DARCY’S LAW APPLICATIONS IN GROUNDWATER FLOW
1 COMPLEX NUMERICAL PROBLEMS
1.1 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
a) The average linear velocity of groundwater flow
b) The time taken for the contaminant to reach the detection point, assuming no
retardation
c) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.2 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
a) The drawdown at a distance of 100 m from the well
b) The radius of influence of the well
c) The time it would take for the drawdown to reach 1 m at a distance of 200 m from the
well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.3 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
a) The flow rate per unit width of the aquifer
b) The hydraulic head at a point 300 m from the higher-head boundary
c) The volume of water flowing through the entire aquifer in one year if its width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.4 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
a) The discharge per unit width of the coastline
b) The position of the water table 200 m from the coast
c) The location where saltwater intrusion would begin, assuming the Ghyben-Herzberg
relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.5 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
a) The aquifer’s hydraulic conductivity if its thickness is 25 m
b) The expected drawdown at the same observation well after 10 days of pumping
c) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.6 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
a) The drawdown at a distance of 200 m after 5 days of pumping
b) The steady-state drawdown at the same location
c) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.7 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.8 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
d) The average linear velocity of groundwater flow
e) The time taken for the contaminant to reach the detection point, assuming no
retardation
f) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.9 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
g) The drawdown at a distance of 100 m from the well
h) The radius of influence of the well
i) The time it would take for the drawdown to reach 1 m at a distance of 200 m from the
well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.10 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
j) The flow rate per unit width of the aquifer
k) The hydraulic head at a point 300 m from the higher-head boundary
l) The volume of water flowing through the entire aquifer in one year if its width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.11 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
m) The discharge per unit width of the coastline
n) The position of the water table 200 m from the coast
o) The location where saltwater intrusion would begin, assuming the Ghyben-Herzberg
relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.12 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
p) The aquifer’s hydraulic conductivity if its thickness is 25 m
q) The expected drawdown at the same observation well after 10 days of pumping
r) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.13 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
s) The drawdown at a distance of 200 m after 5 days of pumping
t) The steady-state drawdown at the same location
u) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.14 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.15 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
v) The average linear velocity of groundwater flow
w) The time taken for the contaminant to reach the detection point, assuming no
retardation
x) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.16 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
y) The drawdown at a distance of 100 m from the well
z) The radius of influence of the well
aa) The time it would take for the drawdown to reach 1 m at a distance of 200 m from the
well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.17 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
bb) The flow rate per unit width of the aquifer
cc) The hydraulic head at a point 300 m from the higher-head boundary
dd) The volume of water flowing through the entire aquifer in one year if its width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.18 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
ee) The discharge per unit width of the coastline
ff) The position of the water table 200 m from the coast
gg) The location where saltwater intrusion would begin, assuming the Ghyben-Herzberg
relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.19 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
hh) The aquifer’s hydraulic conductivity if its thickness is 25 m
ii) The expected drawdown at the same observation well after 10 days of pumping
jj) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.20 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
kk) The drawdown at a distance of 200 m after 5 days of pumping
ll) The steady-state drawdown at the same location
mm) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.21 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.22 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
nn) The average linear velocity of groundwater flow
oo) The time taken for the contaminant to reach the detection point, assuming no
retardation
pp) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.23 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
qq) The drawdown at a distance of 100 m from the well
rr) The radius of influence of the well
ss) The time it would take for the drawdown to reach 1 m at a distance of 200 m from the
well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.24 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
tt) The flow rate per unit width of the aquifer
uu) The hydraulic head at a point 300 m from the higher-head boundary
vv) The volume of water flowing through the entire aquifer in one year if its width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.25 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
ww) The discharge per unit width of the coastline
xx) The position of the water table 200 m from the coast
yy) The location where saltwater intrusion would begin, assuming the Ghyben-Herzberg
relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.26 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
zz) The aquifer’s hydraulic conductivity if its thickness is 25 m
aaa) The expected drawdown at the same observation well after 10 days of pumping
bbb) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.27 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
ccc) The drawdown at a distance of 200 m after 5 days of pumping
ddd) The steady-state drawdown at the same location
eee) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.28 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.29 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
fff) The average linear velocity of groundwater flow
ggg) The time taken for the contaminant to reach the detection point, assuming no
retardation
hhh) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.30 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
iii) The drawdown at a distance of 100 m from the well
jjj) The radius of influence of the well
kkk) The time it would take for the drawdown to reach 1 m at a distance of 200 m from the
well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.31 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
lll) The flow rate per unit width of the aquifer
mmm) The hydraulic head at a point 300 m from the higher-head boundary
nnn) The volume of water flowing through the entire aquifer in one year if its width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.32 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
ooo) The discharge per unit width of the coastline
ppp) The position of the water table 200 m from the coast
