Principles of Geotechnical Engineering (MindTap Course List)
Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781305970939
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
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Chapter 7, Problem 7.1CTP

(a)

To determine

Find the quantity of water flowing through the sample per hour.

(a)

Expert Solution
Check Mark

Answer to Problem 7.1CTP

The quantity of water flowing through the sample per hour is 536.0cm3/hr_.

Explanation of Solution

Given information:

The length of each soil layer (H1,H2,H3) is 200 mm.

The total length of the soil layer H is 600 mm.

The diameter of the cylindrical tube d is 150 mm.

The constant head difference Δh is 470 mm.

The porosity of the soil layer I (n1) is 0.5.

The porosity of the soil layer II (n2) is 0.6.

The porosity of the soil layer III (n3) is 0.33.

The hydraulic conductivity of soil layer I (k1) is 5×103cm/sec.

The hydraulic conductivity of soil layer II (k2) is 4.2×102cm/sec.

The hydraulic conductivity of soil layer III (k3) is 3.9×104cm/sec.

Calculation:

Determine the hydraulic conductivity in the vertical direction using the relation.

kV(eq)=HH1k1+H2k2+H3k3

Substitute 600 mm for H, 200 mm for H1, 5×103cm/sec for k1, 200 mm for H2, 4.2×102cm/sec for k2, 200 mm for H3, and 3.9×104cm/sec for k3.

kV(eq)=600mm×1cm10mm200mm×1cm10mm5×103+200mm×1cm10mm4.2×102+200mm×1cm10mm3.9×104=604,000+476.19+51,282.05=1.076×103cm/sec

Determine the hydraulic gradient using the relation.

i=ΔhL

Here, L is the total length of the soil layer.

Substitute 470 mm for Δh and 600 mm for L.

i=470600=0.783

Determine the area of the cylindrical tube using the relation.

A=πd24

Substitute 150 mm for d.

A=π(150mm×1cm10mm)24=176.71cm2

Determine the rate of seepage per unit length of the dam using the relation.

q=kV(eq)iA

Substitute 1.076×103cm/sec for kV(eq), 0.783 for i, and 176.71cm2 for A.

q=1.076×103×0.783×176.71=0.149cm3/sec×3,600sec1hr=536.0cm3/hr

Therefore, the quantity of water flowing through the sample per hour is 536.0cm3/hr_.

(b)

To determine

Find the elevation head (Z), pressure head (u/γw), and the total head (h) at the entrance and exit of the each soil layer.

(b)

Expert Solution
Check Mark

Answer to Problem 7.1CTP

The elevation head (Z0) at x is 0 mm is 220mm_.

The pressure head (u/γw)0 at x is 0 mm is 690mm_.

The total head h0 at x is 0 mm is 470mm_.

The elevation head (Z200) at x is 200 mm is 220mm_.

The pressure head (u/γw)200 at x is 200 mm is 656.3mm_.

The total head h200 at x is 200 mm is 436.3mm_.

The elevation head (Z400) at x is 400 mm is 220mm_.

The pressure head (u/γw)400 at x is 400 mm is 652.3mm_.

The total head h400 at x is 400 mm is 432.3mm_.

The elevation head (Z600) at x is 600 mm is 220mm_.

The pressure head (u/γw)600 at x is 600 mm is 220mm_.

The total head h600 at x is 600 mm is 0mm_.

Explanation of Solution

Given information:

The length of each soil layer (H1,H2,H3) is 200 mm.

The total length of the soil layer H is 600 mm.

The diameter of the cylindrical tube d is 150 mm.

The constant head difference Δh is 470 mm.

The porosity of the soil layer I (n1) is 0.5.

The porosity of the soil layer II (n2) is 0.6.

The porosity of the soil layer III (n3) is 0.33.

The hydraulic conductivity of soil layer I (k1) is 5×103cm/sec.

The hydraulic conductivity of soil layer II (k2) is 4.2×102cm/sec.

The hydraulic conductivity of soil layer III (k3) is 3.9×104cm/sec.

Calculation:

Determine the elevation head (Z0) at x is 0 mm.

Z0=hX-Y

Here, hX-Y is the depth of water from the Y point to the middle point of X.

Substitute 220 mm for hX-Y.

Z0=220mm

Therefore, the elevation head (Z0) at x is 0 mm is 220mm_.

Determine the pressure head (u/γw)0 at x is 0 mm.

