QUESTIONS
What is the limit between unsaturated and saturated zones in groundwater flow?
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In groundwater flow, the groundwater table, or water table, is the limit
between the unsaturated and saturated zones.
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What factors determine the feasibility of mining a groundwater reservoir? Explain.
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The feasibility of mining or extracting water from a groundwater
reservoir is determined by the following three properties: (1) porosity,
(2) permeability, and (3) replenishment.
Porosity is a measure of the ability of the soil deposit or rock formation to hold water in
sufficiently large quantities.
Permeability describes the rate at which water can pass through a soil deposit or rock formation.
Replenishment relates to the size and extent of a groundwater reservoir and its
interconnection to other neighboring groundwater resources in the region.
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What is an aquifer? An aquiclude? An aquitard?
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An aquifer is a saturated permeable geologic formation which can yield
significant quantities of water to wells and springs.
An aquiclude is a geologic formation that is incapable of yielding significant amounts of
water under ordinary circumstances.
Aquitards are the less permeable geologic formations whose yield may not be sufficient to justify the cost
of drilling wells.
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What are the major differences between confined and unconfined aquifers?
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A confined aquifer is bounded between two relatively impermeable layers
or aquitards. By contrast, in an unconfined aquifer the water table
constitutes its upper boundary.
Unconfined aquifers occur near the ground surface; confined aquifers occur at substantial depths below the
ground surface.
In an unconfined aquifer, the water level rests at the water table. By contrast, in a confined aquifer, the water table may
rise above the top of the aquifer.
Unlike confined aquifers, unconfined aquifers can discharge by vapor diffusion upwards through the soil or
through evapotranspiration by vegetation.
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What is a potentiometric surface? A perched aquifer?
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A potentiometric surface is the loci to which water levels will rise in
wells located in a confined aquifer.
A perched aquifer is a special case of unconfined aquifer which forms on
top of an impermeable layer located well above the water table.
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Explain the processes of recharge and discharge in unconfined and confined aquifers.
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The recharge area is the area of aquifer replenishment with infiltrated
water. The discharge area is the area where the infiltrated water
returns back to the surface.
In unconfined aquifers, recharge takes place by rainfall infiltration reaching the water table; discharge takes
place by vapor diffusion and evapotranspiration, exfiltration to streams
and rivers, and well pumping.
In confined aquifers, recharge takes place through permeable paths; discharge takes place by fast seepage through
permeable paths or slow seepage through aquitards, and also by well
pumping.
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What is specific yield? Specific retention?
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Specific yield is the ratio of free-draining water volume to aquifer
volume.
Specific retention is the ratio of volume of retained water to volume of
aquifer.
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What is field capacity? Permanent wilting point?
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Field capacity is the level of soil moisture which is equivalent to
specific retention, i.e., the amount of water that can be held in the
soil against the action of gravity.
Permanent wilting point is the level of soil moisture at which vegetation
begins to die for lack of sufficient water.
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What is specific discharge? How does hydraulic conductivity differ from intrinsic permeability?
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In flow through porous media, specific discharge is the discharge per
unit area measured perpendicular to the flow direction.
Hydraulic conductivity is a function of the physical properties of fluid
and porous media, including intrinsic permeability, fluid density, and
absolute viscosity. Intrinsic permeability is a function only of the
geometric characteristics of the porous media, including mean grain
diameter, grain size distribution, sphericity and roundness of the
particles, and the nature of their packing. Hydraulic conductivity is
measured in units of velociy (L/T), whereas intrinsic permeability is
measured in units of area (L2).
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Explain the differences between the compressibility of water and that of porous media.
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The compressibility of water changes very little over the range of fluid
pressures normally encountered in practice. The accepted value of the
compressibility of water is β = 4.4 × 10-10 m2/N.
The compressibility of porous media is subject to greater variations
than the compressibility of water. It is a function of the
characteristics of the granular skeleton a te nature of its packing,
varying normally in the range α of 10-6 to 10-11 m2 /N.
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Define specific storage, transmisivity and storativity. How does the concept of storativity differ from that of specific yield?
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The specific storage of a confined aquifer is the volume of water
released per unit volume of aquifer per unit decrease in hydraulic head.
The transmissivity of a confined aquifer is the product of its hydraulic
conductivity and thickness. The storativity of a confined aquifer is the
product of its specific storage and its thickness.
