SME
Hydrology · Problem type
A well’s yield from two observation wells
Use this for “A well pumps a confined aquifer 15 m thick with K = 20 m/day. Observation wells 10 m and 100 m away read heads of 40 m and 41.2 m. Find the pumping rate.”
Diagram
Q
=
2π
Kb(h₂ − h₁)
ln(r₂ ÷ r₁)
Q
=
2π
×
20
×
15
×
1.2
ln(10)
Q = 982.4 m³/day
The cone passes both observed heads (r₁ = 10 m); heads drawn on a stretched scale, solid where Thiem's equation is measured.
This is an example. Type your own values to start a new problem.
Q (m³/day)
=
2π
×
K (m/day)
×
b (m)
×
(
h₂ (m)
−
h₁ (m)
)
÷
ln(
r₂ (m)
÷
r₁ (m)
)
Formulas
Q
=
2π
×
K
×
b
×
h₂
−
h₁
ln(
r₂
÷
r₁
)
982.351
=
2π
×
20
×
15
×
41.2 − 40
ln(100 ÷ 10)
Assumptions
•
A confined aquifer pumped steadily for a long time, so the cone of depression no longer grows.
•
The aquifer is even in thickness and conductivity, and the well goes through all of it.
•
Heads rise toward the far well: h₂ at r₂ is above h₁ at r₁.
Step-by-step
WE KNOW
K = 20 m/day, b = 15 m, h₁ = 40 m, h₂ = 41.2 m, r₁ = 10 m, r₂ = 100 m
FIND
pumping rate (Q)
Step 1 · Find pumping rate (Q)
Use Q = 2π × K × b × (h₂ − h₁) ÷ ln(r₂ ÷ r₁)
Thiem: 2πKb times the head difference, divided by ln of the distance ratio.
Q
=
2π
×
20
×
15
×
41.2 − 40
ln(100 ÷ 10)
100
10
=
10
ln(10) = 2.303
41.2 − 40 = 1.2
Q
=
982.351
m³/day
Check
Put the numbers back in. Both sides should match.
982.351
=
2π
×
20
×
15
×
41.2 − 40
ln(100 ÷ 10)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Groundwater flow (Darcy's law)
Main lesson
Hydraulic head from elevation and pressure
Use this for “A piezometer’s screen is 120 m above sea level, and the water pressure there is 98.1 kPa. Find the pressure head and the total head.”
Aquifer or aquitard
Use this for “A well is drilled into fractured basalt, then into a clay layer. Which is the aquifer and which the aquitard?”