--- title: "Toe-drain earthen dam: discharge and phreatic line" subject: "Hydropower Engineering" chapter: 4 key: "semester_7/hydropower_engineering:earthen-dam-toe-drain-2069" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/earthen-dam-toe-drain-2069 site: BCT Engineering --- # Toe-drain earthen dam: discharge and phreatic line Hydropower Engineering · Chapter 4: Headworks of Storage Plants - Asked in: 2069 Chaitra, 10 marks - Video: [Design of Phreatic line in earthen dam with filter](https://www.youtube.com/watch?v=V5pnA13B3r0) — 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-toe-drain-2069 - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4 ## Question The u/s and d/s slopes of a homogeneous earthen dam with 12 m toe drain are 2:1 and 3:1 (H:V) respectively. The water depth at u/s of the dam is 50 m. The dam has a crest width of 20 m and free board of 5 m. The coefficient of permeability of the dam material is 2.5 cm/hr. Calculate (i) the specific discharge through the body of the dam, (ii) the co-ordinates of the phreatic line. ## Short answer With focus at the upstream end of the toe drain, y0 = 5.79 m, phreatic line y = sqrt(33.52+11.58x). Specific discharge q = k·y0 = 4.021×10⁻⁵ m³/s per m. ## Answer **Given**: u/s slope 2:1, d/s slope 3:1 (H:V), water depth $h=50$ m, free board 5 m so dam height $=55$ m, crest width 20 m, toe drain 12 m long, $k=2.5$ cm/hr $=\dfrac{2.5}{100\times3600}=6.944 \times 10^{-6}$ m/s. ```text 55 m 20 m ________ 2:1 / \ 3:1 50m / dam \ toe drain 12 m ~~~ / \_____[#######] //////////////////////////////// ``` #### Step 1: Geometry $$ \begin{aligned} B &= 2(55)+20+3(55)=295\ \text{m}\\ b &= 2\times50=100\ \text{m},\quad 0.3b=30\ \text{m}\\ d &= B-0.7b-12 = 295-70-12 = 213\ \text{m} \end{aligned} $$ ($d$ is measured from the corrected entry point to the u/s end of the toe drain, which is the focus of the parabola.) #### Step 2: (i) Specific discharge $$ y_0=\sqrt{h^2+d^2}-d=\sqrt{50^2+213^2}-213=5.790\ \text{m} $$ $$ q=k\,y_0=6.944 \times 10^{-6}\times5.790=4.021 \times 10^{-5}\ \text{m}^3/\text{s per m}\quad(\approx3.47\ \text{m}^3/\text{day per m}) $$ #### Step 3: (ii) Co-ordinates of the phreatic line $$ y=\sqrt{y_0^2+2y_0x}=\sqrt{33.52+11.58\,x} $$ with $x$ from the focus (u/s end of the toe drain) towards the u/s side: | x from focus (m) | y (m) | |---|---| | 0.0 | 5.79 | | 10.0 | 12.22 | | 25.0 | 17.97 | | 50.0 | 24.75 | | 75.0 | 30.03 | | 100.0 | 34.52 | | 125.0 | 38.48 | | 150.0 | 42.08 | | 175.0 | 45.39 | | 200.0 | 48.47 | | 213.0 | 50.00 | At $x=213$ m the ordinate is $50$ m (corrected entry). The line is joined to the u/s face at right angles at the entry, and since the line enters the toe drain ($\alpha=180^\circ$) no exit correction is needed ($\Delta a=0$); it meets the drain at about $a=y_0/2=2.89$ m beyond the focus. **Answer**: (i) $q=4.021 \times 10^{-5}$ m³/s per m; (ii) $y=\sqrt{33.52+11.58\,x}$ with $y_0=5.790$ m, as tabulated. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With water depth 50 m and free board 5 m, how are B and b taken? A. Dam height 55 m: B = 2(55)+20+3(55) = 295 m; b = 2 x 50 = 100 m B. Dam height 50 m: B = 2(50)+20+3(50) = 270 m C. Dam height 55 m: b = 2 x 55 = 110 m D. Dam height 45 m: B = 245 m Answer: A. Slopes act over the full dam height (55 m), while b is the wetted u/s horizontal length using the water depth. (Key point: Dam height and geometry) 2. For a 12 m toe drain, d is measured to which point? A. From the corrected entry point to the u/s end of the toe drain, d = 295 - 70 - 12 = 213 m B. From the u/s toe to the d/s toe, 295 m C. From the crest to the drain, 20 m D. From the corrected entry to the d/s toe, 225 m Answer: A. The u/s end of the toe drain is the focus, so the drain length is subtracted: d = B - 0.7b - 12. (Key point: Distance d with toe drain) 3. What is the correct approach for specific discharge? A. Convert k = 2.5 cm/hr to 6.944 x 10^-6 m/s and use q = k y0 with y0 = 5.79 m B. Use k = 2.5 m/s directly C. Use q = k h with h = 50 m D. Use q = k d with d = 213 m Answer: A. k must be in m/s: 2.5/(100 x 3600). Then q = k y0 = 4.02 x 10^-5 m3/s per m. 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