← GW Home
Chapter 3

Potential Groundwater Flow Theory and Flow Net Analysis [8 Hours]

Flow Theory & Flow Nets ~15+ Past Questions

MOST IMPORTANT TOPICS & MARKS DISTRIBUTION

High-Yield Topics (by exam frequency):

  1. Numericals: seepage discharge, flow between canals/rivers with/without recharge — Asked in 2078, 2073, 2072 Ashwin, 2072 Magh, 2071 [7–8 marks]
  2. Water table contour maps — preparation, uses, flow direction — Asked in 2073, 2072 Ashwin, 2072 Magh [4–8 marks]
  3. Flow net — definition, properties, seepage discharge derivation — Asked in 2078, 2071 [6–10 marks]
  4. Confined aquifer steady flow (variable thickness) — Asked in 2072 Ashwin, 2072 Magh [8 marks]

Marks Each Time Asked:

ExamQuestionsTotal
2078 ChaitraQ3a [3+3+4], Q3b [8 num]18
2073 BhadraQ3a [3+4 num], Q3b [7 num]14
2072 AshwinQ4 [6+4 num], Q5 [8 num]18
2072 MaghQ5a [4], Q7 [8 num]12
2071 BhadraQ3 [6], Q4 [3+3+3 num]15

CHAPTER ROADMAP

  1. Stream lines, equipotential lines, and flow nets
  2. Seepage discharge from a flow net
  3. Water table contour maps and direction of flow
  4. Derivation of the Laplace equation for steady groundwater flow
  5. Steady 1-D flow in an unconfined aquifer (Dupuit theory)
  6. Unconfined aquifer between two water bodies with recharge
  7. Steady flow in a confined aquifer of constant and variable thickness
  8. Flow into horizontal galleries
  9. Past question bank for Chapter 3
  10. Worked numerical solutions
  11. Chapter summary and exam formula sheet

3.1 STREAM LINES, EQUIPOTENTIAL LINES, AND FLOW NETS

PAST EXAM QUESTIONS ADDRESSED IN THIS SECTION:

  1. Define flow net with a neat sketch. State applications of flow net in geotechnical engineering and derive the expression for seepage discharge from a flow net. [3+3+4] (2078 Chaitra, Q3a)
  2. Derive an expression for seepage discharge from a flow net for an isotropic aquifer. [6] (2071 Bhadra, Q3)
DEFINITION: A flow net is a graphical representation of two-dimensional groundwater flow consisting of two families of curves — stream lines (flow lines) and equipotential lines — that intersect each other at right angles in isotropic media.

Key Terminology

  • Stream line (flow line): A line that is everywhere tangent to the velocity vector. Water particles move along stream lines. No flow crosses a stream line (Fetter, 2001).
  • Equipotential line: A line connecting points of equal hydraulic head. If a piezometer is installed at any point along an equipotential line, the water would rise to the same elevation. A line along which the velocity potential function \(\phi\) is constant (Todd & Mays, 2005).
  • Stream function (\(\Psi\)): A scalar function whose contour lines define stream lines. In a 2-D isotropic aquifer, streamlines are curves where \(\Psi\) is constant. The distance between two streamlines represents a constant discharge.
  • Velocity potential (\(\phi\)): Defined such that the velocity components are given by \(u = -\partial\phi/\partial x\), \(v = -\partial\phi/\partial y\). For groundwater flow, \(\phi = Kh\) (Raghunath, 2006).
  • Flow channel: The zone between two adjacent stream lines through which a constant fraction of the total flow passes.
  • Potential drop: The difference in head between two successive equipotential lines.
Stream Function (ψ) and Velocity Potential (φ) Stream Function ψ ψ = constant along streamlines Flow between two streamlines: Δq = ψ₂ − ψ₁ vx = ∂ψ/∂y vy = −∂ψ/∂x Exists for any 2D flow Velocity Potential φ φ = constant on equipotentials Head drop between two lines: Δh = (φ₂ − φ₁)/K vx = −∂φ/∂x vy = −∂φ/∂y Exists only for irrotational flow ψ-lines ⊥ φ-lines → Flow Net
Figure 3.0b: Stream function and velocity potential — their contours form the flow net when plotted together.

Orthogonality of Stream Lines and Equipotential Lines

In an isotropic medium, stream lines and equipotential lines always intersect at right angles (90°). This can be proven mathematically (Bear, 1972):

For a streamline, \(\Psi(x, y) = \text{constant}\), so \(d\Psi = 0\):

$$ d\Psi = \frac{\partial\Psi}{\partial x}\,dx + \frac{\partial\Psi}{\partial y}\,dy = 0 \implies \left(\frac{dy}{dx}\right)_\Psi = -\frac{\partial\Psi/\partial x}{\partial\Psi/\partial y} $$

For an equipotential line, \(\phi(x, y) = \text{constant}\), so \(d\phi = 0\):

$$ d\phi = \frac{\partial\phi}{\partial x}\,dx + \frac{\partial\phi}{\partial y}\,dy = 0 \implies \left(\frac{dy}{dx}\right)_\phi = -\frac{\partial\phi/\partial x}{\partial\phi/\partial y} $$

The product of the two slopes equals \(-1\), confirming perpendicularity:

$$ \boxed{\text{Slope of equipotential line} \times \text{Slope of stream function} = -1} $$
DAM h₁ h₂ Impervious bedrock H LEGEND Stream line Equipotential Intersect at 90° Complete flow net: stream lines and equipotential lines intersect at 90° Each curvilinear square carries equal discharge Δq Total seepage: Q = K H (Nf / Nd)
Figure 3.1: A flow net beneath a dam showing stream lines (solid) and equipotential lines (dashed) intersecting at right angles.

