Class Lecture

1. Permeability and Seepage
Course Code: CE 1115
Course Title: Permeability and seepage
Credit Hours: 3
Lecture: 1
Presented by:
Sabiha Ashtara Rekha
Lecturer
Dept. of Civil Engineering
World University of Bangladesh
(Prepared By: Rokhsana Parvin)

2. Permeability and Seepage
Outline of the lecture:
 Permeability
 Factors affecting permeability.
 Darcy’s law
 Determination of co-efficient of permeability
 Constant head permeability test
 Mathematical problems on Constant head
permeability test

3. Permeability and Seepage
Permeability:
Permeability is the property of porous material which permits the passage or seepage of
water through its interconnecting voids. A material having continuous void is called
permeable. Gravels are highly permeable while stiff clay is the least permeable and such a
clay may be termed as impermeable for all practical purposes.

4. Permeability and Seepage
Figure. Process of Permeability

5. Permeability and Seepage
FACTORS AFFECTING
PERMEABILITY:
 Grain size.
 Properties of pore fluid.
 Void ratio of soil.
 Structural arrangement of soil
particle.
 Entrapped air and foreign
matter.
 Absorbed water in clayey soil.

6. Permeability and Seepage
DARCY’S LAW:
“The laminar flow condition in a saturated
soil, the rate of flow or discharge per unit
time is proportional to the hydraulic
gradient”.
Thus,
q= Ki A
q/A = ki -------- (1)
again, q= Av
So, v = q/A = ki
Where,
q= discharge per unit time.
A= total cross-sectional area.
i= hydraulic gradient =
𝒉𝑳
𝑳
K= Darcy’s coefficient of permeability.
V= velocity of flow or discharge velocity.

7. Permeability and Seepage
DETERMINATION OF COEFFICIENT OF PERMEABILITY:
The co-efficient of permeability can be determined by the
following methods.
 Laboratory methods
(1) Constant head permeability test.
(2) Falling head permeability test.
 Field methods
(1) Pumping – out tests.
(2) Pumping – in tests.
 Indirect methods
(1) Computation from grain size or specific surface.
(2) Horizontal capillarity test.
(3) Consolidation test data.

8. Permeability and Seepage
Constant head permeability test
The constant head permeability test is a common laboratory
testing method used to determine the permeability of granular soils like
sands and gravels containing little or no silt.
Falling head permeability test
The falling head permeability test is used to determine the permeability of
fine grained soils with intermediate and low permeability such as silts and
clays.

9. Permeability and Seepage
Constant head permeability test
Figure shows the diagrammatical representation of
constant head test. Water flows from the overhead
tank consisting of three tubes: the inlet tube, the
overflow tube and the outlet tube. The constant
hydraulic gradient ‘i’ causing the flow is the head ‘h’
(i.e. difference in the water levels of the overhead and
bottom tanks) divided by the length, ‘L’ of the sample.
If Q is the total quantity of flow in a time interval t, we
have from Darcy’s law-
q=
𝑸
𝒕
= k i A [ From Darcy’s law, q = Ki A ]
𝑸
𝒕
= k i A [ i =
𝒉
𝑳
]
K =
𝑸
𝒕
.
𝟏
𝑨
=
𝑸
𝒕
.
𝑳
𝒉
.
𝟏
𝑨
=
𝑸𝑳
𝑨𝒉𝒕
Where, A = total cross sectional area of the sample.
q = discharge per unit time.
t = time
i = hydraulic gradient =
𝒉
𝑳
K = Darcy’s coefficient of permeability.
Q = Total quantity of flow in a time interval t
Figure. Constant head permeability test.

10.  PROBLEM:
A sample coarse sand is 15 cm high and 5.5 cm in diameter. It was tested in a constant
head parameter. Water permeated through the soil under a hydrostatic head of 50 cm for
6 seconds. The water was collected and it was found to weigh 500gm. Find the
permeability in cm/hr.
SOLUTION:
We know that, Here , L= 15cm
K =
𝑸𝑳
𝑨𝒉𝒕
d = 5.5 cm
A =
𝝅
𝟒
𝒅𝟐
=
𝟓𝟎𝟎 ×𝟏𝟓
𝟐𝟑.𝟕 ×𝟓𝟎 ×𝟏.𝟔𝟕𝒙𝟏𝟎−𝟑 = (3.1416/4)× (5.5)2 = 23.7 cm2
h = 50 cm
= 3790 cm/hr. Q = 500gm = 500 cc
(Answer). t = 6 seconds
= 6/(60x60) = 1.67x 𝟏𝟎−𝟑
hrs
Permeability and Seepage
Note:1 gram (g) = weight of 1 cubic centimeter (cc, cm3)

11. Permeability and Seepage