qqq) The location where saltwater intrusion would begin, assuming the Ghyben-Herzberg
relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.33 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
rrr) The aquifer’s hydraulic conductivity if its thickness is 25 m
sss) The expected drawdown at the same observation well after 10 days of pumping
ttt) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.34 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
uuu) The drawdown at a distance of 200 m after 5 days of pumping
vvv) The steady-state drawdown at the same location
www) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.35 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.36 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
xxx) The average linear velocity of groundwater flow
yyy) The time taken for the contaminant to reach the detection point, assuming no
retardation
zzz) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.37 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
aaaa) The drawdown at a distance of 100 m from the well
bbbb) The radius of influence of the well
cccc) The time it would take for the drawdown to reach 1 m at a distance of 200 m
from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.38 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
dddd) The flow rate per unit width of the aquifer
eeee) The hydraulic head at a point 300 m from the higher-head boundary
ffff) The volume of water flowing through the entire aquifer in one year if its width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.39 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
gggg) The discharge per unit width of the coastline
hhhh) The position of the water table 200 m from the coast
iiii) The location where saltwater intrusion would begin, assuming the Ghyben-Herzberg
relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.40 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
jjjj) The aquifer’s hydraulic conductivity if its thickness is 25 m
kkkk) The expected drawdown at the same observation well after 10 days of pumping
llll) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.41 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
mmmm) The drawdown at a distance of 200 m after 5 days of pumping
nnnn) The steady-state drawdown at the same location
oooo) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.42 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.43 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
pppp) The average linear velocity of groundwater flow
qqqq) The time taken for the contaminant to reach the detection point, assuming no
retardation
rrrr) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.44 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
ssss) The drawdown at a distance of 100 m from the well
tttt) The radius of influence of the well
uuuu) The time it would take for the drawdown to reach 1 m at a distance of 200 m
from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.45 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
vvvv) The flow rate per unit width of the aquifer
wwww) The hydraulic head at a point 300 m from the higher-head boundary
xxxx) The volume of water flowing through the entire aquifer in one year if its width is
5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.46 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
yyyy) The discharge per unit width of the coastline
zzzz) The position of the water table 200 m from the coast
aaaaa) The location where saltwater intrusion would begin, assuming the Ghyben-
Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.47 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
bbbbb) The aquifer’s hydraulic conductivity if its thickness is 25 m
ccccc) The expected drawdown at the same observation well after 10 days of pumping
ddddd) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.48 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
eeeee) The drawdown at a distance of 200 m after 5 days of pumping
fffff) The steady-state drawdown at the same location
ggggg) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.49 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.50 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
hhhhh) The average linear velocity of groundwater flow
iiiii) The time taken for the contaminant to reach the detection point, assuming no
retardation
jjjjj) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.51 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
kkkkk) The drawdown at a distance of 100 m from the well
lllll) The radius of influence of the well
mmmmm) The time it would take for the drawdown to reach 1 m at a distance of 200 m
from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.52 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
nnnnn) The flow rate per unit width of the aquifer
ooooo) The hydraulic head at a point 300 m from the higher-head boundary
ppppp) The volume of water flowing through the entire aquifer in one year if its width is
5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.53 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
qqqqq) The discharge per unit width of the coastline
rrrrr) The position of the water table 200 m from the coast
sssss) The location where saltwater intrusion would begin, assuming the Ghyben-
Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.54 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
ttttt) The aquifer’s hydraulic conductivity if its thickness is 25 m
uuuuu) The expected drawdown at the same observation well after 10 days of pumping
vvvvv) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.55 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
wwwww) The drawdown at a distance of 200 m after 5 days of pumping
xxxxx) The steady-state drawdown at the same location
yyyyy) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.56 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.57 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
zzzzz) The average linear velocity of groundwater flow
aaaaaa) The time taken for the contaminant to reach the detection point, assuming no
retardation
bbbbbb) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.58 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
cccccc) The drawdown at a distance of 100 m from the well
dddddd) The radius of influence of the well
eeeeee) The time it would take for the drawdown to reach 1 m at a distance of 200 m
from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.59 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
ffffff) The flow rate per unit width of the aquifer
gggggg) The hydraulic head at a point 300 m from the higher-head boundary
hhhhhh) The volume of water flowing through the entire aquifer in one year if its width is
5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.60 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
iiiiii) The discharge per unit width of the coastline
jjjjjj) The position of the water table 200 m from the coast
kkkkkk) The location where saltwater intrusion would begin, assuming the Ghyben-
Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.61 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
llllll) The aquifer’s hydraulic conductivity if its thickness is 25 m
mmmmmm) The expected drawdown at the same observation well after 10 days of pumping
nnnnnn) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.62 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
oooooo) The drawdown at a distance of 200 m after 5 days of pumping
pppppp) The steady-state drawdown at the same location
qqqqqq) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.63 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.64 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
rrrrrr) The average linear velocity of groundwater flow
ssssss) The time taken for the contaminant to reach the detection point, assuming no
