Dupuit's Formula

Hydrology and Water Resources · FE Reference Handbook section

Hydrology and Water Resources

10 formulas

10 exam-style examples

~60 min

All Hydrology and Water Resources lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Dupuit's Formula within Hydrology and Water Resources. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what dupuit's formula describes physically and when it applies.
State every one of the 10 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: acre-in/hr ≈ cfs makes the rational formula work in US units.

Lecture

Why this section exists. Dupuit's Formula is the part of Hydrology and Water Resources that lets you connect a watershed, aquifer or detention facility to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a rainfall-runoff or well-drawdown calculation with one lookup. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. acre-in/hr ≈ cfs makes the rational formula work in US units. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Row of centrifugal pumps and valved steel piping inside a water pumping station. — Photo 1. Where this shows up in practice: dupuit's formula.
Capstone Studio instructional photograph

Hydrology and Water Resources — Dupuit's Formula: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a watershed, aquifer or detention facility. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 10 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

Q	Quantity produced by "Q = flowrate of water drawn from well (cfs)" — read its definition and unit from the handbook line directly above the equation.
K	Quantity produced by "K = coefficient of permeability of soil; hydraulic conductivity (ft/sec)" — read its definition and unit from the handbook line directly above the equation.
h1	Quantity produced by "h1 = height of water surface above bottom of aquifer at perimeter of well (ft)" — read its definition and unit from the handbook line directly above the equation.
h2	Quantity produced by "h2 = height of water surface above bottom of aquifer at distance r2 from well centerline (ft)" — read its definition and unit from the handbook line directly above the equation.
r1	Quantity produced by "r1 = radius to water surface at perimeter of well, i.e., radius of well (ft)" — read its definition and unit from the handbook line directly above the equation.
r2	Quantity produced by "r2 = radius to water surface whose height is h2 above bottom of aquifer (ft)" — read its definition and unit from the handbook line directly above the equation.
ln	Quantity produced by "ln = natural logarithm" — read its definition and unit from the handbook line directly above the equation.
Q/Dw	Quantity produced by "Q/Dw = specific capacity" — read its definition and unit from the handbook line directly above the equation.
Dw	Quantity produced by "Dw = well drawdown (ft)" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

rK `h 22 - h12 j
ln d r2 n
where

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Confined-aquifer drawdown between two observation wells — Dupuit's Formula

A well pumps 262 gpm from a confined aquifer with K = 125.0 ft/day and thickness 34 ft. Two observation wells sit 50 ft and 983 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 262 gpm
K = 125.0 ft/day
b = 34 ft
r₁ = 50 ft
r₂ = 983 ft

Find

Head difference h₂ − h₁

Start with the thinking

Convert gpm to ft³/day (1 gpm = 192.5 ft³/day) so units match K.
Confined (Thiem) flow is logarithmic in radius, so distant wells see little extra drawdown.

Figure for Confined-aquifer drawdown between two observation wells — Dupuit's Formula

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 125.0 \times 34 = 4, 250 f t^{2} / d a y$
Discharge
$Q = 262 g p m \times 192.5 = 50, 435 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (983/50) = 2.979$
Substituting — Δh = 50,435 × 2.979 / (2π × 4,250)
Evaluate — Δh = 5.63 ft

Answer: h₂ − h₁ ≈ 5.63 ft

Why the other options are there

37.13 ft (ratio used instead of its log)
35.35 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Dupuit's Formula

Example 2

Confined-aquifer drawdown between two observation wells — Dupuit's Formula (2)

A well pumps 369 gpm from a confined aquifer with K = 110.0 ft/day and thickness 65 ft. Two observation wells sit 50 ft and 1254 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 369 gpm
K = 110.0 ft/day
b = 65 ft
r₁ = 50 ft
r₂ = 1254 ft

Find

Head difference h₂ − h₁

Start with the thinking

Convert gpm to ft³/day (1 gpm = 192.5 ft³/day) so units match K.
Confined (Thiem) flow is logarithmic in radius, so distant wells see little extra drawdown.

Figure for Confined-aquifer drawdown between two observation wells — Dupuit's Formula (2)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 110.0 \times 65 = 7, 150 f t^{2} / d a y$
Discharge
$Q = 369 g p m \times 192.5 = 71, 033 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (1254/50) = 3.222$
Substituting — Δh = 71,033 × 3.222 / (2π × 7,150)
Evaluate — Δh = 5.09 ft

Answer: h₂ − h₁ ≈ 5.09 ft

Why the other options are there

39.66 ft (ratio used instead of its log)
32.01 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Dupuit's Formula

Example 3

Confined-aquifer drawdown between two observation wells — Dupuit's Formula (3)

A well pumps 794 gpm from a confined aquifer with K = 60 ft/day and thickness 25 ft. Two observation wells sit 50 ft and 949 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 794 gpm
K = 60 ft/day
b = 25 ft
r₁ = 50 ft
r₂ = 949 ft

Find

Head difference h₂ − h₁

Start with the thinking

Convert gpm to ft³/day (1 gpm = 192.5 ft³/day) so units match K.
Confined (Thiem) flow is logarithmic in radius, so distant wells see little extra drawdown.

