CHAPTER 4 — NUMERICALS
Well Hydraulics (Thiem’s Steady-State Equations)

2 Problems — Confined & Unconfined Steady-State Well Discharge

KEY FORMULAE — CHAPTER 4

A. Thiem’s Equation — Confined (Artesian) Aquifer:

$$ Q = \frac{2\pi\,K\,b\,s_w}{\ln(R/r_w)} = \frac{2\pi\,T\,s_w}{\ln(R/r_w)} $$

B. Dupuit’s Equation — Unconfined (Water-Table) Aquifer:

$$ Q = \frac{\pi\,K\,(H^2 - h_w^2)}{\ln(R/r_w)} $$

C. Key relationships:

$$ T = K \cdot b, \qquad s_w = H - h_w $$

Note: Q depends on \(\ln(R/r_w)\), so doubling the well diameter increases Q by only a small percentage — the log function grows very slowly.

SymbolTypical RangeSI Unit
\(K\)10−7–10−2 m/sm/s
\(b\) (confined)5–100 mm
\(T\)10–10,000 m²/daym²/day or m²/s
\(R\) (influence)100–1000 mm
\(r_w\)0.05–0.50 mm
\(s_w\) (drawdown)1–30 mm

DERIVATION HINT — Thiem’s Equation (Confined)

For steady radial flow to a well in a confined aquifer, by Darcy’s law at radial distance r: \(Q = 2\pi r b K \cdot dh/dr\). Separating variables: \(dr/r = (2\pi bK/Q)\,dh\). Integrating from the well (\(r_w, h_w\)) to the radius of influence (\(R, H\)):

$$ \ln(R/r_w) = \frac{2\pi K b}{Q}(H - h_w) = \frac{2\pi T s_w}{Q} $$

Rearranging gives \(Q = 2\pi T s_w / \ln(R/r_w)\). The key insight: \(Q \propto 1/\ln(R/r_w)\), so Q is logarithmically insensitive to both R and rw.

CONFINED vs UNCONFINED — KEY DIFFERENCES

CONFINED WELL (Thiem) Well Original piezo. sw b R rw Q = 2πKbs / ln(R/rw) UNCONFINED WELL (Dupuit) Well Orig. WT H hw R Q = πK(H²−hw²) / ln(R/rw) Q Q Both formulae assume steady-state radial flow (Thiem/Dupuit assumptions)
Left: Confined (artesian) well with drawdown sw. Right: Unconfined (water-table) well with saturated thickness hw.
Problem 11: Thiem’s Confined Well — Maximum Yield & Diameter Effect 2078 Chaitra Q4(b) — 6 marks

Question: A 30 cm diameter well penetrates a 23 m thick artesian aquifer. Determine maximum yield for drawdown 3.5 m (K = 2.8×10−4 m/s, R = 450 m). Also find the percentage change in discharge if well diameter is reduced to 20 cm.

GIVEN DATA

Well diameter30 cm ⇒ \(r_w = 0.15\) m
Aquifer thickness, \(b\)23 m
Drawdown, \(s_w\)3.5 m
Hydraulic conductivity, \(K\)\(2.8 \times 10^{-4}\) m/s
Radius of influence, \(R\)450 m
Modified diameter20 cm ⇒ \(r_w' = 0.10\) m
30cm Q = ? Original piezometric surface sw=3.5 b=23m R=450m rw
Confined well with drawdown cone. Discharge is governed by Thiem’s equation.

SOLUTION — Part (a): Maximum Yield

APPROACH

  1. Identify the aquifer type (confined → Thiem’s equation)
  2. Compute the numerator: \(2\pi K b s_w\)
  3. Compute the denominator: \(\ln(R/r_w)\)
  4. Divide to get Q
  5. For diameter effect: recompute Q with new rw and compare

Step 1 — Apply Thiem’s confined formula. Since the aquifer is artesian (confined) and we seek steady-state discharge:

$$ Q = \frac{2\pi\,K\,b\,s_w}{\ln(R/r_w)} $$

Step 2 — Compute numerator.

\(K \cdot b \cdot s_w = 2.8 \times 10^{-4} \times 23 \times 3.5 = 2.8 \times 10^{-4} \times 80.5 = 0.02254\) m³/s per radian

$$ 2\pi\,K\,b\,s_w = 2\pi \times 0.02254 = 0.14166 \text{ m}^3/\text{s} $$

Step 3 — Compute denominator.