retardation
tttttt) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.65 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
uuuuuu) The drawdown at a distance of 100 m from the well
vvvvvv) The radius of influence of the well
wwwwww) The time it would take for the drawdown to reach 1 m at a distance of 200 m
from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.66 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
xxxxxx) The flow rate per unit width of the aquifer
yyyyyy) The hydraulic head at a point 300 m from the higher-head boundary
zzzzzz) The volume of water flowing through the entire aquifer in one year if its width is
5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.67 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
aaaaaaa) The discharge per unit width of the coastline
bbbbbbb) The position of the water table 200 m from the coast
ccccccc) The location where saltwater intrusion would begin, assuming the Ghyben-
Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.68 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
ddddddd) The aquifer’s hydraulic conductivity if its thickness is 25 m
eeeeeee) The expected drawdown at the same observation well after 10 days of pumping
fffffff) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.69 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
ggggggg) The drawdown at a distance of 200 m after 5 days of pumping
hhhhhhh) The steady-state drawdown at the same location
iiiiiii) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.70 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.71 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
jjjjjjj) The average linear velocity of groundwater flow
kkkkkkk) The time taken for the contaminant to reach the detection point, assuming no
retardation
lllllll) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.72 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
mmmmmmm) The drawdown at a distance of 100 m from the well
nnnnnnn) The radius of influence of the well
ooooooo) The time it would take for the drawdown to reach 1 m at a distance of 200 m
from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.73 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
ppppppp) The flow rate per unit width of the aquifer
qqqqqqq) The hydraulic head at a point 300 m from the higher-head boundary
rrrrrrr) The volume of water flowing through the entire aquifer in one year if its width is
5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.74 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
sssssss) The discharge per unit width of the coastline
ttttttt) The position of the water table 200 m from the coast
uuuuuuu) The location where saltwater intrusion would begin, assuming the Ghyben-
Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.75 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
vvvvvvv) The aquifer’s hydraulic conductivity if its thickness is 25 m
wwwwwww) The expected drawdown at the same observation well after 10 days of
pumping
xxxxxxx) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.76 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
yyyyyyy) The drawdown at a distance of 200 m after 5 days of pumping
zzzzzzz) The steady-state drawdown at the same location
aaaaaaaa) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.77 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.78 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
bbbbbbbb) The average linear velocity of groundwater flow
cccccccc) The time taken for the contaminant to reach the detection point, assuming no
retardation
dddddddd) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.79 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
eeeeeeee) The drawdown at a distance of 100 m from the well
ffffffff) The radius of influence of the well
gggggggg) The time it would take for the drawdown to reach 1 m at a distance of 200 m
from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.80 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
hhhhhhhh) The flow rate per unit width of the aquifer
iiiiiiii) The hydraulic head at a point 300 m from the higher-head boundary
jjjjjjjj) The volume of water flowing through the entire aquifer in one year if its width is
5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.81 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
kkkkkkkk) The discharge per unit width of the coastline
llllllll) The position of the water table 200 m from the coast
mmmmmmmm) The location where saltwater intrusion would begin, assuming the
Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.82 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
nnnnnnnn) The aquifer’s hydraulic conductivity if its thickness is 25 m
oooooooo) The expected drawdown at the same observation well after 10 days of pumping
pppppppp) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.83 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
qqqqqqqq) The drawdown at a distance of 200 m after 5 days of pumping
rrrrrrrr) The steady-state drawdown at the same location
ssssssss) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.84 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.85 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
tttttttt) The average linear velocity of groundwater flow
uuuuuuuu) The time taken for the contaminant to reach the detection point, assuming no
retardation
vvvvvvvv) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.86 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
wwwwwwww) The drawdown at a distance of 100 m from the well
xxxxxxxx) The radius of influence of the well
yyyyyyyy) The time it would take for the drawdown to reach 1 m at a distance of 200 m
from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.87 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
zzzzzzzz) The flow rate per unit width of the aquifer
aaaaaaaaa) The hydraulic head at a point 300 m from the higher-head boundary
bbbbbbbbb) The volume of water flowing through the entire aquifer in one year if its width is
5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.88 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
ccccccccc) The discharge per unit width of the coastline
ddddddddd) The position of the water table 200 m from the coast
eeeeeeeee) The location where saltwater intrusion would begin, assuming the Ghyben-
Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.89 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
fffffffff) The aquifer’s hydraulic conductivity if its thickness is 25 m
ggggggggg) The expected drawdown at the same observation well after 10 days of pumping
hhhhhhhhh) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.90 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
iiiiiiiii) The drawdown at a distance of 200 m after 5 days of pumping
jjjjjjjjj) The steady-state drawdown at the same location
kkkkkkkkk) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.91 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.92 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
lllllllll) The average linear velocity of groundwater flow
mmmmmmmmm) The time taken for the contaminant to reach the detection point,
assuming no retardation
nnnnnnnnn) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.93 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
ooooooooo) The drawdown at a distance of 100 m from the well
ppppppppp) The radius of influence of the well
qqqqqqqqq) The time it would take for the drawdown to reach 1 m at a distance of 200 m
from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.94 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
rrrrrrrrr) The flow rate per unit width of the aquifer
sssssssss) The hydraulic head at a point 300 m from the higher-head boundary
ttttttttt) The volume of water flowing through the entire aquifer in one year if its width is
