Question Bank • QB_IDE_ADV_088 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Groundwater Hydrology

Subtopic: Steady and unsteady flow in confined and unconfined aquifers

Question Statement:

A fully penetrating test well discharges steadily at $Q = 1800.0\text{ m}^3/\text{day}$ from a confined aquifer of uniform thickness $b = 25.0\text{ m}$. Steady-state drawdowns observed in two observation piezometers are $s_1 = 3.60\text{ m}$ at radial distance $r_1 = 15.0\text{ m}$ and $s_2 = 1.20\text{ m}$ at $r_2 = 60.0\text{ m}$. The transmissivity ($T$) of the confined aquifer in square meters per day ($\text{m}^2/\text{day}$) is ________ (round off to one decimal place). Take $\pi = 3.1416$. ``` Pumping Well (Q = 1800 m³/d) | Observation Well 1 (r1=15m, s1=3.6m) | | Observation Well 2 (r2=60m, s2=1.2m) | | | ═══════╪════════════════╪═════════════════╪═══════════════ Static Piezometric Surface | \ s1 | | | `--___ | | | `----o-----------------o────────────── Drawdown Cone | Aquifer Thickness b = 25m ═══════╧═════════════════════════════════════════════════ Impermeable Stratum ```

Answer Key:

165.5

Step-by-Step Solution & Explanation

**Method 1: Thiem Confined Aquifer Equation** $Q = \frac{2 \pi T (s_1 - s_2)}{\ln(r_2 / r_1)} \implies T = \frac{Q \ln(r_2 / r_1)}{2 \pi (s_1 - s_2)}$ 1. Parameters: - $Q = 1800.0\text{ m}^3/\text{day}$ - $s_1 - s_2 = 3.60 - 1.20 = 2.40\text{ m}$ - $\frac{r_2}{r_1} = \frac{60.0}{15.0} = 4.0$ - $\ln(4.0) = 1.386294$ 2. Evaluating Transmissivity $T$: $T = \frac{1800.0 \times 1.386294}{2 \times 3.1416 \times 2.40} = \frac{2495.33}{15.07968} = 165.476\text{ m}^2/\text{day} \approx 165.5\text{ m}^2/\text{day}$ **Method 2: Hydraulic Conductivity Verification** $K = \frac{T}{b} = \frac{165.476}{25.0} = 6.619\text{ m/day}$ $Q = \frac{2 \pi (6.619)(25)(2.4)}{1.3863} = \frac{2495.3}{1.3863} = 1800.0\text{ m}^3/\text{day}$
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