Permeability and Seepage -2

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Permeability and Seepage -2

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Conditions favourable for the formation quick sand Quick sand is not a type of sand but a flow condition occurring within a cohesion-less soil when its effective stress is reduced to zero due to upward flow of water.

Quick sand occurs in nature when water is being forced upward under pressurized conditions.

In this case, the pressure of the escaping water exceeds the weight of the soil and the sand grains are forced apart. The result is that the soil has no capability to support a load.

Why does quick condition or boiling occurs mostly in fine sands or silts?

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Some practical examples of quick conditions

Excavations in granular materials behind cofferdams alongside rivers Any place where artesian pressures exist (i.e. where head of water is greater than the usual static water pressure). -- When a pervious underground structure is continuous and connected to a place where head is higher. Behind river embankments to protect floods

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Some practical examples of quick conditions

Contrary to popular belief, it is not possible to drown in quick sand, because the density of quick sand is much greater than that of water.

ρquicksand >> ρwater

Consequently, it is literally impossible for a person to be sucked into quicksand and disappear. So, a person walking into quicksand would sink to about waist depth and then float.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Boiling condition

• Example • The sand layer of the soil profile shown in figure Water level is under artesian pressure. 4m A trench is to be 6m h 5m excavated in the clay up to a depth of 4m. stiff clay γ = 18kN/m3 Determine the depth of P 3m water h to avoid boiling. Sand

Rock

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Boiling condition • Solution: • At P: σ = hγw + (6-4)γclay • = hγw + 2γclay Water level • At P: u = 5γw 4m ’ • At P: σ = hγw + 2γclay - 5γw 6m h = 1.33 m 5m • σ’ = 0 (boiling condition) stiff clay γ = 18kN/m3 • hγw + 2γclay - 5γw = 0 P • h = (5γ - 2γ )/γ 3m w clay w Sand • = (5 x 9.81 – 2x18)/9.81 Rock • h = 1.33m

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay B Seepage forces (F ) Water flowing past a h s c soil particle exerts a drag Wd force on the particle in the direction of flow.

L The drag force is caused by pressure gradient and by viscous drag.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Seepage forces B

(F ) Direction W s c of flow Wd

FBD of grain FBD of soil

At critical condition; h = hc h  G −1  Gs + e  c s  γ AL = (h + L)Aγ ic = =    1+ e  w c w L  1+ e 

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Considering FBD of grain in the direction of flow at i = ic

Wd = (Fs )c + B  G −1 (F ) = W − B (F ) =  s γ V s c d s c  1+ e  w  Gsγ w  (F ) =  AL −V γ (Fs )c = icγ wV s c  1+ e  s w  G γ  (V −V ) (F ) =  s w AL − V Vγ Seepage pressure s c  1+ e  V w ps = icγw  G γ  (F ) =  s w AL − (1− n)ALγ s c  1+ e  w

If h < hc then seepage force Js is iγwV Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Seepage force Area per unit width of section = ∆l x 1

h1 > h2 Volume affected by ∆h seepage force = ∆l2 x 1 h 1 h2

Direction of ∆l flow

Force applied to sand particles = γwh1 -γwh2 (∆l x 1) ∆l

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Seepage force

Force applied to sand particles = γw (∆h) (∆l x 1) 2 = γw (∆h/∆l) (∆l x 1)

J = i γw V

Seepage pressure ps = Seepage force per unit volume

= i γw

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Critical hydraulic gradient – Quick condition 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. (Shear stresses on the sides of the element are neglected.)  G −1 i =  s  c  υ 

 The critical hydraulic gradient is typically around 1.0 for many soils. Fluidized beds in chemical engineering systems rely on deliberate generation of quick conditions to ensure that the chemical process can occur most efficiently.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Example problem

Determine and plot total h stress, PWP and effective a stress diagrams if: (i) h = 1 m; (ii) h = 4 m; and (iii) h = 2 m. 2 m h b i = 1 m 3 2 γsat =20 kN/m c Case – I; i = 0.5; 1 m d Case – II; i = 2; Datum Case – III; i = 1

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Case - I i = 0.5

Point PH [m] EH [m] TH [m]

a 0 4 4

b 2 2 4

d 5 0 5

c 3.5 1 4.5

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Case - I i = 0.5 h = 1 m σ u σ´ a

2 m b 20 1 m 20 γ =20 kN/m3 c sat 1 m 40 35 5 d 60 50 10 Datum Stress units are in kN/m2

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Case - II i = 2

Point PH [m] EH [m] TH [m]

a 0 4 4

b 2 2 4

d 8 0 8

c 5 1 6

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Case - II Stress units are in kN/m2 h = 4 m σ u σ´ a

2 m b 20 1 m 20 γ =20 kN/m3 c sat -10 1 m 40 50 d -20 60 80 Datum Quick sand condition would have already occurred…

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Case - III i = 1 Point PH [m] EH [m] TH [m]

a 0 4 4

b 2 2 4

d 6 0 6

c 4 1 5

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Case - III Stress units are in kN/m2 h = 2 m σ u σ´ a 0

2 m b 0 20 1 m 20 γ =20 kN/m3 c sat 0 1 m 40 40 d 0 60 60 Datum Just subjected to Quick sand condition… Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Example problem

A large open excavation was made in a stratum of clay with a saturated unit weight of 17.6 kN/m3. When the depth of the excavation reached 7.5 m, the bottom rose, gradually cracked and was flooded from below by a mixture of sand and water. Subsequent borings showed that the clay was underlain by a bed of sand with its surface at a depth of 11 m.