Figure for Confined-aquifer drawdown between two observation wells — Dupuit's Formula (3)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 60 \times 25 = 1, 500 f t^{2} / d a y$
Discharge
$Q = 794 g p m \times 192.5 = 152, 845 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (949/50) = 2.943$
Substituting — Δh = 152,845 × 2.943 / (2π × 1,500)
Evaluate — Δh = 47.73 ft

Answer: h₂ − h₁ ≈ 47.73 ft

Why the other options are there

307.8 ft (ratio used instead of its log)
299.9 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Dupuit's Formula

Example 4

Confined-aquifer drawdown between two observation wells — Dupuit's Formula (4)

A well pumps 163 gpm from a confined aquifer with K = 20 ft/day and thickness 44 ft. Two observation wells sit 50 ft and 623 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 163 gpm
K = 20 ft/day
b = 44 ft
r₁ = 50 ft
r₂ = 623 ft

Find

Head difference h₂ − h₁

Start with the thinking

Convert gpm to ft³/day (1 gpm = 192.5 ft³/day) so units match K.
Confined (Thiem) flow is logarithmic in radius, so distant wells see little extra drawdown.

Figure for Confined-aquifer drawdown between two observation wells — Dupuit's Formula (4)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 20 \times 44 = 880.0 f t^{2} / d a y$
Discharge
$Q = 163 g p m \times 192.5 = 31, 378 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (623/50) = 2.523$
Substituting — Δh = 31,378 × 2.523 / (2π × 880.0)
Evaluate — Δh = 14.31 ft

Answer: h₂ − h₁ ≈ 14.31 ft

Why the other options are there

70.71 ft (ratio used instead of its log)
89.94 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Dupuit's Formula

Example 5

Confined-aquifer drawdown between two observation wells — Dupuit's Formula (5)

A well pumps 846 gpm from a confined aquifer with K = 135.0 ft/day and thickness 45 ft. Two observation wells sit 50 ft and 1395 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 846 gpm
K = 135.0 ft/day
b = 45 ft
r₁ = 50 ft
r₂ = 1395 ft

Find

Head difference h₂ − h₁

Start with the thinking

Convert gpm to ft³/day (1 gpm = 192.5 ft³/day) so units match K.
Confined (Thiem) flow is logarithmic in radius, so distant wells see little extra drawdown.

Figure for Confined-aquifer drawdown between two observation wells — Dupuit's Formula (5)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 135.0 \times 45 = 6, 075 f t^{2} / d a y$
Discharge
$Q = 846 g p m \times 192.5 = 162, 855 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (1395/50) = 3.329$
Substituting — Δh = 162,855 × 3.329 / (2π × 6,075)
Evaluate — Δh = 14.20 ft

Answer: h₂ − h₁ ≈ 14.20 ft

Why the other options are there

119.0 ft (ratio used instead of its log)
89.23 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Dupuit's Formula

Example 6

Confined-aquifer drawdown between two observation wells — Dupuit's Formula (6)

A well pumps 430 gpm from a confined aquifer with K = 170.0 ft/day and thickness 33 ft. Two observation wells sit 50 ft and 454 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 430 gpm
K = 170.0 ft/day
b = 33 ft
r₁ = 50 ft
r₂ = 454 ft

Find

Head difference h₂ − h₁

Start with the thinking

Convert gpm to ft³/day (1 gpm = 192.5 ft³/day) so units match K.
Confined (Thiem) flow is logarithmic in radius, so distant wells see little extra drawdown.

Figure for Confined-aquifer drawdown between two observation wells — Dupuit's Formula (6)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 170.0 \times 33 = 5, 610 f t^{2} / d a y$
Discharge
$Q = 430 g p m \times 192.5 = 82, 775 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (454/50) = 2.206$
Substituting — Δh = 82,775 × 2.206 / (2π × 5,610)
Evaluate — Δh = 5.18 ft

Answer: h₂ − h₁ ≈ 5.18 ft

Why the other options are there

21.32 ft (ratio used instead of its log)
32.55 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Dupuit's Formula

Example 7

Confined-aquifer drawdown between two observation wells — Dupuit's Formula (7)

A well pumps 219 gpm from a confined aquifer with K = 130.0 ft/day and thickness 76 ft. Two observation wells sit 50 ft and 972 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 219 gpm
K = 130.0 ft/day
b = 76 ft
r₁ = 50 ft
r₂ = 972 ft

Find

Head difference h₂ − h₁

Start with the thinking

Convert gpm to ft³/day (1 gpm = 192.5 ft³/day) so units match K.
Confined (Thiem) flow is logarithmic in radius, so distant wells see little extra drawdown.