Properties of a Flow Net

  1. Stream lines and equipotential lines are orthogonal (perpendicular) in isotropic media.
  2. The head drop between successive equipotential lines is constant: \(\Delta h = H / N_d\).
  3. The discharge through each flow channel is the same: \(\Delta q = K \Delta h \cdot b/a\).
  4. For a square flow net (\(a = b\)), the cell dimensions are equal, simplifying computation.
  5. Neither stream lines nor equipotential lines can intersect themselves.
  6. Impervious boundaries are stream lines; constant-head boundaries are equipotential lines.

Applications of Flow Nets in Engineering

  1. Determination of seepage discharge beneath dams, sheet piles, and cofferdams.
  2. Estimation of uplift pressure on the base of hydraulic structures.
  3. Assessment of exit gradient at the downstream toe to check piping safety.
  4. Computation of pore-water pressure distribution for slope stability analysis.
  5. Analysis of dewatering and well-field problems.
  6. Determination of head at any point: \(\Delta h = H/N_d\) gives the head drop between two consecutive equipotential lines, so head at any point can be computed (Todd & Mays, 2005).
Seepage Discharge from a Flow Net Flow Net Parameters Δs≈Δn Curvilinear square Nf = number of flow channels Nd = number of potential drops H = total head difference K = hydraulic conductivity Seepage Formula q = KH(Nf/Nd) per unit width (m³/s per m) For total: Q = q × width Δh per drop = H/Nd | Δq per channel = KΔh = KH/Nd
Figure 3.2b: Seepage discharge from a flow net — total seepage equals K × H × (N_f/N_d) per unit width of the structure.

Construction of a Flow Net

Flow nets are constructed by trial-and-error sketching following these steps (Cedergren, 1989; Das, 2013):

Flow Net Construction — Essential Rules Rule 1 Streamlines ⊥ Equipotential lines (90° intersections) Rule 2 Elements should be curvilinear squares (Δs ≈ Δn) Rule 3 Impervious boundary = streamline No flow crosses it Rule 4 Constant head boundary = equipotential line Equal head along it Rule 5 Water table (unconfined) = top streamline Pressure = atmospheric q = KH(Nf/Nd) where Nf = flow channels, Nd = potential drops
Figure 3.1b: Five essential rules for constructing a valid flow net — orthogonality, curvilinear squares, and boundary conditions.
  1. Identify boundary conditions: Impervious boundaries are stream lines. Constant-head boundaries (e.g., reservoir surface) are equipotential lines. Note the starting directions of flow and equipotential lines at boundaries.
  2. Draw trial flow lines (or equipotential lines) consistent with the boundary conditions.
  3. Sketch the other family of lines ensuring they intersect the first set at 90°.
  4. Adjust to form curvilinear squares: The shapes formed by the intersection should approximate squares — the mean width should be nearly equal to the mean length.
  5. Iterate: Erase and redraw lines until all figures are approximately square. Subdivide as desired for detail and accuracy.

3.2 SEEPAGE DISCHARGE FROM A FLOW NET

PAST EXAM QUESTIONS ADDRESSED IN THIS SECTION:

  1. Derive the expression for seepage discharge from flow net. [Part of Q3a] (2078 Chaitra)
  2. Derive an expression for seepage discharge from a flow net for an isotropic aquifer. [6] (2071 Bhadra, Q3)
a b h h − Δh Δq DERIVATION By Darcy's law for one cell: Δq = K · (Δh / a) · b · 1 For square flow net (a = b): Δq = K · Δh Total seepage per unit width: Q = K H (Nf / Nd)
Figure 3.2: Derivation of seepage from a single curvilinear square of a flow net.

Consider a single cell of the flow net with length \(a\) along the flow direction and width \(b\) perpendicular to flow. By Darcy's law the discharge per unit thickness through this cell is:

$$ \Delta q = K \frac{\Delta h}{a} \cdot b \cdot 1 $$

For a square flow net, \(a = b\), so \(\Delta q = K \Delta h\). The total head loss is \(H\), divided into \(N_d\) drops giving \(\Delta h = H/N_d\). With \(N_f\) flow channels:

$$ \boxed{Q = K H \frac{N_f}{N_d}} $$
Sheet pile H₁ H₂ Impervious base Nf = 2 flow channels (each side) Nd = 4 equipotential drops (each side) Q = K(H₁ − H₂) × Nf/Nd
Figure 3.3: Flow net around a sheet pile showing stream lines (solid), equipotential lines (dashed), and the seepage formula.

ISOTROPIC CASE

\(Q = K H \frac{N_f}{N_d}\), where \(K\) is the uniform hydraulic conductivity.

ANISOTROPIC CASE

Transform domain: \(x' = x\sqrt{K_z/K_x}\). Then \(Q = H \frac{N_f}{N_d}\sqrt{K_x K_z}\).

Example (Neyveli Lignite Project — Raghunath, 2006, Ex. 4.15): A flow net is drawn for flow through a confined aquifer at the Neyveli Lignite mines. The flow net has 30 flow channels and the head drop is 6 m spread over 30 equipotential drops. The aquifer thickness is 30 m. Calculate the transmissivity and hydraulic conductivity.

Solution:

For a plan-view flow net in a confined aquifer of thickness \(b\), the total seepage is:

$$ Q = T \cdot \Delta H \cdot \frac{N_f}{N_d} = T \times 6 \times \frac{30}{30} = 6T $$

Given \(Q = 7200\) m³/day per unit width (from field data):

$$ T = \frac{Q}{6} = \frac{7200}{6} = 1200 \text{ m}^2/\text{day} $$
$$ K = \frac{T}{b} = \frac{1200}{30} = 40 \text{ m/day} $$
SymbolMeaning
\(Q\)Total seepage discharge per unit width of structure (m³/s per m)
\(K\)Hydraulic conductivity (m/s or m/day)
\(H\)Total head loss = upstream head − downstream head
\(N_f\)Number of flow channels (spaces between stream lines)
\(N_d\)Number of equipotential drops

3.3 WATER TABLE CONTOUR MAPS AND DIRECTION OF FLOW

PAST EXAM QUESTIONS ADDRESSED IN THIS SECTION:

  1. State uses of water-table contour. Prepare a contour map and show flow direction from four observation wells. [3+4] (2073 Bhadra, Q3a)
  2. Explain how water table contour maps are prepared, state their uses, and prepare a groundwater contour map. [8] (2072 Ashwin, Q4)
  3. Explain how a water table contour map is prepared. [4] (2072 Magh, Q5a)
DEFINITION: A water table contour map is a plan view showing lines of equal water-table elevation. It is the groundwater equivalent of a topographic contour map.

