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Exercise 6.A pump test is carried out to determine the hydraulic conductivity of a confined aquifer, as shown in Figure P6 on page 128. Show that the equation for k isቀnullቁሺ ሻSolution 6.() ()∫∫ቀnullቁሺ ሻDetermine the pressure head, elevation head, and total head at A, B, and C for the arrangement sh...

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Exercise 6.1 A pump test is carried out to determine the hydraulic conductivity of a confined aquifer, as shown in Figure P6.1 on page 128. Show that the equation for k is

(

( )

)

Solution 6.1

(

)

( ∫

∫

(

( )

)

)

Exercise 6.2: Determine the pressure head, elevation head, and total head at A, B, and C for the arrangement shown in Fig. P6.2. Take the water level at exit as datum. Hint: You need to convert the pressure 10 kPa to head. Solution 6.2:

As we are interested in the heads at A, write Bernoulli’s equation between points B and A. 10 kPa P  0.9 m  A  0.25 m  h L B A 3 9.81 kN / m w

(1)

For hL(B-A), write Bernoulli’s equation from point B to point C:

10 kPa  0.9 m  0.0 m  0.0 m  h L B C  1.02 m  0.9 m  1.92m 9.81 kN / m3

(2)

The head loss is linear in the system. Hence, after moving through half of the length of the soil, half of the head loss has occurred. hL(B-A) = 1.92 m / 2 = 0.96 m. Now, insert this result back into (1):

10 kPa P  0.9 m  A  0.25 m  0.96 m 3 w 9.81 kN / m PA  1.02 m  0.9 m  0.25 m  0.96 m  0.71 m w Present all results in a table: Point A B C (datum)

Elev. Head (m) 0.25 0.90 0.0

Pressure Head (m) 0.71 1.02 0.0

Total Head (m) 0.96 1.92 0.0

Exercise 6.3: The groundwater level in a soil layer 10 m thick is located at 3 m below the surface. (a) Plot the distribution of hydrostatic pressure with depth. (b) If the groundwater were to rise to the surface, plot on the same graph as (a), using a different line type, the distribution of hydrostatic pressure with depth. (c) Repeat (b), but the groundwater is now 2 m above the ground surface (flood condition). Interpret and discuss these plots with respect to the effects of fluctuating groundwater levels. Solution 6.3 Hydrostatic pressure = w x depth

2m

3m

10 m

Increase in groundwater elevation decrease the effective stresses ( =  – u) where u = hydrostatic pressure.

Exercise 6.4: In a constant-head permeability test, a sample of soil 12 cm long and 6 cm in diameter discharged 1.5 × 10−3 m3 of water in 10 minutes. The head difference in two manometers A and B located at 1 cm and 11 cm, respectively, from the bottom of the sample is 2 cm. Determine the hydraulic conductivity of the soil. What is the soil type tested? Solution 6.4 Cross sectional area of soil Although the length of the soil sample is 12cm, the distance over which the head was measured is 11 – 1 = 10 cm Therefore, the hydraulic gradient,

⁄ The constant head permeability test is used to determine the hydraulic conductivity of coarsegrained soils. Soil types: clean sands, clean sand and gravel mixtures.

Exercise 6.5: A constant-head test was conducted on a sample of soil 15 cm long and 60 cm2 in crosssectional area. The quantity of water collected was 50 cm3 in 20 seconds under a head difference of 24 cm. Calculate the hydraulic conductivity. If the porosity of the sand were 55%, calculate the average velocity and the seepage velocity. Estimate the hydraulic conductivity of a similar soil with a porosity of 35% from the results of this test. Solution 6.5:

This test was completed at n = 0.55. For sands, Taylor (1948) presented a relationship between k and the void ratio, e. k 

e

e3 e3 ; or k  (constant) 1  e 1  e

n 0.55  1.22 ; e n   1 1 0.55

Solving for the constant, we get constant = 4.24 x 10-2 cm/sec Hence, for the same soil at another void ratio, the same constant may be used. Hence, the following is calculated for k at a porosity of 0.35 ( e = 0.54): k  (4.24 10 2 cm / sec)

0.543  4.34 10 3 cm/sec. 1  0.54

Exercise 6.6: A falling-head permeability test was carried out on a clay soil of diameter 100 mm and length 150 mm. In 1 hour the head in the standpipe of diameter 5 mm dropped from 680 to 502 mm. Calculate the hydraulic conductivity of this clay. Solution 6.6 Cross sectional area of soil;

Standpipe cross sectional area;

(

)

(

)

⁄

Exercise 6.7: Calculate the equivalent hydraulic conductivity for the soil profile shown in Fig. P6.7. Solution 6.7

Equivalent vertical hydraulic conductivity: ( )

∑

(

∑

)

⁄

(

)

⁄

⁄ ⁄

⁄

You should observe that the flow in this case is controlled (primarily) by the layer with the lowest hydraulic conductivity – the bottom layer. Equivalent horizontal hydraulic conductivity: ∑ ( ) ∑ ( ⁄ )( ) ( ⁄ )( ) ( ( ) ⁄

⁄ )( )

You should observe that the flow is controlled (primarily) by the layer with the highest hydraulic conductivity – the top layer. √( )

( )

√

⁄

Exercise 6.8: A pumping test was carried out to determine the average hydraulic conductivity of a sand deposit 20 m thick overlying impermeable clay. The discharge from the pumping well was 10 × 10 −3 m3/s. Drawdowns in the observation wells located 15 m and 30 m from the centerline of the pumping well were 2.1 m and 1.6 m, respectively. Groundwater table was reached at 3.2 m below the ground surface. Determine the hydraulic conductivity of the sand. Estimate the effective grain size using Hazen’s equation. Solution 6.8 Use simple well formula. r1 = 15 m, r2 = 30 m, h1 = 14.7 m, h2 = 15.2 m (

