Question Bank • QB_SWCE_EXP_070 NAT • 2 Marks

Section 4: Soil and Water Conservation Engineering

Watershed Management

Subtopic: Check dams and farm ponds

Question Statement:

A dugout farm pond has top dimensions $30\text{ m} \times 20\text{ m}$, bottom dimensions $24\text{ m} \times 14\text{ m}$, and a uniform depth of $3.0\text{ m}$. Using the prismoidal formula $V = \frac{d}{6} (A_1 + 4 A_m + A_2)$, calculate the water storage capacity of the pond in $\text{m}^3$.

Answer Key:

1404

Step-by-Step Solution & Explanation

Top area $A_1 = 30 \times 20 = 600\text{ m}^2$. Bottom area $A_2 = 24 \times 14 = 336\text{ m}^2$. Mid-dimensions: length $= \frac{30 + 24}{2} = 27\text{ m}$, width $= \frac{20 + 14}{2} = 17\text{ m}$. Mid-area $A_m = 27 \times 17 = 459\text{ m}^2$. Depth $d = 3.0\text{ m}$. Prismoidal volume: $V = \frac{3.0}{6} [600 + 4(459) + 336] = 0.5 [600 + 1836 + 336] = 0.5 [2772] = 1404\text{ m}^3$
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