Piezometric Head

The piezometric head (or hydraulic head) at a specific point is defined as the sum of the elevation head and pressure head (Freeze & Cherry, 1979):

$$ h = z + \frac{P}{\gamma} $$

where \(z\) = elevation head above datum, \(P\) = pore water pressure, and \(\gamma\) = unit weight of water. It is measured using piezometers (observation wells) penetrating the aquifer and is the driving potential for groundwater movement.

Hydraulic Gradient

The hydraulic gradient is the change in hydraulic head per unit distance in the direction of maximum decrease in head (Fetter, 2001):

$$ i = \frac{dh}{dl} $$

It is directly proportional to the Darcy velocity \(v = -Ki\). Steeper gradients imply faster flow velocities.

Water Table

The water table is the surface in an unconfined aquifer where pore water pressure equals atmospheric pressure (Freeze & Cherry, 1979). It marks the upper boundary of the saturated zone. The water table elevation is influenced by topography, recharge rates, extraction (pumping), and seasonal variations.

How a Water Table Contour Map Is Prepared

  1. Install observation wells (piezometers) at multiple locations across the study area.
  2. Measure static water level in each well at a common reference time.
  3. Convert water levels to elevation above a common datum (e.g., mean sea level).
  4. Plot the well locations on a scaled plan and annotate each with its water-table elevation.
  5. Interpolate contour lines of equal head between measurement points.
76 m Well D 66 m Well C 56 m Well A 51 m Well B 55 m 60 m 65 m 70 m 75 m Flow direction NOTES Contour interval: 5 m Flow: high h → low h ⊥ to contour lines Flow is perpendicular to contours in the direction of decreasing head
Figure 3.4: Water table contour map from four observation wells (51, 56, 66, 76 m) with 5 m contour interval, showing groundwater flow direction.

Uses of Water Table Contour Maps

  1. Determine the direction of groundwater flow at any point.
  2. Estimate the hydraulic gradient from the spacing of contour lines.
  3. Identify recharge and discharge zones.
  4. Assist in well-field planning.
  5. Evaluate potential path of contaminant plumes.
  6. Calculate seepage flow through any cross section using Darcy's law.
Applications of Water Table Contour Maps WT Contour Map Flow direction ⊥ to contours Hydraulic gradient (i) GW-SW interaction Recharge / discharge areas Well siting / pollution track Gaining stream: contours curve upstream (GW → river) Losing stream: contours curve downstream (river → GW) Closely spaced contours → steep gradient → high velocity
Figure 3.3c: Applications of water table contour maps — flow direction, gradient estimation, GW–SW interaction, and contamination tracking.
70 m 65 m 60 m 65 m (interpolated) 65 m contour Flow direction
Figure 3.5: Three-well method for finding groundwater flow direction from a triangular network of piezometers.

Three-Well Triangulation Procedure (Todd & Mays, 2005)

Given three observation wells, the flow direction and hydraulic gradient can be determined using the following steps:

Three-Well Triangulation — Finding Flow Direction A h = 48 m B h = 42 m C h = 36 m D h = 42 m (interpolated) Equipotential h = 42 m Flow direction ⊥ to equipotential ① Find mid-head well (B, 42 m) ② Interpolate D on AC at h=42 ③ BD = equipotential ④ Flow ⊥ BD
Figure 3.3b: Three-well triangulation — interpolate an equal-head point, draw the equipotential, and find the flow direction perpendicular to it.
  1. Step 1: Identify the well with the intermediate water level.
  2. Step 2: Along the straight line between the wells with the highest and lowest head, locate the point with the same head as the intermediate well (by linear interpolation).
  3. Step 3: Draw a straight line between the intermediate well and the point identified in Step 2. This is a segment of the equipotential line along which the head equals the intermediate value.
  4. Step 4: Draw a line perpendicular to the equipotential line passing through the well with the lowest head. This perpendicular indicates the direction of groundwater flow. The hydraulic gradient is the slope of this perpendicular line.

Analysis of Water Table Maps (Raghunath, 2006)

If the flow rates across successive water table contours between two limiting flow lines are known, the following deductions can be made:

  • Transmissivity: From Darcy's law, \(q = T \cdot i \cdot w\), where \(i = \Delta h / l\) (head difference over distance between contours) and \(w\) = width between limiting flow lines (scaled from map). By computing \(q\) at different sections, the spatial distribution of \(T\) can be estimated.
  • Storage coefficient: If there is no leakage, \(q_2 - q_1 = S \cdot \Delta h_t \cdot A_s\), where \(\Delta h_t\) = average rate of water level change, and \(A_s\) = area between the two limiting flow lines and successive contours.
  • Vertical leakage: If vertical leakage occurs through an aquitard, \(P \cdot A_s = (q_2 - q_1) + S \cdot \Delta h_t \cdot A_s\), where \(P\) = rate of vertical leakage per unit area.

Such water table contour maps with flow lines provide useful data for locating new wells, identifying the most favourable permeability zones, and understanding regional groundwater movement patterns. Portions with wide contour spacing (flat gradients) indicate higher permeabilities (Raghunath, 2006).

3.4 DERIVATION OF LAPLACE EQUATION FOR STEADY GROUNDWATER FLOW

OBJECTIVE: The governing PDE for steady-state groundwater flow in a homogeneous isotropic aquifer is the Laplace equation, providing the theoretical foundation for flow-net analysis.

