Solution Manual For Groundwater Science 3rd Edition By Charles R. Fitts chapter 1-12

Contents Chapter 1 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Chapter 3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Chapter 4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Chapter 5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Chapter 6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Chapter 7 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Chapter 8 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Chapter 9 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Chapter 10 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Chapter 11 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Chapter 12 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

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CONTENTS

Chapter 1 Solutions 1. The oceans cover about 71% of earth’s surface, and other than ice-covered areas, there is no limit to the availability of water to evaporate. On the continents, evapotransipration is often limited by lack of water.

2. Low rates occur in polar latitudes because water is frozen much of the year and cold air can hold little water vapor. Low rates in north Africa and the Arabian peninsula are due to low levels of soil moisture in desert areas.

3. Answers vary with time. See the USGS waterwatch web site.

4. High infiltration and recharge are associated with permeable surface materials, flat topography, high precipitation rates, and land use/vegetation with low evapotranspiration rates.

5. Coarse-grained granular deposits have small capillary fringes because there is relatively little mineral surface area to attract water molecules and with larger pores the water molecules, on average, are farther from the mineral surfaces.

6. (a) The lake volume is area × depth = 1.67 × 109 ft3 (4.72 × 107 m3 ). (b) The residence time is the lake volume divided by the average of the inflow and outflow discharges (ft, sec units): Tr

= = =

V Q 1.67 × 109 14.6 1.14 × 108 sec = 3.62 yr

(c) Other possible terms are the net groundwater flow into the reservoir (G), the evaporation from the surface of the reservoir (E), and the rate of change in the volume stored in the reservoir (dV /dt). The balance equation for the reservoir is dV I +G−O−E = dt where I and O are the inlet and outlet discharges. Rearranging the balance equation gives (ft, sec units): I −O

=

dV dt 15.3 − 13.9

=

1.4 ft3 /sec

=

0.0396 m3 /sec

=

1

E−G+

7. The volume of the tank is 1000 gal = 133.7 ft3 . The area of the roof is 25 · 45 = 1125 ft2 . Dividing these two figures gives the depth of rain on the roof needed to fill the tank.

133.7 1125

=

0.119 ft

=

1.43 in (3.6 cm)

By 5:00, 1.21 inches had fallen, and by 6:00 1.70 inches had fallen. Overflow occurs between 5:00 and 6:00. Linear interpolation between these times estimates overflow at 5:27.

8. The discharge is Q=

1000 gal 80 min

=

12.5 gal/min

=

0.028 ft3 /s

=

7.9 × 10−4 m3 /sec

The radius of the hose is r = 0.625 in = 0.0521 ft, so the cross-sectional area of the hose is A = πr2 = 8.52 × 10−3 ft2 The water velocity in the hose is (ft, sec units): v=

0.028 Q = = 3.26 ft/sec A 8.52 × 10−3

which is equal to 0.99 m/sec.

9. Start by converting all given quantities into meter units: the drainage area is A = 6.53 × 105 km2 = 6.53 × 1011 m2 ; the annual discharge is Q = 1.5 × 107 acre-ft/yr = 1.85 × 1010 m3 /yr. The river discharge per drainage area is Q A

=

1.85 × 1010 = 0.0283 m/yr 6.53 × 1011 0.093 ft/yr

=

1.1 in/yr

=

This represents 9% of the annual precipitation of 12 in/yr.

Besides discharging into the Colorado River, water transpires through plants, evaporates from near-surface soils, evaporates from surface waters, and is pumped from aquifers.

10. (a) Overland flow (OF ) plus interflow (IF ) is related to stream flow (SF ) and groundwater discharge to streams (Gs ) as follows: OF + IF = SF − Gs = 13 − 6 = 7 in/yr (18 cm/yr) These values are in units of discharge per drainage basin area. (b) Looking at the saturated zone fluxes, recharge (R) is related to groundwater discharge to streams (Gs ), well discharge (Qw ), and deep transpiration (ETd ) as follows: R = Gs + Qw + ETd = 6 + 2.3 + ETd ≥ 8.3 in/yr (21 cm/yr) 2

The well discharge (Qw = 2.3 in/yr) is calculated as the discharge (1.8 × 109 ft3 /yr) divided by the drainage basin area (340 mi2 ), with the necessary conversions to yield in/yr units. The deep transpiration from the saturated zone is unknown, so we can only conclude that the recharge is at least 8.3 in/yr. If we assume that there is no significant deep transpiration (ETd = 0), then R = 8.3 in/yr. (c) Looking only at fluxes entering or exiting the ground, evapotranspiration (ET ) can be estimated from precipitation (P ), stream flow (SF ), and well discharge (Qp ) as follows: ET = P − SF − Qp = 28 − 13 − 2.3 = 12.7 in/yr (32 cm/yr) 11. (a) Fluxes into the lake are: Precipitation (P ), inlet stream discharge (Si ), overland flow (OF ), interflow (IF ), and groundwater inflow (Gi ). Fluxes out of the lake are: Evaporation (E), outlet stream discharge (So ), and groundwater outflow (Go ). (b) Precipitation’s contribution during June is 1.63 inches times the area of the lake (285 acres). With appropriate unit conversions this works out to P = 1.69×106 ft3 . Evaporation is calculated in a similar manner: E = 3.58 × 106 ft3 . The amount of storage lost from the lake is also calculated as the change in level times the area of the lake, which works out to dV = −4.45 × 106 ft3 . The hydrologic balance equation for this situation is P + Si + OF + IF + Gi − E − So − Go = dV The unknown terms are OF , IF , Gi , and Go . Solving for the sum of these unknowns gives: OF + IF + Gi − Go = dV − P − Si + E + So = 9.4 × 105 ft3 (2.7 × 104 m3 ) (c) The net of the four unknown fluxes is positive (into the lake). If we assume that overland flow and interflow are negligible, then Gi − Go ≈ 9.4 × 105 ft3 , and there is more groundwater inflow than outflow at the lake bottom. So, with these assumptions, the lake is spring-fed to some degree. The net groundwater inflow is smaller than the contribution from precipitation and much smaller than the contribution of the inlet stream. (d) Possibilities include: • Install shallow wells around lake shore – where well levels exceed the lake level, flow is into the lake from the groundwater.

• Measure water temperatures along the bottom of the lake. Where spring discharge is high, the temperature may be anomalous. Summer would be better for this survey, because cold spring discharge will be denser and stay on the bottom.

• Repeat the measurements at a time when OF , E, and P are known to be low (winter in a cool climate would be good).

12. (a) The hydrologic balance equation in this case is P + Si + G − E − So = dV /dt where P is precipitation, Si and So are the inlet and outlet stream discharges, G is net groundwater inflow, E is evaporation, and dV /dt is the rate of change in reservoir volume. (b) A spreadsheet of the solution is shown in Table 1 and the graph is shown in Figure 1.

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Table 1: Problem 12.

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Si (m3 ) 3.90E+06 3.09E+06 5.47E+06 6.48E+06 5.16E+06 3.24E+06 5.37E+06 1.45E+06 7.00E+05 1.73E+06 2.37E+06 2.02E+06

So (m3 ) 4.32E+06 4.02E+06 5.42E+06 6.19E+06 5.28E+06 3.73E+06 5.61E+06 2.07E+06 1.61E+06 2.57E+06 3.39E+06 3.03E+06

E (m3 ) 1.04E+04 1.40E+04 4.19E+04 9.41E+04 1.01E+05 1.50E+05 1.92E+05 1.78E+05 1.18E+05 5.57E+04 3.50E+04 2.09E+04

P (m3 ) 8.37E+04 4.53E+04 1.57E+05 2.33E+05 1.12E+05 9.06E+04 1.60E+05 2.78E+04 1.40E+04 1.50E+05 9.75E+04 5.22E+04

figure file: solutions/ch1res Figure 1: Problem 12.

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dV /dt (m3 ) 1.61E+05 -9.20E+04 3.68E+05 4.83E+05 2.30E+05 2.30E+04 1.61E+05 -1.15E+05 -1.84E+05 4.60E+04 -6.90E+04 -9.20E+04

G (m3 ) 5.08E+05 8.07E+05 2.03E+05 5.36E+04 3.39E+05 5.72E+05 4.32E+05 6.55E+05 8.30E+05 7.92E+05 8.88E+05 8.87E+05

Chapter 2 Solutions 1. The weight of water W equals (ft, lb units): W

= ρw gV = ρw gπr2 H =

(62.4)π(2)2 (6) = 4705 lb

This is equivalent to 20,927 N. 2. Pressure is force per area. Since a Newton = kg m sec−2 , N/m2 = kg m−1 sec−2 .

3. First, calculate the pressure change in going from the lake surface to 130 m deep (m, N units): ∆P

= ρw g∆z =

(9810)(130) = 1.275 × 106 N/m2

Use Eq. 2.1 to calculate the change in density: dρw

=

ρw βdP

=

(1.002)(4.9 × 10−10 )(1.275 × 106 ) = 6.26 × 10−4 g/cm3

Therefore the density of water at the lake bottom is 1.002 + 6.26 × 10−4 ' 1.0026 g/cm3 4. Viscosity is a measure of the resistance to shear within a fluid. The viscosity of water is due to the attraction of polar water molecules for themselves and the weak hydrogen bonds that form between molecules.

5. Rearrange Eq. 2.2 to solve for viscosity (m, N units): µ = =

F dz Adv (0.6)(0.002) = 0.024 N·sec/m2 (0.5)2 (0.2)

This is about 20 times higher than the viscosity of water.

6. In a fine sand there is much more mineral surface area per volume of material and the attraction of pore water for these surfaces causes much lower pore water pressure (well below atmospheric pressure), which is what helps hold the sand castle together.

In dry sand there is no pore water pressure, just atmospheric pressure in the pores.

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Table 2: Problem 10.

Volume (cm3 ) 34.6 Air 205.3 Water 488.3 Solids 728.2 Total

Mass (g) 0.0 205.3 1318.3 1523.6

Table 3: Problem 11.

Volume (cm3 ) 0.0 Air 239.9 Water 488.3 Solids 728.2 Total

Mass (g) 0.0 239.9 1318.3 1558.2

7. Gage pressure is pressure relative to atmospheric pressure. A negative gage pressure is a pressure that is less than atmospheric pressure and a positive gage pressure is a pressure that exceeds atmospheric pressure. Absolute pressure is pressure relative to absolute zero pressure (zero force per area).

8. Porosity tends to be higher in well sorted materials than in poorly sorted materials.

This is because in poorly sorted materials, there is a wide range of grain sizes, and smaller grains can pack between larger grains.

9. The porosity of pumice is quite high due to the large fraction of voids. However, many of these voids are sealed bubbles that do not participate in fluid flow. Therefore the effective porosity of pumice is much smaller than the porosity.

10. (a) The completed phase table is shown in Table 2. (b) n

=

e

=

θ

=

ρb

=

Vv 34.6 + 205.3 = 0.33 = Vt 728.2 Vv 34.6 + 205.3 = = 0.49 Vs 488.3 Vw 205.3 = = 0.282 Vt 728.2 ms 1318.3 = = 1.81 g/cm3 Vt 728.2

11. If the sample were fully saturated, the air in the voids would be replaced by water, and the phase table would be as shown in Table 3. The total mass per total volume, from the phase table, is mt 1558.2 = = 2.140 g/cm3 Vt 728.2 At this density, 1 m3 would have a mass of 2140 kg.

12. The completed phase table is shown in Table 4. The table assumes a mineral density of 2700 kg/m3 (2.70 g/cm3 ). With this assumption, the mass of one cubic meter is 1748 kg.

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Table 4: Problem 12.

Volume (m3 ) 0.44 0.56 1.00

Water Solids Total

Mass (kg) 440 1512 1952

Table 5: Problem 13.

Volume (m3 ) 0.22 0.78 1.00

Water Solids Total

Mass (kg) 220 2067 2287

13. (a) The completed phase table is shown in Table 5. The table assumes a mineral density of 2650 kg/m3 (2.70 g/cm3 ). With this assumption, the mass of one cubic meter is 1952 kg. (b) ρb = ms /Vt = 2067 kg/m3 . (c) ρt = mt /Vt = 2287 kg/m3 .

14. Gravitational potential energy (elevation head), elastic potential energy (pressure head), and kinetic energy (velocity head). Generally the velocity head is insignificant compared to the other two. The units of head are length.

15. We need to first calculate the velocity as the water exits the hose. The velocity in the hose is (cm, sec units): V =

Q A

= =

Q πr2 2000 = 636 cm/sec π(1)2

The velocity is 6.36 m/sec. The hydraulic head is given by Eq. 2.13. At the end of the hose, P = 0 and this equation becomes (m, sec units) h = z+ =

v2 2g

134.5 +

6.262 = 136.6 m (2)(9.81)

At the top of its arc, the water velocity drops to zero, the pressure is zero, and the elevation equals the total head. z = h = 136.6 m 16. Applying Eq. 2.15 results in the following gage pressure (m, sec units): P

=

(depth)ρw g

=

(4)(9810) = 39,200 N/m2

This answer, properly converted, is equal to 819 lb/ft2 .

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figure file: solutions/ch2wellflow Figure 2: Vertical cross-section showing groundwater flow patterns for Problem 17. The well screen is light gray, the solid well casing is dark gray.

Table 6: Problem 21.

Point C D E

h (m) 476.69 472.17 471.70

z (m) 472.74 458.03 466.84

p (N/m2 ) 38,750 138,710 47,677

The absolute pressure is the gage pressure plus atmospheric pressure, which is about 1.01 × 105 N/m2 or 14.7 psi. Adding these to the gage pressure results in absolute pressure = 1.40 × 105 N/m2 = 2940 lb/ft2 .

17. (a) The flow is downward and head always decreases in the direction of flow. Therefore the head increases in the upward direction. (b) The pattern of flow will be like that illustrated in Figure 2. Water will flow into the top end of the well and exit the bottom end of it, because the well is a path of less resistance.

18. The water pressure P = 5.32 lb/in2 = 766 lb/ft2 . The head at the transducer is (ft, lb units): h = =

P +z ρw g 766 + 417.9 = 430.2 ft (131.1 m) 62.4

19. Under a hurricane atmospheric pressure is lower, so the gage pressure is higher.

20. (a) The hydraulic head equals the elevation of the water in the well: h = 267.4 m. (b) The pressure head is P = h − z = 267.4 − 245.0 = 22.4 m ρw g (c) The pressure is calculated from the pressure head (n, N units):

P ρw g P = ρw g = (22.4)(9810) = 219,700 N/m2 21. (a) The head is calculated by subtracting the depth to water from the elevation of the top of casing (TOC). Doing this results in hC = 476.69, hD = 472.17, and hE = 471.70. With heads known, pressures are computed using p = (h − z)ρg.

See Table 6. (b) See Figure 3. (c) There is generally an upward component, as indicated by the head differences between the stream and wells D and E. (d) It enters at the top of the section everywhere except in the immediate vicinity of the wetland and the stream. These two places are where the groundwater exits the subsurface.

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figure file: solutions/ch2sectionflow Figure 3: Vertical cross-section showing hydraulic heads and groundwater flow patterns for Problem 21.

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Chapter 3 Solutions 1. Use a form of Eq. 3.2 (cm, min units): Ks

Qs ds A dh 20.0 3.5 = 0.17 cm/min = − 2 π(1.5) (37.0 − 95.0) = −

This is equivalent to K = 2.8 × 10−3 cm/sec. Such a conductivity is consistent with a fine sand or silty sand.

2. Fourier’s Law governs heat conduction in solids. For one-dimensional conduction in the x direction it is ∂T qcx = −κ ∂x where qcx is the energy conducted per time per unit area normal to the x direction (analogous to specific discharge), κ is the thermal conductivity (analogous to hydraulic conductivity), and T is temperature (analogous to hydraulic head).

