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

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Unit hydrograph: The direct runoff hydrograph that would result from one unit of runoff occurring uniformly in space and 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 — solve for flow rate — Darcy's Law

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 78.0000 ft/day; hydraulic gradient (i) = 0.0660 ft/ft; cross-sectional area (A) = 32.0000 ft², determine the flow rate (Q) in ft³/day.

Given

$hydraulic conductivity (K) = 78.0000 ft/day$
$hydraulic gradient (i) = 0.0660 ft/ft$
$cross-sectional area (A) = 32.0000 ft^{2}$

Find

flow rate (Q), in ft³/day

Start with the thinking

The governing relation printed in this handbook section is Darcy's law.
Everything except Q is given, so isolate Q symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$Q = K i A$
Step 2 — Rearrange the relation so that Q stands alone on the left-hand side.
Step 3 — List the givens: hydraulic conductivity (K) = 78.0000 ft/day, hydraulic gradient (i) = 0.0660 ft/ft, cross-sectional area (A) = 32.0000 ft².
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$Q = 164.7\ \text{ft³/day}$
Step 6 — Check: returning Q = 164.7 ft³/day to
$Q = K i A$
reproduces the given quantities, and both sides carry the same units.

Answer:

Q = 164.7\ \text{ft³/day}

Why the other options are there

329.5 — kept a factor of two that cancels in the correct rearrangement.
82.3680 — dropped that same factor in the other direction.
181.2 — rounded an intermediate value before the final step.

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

Example 2

Darcy's law — solve for hydraulic conductivity — Darcy's Law (2)

A hydrology and water resources problem uses Darcy's law. Given hydraulic gradient (i) = 0.0530 ft/ft; cross-sectional area (A) = 423.0 ft²; flow rate (Q) = 4,451 ft³/day, determine the hydraulic conductivity (K) in ft/day.

Given

$hydraulic gradient (i) = 0.0530 ft/ft$
$cross-sectional area (A) = 423.0 ft^{2}$
$flow rate (Q) = 4,451 ft^{3}/day$

Find

hydraulic conductivity (K), in ft/day

Start with the thinking

The governing relation printed in this handbook section is Darcy's law.
Everything except K is given, so isolate K symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$Q = K i A$
Step 2 — Rearrange the relation so that K stands alone on the left-hand side.
Step 3 — List the givens: hydraulic gradient (i) = 0.0530 ft/ft, cross-sectional area (A) = 423.0 ft², flow rate (Q) = 4,451 ft³/day.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$K = 198.5\ \text{ft/day}$
Step 6 — Check: returning K = 198.5 ft/day to
$Q = K i A$
reproduces the given quantities, and both sides carry the same units.

Answer:

K = 198.5\ \text{ft/day}

Why the other options are there

397.0 — kept a factor of two that cancels in the correct rearrangement.
99.2618 — dropped that same factor in the other direction.
218.4 — rounded an intermediate value before the final step.

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

Example 3

Darcy's law — solve for cross-sectional area — Darcy's Law (3)

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 21.9000 ft/day; hydraulic gradient (i) = 0.0380 ft/ft; flow rate (Q) = 3,051 ft³/day, determine the cross-sectional area (A) in ft².

Given

$hydraulic conductivity (K) = 21.9000 ft/day$
$hydraulic gradient (i) = 0.0380 ft/ft$
$flow rate (Q) = 3,051 ft^{3}/day$

Find

cross-sectional area (A), in ft²

Start with the thinking

The governing relation printed in this handbook section is Darcy's law.
Everything except A is given, so isolate A symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$Q = K i A$
Step 2 — Rearrange the relation so that A stands alone on the left-hand side.
Step 3 — List the givens: hydraulic conductivity (K) = 21.9000 ft/day, hydraulic gradient (i) = 0.0380 ft/ft, flow rate (Q) = 3,051 ft³/day.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$A = 3666\ \text{ft²}$
Step 6 — Check: returning A = 3,666 ft² to
$Q = K i A$
reproduces the given quantities, and both sides carry the same units.