Potential Flow Theory

Potential flow theory assumes that the fluid is incompressible, the flow is irrotational, and the medium is homogeneous and isotropic. In groundwater hydrology, water flows because of differences in hydraulic head. Darcy's law relates velocity to hydraulic head:

$$ \vec{v} = -K\nabla h $$

Since velocity is the gradient of a scalar potential (head), groundwater flow is a potential flow. From potential flow theory: equipotential lines are lines of constant head, flow lines represent the direction of groundwater flow, and they intersect at right angles (90°).

Derivation via Conservation of Mass and Darcy's Law

Under steady-state conditions, inflow equals outflow (no storage change). Using Darcy's law and conservation of mass:

Laplace Equation — Derivation from Continuity Control volume dx × dy × dz qx qx+dx qy qy+dy Continuity (steady, incompressible): ∂qx/∂x + ∂qy/∂y + ∂qz/∂z = 0 Substitute Darcy (q = −K∇h): ∂²h/∂x² + ∂²h/∂y² + ∂²h/∂z² = 0
Figure 3.4b: Laplace equation derived by combining continuity (mass balance) with Darcy's law — governs steady-state groundwater flow in homogeneous isotropic media.
$$ \frac{\partial q_x}{\partial x} + \frac{\partial q_y}{\partial y} + \frac{\partial q_z}{\partial z} = 0 $$

Substituting Darcy's law \(q_i = -K_i \frac{\partial h}{\partial x_i}\):

$$ \frac{\partial}{\partial x}\!\left(K_x \frac{\partial h}{\partial x}\right) + \frac{\partial}{\partial y}\!\left(K_y \frac{\partial h}{\partial y}\right) + \frac{\partial}{\partial z}\!\left(K_z \frac{\partial h}{\partial z}\right) = 0 $$

For a homogeneous isotropic medium (\(K\) = constant):

$$ \boxed{\frac{\partial^2 h}{\partial x^2} + \frac{\partial^2 h}{\partial y^2} + \frac{\partial^2 h}{\partial z^2} = 0 \qquad (\nabla^2 h = 0)} $$
Conservation of Mass ∇ · q = 0 + Darcy's Law q = −K∇h ⇒ General Eqn. ∇·(K∇h) = 0 K = const. Laplace ∇²h = 0 For anisotropic media, the Laplace form requires a coordinate transformation.
Figure 3.6: Derivation pathway from conservation of mass and Darcy's law to the Laplace equation.

Anisotropic Media

When \(K_x \neq K_z\), a coordinate transformation \(x' = x\sqrt{K_z/K_x}\) converts the anisotropic medium into an equivalent isotropic one. The effective conductivity is \(K_e = \sqrt{K_x K_z}\).

Alternative Derivation via Velocity Potential (Raghunath, 2006)

A velocity potential \(\phi\) is a scalar function such that the velocity components are given by:

$$ u = -\frac{\partial\phi}{\partial x}, \quad v = -\frac{\partial\phi}{\partial y}, \quad w = -\frac{\partial\phi}{\partial z} $$

From Darcy's law, \(V = -K\,dh/ds\), and comparing with \(V = -d\phi/ds\):

$$ \phi = Kh + C $$

where \(C\) is a constant (usually taken as zero). Since groundwater is incompressible, the continuity equation for steady flow gives:

$$ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0 $$

Substituting the velocity potential:

$$ \frac{\partial^2\phi}{\partial x^2} + \frac{\partial^2\phi}{\partial y^2} + \frac{\partial^2\phi}{\partial z^2} = 0 \qquad (\nabla^2\phi = 0) $$

Since \(\phi = Kh\) for a homogeneous isotropic medium, this directly gives \(\nabla^2 h = 0\), the Laplace equation.

Anisotropic Laplace Equation

For an anisotropic formation with principal permeabilities \(K_x\), \(K_y\), and \(K_z\), the steady-state flow equation becomes (Raghunath, 2006):

$$ K_x\frac{\partial^2 h}{\partial x^2} + K_y\frac{\partial^2 h}{\partial y^2} + K_z\frac{\partial^2 h}{\partial z^2} = 0 $$

Using the coordinate transformation \(X = x/\sqrt{K_x}\), \(Y = y/\sqrt{K_y}\), \(Z = z/\sqrt{K_z}\), this reduces to:

$$ \frac{\partial^2 h}{\partial X^2} + \frac{\partial^2 h}{\partial Y^2} + \frac{\partial^2 h}{\partial Z^2} = 0 $$

Thus, the Laplace equation is satisfied in the transformed domain, and a flow net can be constructed.

Assumptions and Boundary Conditions

  • Steady state: No change in head over time.
  • Homogeneous: Hydraulic conductivity \(K\) is independent of position.
  • Isotropic: \(K\) is the same in all directions (\(K_x = K_y = K_z\)).
  • Incompressible: Fluid and matrix density are constant.
  • Conservative: No internal sources or sinks (no recharge/pumping).

Boundary conditions are essential to solve the PDE (Wang & Anderson, 1982):

  • Dirichlet (first-type): Constant head (e.g., a lake or river).
  • Neumann (second-type): Constant flux (e.g., an impervious rock layer where flux = 0).

3.5 STEADY ONE-DIMENSIONAL FLOW IN AN UNCONFINED AQUIFER (DUPUIT THEORY)

PAST EXAM QUESTIONS ADDRESSED IN THIS SECTION:

  1. Two rivers are separated by a homogeneous unconfined aquifer. Compute seepage flow per unit river length and the phreatic line. [7] (2073 Bhadra, Q3b)
DUPUIT ASSUMPTIONS: (1) Hydraulic gradient equals the slope of the water table. (2) Flow is essentially horizontal — all stream lines are horizontal and equipotential lines are vertical.
Impervious bedrock h₁ h₂ Phreatic surface (water table) h(x) h₁ h₂ L x Q →
Figure 3.7: Steady unconfined flow between two water bodies on an impervious base (Dupuit parabola).