3. For water to flow through fine-grained material, it must squeeze though many small, tortuous pores. In doing so, it must shear itself, which involves frictional resistance (see the discussion of water viscosity in Section 2.2.2). Flowing a given distance through fine-grained materials requires more shearing and viscous resistance than flowing the same distance through a coarse-grained material.

4. (a) Use Darcy’s law (ft, day units): ∂h A ∂x (25)(0.006)(5280 · 65) = 51,000 ft3 /day (1440 m3 /day)

|Qx | = Kx =

The area A is normal to flow, one mile in horizontal length and 65 ft in the vertical. (b) |qx | = Kx

∂h = (25)(0.006) = 0.15 ft/day (4.6 cm/day) ∂x

(c) vx =

qx 0.15 = = 0.94 ft/day (29 cm/day) ne 0.16

5. (a) The head is merely the ground elevation minus the depth to water. hA = 90.86 m, hB = 88.50 m, hC = 92.46 m.

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figure file: solutions/ch3vectors Figure 4: Problem 5.

(b) Say the x direction is east and the y direction is north. In meter, day units, the specific discharge components are as follows: qx = −Kx

hC − hA xC − xA 92.46 − 90.86 = −5 = −0.0080 m/day 1000

∂h ∂x

qy = −Ky

= −Kx

∂h ∂y

hB − hA yB − yA 88.50 − 90.86 = 0.047 m/day −20 1000

= −Ky =

(c) See Figure 4. (d) |~q |

=

q

qx2 + qy2

=

p

(−0.008)2 + (0.047)2 = 0.048 m/day

6. (a) Apply Darcy’s law to the gradient between A and C (m, day units): ∂h ∂x 44.62 − 44.50 = −3.6 × 10−4 m/day = −(0.3) 100 = −Kx

qx

(b) The total discharge is the specific discharge times the area that the discharge goes through: Qx

= qx A (−3.6 × 10−4 m/day)(1 × 6) = −2.2 × 10−3 m3 /day

=

The minus sign just means that the discharge is in the negative x direction. (c) Apply Darcy’s law to the gradient between B and D, using the average Kx from these points (m, day units): qx

= =

−Kx

∂h ∂x

−(4.5)

42.75 − 42.64 100

= −4.9 × 10−3 m/day

(d) The total discharge is the specific discharge times the area that the discharge goes through: Qx = qx A = (−4.9 × 10−3 m/day)(1 × 15) = −0.074 m3 /day 12

(e) Assume that the heads measured at A and C apply at the top of the silty clay and that the heads measured at B and D apply at the base of the silty clay.

This is reasonable since the head gradients in the permeable sands are much smaller than head gradients in the silty clay. There are two sets of data that could be analyzed: location A-E-B and location C-F -D. One approach is to use average Kz and vertical gradients from the two locations in one calculation of qz . Averaging the Kz values at E and F gives Kz = 0.0025 m/day. The vertical hydraulic gradient at A-E-B is hA − hB 44.50 − 42.64 ∂h = = = 0.372 ∂z 5 5 Similarly, the vertical gradient at C-F -D is hC − hD 44.62 − 42.75 ∂h = = = 0.374 ∂z 5 5 The average of these two gradient values is ∂h/∂z = 0.373. Now the average specific discharge is calculated (m, day units): qz = −Kz

∂h = −(0.0025)(0.373) = −9.3 × 10−4 m/day ∂z

7. Assume that the effective porosity equals the total porosity. The average linear velocity is (m, day units): vs =

qs ne

K ∂h ne ∂s 0.7 (0.012) = 0.026 m/day 0.32

= − =

8. Before calculating the intrinsic permeability, convert the hydraulic conductivity to m/sec units: K = 0.7 m/day = 8.1 × 10−6 m/sec. Now the intrinsic permeability is calculated (m, sec, kg units): k

= =

Kµ ρw g (8.1 × 10−6 )(1.0 × 10−3 ) = 8.3 × 10−13 m2 9810

This is equivalent to k = 8.3 × 10−9 cm2 .

The conductivity for leaded gasoline is (m, sec, kg units) K

= =

kρg µ (8.3 × 10−13 )(730)(9.81) = 1.2 × 10−5 m/sec 5 × 10−4

The conductivity for Arabian medium crude is (m, sec, kg units) K

= =

kρg µ (8.3 × 10−13 )(880)(9.81) = 3.1 × 10−7 m/sec 2.3 × 10−2 13

9. Media with very large pores like karst limestone, porous basalt, or coarse gravel may have turbulent flow. Large pores and high velocity promote turbulent flow. See the definition of Reynolds number; turbulence occurs at high Reynolds numbers.

10. The specific discharge is calculated as |q s | = =

∂h ∂s (1000)(0.05) = 50 m/day K

Calculate the Reynolds number to see if this flow is laminar or turbulent. Convert inputs to the Reynolds number equation to common units (m, sec, kg): d = 1.5 inches = 0.038 m, q = 50 m/day = 0.00058 m/sec.

Re

= =

ρqd µ (1000)(0.00058)(0.0381) = 16 1.4 × 10−3

Since Re > 10, this flow is probably turbulent and therefore not governed by Darcy’s law.

11. In m, day units: (a) qx = v x ne = (13)(0.34) = 4.4 m/day. (b) Q = qx A = (4.4)(95 × 0.6) = 252 m3 /day. (c) Kx = −qx dx/dh = −(4.4)(25)/(−14) = 7.9 m/day.

12. Heterogeneity is when a material property varies with location. Anisotropy is when, at a point, a material property varies with direction. Small scale heterogeneity in the form of thin layers of different K leads to anisotropy of a large-scale package of thin layers, with higher K parallel to the layers and lower K normal to the layers.

13. (a) The horizontal equivalent average conductivity is (m, day units): P K d P xi i Kxe = di (6 × 1.4) + (0.5 × 3.7) + (2.5 × 2.5) = = 2.2 m/day 1.4 + 3.7 + 2.5 The vertical equivalent average conductivity is: P di Kze = P (di /Kzi ) 1.4 + 3.7 + 2.5 = = 0.031 m/day (1.4/0.1) + (3.7/0.02) + (2.5/0.05) (b) The vertical specific discharge is qz

∂h ∂z (101.3 − 96.2) = −0.031 = −0.021 m/day 7.6 = −Kze

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figure file: solutions/Ch3EquivKsProfile Figure 5: Problem 13 (c) Using this specific discharge in an analysis of the top layer gives the head difference across the top layer: qz dz Kz (−0.021)(−1.4) = −0.29 m = − 0.1

dh = −

Therefore, the head at the interface between the top and middle layers is h = 101.3 − 0.29 = 101.0 m. Using the specific discharge in an analysis of the bottom layer gives the head difference across the bottom layer: qz dz Kz (−0.021)(2.5) = 1.04 m = − 0.05

dh = −

Therefore, the head at the interface between the middle layers and bottom is h = 96.2+1.04 = 97.2 m. As a check, a similar analysis will show that the head difference across the middle layer is 3.85 m, which agrees with 101.0−97.2 = 3.8 m. (d) A profile of the heads through this section is shown in Figure 5.

14. Ft, day units were used in the following calculations. (a) T

=

(Kt )1 b1 + (Kt )2 b2

=

(3.0)(75 − 68) + (0.8)(68 − 50) = 35 ft2 /day (3.3 m2 /day)

=

(Kt )1 b1 + (Kt )2 b2

=

(3.0)(85 − 68) + (0.8)(68 − 50) = 65 ft2 /day (6.0 m2 /day)

(b) T

15. Consider first the discharge through a single layer with conductivity Kt = Kx parallel to the layer, and thickness b. Darcy’s law states that the discharge in the x direction is ∂h Qx = −Kt A ∂x where A is the cross-sectional area that the flow goes through. In this case, A = b∆y, so ∂h ∂h Qx = −Kt b∆y = −T ∆y ∂x ∂x If the layer consisted of multiple layers with conductivities (Kt )i and thicknesses bi , then m X ∂h ∂h Qx = (Kt )i bi ∆y = −T ∆y ∂x ∂x i=1 16. (a) Cm, sec units are used in the following calculation.

K = C(d10 )2 = (120)(0.01)2 = 0.012 cm/sec 15

(b) M, sec, kg units are used in the following calculations.

(d50 )2 n3 ρw g K = µ (1 − n)2 180

3 9810 0.37 (2.8 × 10−4 )2 = 1.4 × 10−3 (1 − 0.37)2 180 =

3.9 × 10−4 m/sec = 3.9 × 10−2 cm/sec

(c) This is just like the calculation for (b), but with n = 0.31.

0.313 (2.8 × 10−4 )2 9810 K = 1.4 × 10−3 (1 − 0.31)2 180 =

1.9 × 10−4 m/sec = 1.9 × 10−2 cm/sec

17. Rearrange Eq. 3.33 to solve for t1 − t0 (cm, sec units):

dh0 a L ln t1 − t0 = AK dh1

2 80 π(0.1) (22) ln = 191 sec = π(4)2 (10−4 ) 20 18. Use m, sec, kg units in this calculation.

Qx

= =

ρw gb3 w ∂h 12µ ∂x (9810)(0.0002)3 (15) (0.004) = 2.8 × 10−7 m3 /sec = 0.28 cm3 /sec 12(1.4 × 10−3 )

19. Water is attracted to mineral surfaces and is “pulled” by these forces to form a film over the mineral surfaces. This pulling is like tension in a solid, reducing pore water pressures to below atmospheric pressure.

20. (a) Using m, sec, kg units: h = =

P +z ρw g −6800 + 231.49 = 230.80 m 9810

(b) ∂h 230.80 − 230.54 = = 0.079 ∂z 231.49 − 228.25 (c) The flow is downward, toward the lower head. (d) The pressure head is P −6800 = = −0.69 m = −69 cm ρw g 9810 In the fine sand soil, this corresponds to θ ≈ 0.10.

21. (a) (m, kg, s, N units): h=

P −12800 +z = + 14.19 = −1.30 + 14.19 = 12.89 m ρg (1000)(9.81) 16

figure file: solutions/ch3unsat Figure 6: Problem 23. (b) P/(ρg) = −1.30 m = −130 cm. The water content at this pressure head for the silt loam is about θ = 0.22.

22. Water will have an upward flow component any time the head decreases with increasing elevation. This can only occur if the pressure head decreases with elevation at a great enough rate to offset the increase in elevation head. Therefore, water content must decrease with elevation.

23. The plots for this problem are shown in Figure 6. The plot of θ is given. The plot of P is determined by using the silt loam curve to estimate P/ρw g from θ and then finding P = (P/ρw g) · ρw g. The plot of h is determined using h = (P/ρw g) + z.

Because head decreases in the downward direction, there is a downward component of flow here.

24. To apply Darcy’s law for variable density vertical flow (Eq. 3.40), we need an average water density. Averaging the densities at 500 and 700 m depths gives ρw = (1.037 + 1.042)/2 = 1.0395 gm/cm3 = 1039.5 kg/m3 . The intrinsic permeability is 10−12 cm2 = 10−16 m2 . The vertical specific discharge is then (m, sec, kg units):

kz ∂P + ρw g qz = − µ ∂z

10−16 5.002 × 106 − 7.157 × 106 = − + (1039.5)(9.81) 1.01 × 10−3 200 =

5.7 × 10−11 m/sec = 0.0018 m/yr

Since qz > 0, flow is upward in the positive z direction.

25. The depth to the interface is (g/cm3 units for ρ which cancel, ft units for h, zs ): zs

= =

ρf h ρs − ρf 1.001 2.3 = 88.6 ft 1.027 − 1.001

26. (a) Assuming typical fresh and salt water densities (1.000 and 1.025 gm/cm3 ), the depth to the interface will be 40 times the height of the water table: zs = 40h = 40(0.9) = 36 m (b) I assumed typical fresh and salt water density, which is probably close for Cape Cod. Deviation on these densities could affect the result slightly. I assumed hydrostatic conditions, which below the well is not a good assumption. Below the well there is upward flow, and heads at depth are higher than heads in the well. This deviation will make the actual interface depth greater than what was computed in part (a). (c) The pumping rate may not be sustainable. Storage changes along the interface occur very slowly, since there is so much storage involved with moving the interface. Such a system may not equilibrate for years or decades. Given enough time, the interface may rise to the bottom of the well screen. A more sophisticated transient, 3D analysis could give a better prediction.

17

27. The pressure at the well screen is (m, s, kg units): P

=

(depth)ρwell g

=

(−0.11 − (−74.3))(1009)(9.81) = 7.34 × 105 N/m2

The fresh water head at the well is hf =

P 7.34 × 105 = = 0.558 m ρw g 9810

This head is higher than zero, so water flow toward the shore from the vicinity of the screen.

18

Chapter 4 Solutions 1. See the relevant sections in this chapter. Rotary drilling can be employed in a wide range of circumstances: shallow and deep, unconsolidated and rock. Direct-push probes are limited to shallow depths in finer unconsolidated materials (sand, silt, clay). Some probes can collect pore gas and/or pore water samples. Most probes do not allow soil sampling.

2. The electrical conductivity of frozen ground is low compared to unfrozen ground, so resistivity and electromagnetic methods would probably penetrate far and work well for this application. On the other hand, seismic refraction would not work well. Because the P-wave velocity of frozen ground exceeds that of unfrozen ground, this situation has a high-velocity layer over a low-velocity layer, a condition that is difficult to resolve with the seismic refraction method.

3. At about 24 m depth, about 23 L/min enters the borehole. At about 32 m depth, about 21 L/min exits the borehole. At about 39 m depth, about 4 L/min enters the bore hole. At about 55 m depth, about 6 L/min exits the bore hole.

4. Mud is denser than water, so it can build higher pressures in the bore hole, which helps reduce groundwater flow into the bore hole and helps prevent collapse of uncased zones. Thick drilling mud can create a thin zone of low K materials on the wall of the bore hole, which helps limit flow to or from the bore hole (this can later be undesirable if a well is to be installed). Drilling mud is more viscous than water and returns drill cuttings to the surface more easily.

5. When the aquifer formation is very thin and horizontal, and where the aquifer extends under a surface water like a river.

6. Well development increases the permeability near the well screen and thus increases the yield of a well. It also removes fines which later reduces the suspended fines and turbidity of water pumped from a well.

7. A filter pack is a zone of higher permeability around a well screen, which increases well yield. Properly designed filter packs allow water to move to the screen, but not fine particles from the aquifer. Seals in the upper portion of a well installation prevent surface water or water from higher zones from migrating down to the well screen. These waters are often poor quality compared to groundwater from the screened interval.

8. (a) Interpolating the grain sizes from the table, the estimated d50 of the formation is about 0.28 mm. This should be the screen opening size. (b) The d30 of the formation sand is about 0.17 mm. The sand is very uniform (well sorted), so it would be conservative to use a multiplier of 4 to get the d30 of the filter sand: d30 f ilter = (4)(0.17) = 0.68 mm 19

Assuming that the filter sand d30 /d10 = 1.5, the filter sand d10 = 0.68/1.5 = 0.45 mm. The recommended screen opening size equals the filter pack d10 size, 0.45 mm. (c) The design with a filter pack has a larger screen opening size and has a guaranteed zone of high conductivity filter pack with few fines. It would likely have a higher yield.

9. (a) From the graph the d50 of the sand is about 0.8 mm. This should be the screen opening size. (b) The d30 of the sand is about 0.35 mm. The sand is widely graded (poorly sorted), so a multiplier of about 9 is appropriate to get the d30 of the filter sand: d30 f ilter = (9)(0.35) = 3.1 mm Assuming that the filter sand d30 /d10 = 1.5, the filter sand d10 = 0.68/1.5 = 2.1 mm. The recommended screen opening size equals the filter pack d10 size, 2.1 mm. (c) The design with a filter pack has a larger screen opening size and has a guaranteed zone of high conductivity filter pack with few fines. It would likely have a higher yield.

10. It would descend at a rapid, relatively constant rate until the liner covered the fracture. Below that, the liner would descend slowly, slowing with more depth. To analyze the conductivity of the bedrock matrix, you would analyze the slow descent after the liner covered the fracture.