Answer:

A = 3666\ \text{ft²}

Why the other options are there

7,332 — kept a factor of two that cancels in the correct rearrangement.
1,833 — dropped that same factor in the other direction.
4,033 — rounded an intermediate value before the final step.

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

Example 4

Darcy's law — solve for flow rate (case 2) — Darcy's Law (4)

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 11.2000 ft/day; hydraulic gradient (i) = 0.0950 ft/ft; cross-sectional area (A) = 107.0 ft², determine the flow rate (Q) in ft³/day.

Given

$hydraulic conductivity (K) = 11.2000 ft/day$
$hydraulic gradient (i) = 0.0950 ft/ft$
$cross-sectional area (A) = 107.0 ft^{2}$

Find

flow rate (Q), in ft³/day

Start with the thinking

The governing relation printed in this handbook section is Darcy's law.
Everything except Q is given, so isolate Q symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$Q = K i A$
Step 2 — Rearrange the relation so that Q stands alone on the left-hand side.
Step 3 — List the givens: hydraulic conductivity (K) = 11.2000 ft/day, hydraulic gradient (i) = 0.0950 ft/ft, cross-sectional area (A) = 107.0 ft².
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$Q = 113.8\ \text{ft³/day}$
Step 6 — Check: returning Q = 113.8 ft³/day to
$Q = K i A$
reproduces the given quantities, and both sides carry the same units.

Answer:

Q = 113.8\ \text{ft³/day}

Why the other options are there

227.7 — kept a factor of two that cancels in the correct rearrangement.
56.9240 — dropped that same factor in the other direction.
125.2 — rounded an intermediate value before the final step.

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

Example 5

Darcy's law — solve for hydraulic conductivity (case 2) — Darcy's Law (5)

A hydrology and water resources problem uses Darcy's law. Given hydraulic gradient (i) = 0.0930 ft/ft; cross-sectional area (A) = 321.0 ft²; flow rate (Q) = 331.0 ft³/day, determine the hydraulic conductivity (K) in ft/day.

Given

$hydraulic gradient (i) = 0.0930 ft/ft$
$cross-sectional area (A) = 321.0 ft^{2}$
$flow rate (Q) = 331.0 ft^{3}/day$

Find

hydraulic conductivity (K), in ft/day

Start with the thinking

The governing relation printed in this handbook section is Darcy's law.
Everything except K is given, so isolate K symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$Q = K i A$
Step 2 — Rearrange the relation so that K stands alone on the left-hand side.
Step 3 — List the givens: hydraulic gradient (i) = 0.0930 ft/ft, cross-sectional area (A) = 321.0 ft², flow rate (Q) = 331.0 ft³/day.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$K = 11.0877\ \text{ft/day}$
Step 6 — Check: returning K = 11.0877 ft/day to
$Q = K i A$
reproduces the given quantities, and both sides carry the same units.

Answer:

K = 11.0877\ \text{ft/day}

Why the other options are there

22.1753 — kept a factor of two that cancels in the correct rearrangement.
5.5438 — dropped that same factor in the other direction.
12.1964 — rounded an intermediate value before the final step.

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

Example 6

Darcy's law — solve for cross-sectional area (case 2) — Darcy's Law (6)

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 9.2000 ft/day; hydraulic gradient (i) = 0.0730 ft/ft; flow rate (Q) = 4,751 ft³/day, determine the cross-sectional area (A) in ft².

Given

$hydraulic conductivity (K) = 9.2000 ft/day$
$hydraulic gradient (i) = 0.0730 ft/ft$
$flow rate (Q) = 4,751 ft^{3}/day$

Find

cross-sectional area (A), in ft²

Start with the thinking

The governing relation printed in this handbook section is Darcy's law.
Everything except A is given, so isolate A symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$Q = K i A$
Step 2 — Rearrange the relation so that A stands alone on the left-hand side.
Step 3 — List the givens: hydraulic conductivity (K) = 9.2000 ft/day, hydraulic gradient (i) = 0.0730 ft/ft, flow rate (Q) = 4,751 ft³/day.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$A = 7074\ \text{ft²}$
Step 6 — Check: returning A = 7,074 ft² to
$Q = K i A$
reproduces the given quantities, and both sides carry the same units.

Answer:

A = 7074\ \text{ft²}

Why the other options are there

14,148 — kept a factor of two that cancels in the correct rearrangement.
3,537 — dropped that same factor in the other direction.
7,781 — rounded an intermediate value before the final step.

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

Example 7

Darcy's law — solve for flow rate (case 3) — Darcy's Law (7)

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 90.6000 ft/day; hydraulic gradient (i) = 0.0630 ft/ft; cross-sectional area (A) = 160.0 ft², determine the flow rate (Q) in ft³/day.

Given

$hydraulic conductivity (K) = 90.6000 ft/day$
$hydraulic gradient (i) = 0.0630 ft/ft$
$cross-sectional area (A) = 160.0 ft^{2}$

Find

flow rate (Q), in ft³/day

Start with the thinking

The governing relation printed in this handbook section is Darcy's law.
Everything except Q is given, so isolate Q symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$Q = K i A$
Step 2 — Rearrange the relation so that Q stands alone on the left-hand side.
Step 3 — List the givens: hydraulic conductivity (K) = 90.6000 ft/day, hydraulic gradient (i) = 0.0630 ft/ft, cross-sectional area (A) = 160.0 ft².
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$Q = 913.2\ \text{ft³/day}$
Step 6 — Check: returning Q = 913.2 ft³/day to
$Q = K i A$
reproduces the given quantities, and both sides carry the same units.

Answer:

Q = 913.2\ \text{ft³/day}

Why the other options are there

1,826 — kept a factor of two that cancels in the correct rearrangement.
456.6 — dropped that same factor in the other direction.
1,005 — rounded an intermediate value before the final step.

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

Example 8

Darcy's law — solve for hydraulic conductivity (case 3) — Darcy's Law (8)

A hydrology and water resources problem uses Darcy's law. Given hydraulic gradient (i) = 0.0040 ft/ft; cross-sectional area (A) = 431.0 ft²; flow rate (Q) = 3,716 ft³/day, determine the hydraulic conductivity (K) in ft/day.

Given

$hydraulic gradient (i) = 0.0040 ft/ft$
$cross-sectional area (A) = 431.0 ft^{2}$
$flow rate (Q) = 3,716 ft^{3}/day$

Find

hydraulic conductivity (K), in ft/day

Start with the thinking

The governing relation printed in this handbook section is Darcy's law.
Everything except K is given, so isolate K symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$Q = K i A$
Step 2 — Rearrange the relation so that K stands alone on the left-hand side.
Step 3 — List the givens: hydraulic gradient (i) = 0.0040 ft/ft, cross-sectional area (A) = 431.0 ft², flow rate (Q) = 3,716 ft³/day.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$K = 2155\ \text{ft/day}$
Step 6 — Check: returning K = 2,155 ft/day to
$Q = K i A$
reproduces the given quantities, and both sides carry the same units.

Answer:

K = 2155\ \text{ft/day}

Why the other options are there

4,311 — kept a factor of two that cancels in the correct rearrangement.
1,078 — dropped that same factor in the other direction.
2,371 — rounded an intermediate value before the final step.

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

Example 9

Darcy's law — solve for cross-sectional area (case 3) — Darcy's Law (9)

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 17.4000 ft/day; hydraulic gradient (i) = 0.0450 ft/ft; flow rate (Q) = 1,037 ft³/day, determine the cross-sectional area (A) in ft².