5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.95 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
uuuuuuuuu) The discharge per unit width of the coastline
vvvvvvvvv) The position of the water table 200 m from the coast
wwwwwwwww) The location where saltwater intrusion would begin, assuming the
Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.96 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
xxxxxxxxx) The aquifer’s hydraulic conductivity if its thickness is 25 m
yyyyyyyyy) The expected drawdown at the same observation well after 10 days of pumping
zzzzzzzzz) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.97 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
aaaaaaaaaa) The drawdown at a distance of 200 m after 5 days of pumping
bbbbbbbbbb) The steady-state drawdown at the same location
cccccccccc) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.98 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.99 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
dddddddddd) The average linear velocity of groundwater flow
eeeeeeeeee) The time taken for the contaminant to reach the detection point,
assuming no retardation
ffffffffff) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.100 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
gggggggggg) The drawdown at a distance of 100 m from the well
hhhhhhhhhh) The radius of influence of the well
iiiiiiiiii) The time it would take for the drawdown to reach 1 m at a distance of 200 m
from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.101 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
jjjjjjjjjj) The flow rate per unit width of the aquifer
kkkkkkkkkk) The hydraulic head at a point 300 m from the higher-head boundary
llllllllll) The volume of water flowing through the entire aquifer in one year if its width is
5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.102 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
mmmmmmmmmm) The discharge per unit width of the coastline
nnnnnnnnnn) The position of the water table 200 m from the coast
oooooooooo) The location where saltwater intrusion would begin, assuming the
Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.103 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
pppppppppp) The aquifer’s hydraulic conductivity if its thickness is 25 m
qqqqqqqqqq) The expected drawdown at the same observation well after 10 days of
pumping
rrrrrrrrrr) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.104 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
ssssssssss) The drawdown at a distance of 200 m after 5 days of pumping
tttttttttt) The steady-state drawdown at the same location
uuuuuuuuuu) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.105 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.106 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
vvvvvvvvvv) The average linear velocity of groundwater flow
wwwwwwwwww) The time taken for the contaminant to reach the detection point,
assuming no retardation
xxxxxxxxxx) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.107 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
yyyyyyyyyy) The drawdown at a distance of 100 m from the well
zzzzzzzzzz) The radius of influence of the well
aaaaaaaaaaa) The time it would take for the drawdown to reach 1 m at a distance of
200 m from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.108 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
bbbbbbbbbbb) The flow rate per unit width of the aquifer
ccccccccccc) The hydraulic head at a point 300 m from the higher-head boundary
ddddddddddd) The volume of water flowing through the entire aquifer in one year if its
width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.109 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
eeeeeeeeeee) The discharge per unit width of the coastline
fffffffffff) The position of the water table 200 m from the coast
ggggggggggg) The location where saltwater intrusion would begin, assuming the
Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.110 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
hhhhhhhhhhh) The aquifer’s hydraulic conductivity if its thickness is 25 m
iiiiiiiiiii) The expected drawdown at the same observation well after 10 days of pumping
jjjjjjjjjjj) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.111 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
kkkkkkkkkkk) The drawdown at a distance of 200 m after 5 days of pumping
lllllllllll) The steady-state drawdown at the same location
mmmmmmmmmmm) The percentage of pumped water derived from leakage at steady-
state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.112 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.113 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
nnnnnnnnnnn) The average linear velocity of groundwater flow
ooooooooooo) The time taken for the contaminant to reach the detection point,
assuming no retardation
ppppppppppp) The width of the plume if the longitudinal dispersivity is 10 m and the
transverse dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.114 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
qqqqqqqqqqq) The drawdown at a distance of 100 m from the well
rrrrrrrrrrr) The radius of influence of the well
sssssssssss) The time it would take for the drawdown to reach 1 m at a distance of 200 m
from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.115 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
ttttttttttt) The flow rate per unit width of the aquifer
uuuuuuuuuuu) The hydraulic head at a point 300 m from the higher-head boundary
vvvvvvvvvvv) The volume of water flowing through the entire aquifer in one year if its
width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.116 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
wwwwwwwwwww) The discharge per unit width of the coastline
xxxxxxxxxxx) The position of the water table 200 m from the coast
yyyyyyyyyyy) The location where saltwater intrusion would begin, assuming the
Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.117 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
zzzzzzzzzzz) The aquifer’s hydraulic conductivity if its thickness is 25 m
aaaaaaaaaaaa) The expected drawdown at the same observation well after 10 days of
pumping
bbbbbbbbbbbb) The pumping rate required to maintain a steady-state drawdown of 5 m
at the observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.118 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
cccccccccccc) The drawdown at a distance of 200 m after 5 days of pumping
dddddddddddd) The steady-state drawdown at the same location
eeeeeeeeeeee) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.119 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.120 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
ffffffffffff) The average linear velocity of groundwater flow
gggggggggggg) The time taken for the contaminant to reach the detection point,
assuming no retardation
hhhhhhhhhhhh) The width of the plume if the longitudinal dispersivity is 10 m and the
transverse dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.121 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
iiiiiiiiiiii) The drawdown at a distance of 100 m from the well
jjjjjjjjjjjj) The radius of influence of the well
kkkkkkkkkkkk) The time it would take for the drawdown to reach 1 m at a distance of
200 m from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.122 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
llllllllllll) The flow rate per unit width of the aquifer
mmmmmmmmmmmm) The hydraulic head at a point 300 m from the higher-head
boundary
nnnnnnnnnnnn) The volume of water flowing through the entire aquifer in one year if its
width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.123 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
oooooooooooo) The discharge per unit width of the coastline
pppppppppppp) The position of the water table 200 m from the coast
