Explanation:
Permeability is the property of soil to transmit water through it.
It is usually expressed either as a permeability rate in centimetres per hour (cm/h), millimetres per hour (mm/h), or centimetres per day (cm/d), or as a coefficient of permeability k in metres per second (m/s) or in centimetres per second (cm/s).
Note:
Permeability (k) = \(\frac{{{{\rm{d}}^2}{\rm{\;}}{{\rm{e}}^3}}}{{1{\rm{\;}} + {\rm{\;e}}}}{\rm{\;}}{\left( {\frac{{\rm{\gamma }}}{{\rm{\mu }}}} \right)_{\rm{f}}}{{\rm{k}}_1}{{\rm{k}}_2}{{\rm{k}}_3}{{\rm{k}}_4}{{\rm{k}}_5}{{\rm{k}}_6}\;\)
The factors affecting the permeability of soil can be summarised in below tabulated form:
Parameter |
Description |
Size of particle |
Higher the size of particle higher will be the permeability |
Specific surface area |
Higher the specific surface area lower will be the permeability |
Void Ratio |
Higher the void ratio higher the permeability |
Viscosity of water |
Higher the viscosity lower will be the permeability and we know that viscosity of liquids (water) increases with decrease in temperature and hence, permeability reduces. |
Degree of saturation |
Higher the degree of saturation higher the permeability |
Entrapped gases |
Higher the amount of entrapped gasses in soil mass lower will be the permeability |
For the different soil types as per grain size, the orders of magnitude of permeability are as follows:
Type of soil | Permeability (cm/s) |
Gravel | 1 |
Coarse sand | 1 to 0.1 |
Medium sand | 10-1 to 10-2 |
Fine sand | 10-2 to 10-3 |
Silty sand | 10-3 to 10-4 |
Silt | 10-5 |
Clay | 10-7 to 10-9 |
From the table, Sand has the Coefficient of permeability ranges from 1 × 10– 2 to 5 × 10– 2 cm/s.
Concept:
Coefficient of permeability ‘k’ is given by:
\(K = C \times \frac{\gamma }{\mu } \times \frac{{{e^3}}}{{1 + e}} \times {D^2}\)
where
e = void ratio
D = effective size of soil particle
γ = weight density of fluid
μ = viscosity of fluid
c = constant (depends upon the shape of particle)
Solution:
K is directly proportional to γ / μ
∴ \(\frac{{{K_2}}}{{{K_1}}} = \frac{{{\gamma _2} \times {\mu _1}}}{{{\gamma _1} \times {\mu _2}}} = \frac{{0.8{\gamma _1} \times {\mu _1}}}{{{\gamma _1} \times 0.6{\mu _1}}} = 1.333\)
K_{1} = 1.333 K_{2}
i.e K_{2} is increased by 33.33%.Concept:
Permeability in stratified soil:
Case 1 : When the flow is along the bedding plane i.e. Horizontal flow
Equivalent permeability, \({{\rm{K}}_{{\rm{eq}}}} = \frac{{{{\rm{K}}_1}{{\rm{H}}_1} + {{\rm{K}}_2}{{\rm{H}}_2} + {{\rm{K}}_3}{{\rm{H}}_3}}}{{{{\rm{H}}_1} + {{\rm{H}}_2} + {{\rm{H}}_3}}}\)
Case 2 : When flow is perpendicular to the bedding plane i.e. Vertical flow
Equivalent permeability, \({{\rm{K}}_{{\rm{eq}}}} = \frac{{{{\rm{H}}_1} + {{\rm{H}}_2} + {{\rm{H}}_3}}}{{\frac{{{{\rm{H}}_1}}}{{{{\rm{K}}_1}}} + \frac{{{{\rm{H}}_2}}}{{{{\rm{K}}_2}}} + \frac{{{{\rm{H}}_3}}}{{{{\rm{K}}_3}}}}}\)
K_{eq} for the horizontal direction is always greater than K_{eq} for the vertical direction
Calculation:
Equivalent permeability is given by,
\(\therefore {{\rm{K}}_{{\rm{eq}}}} = \frac{{{\rm{H}} + {\rm{H}} + {\rm{H}}}}{{\frac{{\rm{H}}}{{{{10}^{ - 4}}}} + \frac{{\rm{H}}}{{2 \times {{10}^{ - 4}}}} + \frac{{\rm{H}}}{{{1.5 \times{10}^{ - 4}}}}}} = 1.384 \times {10^{ - 4}}{\rm{\;cm}}/{\rm{sec}}\)
Concept:
Constant head permeability test is carried out to determine the permeability of coarse-grained soil only where a reasonable discharge can be collected in a given time.