The storativity of a confined aquifer is equivalent to the concept of
specific yield of an unconfined aquifer. However, typical values of
specific yield are usually much greater than those of storativity. This
is because releases from an unconfined aquifer represent an actual
dewatering of the pore spaces, whereas releases from a confined aquifer
represent aquifer compaction due to changes in fluid pressure.
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What fluid and aquifer properties are needed in the description of saturated groundwater flow?
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The fluid and aquifer properties needed in the description of saturated groundwater flow are: fluid density, absolute viscosity and aquifer porosity.
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What additional term is added to the equation of continuity in order to account for transient flow conditions?
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The term added to the continuity equation in order to account for transient flow conditions is the time rate of change of porosity (i.e., fluid mass storage).
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What is hydraulic diffusivity in connection with groundwater flow?
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Hydraulic diffusivity is the coefficient (or parameter) in the general
diffusion equation describing flow in porous media. It is defined as the
ratio of hydraulic conductivity to specific storage, or alternatively, as
the ratio of transmissivity to storativity.
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What determines whether streams are classified as effluent or influent?
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Streams are classified as either effluent or influent depending on
whether they serve as discharge or recharge areas for groundwater.
Effluent streams serve as discharge areas. Conversely, influent streams
serve as recharge areas.
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What theories of streamflow generation lie at the two extremes of the spectrum of hillslope hydrologic models?
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On one extreme, there is the classical Hortonian theory of overland flow,
which stresses overland flow as the primary mechanism responsible for
runoff generation. On the other extreme, there is the throughflow
concept, which relies heavily on interflow to explain the process of
runoff generation.
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What is the essential difference between the partial-area and the variable-source-area models of hillslope hydrology?
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Partial-area models assume that the contributing partial areas are constant in time. Variable-source-area models relax the assumption of time invariance, permiting the source areas to expand and contract as a function of the rainfall input and soil-moisture storage.
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What runoff components are justified by current theories of hillslope hydrology? Explain.
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Two runoff components are justified in light of recent theories of hillslope hydrology: (1) quickflow, consisting of fast interflow (subsurface stormflow), overland flow, and rain falling directly on the channels, and (2) baseflow, consisting of slow interflow and groundwater
flow.
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PROBLEMS
What is the volume of free-draining water in an unconfined aquifer of surface area A = 130 mi2, thickness b = 300 ft, and specific yield Sy = 0.05?
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From Eq. 11-1: V = 130 mi2 × 300 ft × 0.05 × 640 ac/mi2
V = 1,248,000 ac-ft. ANSWER.
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What is the specific retention of an unconfined aquifer of surface area A = 125 mi2, thickness b = 80 ft, porosity n = 0.30, which can yield 1,536,000 ac-ft of free-draining water?
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From Eq. 11-2, specific yield is:
Sy = 1,536,000 ac-ft / (125 mi2 × 80 ft × 640 ac/mi2) = 0.24
Specific retention = n - Sy = 0.30 - 0.24 = 0.06. ANSWER.
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What is the flow through a porous media with cross-sectional area A = 1250 m2, hydraulic conductivity K = 0.8 × 10-4 m/s, under a hydraulic gradient i = 0.01?
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Using Eq. 11-5, the flow is:
Q = 0.8 × 10-4 m/s × 0.01 × 1250 m2 = 0.001 m3/s.
Q = 0.001 m3/s × 1000 L/m3 = 1 L/s, or 60 L/min. ANSWER.
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Derive the exact equivalence between darcys and square centimeters.
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From Eq. 11-5: Q / A = Ki.
With Eq. 11-9:
Q / A = k ρ g i / μ
Therefore: k = (Q / A) μ / (ρgi)
By definition: 1 darcy = 1 cm/s × 1 cp / (1 atm/cm)
1 cm/s × 1 cp × 0.001 (N-s/m2)/cp
1 darcy = _______________________________________ = 0.987 × 10-8 cm2
1 atm/cm × 101,320 (N/m2)/atm
1 darcy = 0.987 × 10-8 cm2. ANSWER.
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What is the hydraulic conductivity of a porous medium with fluid density ρ = 1 g/cm3, intrinsic permeability k = 5 darcys, and fluid absolute viscosity μ = 1 cp.
