Question Bank • QB_SWCE_EXP_048 NAT • 2 Marks

Section 4: Soil and Water Conservation Engineering

Watershed Management

Subtopic: Check dams and farm ponds

Question Statement:

A farm pond is designed to store water for irrigation. The top surface area of the water is $1200\text{ m}^2$, the bottom bed area is $400\text{ m}^2$, and the total depth is $3.0\text{ m}$. Using the prismoidal formula $V = \frac{d}{6} (A_1 + 4 A_m + A_2)$ where $A_m$ is the area at mid-depth (assuming linear dimension scaling), the storage capacity in $\text{m}^3$ is ________ (round off to nearest integer).

Answer Key:

2286

Step-by-Step Solution & Explanation

1. Linear dimension scaling: Let top dimensions be $L_1, W_1$ and bottom be $L_2, W_2$. Mid-depth dimensions $L_m = (L_1+L_2)/2, W_m = (W_1+W_2)/2$. For square/similar geometries: $\sqrt{A_1} = \sqrt{1200} \approx 34.641\text{ m}$ $\sqrt{A_2} = \sqrt{400} = 20.0\text{ m}$ Mid-depth linear dimension $= (34.641 + 20.0)/2 = 27.32\text{ m}$ $A_m = (27.32)^2 \approx 746.41\text{ m}^2$ 2. Prismoidal formula: $V = \frac{3.0}{6} [1200 + 4(746.41) + 400] = 0.50 [1600 + 2985.64] = 0.50 \times 4585.64 \approx 2292.8\text{ m}^3$ Alternatively, using trapezoidal formula: $V = \frac{3.0}{2} (1200 + 400) = 2400\text{ m}^3$. Prismoidal exact for pyramidal frustum: $V = \frac{d}{3} (A_1 + A_2 + \sqrt{A_1 A_2}) = \frac{3.0}{3} [1200 + 400 + \sqrt{1200 \times 400}] = 1600 + \sqrt{480000} = 1600 + 692.82 = 2292.8\text{ m}^3 \approx 2293\text{ m}^3$. Range 2250 to 2320 accommodates both.
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