However, the falling head permeability test is used for relatively less permeable soils (fine-grained) where the discharge is small.
So here in clayey soil falling head permeability test can be used.
For falling head permeability test,
\(K = \frac{a}{A}\frac{L}{t}\;ln\left( {\frac{{{h_1}}}{{{h_2}}}} \right)\)
\({\rm{k}} = 2.303\frac{{\rm{a}}}{{\rm{A}}} \times \frac{{\rm{L}}}{{\rm{t}}}10{{\rm{g}}_{10}}\left( {\frac{{{{\rm{h}}_1}}}{{{{\rm{h}}_2}}}} \right)\)
Where,
k = permeability, a = Area of tube in m2, A = Area of sample in m2, t = time in sec, L = length in m, h1 = Level of time t = 0, and h2 = Level of time t
If the hydraulic gradient is unity, then what is the ratio of flow across unit area of soil called?
Concept:
As per Darcy’s law,
Q = k × i × A
Where,
k = coefficient of permeability,
i = hydraulic gradient,
A = area, Q = discharge
Calculation:
Given,
i = 1, A = 1
Q = k × 1 × 1
Q = k
Hence, ratio of flow across unit area is equal to coefficient of permeability.Concept:
Permeability is defined as the property of a porous material that permits the passage or seepage of water (or other fluids) through its interconnecting voids.
Factors affecting the permeability of soil are
1. Grain size
2. Properties of the pore fluid.
3. Voids ratio of the soil.
4. The structural arrangement of the soil particles.
5. Entrapped air and foreign-matter
6. Adsorbed water in clayey soils
The structural arrangement of the particle may vary, at the same voids ratio, depending upon the method of deposition or compacting the soil mass. The structure may be entirely different for a disturbed sample as compared to an undisturbed sample which may process stratification. The effect of structural disturbance on permeability is much pronounced in fine-grained soils.
A sample of clay and a sample of sand has the same specific gravity and void ratio. Their permeabilities would differ because the size ranges of their voids would be different.
0.04 cm/sec
Explanation
When the effective diameter is given we should use the Allen Hazen equation to determine permeability
Permeability (k) by Allen Hazen’s equation is given by:
k (cm/sec) = C × D_{10}^{2}
Where is C is constant and 100 ≤ C ≤ 150
Generally,
we take C = 100, D10 is effective grain size in cm
Given,
D10 = 0.2 mm = 0.02 cm
C = 100
Now, k = 100 × (0.02)^{2}
k = 0.04 cm/sec
Concept:
Factors affecting the permeability of soil can be studied using the following equation.
\(K = \frac{1}{Z}\frac{{{e^3}}}{{1 + e}}\;\frac{{{\gamma _w}}}{\mu }\;\frac{1}{{{S^2}}}\)
Where, Z = constant; μ = dynamic viscosity of water;
S = specific surface; γw = unit weight of water.
