Permeability MCQ Quiz - Objective Question with Answer for Permeability - Download Free PDF

Explanation: 

The average permeability for flow perpendicular to the bedding planes is given by,

\(k_{avg} = \frac{z_1 + z_2}{\frac{z_1}{k_1}+\frac{z_2}{k_2}}\)

The average permeability for flow parallel to the bedding planes is given by,

\(k_{avg} = \frac{k_1z_1 + k_2z_2 + k_3z_3 + ---}{z_1 + z_2 + z_3 + ---}\)

∴ Average permeability for flow perpendicular to the bedding plane is harmonic mean whereas, average permeability for flow parallel to the bedding plane is the arithmetic mean.

Important PointsAs arithmetic mean is always greater than harmonic mean therefore, the average permeability for flow perpendicular to the bedding plane is always lesser than the average permeability for flow parallel to the bedding plane.

Explanation:

Factors Not Affecting Bacterial Efficiency of Chlorine

Chlorine is one of the most widely used disinfectants for water treatment due to its strong oxidizing properties and ability to inactivate a variety of microorganisms, including bacteria. However, the efficiency of chlorine as a disinfectant can be influenced by several factors. These include the water's chemical composition, physical characteristics, and the presence of various compounds. The question asks us to identify the factor that does not affect the bacterial efficiency of chlorine, and the correct answer is Option 2: Coagulants used.

Detailed Explanation:

The addition of coagulants during water treatment is a process primarily aimed at removing suspended particles, turbidity, and colloidal impurities from water. This process does not directly interact with chlorine or its ability to kill bacteria. Coagulation is a separate step in the water treatment process that precedes disinfection. The following reasons explain why the use of coagulants does not affect the bacterial efficiency of chlorine:

• Purpose of Coagulation: Coagulants such as alum (aluminum sulfate) or ferric chloride are used to aggregate fine particles into larger flocs, which can then be removed through sedimentation or filtration. This step is unrelated to the disinfection process where chlorine is applied.
• No Direct Interaction: Coagulants do not chemically react with chlorine to form compounds that would inhibit its disinfectant properties. Therefore, they do not interfere with chlorine's ability to inactivate bacteria.
• Separate Treatment Stages: The coagulation and disinfection stages in water treatment are independent of each other. Coagulation improves the physical quality of water, while disinfection ensures microbial safety.

Explanation:

• Transmissibility (T) is the ability of an aquifer to transmit water through its entire saturated thickness.

• Permeability (K) is the ability of the aquifer material (soil or rock) to transmit water per unit thickness.

• The depth or thickness of the aquifer is represented by  d" id="MathJax-Element-2-Frame" role="presentation" style="position: relative;" tabindex="0">$d$ $d$ .

• The relation T$=K\times d$ means that transmissibility is the product of permeability and the aquifer thickness.

• This formula assumes uniform permeability throughout the depth.

 Additional Information

Permeability (K):

• It is a property of the porous material that measures the ease with which water can flow through its pores.

• Units: typically m/s or m/day.

Transmissibility (T):

• It represents the total volume of water transmitted through the entire thickness of the aquifer per unit width and hydraulic gradient.

• Units: m²/s or m²/day.

• It combines permeability and aquifer thickness into a single term useful for groundwater flow analysis.

Explanation:

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 the flow is perpendicular to the bedding plane i.e. Vertical flow

Equivalent permeability, \({{\rm{K}}_{{\rm{eq}}}} = \frac{{{{\rm{H}}_1} + {{\rm{H}}_2} + {{\rm{H}}_3}}}{{\frac{{{{\rm{H}}_1}}}{{{{\rm{K}}_1}}} + \frac{{{{\rm{H}}_2}}}{{{{\rm{K}}_2}}} + \frac{{{{\rm{H}}_3}}}{{{{\rm{K}}_3}}}}}\)

For the given case, we have three layers with the following properties:

First layer: h1=h,  h_1 = h" id="MathJax-Element-33-Frame" role="presentation" style="position: relative;" tabindex="0"> h1=h, $h_1 = h$ $h_1 = h$  k1=k k_1 = k" id="MathJax-Element-34-Frame" role="presentation" style="position: relative;" tabindex="0"> k1=k$k_1 = k$ $k_1 = k$

Second layer: h2=h h_2 = h" id="MathJax-Element-35-Frame" role="presentation" style="position: relative;" tabindex="0"> h2=h$h_2 = h$ $h_2 = h$ , k2=2k k_2 = 2k" id="MathJax-Element-36-Frame" role="presentation" style="position: relative;" tabindex="0">, k2=2k$k_2 = 2k$ $k_2 = 2k$

Third layer: h3=h,  h_3 = h" id="MathJax-Element-37-Frame" role="presentation" style="position: relative;" tabindex="0"> h3=h, $h_3 = h$ $h_3 = h$  k3=k k_3 = k" id="MathJax-Element-38-Frame" role="presentation" style="position: relative;" tabindex="0"> k3=k$k_3 = k$ $k_3 = k$

 Equivalent permeability, \({{\rm{K}}_{{\rm{eq}}}} = \frac{{{{\rm{H}}_1} + {{\rm{H}}_2} + {{\rm{H}}_3}}}{{\frac{{{{\rm{H}}_1}}}{{{{\rm{K}}_1}}} + \frac{{{{\rm{H}}_2}}}{{{{\rm{K}}_2}}} + \frac{{{{\rm{H}}_3}}}{{{{\rm{K}}_3}}}}}\)

\({{\rm{K}}_{{\rm{eq}}}} = \frac{{{h} + {{h}} + {{\rm{h}}}}}{{\frac{{{{\rm{h}}}}}{{{{\rm{k}}}}} + \frac{{{{\rm{h}}}}}{{{{\rm{2k}}}}} + \frac{{{{\rm{h}}}}}{{{{\rm{k}}}}}}}\)

Keq = 6k/5

Explanation:

1. Darcy's law describes the flow of fluid through porous media, and it is commonly applied in soil flow problems, particularly in hydrogeology and geotechnical engineering.
2. Darcy’s law relates the flow rate of water through soil to the hydraulic gradient and the soil's permeability.

