EMPIRICA L CORRELATION S FOR HYDRAULIC CONDUCTIVIT Y Granular Soil s

SOIL PERMEABILITY AND SEEPAGE

4.11 EMPIRICA L CORRELATION S FOR HYDRAULIC CONDUCTIVIT Y Granular Soil s

Some o f the factors that affect th e permeability are interrelated such as grain size, voi d ratio, etc.

The smaller th e grain size, the smaller the voids which leads t o the reduced siz e of flow channel s and lower permeability .

The average velocity of flow in a pore channel from Eq . (4.2b) is

8// 32/ /

where d is the average diameter o f a pore channel equal to 2R.

(4.29)

Eq. (4.29) expresses fo r a given hydraulic gradient /, the velocity of water in a circular pore channel is proportional to the square of the diameter of the pore channel. The average diamete r of the voids in a soil at a given porosity increases practically in proportion to the grain size, D

Extensive investigations of filter sand s by Hazen (1892) led to the equation

k(m/s) = CD ² (4.30)

where D^e is a characteristic effective grain size which was determined to be equal to D¹⁰ (10% size).

Fig. 4.10 gives a relationship between k and effective grain size D¹⁰ of granular soil which validates Eq. (4.30) . Th e permeabilit y dat a approximates a straight line with a slope equa l t o 2 consistent with Eq. (4.30). These data indicate an average value of C - 10~ ² where k is expressed i n m/s and D¹⁰ in mm. According t o the data in Fig. 4.10, Eq . (4.30) ma y underestimate or overestimate th e permeability o f granular soils by a factor of about 2.

Further investigations on filter sand s were carried ou t by Kenney et al., (1984). The y foun d the effective grain size D5 woul d be a better choice compared t o D}Q. Fig. 4.11 gives relationship s between D⁵ and k. The sand they used in the investigation had a uniformity coefficient ranging from

1.04 to 12.

Hydraulic Conductivit y a s a Functio n o f Voi d Rati o fo r Granula r Soil s Further analysis of hydraulic conductivity in granular soils based on Hagen-Poiseuille's Eq . (4.2b) leads t o interesting relationships between k and void ratio e. Three types of relationships ma y be expressed a s follows.

It can be shown that the hydraulic conductivity k can be expressed a s

k = kF(e) (4.31)

Silt Silty

Sand Sand

Fine Medium | Coarse Gravel

10-

10

~

o 10 "

-Qo_CJ

o

10"

10"

10"

0.002

Hazen equation Jt= 1/100 £)?⁰ m/sec

0.01 0.1 1

D¹⁰ (mm ) 10

Figure 4.10 Haze n equation an d data relatin g hydrauli c conductivit y an d D¹⁰ of granular soil s (after Louden, 1952 )

Soil Permeabilit y an d Seepage 105

Sand

Fine Medium] Coars e Sand

10"

T3

8 io -

-³

•o2

X 10-

10'

C, ,= 1 - 3

10-

10-

^{10° IO1}

D5 (mm)

Figure 4.11 Influenc e o f gradation o n permeability o n granular soil s (after Kenne y et al., 1984 )

where k = a soil constant depending on temperature and void ratio e . F(e) ma y be expressed a s

F(e) = ^o2e

l + e (4.32)

When e = 1, F(e) ~ 1 . Therefore k represent s th e hydraulic conductivity corresponding t o void ratio e - 1 . Since k i s assumed a s a constant, k is a function of e only.

By substituting in F(e), the limiting values, ;c = 0, x = 0.25, an d x = 0.5, we get For J c = 0,

x = 0.25,

(4.33)

(4.34)

x = 0.5 0

F,(e) represents the geometric mean of F.(e) an d F.(

The arithmetic mean of the functions F^e) an d F3(e) is

(4.35)

= e2 (4.36)

1000

u

o o o

Void ratio function

Figure 4.12 Relationshi p between voi d rati o an d permeability fo r coars e grained soils

Best Valu e fo r x fo r Coars e Graine d Soil s

From laboratory tests determine k for various void ratios e of the sample. Then plot curves k versus 2e^2(1+x)/(l + e) fo r values of x = 0, 0.25, 0.5 and k versus e². The plot that fits wel l gives the best value of x. It has been found fro m experimenta l results that the functio n

2e³

l + e (4.37)

gives better agreement than the other functions. However, the function F⁴(e) = e² is sometimes preferred because of its simplicity and its fair degree of agreement with the experimental data. Fig. 4.12 present experimental data in the form of k versus functions o f e.

