13.3 Ground water hydrology

Achievable FE Civil

13. Water resources engineering

Ground water hydrology

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This chapter covers the following:

Darcy’s law
Unconfined aquifer - Dupuit’s formula
Confined aquifer - Theim equation

Exam tip: The Dupuit and Thiem equations are given in the FE Reference Handbook, so you don’t need to memorize them. Instead, focus on identifying $r_{1}$ , $r_{2}$ , $h_{1}$ , and $h_{2}$ correctly from the problem setup and knowing which equation applies to a confined versus unconfined aquifer.

Darcy’s law

Darcy’s law is the core relationship used to describe groundwater flow through saturated porous media. It says that the volumetric flow rate is proportional to the hydraulic gradient, with hydraulic conductivity capturing how easily water moves through the material (based on both fluid properties and the permeability of the soil or rock).

Darcy’s law applies when flow is laminar, which is typical in most groundwater systems. You’ll use it as the starting point for analyzing seepage, aquifer behavior, and contaminant transport.

$Q = - K A (\frac{d h}{d x})$

$Q$ = discharge rate (ft³/sec or m³/s)
$K$ = hydraulic conductivity (ft/sec or m/s)
$h$ = hydraulic head (ft or m)
$A$ = cross-sectional area of flow (ft² or m²)

The negative sign indicates that flow goes from higher head to lower head.

Specific discharge:

$q = - K (\frac{d h}{d x})$

Average seepage velocity:

$v = \frac{q}{n} = \frac{- K}{n} (\frac{d h}{d x})$

$n$ = effective porosity

Example: Darcy discharge

$K = 0.003$ ft/sec, $A = 25$ ft², head drop $4$ ft over $800$ ft.

$Q = - K A (\frac{d h}{d x}) = - (0.003) (25) (\frac{- 4}{800}) = 0.000375 ft^{3} / sec$

Answer: $Q \approx 3.75 \times 1 0^{- 4}$ ft³/sec

Watch out: Keep $K$ , $A$ , $Q$ in one unit system - mixing ft and m is a common error. $\frac{d h}{d x}$ is head change per length, not a bare number. Round only the final answer.

Unconfined aquifer - Dupuit’s formula

Dupuit’s formula is used for steady-state groundwater flow in an unconfined aquifer, where the top of the saturated zone is the water table. The approach relies on the Dupuit assumptions:

Vertical flow components are neglected.
The hydraulic gradient is approximated by the slope of the water table.

These assumptions simplify the flow field, but the result is often accurate enough for shallow, laterally extensive unconfined aquifers. You can use the equation to estimate discharge to a well, drawdown, or the shape of the water table.

$Q = \frac{π K ( h _{2}^{2} - h _{1}^{2} )}{ln ( \frac{r _{2}}{r _{1}} )}$

Where:

$Q$ = flowrate of water drawn from well (cfs)
$K$ = hydraulic conductivity (ft/sec)
$h_{1}$ = head at the well (radius $r_{1}$ ) (ft)
$h_{2}$ = water height at radius $r_{2}$ (ft)
$r_{1}$ = radius of well (ft)
$r_{2}$ = distance from well centerline to $h_{2}$ point (ft)

Example: Unconfined well discharge

$K = 0.002$ ft/sec; $r_{1} = 1$ ft, $h_{1} = 45$ ft; $r_{2} = 500$ ft, $h_{2} = 60$ ft.

$Q = \frac{π K ( h _{2}^{2} - h _{1}^{2} )}{ln ( r _{2} / r _{1} )} = \frac{π ( 0.002 ) ( 6 0 ^{2} - 4 5 ^{2} )}{ln ( 500 )} = \frac{9.90}{6.215} = 1.59 cfs$

Answer: $Q \approx 1.59$ cfs

Confined aquifer - Theim equation

The Thiem equation is a classical steady-state solution for radial flow toward a pumping well in a confined aquifer. It connects the pumping rate to aquifer properties and shows that hydraulic head changes logarithmically with distance from the well.