 Additional Information

1. Hydraulic head: This is the total energy per unit weight of water at a point in the soil, and it consists of the elevation head and the pressure head.

2. Pressure head: This refers to the height of the water column that would exert the same pressure as the water at the point of interest.

3. Datum head: This is the reference level, typically the ground surface or sea level, from which other heads are measured.

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

\(K = \frac{{{V_s} \times n}}{i} = \frac{{1 \times 0.15}}{{\frac{3}{{100}}}} = 5\;m/day\)

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

∴ \(\frac{{{K_2}}}{{{K_1}}} = \frac{{{\gamma _2} \times {\mu _1}}}{{{\gamma _1} \times {\mu _2}}} = \frac{{0.8{\gamma _1} \times {\mu _1}}}{{{\gamma _1} \times 0.6{\mu _1}}} = 1.333\)

K₁ = 1.333 K₂

Concept:

The coefficient of permeability ‘k’ is given by:

\(k = C \times \frac{γ }{μ } \times \frac{{{e^3}}}{{1 + e}} \times {D^2}\)

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)

Calculation:

Given,

Viscosity reduced to 80% i.e μ₂ = 0.8 μ₁

Unit Weight reduced to 90%  γ₂ = 0.9 γ₁  

As 'k' is directly proportional to γ / μ

∴ \(\frac{{{k_2}}}{{{k_1}}} = \frac{{{γ _2} \times {μ _1}}}{{{γ _1} \times {μ _2}}} = \frac{{0.9{γ _1} \times {μ _1}}}{{{γ _1} \times 0.8{μ _1}}} = 1.125\)

k2 = 1.125 k1

Explanation:

Permeability:

It 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:

Different type of soil and their standard permeability rate:

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 ‘

Concept:

The capillarity rise in soil is calculated using empirical equation proposed by Terzaghi:

\({\rm{h}} = \frac{{\rm{C}}}{{{\rm{e}}{{\rm{D}}_{10}}}}\)

Where

 C is constant with value ranges from 0.1 to 0.5 Sq cm.

e is the void ratio.

D₁₀ is the effective grain size in cm.

Calculation:

Given

e_A = 2 e_B

(D₁₀)_A = 1/3 (D₁₀)_B

Now

\(\frac{{{h_A}}}{{{h_B}\;}} = \frac{{{{\left( {e{D_{10}}} \right)}_B}}}{{{{\left( {e{D_{10}}} \right)}_A}}}\)

Or

\(\frac{{{h_A}}}{{{h_B}}} = \frac{1}{2}\; \times \frac{3}{1}\)

\(\frac{{{h_A}}}{{{h_B}\;}} = 1.5\)

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:

(D₁₀)_A = 0.02 mm, (D10)_B = 0.04 mm, (h_c)_A = 60 cm

\(\therefore \frac{{{{\left( {{h_c}} \right)}_A}}}{{{{\left( {{h_c}} \right)}_B}}} = \;\frac{{{{\left( {{D_{10}}} \right)}_B}}}{{{{\left( {{D_{10}}} \right)}_A}}}\)

\(\therefore \frac{{60}}{{{{\left( {{h_c}} \right)}_B}}} = \frac{{0.04}}{{0.02}}\)

\(\Rightarrow {\left( {{h_c}} \right)_B} = 30\;cm\)

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:

For the different soil types as per grain size, the orders of magnitude of permeability are as follows:

From the table, Sand has the Coefficient of permeability ranges from 1 × 10– 2 to 5 × 10– 2 cm/s.

Concepts:

Hydraulic conductivity, also called permeability, is defined as the how much water can flow inside the soil voids. 

Factor affecting the permeability:

1. It depends on size of pores in soil mass as such on increasing the pore size friction losses will be less and hence, water can flow easily inside the soil. So, coarse sand has higher void size and hence, higher the permeability then that of fine sand.

2. Specific surface area: Higher the specific surface area, more will be the friction so water will not flow easily and hence, low will be the permeability.

3. Void ratio: Higher the void ratio, higher will be the permeability.

4. Fluid properties: It is inversely related to dynamic viscosity of fluid being flowed inside the soil and is directly related to unit weight of fluid.

6. Degree of Saturation: It is directly related to degree of saturation of soil i.e. higher the saturation, higher will be permeability.

7. Entrapped gasses: It is inversely proportional yo entrapped gasses i.e. more quantity of entrapped gasses reduces its permeability.

8. Adsorbed Water: Higher the presence of adsorbed water, lower will be permeability because lower the area is available to flow the fluid.

9. Foreign impurities: Higher the impurities lower will be the permeability because foreign substance causes obstruction in flow of water inside soil.

Explanation:

Flow Nets: It is graphical solutions to the Laplace equation for two-dimensional seepage. Two orthogonal sets of curves forming a flow net are:

• Equipotential lines → connecting points of the equal total head
• Flow lines → indicating the direction of seepage down under a hydraulic gradient.

Application of flow net is as follows:

1. Determination of seepage
2. Determination of hydrostatic pressure
3. Determination of seepage pressure
4. Determination of exit gradient