--- title: "Phreatic line and discharge (k = 5.1×10⁻⁴ cm/s)" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:earthen-dam-phreatic-line-2082" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/earthen-dam-phreatic-line-2082 site: BCT Engineering --- # Phreatic line and discharge (k = 5.1×10⁻⁴ cm/s) Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2082 Bhadra, 8 marks - Video: [Problem on Design of Phreatic line in earthen dam without filter](https://www.youtube.com/watch?v=gPWo52zz3OI) — MSR Geo Spatial Solutions - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/earthen-dam-phreatic-line-2082 - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question An earthen dam made of homogeneous material has the following data: coefficient of permeability of dam material = $5.1 \times 10^{-4}$ cm/sec; top level of dam = 290 masl; bed level of river = 235 masl; reservoir water level = 285 masl; top width of dam = 5 m; U/S slope of dam = 3:1; D/S slope of dam = 2:1. Determine the phreatic line for the dam section and the discharge passing through the dam. ## Short answer Using Casagrande’s method: base width 280 m, wetted projection b = 150 m, d = 175 m, y₀ = √(50²+175²) – 175 = 7.00 m. The phreatic parabola is y = √(49+14x) (m) with exit about 35 m up the face. Seepage q = k y₀ = 5.1×10⁻⁶ × 7.00 = 3.57×10⁻⁵ m³/s per metre. ## Answer **Given data and assumptions**: homogeneous dam without any drain; foundation impervious; tail water ignored. Casagrande's method (basic parabola with focus at the d/s toe) is used. - Dam height $= 290-235 = 55$ m; water head $h = 285-235 = 50$ m; free board $= 5$ m. - $k = 5.1\times10^{-4}$ cm/s $= 5.1\times10^{-6}$ m/s. ```text 290 ______5 m______ /| \ 3:1/ | homogeneous \ 2:1 / | dam \ 285 ~~| \ / | \ 235 /////////////////////////// |<--------- 280 m ------->| ``` #### Step 1: Geometry $$ \begin{aligned} \text{Base width } B &= 3(55)+5+2(55) = 280\ \text{m} \\ \text{Wetted horizontal projection } b &= 3\times 50 = 150\ \text{m} \\ \text{Entry correction } 0.3b &= 45.0\ \text{m} \\ d &= B-0.7b = 280-105.0 = 175.0\ \text{m} \end{aligned} $$ Here $d$ is the horizontal distance from the corrected entry point to the d/s toe (the focus $F$). #### Step 2: Basic parabola $$ y_0=\sqrt{h^2+d^2}-d=\sqrt{50^2+175.0^2}-175.0=7.00\ \text{m} $$ $$ y=\sqrt{y_0^2+2y_0x} $$ with $x$ measured from the focus (d/s toe) towards the u/s side: | x from focus (m) | y (m) | |---|---| | 0.0 | 7.00 | | 5.0 | 10.91 | | 10.0 | 13.75 | | 20.0 | 18.14 | | 40.0 | 24.68 | | 60.0 | 29.82 | | 80.0 | 34.20 | | 100.0 | 38.07 | | 125.0 | 42.42 | | 150.0 | 46.37 | | 175.0 | 50.00 | At $x=d=175.0$ m the parabola reaches $y=h=50$ m, which is the corrected entry point. The u/s end of the actual line is drawn to meet the u/s face at right angles. #### Step 3: Exit correction (d/s face) The d/s slope angle is $\alpha=\tan^{-1}(1/2)=26.57^\circ$. The parabola meets the d/s face at a distance $$ a+\Delta a=\frac{y_0}{1-\cos\alpha}=\frac{7.00}{1-\cos 26.57^\circ}=66.3\ \text{m} $$ (measured along the slope from the toe). Casagrande's chart covers $30^\circ\le\alpha\le180^\circ$, so for $\alpha<30^\circ$ the exit point is found from Schaffernak's formula: $$ a=\frac{d}{\cos\alpha}-\sqrt{\frac{d^2}{\cos^2\alpha}-\frac{h^2}{\sin^2\alpha}}=195.66-\sqrt{38,281-12,500}=35.1\ \text{m} $$ So the line leaves the d/s face about 35.1 m along the slope