The equation assumes:

The aquifer is homogeneous and isotropic.
The well fully penetrates the confined aquifer.
Flow is steady (equilibrium conditions).

These conditions make the Thiem equation especially useful for interpreting pumping test data and estimating transmissivity.

$Q = \frac{2 π T ( h _{2} - h _{1} )}{ln ( \frac{r _{2}}{r _{1}} )}$

Where:

$T = K b$ = transmissivity, the product of hydraulic conductivity and the confined aquifer thickness $b$ (ft²/sec)
$b$ = thickness of confined aquifer (ft)
$h_{1}, h_{2}$ = head at radius $r_{1}, r_{2}$ (ft)
$r_{1}, r_{2}$ = radii from pumping well (ft)

Example: Confined well discharge

$T = 0.05$ ft²/sec; $r_{1} = 0.5$ ft, $h_{1} = 100$ ft; $r_{2} = 1, 000$ ft, $h_{2} = 120$ ft.

$Q = \frac{2 π T ( h _{2} - h _{1} )}{ln ( r _{2} / r _{1} )} = \frac{2 π ( 0.05 ) ( 120 - 100 )}{ln ( 2000 )} = \frac{6.28}{7.60} = 0.83 cfs$

Answer: $Q \approx 0.83$ cfs

Storativity or storage coefficient of an aquifer, $S$

Storativity (storage coefficient) is the volume of water taken into or released from storage per unit surface area per unit change in potentiometric (piezometric) head. You won’t need to calculate storativity for the FE exam - just recognize it as the property that describes how much water a confined aquifer releases from storage as head declines, distinct from $T$ ’s role in determining flow rate.

Darcy’s law

Governs groundwater flow in saturated porous media
Key formula: $Q = - K A (\frac{d h}{d x})$
- $Q$ : discharge, $K$ : hydraulic conductivity, $A$ : area, $d h / d x$ : hydraulic gradient
Specific discharge: $q = - K (\frac{d h}{d x})$
Seepage velocity: $v = \frac{q}{n} = \frac{- K}{n} (\frac{d h}{d x})$
- $n$ : effective porosity

Unconfined aquifer - Dupuit’s formula

Used for steady-state flow in unconfined aquifers
Assumes:
- Negligible vertical flow
- Hydraulic gradient ≈ water table slope
Key formula: $Q = \frac{π K ( h _{1}^{2} - h _{2}^{2} )}{l n ( \frac{r _{2}}{r _{1}} )}$
- $h_{1}$ , $h_{2}$ : water heights, $r_{1}$ , $r_{2}$ : radii

Confined aquifer - Thiem equation

Describes steady-state radial flow to a well in confined aquifer
Assumes:
- Homogeneous, isotropic aquifer
- Well fully penetrates aquifer
- Steady flow
Key formula: $Q = \frac{2 π T ( h _{1} - h _{2} )}{l n ( \frac{r _{2}}{r _{1}} )}$
- $T$ : transmissivity, $h_{1}$ , $h_{2}$ : heads, $r_{1}$ , $r_{2}$ : radii

Transmissivity, $T$

$T = K b$ (hydraulic conductivity × aquifer thickness)
Units: $[L^{2} / T]$ (e.g., ft²/sec)

Storativity (storage coefficient), $S$

Volume of water released or stored per unit area per unit head change
Important for aquifer response to pumping

Next | 14.1 Water quality

Ground water hydrology

This chapter covers the following:

Darcy’s law
Unconfined aquifer - Dupuit’s formula
Confined aquifer - Theim equation

Darcy’s law

Darcy’s law applies when flow is laminar, which is typical in most groundwater systems. You’ll use it as the starting point for analyzing seepage, aquifer behavior, and contaminant transport.