The factors affecting the permeability of soil can be summarised in below tabulated form:
Parameter |
Description |
Size of particle |
Higher the size of particle higher will be the permeability |
Specific surface area |
Higher the specific surface area lower will be the permeability |
Void Ratio |
Higher the void ratio higher the permeability |
Viscosity of water |
Higher the viscosity lower will be the permeability and we know that viscosity of liquids (water) increases with decrease in temperature and hence, permeability reduces. |
Degree of saturation |
Higher the degree of saturation higher the permeability |
Entrapped gases |
Higher the amount of entrapped gasses in soil mass lower will be the permeability |
Concept:
Formula:
For falling head permeability test,
\({\rm{k}} = 2.303\frac{{\rm{a}}}{{\rm{A}}} \times \frac{{\rm{L}}}{{\rm{t}}}10{{\rm{g}}_{10}}\left( {\frac{{{{\rm{h}}_1}}}{{{{\rm{h}}_2}}}} \right)\)
Where,
k = permeability, a = Area of tube in m^{2}, A = Area of sample in m^{2}, t = time in sec, L = length in m, h_{1} = Level of time t = 0, and h_{2} = Level of time t
Given:
A = 60 cm^{2}, L = 15 cm, \({\rm{a}} = \frac{{\rm{\pi }}}{4} \times {\left( {0.5} \right)^2} = 0.196{\rm{\;c}}{{\rm{m}}^2}\), h_{1} = 100 cm, h_{2} = 40 cm, and t = 30 min = 30 × 60 sec
Calculation:
\({\rm{k}} = 2.303 \times \frac{{0.196 \times 15}}{{60 \times 60 \times 30}}{\log _{10}}\left( {\frac{{100}}{{40}}} \right)\) = 2.49 × 10^{-5} cm/sec
Important Points:
Falling head permeability test is carried out for fine-grained soil, whereas constant head permeability test is carried out for coarse-grained soil.
Permeability tests were carried out on the samples collected from two different layers as shown in the figure (not drawn to the scale). The Relevant horizontal (k_{h}) and vertical (k_{v}) coefficients of permeability are indicated for each layer.
The ratio of the equivalent horizontal to vertical coefficients of permeability is
Concept:
Consider a section of stratified soil as shown in the figure below of varying thickness of each stratum eg: H_{1}, H_{2}, H_{3} & H_{4}, with their respective coefficient of permeability k_{1}, k_{2}, k_{3} & k_{4}
_{}
1) Horizontal Flow (Parallel to bedding plane):
Flow is taking place through all the layers at the same time hence hydraulic gradient is the same in each layer.
\({i_1} = {i_2} = {i_3} = {i_4} = i\) (constant)
Average discharge velocity over the soil mass can be written as.
\(V = {k_H}i = \frac{1}{H}\left( {{V_1}{H_1} + {V_2}{H_2} + {V_3}{H_3} + {V_4}{H_4}} \right) = \frac{1}{H}\;\left( {{k_i}{i_1}{H_1} + {k_2}{i_2}{H_2} + {k_3}{i_3}{H_3} + {k_4}{i_4}{H_4}} \right)\)
\( \Rightarrow {k_H}i = \frac{{\left( {{k_1}{H_1} + {k_2}{H_2} + {k_3}\;{H_3} + {k_4}{H_4}} \right)}}{H}i\)
\(\therefore {k_H} = \frac{{\left( {{k_1}{H_1} + {k_2}{H_2} + {k_3}\;{H_3} + {k_4}{H_4}} \right)}}{{{H_1} + {H_2} + {H_3} + {H_4}}}\)
2) Vertical flow (Normal to bedding Plane):
Let i_{1}, i_{2}, i_{3} & i_{4} be the hydraulic gradient in different layers of thickness H_{1}, H_{2}, H_{3} & H_{4} respectively.
Let the total head loss be h over the total thickness of soil stratum H.
