Seepage Analysis MCQ Quiz - Objective Question with Answer for Seepage Analysis - Download Free PDF

Explanation:

Zoned Dam

• Zoned darns usually have a central zone of selected soil material, to form a relatively impermeable core, a transition zone along both faces of the core to prevent piping through cracks which may form in the core, and outer zones of more pervious material for stability.
• This construction is widely used in earth dams and is selected whenever suitable materials are available.
• Clay, even though highly impermeable, may not make the best core if it shrinks and swells too much.
• The most satisfactory cores are of clay mixed with sand and fine gravel

Explanation:

• Seepage velocity (vs): This is the velocity at which the fluid actually moves through the voids in the material.

• Superficial velocity (v): This is the velocity of the fluid if it were to move through the entire cross-sectional area (including both the solid and void portions of the material).

The relationship between seepage velocity and superficial velocity is given by the equation: v = Avvs/A

Where:

• A_v $A_v$ ​ is the area of voids in the cross-section.

• A $A$  is the total area of the cross-section (including both solids and voids).

• v_s $v_s$  is the seepage velocity.

• v $v$  is the superficial velocity.

Additional InformationSeepage Velocity

• Definition: Seepage velocity refers to the actual velocity of water or fluid that moves through the void spaces (pores) of a material, such as soil or a porous medium.

• Importance: It is used to understand how quickly fluid travels through the porous medium, which is crucial in applications like groundwater flow, soil permeability tests, and drainage.

• Relation to Porosity: Seepage velocity is dependent on the porosity and the permeability of the material. Higher porosity generally results in higher seepage velocity for the same hydraulic gradient.

Explanation:

Clay:

• Clay is commonly used for the core of embankments because it has very low permeability, which helps minimize seepage.

• The fine particles in clay make it highly effective at reducing the movement of water, making it ideal for controlling seepage in embankment dams.

Additional Information Sand:

• Sand has relatively high permeability, which makes it unsuitable for the core of an embankment where seepage control is critical.

• It would allow water to pass through more easily, undermining the stability of the embankment.

Silty Sand:

• Silty sand has a higher permeability than clay, though not as much as pure sand.

• It may still allow some seepage, making it less ideal for minimizing water movement compared to clay.

Gravel:

• Gravel has very high permeability, allowing significant water flow.

• It is typically used in the outer layers or drainage zones of embankments but not in the core, where seepage control is essential.

Explanation:

• When flow takes place in an upward direction, seepage pressure (ps) also acts in an upward direction and effective stress is reduced.
• If seepage pressure equals the submerged weight of soil mass then effective stress reduces to zero. In such a case, cohesionless soil loses all its shear strength and has the tendency to flow along with water. This phenomenon is known as the quicksand condition.

For quicksand / Piping condition:

Effective stress = γ'z - P_s = 0

γ'z = P_s = i × z × γ_w

Where P_s = Seepage pressure = i z γ_w

γ = i × γ_w

i = \(\left ( \frac{\gamma '}{\gamma _{w}} \right )=\frac{G-1}{1+e}.\gamma _{w}\)

i_c = \(\frac{G-1}{1+e}\)

Note:

The hydraulic gradient at which quicksand condition occurs is termed as the critical hydraulic gradient piping gradient, floating gradient, or bursting gradient.

Important Points

• For piping, the hydraulic gradient should be more than or equals to the critical hydraulic gradient.  
• For most of the soils, the critical hydraulic gradient is close to unity. So when the hydraulic gradient is nearly unity piping will occur.

Explanation:

Quick Sand Condition

Quick sand condition is a state where saturated soil loses its strength and behaves like a liquid due to the upward flow of water. This phenomenon occurs in certain types of soil under specific conditions.

Analyzing the Given Options

1. "Quick sand condition is most likely to occur in silt and fine sand." (Correct)

   • Quick sand condition primarily occurs in fine-grained soils such as silt and fine sand.

   • These soils are susceptible to liquefaction under certain conditions.