qqqqqqqqqqqq) The location where saltwater intrusion would begin, assuming the
Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.124 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
rrrrrrrrrrrr) The aquifer’s hydraulic conductivity if its thickness is 25 m
ssssssssssss) The expected drawdown at the same observation well after 10 days of
pumping
tttttttttttt) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.125 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
uuuuuuuuuuuu) The drawdown at a distance of 200 m after 5 days of pumping
vvvvvvvvvvvv) The steady-state drawdown at the same location
wwwwwwwwwwww) The percentage of pumped water derived from leakage at steady-
state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.126 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so:
ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.127 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
xxxxxxxxxxxx) The average linear velocity of groundwater flow
yyyyyyyyyyyy) The time taken for the contaminant to reach the detection point,
assuming no retardation
zzzzzzzzzzzz) The width of the plume if the longitudinal dispersivity is 10 m and the
transverse dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.128 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
aaaaaaaaaaaaa) The drawdown at a distance of 100 m from the well
bbbbbbbbbbbbb) The radius of influence of the well
ccccccccccccc) The time it would take for the drawdown to reach 1 m at a distance of
200 m from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.129 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
ddddddddddddd) The flow rate per unit width of the aquifer
eeeeeeeeeeeee) The hydraulic head at a point 300 m from the higher-head boundary
fffffffffffff) The volume of water flowing through the entire aquifer in one year if its width is
5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.130 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
ggggggggggggg) The discharge per unit width of the coastline
hhhhhhhhhhhhh) The position of the water table 200 m from the coast
iiiiiiiiiiiii) The location where saltwater intrusion would begin, assuming the Ghyben-
Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.131 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
jjjjjjjjjjjjj) The aquifer’s hydraulic conductivity if its thickness is 25 m
kkkkkkkkkkkkk) The expected drawdown at the same observation well after 10 days of
pumping
lllllllllllll) The pumping rate required to maintain a steady-state drawdown of 5 m at the
observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.132 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
mmmmmmmmmmmmm) The drawdown at a distance of 200 m after 5 days of pumping
nnnnnnnnnnnnn) The steady-state drawdown at the same location
ooooooooooooo) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.133 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so:
ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.134 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
ppppppppppppp) The average linear velocity of groundwater flow
qqqqqqqqqqqqq) The time taken for the contaminant to reach the detection point,
assuming no retardation
rrrrrrrrrrrrr) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.135 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
sssssssssssss) The drawdown at a distance of 100 m from the well
ttttttttttttt) The radius of influence of the well
uuuuuuuuuuuuu) The time it would take for the drawdown to reach 1 m at a distance of
200 m from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.136 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
vvvvvvvvvvvvv) The flow rate per unit width of the aquifer
wwwwwwwwwwwww) The hydraulic head at a point 300 m from the higher-head
boundary
xxxxxxxxxxxxx) The volume of water flowing through the entire aquifer in one year if its
width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.137 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
yyyyyyyyyyyyy) The discharge per unit width of the coastline
zzzzzzzzzzzzz) The position of the water table 200 m from the coast
aaaaaaaaaaaaaa) The location where saltwater intrusion would begin, assuming the
Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.138 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
bbbbbbbbbbbbbb) The aquifer’s hydraulic conductivity if its thickness is 25 m
cccccccccccccc) The expected drawdown at the same observation well after 10 days of
pumping
dddddddddddddd) The pumping rate required to maintain a steady-state drawdown of 5 m
at the observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.139 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
eeeeeeeeeeeeee) The drawdown at a distance of 200 m after 5 days of pumping
ffffffffffffff) The steady-state drawdown at the same location
gggggggggggggg) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.140 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.141 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
hhhhhhhhhhhhhh) The average linear velocity of groundwater flow
iiiiiiiiiiiiii) The time taken for the contaminant to reach the detection point, assuming no
retardation
jjjjjjjjjjjjjj) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.142 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
kkkkkkkkkkkkkk) The drawdown at a distance of 100 m from the well
llllllllllllll) The radius of influence of the well
mmmmmmmmmmmmmm) The time it would take for the drawdown to reach 1 m at a
distance of 200 m from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.143 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
nnnnnnnnnnnnnn) The flow rate per unit width of the aquifer
oooooooooooooo) The hydraulic head at a point 300 m from the higher-head boundary
pppppppppppppp) The volume of water flowing through the entire aquifer in one year if its
width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.144 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
qqqqqqqqqqqqqq) The discharge per unit width of the coastline
rrrrrrrrrrrrrr) The position of the water table 200 m from the coast
ssssssssssssss) The location where saltwater intrusion would begin, assuming the
Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.145 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
tttttttttttttt) The aquifer’s hydraulic conductivity if its thickness is 25 m
uuuuuuuuuuuuuu) The expected drawdown at the same observation well after 10 days of
pumping
vvvvvvvvvvvvvv) The pumping rate required to maintain a steady-state drawdown of 5 m
at the observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.146 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
wwwwwwwwwwwwww) The drawdown at a distance of 200 m after 5 days of pumping
xxxxxxxxxxxxxx) The steady-state drawdown at the same location
yyyyyyyyyyyyyy) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.147 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.148 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
zzzzzzzzzzzzzz) The average linear velocity of groundwater flow
aaaaaaaaaaaaaaa) The time taken for the contaminant to reach the detection point,
assuming no retardation
bbbbbbbbbbbbbbb) The width of the plume if the longitudinal dispersivity is 10 m and the
transverse dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.149 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
ccccccccccccccc) The drawdown at a distance of 100 m from the well
ddddddddddddddd) The radius of influence of the well
eeeeeeeeeeeeeee) The time it would take for the drawdown to reach 1 m at a distance of
200 m from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.150 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
fffffffffffffff) The flow rate per unit width of the aquifer
ggggggggggggggg) The hydraulic head at a point 300 m from the higher-head boundary
