Confined 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 Confined 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 confined 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. Confined 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: confined aquifer.
Capstone Studio instructional photograph

Hydrology and Water Resources — Confined 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 PIEZOMETRIC SURFACE
AQUITARD
H TOP OF AQUIFER
h1 h2
PERMEABLE SOIL b
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 — Confined aquifer

A well pumps 219 gpm from a confined aquifer with K = 115.0 ft/day and thickness 31 ft. Two observation wells sit 50 ft and 672 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 219 gpm
K = 115.0 ft/day
b = 31 ft
r₁ = 50 ft
r₂ = 672 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 — Confined aquifer

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 115.0 \times 31 = 3, 565 f t^{2} / d a y$
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
$ln (672/50) = 2.598$
Substituting — Δh = 42,158 × 2.598 / (2π × 3,565)
Evaluate — Δh = 4.89 ft

Answer: h₂ − h₁ ≈ 4.89 ft

Why the other options are there

25.30 ft (ratio used instead of its log)
30.73 ft (2π omitted)

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

Example 2

Darcy's law seepage flow — Confined aquifer

An aquifer has K = 1.90e-3 ft/s under a gradient of 0.025 across a 2597 ft² cross section. Find the seepage discharge.

Given

K = 1.90e-3 ft/s
i = 0.025
A = 2597 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 — Confined aquifer

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 1.90 e - 3 (0.025) (2597)$
Evaluate
$Q = 1.234 e - 1 f t^{3} / s$
Convert
$Q = 55.3628 g p m$
Darcy velocity
$v = K i = 4.750 e - 5 f t / s$

Answer: Q ≈ 1.23e-1 cfs (55.36 gpm)

Why the other options are there

4.93e+0 cfs (gradient omitted)
7.60e-2 cfs (gradient divided)

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

Example 3

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

A well pumps 628 gpm from a confined aquifer with K = 70 ft/day and thickness 25 ft. Two observation wells sit 50 ft and 1383 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 628 gpm
K = 70 ft/day
b = 25 ft
r₁ = 50 ft
r₂ = 1383 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 — Confined aquifer (2)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 70 \times 25 = 1, 750 f t^{2} / d a y$
Discharge
$Q = 628 g p m \times 192.5 = 120, 890 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (1383/50) = 3.320$
Substituting — Δh = 120,890 × 3.320 / (2π × 1,750)
Evaluate — Δh = 36.50 ft

Answer: h₂ − h₁ ≈ 36.50 ft

Why the other options are there

304.1 ft (ratio used instead of its log)
229.3 ft (2π omitted)

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

Example 4

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

An aquifer has K = 2.70e-3 ft/s under a gradient of 0.020 across a 1053 ft² cross section. Find the seepage discharge.

Given

K = 2.70e-3 ft/s
i = 0.020
A = 1053 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 — Confined aquifer (2)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 2.70 e - 3 (0.020) (1053)$
Evaluate
$Q = 5.686 e - 2 f t^{3} / s$
Convert
$Q = 25.5197 g p m$
Darcy velocity
$v = K i = 5.400 e - 5 f t / s$

Answer: Q ≈ 5.69e-2 cfs (25.52 gpm)

Why the other options are there

2.84e+0 cfs (gradient omitted)
1.35e-1 cfs (gradient divided)

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

Example 5

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

A well pumps 261 gpm from a confined aquifer with K = 95 ft/day and thickness 38 ft. Two observation wells sit 50 ft and 312 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 261 gpm
K = 95 ft/day
b = 38 ft
r₁ = 50 ft
r₂ = 312 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 — Confined aquifer (3)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 95 \times 38 = 3, 610 f t^{2} / d a y$
Discharge
$Q = 261 g p m \times 192.5 = 50, 243 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (312/50) = 1.831$
Substituting — Δh = 50,243 × 1.831 / (2π × 3,610)
Evaluate — Δh = 4.06 ft

Answer: h₂ − h₁ ≈ 4.06 ft

Why the other options are there

13.82 ft (ratio used instead of its log)
25.48 ft (2π omitted)

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

Example 6

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

An aquifer has K = 3.00e-4 ft/s under a gradient of 0.055 across a 1232 ft² cross section. Find the seepage discharge.