(u/γw)0=Δh+hX-Y

Substitute 470 mm for Δh and 220 mm for hX-Y.

(u/γw)0=470+220=690mm

Therefore, the pressure head (u/γw)0 at x is 0 mm is 690mm_.

Determine the total head h0 at x is 0 mm.

h0=(u/γw)0+Z0

Substitute 690 mm for (u/γw)0 and –220 mm for Z0.

h0=690220=470mm

Therefore, the total head h0 at x is 0 mm is 470mm_.

Determine the elevation head (Z200) at x is 200 mm.

Z200=hX-Y

Substitute 220 mm for hX-Y.

Z200=220mm

Therefore, the elevation head (Z200) at x is 200 mm is 220mm_.

Determine the value of ΔH value using the relation.

kV(eq)i=k1i1kV(eq)i=k1ΔHH1 (1)

Substitute 1.076×103cm/sec for kV(eq), 0.783 for i, 5×103cm/sec for k1, and 200 mm for H1 in Equation (1).

1.076×103×0.783=5×103×ΔH200mm×1cm10mm8.425×1042.5×104=ΔHΔH=3.37cm×10mm1cmΔH=33.70mm

Determine the total head h200 at x is 200 mm.

h200=ΔhΔH

Substitute 470 mm for Δh and 33.70 mm for ΔH.

h200=47033.70=436.3mm

Therefore, the total head h200 at x is 200 mm is 436.3mm_.

Determine the pressure head (u/γw)200 at x is 200 mm.

(u/γw)200=h200(Z200)

Substitute 436.3 mm for h200 and –220 mm for Z200.

(u/γw)200=436.3+220=656.3mm

Therefore, the pressure head (u/γw)200 at x is 200 mm is 656.3mm_.

Determine the elevation head (Z400) at x is 400 mm.

Z400=hX-Y

Substitute 220 mm for hX-Y.

Z400=220mm

Therefore, the elevation head (Z400) at x is 400 mm is 220mm_.

Determine the value of ΔH value using the relation.

kV(eq)i=k2i2kV(eq)i=k2ΔHH2 (2)

Substitute 1.076×103cm/sec for kV(eq), 0.783 for i, 4.2×102cm/sec for k2, and 200 mm for H2 in Equation (2).

1.076×103×0.783=4.2×102×ΔH200mm×1cm10mm8.425×1042.1×103=ΔHΔH=0.40cm×10mm1cmΔH=4.0mm

Determine the total head h400 at x is 400 mm.

h400=h200ΔH

Substitute 436.3 mm for h200 and 4.0 mm for ΔH.

h400=436.34.0=432.3mm

The total head h400 at x is 400 mm is 432.3mm_.

Determine the pressure head (u/γw)400 at x is 400 mm.

(u/γw)400=h400(Z400)

Substitute 432.3 mm for h400 and –220 mm for Z400.

(u/γw)400=432.3+220=652.3mm

Therefore, The pressure head (u/γw)400 at x is 400 mm is 652.3mm_.

Determine the elevation head (Z600) at x is 600 mm.

Z600=hX-Y

Substitute 220 mm for hX-Y.

Z600=220mm

Therefore, the elevation head (Z600) at x is 600 mm is 220mm_.

Determine the value of ΔH value using the relation.

kV(eq)i=k3i3kV(eq)i=k3ΔHH3 (3)

Substitute 1.076×103cm/sec for kV(eq), 0.783 for i, 3.9×104cm/sec for k3, and 200 mm for H3 in Equation (3).

1.076×103×0.783=3.9×104×ΔH200mm×1cm10mm8.425×1041.95×105=ΔHΔH=43.21cm×10mm1cmΔH=432.10mm

Determine the total head h600 at x is 600 mm.

h600=h400ΔH

Substitute 432.3 mm for h400 and 432.10 mm for ΔH.

h600=432.3432.10=0.2mm0mm

Therefore, the total head h600 at x is 600 mm is 0mm_.

Determine the pressure head (u/γw)600 at x is 600 mm.

(u/γw)600=h600(Z600)

Substitute 432.3 mm for h600 and –220 mm for Z600.

(u/γw)600=0+220=220mm

Therefore, the pressure head (u/γw)600 at x is 600 mm is 220mm_.

(c)

To determine

Plot the variation of the elevation head, pressure head, and the total head with the horizontal distance along the sample axis (X–X).

