Question

Consider flow in a long and very wide rectangular open channel. Width of the channel can be considered as infinity compared to the depth of flow. Uniform flow depth is 1.0 m. The bed slope of the channel is 0.0001. The Manning roughness coefficient value is 0.02. Acceleration due to gravity, \( g \), can be taken as 9.81 m/s\(^2\).
The critical depth (in m) corresponding to the flow rate resulting from the above conditions is ________ (round off to one decimal place).
 

• A. 0.4
• B. 0.3
• C. 0.6
• D. 0.1

Hint:

When calculating the critical depth in open channel flow, use the Manning's equation for discharge and the formula for critical depth based on the flow rate.

Answer 1

The Correct Option is B

Given:
Bed slope: \( s = 0.0001 \)
Manning’s coefficient: \( n = 0.02 \)
Depth of flow: \( y = 1 \, \text{m} \)
Acceleration due to gravity: \( g = 9.81 \, \text{m/s}^2 \)

For a very wide rectangular channel (\( B \gg y \)):

Step 1: Hydraulic radius
The hydraulic radius for a very wide rectangular channel is given by: \[ R = \frac{A}{P} = \frac{By}{B+2y} \approx y \quad \text{(since \( B \gg y \))} \] Thus, the hydraulic radius \( R \) is approximately equal to the flow depth \( y \). Therefore: \[ R = 1 \, \text{m} \]
Step 2: Critical depth of flow
The critical depth \( y_c \) is given by the formula: \[ y_c = \left( \frac{q^2}{g} \right)^{1/3} \] where \( q \) is the discharge per unit width. Now we calculate \( q \) using Manning's equation for discharge: \[ q = A \times v = B \times \frac{R^{2/3}}{n} \times s^{1/2} \] Simplifying this, we get: \[ q = \frac{1}{n} \times y^{5/3} \times s^{1/2} \] Substituting the values: \[ q = \frac{1}{0.02} \times (1)^{5/3} \times (0.0001)^{1/2} = 0.5 \, \text{m}^3/\text{s/m} \]
Step 3: Calculation of critical depth
Now, using the formula for critical depth: \[ y_c = \left( \frac{(0.5)^2}{9.81} \right)^{1/3} \] Solving this: \[ y_c = 0.294 \, \text{m} \approx 0.3 \, \text{m} \]
Thus, the critical depth is: \[ \boxed{0.3} \, \text{m} \]