Allied & ICAR Exam Bank • ALLIED_0552 NAT • 2 Marks

Section 5: Irrigation and Drainage Engineering

Groundwater Wells and Pumps

Subtopic: Groundwater Wells and Pumps

Question Statement:

In an unconfined aquifer with $K = 20\text{ m/day}$, steady heads at radial distances $r_1 = 10\text{ m}$ and $r_2 = 100\text{ m}$ ($\ln(r_2/r_1) = 2.303$) are $h_1 = 15\text{ m}$ and $h_2 = 17\text{ m}$. Here $h_2^2 - h_1^2 = 289 - 225 = 64\text{ m}^2$. The discharge in $\text{m}^3/\text{day}$ is (rounded off to one decimal place):

Answer Key:

1746.0

Step-by-Step Solution & Explanation

$Q = \frac{3.14159 \times 20 \times 64}{2.302585} = \frac{4021.24}{2.302585} = 1746.36\text{ m}^3/\text{day} \approx 1746.0-1746.4\text{ m}^3/\text{day}$.
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