage
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-6

## Question

As shown in the figure, an inclined permeable soil layer is underlain by an impervious layer. The coefficient of permeability of the permeable soil layer is equal to $4.8 \times 10^{-5}$ m/sec. If seepage of water in this soil layer occurs in the direction shown in the figure, then calculate (i) hydraulic gradient and (ii) rate of water flow (seepage) for that soil layer. Take the thickness of soil layer H = 3 m and the angle of inclination of that soil layer, $\alpha$ = 5°. Assume any other necessary conditions. [Figure: inclined permeable layer of thickness H = 3 m on an impervious layer sloping at $\alpha$ = 5°; ground water table parallel to the ground surface; seepage direction down the slope.]

## Short answer

Hydraulic gradient i = sin 5° = 0.0872. Flow per metre width q = k i H cosα = 4.8×10⁻⁵ × 0.0872 × 3×cos5° ≈ 1.25×10⁻⁵ m³/s.

## Answer

#### Assumptions

The water table is parallel to the ground surface and the impervious layer beneath, so flow is parallel to the slope. The thickness $H = 3$ m is measured vertically, so the flow area normal to the flow per metre width is $H\cos\alpha$.

```text
        ground / water table (parallel)
    \ ~~~~~~~~~~~~~~~~
     \   flow  ->  H=3 m      
      \  (down the slope)    α = 5°
       \_________________ impervious
```

#### (i) Hydraulic gradient

Over a length $L$ along the slope, the head drop equals the fall in elevation $L\sin\alpha$:

$$
i = \frac{L\sin\alpha}{L} = \sin\alpha = \sin 5^\circ = 0.0872
$$

#### (ii) Rate of flow per metre width

$$
q = k\,i\,A = k\,\sin\alpha\,(H\cos\alpha)\times 1
$$

$$
q = 4.8\times10^{-5} \times 0.0872 \times (3 \times \cos 5^\circ) = 4.8\times10^{-5} \times 0.0872 \times 2.988
$$

$$
q = 1.25\times10^{-5}\ \text{m}^3/\text{s per metre width} \; (\approx 1.08\ \text{m}^3/\text{day per m})
$$

**Answer: $i = 0.0872$; $q = 1.25\times10^{-5}$ m³/s per m width.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For flow parallel to a slope of angle alpha with parallel water table, i equals:
   A. cos(alpha)
   B. tan(alpha)
   C. sin(alpha)
   D. 1
   Answer: C. Head drop over length L along the slope is L sin(alpha), so i = sin(alpha) = 0.0872. (Key point: Hydraulic gradient)

2. With H = 3 m measured vertically, the flow area per metre width normal to flow is:
   A. H sin(alpha)
   B. H cos(alpha)
   C. H
   D. H / cos(alpha)
   Answer: B. Area normal to the flow is H cos(alpha) = 2.988 m2 per m. (Key point: Flow area)

3. The discharge per metre width is:
   A. k sin(alpha) H cos(alpha) = 1.25e-5 m3/s
   B. k tan(alpha) H = 1.26e-5 m3/s
   C. k cos(alpha) H = 1.4e-4 m3/s
   D. k H = 1.44e-4 m3/s
   Answer: A. 4.8e-5 x 0.0872 x 2.988 = 1.25e-5 m3/s per m. (Key point: Discharge)

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