---
title: "Constant head test, k at porosity 39%"
subject: "Soil Mechanics"
chapter: 6
key: "semester_4/soil_mechanics:constant-head-permeability-porosity"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/constant-head-permeability-porosity
site: BCT Engineering
---

# Constant head test, k at porosity 39%

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

- Asked in: 2078 Chaitra, 6 marks
- Video: [Constant Head Permeability Test](https://www.youtube.com/watch?v=B3gvVN29u8E) — Elementary Engineering
- 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-permeability-porosity
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-6

## Question

The following data were recorded in a constant head permeability test. Internal diameter of the permeameter = 7.5 cm, porosity of sample = 44%. Quantity of water collected in 60 s = 626 ml and head loss over a sample length of 18 cm = 24.7 cm. Calculate the permeability, flow velocity and seepage velocity. Also calculate the permeability of soil at a porosity of 39%.

## Short answer

Permeability k = 0.172 cm/s, discharge velocity v = 0.236 cm/s, seepage velocity vs = 0.537 cm/s. At porosity 0.39, k = 0.101 cm/s.

## Answer

#### Given

$D = 7.5$ cm, $L = 18$ cm, head loss $h = 24.7$ cm, $Q = 626$ ml in $t = 60$ s, porosity $n = 0.44$.

$$
A = \frac{\pi}{4}(7.5)^2 = 44.18\ \text{cm}^2, \qquad q = \frac{626}{60} = 10.43\ \text{cm}^3/\text{s}, \qquad i = \frac{24.7}{18} = 1.372
$$

#### Permeability

$$
k = \frac{q}{A\,i} = \frac{10.43}{44.18 \times 1.372} = 0.172\ \text{cm/s}
$$

#### Flow (discharge) velocity

$$
v = \frac{q}{A} = \frac{10.43}{44.18} = 0.236\ \text{cm/s} \quad (= ki)
$$

#### Seepage velocity

$$
v_s = \frac{v}{n} = \frac{0.236}{0.44} = 0.537\ \text{cm/s}
$$

#### Permeability at $n = 0.39$

For the same soil, $k \propto \dfrac{e^3}{1+e}$ (Kozeny–Carman type relation).

$$
e_1 = \frac{0.44}{0.56} = 0.786, \qquad e_2 = \frac{0.39}{0.61} = 0.639
$$

$$
\frac{k_2}{k_1} = \frac{e_2^3/(1+e_2)}{e_1^3/(1+e_1)} = \frac{0.1594}{0.2716} = 0.587
$$

$$
k_2 = 0.172 \times 0.587 = 0.101\ \text{cm/s}
$$

**Answer: $k = 0.172$ cm/s; $v = 0.236$ cm/s; $v_s = 0.537$ cm/s; $k$ at $n = 0.39$ is $0.101$ cm/s.**

## Check yourself

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

1. In a constant head test, k is:
   A. q/(A i)
   B. q i/A
   C. q A/i
   D. q L A
   Answer: A. From Darcy, k = q/(A i) with i = h/L. (Key point: Permeability formula)

2. With A = 44.18 cm2, q = 10.43 cm3/s, i = 1.372, k is about:
   A. 0.236 cm/s
   B. 0.172 cm/s
   C. 0.537 cm/s
   D. 1.72 cm/s
   Answer: B. k = 10.43/(44.18 x 1.372) = 0.172 cm/s. (Key point: Value of k)

3. Seepage velocity is obtained from discharge velocity v = 0.236 cm/s as:
   A. v n = 0.104
   B. v / n = 0.537 cm/s
   C. v (1+e) = 0.42
   D. v / (1-n) = 0.42
   Answer: B. v_s = v/n = 0.236/0.44 = 0.537 cm/s. (Key point: Seepage velocity)

4. k at n = 0.39 is found using:
   A. k proportional to n
   B. k proportional to e^3/(1+e), giving about 0.101 cm/s
   C. k independent of porosity
   D. k proportional to 1/e
   Answer: B. k2/k1 = 0.587 using e^3/(1+e), so k2 = 0.172 x 0.587 = 0.101 cm/s. (Key point: Change of porosity)

## Related in Soil Mechanics

- [Constant head permeameter, seepage velocity](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aconstant-head-permeameter-seepage-velocity) — 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
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