2. "Quick sand is not a special type of soil." (Correct)

   • Quick sand condition refers to a state of soil rather than a specific type of soil.

   • It occurs when the soil becomes saturated with water and loses its strength.

3. "Cohesive soil will become quick as it still possesses some strength equal to cohesion intercept." (Incorrect)

   • Cohesive soils have significant cohesion, which prevents them from becoming quick sand.

   • Quick sand condition typically occurs in cohesionless soils that lack significant cohesion.

   • The given statement contradicts this fact, making it the correct answer.

4. "This condition occurs in cohesionless soils only." (Correct)

   • Quick sand condition is common in cohesionless soils like sand and silt.

   • These soils lose their strength when saturated with water and subjected to upward hydraulic pressure.

Given,

Discharge velocity (V) = 6 × 10^-7 m/s

Void ratio = 0.5

Then porosity (n)

\(n = \frac e {1 + e}\)

\(n = \frac {0.5} {1 + 0.5} = \frac {0.5}{1.5} = \frac 1 3\)

seepage velocity (V_s) = ?

seepage velocity V_s \(= \frac V n = \frac {6 × 10^{-7}}{1 / 3}\)

V_s = 18 × 10^-7 m/s

Explanation:

In the process of loading, initially, as we increase the load the pore water pressure increases, due to which pore water escapes out from the voids of the soil.

Hence, the pore water pressure in the soil sample of the consolidometer test is maximum at the center and when the primary consolidation is completed, which leads to the expulsion of pore water from the voids of the soil stops then the excess pore water process developed due to the application of load reduced to zero.

Quicksand condition:

(i) This process in which soil particles are lifted over the soil mass is called quicksand condition. It is also known as 'boiling of sand' as the surface of sand looks it is boiling.

(ii) At quick sand condition, net effective stress is reduce to zero, i.e.

σ̅ = 0

∴ γ'z = ps = 0     {ps = Seepage pressure}

upward seepage pressure in the soil becomes equal to submerged unit weight of the soil

or, γ'z - izγw = 0

i = icr = γ'/γw

(iii) The hydraulic gradient under which quick sand condition occurs is termed as a critical hydraulic gradient. If void ratio and specific gravity of soil is known, then

\(i = {i_{cr}} = \frac{{\gamma '}}{{{\gamma _w}}} = \frac{{\frac{{\left( {G - 1} \right){\gamma _w}}}{{1 + e}}}}{{{\gamma _w}}}\)  

⇒ \(i = \frac{{\left( {G - 1} \right)}}{{1 + e}}\)

(v) quicksand condition can be prevented by lowering the water table at the site before excavation.

• It can also be prevented by increasing the upward flow length by providing a sheet pile wall.
• Another method adopted at sites to avoid quicksand Conditions is the addition of surcharge (i.e additional weight) on the excavation side to increase the net weight of the soil mass.

(i) Quick sand condition is generally observed when excavation is done below the GWT and water is pumped out to keep the excavated area free from it or it is observed when sand is under artesian pressure and overlain by a clay layer. 

Phreatic Line:

The top flow line of a saturated soil mass below which seepage takes place, is called the Phreatic line.

• The hydrostatic pressure on the Phreatic line within a dam section is equal to atmospheric pressure
• This line separates a saturated soil mass from an unsaturated soil mass
• It is not an equipotential line, but a flow line

Concept:

The velocity potential function may be defined as a scalar function of space and time such that its derivative with respect to any direction gives the fluid velocity in that direction.

Velocity potential function (φ) satisfies the Laplace equation.

So \(\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {y^2}}} = 0\) 

The solution of the above equation gives two sets of curves known as equipotential lines and stream lines (or flow lines), mutually orthogonal to each other as shown in figure.

The equipotential lines represent contours of equal head (potential).

The direction of seepage is always perpendicular to the equipotential lines.

The path along which the individual particles of water seep through the soil are called streamlines or flow lines.