hhhhhhhhhhhhhhh) The volume of water flowing through the entire aquifer in one
year if its width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.151 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
iiiiiiiiiiiiiii) The discharge per unit width of the coastline
jjjjjjjjjjjjjjj) The position of the water table 200 m from the coast
kkkkkkkkkkkkkkk) The location where saltwater intrusion would begin, assuming the
Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.152 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
lllllllllllllll) The aquifer’s hydraulic conductivity if its thickness is 25 m
mmmmmmmmmmmmmmm) The expected drawdown at the same observation well
after 10 days of pumping
nnnnnnnnnnnnnnn) The pumping rate required to maintain a steady-state drawdown
of 5 m at the observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.153 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
ooooooooooooooo) The drawdown at a distance of 200 m after 5 days of pumping
ppppppppppppppp) The steady-state drawdown at the same location
qqqqqqqqqqqqqqq) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.154 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.155 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
rrrrrrrrrrrrrrr) The average linear velocity of groundwater flow
sssssssssssssss) The time taken for the contaminant to reach the detection point,
assuming no retardation
ttttttttttttttt) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.156 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
uuuuuuuuuuuuuuu) The drawdown at a distance of 100 m from the well
vvvvvvvvvvvvvvv) The radius of influence of the well
wwwwwwwwwwwwwww) The time it would take for the drawdown to reach 1 m at a
distance of 200 m from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.157 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
xxxxxxxxxxxxxxx) The flow rate per unit width of the aquifer
yyyyyyyyyyyyyyy) The hydraulic head at a point 300 m from the higher-head boundary
zzzzzzzzzzzzzzz) The volume of water flowing through the entire aquifer in one year if its
width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.158 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
aaaaaaaaaaaaaaaa) The discharge per unit width of the coastline
bbbbbbbbbbbbbbbb) The position of the water table 200 m from the coast
cccccccccccccccc) The location where saltwater intrusion would begin, assuming the
Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.159 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
dddddddddddddddd) The aquifer’s hydraulic conductivity if its thickness is 25 m
eeeeeeeeeeeeeeee) The expected drawdown at the same observation well after 10
days of pumping
ffffffffffffffff) The pumping rate required to maintain a steady-state drawdown of 5 m
at the observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.160 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
gggggggggggggggg) The drawdown at a distance of 200 m after 5 days of pumping
hhhhhhhhhhhhhhhh) The steady-state drawdown at the same location
iiiiiiiiiiiiiiii) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.161 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.162 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
jjjjjjjjjjjjjjjj) The average linear velocity of groundwater flow
kkkkkkkkkkkkkkkk) The time taken for the contaminant to reach the detection point,
assuming no retardation
llllllllllllllll) The width of the plume if the longitudinal dispersivity is 10 m and the transverse
dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.163 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
mmmmmmmmmmmmmmmm) The drawdown at a distance of 100 m from the well
nnnnnnnnnnnnnnnn) The radius of influence of the well
oooooooooooooooo) The time it would take for the drawdown to reach 1 m at a
distance of 200 m from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.164 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
pppppppppppppppp) The flow rate per unit width of the aquifer
qqqqqqqqqqqqqqqq) The hydraulic head at a point 300 m from the higher-head
boundary
rrrrrrrrrrrrrrrr) The volume of water flowing through the entire aquifer in one year if its
width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.165 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
ssssssssssssssss) The discharge per unit width of the coastline
tttttttttttttttt) The position of the water table 200 m from the coast
uuuuuuuuuuuuuuuu) The location where saltwater intrusion would begin, assuming the
Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.166 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
vvvvvvvvvvvvvvvv) The aquifer’s hydraulic conductivity if its thickness is 25 m
wwwwwwwwwwwwwwww) The expected drawdown at the same observation well
after 10 days of pumping
xxxxxxxxxxxxxxxx) The pumping rate required to maintain a steady-state drawdown of 5 m
at the observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.167 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
yyyyyyyyyyyyyyyy) The drawdown at a distance of 200 m after 5 days of pumping
zzzzzzzzzzzzzzzz) The steady-state drawdown at the same location
aaaaaaaaaaaaaaaaa) The percentage of pumped water derived from leakage at steady-
state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.168 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.169 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
bbbbbbbbbbbbbbbbb) The average linear velocity of groundwater flow
ccccccccccccccccc) The time taken for the contaminant to reach the detection point,
assuming no retardation
ddddddddddddddddd) The width of the plume if the longitudinal dispersivity is 10 m and
the transverse dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.170 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
eeeeeeeeeeeeeeeee) The drawdown at a distance of 100 m from the well
fffffffffffffffff) The radius of influence of the well
ggggggggggggggggg) The time it would take for the drawdown to reach 1 m at a
distance of 200 m from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.171 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
hhhhhhhhhhhhhhhhh) The flow rate per unit width of the aquifer
iiiiiiiiiiiiiiiii) The hydraulic head at a point 300 m from the higher-head boundary
jjjjjjjjjjjjjjjjj) The volume of water flowing through the entire aquifer in one year if its width is
5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.172 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
kkkkkkkkkkkkkkkkk) The discharge per unit width of the coastline
lllllllllllllllll) The position of the water table 200 m from the coast
mmmmmmmmmmmmmmmmm) The location where saltwater intrusion would begin,
assuming the Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.173 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
nnnnnnnnnnnnnnnnn) The aquifer’s hydraulic conductivity if its thickness is 25 m
ooooooooooooooooo) The expected drawdown at the same observation well after 10
days of pumping
ppppppppppppppppp) The pumping rate required to maintain a steady-state drawdown
of 5 m at the observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.174 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
qqqqqqqqqqqqqqqqq) The drawdown at a distance of 200 m after 5 days of pumping
rrrrrrrrrrrrrrrrr) The steady-state drawdown at the same location
sssssssssssssssss) The percentage of pumped water derived from leakage at steady-state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.175 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.176 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
ttttttttttttttttt) The average linear velocity of groundwater flow
uuuuuuuuuuuuuuuuu) The time taken for the contaminant to reach the detection point,
assuming no retardation
vvvvvvvvvvvvvvvvv) The width of the plume if the longitudinal dispersivity is 10 m and
