---
title: "Phreatic line and discharge (crest 300 masl)"
subject: "Hydropower Engineering"
chapter: 4
key: "semester_7/hydropower_engineering:earthen-dam-phreatic-line-2079-baishakh"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/earthen-dam-phreatic-line-2079-baishakh
site: BCT Engineering
---

# Phreatic line and discharge (crest 300 masl)

Hydropower Engineering · Chapter 4: Headworks of Storage Plants

- Asked in: 2079 Baishakh, 4+4 marks
- 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-2079-baishakh
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-4

## Question

A homogeneous earthen dam has the following data: dam crest level = 300.00 masl; deepest river bed level = 278.00 masl; HFL in the reservoir = 297.50 masl; dam crest width = 4.50 m; dam u/s slope = 3:1; dam d/s slope = 2:1 and coefficient of permeability of the dam material = $5 \times 10^{-4}$ cm/s. Determine the phreatic line of the dam section and the discharge passing through the dam.

## Short answer

Using Casagrande's method, y0 = 2.54 m, the phreatic line is y = sqrt(6.46+5.08x), and the exit lies 12.5 m along the downstream face. Specific discharge q = k·y0 = 1.271×10⁻⁵ m³/s per m.

## Answer

**Given data and assumptions**: homogeneous dam on an impervious foundation, no drain, tail water neglected. Casagrande's basic parabola with focus at the d/s toe.

- Dam height $= 300-278 = 22$ m; water head $h = 297.5-278 = 19.5$ m; $k = 5\times10^{-4}$ cm/s $=5\times10^{-6}$ m/s.

#### Step 1: Geometry
$$
\begin{aligned}
B &= 3(22)+4.5+2(22)=114.5\ \text{m}\\
b &= 3\times 19.5 = 58.5\ \text{m},\quad 0.3b=17.55\ \text{m}\\
d &= B-0.7b = 114.5-40.95 = 73.55\ \text{m}
\end{aligned}
$$

#### Step 2: Basic parabola
$$
y_0=\sqrt{h^2+d^2}-d=\sqrt{19.5^2+73.55^2}-73.55=2.54\ \text{m}
$$
$$
y=\sqrt{y_0^2+2y_0x}=\sqrt{6.46+5.08\,x}
$$
Coordinates ($x$ from the d/s toe, $y$ above the base):

| x from focus (m) | y (m) |
|---|---|
| 0.0 | 2.54 |
| 2.0 | 4.08 |
| 5.0 | 5.65 |
| 10.0 | 7.57 |
| 20.0 | 10.40 |
| 30.0 | 12.61 |
| 40.0 | 14.48 |
| 50.0 | 16.14 |
| 60.0 | 17.65 |
| 73.6 | 19.50 |

At $x=d$, $y=h=19.5$ m (corrected entry point A, located $0.3b=17.55$ m upstream of the point where the water line meets the u/s face).

#### Step 3: Corrections
- Entry: the line starts at right angles to the u/s face and joins the parabola smoothly at A.
- Exit: $\alpha=\tan^{-1}(1/2)=26.57^\circ$, so the parabola meets the d/s face at $a+\Delta a=\dfrac{y_0}{1-\cos\alpha}=24.1$ m from the toe. Since $\alpha<30^\circ$ (outside Casagrande's chart), the exit point is found from Schaffernak's formula:
$$
a=\frac{d}{\cos\alpha}-\sqrt{\frac{d^2}{\cos^2\alpha}-\frac{h^2}{\sin^2\alpha}}=82.23-\sqrt{6,762-1,901}=12.5\ \text{m}
$$
so the line leaves the d/s face about 12.5 m along the slope from the toe (about 5.6 m above the base). (Extrapolating Casagrande's factor 0.36 would give about 15.4 m.)

#### Step 4: Discharge
$$
q=k\,y_0=5\times10^{-6}\times2.54=1.271 \times 10^{-5}\ \text{m}^3/\text{s per m}
$$
(Schaffernak's $q=k\,a\sin\alpha\tan\alpha=5\times10^{-6}\times12.5\times0.447\times0.5=1.40\times10^{-5}$ m³/s per m is of the same order.)

**Answer**: $y_0=2.54$ m, phreatic line $y=\sqrt{6.46+5.08\,x}$, exit point about 12.5 m up the d/s face; discharge $q=1.271 \times 10^{-5}$ m³/s per metre ($\approx1.098$ m³/day per m).

## Check yourself

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

1. What are dam height and water head here?
   A. 22 m and 19.5 m
   B. 19.5 m and 22 m
   C. 22 m and 22 m
   D. 2.5 m and 19.5 m
   Answer: A. H = 300 - 278 = 22 m; h = 297.5 - 278 = 19.5 m. (Key point: Dam height and head)

2. The base width is
   A. 110 m
   B. 114.5 m
   C. 44 m
   D. 66 m
   Answer: B. B = 3(22) + 4.5 + 2(22) = 114.5 m. (Key point: Base width)

3. y0 is about
   A. 2.54 m
   B. 19.5 m
   C. 10 m
   D. 73.55 m
   Answer: A. d = 114.5 - 0.7 x 58.5 = 73.55 m; y0 = sqrt(19.5^2 + 73.55^2) - 73.55 = 2.54 m. (Key point: Parabola ordinate)

4. The discharge q = k y0 equals
   A. 5.0e-6 m^3/s per m
   B. 2.54 m^3/s
   C. 1.271e-5 m^3/s per m
   D. 1.098e-5 m^3/s
   Answer: C. 5e-6 x 2.54 = 1.271e-5 m^3/s per metre. (Key point: Discharge)

## Close variants set in other exams

- [Phreatic line and discharge (k = 5.1×10⁻⁴ cm/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aearthen-dam-phreatic-line-2082) — Hydropower Engineering, chapter 4, asked 2×

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