20

Chapter 5 Solutions 1. The perched aquifer should have a low K layer underneath and some unsaturated zone beneath that. The head in the confined aquifer should be above the top of the aquifer and there should be some confining layer above the aquifer. The water table in the unconfined aquifer should be within the aquifer layer.

2. The flow in the sand layers will be nearly parallel to the layers, and flow in the silt/clay layers will be almost normal to the layers.

3. In an area underlain by low-conductivity materials, the water table will be at or near the ground surface. Where the water table is at the surface there will be streams and wetlands, and these will typically be closely spaced. On the other hand, areas underlain by high conductivity materials will have few, widely spaced streams and wetlands.

4. There are many possibilities here. All it takes is a large aquifer system with long residence times. A couple that come to mind are the Great Artesian Aquifer in Australia and the Dakota aquifer in the northern Great Plains of the U.S. 5. The answers to this question are in a list of items found in Section 5.1.4.

6. The analysis and Q estimate are shown in Table 7.

7. Convert the drainage areas to ft2 units: Tippecanoe River A = 113 mi2 = 3.15 × 109 ft2 ; Wildcat Creek A = 146 mi2 = 4.07 × 109 ft2 . The baseflow/area for the Tippecanoe River is (ft, sec units): Qb 22 = = 6.98 × 10−9 ft/sec A 3.15 × 109 which equals 2.6 in/yr (6.6 cm/yr). The baseflow/area for Wildcat Creek is (ft, sec units): Qb 6.4 = = 1.57 × 10−9 ft/sec A 4.07 × 109

Table 7: Problem 6.

Section No. v (ft/s) width (ft) depth (ft) Q (ft3 /s) Q (m3 /s)

1 0.31 9.4 3.2 9.3

2 0.89 7.3 5.5 35.7

3 1.93 10.7 8.5 175.5

21

4 1.83 11.1 8.2 166.6

5 0.48 5.2 3.4 8.5

Total

396 11.2

figure file: solutions/ch1hydrog Figure 7: Problem 8. which equals 0.59 in/yr (1.50 cm/yr).

The period of drought leaves the baseflow a small fraction of the annual precipitation rate in both basins, but especially so in Wildcat Creek where the baseflow is about 1/4 the baseflow in the Tippecanoe River. The difference is explained by the lower conductivity of subsurface materials in the Wildcat Creek basin.

8. (a) The equation for baseflow is derived by substituting Qq = Qs − Qb into Qs cs = Qb cc + Qq cq and solving for Qb , with the result:

Qb = Qs

cs − cq cb − cq

The quantity in parentheses above is dimensionless, so the units of Qb are the same as the units of Qs , [L3 /T]. (b) The graph shown in Figure 7 was created with a spreadsheet using the above equation for each time step.

9. Scenarios (c) and (d) are most likely to represent long-term average conditions.

Pattern (a) requires a consistent loss mechanism (usually an outlet), and pattern (b) requires a consistent source mechanism (usually an inlet).

10. The exceedance probability is about 0.72. The return period of such a flood is the inverse, 1/0.72 = 1.4 yrs.

11. The peak concentrations are about 11 m below the water table. Assuming the peak concentrations in precipitation occurred in 1963 and the concentration measurements were made in 1986, the vertical average linear velocity here is (m, yr units) vz =

dz −11 = = −0.48 m/yr dt 1986 − 1963

The recharge rate equals the magnitude of vertical specific discharge: R = |qz | = |v z |n = (0.48)(0.29) = 0.14 m/yr 12. There are many contaminant plumes to the southwest of the highest part of the water table, so that is a less desirable quadrant to explore. To the north and southeast are less affected by pollution. It is good to screen wells deep to maximize potential drawdown and yield and to have greater isolation from surface waters which are less clean. Based on the profile, more permeable materials occur deep in the north, just south of the Sandwich moraine. They also occur in the south, just north of the coast.

13. The most conductive materials are coarse sediments deposited high up in alluvial fans near the sides of the basins. These sediments are deposited by mountain streams that slow down as they enter the basin. Finer sediments and evaporite salts are deposited near the center of basins in distal parts of the alluvial fans and in ephemeral lakes that can form in the center of basins. These fine sediments have low conductivity.

22

14. The Dakota Sandstone crops out and is recharged at relatively high elevation in the Black Hills in western South Dakota. It dips down to the east and is confined by shale layers in the east. The elevation of the Dakota Sandstone recharge areas in the west are well above the ground surface elevations in the east, hence the artesian heads. Flow in the aquifer is roughly parallel to the bedding, and flow in the aquitard above is vertically upward normal to the bedding. Flow in the aquifer is to the east in most of S. Dakota. Near the Black Hills dome, flow in the aquifer is radially away from the high recharge areas.

15. First, convert the pressure to lb/ft2 units: 180 lb/in2 = 25,920 lb/ft2 . Then calculate the head at the top of the capped well, which is also the head at the well screen (ft, lb units): h = =

P +z ρw g 25,920 + 0 = 415 ft (126 m) 62.4

If the well casing were extended to an absurd height, the water level would stabilize 415 ft above ground!

16. The travel times are very rapid in karst limestone, so protections would probably be more limiting than in non-karst cases. Surface water and groundwater are tightly linked, so strong protections on surface waters may also be needed.

23

24

Chapter 6 Solutions 1. Rearrange Eq. 6.43 to give (m, day units): T

Q r1 ln 2π(h1 − h0 ) r0 8 600 ln = 2π(430.2 − 427.9) 0.15 = 165 m2 /day =

where h0 , r0 refer to the pumping well, and h1 , r1 refer to the observation well.

2. Write Eq. 6.38 once for point A and once for point B. Then subtract the two equations to get Q rB hB − hA = ln 2πT rA Rearranging this gives (m, day units): 2πT (hB − hA ) ln(rB /rA ) 2π(1200)(134.28 − 134.20) = ln(24/8) 3 = 550 m /day

Q =

3. Use the solution of Eq. 6.44, with D = 1 (horizontal flow is constrained to the x direction, perpendicular to the ditches). The solution is therefore h=−

N 2 x +C 2T

Midway between the two ditches, x = x0 = 0 m and the head is h0 . At the ditches, x = xd = ±125 m and the head is hd . Evaluate the above equation at both xd and x0 , and then subtract to get hd − h0 = −

N 2 (x − x20 ) 2T d

Solve this for N as follows (ft, day units): N

2T (hd − h0 ) x2d − x20 2(35)(1.4) = − 1252 − 02 = −0.0063 ft/day (0.19 cm/day) = −

That N < 0 means that the leakage out the base of the aquifer exceeds the recharge in the top by this amount.

25

4. Use the uniform recharge/leakage solution of Eq. 6.44. The constant D is determined by measuring the ratio ∆y/∆x for one ellipse and plugging that into Eq. 6.48. Doing this yields ∆y/∆x ≈ 0.44 and D ≈ 0.159. If we write two versions of Eq. 6.44 at two points A and B that lie on the x axis (yA = yB = 0) and then subtract these equations, we get N hA − hB = − D(x2A − x2B ) 2T Solve this equation for T , then insert the x coordinate and head at two points A and B. Any two points will do, so long as they are at different x values. In the calculations below, I chose point A on the x axis and the innermost contour, and point B on the x axis and the contour 4 removed from the innermost one (ft, day units): N D(x2A − x2B ) 2(hA − hB ) 0.0025 (0.159)(2192 − 4892 ) = − 2(0.40) = 95 ft2 /day (8.8 m2 /day) = −

T

5. Superposition is when solutions of a linear differential equation are added to each other, and the resulting solution is also a solution. This applies to linear differential equations. Many examples of superposition are described in the chapter, including a well in a uniform flow and wells with image wells near linear and circular boundaries.

6. There is a range of possible answers here. One possibility is to assume that the well is located 5 m ahead of the leading edge of the plume. Assuming that the origin of an x, y coordinate system is at the well, like in Figure 6.16, a point at the upper limit of the capture zone would have approximate coordinates of (−15.4, 4.5). The angle Θ = arctan(y/x) = 2.86 radians. Use Eq. 6.60 to solve for Q (m, day units): 2πT Ay Θ 2π(14 · 8)(−0.0015)(4.5) − 2.86 1.7 m3 /day

Q = − = =

As a check, we could examine the ultimate width of this capture zone far upstream from the well, using Eq. 6.62: y=±

Q 1.7 =± = ±5.1 m 2T A 2(14 · 8)(0.0015)

This seems reasonable, since the y coordinates of the capture zone at the plume are ±4.5.

7. The solution neglects recharge and leakage – all flow is from lateral flow. Close to the well this may be reasonable, but on a larger scale recharge and leakage contribute a significant amount of flow and should not be ignored. The capture zone area is finite when there is finite recharge/leakage into the aquifer. Without accounting for recharge/leakage the area and extent of the capture zone is infinite.

8. (a) Vertical thickness is 30 m (domain top elevation - bottom elevation). K = 1 m/day, T = 30 m2 /day. (b) qx ' 0.0121 m/day.

26

(c) ∂h/∂x ' −0.0121. (d) qx = −K(∂h/∂x) = −(1)(−0.0121) = 0.0121 m/day. (e) Q = qx A = (0.0121)(30)(277) = 100.6 m3 /day (f) The well Q is 100 m3 /day, which essentially the same as computed in the previous step, allowing for inaccuracy in measuring the width of the capture zone.

9. The sketch map would look like Figure 6.20 with a real well and an image well with opposite discharge. The model equation would be

Q r1 2 h = ln +C πK r2 where Q is the well discharge, r1 is the distance from the real well and r2 is the distance from the image well.

10. First, convert quantities to a consistent set of units (ft, day). The discharge Q = 70 gal/min = 13,500 ft3 /day. The radius of the well is rw = 3 in = 0.25 ft. The problem is solved using an image well (Eq. 6.64). Write this equation for a point on the well screen: r1→1 Q ln hw = +C 2πT r2→1 where r1→1 is the distance from the centerline of the real well to a point on its well screen (0.25 ft), and r2→1 is the distance from the imaginary well to the well screen of the real well (100 ft). Because of the symmetry of the problem, r1 = r2 everywhere along the lake shore and there the head is constant hL = C. Subtracting these two heads allows isolation of T : Q r1→1 hL − hw = − ln 2πT r2→1 Q r1→1 T = − ln 2π(hL − hw ) r2→1 13, 500 (0.25) = − ln 2π(6.5) (100) 2 = 1980 ft /day (184 m2 /day) 11. Call the real injection well at (26,37) well 1 and the real pumping well at (13,0) well 2. (a) Each real wells needs a corresponding image well. The image for well 2 is located at (−13,0) with a discharge Q (well 3), and the image for well 1 is located at (−26,37) with a discharge −Q (well 4). The solution is then Q r2 r3 ln +C 2πT r1 r4 Far away from these wells, r1 ≈ r2 ≈ r3 ≈ r4 , heads are unaffected by the pumping and injection, and h ≈ C, so C = 77.3 m. The transmissivity of the aquifer is T = Kb = (0.07)(73.9 − 70.0) = 0.273 m2 /day. (b) Rearrange the solution to solve for the maximum discharge Qm , where h is evaluated at the well screen of well 1 (m, day units): r2 r3 Qm = (h − C)2πT / ln r1 r4 (39.2)(53.7) = (79.0 − 77.3)2π(0.273)/ ln (0.1)(52.0) = 0.49 m3 /day h=

27

Table 8: Problem 11(c).

Location Pumping well 1/4 1/2 3/4 Injection well

h 75.4 76.8 77.2 77.6 79.0

(c) The heads are listed in Table 8. (d) Use the heads at the 1/4 and 1/2 points in Table 8 in a calculation of average linear velocity: vs

= ≈ = =

qs ne −K∆h/∆s ne −(0.07)(76.8 − 77.2)/(9.8) 0.31 0.0092 m/day

(e) To evaluate the head at point M , measure the distances r1 , r2 , etc. and plug them into the solution listed in part (a). Doing that results in hM = 77.5 m.

At the base of the aquifer, P = (hM − z)ρw g = (77.5 − 70.0)9810 = 73,600 N/m2 12. The arrangement of image wells for this solution is shown in Figure 8. The equation for head is Q r1 r4 h= ln +C 2πT r2 r3 where the subscripts refer to the wells as numbered in Figure 8. At any point on the canal (x axis), r1 = r2 and r3 = r4 and the p solution reduces to h = C. Therefore, C = 200 m. At the point (10,40), r1 = (100 − 10)2 + (50 − 40)2 = 90.6 m.

Similarly, r2 = 127.3, r3 = 142.1, and r4 = 110.5 m. Using these values in the solution gives (m, day units): h = =

2500 (90.6)(110.5) ln + 200 2π(600) (127.3)(142.1) 199.6 m

13. This problem is similar to the example worked in Section 6.4.10. The drawdown at the well is given by (m, day units): hd

= = =

Q r1 ln 2πT r2 0.4 600 ln 2π(150) 130 −3.7 m

where r1 is the distance from the centerline of the real well to a point on its screen, and r2 is the distance from the image well to a point on the screen of the real well.

28

figure file: solutions/ch6slurrycanal Figure 8: Arrangement of image wells for Problem 12.

14. First, get all given values in ft, day units: Q = 50 gal/min = 9625 ft3 /day; the well radius is 3 in = 0.25 ft. This problem makes use of imaging to create a circular constant head boundary, and it makes use of the long-term drawdown analysis as discussed in Section 6.4.10. The stream and wetlands boundaries form a rough circle of radius R = 225 ft. The real well is located at d1 = 100 ft from the center of the circle. In accord with Eq. 6.74, the image well has to be located outside the circle at d2 = R2 /d1 = 506 ft from the center of the circle. The equation for the long-term drawdown due to the real well is hd =

r1 Q ln +C 2πT r2

where r1 is the distance from the real well to the point where hd is evaluated, and r2 is the radial distance from the image well to this point. Along the circular boundary, the head is constant so hd = 0. Use this condition to solve for the constant C (ft, day units): C

= = =

Q r1 ln 2πT r2 125 9625 ln 0− 2π(4500) 506 − 225 0.28 ft (8.5 cm) hd −

where the distances r1 and r2 are from the real and imaginary wells, respectively, to a point on the circle directly between the two wells. Now use the solution to predict the drawdown at the screen of the real well: hd

Q r1 ln +C 2πT r2 9625 0.25 = ln + 0.28 2π(4500) (506 − 100) = −2.2 ft (-67 cm) =

The head at the well drops 2.2 ft after a long time of pumping.

15. The solution has the following form: h2 = −

N 2 x + Ax + C K

The constants A and C can be determined by applying the known head conditions at x = 0 and x = 300. At x = 0, h2 = 202 = 400 = C Rearrange the solution to solve for A using the conditions at x = 300 (ft, day units):

1 N h2 − C + x2 A = x K

1 0.007 = 102 − 400 + 3002 300 1.5 = 0.40 29

Table 9: Problems 15 and 16. x 0 25 50 75 100 125 150 175 200 225 250 275 300

h (15) 20.00 20.18 20.21 20.09 19.83 19.42 18.84 18.09 17.13 15.93 14.43 12.53 10.00

h (16) 20.00 19.36 18.71 18.03 17.32 16.58 15.81 15.00 14.14 13.23 12.25 11.18 10.00

figure file: solutions/ch6unconf Figure 9: Problems 15 and 16.

The solution, with the constants A and C determined is h2 = −

N 2 x + 0.40x + 400 K

The solution is tabulated in Table 9 and graphed in Figure 9.

The divide will occur where ∂h/∂x = 0, which occurs where ∂(h2 )/∂x = 0. Differentiating the solution gives ∂(h2 ) N = − 2x + A ∂x K Setting this equal to zero lets us find xd , the x coordinate at the divide: xd =

AK (0.4)(1.5) = = 42.9 ft (13.1 m) 2N 2(0.007)

16. In this case, the solution is of the form h2 = Ax + C The constants A and C can be determined by applying the known head conditions at x = 0 and x = 300. At x = 0, h2 = 202 = 400 = C Rearrange the solution to solve for A using the conditions at x = 300 (ft, day units): A

h2 − C x 102 − 400 = 300 = −1.0 =

30

The solution, with the constants A and C determined is h2 = −x + 400 The solution is tabulated in Table 9 and graphed in Figure 9. Without the recharge, the water table is lower than in the previous problem. There is no divide; everywhere between the canals water flows in the positive x direction.