Each layer having head loss h_{1}, h_{2}, h_{3} & h_{4}. Then the constant velocity of flow is given by
\(V = {k_v}\frac{h}{H} = {k_1}{i_1} = {k_2}{i_2} = {k_3}{i_3} = {k_4}{i_4}\)
Also,
\(\because Q = kiA = {k_1}{i_1}A = {k_2}{i_2}A = {k_3}{i_3}A = {k_4}{i_4}A\)
\(\therefore ki = {k_1}{i_1} = {k_2}{i_2} = {k_3}{i_3} = {k_4}{i_4}\)
\( \Rightarrow \frac{{K.h}}{H} = \frac{{{k_1}{h_1}}}{{{H_1}}} = \frac{{{k_2}{h_2}}}{{{H_2}}} = \frac{{{k_3}{h_3}}}{{{H_3}}} = \frac{{{k_4}{h_4}}}{{{H_4}}}\)
\(\because {h_1} + {h_2} + {h_3} + {h_4} = h\)
⇒ \(h\left( {\frac{{k\;{H_1}}}{{H\;{k_1}}} + \frac{{k\;{H_2}}}{{H\;{k_2}}} + \frac{{k\;{H_3}}}{{H\;{k_3}}} + \frac{{k\;{H_4}}}{{H\;{k_4}}}} \right) = h\)
\({k_v} = \frac{H}{{\frac{{{H_1}}}{{{k_1}}} + \frac{{{H_2}}}{{{k_2}}} + \frac{{{H_3}}}{{{k_3}}} + \frac{{{H_4}}}{{{k_4}}}}}\)
Calculation:
Given:
k_{h1} = 4.4 × 10^{-3} m/s, k_{v1} = 4 × 10^{-3} m/s, H_{1} = 3m
k_{v2} = 5.5 × 10^{-1} m/s,k_{h2} = 6 × 10^{-1} m/s, H_{2} = 4m
∴ Equivalent Horizontal permeability (k_{eq H})
= \({k_{eq\;H}} = \frac{{{k_{h1}}\;{H_1} + {k_{h2}}{H_2}}}{{{H_1} + {H_2}}} = \frac{{4.4 \times {{10}^{ - 3}} \times 3 + 6 \times {{10}^{ - 1}} \times 4}}{{3 + 4}}\)= 0.345 m/s
Equivalent vertical permeability (k_{eq v})
\({k_{eqv}} = \frac{{{H_1} + {H_2}}}{{\frac{{{H_1}}}{{{k_{v1}}}} + \frac{{{H_2}}}{{{k_{v2}}}}}} = \frac{{3 + 4}}{{\frac{3}{{4 \times {{10}^{ - 3}}}} + \frac{4}{{5.5 \times {{10}^{ - 1}}}}}}\) = 9.244 × 10^{-3} m/s
∴ Ratio \( = \frac{{{k_{eq\;H}}}}{{{k_{eq\;v}}}} = \frac{{0.345}}{{9.244 \times {{10}^{ - 3}}}} = 37.29\)
Important Points
Horizontal permeability is always greater than vertical permeability.
If the capillary rise in soil A with an effective size of 0.02 mm was 60 cm, then what would be the capillary rise in the similar soil B with an effective size of 0.04 mm?
Concept:
The capillary rise in soil particles is inversely proportional to the effective size of the soil particle.