Applying Darcy's law with Dupuit assumption (\(q = -Kh\,dh/dx\)) and integrating:

$$ Q = \frac{K}{2L}(h_1^2 - h_2^2) \qquad \text{(Dupuit equation)} $$

The water-table profile (phreatic surface equation) is:

$$ h^2 = h_1^2 - \frac{h_1^2 - h_2^2}{L}\,x $$

This indicates that for a given discharge, the water table is parabolic in form (Dupuit parabola).

Limitations of the Dupuit Equation and the Seepage Face

There is a difference between the water level calculated using the Dupuit equation and the actual water level. In reality, the actual water table sits higher than what the Dupuit equation predicts. This departure is most noticeable near the well or discharge boundary. The reasons are:

Dupuit Assumption vs Actual Flow — Seepage Face Impervious Base Dam h₁ Dupuit (parabolic) Actual phreatic surface Seepage Face Dupuit assumes horizontal flow (∂h/∂z ≈ 0) — cannot predict the seepage face at exit
Figure 3.5b: The Dupuit assumption predicts a parabolic water table but cannot capture the seepage face where flow exits the downstream boundary.
  • Vertical flow component: The Dupuit equation assumes water flows only horizontally. In reality, near a well or discharge boundary, water also moves downward, requiring a greater saturated thickness to convey the same discharge.
  • Slope error: The assumption that \(dh/ds \approx dh/dx\) breaks down near wells where the water table slope is steep.
  • Seepage face: The water table cannot connect directly to the water surface inside the well. It curves and meets the well wall tangentially, forming a seepage face — a zone where water emerges from the aquifer above the well water level.

Despite these errors in predicting the water table shape, the Dupuit equation remains accurate for calculating the total flow rate (q) and the aquifer's hydraulic conductivity (K) (Raghunath, 2006).

Vertical Velocity Component

From the 2-D continuity equation \(\partial u/\partial x + \partial w/\partial z = 0\) and the Dupuit velocity \(u = q/h\), the vertical velocity component can be derived. Since the horizontal velocity varies with \(x\) (through the changing \(h\)), this creates a vertical component by continuity:

$$ \frac{\partial u}{\partial x} = -\frac{q}{h^2}\frac{dh}{dx} = \frac{q^2}{Kh^3} $$

The vertical velocity \(w\) increases linearly with elevation \(z\) (zero at the impervious base, maximum at the water table):

$$ w = -\frac{q^2}{Kh^3} \cdot z $$

This shows that the vertical velocity is zero at the base and increases upward, being greatest at the water table.

3.6 UNCONFINED AQUIFER BETWEEN TWO WATER BODIES WITH RECHARGE

PAST EXAM QUESTIONS ADDRESSED IN THIS SECTION:

  1. Two irrigation canals with recharge — determine q1, q2, phreatic line, divide. [8] (2078 Chaitra, Q3b)
  2. Rain on a land strip between canals — q1, q2, divide, and no-rain case. [8] (2071 Bhadra, Q4)

When uniform recharge \(p\) is applied, the governing 1-D Dupuit equation becomes:

$$ \frac{d}{dx}\!\left(Kh\frac{dh}{dx}\right) + p = 0 $$
Rainfall, p (mm/hr) Ground h₁ h₂ Water table with recharge Water divide (d) q₁ q₂ Impervious stratum h₁ h₂ L d
Figure 3.8: Unconfined aquifer between two canals with uniform recharge, showing water table divide and seepage components q1 and q2.

Phreatic Surface Equation

$$ h^2 = h_1^2 - \frac{h_1^2 - h_2^2}{L}\,x + \frac{p}{K}\,x(L - x) $$

Seepage to Each Canal (per unit length)

$$ q_1 = \frac{pL}{2} - \frac{K(h_1^2 - h_2^2)}{2L} \qquad \text{(into the deeper canal, at } x=0\text{)} $$
$$ q_2 = \frac{pL}{2} + \frac{K(h_1^2 - h_2^2)}{2L} \qquad \text{(into the shallower canal, at } x=L\text{)} $$

Verify: \(q_1 + q_2 = pL\) (total recharge on the strip).

Location of the Water Table Divide

$$ d = \frac{L}{2} - \frac{K(h_1^2 - h_2^2)}{2pL} $$
Water Table Divide with Uniform Recharge Impervious base h₁ h₂ Water table (parabolic) Recharge W (uniform) Water divide (d) q₁→ →q₂ d = L/2 − K(h₁²−h₂²)/(2WL)
Figure 3.6b: Water table divide — the point where flow splits toward the two canals, shifted from the midpoint when h₁ ≠ h₂.
Distance x (from left canal →) Discharge q(x) q = 0 d (water divide) +q (toward left canal) −q (toward right canal) q₁ q₂
Figure 3.9: Darcy flux varies linearly with distance. The zero-crossing locates the water table divide.

SPECIAL CASE: NO RECHARGE (p = 0)

Equations reduce to the simple Dupuit case:

$$ q = \frac{K(h_1^2 - h_2^2)}{2L} $$

All flow moves toward the canal with lower head. No water divide exists.

3.7 STEADY FLOW IN A CONFINED AQUIFER

PAST EXAM QUESTIONS ADDRESSED IN THIS SECTION:

  1. Determine flow in the confined aquifer (variable thickness). [8] (2072 Ashwin, Q5)
  2. Determine flow in the confined aquifer (variable thickness). [8] (2072 Magh, Q7)

A. Constant Thickness

$$ Q = T \frac{h_1 - h_2}{L} = Kb \frac{h_1 - h_2}{L} $$
Steady Flow in Confined Aquifer — Linear Head Confined aquifer (K, b) Confining layer (aquitard) h₁ h₂ Piezometric surface (linear) Flow Q → b h(x) = h₁ − (h₁−h₂)x/L (linear head distribution) Q = Kb(h₁−h₂)/L = T(h₁−h₂)/L per unit width
Figure 3.7c: Steady flow in a confined aquifer — head varies linearly with distance, and discharge per unit width equals T(h₁−h₂)/L.
Confining layer (impervious) Confined Aquifer (thickness b) Impervious base h₁ h₂ Piezometric surface Q b L
Figure 3.10: Steady one-dimensional flow in a confined aquifer of constant thickness b.

