---
title: "Constant head permeameter, seepage velocity"
subject: "Soil Mechanics"
chapter: 6
key: "semester_4/soil_mechanics:constant-head-permeameter-seepage-velocity"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/constant-head-permeameter-seepage-velocity
site: BCT Engineering
---

# Constant head permeameter, seepage velocity

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

- Asked in: 2077 Chaitra, 7 marks
- Video: [constant head permeability | Numerical on Permeability of Soil| Discharge & Seepage velocity of soil](https://www.youtube.com/watch?v=6JYWnrnddLw) — Civil Engineering Exam
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/constant-head-permeameter-seepage-velocity
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-6

## Question

The discharge of water collected from a constant head permeameter in a period of 15 minutes is 500 ml. The internal diameter of the permeameter is 5 cm and the measured difference in head between two gauging points 15 cm vertically apart is 40 cm. Calculate the coefficient of permeability. If the dry weight of the 15 cm long sample is 4.86 N and the specific gravity of the solids is 2.65, calculate the seepage velocity.

## Short answer

Coefficient of permeability is 0.0106 cm/s. Porosity is 0.365, giving a seepage velocity of 0.0775 cm/s.

## Answer

#### Given

$Q = 500$ ml in $t = 15$ min $= 900$ s, $D = 5$ cm, head loss $h = 40$ cm between two points $L = 15$ cm apart. Dry weight of sample $W_d = 4.86$ N, $G = 2.65$, $\gamma_w = 9.81$ kN/m³.

$$
A = \frac{\pi}{4}(5)^2 = 19.63\ \text{cm}^2, \qquad q = \frac{500}{900} = 0.5556\ \text{cm}^3/\text{s}, \qquad i = \frac{h}{L} = \frac{40}{15} = 2.667
$$

#### Coefficient of permeability

$$
k = \frac{q}{A\,i} = \frac{0.5556}{19.63 \times 2.667} = 0.0106\ \text{cm/s} = 1.06\times10^{-4}\ \text{m/s}
$$

#### Seepage velocity

Discharge velocity:

$$
v = \frac{q}{A} = \frac{0.5556}{19.63} = 0.0283\ \text{cm/s}
$$

Volume of sample $= 19.63 \times 15 = 294.5\ \text{cm}^3 = 2.945\times10^{-4}\ \text{m}^3$.

$$
\gamma_d = \frac{4.86\times10^{-3}\ \text{kN}}{2.945\times10^{-4}\ \text{m}^3} = 16.50\ \text{kN/m}^3
$$

$$
e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.65 \times 9.81}{16.50} - 1 = 0.575, \qquad n = \frac{e}{1+e} = 0.365
$$

$$
v_s = \frac{v}{n} = \frac{0.0283}{0.365} = 0.0775\ \text{cm/s}
$$

**Answer: $k = 0.0106$ cm/s; $v_s = 0.0775$ cm/s ($n = 0.365$).**

## Check yourself

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

1. With head difference 40 cm over 15 cm between gauging points, i is:
   A. 0.375
   B. 2.667
   C. 15
   D. 0.4
   Answer: B. i = h/L = 40/15 = 2.667. (Key point: Hydraulic gradient)

2. With q = 0.5556 cm3/s, A = 19.63 cm2, i = 2.667, k is about:
   A. 0.0283 cm/s
   B. 0.106 cm/s
   C. 0.0106 cm/s
   D. 0.0775 cm/s
   Answer: C. k = q/(A i) = 0.5556/(19.63 x 2.667) = 0.0106 cm/s. (Key point: Permeability)

3. Void ratio is found from gamma_d = 16.5 kN/m3 using:
   A. e = G gamma_w/gamma_d - 1
   B. e = gamma_d/(G gamma_w)
   C. e = 1 - gamma_d/gamma_w
   D. e = G - gamma_d
   Answer: A. gamma_d = G gamma_w/(1+e), so e = 2.65 x 9.81/16.5 - 1 = 0.575. (Key point: Void ratio from dry weight)

4. With v = 0.0283 cm/s and n = 0.365, v_s is about:
   A. 0.0103 cm/s
   B. 0.0775 cm/s
   C. 0.0283 cm/s
   D. 0.0447 cm/s
   Answer: B. v_s = v/n = 0.0283/0.365 = 0.0775 cm/s. (Key point: Seepage velocity)

## Related in Soil Mechanics

- [Constant head test, k at porosity 39%](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aconstant-head-permeability-porosity) — Soil Mechanics, chapter 6
- [Discharge velocity vs seepage velocity](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Adischarge-vs-seepage-velocity) — Soil Mechanics, chapter 6, asked 2×
- [Impervious dam seepage discharge](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aimpervious-dam-seepage-discharge-numerical) — Soil Mechanics, chapter 7
- [Seepage and superficial velocity relation](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aseepage-superficial-velocity-relation) — Soil Mechanics, chapter 6
- [Variable head test, two soils](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Avariable-head-two-soils-permeability) — Soil Mechanics, chapter 6
- [Upward seepage stress variation diagram](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aupward-seepage-stress-diagram) — Soil Mechanics, chapter 7

## Related in other subjects

- [MOC numerical: discharge and head at P (30 cm pipe)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Amoc-pipe-discharge-head-pc-30cm) — Computational Techniques in Civil Engineering, chapter 6
- [Velocity, head loss, discharge in loop pipes BC, CD, BF, FD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Aloop-a76) — Water Supply Engineering, chapter 7
- [Internal discharge voltage in 1 cm specimen](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Avoid-80kv-specimen-discharge) — High Voltage Engineering, chapter 6
- [Velocity, head loss, discharge in pipes BC, CD, BF, FD (0.1 m)](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Aloop-c72) — Water Supply Engineering, chapter 7