The rate of flow of water through a circular channel of diameter 0.8 m is 200 litres/s. Find the slope of the bed of the channel for maximum velocity.

Circular Channel Slope for Maximum Velocity

Problem Statement

The rate of flow of water through a circular channel of diameter 0.8 m is 200 litres/s. Find the slope of the bed of the channel for maximum velocity. Take C = 50.

Given Data & Constants

Diameter of channel, $D = 0.8 \, \text{m}$ (Radius, $R = 0.4 \, \text{m}$)
Discharge, $Q = 200 \, \text{L/s} = 0.2 \, \text{m}^3/\text{s}$
Chezy's constant, $C = 50$

Solution

1. Conditions for Maximum Velocity in a Circular Channel

For the velocity to be maximum in a circular channel, the following hydraulic conditions must be met:

The depth of flow is $d = 0.81 \times D$
The hydraulic mean depth is $m = 0.3 \times D$

2. Calculate Geometric Properties for Maximum Velocity

Using the given diameter, we can find the specific depth, area, and hydraulic mean depth for this condition.

$$ \text{Depth of flow, } d = 0.81 \times 0.8 = 0.648 \, \text{m} $$ $$ \text{Half-angle of flow, } \cos(\alpha) = \frac{R - d}{R} = \frac{0.4 - 0.648}{0.4} = -0.62 $$ $$ \alpha = \arccos(-0.62) \approx 2.24 \, \text{radians} $$ $$ \text{Area of flow, } A = R^2 (\alpha - \sin\alpha \cos\alpha) = (0.4)^2 (2.24 - \sin(2.24)\cos(2.24)) $$ $$ A = 0.16 \times (2.24 - (0.784 \times -0.62)) = 0.16 \times (2.24 + 0.486) \approx 0.436 \, \text{m}^2 $$ $$ \text{Hydraulic Mean Depth, } m = 0.3 \times D = 0.3 \times 0.8 = 0.24 \, \text{m} $$

3. Calculate the Required Bed Slope ($i$)

First, find the actual velocity of flow from the given discharge and the calculated area.

$$ V = \frac{Q}{A} = \frac{0.2 \, \text{m}^3/\text{s}}{0.436 \, \text{m}^2} \approx 0.4587 \, \text{m/s} $$

Now, rearrange Chezy's formula to solve for the slope ($i$).

$$ V = C \sqrt{m \cdot i} \implies i = \frac{V^2}{C^2 \cdot m} $$ $$ i = \frac{(0.4587)^2}{50^2 \times 0.24} = \frac{0.2104}{2500 \times 0.24} = \frac{0.2104}{600} \approx 0.0003507 $$ $$ \text{Slope} = 1 \text{ in } \frac{1}{0.0003507} \approx 1 \text{ in } 2851 $$

Final Result:

The required bed slope for maximum velocity is approximately 1 in 2851.

Explanation of Maximum Velocity Condition

In a circular channel, the velocity of flow is not necessarily highest when the pipe is full. Due to the shape of the pipe, the wetted perimeter increases very rapidly as the pipe approaches full, which increases frictional drag. The maximum velocity is actually achieved when the pipe is about **81% full** ($d = 0.81D$).

This problem requires finding the specific bed slope that will produce the given discharge (200 L/s) when the water is flowing at this optimal depth for maximum velocity. The solution involves first calculating all the geometric properties of the channel at this specific depth and then using the known discharge to find the velocity. Finally, Chezy's formula is used to solve for the unknown slope that corresponds to these conditions.