Figure for Confined-aquifer drawdown between two observation wells — Dupuit's Formula (7)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 130.0 \times 76 = 9, 880 f t^{2} / d a y$
Discharge
$Q = 219 g p m \times 192.5 = 42, 158 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (972/50) = 2.967$
Substituting — Δh = 42,158 × 2.967 / (2π × 9,880)
Evaluate — Δh = 2.02 ft

Answer: h₂ − h₁ ≈ 2.02 ft

Why the other options are there

13.20 ft (ratio used instead of its log)
12.66 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Dupuit's Formula

Example 8

Confined-aquifer drawdown between two observation wells — Dupuit's Formula (8)

A well pumps 257 gpm from a confined aquifer with K = 85 ft/day and thickness 67 ft. Two observation wells sit 50 ft and 1050 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 257 gpm
K = 85 ft/day
b = 67 ft
r₁ = 50 ft
r₂ = 1050 ft

Find

Head difference h₂ − h₁

Start with the thinking

Convert gpm to ft³/day (1 gpm = 192.5 ft³/day) so units match K.
Confined (Thiem) flow is logarithmic in radius, so distant wells see little extra drawdown.

Figure for Confined-aquifer drawdown between two observation wells — Dupuit's Formula (8)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 85 \times 67 = 5, 695 f t^{2} / d a y$
Discharge
$Q = 257 g p m \times 192.5 = 49, 473 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (1050/50) = 3.045$
Substituting — Δh = 49,473 × 3.045 / (2π × 5,695)
Evaluate — Δh = 4.21 ft

Answer: h₂ − h₁ ≈ 4.21 ft

Why the other options are there

29.03 ft (ratio used instead of its log)
26.45 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Dupuit's Formula

Example 9

Confined-aquifer drawdown between two observation wells — Dupuit's Formula (9)

A well pumps 356 gpm from a confined aquifer with K = 70 ft/day and thickness 83 ft. Two observation wells sit 50 ft and 1217 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 356 gpm
K = 70 ft/day
b = 83 ft
r₁ = 50 ft
r₂ = 1217 ft

Find

Head difference h₂ − h₁

Start with the thinking

Convert gpm to ft³/day (1 gpm = 192.5 ft³/day) so units match K.
Confined (Thiem) flow is logarithmic in radius, so distant wells see little extra drawdown.

Figure for Confined-aquifer drawdown between two observation wells — Dupuit's Formula (9)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 70 \times 83 = 5, 810 f t^{2} / d a y$
Discharge
$Q = 356 g p m \times 192.5 = 68, 530 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (1217/50) = 3.192$
Substituting — Δh = 68,530 × 3.192 / (2π × 5,810)
Evaluate — Δh = 5.99 ft

Answer: h₂ − h₁ ≈ 5.99 ft

Why the other options are there

45.69 ft (ratio used instead of its log)
37.65 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Dupuit's Formula

Example 10

Confined-aquifer drawdown between two observation wells — Dupuit's Formula (10)

A well pumps 539 gpm from a confined aquifer with K = 145.0 ft/day and thickness 65 ft. Two observation wells sit 50 ft and 792 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 539 gpm
K = 145.0 ft/day
b = 65 ft
r₁ = 50 ft
r₂ = 792 ft

Find

Head difference h₂ − h₁

Start with the thinking

Convert gpm to ft³/day (1 gpm = 192.5 ft³/day) so units match K.
Confined (Thiem) flow is logarithmic in radius, so distant wells see little extra drawdown.

Figure for Confined-aquifer drawdown between two observation wells — Dupuit's Formula (10)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 145.0 \times 65 = 9, 425 f t^{2} / d a y$
Discharge
$Q = 539 g p m \times 192.5 = 103, 758 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (792/50) = 2.763$
Substituting — Δh = 103,758 × 2.763 / (2π × 9,425)
Evaluate — Δh = 4.84 ft

Answer: h₂ − h₁ ≈ 4.84 ft

Why the other options are there

27.75 ft (ratio used instead of its log)
30.41 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Dupuit's Formula

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a watershed, aquifer or detention facility, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Dupuit's Formula contains 10 relations; you must be able to find this page in under 15 seconds.
Exam style: a rainfall-runoff or well-drawdown calculation with one lookup.
Unit rule: acre-in/hr ≈ cfs makes the rational formula work in US units.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

acre-in/hr ≈ cfs makes the rational formula work in US units
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Dupuit's Formula

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

Q=Needs technical review

Q=flowrateofwaterdrawnfromwell(cfs)

K=coefficientofpermeabilityofsoil;hydraulicconductivity(ft/sec)

h1​=heightofwatersurfaceabovebottomofaquiferatperimeterofwell(ft)

h2​=heightofwatersurfaceabovebottomofaquiferatdistancer2​fromwellcenterline(ft)

r1​=radiustowatersurfaceatperimeterofwell,i.e.,radiusofwell(ft)

r2​=radiustowatersurfacewhoseheightish2​abovebottomofaquifer(ft)

ln=naturallogarithm

Q/Dw=specificcapacity

Dw=welldrawdown(ft)

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section