CHAPTER 3 — NUMERICALS
Potential Groundwater Flow Theory & Flow Net Analysis

7 Problems — Dupuit Equation, Recharge Formula, Variable-Thickness Confined Flow, Contour Maps

KEY FORMULAE — CHAPTER 3

A. Dupuit Equation (Unconfined, No Recharge):

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

B. Unconfined Aquifer with Uniform Recharge \(p\):

$$ h^2 = h_1^2 - \frac{(h_1^2-h_2^2)}{L}\,x + \frac{p}{K}\,x(L-x) $$
$$ q_1 = \frac{K(h_1^2-h_2^2)}{2L} + \frac{pL}{2}, \qquad q_2 = \frac{pL}{2} - \frac{K(h_1^2-h_2^2)}{2L} $$
$$ d = \frac{L}{2} - \frac{K(h_1^2-h_2^2)}{2pL} \quad \text{(water divide from deeper canal)} $$

C. Confined Aquifer with Variable (Wedge) Thickness:

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

D. Horizontal Gallery (Unconfined):

$$ q = \frac{K}{2L}(H^2 - h_1^2), \qquad Q_{\text{total}} = 2ql $$

E. Water-Table Contour Map:

SymbolTypical RangeSI Unit
\(K\)0.01–100 m/daym/day or m/s
\(p\) (recharge)0–50 mm/hrm/day (convert!)
\(h_1, h_2\)5–50 m (unconfined)m
\(L\)100–10,000 mm

DERIVATION HINT — Dupuit Equation (No Recharge)

Start from the Dupuit assumption: flow is horizontal and the hydraulic gradient equals the slope of the water table. For a unit-width strip: \(Q = -Kh\,dh/dx\). Separate and integrate from \(x=0\) (where \(h=h_1\)) to \(x=L\) (where \(h=h_2\)):

$$ Q\,dx = -Kh\,dh \quad \Rightarrow \quad Q \cdot L = \frac{K(h_1^2 - h_2^2)}{2} $$

This gives the Dupuit parabola: \(h^2\) varies linearly with distance \(x\).

DERIVATION HINT — Recharge Formula

With uniform recharge \(p\) (m/day) over a strip of width \(L\), the flow rate at any section \(x\) is not constant — it increases linearly from the water divide toward each boundary. By continuity: \(dQ/dx = p\). This modifies the ODE to: \(d(q)/dx = p\), where \(q = -Kh\,dh/dx\). Integrating twice with boundary conditions \(h=h_1\) at \(x=0\) and \(h=h_2\) at \(x=L\) yields the complete phreatic surface and the discharge formulas.

WHEN TO USE WHICH FORMULA?

Problem 4: Unconfined Aquifer with Recharge (q1, q2, divide) 2078 Chaitra Q3(b) — 8 marks

Question: Two irrigation canals are separated by 1.57 km of land underlain by an impervious layer. Rainfall at 25 mm/hr fully infiltrates. Canal depths: 12 m and 7 m. K = 12 m/day. Find seepage into both canals, the phreatic surface equation, and the water-table divide.

GIVEN DATA

Strip width, \(L\)1570 m
Rainfall, \(p\)25 mm/hr
Canal depths\(h_1 = 12\) m, \(h_2 = 7\) m
Permeability, \(K\)12 m/day
Rainfall p = 25 mm/hr = 0.6 m/day h₁=12 h₂=7 Phreatic surface d = ? q₁ q₂ Impervious stratum L = 1570 m K = 12 m/day
Two canals separated by a land strip with uniform rainfall infiltration. The water divide occurs where horizontal flow reverses direction.

SOLUTION

Unit conversion first! Rainfall is given as 25 mm/hr. Convert to m/day: \(25 \times 24/1000 = 0.6\) m/day.

Step 1 — Convert rainfall. \(p = 25 \text{ mm/hr} \times 24 \text{ hr/day} \div 1000 = 0.6\) m/day.

Step 2 — Dupuit gradient term. This term accounts for flow driven purely by the head difference between the two canals (it would be the only term if there were no rainfall):

$$ \frac{K(h_1^2-h_2^2)}{2L} = \frac{12(144-49)}{3140} = \frac{12 \times 95}{3140} = \frac{1140}{3140} = 0.363 \text{ m}^2/\text{day} $$

Step 3 — Recharge term. This represents half the total rainfall volume per unit length of strip:

$$ \frac{pL}{2} = \frac{0.6 \times 1570}{2} = 471.0 \text{ m}^2/\text{day} $$

Step 4 — Discharge into deeper canal (canal 1): It receives both the gravity-driven flow AND the recharge contribution from its side of the divide:

\( q_1 = 0.363 + 471.0 = 471.36 \text{ m}^2/\text{day per m} \)