B. Variable Thickness (Wedge-Shaped Aquifer)

When aquifer thickness varies linearly from \(b_1\) at \(x = 0\) to \(b_2\) at \(x = L\), applying Darcy's law and integrating:

$$ Q = K \frac{(h_1 - h_2)(b_2 - b_1)}{L \ln(b_2/b_1)} $$
b₁ b₂ h₁ h₂ Confined Aquifer (variable thickness) Q L
Figure 3.11: Steady flow in a confined aquifer of variable (wedge-shaped) thickness.

3.7b FLOW INTO HORIZONTAL GALLERIES

A horizontal gallery (or infiltration gallery) is a horizontal conduit laid beneath the water table to collect groundwater by gravity flow. The analysis is analogous to the Dupuit case for unconfined flow between two boundaries.

Discharge Equation

Consider a horizontal gallery of length \(l\) (perpendicular to flow) receiving flow from both sides. The water table stands at height \(H\) at a distance \(L\) from the gallery, and the water level in the gallery is \(h_1\). Applying the Dupuit assumptions, the discharge per unit length from one side is:

Infiltration Gallery — Cross Section Impervious layer Water table Ground surface Gallery Seepage Seepage H s q = K(H² − h²)/(2L) per unit length per side
Figure 3.7b: Infiltration gallery — a horizontal perforated pipe below the water table collects water by gravity flow from both sides.
$$ q = \frac{K}{2L}(H^2 - h_1^2) $$

The total discharge collected from both sides over gallery length \(l\) is:

$$ Q = 2ql = \frac{Kl}{L}(H^2 - h_1^2) $$

Phreatic Surface Profile

The water table (depression curve) between the undisturbed level and the gallery follows a parabolic profile:

$$ h^2 = h_1^2 + \frac{H^2 - h_1^2}{L}\,x $$

where \(x\) is measured from the gallery towards the undisturbed water table.

Worked Example

Problem: A horizontal gallery of length \(l\) taps an unconfined aquifer with \(K = 60\) m/day. The undisturbed saturated thickness is \(H = 7\) m at a distance \(L = 400\) m from the gallery where the water depth is \(h_1 = 2\) m. Find the discharge per unit length from one side.

Solution:

$$ q = \frac{K}{2L}(H^2 - h_1^2) = \frac{60}{2 \times 400}(7^2 - 2^2) = \frac{60}{800}(49 - 4) = 0.075 \times 45 $$
$$ q = 3.375 \text{ m}^3/\text{day per metre length} $$

3.8 PAST QUESTION BANK FOR CHAPTER 3

A. Theory and Definition Questions

2078 ChaitraQ3(a)Theory + Derivation

Define flow net with a neat sketch. State applications of flow net in geotechnical engineering and derive the expression for seepage discharge from a flow net. [3+3+4]

2073 BhadraQ3(a)Numerical

State the uses of water table contour. Prepare a water table contour map and show the direction of groundwater flow from four observation wells. [3+4]

Water table elevations at rectangle vertices = 51 m, 56 m, 76 m, 66 m
Contour interval = 5 m
2072 AshwinQ4Theory + Numerical

Explain how water table contour maps are prepared, state their uses, and prepare a groundwater contour map with flow direction from the given figure. [8]

2072 MaghQ5(a)Theory

Explain how a water table contour map is prepared. [4]

2071 BhadraQ3Derivation

Derive an expression for seepage discharge from a flow net for an isotropic aquifer. [6]

B. Numerical and Applied Questions

2078 ChaitraQ3(b)Numerical

Two irrigation canals with recharge — determine q1, q2, phreatic line, water divide. [8]

L = 1.57 km = 1570 m,  Rainfall p = 25 mm/hr
Canal depths h₁ = 12 m, h₂ = 7 m,  K = 12 m/day
2073 BhadraQ3(b)Numerical

Two rivers separated by unconfined aquifer — seepage and phreatic line. [7]

L = 5 km = 5000 m,  h₁ = 20 m, h₂ = 11 m,  K = 18 m/day
2072 AshwinQ5Numerical

Determine flow in variable-thickness confined aquifer. [8]

K = 0.4 m/day,  L = 800 m
b₁ = 8 m, b₂ = 6 m,  h₁ = 10 m, h₂ = 8 m
2072 MaghQ7Numerical

Determine flow in variable-thickness confined aquifer. [8]

L = 950 m,  K = 0.56 m/day
h₁ = 12 m, b₁ = 7 m,  h₂ = 9 m, b₂ = 5 m
2071 BhadraQ4Numerical

Rain on a land strip between canals — q1, q2, divide, and no-rain case. [8]

p = 12 mm/hr,  L = 1200 m
h₁ = 10 m, h₂ = 8 m,  K = 10 m/day

C. Quick Revision Matrix

ThemeYears askedWhat to prepare
Flow net definition and seepage2078, 2071Sketch, definition, derivation of Q = KH(Nf/Nd), applications
Water table contour maps2073, 2072 Ashwin, 2072 MaghPreparation steps, uses, 3-well or 4-well interpolation, flow direction
Unconfined seepage with recharge2078, 2071Phreatic surface equation, q1 and q2, water table divide d
Unconfined seepage without recharge2073Dupuit equation: Q = K(h₁² − h₂²) / 2L
Confined aquifer flow (constant/variable)2072 Ashwin, 2072 MaghQ = Kb(h₁−h₂)/L and variable-thickness form

3.9 WORKED NUMERICAL SOLUTIONS

SOLUTION 1: 2078 Chaitra Q3(b) — Unconfined Aquifer with Recharge

Given: \(L = 1570\) m, \(p = 25\) mm/hr \(= \frac{25}{1000 \times 24} = 0.6\) m/day, \(h_1 = 12\) m, \(h_2 = 7\) m, \(K = 12\) m/day.