Given

K = 3.00e-4 ft/s
i = 0.055
A = 1232 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 — Confined aquifer (3)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 3.00 e - 4 (0.055) (1232)$
Evaluate
$Q = 2.033 e - 2 f t^{3} / s$
Convert
$Q = 9.1232 g p m$
Darcy velocity
$v = K i = 1.650 e - 5 f t / s$

Answer: Q ≈ 2.03e-2 cfs (9.12 gpm)

Why the other options are there

3.70e-1 cfs (gradient omitted)
5.45e-3 cfs (gradient divided)

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

Example 7

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

A well pumps 261 gpm from a confined aquifer with K = 95 ft/day and thickness 75 ft. Two observation wells sit 50 ft and 512 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 261 gpm
K = 95 ft/day
b = 75 ft
r₁ = 50 ft
r₂ = 512 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 — Confined aquifer (4)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 95 \times 75 = 7, 125 f t^{2} / d a y$
Discharge
$Q = 261 g p m \times 192.5 = 50, 243 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (512/50) = 2.326$
Substituting — Δh = 50,243 × 2.326 / (2π × 7,125)
Evaluate — Δh = 2.61 ft

Answer: h₂ − h₁ ≈ 2.61 ft

Why the other options are there

11.49 ft (ratio used instead of its log)
16.40 ft (2π omitted)

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

Example 8

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

An aquifer has K = 4.10e-3 ft/s under a gradient of 0.010 across a 992 ft² cross section. Find the seepage discharge.

Given

K = 4.10e-3 ft/s
i = 0.010
A = 992 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 — Confined aquifer (4)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 4.10 e - 3 (0.010) (992)$
Evaluate
$Q = 4.067 e - 2 f t^{3} / s$
Convert
$Q = 18.2536 g p m$
Darcy velocity
$v = K i = 4.100 e - 5 f t / s$

Answer: Q ≈ 4.07e-2 cfs (18.25 gpm)

Why the other options are there

4.07e+0 cfs (gradient omitted)
4.10e-1 cfs (gradient divided)

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

Example 9

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

A well pumps 333 gpm from a confined aquifer with K = 155.0 ft/day and thickness 83 ft. Two observation wells sit 50 ft and 1435 ft from the pumping well. Find the head difference between them at steady state.

Given

Q = 333 gpm
K = 155.0 ft/day
b = 83 ft
r₁ = 50 ft
r₂ = 1435 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 — Confined aquifer (5)

Step-by-step solution

Transmissivity
$T = K b$
Substituting
$T = 155.0 \times 83 = 12, 865 f t^{2} / d a y$
Discharge
$Q = 333 g p m \times 192.5 = 64, 103 f t^{3} / d a y$
Thiem equation — h₂ − h₁ = Q ln(r₂/r₁) / (2πT)
Log term
$ln (1435/50) = 3.357$
Substituting — Δh = 64,103 × 3.357 / (2π × 12,865)
Evaluate — Δh = 2.66 ft

Answer: h₂ − h₁ ≈ 2.66 ft

Why the other options are there

22.76 ft (ratio used instead of its log)
16.73 ft (2π omitted)

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

Example 10

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

An aquifer has K = 3.70e-3 ft/s under a gradient of 0.050 across a 1963 ft² cross section. Find the seepage discharge.

Given

K = 3.70e-3 ft/s
i = 0.050
A = 1963 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 — Confined aquifer (5)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 3.70 e - 3 (0.050) (1963)$
Evaluate
$Q = 3.632 e - 1 f t^{3} / s$
Convert
$Q = 163.0 g p m$
Darcy velocity
$v = K i = 1.850 e - 4 f t / s$

Answer: Q ≈ 3.63e-1 cfs (163.0 gpm)

Why the other options are there

7.26e+0 cfs (gradient omitted)
7.40e-2 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Confined 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

Confined 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.