Step 5 — Discharge into shallower canal (canal 2): It gets the recharge contribution MINUS the gravity-driven flow toward canal 1:

\( q_2 = 471.0 - 0.363 = 470.64 \text{ m}^2/\text{day per m} \)

Verification: Total rainfall on the strip = \(pL = 0.6 \times 1570 = 942.0\) m²/day. Sum of discharges: \(q_1 + q_2 = 471.36 + 470.64 = 942.0\). ✓ (Conservation of mass satisfied)

Step 6 — Phreatic surface equation. The general form is:

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

Substituting: \(h_1^2 = 144\), \((h_1^2-h_2^2)/L = 95/1570 = 0.0605\), \(p/K = 0.6/12 = 0.05\):

\( h^2 = 144 - 0.0605\,x + 0.05\,x(1570 - x) \)

Step 7 — Water divide (ground-water divide). The position d (from the deeper canal) where the horizontal flow rate is zero:

$$ d = \frac{L}{2} - \frac{K(h_1^2-h_2^2)}{2pL} = 785 - \frac{12 \times 95}{2 \times 0.6 \times 1570} = 785 - \frac{1140}{1884} $$
\( d = 785 - 0.605 = 784.4 \text{ m from the deeper canal} \)
Physical insight: The divide is nearly at the midpoint (784.4 vs 785.0 m). This is because the intense rainfall (0.6 m/day) produces a recharge term (471 m²/day) that dwarfs the Dupuit gradient term (0.363 m²/day) by a factor of 1300. Since the rainfall overwhelms the head difference, the water table is nearly symmetric and almost equal volumes drain to each canal. In dry conditions, the divide would shift significantly toward the shallower canal.
This problem appeared in 2078 Chaitra Q3(b) for 8 marks. The recharge formula is among the most important in the course.
Problem 5: Water-Table Contour Map (4 Wells, Rectangle) 2073 Bhadra Q3(a) — 3+4 marks

Question: Prepare a water-table contour map and show flow direction from 4 wells at rectangle vertices (elevations: 51, 56, 76, 66 m). Contour interval = 5 m.

GIVEN DATA

Well A (bottom-left)56 m
Well B (bottom-right)51 m
Well C (top-right)66 m
Well D (top-left)76 m
Contour interval5 m
D: 76 m C: 66 m A: 56 m B: 51 m 60 65 70 75 55 55 60 65 70 75 75 70 65 60 55 Flow direction N Flow: D (76 m) → B (51 m) along the rectangle diagonal Flow is always perpendicular to contour lines, toward decreasing head
Completed water-table contour map with 5 m interval. Equal-head points are linearly interpolated on rectangle edges.

SOLUTION

Step 1 — Linear interpolation on each side:

Step 2: Join equal-head points → contour lines (55, 60, 65, 70, 75 m).

Step 3: Flow is perpendicular to contours toward lower head.

General flow: from D (76 m) toward B (51 m) along the rectangle diagonal. Hydraulic gradient ≈ 5 m / (contour spacing).

Uses of Water-Table Contour Maps

  1. Determine groundwater flow direction at any point
  2. Estimate hydraulic gradient from contour spacing
  3. Identify recharge and discharge zones
  4. Plan well-field locations and track contamination plumes
  5. Calculate seepage flow through any cross section using Darcy’s law
Problem 6: Dupuit Equation — Unconfined Flow Without Recharge 2073 Bhadra Q3(b) — 7 marks

Question: Two rivers 5 km apart through a homogeneous unconfined aquifer. Depths: 20 m and 11 m. K = 18 m/day. Compute seepage per unit length and the phreatic line equation.

GIVEN DATA

\(L\)5000 m
\(h_1, h_2\)20 m, 11 m
\(K\)18 m/day
Impervious bedrock h₁=20m h₂=11m Phreatic surface h(x) Q → L = 5 km
Unconfined aquifer between two rivers (no recharge). The phreatic surface forms a Dupuit parabola.

SOLUTION

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

Phreatic surface:

\( h^2 = 400 - 0.0558\,x \quad \Rightarrow \quad h = \sqrt{400 - 0.0558\,x} \)
Physical insight: With no recharge, flow is one-directional (left to right). The parabolic profile drops more steeply near the discharge boundary because the same flow passes through a progressively thinner saturated section.
Problem 7: Water-Table Contour Map (3 Wells, Triangle) 2072 Ashwin Q4 — 6+4 marks

Question: Prepare a groundwater contour map at 2 m interval and show flow direction from 3 test wells arranged in a triangle with heads 70, 65, and 60 m.

