PRIN.OF GEOTECHNICAL...-MINDTAP(2 SEM)
PRIN.OF GEOTECHNICAL...-MINDTAP(2 SEM)
9th Edition
ISBN: 9781305971271
Author: Das
Publisher: CENGAGE L
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Chapter 7, Problem 7.9P

(a)

To determine

Find the absolute permeability of the soil.

(a)

Expert Solution
Check Mark

Answer to Problem 7.9P

The absolute permeability of the soil is 8.4×1013m2_.

Explanation of Solution

Given information:

The length of the soil sample L is 400 mm.

The area of the sample A is 7,854mm2.

The diameter of the standpipe (d) is 11 mm.

The head difference (h1) at t=0 is 450 mm.

The head difference  (h2) at t=8min is 200 mm.

The unit weight of water (γw) is 9.789kN/m3.

The dynamic viscosity of water (η) is 1.005×103Ns/m2.

Calculation:

Determine the area of the standpipe a using the relation.

a=πd24

Substitute 11 mm for d.

a=π(11)24=95.03mm2

Determine the hydraulic conductivity k using the relation.

k=2.303(aLAt)log10(h1h2)

Substitute 95.03mm2 for a, 400 mm for L, 7,854mm2 for A, 8 min for t, 450 mm for h1, and 200 mm for h2.

k=2.303(95.03×4007,854×8min×60sec1min)log10(450200)=8.18×103mm/s×1cm10mm=8.18×104cm/sec

Determine the absolute permeability of the soil using the relation.

k=γwηK¯

Substitute 8.18×104cm/sec for k, 9.789kN/m3 for γw, and 1.005×103Ns/m2 for η.

8.18×104cm/sec×1m100cm=9.789×1031.005×103×K¯8.18×106×1.005×1039.789×103=K¯K¯=8.4×1013m2

Therefore, the absolute permeability of the soil is 8.4×1013m2_.

(b)

To determine

Find the head difference at 4 min time duration.

(b)

Expert Solution
Check Mark

Answer to Problem 7.9P

The head difference at 4 min time duration is 30cm_.

Explanation of Solution

Given information:

The length of the soil sample L is 400 mm.

The area of the sample A is 7,854mm2.

The diameter of the standpipe (d) is 11 mm.

The head difference (h1) at t=0 is 450 mm.

The head difference (h2) at t=8min is 200 mm.

The unit weight of water (γw) is 9.789kN/m3.

The dynamic viscosity of water (η) is 1.005×103Ns/m2.

Calculation:

Determine the head difference at 4 min time duration using the relation.

k=2.303(aLAt)log10(h1h2)

Substitute 8.18×104cm/sec for k, 95.03mm2 for a, 400 mm for L, 7,854mm2 for A, 8 min for t, and 450 mm for h1.

8.18×104cm/sec×10mm1cm=2.303(95.03×4007,854×4min×60sec1min)log10(450h2)8.18×1030.0464=log10(450h2)10(0.176)=(450h2)h2=4501.5

h2=300mm×1cm10mmh2=30cm

Therefore, the head difference at 4 min time duration is 30cm_.

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