17. (a) For this problem, superpose two radial flow solutions, one for the pumping well (well 1) and one for the injection well (well 2): h2 =

r1 Q ln +C πK r2

Write this equation once for a point on the screen of the pumping well (h1 ) and once for the injection well (h2 ) and then subtract to get

Q r1→2 r1→1 2 2 h2 − h1 = ln − ln πK r2→2 r2→1 where r1→2 = r2→1 = 30 m, r1→1 = 0.1 m is the distance from the centerline of well 1 to its screen, and r2→2 = 0.2 m is the distance from the centerline of well 2 to its screen. Isolating K from the above and combining logarithms gives (m, day units): K

= = =

Q (r1→2 )(r2→1 ) ln 2 2 π(h2 − h1 ) (r2→2 )(r1→1 ) (30)(30) 125 ln 2 2 π(14.9 − 8.6 ) (0.2)(0.1) 2.88 m/day

(b) Far from the well pair, the pumping has no effect since there is no net discharge.

Mathematically, far from the well pair, r1 /r2 → 1, and h2 → C. To find the head far from the wells, solve the solution for C using a point where h is known, such as at the pumping well: h20 = C

Q r1→1 ln πK r2→1 125 0.1 = 8.62 − ln π(2.88) 30 2 = 152.8 m =

h21 −

Taking the square root of this gives the head far from the wells; h0 = 12.4 m.

√

152.8 =

18. This depends on how the model was constructed and in particular what far-field boundary conditions were used. I created a model with head-specified boundaries far away from the area of the wells, and set the heads at these to 12.36 m, the computed far-field head. The model matches the solution computed by hand very closely.

19. Only cases (b) and (d) are appropriate for flow net analysis.

20. Use Eq. 6.96 to estimate the total discharge under the dam (m, day units): ns |Q| = Kb|∆ht | nh 3.5 = (0.06)(312)(86.0 − 62.0) 10 = 160 m3 /day 31

p 21. First draw the domain in X, Z coordinates, where Z = 0.09/0.01 z = 3z. Then draw a flow net in the X, Z domain. Doing that, I get an ns /nh ratio of about 0.52. √ K K The equivalent isotropic conductivity for discharge calculations is K = x z = p (0.09)(0.01) = 0.03 m/day. The estimated total discharge under the dam in this case is (m, day units):

=

ns nh (0.03)(312)(86.0 − 62.0)(0.52)

=

120 m3 /day

|Q| =

Kb|∆ht |

22. (a) Most of the head loss is in the clayey till, because its conductivity is about 67 times lower than that of the fine sand. (b) Draw a flow net for the clayey till, assuming that the head at the upstream interface with the fine sand equals the reservoir head and the head at the downstream interface with the fine sand equals the tailwater head. The interfaces with the concrete dam and the granite can be assumed impermeable. I drew a flow net with 2 head drops and 6.5 stream tubes. The discharge is (ft, day units): |Q| = Kb|∆ht |

ns nh

=

(0.003)(1125)(163.0 − 142.1)

=

230 ft3 /day (6.5 m3 /day)

6.5 2

(c) The flow rates will be the highest at the base of the clayey till near the granite interface. Looking at the flow net, this is where the boxes are smallest and the hydraulic gradients are steepest. At the base, the estimated v is (142.1−163.0)

vs =

−(0.003) −K ∂h (4.0) ∂s = ne 0.17

= 0.09 ft/day (2.7 cm/day)

(d) Assumptions made: (1) the clayey till is homogeneous with isotropic K, (2) the other materials are also homogeneous and isotropic with the conductivities listed, (3) flow is two-dimensional in the vertical plane. Item (1) is probably the most important source of uncertainty. To reduce the uncertainty associated with (1), you could conduct K tests on the clayey till compacted as it will be when constructed. Horizontal K values would be most appropriate for this analysis since flow is mostly horizontal. To reduce uncertainty associated with (2), K tests of these materials could be made. To reduce uncertainty associated with (3), investigate hydrogeologic conditions at the two abutments at the ends of the dam, and check the continuity of conditions along the dam.

23. (a) The discharge per unit length of dam is Q = 0.67 m3 /day. (b) The discharge per unit length of dam is Q = 1.85 m3 /day. (c) The discharge per unit length of dam is Q = 0.84 m3 /day. (d) With the model of part (c), the flow originates closer to the dam in the reservoir and exits closer to the dam in the tailwater, because there is little resistance to vertical flow. The model of part (b) has flow originating and exiting farther from the dam because the vertical resistance to flow exceeds the horizontal resistance.

32

(e) With the geometry of this situation, there is more horizontal than vertical flow.

Thus, changing the horizontal K has a greater impact on the Q than changing the vertical K does.

24. Steady flow will occur when the boundary conditions are steady — when the reservoir and tailwater levels are kept steady. Transient flow occurs when these levels fluctuate, such as when the reservoir is drawn down rapidly or when it rises rapidly in a flood.

33

34

Chapter 7 Solutions 1. (a) The pore water pressure is (ft, lb units): P = ρw g(h − z) = 62.4(16.0 − 12.2) = 237 lb/ft2 (11300 N/m2 ) The total vertical stress is (ft, lb units): X σvt = ρi gdi =

(115)(8.5 − 0.0) + (121)(12.0 − 8.5) + (129)(16.0 − 12.0)

=

1917 lb/ft2 (91800 N/m2 )

where ρ1 is the density of soil layer i, and di is its thickness. The vertical effective stress is (ft, lb units): σve = σvt − P = 1917 − 237 = 1680 lb/ft2 (80400 N/m2 ) (b) The pore water pressure is (ft, lb units): P = ρw g(h − z) = 62.4(16.0 − 13.2) = 175 lb/ft2 (8380 N/m2 ) The total vertical stress is (ft, lb units): X σvt = ρi gdi =

(115)(8.5 − 0.0) + (121)(13.0 − 8.5) + (129)(16.0 − 13.0)

=

1909 lb/ft2 (91400 N/m2 )

The vertical effective stress is (ft, lb units): σve = σvt − P = 1909 − 175 = 1734 lb/ft2 (82000 N/m2 ) (c) It would compress because of the increase in σve .

2. Immediately after starting the pump, the local pore water pressure drops, which does not change the total stress, but it does increase the effective stress. With more time, there is some drainage at the water table, which does decrease total stress some.

3. The water level in the deep, confined well will decline under the high atmospheric pressure. This is because the pressure in the well bore increases in lock-step with the atmospheric pressure, but the pore water pressure in the aquifer increases somewhat less than the atmospheric pressure does. This pressure difference causes water to flow from the well into the aquifer. In the unconfined well, the pressure change in the atmosphere, well bore, and aquifer are all the same, so there is no induced flow or water level change.

35

4. The figure shows an atmospheric pressure change of dPa ' −1.4 inches Hg, which is equivalent to −36 mm Hg or −4725 N/m2 . The head in the well rose by about dh ' 0.81 ft (0.25 m). For this event the barometric efficiency is (m, sec, kg units) BE

ρw gdh dPa (9810)(0.25) = − = 0.51 −4725 = −

5. Precipitation generally increases saturation in the unsaturated zone and increases the height of the water table, both of which contribute to increased total stress. Pore water pressures tend to increase as water levels rise. Effective stress can increase or decrease depending on the magnitude of the total stress and pore water pressure changes. Slopes become less stable because increases in saturation increase total stress, and increases in pore pressure cause reductions in shear strength.

6. We could inject water under high pressure into a fault zone, which would increase pore pressure, decrease effective stress, and decrease shear strength. This would increase the chance of triggering earthquakes.

7. The sand layer will add to the total stress and ultimately to the effective stress an amount equal to (ft, lb units): dσve = ρgb = (120)(10) = 1200 lb/ft2 (57500 N/m2 ) Solving Eq. 7.19 for the change in the clay layer thickness gives (ft, lb units): db = −α dσve b0 = −(4.8 × 10−6 )(1200)(25) = −0.14 ft = −1.7 in (-4.3 cm) 8. The localized zones of greatest subsidence probably correspond to locations where there is more pumping from groundwater wells. These locations tend to be where the underlying sediments are thick in the central parts of the valley, not near the margins where the sediments are thinner. The pumping centers may correspond to locations where certain water-needy crops are grown and/or where the aquifer is more permeable.

9. Drawing a line parallel to the unload–reload cycles, the vertical strain per ∆h = −100 ft is about ∆b/b0 = −3.2 × 10−4 (there is a minus sign because b shrinks as head drops). The compressibility is then calculated as (ft, lb units): α

db/b0 db/b0 db/b0 = = dσve dP ρw g d(P/ρw g) db/b0 ρw g dh −3.2 × 10−4 = 5.1 × 10−8 ft2 /lb (62.4)(−100)

= − = =

which equals α = 1.1 × 10−9 m2 /N. 10. The total strain over the time period is about db/b0 = −1.05 × 10−3 . Therefore (ft units),

db db = b0 = (578)(−1.05 × 10−3 ) = −0.61 ft (-0.19 m) b0 36

figure file: solutions/ch5claydrain Figure 10: Problem 12.

11. Consolidation of the clay layers in the sequence proceeds slowly. Because of this, pore pressures in the clayey layers will exceed the pore pressures in sandy layers for a long time, and the clay layers will continue to drain and compress long after the sandy layers have drained and compressed under new loads. The heads observed in wells reflect the pore pressures in the sandy layers, not in the clayey layers.

12. See Figure 10.

13. (a) The ultimate subsidence after consolidation is about 6.5 cm = 0.065 m. The compressibility is (m, N units): α

db/b0 dσve db/b0 ρw g dh (−0.065)/(5.0) = 1.9 × 10−7 m2 /N (9810)(−7)

−

= = =

(b) About half of the total consolidation (subsidence) occurred by t ≈ 45 days.

From Figure 7.19, 50% of consolidation corresponds to a time factor of Ft = 0.20. Using Eq. 7.21, the vertical conductivity is (m, N, day units): Kz

=

Ft ρw gα(b/2)2 t (0.20)(9810)(1.9 × 10−7 )(5/2)2 45 −5 5.2 × 10 m/d

=

6.0 × 10−8 cm/sec

= =

14. (a) At 50% of total compression the time factor Tf ' 0.20 (Figure 7.19). Solve Eq. 7.21 for Kz (m, sec, kg units): Kz

= = =

Ft ρw gα(b/2)2 t (0.20)(9810)(3 × 10−7 )(6/2)2 (55)(60)(60)(24) 1.1 × 10−9 m/s = 1.1 × 10−7 cm/s

(b) When using Eq. 7.21 in this case, we would replace (b/2)2 with b2 , since the clay can only drain to its lower boundary. Therefore, the time to achieve 50% compression will be t = (55)(2)2 = 220 days.

15. (a) The increase in effective stress is (m, sec, kg units): dσve

= −ρw gdh = −(9810)(−40) = 392400 N/m2 37

(b) The aquifer compression will be dominated by compression of the silt/clay layers, which have a total thickness of 55(0.10) = 5.5 m. Assuming virgin compression, the total silt/clay compression will be db = −α dσve b0 = −(1.6 × 10−7 )(392400)(5.5) = −0.35 m The land surface will go down. (c) It could be more if the aquitards above and below the aquifer also compress.

It could be less if the compression of clay/silt layers in the aquifer is governed by recompression instead of virgin compression (perhaps the aquifer has experienced such high effective stresses before). (d) In this case, the aquifer will swell and the land will come up. The effective stress decreases by the same amount it increased originally, and the correct compressibility is the unload/reload compressibility: db = −α dσve b0 = −(5 × 10−8 )(−392400)(5.5) =

0.11 m

(e) The storativity is S

' bρw gα =

(5.5)(9810)(5 × 10−8 ) = 2.7 × 10−3

16. The critical excavation depth is where, at the clay/sand interface, σvt = P . The pore pressure at the interface is (ft, lb units): P = (h − z)ρw g = (18 − 4)(62.4) = 874 lb/ft2 (41800 N/m2 ) If the depth of the excavation is d, the total vertical stress at the clay/sand interface under the excavation is σvt = (18 − d)ρclay g Solving this for the maximum excavation depth d gives (ft, lb units): σt d = 18 − ρclay g P = 18 − ρclay g 874 = 18 − = 10.9 ft (3.3 m) 123 17. The confined aquifer will experience greater drawdown because its storativity is much smaller than the unconfined aquifer’s specific yield. A given amount of drawdown liberates much more water from storage in an unconfined aquifer than in a confined aquifer. Not as much drawdown is needed in the unconfined aquifer to draw the well’s discharge from storage.

18. (a) The specific storage is (m, N, sec units): Ss

= =

ρw g(nβ + α)

9810 (0.26)(4.5 × 10−10 ) + 2.0 × 10−9 = 2.1 × 10−5 m−1

The storativity is S = Ss b = (2.1 × 10−5 )(20) = 4.1 × 10−4 38

(b) dVw = −SAdh = −(4.1 × 10−4 )(1000)2 (−2) = 830 m3 19. (a) The increase in effective stress is (m, sec, kg units): dσve

=

−ρw gdh

=

−(9810)(−4) = 39240 N/m2

(b) The aquifer compression will be db =

−α dσve b0

=

−(6 × 10−8 )(39240)(45)

=

−0.11 m

(c) Specific storage is Ss

'

ρw gα

=

(9810)(6 × 10−8 ) = 5.9 × 10−4 m−1

Storativity is S

= Ss b =

(5.9 × 10−4 )(45) = 2.6 × 10−2

(d) The volume of extracted water is dVw

= −SAdh = −(2.6 × 10−2 )(1000)2 (−4) = 1.1 × 105 m3

20. The elastic storativity in this case is (m, N, sec units): Ss b ≈ ρw gαb = (9810)(3.0 × 10−9 )(20) = 0.0059 The ratio of water table storage to elastic storage in this aquifer is 0.13 Sy = = 22 Ss b 0.0059 21. Use a rearrangement of Eq. 7.31 (m units): dh = −

dVw 650,000 =− = −1.7 m Sy A (0.15)(2.5 × 106 )

Heads would have to drop 1.7 m, on average.

39

40

Chapter 8 Solutions 1. Analyze this with the Cooper et al.(1967) method. Using the slug test spreadsheet supplied with the text, the best-fit values are about Kr = 0.78 m/day and Ss = 1.8 × 10−5 m−1 . The ε = 10−4 curve was selected as the closest fit. With a higher K, reasonable fit can be had with Kr = 1.0 m/day and Ss = 1.8 × 10−6 m−1 . With a lower K, reasonable fit can be had with Kr = 0.56 m/day and Ss = 1.8 × 10−4 m−1 .

2. Analyze this with the Cooper et al.(1967) method. Using the slug test spreadsheet supplied with the text, the best-fit values are about Kr = 4.3 m/day and Ss = 3.6 × 10−4 m−1 . The ε = 10−2 curve was selected as the closest fit.

3. Using the Dagan method in the slug test spreadsheet supplied with the text, the best-fit value is about Kr = 0.56 m/day. Computed intermediate values include S ∗ = 3.3 × 10−3 m−1 , Υ = 6.3 × 10−3 , b/B = 0.27, and P = 0.23.

4. The Theis solution assumes that an aquifer is infinite with no lateral boundary conditions and that it is perfectly nonleaky. The only water that can supply the well must come from storage. Real aquifers have finite extent, boundaries where they connect to surface waters, and some leakage. The leakage and boundaries supply water that is not accounted for with the Theis solution. These factors are more important with greater distance from the well and with longer times of pumping.