Compute the elevation for which the water would have risen from the sand surface into a drill hole before the excavation was started.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Solution

11m

3 γsat = 17.6 kN/m

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Solution

Elevation for which the water would have risen 7.5 m from the sand surface into a drill hole before the excavation was h started. h = 6.16 m A

For σ´= 0 at A: σ´= 17.6 x (11 – 7.5) - 10 x h = 0

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Example problem Determine and plot the total vertical stress, pore water and effective vertical stress distribution at levels A, B and C. 4 m

6 m B 3 m

Silty clay; Gs = 2.70; w = 45.2 % C 3 m Coarse sand; Gs = 2.65; e = 0.833

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Example problem An artesian pressure 1 m exists in the lower 1 m γ = 18.5 kN/m3 sand layer. d

Determine σ, u, and 3 2 m γsat = 19 kN/m σ ′ at A, B, and C. A

(TH)A = 2 + 3 = 5m 1.33 m 2 m

(TH) = 7 + 0 = 7m 3 B C γclay = 17 kN/m 2 m 1 m B DATUM

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Solution At point B: σ = 18.5 x 1 + 19 x 2 + 17 x 3 = 107.5 kN/m2 u = 10 x 7 = 70 kN/m2 σ′ = 107.5 – 70 = 37.5 kN/m2

At point A: σ = 18.5 x 1 + 19 x 2 = 56.5 kN/m2 u = 10 x 2 = 20 kN/m2 σ′ = 56.5 – 20 = 36.5 kN/m2

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Solution At point c: σ = 18.5 x 1 + 19 x 2 + 17 x 2 = 90.5 kN/m2

2 u = 10 x (PH)c = 10 x 5.33 = 53.3 kN/m σ′ = 107.5 – 53.3 = 38.2 kN/m2

(TH)C = (TH)B - i Z = 7 – (2/3) x 1 = 6.33 m

(PH)C = (TH)C – Zc = 6.33 – 1 = 5.33 m

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay 1 m

1 m

2 m A 56.5 20 36.5 2 m

C 90.5 38.2 1 m 53.3 B DATUM 107.5 70 37.5 Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Measurement of soil permeabilities The rate of flow of water q (volume/time) through cross-sectional area A is found to be proportional to hydraulic gradient i according to Darcy’s law: q h v = = ki i = A L where v is flow velocity and k is coefficient of permeability with dimensions of velocity (length/time).  The coefficient of permeability of a soil is a measure of the conductance (i.e. the reciprocal of the resistance) that it provides to the flow of water through its pores.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Measurement of soil permeabilities

The value of the coefficient of permeability k depends on:

(i) Average size of the pores and is related to the

particle sizes and their packing, d = ed10 (ii) Particle shape, and d d = 10 (iii) Soil structure. 5 The ratio of permeabilities of typical sands/gravels to those of typical clays is of the order of 106. A small proportion of fine material in a coarse-grained soil can lead to a significant reduction in permeability.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Measurement of soil permeabilities A number of tests can be used to measure or estimate the permeability of soils Laboratory methods: The constant head test (used for highly permeable soils) The falling head test (used for relatively impermeable soils) Indirect methods: computation from grain size distribution and During Oedometer test Field methods: 1. Pumping tests and 2. Borehole tests Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Laboratory measurement of permeability In determining permeability of coarse grained material, large quantities of flow occur in short periods of time and small quantities of flow occur over long periods of time for fine grained soils. Two aspects that need careful attention for all types of soils are: (i) To ensure that flow occurs only through the soil and not at the interface between soil and the mould in which the soil is contained. (ii) The soil sample is fully saturated before recording observations

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Determination of coefficient of permeability in the laboratory Water Constant supply water level Constant Head test  Q = Avt A = area of h C/S of the h i = L L specimen  h  Q = Ak t  L  QL k = Aht

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Main features of constant head test

It is suitable for soils having a coefficient of permeability in the range of 10-2 m/s to 10-5 m/s, which applies to clean sand and sand-gravel mixtures with less than 10 % fines.

It can be suitable for soils when used in their completely disturbed or remolded states such as for drainage materials and filters to confirm that their performance will be adequate.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay dh Falling head test or Variable head test

Cross-sectional area of stand pipe = a h1 h h2 Cross-sectional area of specimen = A

At time to head h1

L At time t1 head h2

Let h be the head of water at any time t. Let in time dt the head drop by amount dh

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Falling head test Quantity of water flowing through the sample in time dt from Darcy’s law: dQ = kiA(dt) Quantity of discharge h can also be = k A(dt) dQ = −a(dh) L expressed as:

h2 kA t1 As the time − a dh = dt increases ∫ ∫ = h1 L t t0 head decreases!! aL  h  =  1  k 2.303 log10   At  h2 

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Main features of constant head and falling head tests In the constant head test, permeability is computed on the basis of fluid that passes through the soil sample.

While in the falling head test, k is computed on the basis of fluid flowing into the sample. With the constant head test, time is required to accumulate the fluid volume necessary to perform computation. Extreme care would be required to prevent leaks in the apparatus and evaporation of discharge water.

With the falling head test, the duration of the test is shortened and care is required to prevent evaporation of water in the inlet tube. Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay Problem Determine and plot the total vertical stress, pore water and effective vertical stress distribution at levels A, B and C. 4 m A

6 m B 3 m

Silty clay; Gs = 2.70; w = 45.2 % C 3 m Coarse sand; Gs = 2.65; e = 0.833

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay