Darcy's Law

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 Darcy's Law 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 darcy's law 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. Darcy's Law 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: darcy's law.
Capstone Studio instructional photograph

Hydrology and Water Resources — Darcy's Law: 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 = - KA dx" — read its definition and unit from the handbook line directly above the equation.
K	Quantity produced by "K = hydraulic conductivity (ft/sec or m/s)" — read its definition and unit from the handbook line directly above the equation.
h	Quantity produced by "h = hydraulic head (ft or m)" — read its definition and unit from the handbook line directly above the equation.
A	Quantity produced by "A = cross-sectional area of flow (ft2 or m2)" — read its definition and unit from the handbook line directly above the equation.
q	Quantity produced by "q = - K dx" — read its definition and unit from the handbook line directly above the equation.
v	Quantity produced by "v = n = n dx" — read its definition and unit from the handbook line directly above the equation.
n	Quantity produced by "n = effective porosity" — 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.

where
where
q − K dh
where
Unit hydrograph: The direct runoff hydrograph that would result from one unit of runoff occurring uniformly in space and time
over a specified period of time.
Transmissivity, T: The product of hydraulic conductivity and thickness, b, of the aquifer (L2T –1).
Storativity or storage coefficient of an aquifer, S: The volume of water taken into or released from storage per unit surface area
per unit change in potentiometric (piezometric) head.

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

Darcy's law seepage flow — Darcy's Law

An aquifer has K = 4.20e-3 ft/s under a gradient of 0.040 across a 1697 ft² cross section. Find the seepage discharge.

Given

K = 4.20e-3 ft/s
i = 0.040
A = 1697 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 — Darcy's Law

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 4.20 e - 3 (0.040) (1697)$
Evaluate
$Q = 2.851 e - 1 f t^{3} / s$
Convert
$Q = 128.0 g p m$
Darcy velocity
$v = K i = 1.680 e - 4 f t / s$

Answer: Q ≈ 2.85e-1 cfs (128.0 gpm)

Why the other options are there

7.13e+0 cfs (gradient omitted)
1.05e-1 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Darcy's Law

Example 2

Darcy's law seepage flow — Darcy's Law (2)

An aquifer has K = 6.00e-4 ft/s under a gradient of 0.015 across a 394 ft² cross section. Find the seepage discharge.

Given

K = 6.00e-4 ft/s
i = 0.015
A = 394 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 — Darcy's Law (2)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 6.00 e - 4 (0.015) (394)$
Evaluate
$Q = 3.546 e - 3 f t^{3} / s$
Convert
$Q = 1.5914 g p m$
Darcy velocity
$v = K i = 9.000 e - 6 f t / s$

Answer: Q ≈ 3.55e-3 cfs (1.59 gpm)

Why the other options are there

2.36e-1 cfs (gradient omitted)
4.00e-2 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Darcy's Law

Example 3

Darcy's law seepage flow — Darcy's Law (3)

An aquifer has K = 4.60e-3 ft/s under a gradient of 0.040 across a 1594 ft² cross section. Find the seepage discharge.

Given

K = 4.60e-3 ft/s
i = 0.040
A = 1594 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 — Darcy's Law (3)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 4.60 e - 3 (0.040) (1594)$
Evaluate
$Q = 2.933 e - 1 f t^{3} / s$
Convert
$Q = 131.6 g p m$
Darcy velocity
$v = K i = 1.840 e - 4 f t / s$

Answer: Q ≈ 2.93e-1 cfs (131.6 gpm)

Why the other options are there

7.33e+0 cfs (gradient omitted)
1.15e-1 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Darcy's Law

Example 4

Darcy's law seepage flow — Darcy's Law (4)

An aquifer has K = 1.70e-3 ft/s under a gradient of 0.045 across a 2952 ft² cross section. Find the seepage discharge.

Given

K = 1.70e-3 ft/s
i = 0.045
A = 2952 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 — Darcy's Law (4)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 1.70 e - 3 (0.045) (2952)$
Evaluate
$Q = 2.258 e - 1 f t^{3} / s$
Convert
$Q = 101.4 g p m$
Darcy velocity
$v = K i = 7.650 e - 5 f t / s$

Answer: Q ≈ 2.26e-1 cfs (101.4 gpm)

Why the other options are there

5.02e+0 cfs (gradient omitted)
3.78e-2 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Darcy's Law

Example 5

Darcy's law seepage flow — Darcy's Law (5)

An aquifer has K = 2.90e-3 ft/s under a gradient of 0.005 across a 1515 ft² cross section. Find the seepage discharge.

