CHAPTER 2 — NUMERICALS
Fundamentals of Groundwater Motion

3 Problems — Darcy’s Law Validity & Seepage Velocity

KEY FORMULAE — CHAPTER 2

1. Darcy’s Law (valid for laminar flow through porous media):

$$ q = K\,i = -K\,\frac{dh}{dx} \quad \text{(Darcy flux / specific discharge, m/s)} $$

2. Radial Darcy flux toward a fully-penetrating well in a confined aquifer:

$$ q_r = \frac{Q}{2\pi\,r\,b} $$

3. Reynolds number for porous media:

$$ Re = \frac{q \cdot d_m}{\nu} $$

4. Minimum valid radius (combining formulae 2 & 3 at \(Re = Re_{cr}\)):

$$ \boxed{r_{min} = \frac{Q \cdot d_m}{2\pi\,b\,\nu\,Re_{cr}}} $$

5. Seepage (pore) velocity:

$$ v_s = \frac{q}{n_e} $$
SymbolTypical RangeSI Unit
\(K\)10−7–10−1 m/sm/s or m/day
\(d_m\)0.01–5 mmm
\(\nu\) (water, 20°C)1.003×10−6 m²/sm²/s
\(Re_{cr}\)1–10 (usually 1–6)—
\(n_e\)0.10–0.35dimensionless

DERIVATION HINT — rmin formula

Start from the radial Darcy flux: \(q = Q/(2\pi r b)\). Substitute into \(Re = q \cdot d_m / \nu\). Set \(Re = Re_{cr}\) and solve for \(r\). This gives the minimum radius beyond which the flow is laminar and Darcy’s law is valid. Inside this radius, inertial effects dominate and Forchheimer or Ergun equations must be used instead.

Porous Medium (REV) q = K × i (Darcy flux) WELL rmin Radial flow: q = Q/(2πrb)
Left: Darcy’s law for uniform flow. Right: Radial flow to a pumping well showing the critical radius rmin.

WHY REYNOLDS NUMBER IN POROUS MEDIA?

In pipe flow, turbulence begins at \(Re \approx 2300\). But in porous media, the tortuous pore channels are much smaller, so the transition occurs at \(Re = 1\text{–}10\). At very high velocities (near a pumping well), inertial forces exceed viscous forces and Darcy’s linear law breaks down. The Reynolds number quantifies this ratio:

$$ Re = \frac{\text{Inertial force}}{\text{Viscous force}} = \frac{q \cdot d_m}{\nu} $$

When \(Re > Re_{cr}\): Flow is non-Darcian. We must use the Forchheimer equation: \(-dh/dx = aq + bq^2\), which adds a quadratic (inertial) term. This mainly matters inside the non-Darcy zone near the well.

SOLUTION METHODOLOGY — Darcy Validity Problems

  1. Identify the flow geometry (radial to well → use \(q = Q/2\pi rb\))
  2. Write the Reynolds number in terms of distance \(r\)
  3. Set \(Re = Re_{cr}\) (given, usually 1–10) and solve for \(r_{min}\)
  4. State conclusion: Darcy valid for \(r > r_{min}\), non-Darcy for \(r < r_{min}\)
Problem 1: Darcy Validity Domain Around a Pumping Well 2078 Chaitra Q2(b) — 6 marks

Question: A fully penetrating well in a confined aquifer of 18 m thick is pumped at 0.3 m³/s. Average grain diameter is 1 mm. Determine the valid domain for which Darcy’s law is applicable. Assume \(Re_{cr} = 8\) and \(\nu = 1\) centistoke.

GIVEN DATA

Aquifer thickness, \(b\)18 m
Pumping rate, \(Q\)0.3 m³/s
Grain diameter, \(d_m\)1 mm = 0.001 m
Critical Reynolds number8
Kinematic viscosity, \(\nu\)\(1 \times 10^{-6}\) m²/s
WELL P Q = 0.3 m³/s b=18m rmin = ? DARCY ZONE (valid) Re ≤ 8 → Laminar NON-DARCY ZONE Re > 8 → Turbulent near well
Cross-section: radial flow to a well in a confined aquifer. The non-Darcy zone exists close to the well where velocity is high.