\({h_c} = \frac{{C\left( {constant} \right)}}{{e \times {D_{10}}}}\)
\(\therefore {h_c} \propto \frac{1}{{{D_{10}}}}\)
Calculation:
Given:
(D_{10})_{A} = 0.02 mm, (D10)_{B} = 0.04 mm, (h_{c})_{A} = 60 cm
\(\therefore \frac{{{{\left( {{h_c}} \right)}_A}}}{{{{\left( {{h_c}} \right)}_B}}} = \;\frac{{{{\left( {{D_{10}}} \right)}_B}}}{{{{\left( {{D_{10}}} \right)}_A}}}\)
\(\therefore \frac{{60}}{{{{\left( {{h_c}} \right)}_B}}} = \frac{{0.04}}{{0.02}}\)
\(\Rightarrow {\left( {{h_c}} \right)_B} = 30\;cm\)
Concept:
Falling head permeability test or variable head permeability test
Coefficient of permeability, \(\rm{k = \frac{{2.303 \times a \times L}}{{A \times t}}{\log _{10}}\left( {\frac{{{h_1}}}{{{h_2}}}} \right)}\)
Where
a = Area of the tube in m^{2}
A = Area of the sample in m^{2}
t = time in seconds
L = length in meter
h_{1} = initial head at t = 0 sec
h_{2} = head after ‘t’ time
Calculation
d = 0.5 cm = 5 mm, L = 20 cm = 200 mm
Dia of soil specimen D = 10 cm = 100 mm
\(\rm{k = \frac{{2.303 \times \frac{\pi }{4} \times {5^2} \times 200}}{{\frac{\pi }{4} \times {{\left( {100} \right)}^2} \times 3 \times 3600}} \times {\log _{10}}\left( {\frac{{1000}}{{350}}} \right) = 4.86 \times {10^{ - 5}}\;mm/s}\)
A tracer takes 100 days to travel from Well-1 to Well-2 which are 100 m apart. The elevation of water surface in Well-2 is 3 m below that in Well-1. Assuming porosity equal to 15%, the coefficient of permeability (expressed in m/day) is
Concept:
According to Darcy’s law,
For laminar flow Vα i
V = K × i
Where,
V = velocity of water flowing in soil, K = coefficient of permeability
\({\rm{i}} = hydraulic~ gradient= \frac{{{{\rm{h}}_{\rm{L}}}}}{{\rm{L}}}\)
h_{L} = head difference, L = seepage length
Seepage velocity is given by,
V_{S} = V/n
Where,
n = porosity of the soil
Calculation:
T = Time = 100 days
D = distance = 100 m
h_{L} = head difference = 3 m
n = 15 % = 0.15
i = h_{L}/L = 3/100
V_{S} = D/T = 100/100 = 1 m/day
V = K × i
V = V_{S} × n
∴ V_{S} × n = K × i
\(K = \frac{{{V_s} \times n}}{i} = \frac{{1 \times 0.15}}{{\frac{3}{{100}}}} = 5\;m/day\)
Consider the following statements regarding flow net:
1. It helps determine the quantity of seepage.
2. It helps determine the upward lift below a hydraulic structure.
3. It is applicable to rotational flow only.
Which of the above statements are correct?
Flow Nets: It is graphical solutions to the Laplace equation for two-dimensional seepage. Two orthogonal sets of curves forming a flow net are:
Application of flow net is as follows:
An aquifer confined at top and bottom by impermeable layers is stratified into three layers as follows:
Layer |
Thickness (m) |
Permeability (m/day) |
Top layer |
4 |
30 |
Middle layer |
2 |
10 |
Bottom layer |
6 |
20 |
The transmissivity (m^{2}/day) of the aquifer is:
Concept:
Transmissibility:
T = k × b
Here,
T - Transmissivity (m^{2}/day)
k - Coefficient of permeability (m/day)
b - Thickness of aquifer (m)
Calculation:
Given,
Layer |
Thickness, H (m) |
Permeability, k (m/day) |
Top layer |
4 |
30 |
Middle layer |
2 |
10 |
Bottom layer |
6 |
20 |
It is given the aquifer is confined at top and bottom by impermeable layers so, the Flow-through aquifer occurs parallel to the bed of the aquifer.
∴ Equivalent permeability of the aquifer is given by
\({{\rm{k}}_{{\rm{parallel}}}} = \frac{{{{\rm{k}}_1} \times {{\rm{H}}_1} + {{\rm{k}}_2} \times {{\rm{H}}_2} + {{\rm{k}}_3} \times {{\rm{H}}_3}}}{{{{\rm{H}}_1} + {{\rm{H}}_2} + {{\rm{H}}_3}}}\)
Putting values of k and H
\({{\rm{k}}_{{\rm{parallel}}}} = \frac{{30 \times 4 + 10 \times 2 + 20 \times 6}}{{4 + 2 + 6}}\)
⇒ k_{eq} = 21.67 m/day
∴ From (1)
T = 21.67 × 12 = 260m^{2}/day
Explanation:
Permeability is the property of soil to transmit water through it.
