Unconfined aquifer

Hydrology and Water Resources · FE Reference Handbook section

Hydrology and Water Resources

0 formulas

10 exam-style examples

~45 min

All Hydrology and Water Resources lectures

Learning objectives

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

This chapter section covers Unconfined aquifer 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 unconfined aquifer describes physically and when it applies.
State every one of the 0 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. Unconfined aquifer 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: unconfined aquifer.
Capstone Studio instructional photograph

Hydrology and Water Resources — Unconfined aquifer: 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 0 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.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

ORIGINAL Q r2
GROUNDWATER
LEVEL
PERMEABLE SOIL
BOTTOM OF AQUIFER
IMPERMEABLE

Core formulas for this FE topic

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

This section is conceptual; there are no equations to memorise.

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 — Unconfined aquifer

A well pumps 189 gpm from a confined aquifer with K = 110.0 ft/day and thickness 26 ft. Two observation wells sit 50 ft and 1459 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 189 gpm
K = 110.0 ft/day
b = 26 ft
r₁ = 50 ft
r₂ = 1459 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 — Unconfined aquifer

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 110.0 \times 26 = 2, 860 f t^{2} / d a y$
Discharge
$Q = 189 g p m \times 192.5 = 36, 383 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (1459/50) = 3.373$
Substituting — Δh = 36,383 × 3.373 / (2π × 2,860)
Evaluate — Δh = 6.83 ft

Answer: h₂ − h₁ ≈ 6.83 ft

Why the other options are there

59.08 ft (ratio used instead of its log)
42.91 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Unconfined aquifer

Example 2

Darcy's law seepage flow — Unconfined aquifer

An aquifer has K = 3.30e-3 ft/s under a gradient of 0.045 across a 1691 ft² cross section. Find the seepage discharge.

Given

K = 3.30e-3 ft/s
i = 0.045
A = 1691 ft²

Find

Seepage Q

Start with the thinking

Darcy velocity is superficial — it is not the pore velocity.
Gradient is dimensionless head loss per length.

Figure for Darcy's law seepage flow — Unconfined aquifer

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 3.30 e - 3 (0.045) (1691)$
Evaluate
$Q = 2.511 e - 1 f t^{3} / s$
Convert
$Q = 112.7 g p m$
Darcy velocity
$v = K i = 1.485 e - 4 f t / s$

Answer: Q ≈ 2.51e-1 cfs (112.7 gpm)

Why the other options are there

5.58e+0 cfs (gradient omitted)
7.33e-2 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Unconfined aquifer

Example 3

Confined-aquifer drawdown between two observation wells — Unconfined aquifer (2)

A well pumps 526 gpm from a confined aquifer with K = 175.0 ft/day and thickness 35 ft. Two observation wells sit 50 ft and 1467 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 526 gpm
K = 175.0 ft/day
b = 35 ft
r₁ = 50 ft
r₂ = 1467 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 — Unconfined aquifer (2)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 175.0 \times 35 = 6, 125 f t^{2} / d a y$
Discharge
$Q = 526 g p m \times 192.5 = 101, 255 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (1467/50) = 3.379$
Substituting — Δh = 101,255 × 3.379 / (2π × 6,125)
Evaluate — Δh = 8.89 ft

Answer: h₂ − h₁ ≈ 8.89 ft

Why the other options are there

77.20 ft (ratio used instead of its log)
55.86 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Unconfined aquifer

Example 4

Darcy's law seepage flow — Unconfined aquifer (2)

An aquifer has K = 4.30e-3 ft/s under a gradient of 0.030 across a 2223 ft² cross section. Find the seepage discharge.

Given

K = 4.30e-3 ft/s
i = 0.030
A = 2223 ft²

Find

Seepage Q

Start with the thinking

Darcy velocity is superficial — it is not the pore velocity.
Gradient is dimensionless head loss per length.

Figure for Darcy's law seepage flow — Unconfined aquifer (2)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 4.30 e - 3 (0.030) (2223)$
Evaluate
$Q = 2.868 e - 1 f t^{3} / s$
Convert
$Q = 128.7 g p m$
Darcy velocity
$v = K i = 1.290 e - 4 f t / s$

Answer: Q ≈ 2.87e-1 cfs (128.7 gpm)

Why the other options are there

9.56e+0 cfs (gradient omitted)
1.43e-1 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Unconfined aquifer

Example 5

Confined-aquifer drawdown between two observation wells — Unconfined aquifer (3)

A well pumps 766 gpm from a confined aquifer with K = 155.0 ft/day and thickness 57 ft. Two observation wells sit 50 ft and 441 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 766 gpm
K = 155.0 ft/day
b = 57 ft
r₁ = 50 ft
r₂ = 441 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 — Unconfined aquifer (3)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 155.0 \times 57 = 8, 835 f t^{2} / d a y$
Discharge
$Q = 766 g p m \times 192.5 = 147, 455 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (441/50) = 2.177$
Substituting — Δh = 147,455 × 2.177 / (2π × 8,835)
Evaluate — Δh = 5.78 ft

Answer: h₂ − h₁ ≈ 5.78 ft

Why the other options are there

23.43 ft (ratio used instead of its log)
36.33 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Unconfined aquifer

Example 6

Darcy's law seepage flow — Unconfined aquifer (3)

An aquifer has K = 7.00e-4 ft/s under a gradient of 0.045 across a 611 ft² cross section. Find the seepage discharge.