(c)

Expert Solution
Check Mark

Explanation of Solution

Given information:

The length of each soil layer (H1,H2,H3) is 200 mm.

The total length of the soil layer H is 600 mm.

The diameter of the cylindrical tube d is 150 mm.

The constant head difference Δh is 470 mm.

The porosity of the soil layer I (n1) is 0.5.

The porosity of the soil layer II (n2) is 0.6.

The porosity of the soil layer III (n3) is 0.33.

The hydraulic conductivity of soil layer I (k1) is 5×103cm/sec.

The hydraulic conductivity of soil layer II (k2) is 4.2×102cm/sec.

The hydraulic conductivity of soil layer III (k3) is 3.9×104cm/sec.

Calculation:

Refer Part b)

Draw the graph between the elevation head pressure head, and the total head with the horizontal distance along the sample axis (X–X) as in Figure (1).

Principles of Geotechnical Engineering (MindTap Course List), Chapter 7, Problem 7.1CTP

(d)

To determine

Plot the variation of the discharge velocity and the seepage velocity along the sample axis (X–X).

(d)

Expert Solution
Check Mark

Explanation of Solution

Given information:

The length of each soil layer (H1,H2,H3) is 200 mm.

The total length of the soil layer H is 600 mm.

The diameter of the cylindrical tube d is 150 mm.

The constant head difference Δh is 470 mm.

The porosity of the soil layer I (n1) is 0.5.

The porosity of the soil layer II (n2) is 0.6.

The porosity of the soil layer III (n3) is 0.33.

The hydraulic conductivity of soil layer I (k1) is 5×103cm/sec.

The hydraulic conductivity of soil layer II (k2) is 4.2×102cm/sec.

The hydraulic conductivity of soil layer III (k3) is 3.9×104cm/sec.

Calculation:

Determine the discharge velocity v using the relation.

v=kV(eq)i

Substitute 1.076×103cm/sec for kV(eq) and 0.783 for i.

v=1.076×103×0.783=0.000843cm/sec

Determine the seepage velocity of soil I using the relation.

vs1=vn1

Here, n1 is the porosity of the soil I.

Substitute 0.000843 cm/sec for v and 0.5 for n1.

vs1=0.0008430.5=0.00168cm/sec

Determine the seepage velocity of soil II using the relation.

vs2=vn2

Here, n2 is the porosity of the soil II.

Substitute 0.000843 cm/sec for v and 0.6 for n2.

vs2=0.0008430.6=0.0014cm/sec

Determine the seepage velocity of soil III using the relation.

vs3=vn3

Here, n3 is the porosity of the soil III.

Substitute 0.000843 cm/sec for v and 0.33 for n3.

vs3=0.0008430.33=0.00255cm/sec

Draw graph of variation of the discharge velocity and the seepage velocity along the sample axis (X–X).

Refer Figure (1) in Part (c).

(e)

To determine

Find the height of the vertical columns of water inside piezometers A and B installed on the sample axis.

(e)

Expert Solution
Check Mark

Answer to Problem 7.1CTP

The height of the vertical columns of water at point A is 656.3mm_.

The height of the vertical columns of water at point B is 652.3mm_.

Explanation of Solution

Given information:

The length of each soil layer (H1,H2,H3) is 200 mm.

The total length of the soil layer H is 600 mm.

The diameter of the cylindrical tube d is 150 mm.

The constant head difference Δh is 470 mm.

The porosity of the soil layer I (n1) is 0.5.

The porosity of the soil layer II (n2) is 0.6.

The porosity of the soil layer III (n3) is 0.33.

The hydraulic conductivity of soil layer I (k1) is 5×103cm/sec.

The hydraulic conductivity of soil layer II (k2) is 4.2×102cm/sec.

The hydraulic conductivity of soil layer III (k3) is 3.9×104cm/sec.

Calculation:

The height of water column is equal to the Piezometric or pressure head at a point.

Determine the height of water in point A.

HA=(u/γw)200

Substitute 656.3 mm for (u/γw)200.

HA=656.3mm

Therefore, the height of the vertical columns of water at point A is 656.3mm_.

Determine the height of water in point B.

HB=(u/γw)400

Substitute 652.3 mm for (u/γw)400.

HB=652.3mm

Therefore, the height of the vertical columns of water at point B is 652.3mm_.

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