Given

$hydraulic conductivity (K) = 17.4000 ft/day$
$hydraulic gradient (i) = 0.0450 ft/ft$
$flow rate (Q) = 1,037 ft^{3}/day$

Find

cross-sectional area (A), in ft²

Start with the thinking

The governing relation printed in this handbook section is Darcy's law.
Everything except A is given, so isolate A symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$Q = K i A$
Step 2 — Rearrange the relation so that A stands alone on the left-hand side.
Step 3 — List the givens: hydraulic conductivity (K) = 17.4000 ft/day, hydraulic gradient (i) = 0.0450 ft/ft, flow rate (Q) = 1,037 ft³/day.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$A = 1324\ \text{ft²}$
Step 6 — Check: returning A = 1,324 ft² to
$Q = K i A$
reproduces the given quantities, and both sides carry the same units.

Answer:

A = 1324\ \text{ft²}

Why the other options are there

2,648 — kept a factor of two that cancels in the correct rearrangement.
661.9 — dropped that same factor in the other direction.
1,456 — rounded an intermediate value before the final step.

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

Example 10

Darcy's law — solve for flow rate (case 4) — Darcy's Law (10)

A hydrology and water resources problem uses Darcy's law. Given hydraulic conductivity (K) = 98.9000 ft/day; hydraulic gradient (i) = 0.0180 ft/ft; cross-sectional area (A) = 279.0 ft², determine the flow rate (Q) in ft³/day.

Given

$hydraulic conductivity (K) = 98.9000 ft/day$
$hydraulic gradient (i) = 0.0180 ft/ft$
$cross-sectional area (A) = 279.0 ft^{2}$

Find

flow rate (Q), in ft³/day

Start with the thinking

The governing relation printed in this handbook section is Darcy's law.
Everything except Q is given, so isolate Q symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Hydrology and Water Resources items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$Q = K i A$
Step 2 — Rearrange the relation so that Q stands alone on the left-hand side.
Step 3 — List the givens: hydraulic conductivity (K) = 98.9000 ft/day, hydraulic gradient (i) = 0.0180 ft/ft, cross-sectional area (A) = 279.0 ft².
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$Q = 496.7\ \text{ft³/day}$
Step 6 — Check: returning Q = 496.7 ft³/day to
$Q = K i A$
reproduces the given quantities, and both sides carry the same units.

Answer:

Q = 496.7\ \text{ft³/day}

Why the other options are there

993.4 — kept a factor of two that cancels in the correct rearrangement.
248.3 — dropped that same factor in the other direction.
546.3 — rounded an intermediate value before the final step.

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

Darcy's Law

Handbook notes for this section

Core formulas for this FE topic

Q=−KAdxQ = - KA dxQ=−KAdx

Q=dischargerate(ft3/sec⁡orm3/s)Q = discharge rate (ft_{3}/\sec or m_{3}/s)Q=dischargerate(ft3​/secorm3​/s)

K=hydraulicconductivity(ft/sec⁡orm/s)K = hydraulic conductivity (ft/\sec or m/s)K=hydraulicconductivity(ft/secorm/s)

h=hydraulichead(ftorm)h = hydraulic head (ft or m)h=hydraulichead(ftorm)

A=cross−sectionalareaofflow(ft2orm2)A = cross-sectional area of flow (ft_{2} or m_{2})A=cross−sectionalareaofflow(ft2​orm2​)

q=−Kdxq = - K dxq=−Kdx

q=specificdischarge(alsocalledDarcyvelocityorsuperficialvelocity)q = specific discharge (also called Darcy velocity or superficial velocity)q=specificdischarge(alsocalledDarcyvelocityorsuperficialvelocity)

v=n=ndxv = n = n dxv=n=ndx

v=averageseepagevelocityv = average seepage velocityv=averageseepagevelocity

n=effectiveporosityn = effective porosityn=effectiveporosity

Worked exam-style examples