5. This problem involves straightforward use of the Theis solution, Eqs. 8.31 and 8.32.

1/u is calculated with Eq. 8.31 and then W is determined from Figure 8.9. The results of this problem and the next are tabulated in Table 10.

6. This problem applies the Hantush–Jacob solution. 1/u is calculated with Eq. 8.31 and then W is determined from Figure 8.14 with Λ = 0.5 as calculated below (ft, day units): r Kz0 Λ = r T b0

Table 10: Problems 5, 6, and 7.

Prob. 5 t(days) 0.0069 0.0417 0.2083 1.0000

1/u 2.78 16.67 83.33 400.00

W 0.75 2.2 3.9 5.5

h0 − h (ft) 0.45 1.31 2.33 3.28

Prob. 6 W 0.75 1.6 1.9 1.9

41

h0 − h (ft) 0.45 0.95 1.13 1.13

Prob. 7 W 0.27 0.90

h0 − h (ft) 0.16 0.54

Table 11: Problem 9. t(sec) 0.0 0.1 1.0 10.0 10.1 11.0 20.0

u1 ∞ 0.185 0.0185 0.00185 0.0183 0.00168 0.00924

W (u1 ) 0.0 1.3 3.4 5.7 5.7 5.9 6.3

u2

W (u2 )

∞ 0.185 0.0185 0.00185

s =

100

=

0.5

0.0 1.3 3.4 5.7

h0 − h (m) 0.00 0.21 0.56 0.93 0.72 0.41 0.10

0.05 (200)(10)

The results of this problem and the previous one are tabulated in Table 10.

7. (a) The solution is valid up to time (ft, day units): t = =

b0 S 0 10Kz0 (10)(0.003) = 0.06 days = 1.44 hours 10(0.05)

(b) See Table 10. (c) This solution predicts less drawdown than both Theis (non-leaky) and HantushJacob (leaky, but no storage in aquitard). Hantush (1960) is like HantushJacob, but accounts for water removed from storage in the aquitard, which is significant. Without that water, there needs to be greater drawdown in the aquifer to extract water from storage there.

8. With the Hantush-Jacob solution, drawdown stabilizes because a steady-state solution develops where all of the well discharge is supplied by leakage through the aquitard. The Theis solution never stabilizes because it can only supply the well discharge by removing water from storage in the aquifer.

9. For t < 10 sec, this is an application of the Theis solution. For t > 10 sec, use superposition in time as shown with Eqs. 8.67–8.69. Using the notation of these equations, t1 = 0, t2 = 10 sec, Q1 = 150 m3 /day, and Q2 = 0. The results for this problem are listed in Table 11.

10. This problem is an application of the Hantush–Jacob solution (Section 8.4.2). In this problem the factor Λ is (m, day units): r Kz0 Λ = r T b0 s 1.5 = 0.5 (73)(1) =

0.072

The results for this problem are listed in Table 12.

42

Table 12: Problem 10. t(sec) 0.0 0.1 1.0 10.0 100.0 1000.0

u ∞ 0.185 0.0185 0.00185 0.000185 0.0000185

W (u, Λ) 0.0 1.3 3.4 5.1 5.4 5.4

h0 − h (m) 0.00 0.21 0.56 0.83 0.88 0.88

Table 13: Problem 11. t(min) 0 1 10 100 1000 104 105 106

1/uA ∞ 7.7 × 10−1 7.7 × 100 7.7 × 101

W (uA , η) 0.0 0.048 0.10 0.10

1/uB

1.7 × 10−2 1.7 × 10−1 1.7 × 100 1.7 × 101 1.7 × 102

W (uB , η)

0.10 0.14 0.51 2.20 4.50

h0 − h (ft) 0.00 0.06 0.11 0.11 0.16 0.58 2.50 5.10

11. This problem is an application of the Neuman solution for an unconfined aquifer (Section 8.4.4). The elastic storativity, which applies at early times is (ft, day units): S = Ss b = (7 × 10−7 )(45) = 3.1 × 10−5 The dimensionless parameter η is η=

r2 Kz (250)2 (1.0) = 2.6 = b2 Kr (45)2 (12)

Before using the equations, all parameters must be in a common set of units. Using ft, min units, the discharge is Q = 40 gal/min = 5.35 ft3 /min and the transmissivity is T = Kb = (12)(45) = 540 ft2 /day = 0.375 ft2 /min (0.0348 m2 /min) The results are tabulated in Table 13. The early-time solution applies up until about t = 100 min, at which point both the early and late time solutions apply. For t > 100 min, the late-time solution should be employed.

12. Time will be higher for the high end of the Sy range. Higher Sy means that more water is removed from storage per decline in head. The dewatering project needs to reduce heads to a certain level. If Sy is higher, more water will need to be removed to reduce heads to that level.

13. The drawdown could stabilize for a variety of reasons. It could stabilize as a steady state develops due to some nearby constant head boundary supplying water (a river or lake, for example). It could stabilize as a steady state develops due to leakage through bounding aquitards. It could appear to stabilize due to some other transient effect (e.g. precipitation/recharge) that is counter-acting the well’s impact.

43

14. Using the match point data given in Figure 8.19, the transmissivity is (m, min units): T

= =

Q Wp 4π(h0 − h)p 0.17 0.53 = 0.0072 m2 /min 4π(1.0)

This is equivalent to T = 10.3 m2 /day. The storativity is calculated as S

= =

4T (t − t0 )p (1/u)p r2 4(0.0072)(1.0) = 9 × 10−4 (5.1)(2.5)2

15. The t − t0 vs. h0 − h data are consistent with a nonleaky confined aquifer, so use the Theis solution with the log–log curve-matching procedure. One possible match point is Wp = (1/u)p = 1.0, (h0 − h)p = 0.12 m, and (t − t0 )p = 0.20 min. With these values, the transmissivity is (m, min units): T

= =

Q Wp 4π(h0 − h)p 1.3 1.0 = 0.86 m2 /min 4π(0.12)

This is equivalent to T = 1240 m2 /day. The storativity is calculated as S

= =

4T (t − t0 )p (1/u)p r2 4(0.86)(0.20) = 7.6 × 10−5 (1.0)(95)2

16. Using the method of Section 8.5.2, the results are (m, min units): T

= =

ln(10)Q 4π∆(h0 − h) ln(10)(1.3) = 0.85 m2 /min 4π(0.28)

and S

= =

4T D eγ r 2 4(0.85)(0.31) = 6.6 × 10−5 eγ (95)2

These results are similar to those using the log–log method (see previous problem).

17. The t − t0 vs. h0 − h data are consistent with a leaky confined aquifer, so use the Hantush–Jacob solution with the log–log curve-matching procedure. The leaky layer is the clayey silt, which provides leakage from the shallower sand to the deeper sand where the pumping occurs. The best-fit curve is the one with Λ = 0.6. One possible match point with this curve is Wp = (1/u)p = 1.0, (h0 − h)p = 5.4 ft, and (t − t0 )p = 0.90 min. Before calculating T , convert the discharge to ft, min units: 44

Q = 35 gal/min = 4.7 ft3 /min. Using the match point values, the transmissivity is (ft, min units): T

= =

Q Wp 4π(h0 − h)p 4.7 1.0 = 0.069 ft2 /min 4π(5.4)

This is equivalent to T = 99 ft2 /day. The storativity is calculated as S

= =

4T (t − t0 )p (1/u)p r2 4(0.069)(0.9) = 2.2 × 10−5 (1.0)(107)2

The vertical conductivity of the leaky clayey silt layer can be estimated using a form of Eq. 8.39 (ft, min units): Kz0

2 Λ = Tb r 0

0.6 107

2

=

(0.069)(15)

=

3.3 × 10−5 ft/min (1.0 × 10−5 m/min )

which is equivalent to 4.7 × 10−2 ft/day.

18. Using the pumping test spreadsheet that is supplied with the text, I get the following results: T = 0.0125 m2 /min, S = 0.005, Sy = 0.09, Kz /Kr = 0.3. The data was matched to the type curve for η = 0.06.

19. I used two transient well solutions in the TWODAN software to simulate this one, with the real well (Q = 55 m3 /day) at (35,0) and the image well (Q = −55 m3 /day) at (−35,0). This creates a constant-head, zero-drawdown boundary along the y axis.

With the model, examine h0 − h vs. t − t0 at a point on the real well’s screen, such as (35, 0.4), and adjust the aquifer T and S until the modeled h0 − h vs. t − t0 matches the observed h0 − h vs. t − t0 . Doing this results in T = 1.7 m2 /day and S = 7.6 × 10−6 . The horizontal conductivity is (m, day units): K=

T 1.7 = = 0.28 m/day b 6

20.

21. The computation methods are the same as in the example problem. Results are shown in Table 14. These results show somewhat higher mounding than the example problem. Reducing K and reducing Sy both tend to make mounding larger, all other factors being equal.

45

Table 14: Problem 21 t (day) 1 10 100

ζ (m2 ) 46.7 467 4667

a 4.4 1.39 0.44

b 1.4 0.46 0.15

46

F 0.94 0.67 0.21

h2 (m2 ) 37.6 47.2 71.0

h − h0 (m) 0.13 0.87 2.4

Chapter 9 Solutions 1. Figure 11 shows a sketch of a function y(x). The derivative dy/dx at point a is the slope of the tangent line (dotted) at that point. A finite difference approximation of the derivative at a is the slope of a line connecting two points on the curve near a.

The slope ∆y/∆x of the line from b to c (dashed) is a finite difference approximation of the derivative at a. A derivative is a measure at a point, but a finite difference is a measure over a finite interval.

2. First, examine how the expressions for the coefficients C(x+) , C(x−) , . . . simplify, given the assumptions of homogeneous, isotropic K and uniform grid spacing. Consider C(x+) : Kx(→x+) ∆y∆z = K∆ C(x+) = ∆x where ∆ = ∆x = ∆y = ∆z is the uniform grid spacing and K is the isotropic conductivity. All of the other C coefficients reduce to the same; C(x−) = K∆, C(y+) = K∆, . . . With these values for the C coefficients, and ∂h/∂t = 0, Eq. 9.5 becomes

K∆ (h(x+) − h) + (h(x−) − h) + (h(y+) − h)

+ (h(y−) − h) + (h(z+) − h) + (h(z−) − h) + Qs = 0 Dividing by K∆ and rearranging gives

1 Qs h(x+) + h(x−) + h(y+) + h(y−) + h(z+) + h(z−) + h= 6 K∆ If there is no source/sink in the cell, Qs = 0 and the equation becomes h=

1 h(x+) + h(x−) + h(y+) + h(y−) + h(z+) + h(z−) 6

This states that the head at the central node equals the average of heads at the six neighboring nodes. This result is consistent with the three-dimensional mean value theorem, which applies to solutions of Laplace’s equation.

3. Substitute finite difference representations of the second derivatives, as shown in the text for this problem. ∇2 h = =

∂2h ∂2h ∂2h + 2 + 2 ∂x2 ∂y ∂z ((hx+ − h)/∆) − ((h − hx− )/∆) ∆

figure file: solutions/ch9derivative Figure 11: Problem 1.

47

((hy+ − h)/∆) − ((h − hy− )/∆) ∆ ((hz+ − h)/∆) − ((h − hz− )/∆) + ∆ hx+ + hx− + hy+ + hy− + hz+ + hz− − 6h ∆2 +

=

where ∆ = ∆x = ∆y = ∆z: Since for solutions of Laplace’s equation, ∇2 h = 0, this equation can be rearranged to give the same result as in the previous problem. h=

1 h(x+) + h(x−) + h(y+) + h(y−) + h(z+) + h(z−) 6

This equation states that the head at a node equals the average of heads at the six neighboring nodes. This is consistent with the mean value theorem, which applies to solutions of Laplace’s equation and states that h at a point equals the average of h values on a sphere centered at the point.

4. Modifying Eq. 9.5 for the case of one-dimensional steady flow with no sources/sinks gives C(x+) (h(x+) − h) + C(x−) (h(x−) − h) = 0 C(x−) h(x−) − (C(x−) + C(x+) )h + C(x+) h(x+) = 0 Writing this equation for node 2 gives C(1−2) h1 − (C(1−2) + C(2−3) )h2 + C(2−3) h3 = 0 where, for example, C(1−2) is the conductance factor that applies between nodes 1 and 2. Doing the same for node 3 gives C(2−3) h2 − (C(2−3) + C(3−4) )h3 + C(3−4) h4 = 0 Now examine the conductance factors. The general formulas for conductance factors are Eqs. 9.8 and 9.10, but with uniform grid spacing ∆ = ∆x = ∆y = ∆z these simplify to, for example: C(x+) = Kx(→x+) ∆ K is the same between blocks 1 and 2 and between blocks 3 and 4, but between nodes 2 and 3 K changes and K(2−3) is calculated with Eq. 9.9: K(2−3)

= =

2 (1/K2 ) + (1/K3 ) 2 = 5.217 ft/day (1.59 m/day) (1/3) + (1/20)

Substituting the definitions of the conductance factors into the finite difference equation for node 2 gives: K(1−2) h1 − (K(1−2) + K(2−3) )h2 + K(2−3) h3 = 0 3h1 − 8.217h2 + 5.217h3 = 0

Doing the same for node 3 gives: K(2−3) h2 − (K(2−3) + K(3−4) )h3 + K(3−4) h4 = 0 5.217h2 − 25.217h3 + 20h4 = 0

48

These two previous equations have two unknowns, h2 and h3 , and can be solved by substitution or subtraction. One way to solve by subtraction is as follows: 5.217 8.217 (3h1 − 8.217h2 +

5.217h3 = 0)

+ 5.217h2 − 25.217h3 + 20h4 = 0 which results in 1.905h1 − 21.905h3 + 20h4 = 0 Solving this for the only unknown h3 gives: h3

= =

1 (1.905h1 + 20h4 ) 21.905 1 (1.905(12.0) + 20(11.0)) = 11.09 ft (3.38 m) 21.905

Solving a previous equation for h2 and substituting the value for h3 lets us determine h2 : h2

= =

1 (3h1 + 5.217h3 ) 8.217 1 (3(12.0) + 5.217(11.09)) = 11.42 ft (3.48 m) 8.217

5. The model-predicted head will likely be lower than actual head. Generally model cells are much larger than the well radius. The finite difference method assumes a linear variation in head (constant gradient) between nodes, but near a well the variation in head is actually logarithmic, with steeper gradients closer to the well.

The linear variation in a finite difference model misses this behavior, underestimating the drawdown in a pumping well and draw-up in an injection well.

6. In this case, the finite difference equation (Eq. 9.5) simplifies to

1 Qs h = h(x+) + h(x−) + h(y+) + h(y−) + 4 K∆z

Qs 1 = h(x+) + h(x−) + h(y+) + h(y−) + 4 T The development of this equation directly parallels the development shown for Problem 1 of this chapter, but without contributions from the z direction. Substitute into the previous equation the node discharge and the heads at the four neighboring nodes to get the head at the central node (m, day units):

1 −150 h= 4(382.3) + = 380.42 m 4 20 For comparison, use the analytic solution for radial flow with a known head h0 at radius r0 (see Eq. 6.43): Q rw ln + h0 hw = 2πT r0 where hw is the head at the well radius rw , r0 = 15 m, and h0 = 382.3 m. If rw = 0.1 m, hw = 376.3 m. If rw = 1.0 m, hw = 379.1 m. The result is relatively sensitive to the radius of the well. The finite difference approximation is too high, which is generally the case when grid blocks are much larger than the well radius.

The reason is that the actual solution for h varies logarithmically near a well, but the finite difference solution assumes a linear variation.

49

figure file: solutions/ch82dmodel Figure 12: Head distribution for Problems 8 and 9. figure file: solutions/ch8theis Figure 13: Head vs. time at r = 100 ft for Problem 10.

