Question Bank • QB_SWCE_ADV_153 NAT • 2 Marks

Section 4: Soil and Water Conservation Engineering

Soil and Water Erosion

Subtopic: Vegetative waterways

Question Statement:

A parabolic grassed waterway has a design top water surface width of $T = 6.0\text{ m}$ and a maximum central depth of $d = 0.60\text{ m}$. The bed slope is $S = 0.015$ ($1.50\%$) and Manning's roughness coefficient for the grass cover is $n = 0.040$. For this parabolic geometry, flow area is $A = \frac{2}{3} T d$ and wetted perimeter is $P = T + \frac{8 d^2}{3 T}$. The discharge capacity $Q$ of the waterway in cubic meters per second ($\text{m}^3/\text{s}$) is ________ (round off to two decimal places).

Answer Key:

3.92

Step-by-Step Solution & Explanation

**Method 1: Parabolic Geometric Hydraulic Radius Formulation** 1. Cross-sectional flow area $A$: $A = \frac{2}{3} T d = \frac{2}{3} \times 6.0 \times 0.60 = 2.40\text{ m}^2$ 2. Wetted perimeter $P$: $P = T + \frac{8 d^2}{3 T} = 6.0 + \frac{8 (0.60)^2}{3 (6.0)} = 6.0 + \frac{8 \times 0.36}{18.0} = 6.0 + \frac{2.88}{18.0} = 6.0 + 0.160 = 6.160\text{ m}$ 3. Hydraulic radius $R$: $R = \frac{A}{P} = \frac{2.40}{6.160} = 0.38961\text{ m}$ $R^{2/3} = (0.38961)^{2/3} = 0.53358$ 4. Bed slope factor $\sqrt{S}$: $\sqrt{0.015} = 0.122474$ 5. Discharge via Manning's equation: $Q = \frac{1}{n} A R^{2/3} S^{1/2} = \frac{1}{0.040} \times 2.40 \times 0.53358 \times 0.122474$ $Q = 60.0 \times 0.53358 \times 0.122474 = 3.921\text{ m}^3/\text{s} \approx 3.92\text{ m}^3/\text{s}$ **Method 2: Velocity Check** $V = \frac{Q}{A} = \frac{3.921}{2.40} = 1.634\text{ m/s}$ $V = \frac{1}{0.04} (0.38961)^{2/3} (0.015)^{1/2} = 1.634\text{ m/s}$ Within permissible vegetative limits ($1.5 - 1.8\text{ m/s}$).
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