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Using Eq. 11-9: K = kρg / μ
K = 5 darcys × 1 g/cm3 × 981 cm/s2 / 1 cp
K = 4905 darcy-g/(cm2-s2-cp)
Using the conversion factors: 1 darcy = 0.987 × 10-8 cm2;
1 cp = 0.001 N-s/m2; 1 N = 1000 g × 100 cm/s2; and 1 m2 = 104 cm2
Then: K = 0.00484 cm/s. ANSWER.
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Show that specific storage Ss has units L-1.
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From Eq. 11-12: Ss = ρg (α + n β)
With M = mass, F = force, L = length, and T = time, the units of the right-hand side terms (density ρ, gravitational acceleration g, and compressibility α + n β) are:
[M / L3] × [L2 / T] × [L2 / F] = L-1. ANSWER.
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Calculate the storativity of a confined aquifer of thickness b = 50 m, porosity n = 0.05, and compressibility α = 1.0 × 10-8 Pa-1.
Assume fluid density ρ = 1 g/cm3 and compressibility β = 4.4 × 10-10 Pa-1
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From Eqs. 11-12 and 11-14:
S = ρg (α + n β) b
S = 1 g/cm3 × 981 cm/s2 × [1.0 × 10-8 + (0.05 × 4.4 × 10-10)] Pa-1 × 50 m
S = 981 g/(cm2-s2) × (100.22 × 10-10) Pa-1 × 50 m
Since 1 Pa-1 = 1 m2/N: S = 0.0004916 [g-m3/(cm2-s2)] / N
Since 1 m3 = 106 cm3: S = 491.6 (g-cm/s2) / N
Since 1 g-cm/s2 = 10-5 N: S = 0.004916. ANSWER.
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Calculate the hydraulic diffusivity of an aquifer with the following properties: intrinsic permeability k = 1 darcy, fluid absolute viscosity μ = 1 cp, compressibility α = 1.0 × 10-8 Pa-1, porosity n = 0.08.
Assume fluid compressibility β = 4.4 × 10-10 Pa-1
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From Eq. 11-21, the hydraulic diffusivity of the aquifer is:
vh = K / Ss
From Eqs. 11-9 and 11-12:
vh = [(kρg)/μ] / [ρg (α + n β)] = k / [μ (α + n β)]
vh = 1 darcy / [1 cp × (100 + 0.08 × 4.4) × 10 Pa-1]
vh = 1 darcy / (1 cp × 100.352 × 10-10 m2/N)
vh = 0.9965 × 108 (darcy-N)/(cp-m2)
With 1 darcy = 0.987 × 10-8 cm2; and 1 cp = 0.001 N-s/m2:
vh = 983.5 cm2/s. ANSWER.
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Calculate the hydraulic diffusivity of an aquifer with storativity S = 0.002, hydraulic conductivity K = 1.0 × 10-5 cm/s, and thickness b = 100 m.
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From Eq. 11-22, the hydraulic diffusivity of the aquifer is:
vh = T/S
From Eq. 11-13:
T = K b = 1.0 × 10-5 cm/s × 100 m × 100 cm/m = 0.1 cm2/s.
Therefore: vh = T/S = 0.1 cm2/s / (0.002) = 50 cm2/s. ANSWER.
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A well is pumped at the constant rate of Q = 0.004 m3/s in an aquifer of transmissivity T = 0.004 m2/s and storativity S = 0.0005.
Calculate the drawdown 24 h after the start of pumping in an observational well located at a distance of 250 m from the pumped well.
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Using Eq. 11-28: u = r 2S / (4Tt )
u = (250)2 m2 × 0.0005 /(4 × 0.004 m/s × 24 h × 3600 s/h) = 0.0226.
From Table 11-1, by interpolation, for u = 0.0226: W(u) = 3.25.
Using Eq. 11-29:
Z = Q W(u)/(4 πT )
Z = (0.004 m3/s × 3.25) / (4 × 3.1416 × 0.004 m2/s) = 0.259 m.
The drawdown in the observational well 24 h after the start of pumping is: Z = 0.259 m. ANSWER.
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A well is pumped at the constant rate of Q = 0.008 m3/s.
A match of the well function (W(u) versus 1/u) with drawdown versus time data from an observational well located at a distance of 430 m from the pumped well has produced the following matching values: W(u) = 1, u = 1, Z1 = 0.21 m, and t1 = 1.5 h.