Figure 4.13 I n situ permeabilit y o f sof t clay s in relation t o initia l void ratio , e^o; clay fraction; CF; an d activity A (Afte r Mesr i et al. , 1994 )

Soil Permeabilit y an d Seepage 107 3.5

3.0 2.5

•|2.0

1.0 0.5 0

O BatisconClay

A Berthierville D St . Hilaire V Vosb y

• Bosto n blue

10- 10" 10^,-8

Figure 4.14 Result s of falling-head and constant-head permeabilit y test s on undisturbed samples o f sof t clay s (Terzaghi , Pec k and Mesri, 1996 )

Fine Graine d Soil s

Laboratory experiment s hav e shown that hydraulic conductivity of very fin e graine d soil s ar e not strictly a function o f void ratio since there is a rapid decrease i n the value of k for clays below the plastic limit. This is mostly due to the much higher viscosity of water in the normal channels which results from the fact that a considerable portion of water is exposed t o large molecular attractions by the closely adjacen t solid matter. It also depends upon the fabric of clays especially thos e of marine origin whic h ar e ofte n flocculated. Fig . 4.13 show s that the hydraulic conductivity in the vertica l direction, at in situ void ratio e^Q, is correlated wit h clay fraction (CF) finer than 0.002 mm an d with the activity A (= Ip/CF).

Consolidation o f soft clay s may involve a significant decrease i n void ratio and therefore of permeability. The relationships between e and k (log-scale) fo r a number of soft clays are shown in Fig. 4.14 (Terzaghi, Peck, an d Mesri 1996) .

Example 4. 5

A pumping test was carried ou t for determining the hydraulic conductivity of soil in place. A well of diamete r 4 0 c m wa s drilled dow n t o an impermeabl e stratum . The dept h o f wate r abov e th e bearing stratu m was 8 m. The yield fro m th e well was 4 mVmin at a steady drawdow n of 4.5 m.

Determine th e hydraulic conductivity of the soil in m/day if the observed radiu s of influence was 150m.

Solution

The formula for determining k is [Eq. (4.18)]

k = 2.3 q

xD⁰(2H-D⁰) r 0

q = 4 m³/min = 4 x 60 x 24 m³/day

D⁰ = 4.5 m, H = 8 m, R . = 150 m, r ^Q = 0.2 m

k = 2.3x4x60x24

3.14x4.5(2x8-4.5) 0.log =2 234.4 m/day Example 4. 6

A pumping test was made in pervious gravels and sands extending to a depth of 50 ft, where a bed of cla y wa s encountered. The norma l groun d water leve l wa s a t the ground surface . Observatio n wells were located a t distances of 10 and 25 ft from th e pumping well. At a discharge o f 761 ft³ pe r minute fro m th e pumpin g well, a stead y stat e wa s attaine d in abou t 2 4 hr . The draw-dow n a t a distance of 1 0 ft was 5. 5 ft and a t 25 ft was 1.2 1 ft. Compute the hydraulic conductivity in ft/sec.

Solution

Use Eq. (4.16) where

where = —

60 = 12.683 ft³/sec

= 10 ft, r = 25 ft , h = 50 - 1.21 = 48.79 ft, h = 50 - 5. 5 = 44.5 ft k = 2.3x12.683 ,log — = 9.2 x 10 ⁿ 25 ^{o 1 A J}_ ft/sec .^{r /}

3.14(48.79²-44.5²) 1 0

Example 4. 7

A fiel d pumpin g tes t wa s conducte d fro m a n aquife r o f sand y soi l o f 4 m thicknes s confine d between two impervious strata. When equilibrium was established, 9 0 liters of water was pumped

Test well Observation wells

1 2

Impermeable stratum

} i _y

^.T

-^ — — r , = 3 m

^^

•^-

,..1 — " •

.

T

/i, =2.1 m

1 / / /

^

— T

T

_{/i? = 2.7 m}

1

Confined aquife r \

Impermeable

Figure Ex . 4.7

Soil Permeability and Seepage 10 9 out per hour. The water elevation in an observation wel l 3.0 m away from th e test well was 2.1 m and another 6.0 m away was 2.7 m from the roof level of the impervious stratum of the aquifer. Find the value of k of the soil in m/sec. (Fig. Ex. 4.7)

Solution Use Eq. (4.24a )

, ²³ <7 , ^r 2

k = - - - log—

q = 90 x 10³ cm³/hr = 25 x KT⁶ m³/sec 2.3x25xlO~⁶ , 6 ^{11/1 0 in} _⁶ .

k = - log— = 1.148 x 10 ⁶ m/sec

2x3.14x4(2.7-2.1) 3 Example 4. 8

Calculate th e yiel d pe r hou r fro m a well drive n int o a confined aquifer . The followin g data are available:

height of original piezometric leve l from th e bed of the aquifer, H = 29.53 ft, thickness of aquifer, Ha - 16.4 1 ft,

the depth of water in the well at steady state, hQ = 18.05 ft , hydraulic conductivity of soil = 0.079 ft/min ,

radius of well, rQ = 3.94 in . (0.3283 ft), radius of influence, R. = 574.2 ft.

Solution

Since hQ is greater tha n HQ th e equation for q (refer to Fig 4.7) is Eq. (4.24b )

where k = 0.079 ft/min = 4.74 ft/h r

2x3.14x4.74x16.41(29.53-18.05) ^ ^{1 M r},ⁿ ^ ^ ⁿ

Now q = - - - - = 75 1.87 ft3/hour « 752 ft3/hour

2.3 log(574.2 70.3283) Example 4. 9

A sand deposit contains three distinct horizontal layers of equal thickness (Fig. 4.9). The hydraulic conductivity o f the uppe r and lowe r layer s i s 10~3 cm/sec and tha t of the middl e i s 10~2 cm/sec.

