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Concept explainers
(a)
Find the absolute permeability of the soil.
(a)
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Answer to Problem 7.9P
The absolute permeability of the soil is 8.4×10−13 m2_.
Explanation of Solution
Given information:
The length of the soil sample L is 400 mm.
The area of the sample A is 7,854 mm2.
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=8 min is 200 mm.
The unit weight of water (γw) is 9.789 kN/m3.
The dynamic viscosity of water (η) is 1.005×10−3 N⋅s/m2.
Calculation:
Determine the area of the standpipe a using the relation.
a=πd24
Substitute 11 mm for d.
a=π(11)24=95.03 mm2
Determine the hydraulic conductivity k using the relation.
k=2.303(aLAt)log10(h1h2)
Substitute 95.03 mm2 for a, 400 mm for L, 7,854 mm2 for A, 8 min for t, 450 mm for h1, and 200 mm for h2.
k=2.303(95.03×4007,854×8 min×60 sec1 min)log10(450200)=8.18×10−3 mm/s×1 cm10 mm=8.18×10−4 cm/sec
Determine the absolute permeability of the soil using the relation.
k=γwηˉK
Substitute 8.18×10−4 cm/sec for k, 9.789 kN/m3 for γw, and 1.005×10−3 N⋅s/m2 for η.
8.18×10−4 cm/sec×1 m100 cm=9.789×1031.005×10−3×ˉK8.18×10−6×1.005×10−39.789×103=ˉKˉK=8.4×10−13 m2
Therefore, the absolute permeability of the soil is 8.4×10−13 m2_.
(b)
Find the head difference at 4 min time duration.
(b)
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Answer to Problem 7.9P
The head difference at 4 min time duration is 30 cm_.
Explanation of Solution
Given information:
The length of the soil sample L is 400 mm.
The area of the sample A is 7,854 mm2.
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=8 min is 200 mm.
The unit weight of water (γw) is 9.789 kN/m3.
The dynamic viscosity of water (η) is 1.005×10−3 N⋅s/m2.
Calculation:
Determine the head difference at 4 min time duration using the relation.
k=2.303(aLAt)log10(h1h2)
Substitute 8.18×10−4 cm/sec for k, 95.03 mm2 for a, 400 mm for L, 7,854 mm2 for A, 8 min for t, and 450 mm for h1.
8.18×10−4 cm/sec×10 mm1 cm=2.303(95.03×4007,854×4 min×60 sec1 min)log10(450h2)8.18×10−30.0464=log10(450h2)10(0.176)=(450h2)h2=4501.5
h2=300 mm×1 cm10 mmh2=30 cm
Therefore, the head difference at 4 min time duration is 30 cm_.
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Chapter 7 Solutions
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