Calculate the aquifer transmissivity and storativity.
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From Eq. 11-31:
T = Q / (4π Z1)
T = 0.008 m3/s / (4 × 3.1416 × 0.21 m) = 0.00303 m2/s. ANSWER.
From Eq. 11-32:
S = (4 T t1) / r 2
S = (4 × 0.00303 m/s × 1.5 h × 3600 s/h) / (430)2 m2
S = 0.000354. ANSWER.
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A well is pumped at a constant rate Q = 0.005 m3/s in an aquifer of transmissivity T = 0.0015 m2/s and storativity S = 0.0005.
Calculate the drawdown 24 h after the start of pumping in an observational well located 1000 m from the pumped well.
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Using Eq. 11-28: u = r 2S / (4 Tt)
u = (1000)2 m2 × 0.0005 /(4 × 0.0015 m/s × 24 h × 3600 s/h) = 0.964
From Table 11-1, by interpolation, for u = 0.964: W(u) = 0.234
Using Eq. 11-29:
Z = Q W(u)/ (4πT)
Z = (0.005 m3/s × 0.234) / (4 × 3.1416 × 0.0015 m2/s) = 0.062 m
The drawdown in the observational well 24 h after the start of pumping is: Z = 0.062 m. ANSWER.
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A well is pumped at a constant rate of 4 L/s.
The drawdown in an observational well located at a distance of 150 m from the pumped well has been measured as follows:
Time (min) |
0 |
10 |
15 |
30 |
60 |
90 |
120 |
Drawdown (m) |
0 |
0.16 |
0.25 |
0.42 |
0.62 |
0.73 |
0.85 |
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The well function W(u) vs u (Table 11-1) is plotted on log paper. The drawdown vs time (Z vs t) data is also plotted on log paper. The matching of both graphs in the manner shown in Fig. 11-5 leads to:
Z1 = 0.33 m, and t1 = 330 s.
Using Eq. 11-31:
T = Q / (4π Z1) = (4 L/s × 0.001 m3/L) / (4 × 3.1416 × 0.33 m)
T = 0.000965 m2/s. ANSWER.
Using Eq. 11-32:
S = 4T t1 /r 2 = (4 × 0.000965 m2/s × 330 s) / (150)2 m2
S = 0.0000566. ANSWER.
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- Given the following measured recession flows in a certain river:
Time (h) |
0 |
12 |
24 |
36 |
48 |
Discharge (m3/s) |
1000 |
882 |
779 |
687 |
606 |
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Calculate: (a) the time of storage ts, (b) the volume released from storage in the first 24 h, (c) the volume released from storage in the second 24 h, and (d) the total volume released from storage assuming that the flow eventually recedes back to zero.
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(a) From Eq. 11-33: ts = t / (ln Qo - ln Q)
Then: ts = 12 h / (ln 1000 - ln 882) = 95.57 h
Also: ts = 24 h / (ln 1000 - ln 779) = 96.09 h
Also: ts = 36 h / (ln 1000 - ln 687) = 95.89 h
Also: ts = 48 h / (ln 1000 - ln 606) = 95.83 h
The average of these four values is: ts = 95.85 h. ANSWER.
(b) Using Eq. 11-41: S = (1000 - 779) m3/s × 95.85 h × 3600 s/h
S = 76,258,260 m3 = 76.26 hm3. ANSWER.
(c) Using Eq. 11-41: S = (779 - 606) m3/s × 95.85 h × 3600 s/h
S = 59,695,380 m3 = 59.70 hm3. ANSWER.
(d) Using Eq. 11-41: S = (1000 - 0) m3/s × 95.85 h × 3600 s/h
S = 345,060,000 m3 = 345.06 hm3. ANSWER.
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- Given the following measured recession flows in a certain stream:
Time (h) |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
Discharge (ft3/s) |
1000 |
920 |
846 |
779 |
716 |
659 |
606 |
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Calculate:
(a) the time of storage ts, (b) the recession constant K, (c) the volume released from storage between 3 h and 6 h, using both ts (Eq. 11-41) and K (Eq. 11-40), and (d) the total volume released from storage assuming that the flow eventually recedes back to zero, using both Eqs. 11-40 and 11-41.