7. For the case that ∆x = ∆y and ∆z = constant, the conductance factors (Eqs. 9.8 and 9.10) are of the form C(x+) = Kx(→x+) ∆z The general finite difference equation for a node (Eq. 9.5) under these conditions becomes: C(x+) (h(x+) − h) + C(x−) (h(x−) − h) + C(y+) (h(y+) − h) + C(y−) (h(y−) − h) + Qs = 0 Substituting the conductance factor definitions into the above and dividing through by ∆z gives K(x+) (h(x+) − h) + K(x−) (h(x−) − h) + K(y+) (h(y+) − h) + K(y−) (h(y−) − h) + Qs = 0 Solving this for h results in the following nodal equation: h=

K(x+) h(x+) + K(x−) h(x−) + K(y+) h(y+) + K(y−) h(y−) + Qs K(x+) + K(x−) + K(y+) + K(y−)

The values of K(x+) , K(x−) , K(y+) , and K(y−) are calculated based on equations like Eq. 9.9.

8. Assuming a 15 × 15 grid, the spreadsheet should have multiple arrays of 15 × 15 cells each to hold values of h, K, K(x+) , K(x−) , K(y+) , and K(y−) . The nodal equations in the h array refer to the appropriate cells these K arrays. In the cells along the no-flow boundaries, assign K = 0. The results of an analytic flow model (AnAqSim) are shown in Figure 12.

9. See Figure 12.

10. The analytic solution for an infinite domain (Theis assumption) is shown in Figure 13. A model with constant head boundaries should show a similar curve, but at large times the drawdown should be less than that shown in Figure 13. A model with no-flow boundaries should deviate at large times with more drawdown than that shown in Figure 13.

11. See the USGS Open File Report 00-466. In MODFLOW ET is modeled as a source/sink in a cell in the uppermost model layer (part of Qs for that cell). With the older EVT package, the rate of evapotranspiration varies from zero when the head in the cell is below a user-specified level (call it hmin ), to a user-specified maximum value (call this ETmax ) when the head in the cell rises to a user-specified level (call it hmax ); the rate of evapotranspiration increases linearly from zero to ETmax as head increases from hmin to hmax . With the ETS1 package, The ET rate also goes from zero to ETmax , but it may vary with head according to a multi-segmented nonlinear polyline function that is defined by the user. Evapotranspiration rates are specified with length/time units, which is really volume/time/area.

50

12. Use Darcy’s law to calculate qb2 (m, day units): qb2

∂h ∂x 42.00 − 42.35 = −(2.5) = 0.0109 m/day 200 − 120

= −Kx

The discharge across the boundary of element 2 between node n and node n + 1 is Q = qb2 A =

(0.0109)(13 × 95) = 13.5 m3 /day

where A is the cross-sectional area of this side of element 2.

13. Since this is an interior node, the boundary integral in Eq. 9.17 is zero (Eq. 9.20).

With the area integral written as a summation (see Eq. 9.18), Eq. 9.17 becomes the following in this case: " # " #

! ∂ ĥ ∂fn ∂fn ∂ ĥ + A1 ∂x ∂x 1 ∂y ∂y 1 1 1 " # " #

! ∂ ĥ ∂fn ∂fn ∂ ĥ + A2 + ∂x ∂x 2 ∂y ∂y 2 " #2 " #2

! ∂fn ∂ ĥ ∂fn ∂ ĥ + + A3 ∂x ∂x 3 ∂y ∂y 3 3 " # " #3

! ∂ ĥ ∂fn ∂ ĥ ∂fn + A4 + = 0 ∂x ∂x 4 ∂y ∂y 4 4

4

For this case, A1 = A2 = A3 = A4 , so that factor is common to all four terms above.

The gradients for each element can then be substituted to give the following, where the order is the same as in the previous equation:

ho − hn −1 hp − hn −1 + ∆ ∆ ∆ ∆

hn − hq 1 hp − hn −1 + + ∆ ∆ ∆ ∆

hn − hq 1 hn − hr 1 + + ∆ ∆ ∆ ∆

ho − hn −1 hn − hr 1 + + = 0 ∆ ∆ ∆ ∆ where the subscripts on the h terms denote the node. Multiplying the above by ∆2 and collecting like terms yields 8hn − 2(ho + hp + hq + hr ) = 0 Solving the above for the head at the central node gives hn =

1 (ho + hp + hq + hr ) 4

This is the same result as we had with the finite difference method with a uniform grid and Laplace equation conditions (see Problems 2 and 4 in this chapter). The head at a node equals the average of heads at the neighboring nodes.

51

14. (a) Use the definition of transmissivity, Eqs. 3.24 and 3.25, which results in (h ≤ b1 )

T

= K1 h

T

= K1 b1 + K2 (h − b1 )

T

= K1 b1 + K2 b2

(b1 ≤ h ≤ (b1 + b2 ))

(h ≥ (b1 + b2 ))

(b) The potentials are derived by integrating Eq. 9.27: Z Z Φ = dΦ = T dh For the three cases, this results in Z 1 Φ = K1 h dh = K1 h2 + C1 2 Z Φ = (K1 b1 + K2 (h − b1 )) dh

(h ≤ b1 )

1 (b1 ≤ h ≤ (b1 + b2 )) (K1 − K2 )b1 h + K2 h2 + C2 2 Z = (K1 b1 + K2 b2 ) dh = (K1 b1 + K2 b2 )h + C3 (h ≥ (b1 + b2 )) =

Φ

where C1 , C2 , and C3 are constants. If we set C1 = 0, then when h = b1 the first two expressions for Φ give Φ= and

Φ=

1 K1 b21 2

(h = b1 )

1 K1 − K2 b21 + C2 2

(h = b1 )

For Φ to be continuous at h = b1 , the above two expressions must be equal.

Setting them equal allows solving for C2 :

1 1 1 C2 = K1 b21 − K1 − K2 b21 = (K2 − K1 ) b21 2 2 2 Similarly, when h = b1 + b2 , the second and third expressions for Φ become 1 1 (K1 − K2 )b1 (b1 + b2 ) + K2 (b1 + b2 )2 + (K2 − K1 ) b21 2 2 1 1 = K1 b21 + K1 b1 b2 + K2 b22 (h = b1 + b2 ) 2 2

Φ

=

Φ

=

and (K1 b1 + K2 b2 )(b1 + b2 ) + C3

= K1 b21 + K1 b1 b2 + K2 b1 b2 + K2 b22 + C3

(h = b1 + b2 )

For continuity, the previous two expressions must be equal, so C3

1 1 K1 b21 + K1 b1 b2 + K2 b22 − K1 b21 − K1 b1 b2 − K2 b1 b2 − K2 b22 2 2 1 1 2 = − K1 b1 − K2 b1 b2 − K2 b22 2 2 =

52

15. Eq. 6.31 can be written as

1 K ∇2 h2 = −N 2 Since K/2 is a constant, it can be moved under the Laplacian operator to give

1 2 2 Kh = ∇2 Φ = −N ∇ 2 where Φ is defined as in Eq. 9.34.

16. There are two wells, both extracting water from the aquifer. There is a string of three impermeable barrier line elements in the center of the plot, and a string of three line-sinks in the lower right of the plot. In addition, there is a uniform flow from upper left to lower right.

17. When h = b, Eq. 9.32 gives Φ = T b + Cc = Kb2 + Cc and Eq. 9.34 with Cu = 0 gives 1 2 Kb 2 For continuity, these two expressions must be equal, so Φ=

1 Cc = − Kb2 2 So when the head is high enough that aquifer is confined, 1 Φ = T h − Kb2 2 18. The Φ(h) definitions for this problem are Eq. 9.32 with Cc as defined in the previous problem where confined, and Eq. 9.34 with Cu = 0 where unconfined. The Φ(x, y) in this case involves two superposed radial flow solutions: Φ=

Q rp ln + Φ0 2π ri

where rp is the radial distance from the pumping well to the point where Φ is evaluated, ri is the distance from the injection well, and Φ0 is a constant. Far from the two wells, rp /ri → 1 and Φ → Φ0 . Since far from the wells h → h0 = 39 ft (confined condition), 1 Φ0 = Kbh0 − Kb2 2 Use the known head at the pumping well to solve for K. The discharge potential at the pumping well is Φp =

1 Kh2p 2

= =

Q rp→p ln + Φ0 2π ri→p Q rp→p 1 ln + Kbh0 − Kb2 2π ri→p 2

where Q = 12 gal/min = 2310 ft3 /day is the well discharge rate, rp→p is the distance from the centerline of the pumping well to its screen (its radius), and ri→p is the 53

distance from the injection well to the pumping well. Solving the above equation for K gives K

= =

Q rp→p ln 2π ri→p 2310 0.25 1

ln = 3.8 ft/day (1.2 m/day) 1 2 − (35)(39) + 1 352 2π 118 18 2 2 1

1 2 1 2 2 hp − bh0 + 2 b

19. Answers vary with article chosen.

54

Chapter 10 Solutions 1. Low-TDS groundwater will occur where the water has had a short residence time since precipitation and is in contact with minerals of low solubility. An example would be a recharge area in an area underlain by granite or some other rock containing just silicate minerals. High-TDS groundwater will occur where there are long residence times and contact with more soluble minerals. An example of this would be deep in a sedimentary basin where there are evaporite minerals like gypsum and halite.

2. In going from precipitation → rivers → groundwater, the pH generally increases as does the TDS. This is due to longer contact with basic minerals. The major anion in − river water and groundwater tends to be HCO− 3 , but in precipitation Cl (coastal) 2− and SO4 (inland) dominate. The CO2 content of the atmosphere is small, limiting the amount of carbonates in precipitation.

3. See section 10.4.4. The highest concentrations occur within organic-rich surface soils.

4. Calculate hardness with Eq. 10.6. The hardness of samples 6–9 are calculated below: H6

=

2.5(10) + 4.1(1.5) = 31 mg/L

H7

=

2.5(24.5) + 4.1(10.7) = 105 mg/L

H8

=

2.5(69) + 4.1(29) = 291 mg/L

H9

=

2.5(21) + 4.1(3.1) = 65 mg/L

With these values, samples 6 and 9 would be considered soft (H < 75) and sample 8 would be considered hard (H > 150).

5. To start, calculate the ion concentrations in meq/L and then their percents of the total, as shown in Table 15. Figure 14 shows the plot.

6. Activity is a thermodynamically-effective concentration. In very dilute solutions, it equals the magnitude of molar concentration. In more concentrated solutions, interaction of ions causes ion behavior to deviate from that predicted by perfectly random motions of molecules. Since equilibrium computations are based on this concept of random motion, concentrations used in equilibrium calculations are corrected for the effects of ion interaction. Activity is the corrected concentration that is used in equilibrium calculations. figure file: solutions/ch9trilin Figure 14: Problem 5.

55

Table 15: Problem 5.

Sample 7 meq/L %

Sample 9 meq/L %

Cations: Ca Mg Na K Total

1.223 0.880 1.083 0.120 3.306

37 27 33 4 100

1.048 0.255 7.395 0.215 8.912

12 3 83 2 100

Anions: HCO3 SO4 Cl Total

2.786 0.454 0.200 3.440

81 13 6 100

6.555 0.250 2.398 9.203

71 3 26 100

Table 16: Problem 8. z F.W. mg/L mol/L meq/L Ca 2 40.1 10.0 2.50E-04 0.50 Mg 2 24.3 1.5 6.17E-05 0.12 Na 1 23.0 5.0 2.17E-04 0.22 K 1 39.1 0.8 2.05E-05 0.02 HCO3 −1 61.0 19.0 3.11E-04 0.31 SO4 −2 96.1 5.5 5.73E-05 0.11 Cl −1 35.5 11.0 3.10E-04 0.31 z=charge, F.W.=formula weight, [A]=activity

56

Mz 2 9.98E-04 2.47E-04 2.17E-04 2.05E-05 3.11E-04 2.29E-04 3.10E-04

γ 0.86 0.86 0.96 0.96 0.96 0.86 0.96

[A] 2.15E-04 5.31E-05 2.09E-04 1.97E-05 3.00E-04 4.93E-05 2.99E-04

7. Ion activity product is the product of ion activities on the right side of a mineral dissolution equilibrium equation. When the solution is at equilibrium with the mineral, the IAP equals the solubility product. When the solution is undersaturated, IAP is less than the solubility product. Because precipitation has very low dissolved solids and mineral dissolution is a slow process, fresh infiltration into limestone will be undersaturated and IAP would be less than the solubility product.

8. (a) The calculations of concentrations and activities were done in a spreadsheet and are summarized in Table 16. Calculate the CO2− 3 activity using Eq. 10.47 and the activities of [H+ ] and [HCO− ]: 3 [CO2− 3 ]

= = =

KHCO− [HCO− 3] 3

[H+ ] (10−10.38 )(3.00 × 10−4 ) 10−6.9 −8 9.9 × 10

To get molar concentration from this activity, we need to calculate the ionic strength and then get the activity coefficient. The ionic strength, calculated with Eq. 10.16 and the Mz 2 values from Table 16 is I = 1.2 × 10−3 . Using this ionic strength in Figure 10.7, we estimate γ ' 0.86. Therefore the molar concentration of carbonate is 2− −8 (CO2− /(0.86) = 1.16 × 10−7 mol/L 3 ) = [CO3 ]/γ = 9.9 × 10

(b) Use Eq. 10.28 to calculate the saturation indexes:

[Ca][CO3 ] SIcalcite = log Kso

[2.15 × 10−4 ][9.9 × 10−8 ] = log = −2.2 10−8.48

[Ca][Mg][CO3 ]2 SIdolomite = log Kso

[2.15 × 10−4 ][5.31 × 10−5 ][9.9 × 10−8 ]2 = log = −4.8 10−17.1

[Ca][SO4 ] SIgypsum = log Kso

[2.15 × 10−4 ][4.93 × 10−5 ] = log = −3.4 10−4.58 (c) The water is greatly undersaturated with respect to all three minerals; either these minerals are not present in the aquifer or the water has had a short residence time.

9. The answers to parts a–c are based on the computations shown in Table 17. (a) The sum of the meq/L for the cations is 3.31 meq/L. The sum of the meq/L for the anions is 3.44 meq/L. This amounts to a charge imbalance error of 4.0%. (b) The ionic strength is computed by summing the values in the Mz 2 column and taking half this sum (see Eq. 10.16). Doing this results in I = 4.65 × 10−3 . (c) See Table 17.

57

Table 17: Problem 9. z F.W. mg/L mol/L meq/L Ca 2 40.1 24.5 6.11E-04 1.22 Mg 2 24.3 10.7 4.40E-04 0.88 Na 1 23.0 24.9 1.08E-03 1.08 K 1 39.1 4.7 1.20E-04 0.12 HCO3 −1 61.0 170.0 2.79E-03 2.79 SO−2 −2 96.1 21.8 2.27E-04 0.45 4 Cl −1 35.5 7.1 2.01E-04 0.20 z=charge, F.W.=formula weight, [A]=activity

Mz 2 2.45E-03 1.76E-03 1.08E-03 1.20E-04 2.79E-03 9.08E-04 2.01E-04

γ 0.76 0.76 0.93 0.93 0.93 0.76 0.93

[A] 4.65E-04 3.35E-04 1.01E-03 1.12E-04 2.95E-03 1.72E-04 1.87E-04

Mz 2 6.89E-03 4.77E-03 1.52E-04 2.81E-05 4.87E-03 1.54E-03 2.65E-04

γ 0.68 0.68 0.91 0.91 0.91 0.68 0.91

[A] 1.16E-03 8.07E-04 1.38E-04 2.55E-05 4.41E-03 2.60E-04 2.40E-04

Table 18: Problem 10. z F.W. mg/L mol/L meq/L Ca 2 40.1 69.0 1.72E-03 3.44 Mg 2 24.3 29.0 1.19E-03 2.39 Na 1 23.0 3.5 1.52E-04 0.15 K 1 39.1 1.1 2.81E-05 0.03 HCO3 −1 61.0 297.0 4.87E-03 4.87 SO−2 −2 96.1 37.0 3.85E-04 0.77 4 Cl −1 35.5 9.4 2.65E-04 0.27 z=charge, F.W.=formula weight, [A]=activity

(d) For anhydrite, IAP

=

[Ca2+ ][SO2− 4 ]

=

(4.65 × 10−4 )(1.72 × 10−4 ) = 8.0 × 10−8

The saturation index is SI

= =

log(IAP/Kso )

8.0 × 10−8 log = −2.7 10−4.36

SI is negative, IAP is far less than Kso , so the water is well below saturation with respect to anhydrite. It is unlikely that the rock contains anhydrite.