Given

K = 7.00e-4 ft/s
i = 0.045
A = 611 ft²

Find

Seepage Q

Start with the thinking

Darcy velocity is superficial — it is not the pore velocity.
Gradient is dimensionless head loss per length.

Figure for Darcy's law seepage flow — Unconfined aquifer (3)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 7.00 e - 4 (0.045) (611)$
Evaluate
$Q = 1.925 e - 2 f t^{3} / s$
Convert
$Q = 8.6378 g p m$
Darcy velocity
$v = K i = 3.150 e - 5 f t / s$

Answer: Q ≈ 1.92e-2 cfs (8.64 gpm)

Why the other options are there

4.28e-1 cfs (gradient omitted)
1.56e-2 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Unconfined aquifer

Example 7

Confined-aquifer drawdown between two observation wells — Unconfined aquifer (4)

A well pumps 797 gpm from a confined aquifer with K = 60 ft/day and thickness 84 ft. Two observation wells sit 50 ft and 980 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 797 gpm
K = 60 ft/day
b = 84 ft
r₁ = 50 ft
r₂ = 980 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 — Unconfined aquifer (4)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 60 \times 84 = 5, 040 f t^{2} / d a y$
Discharge
$Q = 797 g p m \times 192.5 = 153, 423 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (980/50) = 2.976$
Substituting — Δh = 153,423 × 2.976 / (2π × 5,040)
Evaluate — Δh = 14.42 ft

Answer: h₂ − h₁ ≈ 14.42 ft

Why the other options are there

94.96 ft (ratio used instead of its log)
90.58 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Unconfined aquifer

Example 8

Darcy's law seepage flow — Unconfined aquifer (4)

An aquifer has K = 3.00e-3 ft/s under a gradient of 0.040 across a 2083 ft² cross section. Find the seepage discharge.

Given

K = 3.00e-3 ft/s
i = 0.040
A = 2083 ft²

Find

Seepage Q

Start with the thinking

Darcy velocity is superficial — it is not the pore velocity.
Gradient is dimensionless head loss per length.

Figure for Darcy's law seepage flow — Unconfined aquifer (4)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 3.00 e - 3 (0.040) (2083)$
Evaluate
$Q = 2.500 e - 1 f t^{3} / s$
Convert
$Q = 112.2 g p m$
Darcy velocity
$v = K i = 1.200 e - 4 f t / s$

Answer: Q ≈ 2.50e-1 cfs (112.2 gpm)

Why the other options are there

6.25e+0 cfs (gradient omitted)
7.50e-2 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Unconfined aquifer

Example 9

Confined-aquifer drawdown between two observation wells — Unconfined aquifer (5)

A well pumps 299 gpm from a confined aquifer with K = 175.0 ft/day and thickness 67 ft. Two observation wells sit 50 ft and 1013 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 299 gpm
K = 175.0 ft/day
b = 67 ft
r₁ = 50 ft
r₂ = 1013 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 — Unconfined aquifer (5)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 175.0 \times 67 = 11, 725 f t^{2} / d a y$
Discharge
$Q = 299 g p m \times 192.5 = 57, 558 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (1013/50) = 3.009$
Substituting — Δh = 57,558 × 3.009 / (2π × 11,725)
Evaluate — Δh = 2.35 ft

Answer: h₂ − h₁ ≈ 2.35 ft

Why the other options are there

15.83 ft (ratio used instead of its log)
14.77 ft (2π omitted)

Reference: FE Reference Handbook — Hydrology and Water Resources → Unconfined aquifer

Example 10

Darcy's law seepage flow — Unconfined aquifer (5)

An aquifer has K = 3.80e-3 ft/s under a gradient of 0.035 across a 2561 ft² cross section. Find the seepage discharge.

Given

K = 3.80e-3 ft/s
i = 0.035
A = 2561 ft²

Find

Seepage Q

Start with the thinking

Darcy velocity is superficial — it is not the pore velocity.
Gradient is dimensionless head loss per length.

Figure for Darcy's law seepage flow — Unconfined aquifer (5)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 3.80 e - 3 (0.035) (2561)$
Evaluate
$Q = 3.406 e - 1 f t^{3} / s$
Convert
$Q = 152.9 g p m$
Darcy velocity
$v = K i = 1.330 e - 4 f t / s$

Answer: Q ≈ 3.41e-1 cfs (152.9 gpm)

Why the other options are there

9.73e+0 cfs (gradient omitted)
1.09e-1 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Unconfined aquifer

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

Unconfined aquifer contains 0 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.