(

)

)

⁄

Effective grain size (D10) Assume C = 1 √

Exercise 6.9: An excavation is proposed for a square area near the bend of a river as shown in Fig. P6.9. It is expected that the flow of water into the excavation will come through the silt layer. Pumping tests reveal an average horizontal hydraulic conductivity of 2 × 10−5 cm/s in the silt layer. The excavation has to be kept dry. Determine the flow rate (qi) into the excavation. Solution 6.9 12.8 m

8m Average width = 10.4 m

Flow will occur from two sides of excavation The average width is 8 + 2 (1.2 x 1) = 10.4 m Area =(10.4 x 2 ) x 2 = 41.6 m 2 ; k = 2 × 10−7 m/s ⁄

⁄

Exercise 6.10: Groundwater is pumped for domestic use from an unconfined aquifer sand layer. The thickness of the clay layer above the sand layer is 20 m and its initial porosity is 40%. After 10 years of pumping, the porosity is reduced to 30%. Determine the subsidence of the clay surface. Solution 6.10 The subsidence is due to the change in void ratio and since the volume of solids remains constant, we can write:

hf

h  (h v ) i  s  hi hs  ( hv ) f

1

(h v)

f

1  ef hs  (h ) 1  ei 1 v i hs

where e is the void ratio, h is the thickness of the layer, i = initial, f = final , s = solids and v is the voids.

ni 0.4 2   1 ni 1  0.4 3 nf 0.3 3 ef    1  n f 1  0.3 7 3 1  e f hf 1 7 h f    1 e i h i 1  2 20 3 h f  0.857  20  17 .14 m ei 

subsidence  20  17 .14  2.86 m Alternatively,

1 n i h f  1 nf hi 1  0.4 hf  1  0.3 20 h f  17 .14 m subsidence  20  17 .14  2.86 m

Exercise 6.11 A canal is dug parallel to a river as shown in Fig. P6.11. A sandy-silt seam of average thickness 0.5 m cuts across the otherwise impermeable clay. The average vertical and horizontal hydraulic conductivities are 1.5 × 10 −5 cm/s and 15 × 10−5 cm/s respectively. Assume a 1 m length of canal, determine the flow rate of water from the canal to the river. Solution 6.11

A

B

Equivalent hydraulic conductivity: √( √ Select B as Datum

)(

)

⁄

Heads at B Elevation head: hZB = 0 Pressure head: hPB = 98 - 96.48 = 1.52 m Total head at B = 1.52 m Heads at A Elevation head : hZA = 99 - 96.48 = 2.52 m Pressure head: hPA = 0.5 m Total head at A = 3.02 m Head loss from A to B = 3.02 – 1.52 = 1.5 m L  2.52 2  30 2  30.1m H 1.5 i   0.0498 L 30.1

L 2.52 m 30m

A (cross-sectional area) per 1 m length = 0.5  1  0.5m

2

3

q (flow rate) = kiA =  4.74 105 /100  0.0498  0.5  1.18 108 m

sec

Exercise 6.12 An excavation is made for a canal that is fed by a stream, as shown in Figure P6.12. The measured flow into the canal is 0.25 x 10-4 m3/s per unit area. Two porewater pressure transducers, A and B, placed along a line parallel to the slope and approximately at the canal mid height gave readings of 3 kPa and 2.5 kPa. Assuming flow parallel to the slope, estimate the hydraulic conductivity.

Solution 6.12 hp = Pressure head difference = p/w = (3-2.5)/9.8 = 0.051m Let  be the slope angle in radians

q = Aki ; A = 1 m2

Slope = 20°

Exercise 6.13: A well, 0.1 m radius, is part of a wellpoint network to keep an excavation dry (Fig. P6.13). The groundwater at the far edge of the excavation must be 0.5 m below the base. a) Calculate the radius of influence. b) Calculate the maximum drawdown. c) Plot the drawdown curve. d) For the radius of influence in (a), (i) calculate the discharge if the well radius increases to 0.2 m and (ii) compare it to the discharge for the 0.1 m radius well.

Solution 6.13 a) Radius of influence The drawdown at a radius of 9.1 m is 2.5 m q w ln(R ) 2 r d H H  k 2

2.5  8  8 

13.2 10 4 ln( R 9.1) 5.8 105 

R  436 m b) Maximum Drawdown occurs at well face, i.e. at r  ro  0.1m √

(

)

√

(

)

(c) Drawdown curve H

8

qw R k

0.00132 436 0.000058

R/r r d

4360 0.1 6.2

0

2

m m3/s m m/s 2180 0.2 5.1

1090 0.4 4.3 radius (m) 6

4

545 0.8 3.7

8

272.5 1.6 3.2

10

136.25 3.2 2.7

12

0.0

Drawdown (m)

1.0 2.0 3.0 4.0 5.0 6.0 7.0

d) For the same radium of influence and a well radius on 0.2 m Assume the drawdown at the well radius is the same. (i)

dmax

qw ln( R ) ro H H  k 2

2

6.18  8  8 

qw ln( 436

) 0.2

5.8  10 5  qw  143 .7 10 5 m 3 / s

(ii) The ratio of the flow is 143.7 x 10-5/13.2 x 10-4 = 1.1

14

68.125 34.0625 6.4 12.8 2.2 1.8

m m...