Learn: Steady radial flow — unconfined aquifer
The unconfined Thiem-type well relation connects steady radial flow with heads measured at two distances from a well. The radius order and head difference determine the sign of the result.
What the formula is saying
Under the Dupuit approximation, thickness varies with saturated head; integration gives πk(h₁² − h₂²)/ln(r₂/r₁). Here r₁ is the inner radius. The stored formula defines positive Q as outward radial flow; pumping toward the well normally gives h₁ < h₂ and a negative Q.
Read the symbols in plain language
- k
- Hydraulic conductivity
The soil/aquifer permeability parameter relating Darcy discharge velocity to hydraulic gradient under the stated flow conditions.
m/sUse m/s as the base unit shown here. k is hydraulic conductivity in m/s; radii and saturated thicknesses h₁, h₂ above the horizontal impermeable base use m. ln is the natural logarithm of the dimensionless radius ratio. Q is m³/s.
- h₁
- Saturated thickness/head 1
Saturated thickness measured above the common horizontal impermeable base, not an arbitrary survey elevation.
mMetres measure length; 1 m = 1000 mm.
- h₂
- Saturated thickness/head 2
Saturated thickness measured above the common horizontal impermeable base, not an arbitrary survey elevation.
mMetres measure length; 1 m = 1000 mm.
- r₁
- Radius 1
Radial distance from the same well centre to the observation location; r2 must be greater than r1.
mMetres measure length; 1 m = 1000 mm.
- r₂
- Radius 2
Radial distance from the same well centre to the observation location; r2 must be greater than r1.
mMetres measure length; 1 m = 1000 mm.
- Q
- Result to find
Steady radial flow — unconfined aquifer. Combine conductivity, aquifer geometry and driving-head difference without hiding the flow sign.
m³/s
Sort out the units first
k is hydraulic conductivity in m/s; radii and saturated thicknesses h₁, h₂ above the horizontal impermeable base use m. ln is the natural logarithm of the dimensionless radius ratio. Q is m³/s.
Assumptions before calculating
Assume steady radial flow in a homogeneous isotropic aquifer around a fully penetrating well, with no recharge between the two observation radii and negligible well losses. Require 0 < r₁ < r₂ and compatible head observations.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find Q and explain the result in the stated output unit.
- k · Hydraulic conductivity
- 0.0001 m/s
- h₁ · Saturated thickness/head 1
- 20 m
- h₂ · Saturated thickness/head 2
- 18 m
- r₁ · Radius 1
- 1 m
- r₂ · Radius 2
- 50 m
Take the natural logarithm of the radius ratio
The outer radius divided by the inner radius is greater than one, giving a positive logarithm.
ln((50) ÷ (1)) ≈ 3.912023005Apply the stated outward-positive convention
Keep inner minus outer head in the stored formula; the sign reveals the modeled radial-flow direction.
(20)^2-(18)^2 = 76 m²Calculate the signed radial discharge
Combine conductivity, aquifer geometry and driving-head difference without hiding the flow sign.
π × (0.0001) × (76) ÷ (3.912023005) ≈ 0.00610326272 m³/s
Signed radial discharge, positive outward from the well and negative inward toward it.
Does this worked answer make sense?
Equal heads give zero radial flow. With r₂ > r₁, the logarithm is positive, so Q must follow the sign of h₁² − h₂². For a pumping-rate magnitude, report −Q only when the calculated Q is negative and explain the changed convention.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- k · Hydraulic conductivity
- 0.00015 m/s
- h₁ · Saturated thickness/head 1
- 8 m
- h₂ · Saturated thickness/head 2
- 10 m
- r₁ · Radius 1
- 2 m
- r₂ · Radius 2
- 40 m
Take the natural logarithm of the radius ratio
The outer radius divided by the inner radius is greater than one, giving a positive logarithm.
ln((40) ÷ (2)) ≈ 2.995732274Apply the stated outward-positive convention
Keep inner minus outer head in the stored formula; the sign reveals the modeled radial-flow direction.
(8)^2-(10)^2 = -36 m²Calculate the signed radial discharge
Combine conductivity, aquifer geometry and driving-head difference without hiding the flow sign.
π × (0.00015) × (-36) ÷ (2.995732274) ≈ -0.005662922711 m³/s
Signed radial discharge, positive outward from the well and negative inward toward it.
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- k · Hydraulic conductivity
- 0.00008 m/s
- h₁ · Saturated thickness/head 1
- 12 m
- h₂ · Saturated thickness/head 2
- 15 m
- r₁ · Radius 1
- 1.5 m
- r₂ · Radius 2
- 60 m
Find: Learn: Steady radial flow — unconfined aquifer
A hint, not the answer
Under the Dupuit approximation, thickness varies with saturated head; integration gives πk(h₁² − h₂²)/ln(r₂/r₁). Here r₁ is the inner radius. The stored formula defines positive Q as outward radial flow; pumping toward the well normally gives h₁ < h₂ and a negative Q.
k is hydraulic conductivity in m/s; radii and saturated thicknesses h₁, h₂ above the horizontal impermeable base use m. ln is the natural logarithm of the dimensionless radius ratio. Q is m³/s.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Take the natural logarithm of the radius ratio
The outer radius divided by the inner radius is greater than one, giving a positive logarithm.
ln((60) ÷ (1.5)) ≈ 3.688879454Apply the stated outward-positive convention
Keep inner minus outer head in the stored formula; the sign reveals the modeled radial-flow direction.
(12)^2-(15)^2 = -81 m²Calculate the signed radial discharge
Combine conductivity, aquifer geometry and driving-head difference without hiding the flow sign.
π × (0.00008) × (-81) ÷ (3.688879454) ≈ -0.005518619041 m³/s
Signed radial discharge, positive outward from the well and negative inward toward it.
Avoid the common trap
Do not silently reverse the head difference just to force a positive answer. Square each saturated thickness before subtracting; (h₁ − h₂)² is different. Do not use log10 instead of ln.
When this method applies — and when it does not
Assume a horizontal base and predominantly horizontal flow; h₁ and h₂ must be positive saturated thicknesses, not arbitrary elevation heads. Vertical flow near the well may violate the approximation. This is not a transient pumping-test solution and does not handle boundaries, partial penetration, anisotropy or changing storage automatically.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Steady radial flow — unconfined aquifer. Steady radial well-flow / Thiem relationships. Match confined versus unconfined assumptions, radii, head datum and the lesson’s outward-positive sign convention.
- U.S. Geological Survey — Shortcuts and Special Problems in Aquifer Tests (1964, Water-Supply Paper 1545-C)
- Dawei Han, University of Bristol — Concise Hydraulics (2008, Ventus Publishing)
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.