GIVEN DATA

Well A70 m
Well B65 m
Well C60 m
Contour interval2 m
A: 70 m B: 65 m C: 60 m 68 66 68 66 64 62 64 62 68 66 64 62 Flow ⊥ contours Flow: from A (70 m) toward C (60 m), perpendicular to contour lines
Triangular well arrangement with 2 m contour intervals. Contours are nearly parallel, indicating uniform regional flow.

SOLUTION

Step 1: Interpolate on each side at 2 m intervals using linear proportion: position = (target − start)/(end − start).

Step 2: Connect equal-head points across the triangle to form contour lines (62, 64, 66, 68 m).

Step 3: Flow is perpendicular to contours toward lower head.

Flow direction: from A (70 m) toward C (60 m), roughly perpendicular to the contour lines.
Three-point method: With exactly 3 wells, you can uniquely determine the regional gradient direction and magnitude. This is the minimum data needed to define a planar water table.
Problem 8: Confined Aquifer — Variable Thickness 2072 Ashwin Q5 — 8 marks

Question: Determine the flow in the confined aquifer shown (K = 0.4 m/day, L = 800 m, b1 = 8 m, b2 = 6 m, h1 = 10 m, h2 = 8 m).

GIVEN DATA

\(K\)0.4 m/day
\(L\)800 m
\(b_1, b_2\)8 m, 6 m
\(h_1, h_2\)10 m, 8 m
b₁=8 b₂=6 h₁=10 h₂=8 Q = ? L=800m
Confined aquifer with linearly varying (wedge) thickness. Thickness tapers from b1 to b2.

SOLUTION

Step 1 — Identify the aquifer type. The aquifer thickness varies linearly from \(b_1 = 8\) m to \(b_2 = 6\) m. This is a “wedge” aquifer. We cannot use the standard formula \(Q = Kb\Delta h/L\) because \(b\) is not constant. Instead, we integrate \(Q = Kb(x)\,dh/dx\) with \(b(x) = b_1 + (b_2-b_1)x/L\).

Step 2 — Apply the variable-thickness formula.

$$ Q = K \cdot \frac{(h_1-h_2)(b_2-b_1)}{L\,\ln(b_2/b_1)} = 0.4 \times \frac{(10-8)(6-8)}{800\,\ln(6/8)} $$

Step 3 — Evaluate each component.

Head difference: \(h_1 - h_2 = 10 - 8 = 2\) m

Thickness change: \(b_2 - b_1 = 6 - 8 = -2\) m

Natural log: \(\ln(6/8) = \ln(0.75) = -0.2877\)

Denominator: \(L \cdot \ln(b_2/b_1) = 800 \times (-0.2877) = -230.16\)

Numerator: \((h_1-h_2)(b_2-b_1) = 2 \times (-2) = -4\)

Step 4 — Compute Q.

$$ Q = 0.4 \times \frac{-4}{-230.16} = 0.4 \times 0.01738 $$
\( Q = 6.95 \times 10^{-3} \text{ m}^2/\text{day per m width} \)
Why a special formula? Since b(x) varies linearly, we cannot simply use Q = Kb(Δh/L). We must integrate \(Q = Kb(x)\,dh/dx\) across the aquifer, which yields the logarithmic formula above. If the aquifer had constant thickness, we would simply use \(Q = K \cdot b \cdot (h_1-h_2)/L\), which gives slightly different results. The log formula correctly accounts for the varying cross-sectional area available for flow along the aquifer.

WHAT IF THE THICKNESS WERE CONSTANT?

If we used the average thickness \(\bar{b} = (8+6)/2 = 7\) m:

$$ Q_{approx} = K\bar{b}\frac{h_1-h_2}{L} = 0.4 \times 7 \times \frac{2}{800} = 7.0 \times 10^{-3} \text{ m}^2/\text{day} $$

This gives 7.0×10−3 vs the exact 6.95×10−3 — only 0.7% difference in this case. The log formula matters more when the thickness ratio \(b_2/b_1\) deviates significantly from 1.

Problem 9: Confined Aquifer — Variable Thickness 2072 Magh Q7 — 8 marks

Question: Same type as Problem 8 with: L = 950 m, K = 0.56 m/day, h1 = 12, b1 = 7, h2 = 9, b2 = 5.

GIVEN DATA

\(K\)0.56 m/day
\(L\)950 m
\(h_1, h_2\)12 m, 9 m
\(b_1, b_2\)7 m, 5 m
b₁=7 b₂=5 h₁=12 h₂=9 Q = ? L=950m K = 0.56 m/day
Problem 9: Wedge-shaped confined aquifer tapering from b1=7 m to b2=5 m over 950 m.

SOLUTION

Step 1 — Identify aquifer type. This is a confined aquifer with linearly varying thickness (wedge). We cannot use the simple \(Q = Kb \cdot \Delta h/L\) formula because \(b\) changes along the flow path. We must use the logarithmic formula derived by integrating \(Q = Kb(x)\,dh/dx\).