the transverse dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.177 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
wwwwwwwwwwwwwwwww) The drawdown at a distance of 100 m from the well
xxxxxxxxxxxxxxxxx) The radius of influence of the well
yyyyyyyyyyyyyyyyy) The time it would take for the drawdown to reach 1 m at a
distance of 200 m from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.178 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
zzzzzzzzzzzzzzzzz) The flow rate per unit width of the aquifer
aaaaaaaaaaaaaaaaaa) The hydraulic head at a point 300 m from the higher-head
boundary
bbbbbbbbbbbbbbbbbb) The volume of water flowing through the entire aquifer in one
year if its width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.179 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
cccccccccccccccccc) The discharge per unit width of the coastline
dddddddddddddddddd) The position of the water table 200 m from the coast
eeeeeeeeeeeeeeeeee) The location where saltwater intrusion would begin, assuming the
Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.180 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
ffffffffffffffffff) The aquifer’s hydraulic conductivity if its thickness is 25 m
gggggggggggggggggg) The expected drawdown at the same observation well after 10
days of pumping
hhhhhhhhhhhhhhhhhh) The pumping rate required to maintain a steady-state drawdown
of 5 m at the observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.181 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
iiiiiiiiiiiiiiiiii) The drawdown at a distance of 200 m after 5 days of pumping
jjjjjjjjjjjjjjjjjj)The steady-state drawdown at the same location
kkkkkkkkkkkkkkkkkk) The percentage of pumped water derived from leakage at steady-
state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.182 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.183 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
llllllllllllllllll) The average linear velocity of groundwater flow
mmmmmmmmmmmmmmmmmm) The time taken for the contaminant to reach the
detection point, assuming no retardation
nnnnnnnnnnnnnnnnnn) The width of the plume if the longitudinal dispersivity is 10 m and
the transverse dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.184 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
oooooooooooooooooo) The drawdown at a distance of 100 m from the well
pppppppppppppppppp) The radius of influence of the well
qqqqqqqqqqqqqqqqqq) The time it would take for the drawdown to reach 1 m at a
distance of 200 m from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.185 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
rrrrrrrrrrrrrrrrrr) The flow rate per unit width of the aquifer
ssssssssssssssssss) The hydraulic head at a point 300 m from the higher-head boundary
tttttttttttttttttt) The volume of water flowing through the entire aquifer in one year if its
width is 5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.186 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
uuuuuuuuuuuuuuuuuu) The discharge per unit width of the coastline
vvvvvvvvvvvvvvvvvv) The position of the water table 200 m from the coast
wwwwwwwwwwwwwwwwww) The location where saltwater intrusion would begin,
assuming the Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.187 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
xxxxxxxxxxxxxxxxxx) The aquifer’s hydraulic conductivity if its thickness is 25 m
yyyyyyyyyyyyyyyyyy) The expected drawdown at the same observation well after 10
days of pumping
zzzzzzzzzzzzzzzzzz) The pumping rate required to maintain a steady-state drawdown of 5 m
at the observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.188 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
aaaaaaaaaaaaaaaaaaa) The drawdown at a distance of 200 m after 5 days of pumping
bbbbbbbbbbbbbbbbbbb) The steady-state drawdown at the same location
ccccccccccccccccccc) The percentage of pumped water derived from leakage at steady-
state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.189 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so: ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
1.190 PROBLEM 1
An unconfined aquifer has a hydraulic conductivity of 25 m/day and a porosity of 0.3. The water
table slopes at 0.005 m/m. A contamination plume is detected 500 m downstream from a point
source. Estimate:
ddddddddddddddddddd) The average linear velocity of groundwater flow
eeeeeeeeeeeeeeeeeee) The time taken for the contaminant to reach the detection point,
assuming no retardation
fffffffffffffffffff) The width of the plume if the longitudinal dispersivity is 10 m and the
transverse dispersivity is 1 m
Solution: a) Average linear velocity:
𝑣 =−𝐾
𝑛𝑒𝑑ℎ
𝑑𝑙
=−25 m/day
0.3 ×(−0.005 m/m)
=0.417 m/day
b) Time to reach detection point:
𝑡 =distance
𝑣
=500 m
0.417 m/day
=1199 days ≈3.3 years
c) Plume width: Using the relationship 𝜎𝑦
2=2𝛼𝑇𝑥, where 𝜎𝑦 is the transverse standard
deviation and 𝑥 is the travel distance:
𝜎𝑦=√2×1 m ×500 m
=31.6 m
Assuming a 95
1.191 PROBLEM 2
A confined aquifer has a transmissivity of 200 m²/day and a storage coefficient of 0.0001. A
well pumping at 1000 m³/day has been operating for 5 days. Calculate:
ggggggggggggggggggg) The drawdown at a distance of 100 m from the well
hhhhhhhhhhhhhhhhhhh) The radius of influence of the well
iiiiiiiiiiiiiiiiiii) The time it would take for the drawdown to reach 1 m at a distance of 200 m
from the well
Solution: a) Drawdown at 100 m: Using the Theis equation: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢), where 𝑢=𝑟2𝑆
4𝑇𝑡
𝑢 = (100 m)2×0.0001
4×200 m²/day ×5 days =0.0025
𝑊(𝑢)≈5.84 (from tables or software)
𝑠 = 1000 m³/day
4𝜋×200 m²/day ×5.84
=1.16 m
b) Radius of influence: Using the empirical formula 𝑅=1.5√𝑇𝑡
𝑆:
𝑅 =1.5√200 m²/day ×5 days
0.0001
=1500 m
c) Time for 1 m drawdown at 200 m: Rearranging the Theis equation:
1 m =1000 m³/day
4𝜋×200 m²/day 𝑊(𝑢)
𝑊(𝑢)=5.03
𝑢 ≈0.0049 (from tables or software)
𝑡 = (200 m)2×0.0001
4×200 m²/day ×0.0049
=10.2 days
1.192 PROBLEM 3
A horizontal confined aquifer is bounded by two constant-head boundaries 1000 m apart. The
hydraulic conductivity is 5 m/day, and the aquifer thickness is 20 m. If the hydraulic heads at
the boundaries are 50 m and 40 m, calculate:
jjjjjjjjjjjjjjjjjjj) The flow rate per unit width of the aquifer
kkkkkkkkkkkkkkkkkkk) The hydraulic head at a point 300 m from the higher-head
boundary
lllllllllllllllllll) The volume of water flowing through the entire aquifer in one year if its width is
5 km
Solution: a) Flow rate per unit width: Using Darcy’s Law: 𝑞=−𝐾𝑑ℎ
𝑑𝑥𝑏
𝑞 =−5 m/day ×40 m −50 m
1000 m ×20 m
=1 m²/day
b) Hydraulic head at 300 m: The head distribution is linear, so:
ℎ =50 m −300 m
1000 m ×(50 m −40 m)
=47 m
c) Annual flow volume: 𝑉 =𝑞×width ×time
=1 m²/day ×5000 m ×365 days
=1,825,000 m³/year
1.193 PROBLEM 4
An unconfined coastal aquifer extends 500 m inland. The water table at the inland boundary is
10 m above sea level. The aquifer base is horizontal and impermeable, located 5 m below sea
level. If the hydraulic conductivity is 15 m/day, calculate:
mmmmmmmmmmmmmmmmmmm) The discharge per unit width of the coastline
nnnnnnnnnnnnnnnnnnn) The position of the water table 200 m from the coast
ooooooooooooooooooo) The location where saltwater intrusion would begin, assuming the
Ghyben-Herzberg relation
Solution: a) Discharge per unit width: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2)
𝑞 = 15 m/day
2×500 m (102−52)
=0.675 m²/day
b) Water table position at 200 m: Using the parabolic equation: ℎ2=ℎ2
2+𝑥
𝐿(ℎ1
2−ℎ2
2)
ℎ2=52+200
500(102−52)
ℎ =8.06 m above sea level
c) Saltwater intrusion location: Using the Ghyben-Herzberg relation: 𝑧=40ℎ, where ℎ is
freshwater head above sea level At the toe of the interface, 𝑧=5 m (aquifer depth), so ℎ=
0.125 m Using the parabolic equation again:
0.1252=52+𝑥
500(102−52)
𝑥 =499.7 m from the inland boundary
The intrusion would begin 0.3 m from the coastline.