10. (a) The calculations of concentrations and activities were done in a spreadsheet and are summarized in Table 18. The ionic strength, calculated with Eq. 10.16 and the Mz 2 values from Table 16 is I = 9.3 × 10−3 . (b) Calculate the CO2− activity using Eq. 10.47 and the activities of [H+ ] and 3 − [HCO3 ]: [CO2− 3 ]

= = =

KHCO− [HCO− 3] 3

[H+ ] (10−10.38 )(4.41 × 10−3 ) 10−7.5 −6 5.8 × 10 58

Table 19: Problem 11. z F.W. mg/L mol/L meq/L Ca 2 40.1 21.0 5.24E-04 1.05 Mg 2 24.3 3.1 1.28E-04 0.26 Na 1 23.0 170.0 7.39E-03 7.39 K 1 39.1 8.4 2.15E-04 0.21 HCO3 −1 61.0 400.0 6.56E-03 6.56 SO4 −2 96.1 12.0 1.25E-04 0.25 Cl −1 35.5 85.0 2.40E-03 2.40 z=charge, F.W.=formula weight, [A]=activity

Mz 2 2.10E-03 5.10E-04 7.39E-03 2.15E-04 6.56E-03 5.00E-04 2.40E-03

γ 0.67 0.67 0.90 0.90 0.90 0.67 0.90

[A] 3.51E-04 8.53E-05 6.69E-03 1.94E-04 5.93E-03 8.36E-05 2.17E-03

(c) Use Eq. 10.28 to calculate the saturation indexes:

[Ca][CO3 ] SIcalcite = log Kso

[1.16 × 10−3 ][5.8 × 10−6 ] = log = 0.31 10−8.48

[Ca][Mg][CO3 ]2 SIdolomite = log Kso

[1.16 × 10−3 ][8.07 × 10−4 ][5.8 × 10−6 ]2 = log = 0.60 10−17.1

[Ca][SO4 ] SIgypsum = log Kso

[1.16 × 10−3 ][2.60 × 10−4 ] = log = −1.9 10−4.58 (d) The water is oversaturated with respect to calcite and dolomite, and undersaturated with respect to gypsum. The carbonates are probably present in the aquifer but gypsum is probably absent.

11. (a) See Table 19. (b) Summing the meq/L of cations gives a total of 8.91 meq/L, and doing the same for anions gives 9.20 meq/L. The percent error is 100 ×

9.20 − 8.91 = 3.2% 8.91

(c) The ionic strength, calculated with Eq. 10.16 and the Mz 2 values from Table 19 is I = 9.8 × 10−3 . (d) See Table 19. (e) Calculate the H2 CO∗3 activity using Eq. 10.46 and the activities of [H+ ] and [HCO− 3 ]: [H2 CO∗3 ]

= = = 59

[H+ ][HCO− 3] KH2 CO3∗ (10−6.9 )(5.93 × 10−3 ) 10−6.35 1.7 × 10−3

Table 20: Problem 12. z F.W. mg/L mol/L meq/L Ca 2 40.1 24.6 6.14E-04 1.23 Mg 2 24.3 6.3 2.59E-04 0.52 Na 1 23.0 2.6 1.13E-04 0.11 K 1 39.1 0.3 7.67E-06 0.01 HCO3 −1 61.0 69.1 1.13E-03 1.13 SO4 −2 96.1 30.1 3.13E-04 0.63 Cl −1 35.5 3.8 1.07E-04 0.11 z=charge, F.W.=formula weight, [A]=activity

Mz 2 2.46E-03 1.04E-03 1.13E-04 7.67E-06 1.13E-03 1.25E-03 1.07E-04

γ 0.79 0.79 0.94 0.94 0.94 0.79 0.94

[A] 4.84E-04 2.04E-04 1.07E-04 7.23E-06 1.07E-03 2.47E-04 1.01E-04

Calculate the CO2− activity using Eq. 10.47 and the activities of [H+ ] and 3 − [HCO3 ]: [CO2− 3 ]

KHCO− [HCO− 3] 3

=

[H+ ] (10−10.38 )(5.93 × 10−3 ) 10−6.9 −6 2.0 × 10

= = (f) The ion activity products are: IAPcalcite IAPdolomite

=

[Ca][CO3 ]

=

[3.51 × 10−4 ][2.0 × 10−6 ] = 10−9.1

=

[Ca][Mg][CO3 ]2

=

[3.51 × 10−4 ][8.53 × 10−5 ][2.0 × 10−6 ]2 = 10−18.9

For both minerals, IAP < Kso , so the water is undersaturated with respect to both. (g) The concentration of CO2 in the atmosphere is about 360 ppm, which is equivalent to 3.6 × 10−4 atm of partial pressure. Using Henry’s law, the aqueous concentration at equilibrium with this gas partial pressure is (CO2 )(aq) = KH (CO2 )(gas) = (10−1.47 )(3.6 × 10−4 ) = 1.2 × 10−5 mol/L This is about two orders of magnitude less than the activity of H2 CO∗3 in this water. By dissolving carbonate minerals, groundwater picks up more carbonate species than precipitation has.

12. (a) Use charge balance to figure out the concentration of SO4 . Referring to the data of Table 20, the sum of the meq/L of cations is 1.87 meq/L. Assuming that the anions also sum to 1.87 meq/L (perfect charge balance), there must be 0.25 meq/L of SO4 . Given this value, the others in the table may be calculated. (b) The ionic strength, calculated with Eq. 10.16 and the Mz 2 values from Table 20 is I = 3.1 × 10−3 . (c) See Table 20. (d) IAPanhydrite = [Ca][SO4 ] = [4.84 × 10−4 ][2.47 × 10−4 ] = 10−6.9 Kso = 10−4.36 for anhydrite, so this water is undersaturated by several orders of magnitude. Anhydrite would dissolve if it were in contact with this water.

60

(e) Calculate the H2 CO∗3 activity using Eq. 10.46 and the activities of [H+ ] and [HCO− 3 ]: [H2 CO∗3 ]

[H+ ][HCO− 3] KH2 CO3∗

=

(10−6.8 )(1.07 × 10−3 ) 10−6.35 3.8 × 10−4

= =

Since H2 CO∗3 is a neutral molecule, its activity coefficient is approximately 1 and the molar concentration is also 3.8 × 10−4 mol/L. Calculate the CO2− activity using Eq. 10.47 and the activities of [H+ ] and 3 − [HCO3 ]: [CO2− 3 ]

KHCO− [HCO− 3] 3

=

[H+ ] (10−10.38 )(1.07 × 10−3 ) 10−6.8 −7 2.8 × 10

= =

At this ionic strength and with a charge of 2, the activity coefficient from Figure 10.7 is γ ' 0.79. The concentration of CO2− 3 is 2− −7 /0.79 = 3.6 × 10−7 mol/L (CO2− 3 ) = [CO3 ]/γ = 2.8 × 10

13. See Table 21. (a) The ionic strength is computed by summing the values in the Mz 2 column and taking half this sum (see Eq. 10.16). Doing this results in I = 1.61 × 10−2 . ∗ (b) The activity [HCO− 3 ] is given in Table 21, and [H2 CO3 ] is computed as

[H2 CO∗3 ]

= = =

[H+ ][HCO− 3] KH2 CO3∗ (10−8.5 )(1.95 × 10−3 ) 10−6.35 1.4 × 10−5

(c) At equilibrium between gas and aqueous phase CO2 , H2 CO∗3 ] = 10−1.47 CO2 (g) (see Table 10.8). With atmospheric CO2 at 380 ppm, the above ratio for the Colorado River sample works out to [H2 CO∗3 ] 1.4 × 10−5 = = 0.036 = 10−1.44 CO2 (g) 380 × 10−6 This is roughly at equilibrium with the atmosphere.

14. The concentration of CO2 in the atmosphere is about 360 ppm, which is equivalent to 3.6 × 10−4 atm of partial pressure. Using Henry’s law, the aqueous concentration at equilibrium with this gas partial pressure is (CO2 )(aq) = KH (CO2 )(gas) = (10−1.47 )(3.6 × 10−4 ) = 1.2 × 10−5 mol/L 61

Table 21: Problem 13.

Mz 2 8.28E-03 3.95E-03 4.13E-03 1.28E-04 2.21E-03 1.12E-02 2.32E-03

z F.W. mg/L mol/L meq/L Ca 2 40.1 83 2.07E-03 4.14 Mg 2 24.3 24 9.87E-04 1.97 Na 1 23.0 95 4.13E-03 4.13 K 1 39.1 5 1.28E-04 0.13 HCO3 −1 61.0 135 2.21E-03 2.21 SO4 −2 96.1 270 2.81E-03 5.62 Cl −1 35.5 82 2.32E-03 2.32 z=charge, F.W.=formula weight, [A]=activity

γ 0.63 0.63 0.88 0.88 0.88 0.63 0.88

[A] 1.30E-03 6.22E-04 3.64E-03 1.13E-04 1.95E-03 1.77E-03 2.04E-03

From Eq. 10.46, [HCO− 3]

= = =

[H2 CO∗3 ]KH2 CO3∗ [H+ ] (1.2 × 10−5 )(10−6.35 ) 10−7.8 3.4 × 10−4

At this ionic strength, the activity coefficient for bicarbonate is γ ' 0.84, so − −4 (HCO− /0.84 = 4.0 × 10−4 mol/L 3 ) = [HCO3 ]/γ = 3.4 × 10 − + Calculate the CO2− 3 activity using Eq. 10.47 and the activities of [H ] and [HCO3 ]:

[CO2− 3 ]

= = =

KHCO− [HCO− 3] 3

[H+ ] (10−10.38 )(3.4 × 10−4 ) 10−7.8 −7 8.9 × 10

At this ionic strength, the activity coefficient for carbonate is γ ' 0.46, so 2− −7 (CO2− /0.46 = 1.9 × 10−6 mol/L 3 ) = [CO3 ]/γ = 8.9 × 10

15. Using Henry’s law gives the aqueous concentration: (TCE)(aq) = KH (TCE)(gas) = (0.10)(3400 × 10−6 ) = 3.4 × 10−4 mol/L The formula weight for TCE is 131.4 g/mol = 131,000 mg/mol. With this factor, the aqueous concentration of 3.4 × 10−4 M = 45 mg/L. 16. Use Eq. 10.47 to calculate this ratio: [HCO− 3] 2− [CO3 ]

=

[H+ ] KHCO−

=

10−8.5 = 76 10−10.38

3

62

figure file: solutions/ch9isotope Figure 15: Problem 23.

Likewise, use Eq. 10.46 to calculate this ratio: [HCO− 3] [H2 CO∗3 ]

= =

KH2 CO3∗ [H+ ] 10−6.35 = 141 10−8.5

These ratios could also be approximated from Figure 10.8.

17. Refer to Figure 10.8 for this question. With pH < 6.35, the rain samples have H2 CO∗3 as the dominant carbonate species. With 6.35 < pH < 10.38, the groundwater samples have HCO− 3 as the dominant carbonate species.

18. Use Eq. 10.48 for sample 6: Alk

=

− + 2− (HCO− 3 ) + 2(CO3 ) + (OH ) − (H )

−4 eq/L ' (HCO− 3 ) = 3.1 × 10

because (HCO− 3 ) is orders of magnitude larger than the other terms. Similarly, for sample 8: −3 eq/L Alk ' (HCO− 3 ) = 4.9 × 10 19. Sulfur is (−II) on the left and (+VI) on the right; it is oxidized. Oxygen goes from (0) on the left to (−II) on the right; it is reduced.

20. Answer varies with article.

21. For metals, the answer lies in an equilibrium equation like Eq. 10.68. As pH rises and [H+ ] falls, the ratio [≡XOCu+ ]/[Cu2+ ] must increase to maintain equilibrium (more Cu is sorbed). For ligands, consider an equation like Eq. 10.70. As pH rises, [OH− ] rises, and the ratio [≡XF]/[F− ] must decrease to maintain equilibrium (less F is sorbed).

22. (a) Use Eq. 10.74: cad = Kd caq = (0.08 L/kg)(80 mg/L) = 6.4 mg/kg (b) The bulk density of the aquifer is 2.28 kg/L = 2280 kg/m3 . sorbed mass = cad ρb = (6.4 mg/kg)(2280 kg/m3 ) = 14, 400 mg/m3 m3 (c) dissolved mass = caq n = (80 mg/L)(0.26) = 20.8 mg/L = 20, 800 mg/m3 m3 23. See Figure 15.

24. Using Eq. 10.88, we can determine the decay constant for tritium: γ=

ln 2 ln 2 = = 0.0559 yr−1 t1/2 12.4 yr 63

Table 22: Problem 24.

Year 1964 1970 1980 1990 2000 2100

t (yrs) 0 6 16 26 36 136

Tritium conc. (TU) 1200 858 491 281 160 6

The tritium concentration can be predicted with a form of Eq. 10.86: c = c0 e−γt where c0 = 1200 TU is the initial concentration at time t = 0 (1964). The results are tabulated in Table 22.

25. Answer varies with article.

64

Chapter 11 Solutions 1. A dense liquid (ρ > 1.00) that has low solubility would sink into hard-to-remove areas and persist. Higher solubility will cause rapid, high levels of contamination, but will lessen persistence. It should have toxicity at much lower concentrations (low MCL) compared to its solubility. It should be resistant to microbiologic decay and relatively nonreactive, so it persists. It should have low sorption (low Kow ), so it can travel far. It should have a low vapor pressure, so it cannot be remediated easily with vapor extraction. Sounds like a recipe for 1,2-DCA, TCE, and PCE! 2. See Figure 16. These are, in fact, DNAPLs. In general, chlorinated hydrocarbon NAPLs tend to be denser than water due to the heavy chlorine atoms.

3. See Figure 17. There is only one isomer because there is only one chlorine atom and no matter which of the four bond positions it is located at, the same molecule results.

4. Apply Eq. 11.1, but before doing so, we need to calculate the mole fraction X of each constituent (mol constituent/mol molecules in gasoline mixture). We are given weight percent W P (mass constituent/mass gasoline mixture). To convert, use: X = WP ×

ave. molecular wt. gasoline molecular wt. constituent

Combining the above with Eq. 11.1 gives, with the benzene values inserted as an example of the calculation:

ave. molecular wt. gasoline caq = W P S molecular wt. constituent

102 = (0.02) 1750 = 46 mg/L 78.1 Doing the same for ethylbenzene (molecular weight = 106.2 g/mol, S = 180 mg/L) results in caq = 3.5 mg/L, and for MTBE (molecular weight = 88.2 g/mol, S = 45,000 mg/L) results in caq = 1,040 mg/L. 5. The results are shown in Table 23. The higher this ratio, the greater potential risk it poses to humans. A high ratio means that concentrations in water can be much higher than the concentration where health risks become significant. High solubility will make the contaminant less persistent as a NAPL, because it will dissolve into passing groundwater more rapidly. figure file: solutions/ch10isomers Figure 16: Problem 2.

65

figure file: solutions/ch10vinylchloride Figure 17: Problem 3.

Table 23: Problem 5.

Contaminant Benzene Ethylbenzene Toluene Xylenes Benzo(a)pyrene PCB-1248 1,2-DCA 1,1,1-TCA TCE PCE DCM MTBE

S/MCL 350,000 260 520 ≈ 50 15 110 1,700,000 6,500 220,000 30,000 2,800,000 ≈ 1,500,000

6. Eq. 11.3, with proper unit conversions, is S

= KH × vapor pressure

mol/L 1 atm 131.3 g 1000 mg = 0.11 (70 mm Hg) atm 760 mm Hg mol g = 1330 mg/L

This is close to the listed solubility of 1100 mg/L. Note that there is easily this much variability in the experimental S, KH , and vapor pressure data reported in the literature.

