Section 4: Soil and Water Conservation Engineering
Fluid Mechanics
Subtopic: Flow in open channels
Question Statement:
A most hydraulically efficient rectangular canal is to be excavated to convey a uniform discharge of $Q = 6.0\text{ m}^3/\text{s}$ on a longitudinal bed slope of $S_0 = 0.0016$. Manning's roughness coefficient is $n = 0.015$. The required design flow depth $y$ in meters is ________ (round off to two decimal places).
Answer Key:
1.24
Step-by-Step Solution & Explanation
**Method 1: Optimum Hydraulic Proportions**
1. For the most efficient rectangular channel:
$\text{Bed width } B = 2 y, \quad \text{Wetted perimeter } P = B + 2 y = 4 y$
$\text{Flow area } A = B y = 2 y^2$
$\text{Hydraulic radius } R = \frac{A}{P} = \frac{2 y^2}{4 y} = \frac{y}{2}$
2. Manning's equation:
$Q = \frac{1}{n} A R^{2/3} S_0^{1/2}$
$6.0 = \frac{1}{0.015} (2 y^2) \left( \frac{y}{2} \right)^{2/3} \sqrt{0.0016}$
$\sqrt{0.0016} = 0.040, \quad \left( \frac{1}{2} \right)^{2/3} \approx 0.62996$
$6.0 = \frac{1}{0.015} \times 2 \times 0.62996 \times 0.040 \times y^{8/3}$
$6.0 = 66.667 \times 0.050397 \times y^{8/3} = 3.3598 \times y^{8/3}$
$y^{8/3} = \frac{6.0}{3.3598} = 1.7858$
$y = (1.7858)^{3/8} = (1.7858)^{0.375} = 1.241\text{ m} \approx 1.24\text{ m}$
**Method 2: Verification of Flow Parameters**
At $y = 1.241\text{ m}$:
$A = 2(1.241)^2 = 3.080\text{ m}^2$, $R = 1.241 / 2 = 0.6205\text{ m}$.
$V = \frac{1}{0.015} (0.6205)^{2/3} (0.04) = 66.667 \times 0.7275 \times 0.04 = 1.940\text{ m/s}$.
$Q = A V = 3.080 \times 1.940 = 5.975 \approx 6.0\text{ m}^3/\text{s}$.