$Q = - K A (\frac{d h}{d x})$

$Q$ = discharge rate (ft³/sec or m³/s)
$K$ = hydraulic conductivity (ft/sec or m/s)
$h$ = hydraulic head (ft or m)
$A$ = cross-sectional area of flow (ft² or m²)

The negative sign indicates that flow goes from higher head to lower head.

Specific discharge:

$q = - K (\frac{d h}{d x})$

Average seepage velocity:

$v = \frac{q}{n} = \frac{- K}{n} (\frac{d h}{d x})$

$n$ = effective porosity

Example: Darcy discharge

$K = 0.003$ ft/sec, $A = 25$ ft², head drop $4$ ft over $800$ ft.

$Q = - K A (\frac{d h}{d x}) = - (0.003) (25) (\frac{- 4}{800}) = 0.000375 ft^{3} / sec$

Answer: $Q \approx 3.75 \times 1 0^{- 4}$ ft³/sec

Watch out: Keep $K$ , $A$ , $Q$ in one unit system - mixing ft and m is a common error. $\frac{d h}{d x}$ is head change per length, not a bare number. Round only the final answer.

Unconfined aquifer - Dupuit’s formula

Dupuit’s formula is used for steady-state groundwater flow in an unconfined aquifer, where the top of the saturated zone is the water table. The approach relies on the Dupuit assumptions:

Vertical flow components are neglected.
The hydraulic gradient is approximated by the slope of the water table.

$Q = \frac{π K ( h _{2}^{2} - h _{1}^{2} )}{ln ( \frac{r _{2}}{r _{1}} )}$

Where:

$Q$ = flowrate of water drawn from well (cfs)
$K$ = hydraulic conductivity (ft/sec)
$h_{1}$ = head at the well (radius $r_{1}$ ) (ft)
$h_{2}$ = water height at radius $r_{2}$ (ft)
$r_{1}$ = radius of well (ft)
$r_{2}$ = distance from well centerline to $h_{2}$ point (ft)

Example: Unconfined well discharge

$K = 0.002$ ft/sec; $r_{1} = 1$ ft, $h_{1} = 45$ ft; $r_{2} = 500$ ft, $h_{2} = 60$ ft.

$Q = \frac{π K ( h _{2}^{2} - h _{1}^{2} )}{ln ( r _{2} / r _{1} )} = \frac{π ( 0.002 ) ( 6 0 ^{2} - 4 5 ^{2} )}{ln ( 500 )} = \frac{9.90}{6.215} = 1.59 cfs$

Answer: $Q \approx 1.59$ cfs

Confined aquifer - Theim equation

The equation assumes:

The aquifer is homogeneous and isotropic.
The well fully penetrates the confined aquifer.
Flow is steady (equilibrium conditions).

These conditions make the Thiem equation especially useful for interpreting pumping test data and estimating transmissivity.

$Q = \frac{2 π T ( h _{2} - h _{1} )}{ln ( \frac{r _{2}}{r _{1}} )}$

Where:

$T = K b$ = transmissivity, the product of hydraulic conductivity and the confined aquifer thickness $b$ (ft²/sec)
$b$ = thickness of confined aquifer (ft)
$h_{1}, h_{2}$ = head at radius $r_{1}, r_{2}$ (ft)
$r_{1}, r_{2}$ = radii from pumping well (ft)

Example: Confined well discharge

$T = 0.05$ ft²/sec; $r_{1} = 0.5$ ft, $h_{1} = 100$ ft; $r_{2} = 1, 000$ ft, $h_{2} = 120$ ft.

$Q = \frac{2 π T ( h _{2} - h _{1} )}{ln ( r _{2} / r _{1} )} = \frac{2 π ( 0.05 ) ( 120 - 100 )}{ln ( 2000 )} = \frac{6.28}{7.60} = 0.83 cfs$

Answer: $Q \approx 0.83$ cfs

Storativity or storage coefficient of an aquifer, $S$

Achievable FE Civil

13. Water resources engineering