Allied & ICAR Exam Bank • ALLIED_0922 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Well Hydraulics and Groundwater

Subtopic: Well Hydraulics and Groundwater

Question Statement:

In a Theis recovery test, well discharge before shutdown was $Q = 0.02\text{ m}^3/\text{s}$. The residual drawdown change over one log cycle of $t/t'$ is $\Delta s' = 0.60\text{ m}$. Using Cooper-Jacob recovery relation $T = \frac{2.303 Q}{4 \pi \Delta s'}$ with $2.303 \times 0.02 = 0.04606$ and $4 \pi \times 0.60 = 7.5398$, transmissivity $T = \frac{0.04606}{7.5398}$ in $10^{-3}\text{ m}^2/\text{s}$ is (rounded off to two decimal places):

Answer Key:

6.11

Step-by-Step Solution & Explanation

$T = 0.04606 / 7.5398 = 0.006109\text{ m}^2/\text{s} = 6.11 \times 10^{-3}\text{ m}^2/\text{s}$.
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