\(R/r_w = 450/0.15 = 3000\)

$$ \ln(3000) = \ln(3 \times 10^3) = \ln 3 + 3\ln 10 = 1.099 + 6.908 = 8.006 $$

Step 4 — Divide.

$$ Q = \frac{0.14166}{8.006} $$
\( Q = 0.01769 \text{ m}^3/\text{s} = 17.69 \text{ lit/s} \approx 63.7 \text{ m}^3/\text{hr} \)

Part (b): Effect of Reducing Diameter to 20 cm

Step 5 — New well radius: \(r_w' = 0.10\) m.

$$ \ln\!\left(\frac{450}{0.10}\right) = \ln(4500) = 8.412 $$
$$ Q' = \frac{0.14166}{8.412} = 0.01684 \text{ m}^3/\text{s} $$

Step 6 — Percentage change.

$$ \Delta\% = \frac{Q' - Q}{Q} \times 100 = \frac{0.01684 - 0.01769}{0.01769} \times 100 $$
\( \Delta\% = -4.8\% \) — Discharge decreases by about 4.8% when diameter is halved from 30 cm to 20 cm.
Physical insight: Well discharge depends on \(\ln(R/r_w)\). Since ln grows very slowly, even halving the well diameter only changes the denominator from 8.006 to 8.412 — a mere 5% increase. This is why increasing well diameter is an inefficient way to increase yield. Instead, increasing aquifer contact (deeper penetration, longer screens) is far more effective.
D = 30 cm → Q = 17.69 L/s D = 20 cm → Q = 16.84 L/s −4.8% Halving diameter reduces discharge by only ~5% because Q ∝ 1/ln(R/rw)
Comparison: halving well diameter causes only a minor (4.8%) reduction in yield.
Problem 12: Dupuit Unconfined Well — Steady-State Discharge 2073 Bhadra Q4(b) — 7 marks

Question: A well of 0.5 m radius fully penetrates an unconfined aquifer 50 m thick (K = 30 m/day). Pumping holds the water level at 40 m above the bottom. Pumping has no effect at 500 m distance. Compute steady-state discharge.

GIVEN DATA

Well radius, \(r_w\)0.5 m
Saturated thickness, \(H\)50 m
Hydraulic conductivity, \(K\)30 m/day
Water level in well, \(h_w\)40 m
Radius of influence, \(R\)500 m
Well Original WT (H=50m) Q = ? H=50 hw=40 sw=10 R=500m
Unconfined well: water table drops from H = 50 m to hw = 40 m at the well face.

SOLUTION

Step 1 — Apply Dupuit unconfined well formula. Since this is a water-table (unconfined) aquifer, the saturated thickness varies with drawdown. We use the Dupuit formula which accounts for this by using squared heads:

$$ Q = \frac{\pi\,K\,(H^2 - h_w^2)}{\ln(R/r_w)} $$

Step 2 — Compute numerator.

\(H^2 = 50^2 = 2500\) m², \(h_w^2 = 40^2 = 1600\) m²

\(H^2 - h_w^2 = 2500 - 1600 = 900\) m²

$$ \pi\,K\,(H^2 - h_w^2) = \pi \times 30 \times 900 = 84{,}823 \text{ m}^3/\text{day} $$

Step 3 — Compute denominator.

\(R/r_w = 500/0.5 = 1000\)

$$ \ln(1000) = 3\ln(10) = 3 \times 2.3026 = 6.908 $$

Step 4 — Divide and convert units.

$$ Q = \frac{84{,}823}{6.908} $$
\( Q = 12{,}280 \text{ m}^3/\text{day} \approx 511.6 \text{ m}^3/\text{hr} \approx 142.1 \text{ L/s} \)
Physical insight: The large yield (12,280 m³/day) is due to three favourable factors:
  • High permeability (K = 30 m/day — typical of very coarse sand/gravel aquifers)
  • Large drawdown (10 m out of 50 m — 20% dewatering, within the Dupuit s/H < 0.5 limit)
  • Thick aquifer (50 m — provides a large saturated cross-section for radial flow)
In practice, this would be an excellent production well. For comparison:
  • Typical municipal well: 500–3000 m³/day (in less permeable formations)
  • Agricultural well in Nepal’s Terai: 200–1000 m³/day
  • This well: 12,280 m³/day — a high-capacity industrial/municipal source
Dupuit validity check: \(s_w/H = 10/50 = 0.2 < 0.5\). The Dupuit approximation is valid. For drawdowns exceeding 50% of the saturated thickness, a correction factor should be applied, or the formula should be modified.
This problem appeared in 2073 Bhadra Q4(b) for 7 marks. Always state the formula, show unit conversions, and mention whether the Dupuit assumption is valid.
DUPUIT ASSUMPTIONS (for validity of this solution) 1. Hydraulic gradient = slope of water table (flow is horizontal) 2. Streamlines are horizontal, equipotentials are vertical 3. Seepage face at the well is neglected (small error if s/H < 0.5) These are reasonably valid for this problem since sw/H = 10/50 = 0.2 < 0.5
Dupuit assumptions underlying the unconfined well equation and their applicability to this problem.