--- title: "Seepage with impervious foundation: derivation" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:seepage-impervious-foundation-tailwater" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/seepage-impervious-foundation-tailwater site: BCT Engineering --- # Seepage with impervious foundation: derivation Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2082 Baishakh, 4 marks - Video: [Seepage though Earthen Dam || Soil Mechanics | Chapter 7 | Most important theory](https://www.youtube.com/watch?v=wmVYDaAoP7s) — Sudip Dhakal - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/seepage-impervious-foundation-tailwater - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question Derive the expression for the specific discharge through a homogeneous earthen dam section with an impervious foundation where the phreatic line emerges at the downstream tailwater. ## Short answer Under Dupuit’s assumptions and Darcy’s law, q dx = –k y dy integrates from headwater h₁ to tailwater h₂ over horizontal distance L to give specific discharge q = k(h₁² – h₂²) / (2L) and the phreatic surface y = √[h₁² – (h₁² – h₂²)x/L]. ## Answer **Dupuit's assumptions**: (i) the hydraulic gradient at any point on the phreatic line equals the slope of the line, $i=\dfrac{dy}{dx}$; (ii) the gradient is constant over the whole vertical section; (iii) flow is horizontal and Darcy's law holds. ```text h1 ___ \ \ phreatic line y \ \____ | \ \____ h2 ........| tail water ______ \ dam \_____ | /////////// impervious base ////|/// |<--------- L ---------------->| ``` #### Derivation Take $x$ along the flow from the u/s end, and $y$ as the height of the phreatic line above the impervious base at distance $x$. The line falls in the flow direction, so $dy/dx$ is negative. Per unit length of dam, the discharge through the vertical section of height $y$ is $$ q = k\, i\, A = -k\,\frac{dy}{dx}\,(y\times 1) $$ Hence $$ q\,dx = -k\,y\,dy $$ Integrate from the u/s end ($x=0$, $y=h_1$) to the exit at tail water ($x=L$, $y=h_2$), where $h_1$ and $h_2$ are the water depths above the base on the u/s and d/s side, and $L$ is the horizontal seepage length: $$ \begin{aligned} q\int_0^L dx &= -k\int_{h_1}^{h_2} y\,dy \\ q\,L &= \frac{k}{2}\left(h_1^2-h_2^2\right) \end{aligned} $$ $$ \boxed{\,q=\frac{k\left(h_1^{2}-h_2^{2}\right)}{2L}\,} $$ The phreatic line is a parabola: integrating only up to the section $x$ gives $q\,x = \frac{k}{2}(h_1^2-y^2)$, so $$ y=\sqrt{h_1^{2}-\frac{(h_1^{2}-h_2^{2})}{L}\,x} $$ Here $L$ is measured from the corrected entry point (0.3 times the horizontal projection of the wetted u/s slope inside the water line) to the point where the line meets the tail water. If the line emerges above tail water on the d/s face, the exit zone is treated by Schaffernak's or Casagrande's correction. For $h_2=0$ this becomes $q=\dfrac{k h_1^2}{2L}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which is a Dupuit assumption? A. Gradient varies with depth B. The hydraulic gradient equals the slope of the phreatic line and is constant over the vertical section C. Flow is vertical D. Darcy's law does not apply Answer: B. i = dy/dx, constant over the depth, with horizontal flow. (Key point: Dupuit assumptions) 2. The discharge per unit length through a section of height y is A. q = k y B. q = k y^2 C. q = -k (dy/dx) y D. q = k dy/dx Answer: C. q = k i A with i = -dy/dx and A = y x 1, giving q dx = -k y dy. (Key point: Differential equation) 3. Integrating between (x=0, y=h1) and (x=L, y=h2) gives A. q = k(h1^2 - h2^2)/(2L) B. q = k(h1 - h2)/L C. q = k(h1^2 + h2^2)/(2L) D. q = k(h1^2 - h2^2)/L Answer: A. q L = (k/2)(h1^2 - h2^2). (Key point: Result) 4. The phreatic line obtained is A. A straight line B. A circle C. A hyperbola D. A parabola Answer: D. y = sqrt(h1^2 - (h1^2-h2^2)x/L) is a parabola. 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