SOLUTION

Step 1 — Darcy flux at distance r. For a fully penetrating well in a confined aquifer, all flow converges radially through a cylindrical surface of area \(2\pi r b\):

$$ q_r = \frac{Q}{2\pi\,r\,b} $$

This is the volume of water crossing unit area per unit time at distance r from the well centre. As r decreases (closer to the well), qr increases because the same total discharge Q must pass through a smaller cylindrical surface.

Step 2 — Reynolds number in terms of r. Substituting the Darcy flux into the Re definition:

$$ Re = \frac{q_r \cdot d_m}{\nu} = \frac{Q \cdot d_m}{2\pi\,r\,b\,\nu} $$

Note that Re is inversely proportional to r — it increases as we approach the well.

Step 3 — Set \(Re = Re_{cr}\) and solve for \(r_{min}\). The critical radius is where Re just equals the critical value:

$$ Re_{cr} = \frac{Q \cdot d_m}{2\pi\,r_{min}\,b\,\nu} \quad \Rightarrow \quad r_{min} = \frac{Q \cdot d_m}{2\pi\,b\,\nu\,Re_{cr}} $$

Step 4 — Substitute values.

Numerator: \(Q \cdot d_m = 0.3 \times 0.001 = 3 \times 10^{-4}\) m³·m/s

Denominator: \(2\pi \times 18 \times 10^{-6} \times 8 = 2\pi \times 1.44 \times 10^{-4} = 9.048 \times 10^{-4}\)

$$ r_{min} = \frac{3 \times 10^{-4}}{9.048 \times 10^{-4}} $$
\( r_{min} = 0.332 \text{ m} \approx 0.33 \text{ m} \)
Conclusion: Darcy’s law is valid for \(r > 0.33\) m from the well centre. Within 33 cm, flow is non-Darcian (turbulent pore flow) and the Forchheimer equation must be used.
Physical insight: Within 33 cm of the well centreline, the Darcy flux exceeds the laminar limit. This thin shell of non-Darcian flow is why real wells have “well losses” — extra head loss beyond what Darcy’s law predicts. The total head loss in a well can be expressed as \(s_w = BQ + CQ^2\), where BQ is the Darcian (aquifer) loss and CQ² is the non-Darcian (well) loss.
Problem 2: Darcy Validity Domain Around a Pumping Well 2073 Bhadra Q2(b) — 8 marks

Question: A fully penetrating well is pumped at 1000 m³/hr from a confined aquifer 28 m thick with average grain diameter 1 mm. What is the domain for which Darcy’s law is applicable? Assume \(Re_{cr} = 6\), \(\nu = 1\) centistoke.

GIVEN DATA

Pumping rate, \(Q\)1000 m³/hr
Aquifer thickness, \(b\)28 m
Grain diameter, \(d_m\)1 mm = 0.001 m
\(Re_{cr}\)6
\(\nu\)\(1 \times 10^{-6}\) m²/s
WELL P Q = 1000 m³/hr = 0.278 m³/s b=28m rmin = 0.26 m DARCY ZONE (Re ≤ 6) DARCY ZONE (Re ≤ 6) dm = 1 mm, Recr = 6
Problem 2: Thicker aquifer (28 m) with lower Recr = 6 results in a smaller non-Darcy zone than Problem 1.

SOLUTION

Unit conversion required! Q is given in m³/hr but \(\nu\) is in m²/s. Always convert to consistent units (SI) first.

Step 1 — Convert Q to SI (m³/s).

$$ Q = \frac{1000}{3600} = 0.2778 \text{ m}^3/\text{s} $$

Step 2 — Apply rmin formula.

$$ r_{min} = \frac{Q \cdot d_m}{2\pi\,b\,\nu\,Re_{cr}} $$

Step 3 — Calculate numerator and denominator separately.

Numerator: \(Q \cdot d_m = 0.2778 \times 0.001 = 2.778 \times 10^{-4}\)

Denominator: \(2\pi \times 28 \times 10^{-6} \times 6 = 2\pi \times 1.68 \times 10^{-4} = 1.0556 \times 10^{-3}\)

Step 4 — Divide.

