Question

A wide unlined channel carries sediment-free water. The depth of water is 1 m. The specific weight of water is 10 kN/m\(^3\). To prevent scouring, the maximum permissible tractive stress on bed is 10 N/m\(^2\). The maximum slope of the channel bed to prevent scouring is 1 in \(n\). The value of \(n\) is (in integer).

Hint:

The tractive stress on the bed of a wide channel is the specific weight of water times depth times bed slope; set this equal to the permissible stress and solve for the slope.

Answer 1

Correct Answer: 1000

Step 1: Understanding the Question.
Water flowing in a wide unlined channel exerts a drag (tractive) force on the bed. If this drag force is too high, it will pick up and move the bed material, which is called scouring. We are told the maximum shear the bed can take without scouring, and we need to find the steepest slope the channel bed can have.

Step 2: Key formula for tractive stress on the bed of a wide channel.
For a wide channel, the hydraulic radius is approximately equal to the flow depth, since the wetted perimeter is dominated by the wide bed rather than the side walls. The average shear stress (tractive force per unit area) on the bed is
\[ \tau_0 = \gamma_w\, y\, S \]
where \(\gamma_w\) is the specific weight of water, \(y\) is the flow depth, and \(S\) is the bed slope.

Step 3: Substitute the known values.
\(\tau_0 = 10\ \text{N/m}^2\) (given maximum permissible tractive stress), \(\gamma_w = 10\ \text{kN/m}^3 = 10000\ \text{N/m}^3\), \(y = 1\) m.
\[ 10 = 10000\times1\times S \]

Step 4: Solve for the slope.
\[ S = \frac{10}{10000} = 0.001 = \frac{1}{1000} \]
So the maximum slope is \(1\) in \(1000\), which means \(n = 1000\).

Final Answer:
\[ \boxed{n = 1000} \]