Required: \(q_1\), \(q_2\), phreatic surface equation, and water table divide location.

Step 1: Convert rainfall to m/day

$$ p = 25 \text{ mm/hr} = 25 \times \frac{24}{1000} = 0.6 \text{ m/day} $$

Step 2: Seepage into the deeper canal (left, at x = 0)

$$ q_1 = \frac{pL}{2} - \frac{K(h_1^2 - h_2^2)}{2L} $$
$$ q_1 = \frac{0.6 \times 1570}{2} - \frac{12(144 - 49)}{2 \times 1570} = \frac{942}{2} - \frac{12 \times 95}{3140} $$
$$ \boxed{q_1 = 471.0 - 0.363 = 470.64 \text{ m}^2/\text{day per m of canal}} $$

Step 3: Seepage into the shallower canal (right, at x = L)

$$ q_2 = \frac{pL}{2} + \frac{K(h_1^2 - h_2^2)}{2L} = 471.0 + 0.363 $$
$$ \boxed{q_2 = 471.36 \text{ m}^2/\text{day per m of canal}} $$

Check: \(q_1 + q_2 = 470.64 + 471.36 = 942.0 = pL = 0.6 \times 1570\). ✓

Step 4: Phreatic surface equation

$$ h^2 = h_1^2 - \frac{h_1^2 - h_2^2}{L}\,x + \frac{p}{K}\,x(L - x) $$
$$ h^2 = 144 - \frac{95}{1570}\,x + \frac{0.6}{12}\,x(1570 - x) $$
$$ \boxed{h^2 = 144 - 0.0605\,x + 0.05\,x(1570 - x)} $$

Expanding: \(h^2 = 144 + 78.44\,x - 0.05\,x^2\).

Step 5: Water table divide

$$ d = \frac{L}{2} - \frac{K(h_1^2 - h_2^2)}{2pL} = \frac{1570}{2} - \frac{12 \times 95}{2 \times 0.6 \times 1570} $$
$$ d = 785 - \frac{1140}{1884} = 785 - 0.605 $$
$$ \boxed{d = 784.4 \text{ m from the deeper canal}} $$

The water divide is nearly at the midpoint because the intense rainfall dominates over the Dupuit gradient term.

SOLUTION 2: 2073 Bhadra Q3(b) — Dupuit Equation (No Recharge)

Given: \(L = 5000\) m, \(h_1 = 20\) m, \(h_2 = 11\) m, \(K = 18\) m/day.

Step 1: Seepage discharge per unit river length

$$ Q = \frac{K(h_1^2 - h_2^2)}{2L} = \frac{18(400 - 121)}{2 \times 5000} = \frac{18 \times 279}{10000} $$
$$ \boxed{Q = 0.502 \text{ m}^2/\text{day per m of river length}} $$

Step 2: Phreatic surface equation

$$ h^2 = h_1^2 - \frac{h_1^2 - h_2^2}{L}\,x = 400 - \frac{279}{5000}\,x $$
$$ \boxed{h^2 = 400 - 0.0558\,x} $$

Or equivalently: \(h = \sqrt{400 - 0.0558\,x}\). This is a parabolic profile that drops more steeply near the discharge boundary.

SOLUTION 3: 2072 Ashwin Q5 — Confined Aquifer (Variable Thickness)

Given: \(K = 0.4\) m/day, \(L = 800\) m, \(b_1 = 8\) m, \(b_2 = 6\) m, \(h_1 = 10\) m, \(h_2 = 8\) m.

Step 1: Identify the formula for variable (wedge-shaped) thickness

Since the thickness changes linearly from \(b_1\) to \(b_2\), we use:

$$ Q = K \frac{(h_1 - h_2)(b_2 - b_1)}{L \ln(b_2 / b_1)} $$

Step 2: Substitute values

$$ Q = 0.4 \times \frac{(10 - 8)(6 - 8)}{800 \times \ln(6/8)} = 0.4 \times \frac{2 \times (-2)}{800 \times \ln(0.75)} $$
$$ Q = 0.4 \times \frac{-4}{800 \times (-0.2877)} = 0.4 \times \frac{-4}{-230.16} $$
$$ Q = 0.4 \times 0.01738 $$
$$ \boxed{Q = 6.95 \times 10^{-3} \text{ m}^2/\text{day per m width}} $$

Note: The positive result confirms flow is from \(h_1 = 10\) m toward \(h_2 = 8\) m (in the direction of decreasing head).

SOLUTION 4: 2072 Magh Q7 — Confined Aquifer (Variable Thickness)

Given: \(L = 950\) m, \(K = 0.56\) m/day, \(h_1 = 12\) m, \(b_1 = 7\) m, \(h_2 = 9\) m, \(b_2 = 5\) m.

Step 1: Apply the variable-thickness formula

$$ Q = K \frac{(h_1 - h_2)(b_2 - b_1)}{L \ln(b_2/b_1)} $$

Step 2: Substitute

$$ Q = 0.56 \times \frac{(12 - 9)(5 - 7)}{950 \times \ln(5/7)} = 0.56 \times \frac{3 \times (-2)}{950 \times \ln(0.7143)} $$
$$ Q = 0.56 \times \frac{-6}{950 \times (-0.3365)} = 0.56 \times \frac{-6}{-319.68} $$
$$ Q = 0.56 \times 0.01877 $$
$$ \boxed{Q = 1.051 \times 10^{-2} \text{ m}^2/\text{day per m width}} $$

SOLUTION 5: 2071 Bhadra Q4 — Unconfined Aquifer with Recharge

Given: \(p = 12\) mm/hr, \(L = 1200\) m, \(h_1 = 10\) m, \(h_2 = 8\) m, \(K = 10\) m/day.