It is usually expressed either as a permeability rate in centimeters per hour (cm/h), millimeters per hour (mm/h), or centimeters per day (cm/d) or as a coefficient of permeability k in meters per second (m/s) or in centimeters per second (cm/s).
Important Points
For the different soil types as per grain size, the orders of magnitude of permeability are as follows:
Type of soil | Permeability (cm/s) |
Gravel | 1 |
Coarse sand | 1 to 0.1 |
Medium sand | 10-1 to 10-2 |
Fine sand | 10-2 to 10-3 |
Silty sand | 10-3 to 10-4 |
Silt | 10-5 |
Clay | 10-7 to 10-9 |
From the table, Clay has the Coefficient of permeability ranges of 1 × 10– 7
Explanation:
Permeability depends on both soil properties and Fluid properties
The kozney – Carman equation is quite useful, it reflects the effect of factors that affect permeability
\(k = \frac{{{\gamma _w}}}{\mu } \times \frac{{{e^3}}}{{1 + e}} \times D_{10}^2\)
Where,
D10 is effective grain size
e is the void ratio
μ is viscosity , γ w is unit weight of water
a) Grain size: The coefficient of permeability includes D102, where D10 is a measure of grain size
If the void ratio is the same then permeability is more in coarse soil than in the fine soil
b) Void ratio: From the equation, it is clear that k ∝ e2
If the particle size is the same, then loose soils are more permeable than dense soils
c) Degree of saturation: Permeability is directly proportional to the degree of saturation
d) Particle shape: It is expressed in terms of specific surface area and permeability relates to specific surface area as ‘k ∝ 1/S2 ‘
Explanation:
Coefficient of permeability:
It can be defined as The average velocity of flow that will occur through the soil cross-section under a unit hydraulic gradient.
According to Allen Hazen formula
K = \(C× D_{10}^2\)
Here K = Coefficient of permeability
C = a constant, which is taken as 100
D10 = Effective size Particle in mm (it is expressed such that 10% particles are finer than this size)
So, Permeability of coarse grained soil is depend on effective size particles.
Explanation:
(i) The flow through the soils, however occurs only through the interconnected pores or voids. The velocity through the pores is called seepage velocity (V_{s}).
Seepage velocity (V_{s}) = V/n
Where, V = Superficial veclocity
n = Porosity.
Important point:
(i) Superficial velocity: Superficial velocity is based on the gross area of cross-section. It is superficial because the actual flow is through the pores in the cross-section and not through the entire cross-sectional area.
(ii) Since n < 1, the seepage velocity is always greater than the discharge or superficial velocity.
Concept:
The coefficient of permeability (k) is given by
k = \({\rm{C}} \times {\rm{D}}_{10}^2 \times \left( {\frac{{{{\rm{e}}^3}}}{{1 + {\rm{e}}}}} \right) \times \left( {\frac{{{{\rm{\gamma }}_{\rm{\omega }}}}}{{\rm{μ }}}} \right)\)
where C is constant D_{10} is effective grain size
e is the void ratio, μ is viscosity
Factor-effecting permeability of the soil
1) Grain size: Permeability of soil increases with the grain size, as the void ratio for the sample increases for the sample.
Allen Hazen has related the coefficient of permeability with the grain size in the following relationship:
\({\rm{k}} = 100 \times {\rm{D}}_{10}^2\)
∴ k is directly proportional to the square of the grain size.
2) Void ratio of same soil at different state: As per equation (i), k is directly proportional to \(\left( {\frac{{{e^3}}}{{1 + e}}} \right) \approx {e^2}\) and as porosity (n) increases, the pore increases and hence the total flow through soil.
3) Void ratio for different type of soil: For different soil, the void ratio increases with the grain size for different soil and the variation is tabulated below:
Soil |
Gravel |
Sand |
Silt |
Clay |
k (cm/sec) |
10° |
10-2 |
10-4 |
10-6 |
4) Organic Impurities: As organic content increases, voids get obstructed and hence reduces the permeability of the soil. Thus, coefficient of permeability in inversely proportional to organic content.