7. To invade a smaller pore requires a smaller radius of curvature for the NAPL-water interface, which requires larger capillary pressure (see Eqs. 11.6 and 11.7). At first, there is not sufficient capillary pressure to enter the finer pores, but as NAPL piles up on the interface, capillary pressure increases and eventually the NAPL does invade the finer pores.

8. Start with Eq. 11.5 and apply hydrostatic principles (Eq. 2.15): Pc

= Pn − Pw = ρn gbg − ρw gbg =

(ρn − ρw )gbg

The greater ρn is, the larger Pc is, and the more likely it is that the capillary pressure is high enough for DNAPL to invade pores and continue moving. Therefore, denser DNAPLs are more mobile.

9. At a given capillary pressure, water saturation is lower for a higher K material. The higher K material has larger pores and at a given Pc , more NAPL can invade the pores, displacing more water in the process.

10. The advective flux is given by Eq. 11.9 (m, day, mg units): Fax = qx c = −Kx

∂h c ∂x 66

Table 24: Problem 11.

Solute Chloride Carbon tet.

Bromoform PCE

∆s (m) 32.0 15.9 14.8 7.9

0–350 days R Kd (L/kg) 1.0 0.0 2.0 0.18 2.2 0.21 4.1 0.56

350–650 days ∆s (m) R Kd (L/kg) 25.9 1.0 0.0 8.8 2.9 0.36 7.3 3.6 0.47 4.1 6.3 0.97

Table 25: Problem 12.

Solute Chloride Carbon tet.

PCE

∆t (days) 63 111 182

R 1.0 1.8 2.9

= −(2)(0.003)(20,000 mg/m3 ) = −120 mg/day/m2 The flux through 15 m2 at this rate is 120 × 15 = −1800 mg/day. The negative sign indicates flux in the negative x direction.

11. In this case, the retardation factor (see Eq. 11.11) for a reactive solute may be calculated with: ∆s chloride R= ∆s reactive solute where ∆s is the distance traveled by the center of mass of the solute cloud. By manipulating Eq. 11.12, Kd may be calculated from R, n, and ρb : Kd =

(R − 1)n ρb

Using these two equations with the data of Figure 11.31 gives the results listed in Table 24.

Possible reasons for the differences include • Diffusion-limited sorption that has greater affect at larger times, and • Heterogeneity in the adsorptive properties of the materials such that the materials are more sorptive where the plume is after 350 days.

12. For this analysis, the retardation factor (see Eq. 11.11) for a reactive solute may be calculated with: ∆t reactive solute R= ∆t chloride where ∆t is the time of the peak arrival. Using this equation with the data of Figure 11.29 gives the results listed in Table 25. These R values are smaller than those calculated in the previous problem. This is probably due to the fact that these data are closer to the source, which limits the effects of intragranular diffusion and introduces some error because the solute pulse occupied a finite volume, not a point, at its source.

67

Table 26: Problem 14.

Solute Carbon tet.

Bromoform PCE

D at 500 days 0.40 0.16 0.18

Kd (L/kg) 0.27 0.95 0.83

13. It would be much higher in the peat because the fraction of organic carbon is so high. Organic carbon typically provides the bulk of sorption sites for nonpolar organic molecules (see section 10.9.2).

14. Use the following form of Eq. 10.77 to calculate Kd :

mg sorbed chemical ρb Kd = mg dissolved chemical θ The ratio in parentheses can be inferred from the data of Figure 11.31, assuming that the sorbed mass = injected mass − mass in solution. Thus, 1−D mg sorbed chemical = mg dissolved chemical D where D is the ratio plotted in Figure 11.31. Using the two previous equations to calculate Kd at t = 500 days gives the results in Table 26. The Kd values calculated this way agree well with those calculated in Problem 11 for carbon tetrachloride and PCE, but not for bromoform. The bromoform Kd calculated this way is 2–4 times higher than the result calculated in Problem 11. This is most likely because bromoform was biodegraded significantly, and this lost mass is incorrectly attributed to sorption in this analysis.

15. Molecular diffusion can make a minor contribution to overall solute transport in a material where advective flux is much larger than diffusive flux. Examples include coarse gravels and porous basalts. However, molecular diffusion can still be an important transient solute transport process in materials with large advective flux, by moving solute mass in and out of low-K lenses, dead-end pores, etc.

16. (a) Apply Eq. 11.10 to the concentrations at the leading edge of the plume, assuming Tx∗ = 0.7 (reasonable for sands) and D = 1.6 × 10−5 cm2 /sec = 1.6 × 10−9 m2 /sec (reasonable for Cl− , see Table 11.5). To keep all units in meters, the concentrations of c = 10 and 300 mg/L are written as 10,000 and 300,000 mg/m3 .

Fdx

= −nTx∗ D

∂c ∂x −9

= −(0.33)(0.7)(1.6 × 10 =

)

10,000 − 300,000 62 − 48

7.7 × 10−6 mg/m2 /sec = 0.66 mg/m2 /day

(b) The advective flux is given by Eq. 11.9. The concentration c = 150 mg/L = 150,000 mg/m3 . The average linear velocity of the tracer cloud is v x = 0.090 m/day (see Figure 11.30).

Fax = qx c = v x ne c =

(0.090)(0.33)(150,000) = 4500 mg/m2 /day 68

(c) The ratio Fax /Fdx ≈ 6700; the advective flux is much larger than the molecular diffusion flux.

17. Combining Eqs. 11.13 and 11.15 gives (mg, m, day units): Fmx

= −nαx |v|

∂c ∂x

= −(0.33)(0.36)(0.090) =

10,000 − 300,000 62 − 48

220 mg/m2 /day

Fmx /Fdx ≈ 330, Fax /Fmx ≈ 20. This transport is occurring in a sand where the dominant transport mechanisms are advection and mechanical dispersion caused by velocity variations (macrodispersion).

18. A zone of depressed O2 levels coincides with the core of the plume where there is the most active aerobic decomposition (respiration) of hydrocarbon solutes. There is elevated CO2 in the core of the plume near the source, probably the byproduct of respiration. Unlike the pattern of depressed O2 , the pattern of elevated CO2 does not persist in the downgradient portion of the plume. The CO2 is probably reacting with water to form bicarbonate (section 10.8.2).

19. With v = 0, the exponential function in this equation reduces to 1, and the equation simplifies to ! x p c = c0 erfc 2 Dmx t/R which, with R = 1, is the solution for one-dimensional diffusion from a sharp boundary where at t = 0, c = 0 for x > 0 and c = c0 for x < 0, and as t increases diffusion blurs the sharp boundary at x = 0 (See Crank, J., 1956, The Mathematics of Diffusion, Oxford University Press). In this case, there is only molecular diffusion and Dmx = Tx∗ D. 20. The modeled plume should resemble the actual plume in a “macro” sense, but the model will give a smoother, more regular concentration pattern. Doubling the longitudinal dispersivity will make for a more dilute plume that is longer in the direction of flow. Doubling the vertical transverse dispersivity will also make for a more dilute plume, but spread more in the vertical direction.

21. On the centerline of the plume, y = z = λ = 0, so with R = 1 the solution simplifies to

M (x − vt)2 p c= exp − 4Dmx t 8n (πt)3 Dmx Dmy Dmz The average linear velocity of the chloride solute cloud from Figure 11.30 is v = 0.09 m/day. The dispersion coefficients are (m, day units): Dmx = αl |v| = (0.36)(0.09) = 0.0324 m2 /day Dmy = αth |v| = (0.039)(0.09) = 0.0035 m2 /day Dmz = αtv |v| = (0.023)(0.09) = 0.0021 m2 /day The injected mass of chloride is M = 10, 700 g and the time is t = 462 days. Using these input parameters in the above equation for c gives the results shown in Figure 18. The units used here for the input parameters are g/m3 , which is equivalent to mg/L. 69

figure file: solutions/ch10charbeneau Figure 18: Problem 21.

22. First, calculate the volume and mass of TCE NAPL in the zone (m units): VT CE

=

πr2 bnSn

=

π(2.5)2 (3)(0.30)(0.08) = 1.41 m3

At a density of 1.46 g/cm3 (kg/L), this much TCE has a mass of 2064 kg.

Now calculate the mass flux F of TCE that can be carried by groundwater flowing through the volume, assuming that water is at the solubility concentration for TCE (mg, L, day units): F = Qc = (250)(1100) = 2.75 × 105 mg/day which equals 0.275 kg/day. At this rate, it will take (kg, day units) t=

2064 m = = 7500 days = 21 yrs F 0.275

to dissolve away all the TCE. In an actual situation, the water flowing through the cylinder will probably leave with concentrations somewhat below the solubility concentration due to kinetic effects and irregular NAPL distributions, so F will be smaller than calculated and t will be larger than calculated.

23. (a) The mass of NAPL per square meter is Mn

= ρgasoline (Area)(depth)nSn =

(800 kg/m3 )(1 m2 )(2 m)(0.33)(0.07) = 37.0 kg

Since benzene is 1.6% by weight in the gasoline, the mass of benzene per square meter is (37.0)(0.016) = 0.591 kg = 591 g. (b) The mole fraction of Benzene in the NAPL is

mole benzene/78.11 g benzene 0.021 mole benzene 0.016 g benzene = X= g gasoline mole gasoline/102 g gasoline mole gasoline Use Raoult’s law to get the aqueous concentration of benzene at equilibrium with the NAPL: caq = XS = (0.021)(1780) = 37.2 mg/L (c) The advective mass flux per square meter is F = qz Ac = (0.5 m/yr)(1 m2 )(37200 mg/m3 ) = 18600 mg/yr = 18.6 g/yr (d) t=

591 g = 31.8 yr 18.6 g/yr

(e) Benzene would persist longer than MTBE because MTBE’s solubility is about 25 times larger than benzene’s, which overwhelms the fact that there is about 4 times as much MTBE as benzene in the NAPL. 70

(f) The calculation ignores biodegradation, which would accelerate disappearance of benzene. It also assumes that the mole fraction of benzene remains constant through time. The mole fraction actually evolves with time as NAPL constituents dissolve at different rates and degrade at different rates.

24. (a) The balanced chemical equation for oxidation of benzene is C6 H6 + 7.5O2 −→ 6CO2 + 3H2 O For a mole of benzene in the above equation, the mass of benzene is 6 × 12.0 + 6 × 1.0 = 78 g, and the mass of oxygen is 7.5 × 2 × 16.0 = 240 g. The ratio of oxygen mass to benzene mass is 240/78 = 3.1 g O2 /g contaminant. (b) The mass of contaminant within the cylinder is (kg, m units) m = V cρb

= πr2 bcρb = π(2)2 (1)(0.007)(2300) = 202 kg

where the total hydrocarbon concentration is c = 0.007 kg contaminant/kg soil, and the soil dry bulk density is 2300 kg/m3 . The total amount of O2 needed to oxidize this amount of hydrocarbon is 3.1 × 202 = 620 kg.

The rate that oxygen flushes through the cylinder in the passing groundwater is (m, kg, day units) F = qAc = K =

∂h Ac ∂x

(1)(0.006)(4 × 1)(0.005) = 1.2 × 10−4 kg/day

where the concentration of oxygen in the water is c = 0.005 kg/m3 . At the above rate, this mass of contaminant will be oxidized in (kg, day units) t=

m 620 = = 5.2 × 106 days = 14,000 yr F 1.2 × 10−4

71

72

Chapter 12 Solutions 1. Portions of section 12.2 explain heat, thermal energy, and enthalpy. Change in thermal energy and change in enthalpy are about the same when the magnitude of the change thermal energy ∆Uk is much larger than the magnitude of the change in potential energy ∆Up plus the change in pressure/density. |∆Uk | |∆Up + ∆(P/ρ)| 2.

100 kW hr

3600 sec hr

1000 J/sec kW

= 3.6 × 108 J

3. For the saturated case (m, sec, ◦ C, kg units): κ = =

(κm )1−n (κf )n (2.9)0.52 (0.6)0.48 = 1.4 W/(m ◦ C)

For the unsaturated case there are two fluids and the geometric mean equation becomes: κ = =

(κm )1−n (κf )θ (κa )n−θ (2.9)0.52 (0.6)0.27 (0.024)0.21 = 0.69 W/(m ◦ C)

where κf and κa are the thermal conductivities of water and air.

4. According to Fourier’s Law, the heat conducted is (m, sec, ◦ C, kg units): qcz

= −κ

∂T ∂z

= −(3.6)

−22 = 0.079 W/m2 1000

The z coordinate is positive upward and heat flows upward. The Basin and Range region has similar rates.

5. Differentiating Eq. 12.13 with respect to z results in

∂T TL − T0 Ph (Ph z/L) = e P h ∂z e −1 L Taking the second derivative results in

2 ∂2T TL − T0 Ph (Ph z/L) = e ∂z 2 eP h − 1 L

∂T Ph = ∂z L 73

figure file: solutions/Ch12climatechangeprofilesProblem Figure 19: Problem 8. figure file: solutions/Ch12MineprofilesProblem Figure 20: Problem 12.

Substituting Eq. 12.14 into the above gives

∂T ρw cw qzw L ∂2T = ∂z 2 ∂z κL Canceling the L in the fraction and multiplying both sides by κ gives the general equation, Eq. 12.12.

6. With very low K, the specific discharge qzw will be near zero, making Ph near zero.

Heat flow through the layer will be conduction-dominated and the temperature profile will be nearly linear, like the Ph = 0 line in the Figure.

7. Estimate κ for the peat as follows (m, sec, ◦ C, kg units): κ = =

(κm )1−n (κf )n (0.25)0.25 (0.6)0.75 = 0.48 W/(m ◦ C)

Using this value in the spreadsheet, the data are well-matched using a value of qzw = −1.6 × 10−7 m/sec. Heat is flowing down (qzw < 0) toward the low temperature at the base of the peat.

8. See Figure 19 for sketches of profiles for cases (a)–(e). Changes in climate occur over long time periods compared to seasonal changes, so the perturbation of the profiles extends much deeper than the seasonal perturbations do.

9. High ct means shallower penetration of surface temperature variations. Increasing ct increases the capacity of the subsurface to absorb heat, so more of the heat change imparted at the surface goes into changing the heat content and less goes into heat flow to deeper levels. Raising κ makes it easier for heat to conduct, so the oscillations at the surface penetrate deeper. Lowering κ reduces heat conduction, so the oscillations at the surface do not penetrate as deep.

10. Using the spreadsheet, a fair match to the envelope of temperature profiles can be made with ∆T = 15◦ Cand qz = 0.002 m/day = 2.3 × 10−8 m/sec. Positive qz indicates upward flow.

11. One possibility is that the surface climate has recently changed toward cooler temperatures. A second possibility is localized downward advection of cool water from the surface at this location. A third possibility is if there is steady heat conduction here, but the medium near the surface has a lower thermal conductivity κ, so a steeper gradient is required to conduct the same heat.

12. See Figure 20 for sketches of temperature profiles. The surface temperature is pinned by the average air temperature, which is cooler than the pre-mining temperature at 100 m depth. After a very long time, assuming constant climate, the profile would be parallel to the original one but shifted to the left to match the surface temperature.

13. See Figure 21 for a sketch of how the state of water changes through phases 1–5.

74

figure file: solutions/Ch12GeyserPEDiagram Figure 21: Problem 13.

14. At present, water-based heat pumps are more efficient than air-based heat pumps.

The coefficient of performance (COP) for water systems is about 3–5, and for air systems COP is about 2–4 (COP for heating is lower for colder air temps). Water heat pumps need liquid water not frozen water, and can work in any air temperature so long as water surrounding the heat exchanger stays liquid. Air heat pumps work best at temperatures above 0◦ C. There is ongoing research into air systems that can work better at lower temperatures.

75