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: Derive the tractive stress from first principles instead of quoting the formula directly.
Consider a column of water of unit plan area (1 m$^2$) and depth $y$, resting on the channel bed, with the bed sloping at angle $\theta$ (where $\tan\theta = S$ for a mild slope, so $\sin\theta \approx S$).
The weight of this column, per unit plan area, is $W = \gamma_w \times y \times 1$.

Step 2: Resolve the weight along the slope direction.
For steady, uniform flow, the component of weight acting along the slope, which drives the flow, must be balanced by the shear resistance offered by the bed:
\[ \tau_0 = W\sin\theta \approx \gamma_w y S \]
using the small-angle approximation that is valid for the mild slopes used in channel design. This reproduces the standard tractive force formula, just derived from a direct force balance instead of citing it.

Step 3: Plug in the numbers.
$\gamma_w = 10\text{ kN/m}^3 = 10000\text{ N/m}^3$, $y = 1$ m, and the bed can take at most $\tau_0 = 10\text{ N/m}^2$ before it scours:
\[ 10 = 10000 \times 1 \times S \]
\[ S = 0.001 \]

Step 4: Express as 1 in n.
\[ S = \frac{1}{n} \implies n = \frac{1}{0.001} = 1000 \]

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