Step 2 — Apply variable-thickness formula.

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

Step 3 — Substitute values.

$$ Q = 0.56 \times \frac{(12-9)(5-7)}{950\,\ln(5/7)} = 0.56 \times \frac{3 \times (-2)}{950 \times (-0.3365)} $$

Numerator of fraction: \(3 \times (-2) = -6\)

Denominator of fraction: \(950 \times (-0.3365) = -319.68\)

$$ = 0.56 \times \frac{-6}{-319.68} = 0.56 \times 0.01877 $$
\( Q = 1.051 \times 10^{-2} \text{ m}^2/\text{day per m width} \)
Comparison with Problem 8: Despite a longer aquifer (950 vs 800 m), the larger K (0.56 vs 0.40 m/day) and bigger head difference (3 vs 2 m) result in 50% greater discharge. This shows that head difference and K are more influential than aquifer length.
Note on signs: Both \(b_2 - b_1\) and \(\ln(b_2/b_1)\) are negative when the aquifer tapers. The two negatives cancel, giving a positive Q (flow from high to low head).
Problem 10: Unconfined Aquifer + Recharge + No-Rain Case 2071 Bhadra Q4 — 3+3+3 marks

Question: Rain at 12 mm/hr on 1200 m strip between two canals (depths 10 & 8 m). K = 10 m/day. Find q1, q2, phreatic equation, divide, and seepage when p = 0.

GIVEN DATA

Rainfall, \(p\)12 mm/hr
Strip width, \(L\)1200 m
\(h_1, h_2\)10 m, 8 m
\(K\)10 m/day
p = 12 mm/hr = 0.288 m/day h₁=10 h₂=8 d = ? q₁ q₂ L=1200m
Recharge between two canals of different depth.

SOLUTION

Unit conversion first! \(p = 12\) mm/hr. Convert: \(12 \times 24/1000 = 0.288\) m/day. All other data (K = 10 m/day, h in m, L in m) is already in m/day-compatible units.

Step 1: \(p = 12 \times 24/1000 = 0.288\) m/day.

Step 2 — Compute the two terms.

Dupuit gradient term (driven by head difference between canals):

$$ \frac{K(h_1^2-h_2^2)}{2L} = \frac{10(100-64)}{2400} = \frac{10 \times 36}{2400} = \frac{360}{2400} = 0.15 \text{ m}^2/\text{day} $$

Recharge term (driven by rainfall infiltration):

$$ \frac{pL}{2} = \frac{0.288 \times 1200}{2} = 172.8 \text{ m}^2/\text{day} $$

Step 3 — Seepage into canals.

\( q_1 = 172.8 + 0.15 = 172.95 \text{ m}^2/\text{day} \) ← into deeper canal
\( q_2 = 172.8 - 0.15 = 172.65 \text{ m}^2/\text{day} \) ← into shallower canal

Verification: \(q_1+q_2 = 345.6 = pL = 0.288 \times 1200\). ✓

Step 4 — Phreatic surface equation.

Using \(h^2 = h_1^2 - [(h_1^2-h_2^2)/L]\,x + (p/K)\,x(L-x)\):

\((h_1^2-h_2^2)/L = 36/1200 = 0.03\), \(p/K = 0.288/10 = 0.0288\)

\( h^2 = 100 - 0.03\,x + 0.0288\,x(1200 - x) \)

Step 5 — Water divide location.

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

Step 6 — No rainfall case (p = 0).

When there is no recharge, the only driving force is the head difference. The recharge formula reduces to the simple Dupuit equation:

\( q_{p=0} = \dfrac{K(h_1^2-h_2^2)}{2L} = \frac{10 \times 36}{2400} = 0.15 \text{ m}^2/\text{day} \)

All flow goes from the deeper canal side towards the shallower canal. There is no water divide because the flow is unidirectional.

Physical insight: The contrast is dramatic: with rainfall, total flow = 345.6 m²/day; without rainfall, only 0.15 m²/day. Rainfall contributes 99.96% of the total flow. This illustrates why seasonal rainfall patterns completely dominate shallow groundwater behaviour.
DISCHARGE COMPARISON: With vs Without Rainfall With rain: q₁+q₂ = 345.6 m²/day Without rain: q = 0.15 m²/day Rainfall contribution: 99.96% of total flow Recharge (172.8×2) dominates head-difference flow (0.15) by a factor of 2300
Rainfall completely dominates the seepage budget. Without rain, discharge drops to nearly zero.
This is a high-marks problem (3+3+3 = 9 marks). Common sub-parts: find q1, q2, phreatic equation, divide location, and compare with the no-rain case.