Determine the maximum discharge of water through a circular channel of diameter 2.0 m when the bed slope of the channel is 1 in 1500.

Circular Channel Maximum Discharge Calculation

Problem Statement

Determine the maximum discharge of water through a circular channel of diameter 2.0 m when the bed slope of the channel is 1 in 1500. Take C = 50.

Given Data & Constants

Diameter of channel, $D = 2.0 \, \text{m}$ (Radius, $R = 1.0 \, \text{m}$)
Bed slope, $i = 1 \text{ in } 1500 = \frac{1}{1500}$
Chezy's constant, $C = 50$

Solution

1. Conditions for Maximum Discharge in a Circular Channel

For the discharge to be maximum in a circular channel, the depth of flow must be $d = 0.95 \times D$.

$$ \text{Depth of flow for max discharge, } d = 0.95 \times 2.0 = 1.9 \, \text{m} $$

2. Calculate Geometric Properties for Maximum Discharge

First, we find the half-angle ($\alpha$) subtended by the free surface at the center.

$$ \cos(\alpha) = \frac{R - d}{R} = \frac{1.0 - 1.9}{1.0} = -0.9 $$ $$ \alpha = \arccos(-0.9) \approx 2.69 \, \text{radians} \quad (154.16^\circ) $$

Now we calculate the wetted area (A) and wetted perimeter (P).

$$ \text{Area of flow, } A = R^2 (\alpha - \sin\alpha \cos\alpha) $$ $$ A = (1.0)^2 \times (2.69 - \sin(2.69) \times (-0.9)) $$ $$ A = 1 \times (2.69 - (0.436 \times -0.9)) = 2.69 + 0.3924 \approx 3.082 \, \text{m}^2 $$ $$ \text{Wetted Perimeter, } P = 2 R \alpha = 2 \times 1.0 \times 2.69 = 5.38 \, \text{m} $$ $$ \text{Hydraulic Mean Depth, } m = \frac{A}{P} = \frac{3.082}{5.38} \approx 0.573 \, \text{m} $$

3. Calculate Velocity and Maximum Discharge

We use Chezy's formula to find the velocity, and then the discharge.

$$ V = C \sqrt{m \cdot i} $$ $$ V = 50 \times \sqrt{0.573 \times \frac{1}{1500}} = 50 \times \sqrt{0.000382} $$ $$ V = 50 \times 0.01954 \approx 0.977 \, \text{m/s} $$ $$ Q_{max} = A \times V = 3.082 \, \text{m}^2 \times 0.977 \, \text{m/s} \approx 3.01 \, \text{m}^3/\text{s} $$

Final Result:

The maximum discharge through the circular channel is approximately $3.01 \, \text{m}^3/\text{s}$.

Explanation of Maximum Discharge Condition

In a circular channel, the maximum discharge does not occur when the pipe is flowing full. As the water level rises past a certain point, the wetted perimeter increases faster than the flow area. This increase in friction (drag) actually causes the discharge to decrease slightly.

The theoretical maximum discharge is achieved when the depth of flow is approximately **95% of the pipe's diameter** ($d = 0.95D$). This problem requires finding the discharge under this specific optimal condition. The solution involves calculating the geometric properties (Area, Wetted Perimeter, Hydraulic Mean Depth) for this 95% full state and then applying Chezy's formula to find the resulting flow rate.