Step 1: Convert rainfall

$$ p = 12 \text{ mm/hr} = \frac{12 \times 24}{1000} = 0.288 \text{ m/day} $$

Step 2: Seepage into the deeper canal (q1)

$$ q_1 = \frac{pL}{2} - \frac{K(h_1^2 - h_2^2)}{2L} = \frac{0.288 \times 1200}{2} - \frac{10(100 - 64)}{2 \times 1200} $$
$$ q_1 = \frac{345.6}{2} - \frac{360}{2400} = 172.8 - 0.15 $$
$$ \boxed{q_1 = 172.65 \text{ m}^2/\text{day}} $$

Step 3: Seepage into the shallower canal (q2)

$$ q_2 = \frac{pL}{2} + \frac{K(h_1^2 - h_2^2)}{2L} = 172.8 + 0.15 $$
$$ \boxed{q_2 = 172.95 \text{ m}^2/\text{day}} $$

Check: \(q_1 + q_2 = 172.65 + 172.95 = 345.6 = pL = 0.288 \times 1200\). ✓

Step 4: Phreatic surface equation

$$ h^2 = 100 - \frac{36}{1200}\,x + \frac{0.288}{10}\,x(1200 - x) $$
$$ \boxed{h^2 = 100 - 0.03\,x + 0.0288\,x(1200 - x)} $$

Step 5: Water table divide

$$ d = \frac{L}{2} - \frac{K(h_1^2 - h_2^2)}{2pL} = 600 - \frac{10 \times 36}{2 \times 0.288 \times 1200} $$
$$ d = 600 - \frac{360}{691.2} = 600 - 0.521 $$
$$ \boxed{d = 599.5 \text{ m from the deeper canal}} $$

Step 6: Seepage when rainfall is absent (p = 0)

$$ q = \frac{K(h_1^2 - h_2^2)}{2L} = \frac{10 \times 36}{2 \times 1200} $$
$$ \boxed{q = 0.15 \text{ m}^2/\text{day (all flow toward the lower canal)}} $$

Without rainfall, the seepage drops from 345.6 m²/day (total) to just 0.15 m²/day, and there is no water divide.

SOLUTION 6: 2073 Bhadra Q3(a) — Water Table Contour Map

Given: Four wells at rectangle vertices with water-table elevations 51 m, 56 m, 76 m, 66 m. Contour interval = 5 m.

Step 1: Assign well positions

Let the rectangle have corners: A (bottom-left) = 56 m, B (bottom-right) = 51 m, C (top-right) = 66 m, D (top-left) = 76 m.

Step 2: Interpolate contour crossings on each side

  • Side AD (56 m to 76 m): The 60, 65, 70, 75 m contours cross this side. For example, the 60 m contour crosses at \(\frac{60-56}{76-56} = 0.2\) (20% from A).
  • Side BC (51 m to 66 m): The 55, 60, 65 m contours cross. The 55 m contour at \(\frac{55-51}{66-51} = 0.267\) (26.7% from B).
  • Side AB (56 m to 51 m): The 55 m contour crosses at \(\frac{56-55}{56-51} = 0.2\) (20% from A).
  • Side DC (76 m to 66 m): The 70 m contour crosses at \(\frac{76-70}{76-66} = 0.6\) (60% from D).

Step 3: Draw contour lines

Connect interpolated points of equal elevation across the rectangle to draw the 55, 60, 65, 70, and 75 m contour lines (see Figure 3.4 for the completed map).

Step 4: Determine flow direction

Groundwater flows perpendicular to the contour lines in the direction of decreasing head. From the map, the general flow direction is from the 76 m corner (highest) toward the 51 m corner (lowest), roughly along the diagonal of the rectangle.

Step 5: Estimate hydraulic gradient

The gradient can be estimated between any two contours:

$$ i = \frac{\Delta h}{\text{distance between contours}} $$

Closely spaced contours indicate steeper gradients and faster flow; widely spaced contours indicate gentle gradients.

CHAPTER SUMMARY AND EXAM FORMULA SHEET

ConceptKey relation / statement
Flow netOrthogonal stream lines and equipotential lines in 2-D steady flow
Seepage from flow net\( Q = KH \frac{N_f}{N_d} \) (isotropic); \( Q = H\frac{N_f}{N_d}\sqrt{K_x K_z} \) (anisotropic)
Laplace equation\(\nabla^2 h = 0\) for steady, homogeneous, isotropic flow
Dupuit equation (no recharge)\( Q = \frac{K}{2L}(h_1^2 - h_2^2) \)
Phreatic surface (no recharge)\( h^2 = h_1^2 - \frac{h_1^2 - h_2^2}{L}\,x \)
Phreatic surface (with recharge \(p\))\( h^2 = h_1^2 - \frac{h_1^2 - h_2^2}{L}\,x + \frac{p}{K}\,x(L-x) \)
Seepage to left canal\( q_1 = \frac{pL}{2} - \frac{K(h_1^2-h_2^2)}{2L} \)
Seepage to right canal\( q_2 = \frac{pL}{2} + \frac{K(h_1^2-h_2^2)}{2L} \)
Water table divide\( d = \frac{L}{2} - \frac{K(h_1^2-h_2^2)}{2pL} \)
Confined flow (constant thickness)\( Q = Kb\frac{h_1-h_2}{L} = T\frac{h_1-h_2}{L} \)
Confined flow (variable thickness)\( Q = K\frac{(h_1-h_2)(b_2-b_1)}{L\ln(b_2/b_1)} \)
Anisotropic transformation\( x' = x\sqrt{K_z/K_x} \), effective \( K_e = \sqrt{K_x K_z} \)

LAST-MINUTE EXAM FOCUS

  1. Sketch and label a flow net including boundary conditions.
  2. Memorize \(Q = KH N_f/N_d\) and know how to count \(N_f\) and \(N_d\).
  3. Know the steps to prepare a water table contour map from piezometer data.
  4. Derive the Laplace equation from continuity + Darcy's law.
  5. Practice the unconfined aquifer with recharge problem — most frequently asked numerical.
  6. For confined problems, check carefully whether the aquifer is constant or variable thickness.
  7. All 6 numerical solutions above cover the complete set of Chapter 3 past question patterns.