from the toe (about 15.7 m above the base). (Extrapolating Casagrande's factor 0.36 would give about 42.5 m.) The last part of the line is drawn tangent to the parabola and blended into this exit point. #### Step 4: Discharge $$ q=k\,y_0=5.1\times10^{-6}\times 7.00=3.571 \times 10^{-5}\ \text{m}^3/\text{s per m} $$ (Schaffernak's $q=k\,a\sin\alpha\tan\alpha=5.1\times10^{-6}\times35.1\times0.447\times0.5=4.0\times10^{-5}$ m³/s per m gives a similar order of value.) **Answer**: the phreatic line is the parabola $y=\sqrt{49.04+14.01\,x}$ (x from the d/s toe, with entry and exit corrections), $y_0=7.00$ m, exit point about 35 m up the d/s face; seepage $q=3.571 \times 10^{-5}\ \text{m}^3/\text{s}$ per metre length of dam ($\approx 3.09$ m³/day per m). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is d computed in Casagrande's method for this dam? A. d = B - 0.7b, from base width and wetted horizontal projection b B. d = B - b C. d = h + b D. d = 0.3b Answer: A. B = 280, b = 150, d = 280 - 105 = 175 m. (Key point: Geometry) 2. y0 for h = 50 m and d = 175 m is A. 50 m B. 25 m C. 175 m D. about 7.0 m Answer: D. y0 = sqrt(50^2 + 175^2) - 175 = 7.00 m. (Key point: Basic parabola) 3. Discharge per metre is found from A. q = k y0 with k in m/s B. q = k h^2 C. q = k d D. q = y0/k Answer: A. q = 5.1e-6 x 7.00 = 3.571e-5 m^3/s per m. (Key point: Discharge) 4. The exit correction uses a + delta a = y0/(1 - cos alpha). What is alpha? A. The u/s slope angle B. The d/s slope angle, tan^-1(1/2) = 26.57 degrees C. 90 degrees D. The angle of the crest Answer: B. alpha is the d/s face angle; C is read from Casagrande's chart (about 0.36 near 30 degrees). (Key point: Exit correction) ## Close variants set in other exams - [Phreatic line and discharge (crest 300 masl)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aearthen-dam-phreatic-line-2079-baishakh) — Hydropower Engineering, chapter 4, asked 2× ## Related in Hydropower Engineering - [Toe-drain earthen dam: discharge and phreatic line](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aearthen-dam-toe-drain-2069) — Hydropower Engineering, chapter 4 - [Seepage line for 22 m high earthen dam](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aearthen-dam-seepage-line-22m-45deg) — Hydropower Engineering, chapter 4 - [Earth dam with drain pipe: phreatic coordinates](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aearthen-dam-drain-pipe-phreatic) — Hydropower Engineering, chapter 4 - [Seepage discharge through dam body (k = 1×10⁻⁶ m/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aearthen-dam-seepage-gradient-figure-2072k) — Hydropower Engineering, chapter 4 - [Gravity dam equilibrium and overtopping discharge coefficient](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Agravity-dam-reservoir-capacity-curve-hrwl) — Hydropower Engineering, chapter 4 - [Filter material in earthen dam](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aearthen-dam-filter-purpose-design) — Hydropower Engineering, chapter 4 ## Related in other subjects - [Top flow line, dam with horizontal filter](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Atop-flow-line-horizontal-filter) — Soil Mechanics, chapter 7 - [Reservoir level change, water balance, 15.5 km2](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Areservoir-level-change-15-5km2) — Engineering Hydrology, chapter 1, asked 2×