Concept:

Critical hydraulic gradient (ic): 

The quick condition occurs at a critical upward hydraulic gradient ic, when the seepage force just balances the buoyant weight of an element of soil. The critical hydraulic gradient is typically around 1.0 for many soils.

At the critical conditions, the effective stress is equal to zero.

The critical hydraulic gradient (i_c) is given by,

\({{\rm{i}}_{\rm{c}}} = \frac{{{\rm{G}} - 1}}{{1 + e}} \)

where,

G - Specific gravity of soil

e - Void ratio of soil

Calculation:

Given: G = 2.65, e = 0.65, h = 1.8 m

Critical hydraulic gradient,

\({{\rm{i}}_{\rm{c}}} = \frac{{{\rm{G}} - 1}}{{1 + e}} = \frac{{2.65 - 1}}{{1 + 0.65}} = 1\)

Head required

= i_c × thickness of sand stratum

Concept:

Flow net: It is a graphical representation of two dimensional steady state flow consisting of two sets of lines – the flow and the equipotential lines.

In case of isotropic soil these lines are orthogonal to each other but need not be orthogonal in anisotropic soil.

Various methods of obtaining flow nets are:

1) Graphical method

2) Electrical flow Analogy method

3) Sand Model

4) Capillary flow Model

5) Solution of Laplace equation/Analytical method.

Concept:

Critical hydraulic gradient (ic): The quicksand / boiling condition occurs at a critical upward hydraulic gradient typically around 1.0 for many soils, when the seepage force just balances the buoyant weight of an element of soil.

At the critical conditions, the effective stress is equal to zero.

\({i_c} = \frac{{G - 1}}{{1 + e}}\;{\rm{OR\;}}\left( {{\rm{G}} - 1} \right) \times \left( {1 - {\rm{n}}} \right)\)

Calculations:

Critical hydraulic gradient

\(\begin{array}{*{20}{c}} {{i_c} = \left( {{\rm{G}} - 1} \right) \times \left( {1 - {\rm{n}}} \right) = 1.65 \times \left( {1 - 0.45} \right) = 0.9075} \end{array}\)

i_s = i_c/4 = 0.9075/4 = 0.225

Explanation:

Flow net:

(i) The entire pattern of flow lines and equipotential lines is referred as a flow net. It is the solution of Laplace's equation for relevant boundary conditions. Thus, a flow net is a graphical representation of the head and direction of seepage at every point.

(ii) Flow net is a function of only boundary condition and is independent of permeability, head loss, or number of flow lines and equipotential lines. Hence the flow length will alter if permeameter length is changed.

Additional Information

Properties of flow net:

(i) Flow lines and equipotential lines meet each other orthogonally.

(ii) There can be no flow across the flow line and velocity of flow is always perpendicular to equipotential lines.

(iii) Area bounded between two adjacent flow lines is called flow channel or flow path and quantity of water flowing through each channel is same.

(iv) Area bounded between two adjacent equipotential lines and adjacent flow lines are referred to as a flow field.

(v) For isotropic medium, flow field is rectangular which may either be linear or curvilinear.

(vi) Loss of head between two equipotential lines i.e. equipotential drop remain constant for all the equipotential lines. 

Concept:

Determination of seepage discharge when the medium is isotropic:

Seepage discharge is given by,

 \(q = kh\frac{{{N_f}}}{{{N_d}}}\)

Where h = Hydraulic head or head difference between upstream and downstream level or head loss through the soil

Nf = Total number of flow channels = Floe lines - 1

Nd = Total number of equipotential drops = Equipotential lines -1

k = Coefficient of permeability

Calculations:

Given data

Nf = 5 - 1 = 4, Nd = 9 - 1 = 8, h = 20 m = 2000 cm

k = 2.6 × 10-6 cm/sec

∵ seepage discharge(q),

 \(q = kh\frac{{{N_f}}}{{{N_d}}}\)

\(q = 2.6 × {10^{ - 6}} × 2000 × \frac{4}{{8}} \)

q = 2.60 × 10-3  cm2/sec