1.194 PROBLEM 5
A pumping well fully penetrates a confined aquifer. The aquifer has a transmissivity of 250
m²/day and a storage coefficient of 0.0005. After pumping at a constant rate of 1500 m³/day
for 2 days, the drawdown in an observation well 100 m away is 2.5 m. Determine:
ppppppppppppppppppp) The aquifer’s hydraulic conductivity if its thickness is 25 m
qqqqqqqqqqqqqqqqqqq) The expected drawdown at the same observation well after 10
days of pumping
rrrrrrrrrrrrrrrrrrr) The pumping rate required to maintain a steady-state drawdown of 5 m
at the observation well
Solution: a) Hydraulic conductivity: 𝐾=𝑇
𝑏=250 m²/day
25 m =10 m/day
b) Drawdown after 10 days: Using the Cooper-Jacob approximation: 𝑠=2.3𝑄
4𝜋𝑇log2.25𝑇𝑡
𝑟2𝑆
𝑠 =2.3×1500 m³/day
4𝜋×250 m²/day log2.25×250 m²/day ×10 days
(100 m)2×0.0005
=3.2 m
c) Pumping rate for 5 m steady-state drawdown: Using the Thiem equation: 𝑄= 2𝜋𝑇𝑠
ln(𝑅/𝑟), where
R is the radius of influence (assume 1000 m)
𝑄 =2𝜋×250 m²/day ×5 m
ln(1000 m/100 m)
=2726 m³/day
1.195 PROBLEM 6
An artesian aquifer is overlain by a leaky confining layer. The aquifer has a transmissivity of
1000 m²/day and a storage coefficient of 0.0002. The confining layer has a vertical hydraulic
conductivity of 0.001 m/day and a thickness of 10 m. A well is pumping at 2000 m³/day.
Calculate:
sssssssssssssssssss) The drawdown at a distance of 200 m after 5 days of pumping
ttttttttttttttttttt) The steady-state drawdown at the same location
uuuuuuuuuuuuuuuuuuu) The percentage of pumped water derived from leakage at steady-
state
Solution: a) Drawdown after 5 days: Using the Hantush-Jacob solution: 𝑠= 𝑄
4𝜋𝑇𝑊(𝑢,𝑟/𝐵)
where 𝐵=√𝑇𝑏′
𝐾′ =√1000 m²/day×10 m
0.001 m/day =1000 m and 𝑢=𝑟2𝑆
4𝑇𝑡=(200 m)2×0.0002
4×1000 m²/day×5 days =0.0002
Using tables or software to find 𝑊(𝑢,𝑟/𝐵)≈4.03
𝑠= 2000 m³/day
4𝜋×1000 m²/day ×4.03=0.64 m
b) Steady-state drawdown: Using the modified Thiem equation: 𝑠= 𝑄
2𝜋𝑇𝐾0(𝑟
𝐵) where 𝐾0 is the
modified Bessel function of the second kind, order zero
𝐾0(200 m
1000 m)≈1.75 (from tables or software)
𝑠= 2000 m³/day
2𝜋×1000 m²/day ×1.75=0.56 m
c) Percentage of water from leakage: At steady-state, all pumped water comes from leakage.
The percentage is 100
1.196 PROBLEM 7
A semi-confined aquifer system consists of an upper unconfined aquifer and a lower confined
aquifer separated by an aquitard. The upper aquifer has a hydraulic conductivity of 5 m/day
and a specific yield of 0.2. The lower aquifer has a transmissivity of 200 m²/day and a storage
coefficient of 0.0001. The aquitard has a vertical hydraulic conductivity of 0.0005 m/day and a
thickness of 5 m. A well fully penetrating the lower aquifer is pumped at 1000 m³/day.
Calculate:
Solution: a) Current freshwater discharge: Using the Dupuit formula: 𝑞= 𝐾
2𝐿(ℎ1
2−ℎ2
2) where
ℎ1=2 m and ℎ2=0 m (sea level)
𝑞= 20 m/day
2×2000 m (22−02)=0.02 m2/day
b) Injection rate for groundwater divide: To create a divide at the coast, we need to balance
the flow from both directions. The injection point will be a new water table high point.
Let ℎ be the water table elevation at the injection point. We can set up two equations:
From injection to coast: 𝑞1=20 m/day
2×500 m (ℎ2−02) From injection to inland: 𝑞2=20 m/day
2×1500 m (ℎ2−22)
For a divide, 𝑞1=𝑞2, so:
ℎ2
500=ℎ2−4
1500
Solving this: ℎ=2.31 m
The injection rate will be the sum of 𝑞1 and 𝑞2:
𝑞inject =20
2×500(2.312)+20
2×1500(2.312−22)=0.107 m2/day
c) New water table elevation at inland boundary: We can use the parabolic equation from the
injection point to the inland boundary:
ℎnew
2=2.312−1500
1500(2.312−ℎnew
2)
Solving this: ℎnew =2.16 m
The new water table elevation at the inland boundary will be 2.16 m above sea level.