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(a) From Eq. 11-33: ts = t /(ln Qo - ln Q)
Then: ts = 1 h / (ln 1000 - ln 920) = 11.99 h
Also: ts = 2 h / (ln 1000 - ln 846) = 11.96 h
Also: ts = 3 h / (ln 1000 - ln 779) = 12.01 h
Also: ts = 4 h / (ln 1000 - ln 716) = 11.97 h
Also: ts = 5 h / (ln 1000 - in 659) = 11.99 h
Also: ts = 6 h / (ln 1000 - ln 606) = 11.98 h
The average of these six values is: ts = 11.98 h, say 12 h. ANSWER.
(b) From Eq. 11-38: ln K = - (1/ts) = -0.0833333 h-1
Then: K = 0.92 h-1 . ANSWER.
(c) Using Eq. 11-40:
S = [(606 - 779) ft3/s × 3600 s/h] / (-0.0833333 h-1 × 43,560 ft2/ac)
S = 171.57 ac-ft. ANSWER.
Using Eq. 11-41:
S = [(779 - 606) ft3/s × 12 h × 3600 s/h] / (43,560 ft2/ac)
S = 171.57 ac-ft. ANSWER.
(d) Using Eq. 11-40:
S = (-1000 ft3/s) / (ln K) = (-1000 ft3/s) / (-0.0833333 h-1)
S = (-1000 ft3/s × 3600 s/h) / (-0.0833333 h-1 × 43,560 ft2/ac)
S = 991.7 ac-ft. ANSWER.
Using Eq. 11-41:
S = (1000 ft3/s × 12 h × 3600 s/h) / (43,560 ft2/ac)
S = 991.7 ac-ft. ANSWER.
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- Given the following flows measured in the receding limb of a flood hydrograph:
Time (h) |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
Flow (m3/s) |
32.0 |
25.2 |
20.5 |
17.9 |
16.1 |
14.5 |
13.5 |
12.9 |
12.6 |
12.3 |
12.0 |
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Calculate: (a) the time of storage for baseflow, (b) the baseflow and quickflow components, (c) the volume released from baseflow in the elapsed 10-h period, and (d) the volume released from baseflow from t = 0 to t = ∞ , assuming that the flow eventually recedes back to zero.
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(a) A relatively constant time of storage is an indication of baseflow recession. Time of storage for baseflow is evaluated at the tail of the flood hydrograph.
From Eq. 11-33: ts = t / (ln Qo - ln Q). Then:
ts = 1 h / (ln 12.3 - ln 12.0) = 40.50 h
Also: ts = 2 h / (ln 12.6 - ln 12.0) = 40.99 h
Also: ts = 3 h / (ln 12.9 - ln 12.0) = 41.48 h
Therefore, the tail of the flood hydrograph is part of the baseflow recession, and the time of storage for baseflow can be taken as the average of these three values:
ts = 40.99, say 41 h. ANSWER.
(b) Using Eq. 11-33, for a 1-h time interval:
Qo = Q et/ts = Q e1/41 = 1.0247 Q
The first baseflow value is: Q = 12.0. Then, the value an hour earlier is: Qo = 1.0247 Q, and so on.
The calculated baseflow values are summarized in the following table. Quickflow is the difference between flow and baseflow.
Time
(h) |
Flow
(m3/s) |
Baseflow
(m3/s) |
Quickflow
(m3/s) |
| 0 |
32.0 |
15.3 |
16.7 |
| 1 |
25.2 |
14.9 |
10.3 |
| 2 |
20.5 |
14.6 |
5.9 |
| 3 |
17.9 |
14.2 |
3.7 |
| 4 |
16.1 |
13.9 |
2.2 |
| 5 |
14.5 |
13.6 |
0.9 |
| 6 |
13.5 |
13.2 |
0.3 |
| 7 |
12.9 |
12.9 |
0.0 |
| 8 |
12.6 |
12.6 |
0.0 |
| 9 |
12.3 |
12.3 |
0.0 |
| 10 |
12.0 |
12.0 |
0.0 |
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(c) Using Eq. 11-41: S = (Qo - Q)ts
S = (15.3 - 12.0) m3/s × 41 h × 3600 s/h = 487,080 m3. ANSWER.
(d) Using Eq. 11-41 for Q = 0: S = Qots
S = (15.3 m3/s × 41 h × 3600 s/h = 2,258,280 m3. ANSWER.
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150908 10:15 |
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