What are the equivalent values of the horizontal and vertical hydraulic conductivities of the three layers, and what is their ratio?

Solution Horizontal flow

~(ki+k² +k³) sinc e z \ = Z² = £3

kh = -(10-³ +10-2 + 10-³) = -(2x 10~³ +10-2) = 4x 10-³ cm/se c Vertical flow

Z 3 3

i L i l i l _ L _ L _ L . 1 _ L

3kik, 3xlO~ _ _³xlO~2 1 / l < = 1.43 x 10 cm/se ca i n _³ . 2k² + ki

kh _ 4xlO~ ³ kv 1.43x10 " = 2.8

Example 4.1 0

The followin g details refe r to a test to determine the value of A ; of a soil sample : sampl e thickness

= 2.5 cm, diameter of soil sample = 7. 5 cm, diameter of stand pipe = 10mm, initial head of water in the stand pipe =100 cm, water level in the stand pipe after 3 h 20 min = 80 cm. Determine th e value of k if e = 0.75. What is the value of k of the same soil at a void ratio e = 0.90?

Solution

Use Eq. (4 . 1 3) where, k = ' ²'3aL lo g

4 (I)2 -0.785 cm2

A = — (7.5)3 14 ² = 44.1 6 cm² t= 1200 0 sec

By substituting the value of k for e{ = 0.75 , , 2.3x0.785x2. 5 , 10 0 ^rtcv ^ ^{i n} , ,

k = k,= - x log - = 0.826 x 10~⁶ cm/se c

1 44.16x1200 0 8 0

For determining k at any other void ratio, use Eq. (4.35 ) i ^e l

Now, k ² = - - x — x k^{ e

For e2 = 0.90 1.75 (0.9 V

= 190 ^X " ^X °'^{826 X 10 = l3146 X}

Soil Permeabilit y an d Seepage 111

Example 4.1 1

In a falling hea d permeameter, th e sample used i s 20 cm long having a cross-sectional are a of 2 4 cm2. Calculat e the tim e require d fo r a dro p o f hea d fro m 2 5 t o 1 2 cm i f th e cross - sectional are a o f th e stan d pip e i s 2 cm2. The sampl e o f soi l i s mad e o f thre e layers . Th e thickness of the firs t laye r from th e top is 8 cm and has a value of k{ = 2 x 10"4 cm/sec, th e second laye r o f thicknes s 8 cm has k2 = 5 x 10~4 cm/se c an d the botto m laye r o f thickness 4 cm ha s &3 = 7 x 10~4 cm/sec . Assume tha t th e flo w i s takin g plac e perpendicula r t o th e layers (Fig. Ex. 4.11).

Solution

Use Eq. (4.28)

k = 20

- + —+ • ^{O O} _l_ ______^^^^^ _ I _ ^{< 4}

2xlO~4 5xlO- 4 7xlO- 4

= 3.24xlO~4 cm/se c

Now from Eq . (4.13) , 2.3aL , h n

log—

or 2.3aL, hlog— = -logQ 2.3x2x2 0 , 2 5— Ak /i , 24x3.24xlO~ ⁴ 1 2

= 3771 sec = 62.9 minute s

8cm Laye r 1 I ^ = 2 x 10"4 cm/sec

8cm Laye r 2 1 ^ 2 = 5 x 10"1 cm/sec

£3 = 7xl0^cm/sec

Figure Ex . 4.11

Example 4.12

The data given below relate to two falling head permeameter tests performed on two different soi l samples:

(a) stan d pip e are a = 4 cm2, (b ) sampl e are a = 2 8 cm2, (c ) sampl e heigh t = 5 cm , (d) initial head in the stand pipe =100 cm, (e) final head = 20 cm, (f) time required for the fall of water level in test 1, t = 500 sec, (g) for test 2, t = 15 sec.

Determine the values of k for each of the samples. If these two types of soils form adjacent layers i n a natura l stat e wit h flow (a ) i n th e horizonta l direction, an d (b ) flo w i n th e vertical

direction, determine the equivalent permeability for both the cases b y assuming that the thickness of each laye r is equal to 15 0 cm.

Solution Use Eq. (4.13)

. 23aL h

k, =— -lo g = 2.3x10 ³ cm/sec

1 28x50 0 2 0 For test 2

2.3x4x5, 10 0

28x15 2 0

F/ovv in the horizontal direction Use Eq. (4.27)

= 76.7xlO~3³ cm/se c

= — -(2.3 x 150 + 76.7 x 150) x ID"³ =39.5xlQ-³ cm/se c

,jL/w

F/ow in the vertical direction Use Eq. (4.28)

Z² 15 0 15 0

fcT 2.3x10- 3 ⁺ 76.7x10-3

300 = 4.46 x 10"³ cm/sec

4.12 HYDRAULI C CONDUCTIVITY O F ROCKS B Y PACKER METHO D