Given

K = 2.90e-3 ft/s
i = 0.005
A = 1515 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 — Darcy's Law (5)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 2.90 e - 3 (0.005) (1515)$
Evaluate
$Q = 2.197 e - 2 f t^{3} / s$
Convert
$Q = 9.8590 g p m$
Darcy velocity
$v = K i = 1.450 e - 5 f t / s$

Answer: Q ≈ 2.20e-2 cfs (9.86 gpm)

Why the other options are there

4.39e+0 cfs (gradient omitted)
5.80e-1 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Darcy's Law

Example 6

Darcy's law seepage flow — Darcy's Law (6)

An aquifer has K = 5.00e-4 ft/s under a gradient of 0.025 across a 2842 ft² cross section. Find the seepage discharge.

Given

K = 5.00e-4 ft/s
i = 0.025
A = 2842 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 — Darcy's Law (6)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 5.00 e - 4 (0.025) (2842)$
Evaluate
$Q = 3.553 e - 2 f t^{3} / s$
Convert
$Q = 15.9436 g p m$
Darcy velocity
$v = K i = 1.250 e - 5 f t / s$

Answer: Q ≈ 3.55e-2 cfs (15.94 gpm)

Why the other options are there

1.42e+0 cfs (gradient omitted)
2.00e-2 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Darcy's Law

Example 7

Darcy's law seepage flow — Darcy's Law (7)

An aquifer has K = 1.90e-3 ft/s under a gradient of 0.025 across a 275 ft² cross section. Find the seepage discharge.

Given

K = 1.90e-3 ft/s
i = 0.025
A = 275 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 — Darcy's Law (7)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 1.90 e - 3 (0.025) (275)$
Evaluate
$Q = 1.306 e - 2 f t^{3} / s$
Convert
$Q = 5.8625 g p m$
Darcy velocity
$v = K i = 4.750 e - 5 f t / s$

Answer: Q ≈ 1.31e-2 cfs (5.86 gpm)

Why the other options are there

5.22e-1 cfs (gradient omitted)
7.60e-2 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Darcy's Law

Example 8

Darcy's law seepage flow — Darcy's Law (8)

An aquifer has K = 1.00e-4 ft/s under a gradient of 0.045 across a 2842 ft² cross section. Find the seepage discharge.

Given

K = 1.00e-4 ft/s
i = 0.045
A = 2842 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 — Darcy's Law (8)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 1.00 e - 4 (0.045) (2842)$
Evaluate
$Q = 1.279 e - 2 f t^{3} / s$
Convert
$Q = 5.7397 g p m$
Darcy velocity
$v = K i = 4.500 e - 6 f t / s$

Answer: Q ≈ 1.28e-2 cfs (5.74 gpm)

Why the other options are there

2.84e-1 cfs (gradient omitted)
2.22e-3 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Darcy's Law

Example 9

Darcy's law seepage flow — Darcy's Law (9)

An aquifer has K = 3.40e-3 ft/s under a gradient of 0.050 across a 2405 ft² cross section. Find the seepage discharge.

Given

K = 3.40e-3 ft/s
i = 0.050
A = 2405 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 — Darcy's Law (9)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 3.40 e - 3 (0.050) (2405)$
Evaluate
$Q = 4.089 e - 1 f t^{3} / s$
Convert
$Q = 183.5 g p m$
Darcy velocity
$v = K i = 1.700 e - 4 f t / s$

Answer: Q ≈ 4.09e-1 cfs (183.5 gpm)

Why the other options are there

8.18e+0 cfs (gradient omitted)
6.80e-2 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Darcy's Law

Example 10

Darcy's law seepage flow — Darcy's Law (10)

An aquifer has K = 3.20e-3 ft/s under a gradient of 0.050 across a 2120 ft² cross section. Find the seepage discharge.

Given

K = 3.20e-3 ft/s
i = 0.050
A = 2120 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 — Darcy's Law (10)

Step-by-step solution

Darcy
$Q = K i A$
Substituting
$Q = 3.20 e - 3 (0.050) (2120)$
Evaluate
$Q = 3.392 e - 1 f t^{3} / s$
Convert
$Q = 152.2 g p m$
Darcy velocity
$v = K i = 1.600 e - 4 f t / s$

Answer: Q ≈ 3.39e-1 cfs (152.2 gpm)

Why the other options are there

6.78e+0 cfs (gradient omitted)
6.40e-2 cfs (gradient divided)

Reference: FE Reference Handbook — Hydrology and Water Resources → Darcy's Law

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

Darcy's Law 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.

Darcy's Law

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

Q=−KAdx

Q=dischargerate(ft3​/secorm3​/s)

K=hydraulicconductivity(ft/secorm/s)

h=hydraulichead(ftorm)

A=cross−sectionalareaofflow(ft2​orm2​)

q=−Kdx

q=specificdischarge(alsocalledDarcyvelocityorsuperficialvelocity)

v=n=ndx

v=averageseepagevelocity

n=effectiveporosity

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section