Question Bank • QB_IDE_EXP_034 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Groundwater Hydrology

Subtopic: Steady and unsteady flow in confined and unconfined aquifers

Question Statement:

A confined aquifer of thickness $15.0\text{ m}$ is pumped at a constant discharge of $0.03\text{ m}^3\text{/s}$. Two observation wells located at radial distances of $20\text{ m}$ and $80\text{ m}$ record drawdowns of $2.5\text{ m}$ and $1.0\text{ m}$ respectively under steady state conditions. The hydraulic conductivity $K$ of the aquifer in $\text{m/day}$ is ________ (round off to 1 decimal place).

Answer Key:

14.8

Step-by-Step Solution & Explanation

Thiem's formula for steady flow in a confined aquifer: $Q = \frac{2 \pi T (s_1 - s_2)}{\ln(r_2 / r_1)}$ Transmissivity $T = K \times b$ where $b = 15.0\text{ m}$. $T = \frac{Q \ln(r_2 / r_1)}{2 \pi (s_1 - s_2)}$ Given: • $Q = 0.03\text{ m}^3\text{/s} = 0.03 \times 86400 = 2592\text{ m}^3\text{/day}$ • $r_1 = 20\text{ m}, r_2 = 80\text{ m} \implies \ln(80/20) = \ln(4) \approx 1.3863$ • $s_1 - s_2 = 2.5 - 1.0 = 1.5\text{ m}$ $T = \frac{2592 \times 1.3863}{2 \pi \times 1.5} = \frac{3593.29}{9.4248} \approx 381.26\text{ m}^2\text{/day}$ $K = \frac{T}{b} = \frac{381.26}{15.0} = 25.42\text{ m/day}$ Wait, in $\text{m/s}$: $T = \frac{0.03 \times 1.3863}{2 \pi \times 1.5} = 0.004412\text{ m}^2\text{/s} \implies K = 0.000294\text{ m/s} = 25.4\text{ m/day}$. Let's check if $K = 25.4\text{ m/day}$, correct_answer: "25.4", range min: 24.5, max: 26.2.
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