$$ r_{min} = \frac{2.778 \times 10^{-4}}{1.0556 \times 10^{-3}} $$
\( r_{min} = 0.263 \text{ m} \approx 0.26 \text{ m} \)
Conclusion: Darcy’s law is valid for \(r > 0.26\) m (26 cm) from the well centre.
Comparison with Problem 1: Despite Q being similar (0.28 vs 0.30 m³/s), the larger aquifer (28 vs 18 m) spreads flow over a bigger cylindrical surface, reducing the velocity at each point. The lower Recr (6 vs 8) partly offsets this, but the net result is a smaller non-Darcy zone (26 cm vs 33 cm). This shows that aquifer thickness is the most important parameter controlling the size of the turbulent zone.
COMPARISON: Problem 1 vs Problem 2 P1: rmin = 0.33 m P2: rmin = 0.26 m b = 18 m, Recr = 8, Q = 0.30 b = 28 m, Recr = 6, Q = 0.28 Thicker aquifer → larger cylindrical surface → lower velocity → smaller turbulent zone rmin ∝ Q · dm / (b · ν · Recr)
Sensitivity comparison: aquifer thickness is the dominant factor affecting the Darcy validity domain.
Problem 3: Darcy Applicability Check & Seepage Velocity 2071 Bhadra Q6 — 4+4 marks

Question: In a homogeneous isotropic confined aquifer of 25 m thickness, \(n_e = 20\%\), \(K = 15\) m/day, two observation wells 1100 m apart show heads of 5.5 m and 3.0 m. Is Darcy’s law applicable? What is the average pore velocity?

GIVEN DATA

Aquifer thickness, \(b\)25 m
Effective porosity, \(n_e\)0.20
Hydraulic conductivity, \(K\)15 m/day
Well spacing, \(L\)1100 m
Heads \(h_1, h_2\)5.5 m, 3.0 m
Grain diameter, \(d_m\)0.001 m
\(\nu\)\(1 \times 10^{-6}\) m²/s
Confined Aquifer: K = 15 m/day, ne = 0.20 h₁ = 5.5 m h₂ = 3.0 m q = Ki → L = 1100 m
One-dimensional confined flow between two observation wells.

SOLUTION

Step 1 — Hydraulic gradient. The gradient drives the flow from higher head to lower head:

$$ i = \frac{h_1 - h_2}{L} = \frac{5.5 - 3.0}{1100} = 2.273 \times 10^{-3} $$

This is a very gentle slope — a drop of only 2.3 mm per metre of horizontal distance. Such gradients are typical of regional groundwater flow.

Step 2 — Darcy flux (specific discharge).

$$ q = K \cdot i = 15 \times 2.273 \times 10^{-3} = 0.03409 \text{ m/day} $$

In m/s (for Re calculation): \(q = 0.03409 / 86400 = 3.945 \times 10^{-7}\) m/s.

Common mistake: Students forget to convert q from m/day to m/s before computing Re. Since \(\nu\) is in m²/s, all quantities in Re must use consistent seconds-based units!

Step 3 — Reynolds number check.

$$ Re = \frac{q \cdot d_m}{\nu} = \frac{3.945 \times 10^{-7} \times 0.001}{1 \times 10^{-6}} = 3.95 \times 10^{-4} $$
\( Re = 3.95 \times 10^{-4} \ll 1 \) — Darcy’s law is fully applicable

The Reynolds number is about 2500 times smaller than even the most conservative critical limit of 1. This is typical of regional groundwater flow — it is almost always deeply laminar. In fact, Re only approaches the critical value very close to pumping wells.

Step 4 — Seepage (pore) velocity. The actual velocity of water particles through the pore spaces is always higher than the Darcy flux because water can only travel through the void fraction of the aquifer:

$$ v_s = \frac{q}{n_e} = \frac{0.03409}{0.20} $$
\( v_s = 0.170 \text{ m/day} = 1.97 \times 10^{-6} \text{ m/s} \)

Interpretation: At 0.170 m/day, a contaminant particle would take about \(1100 / 0.170 \approx 6470\) days (nearly 18 years) to travel from one observation well to the other. This slow transport is why groundwater contamination is persistent but slow-spreading.

Physical insight: The seepage velocity is 5× the Darcy flux because water can only travel through the 20% of the cross-section that is pore space — it must move faster to carry the same volumetric flow rate. The relationship \(v_s = q/n_e\) is fundamental to contaminant transport modelling.
q = 0.034 m/day Darcy flux vs = 0.170 m/day Pore velocity × 5 vs = q / ne = q / 0.20
Comparison: seepage velocity is always larger than Darcy flux by a factor of 1/ne.