Problem Statement
Find the discharge through a circular pipe of diameter 4.0 m, if the depth of water in the pipe is 1.33 m and pipe is laid at a slope of 1 in 1500. Take the value of Chezy's constant = 60.
Given Data & Constants
- Diameter of pipe, \(D = 4.0 \, \text{m}\) (Radius, \(R = 2.0 \, \text{m}\))
- Depth of water, \(d = 1.33 \, \text{m}\)
- Bed slope, \(i = 1 \text{ in } 1500 = \frac{1}{1500}\)
- Chezy's constant, \(C = 60\)
Solution
1. Calculate Geometric Properties
First, we find the half-angle (\(\alpha\)) subtended by the free surface of the water at the center of the pipe.
$$ \cos(\alpha) = \frac{R - d}{R} = \frac{2.0 - 1.33}{2.0} = \frac{0.67}{2.0} = 0.335 $$
$$ \alpha = \arccos(0.335) \approx 1.229 \, \text{radians} $$
Now we calculate the wetted area (A) and the wetted perimeter (P) using the angle in radians.
$$ \text{Area of flow, } A = R^2 (\alpha - \sin\alpha \cos\alpha) $$
$$ A = (2.0)^2 \times (1.229 - \sin(1.229) \cos(1.229)) $$
$$ A = 4 \times (1.229 - 0.942 \times 0.335) = 4 \times (1.229 - 0.3156) \approx 3.654 \, \text{m}^2 $$
$$ \text{Wetted Perimeter, } P = 2 R \alpha = 2 \times 2.0 \times 1.229 \approx 4.916 \, \text{m} $$
$$ \text{Hydraulic Mean Depth, } m = \frac{A}{P} = \frac{3.654}{4.916} \approx 0.743 \, \text{m} $$
2. Calculate Velocity and Discharge
We use Chezy's formula to find the velocity, and then the discharge.
$$ V = C \sqrt{m \cdot i} $$
$$ V = 60 \times \sqrt{0.743 \times \frac{1}{1500}} = 60 \times \sqrt{0.0004953} $$
$$ V = 60 \times 0.02225 \approx 1.335 \, \text{m/s} $$
$$ Q = A \times V = 3.654 \, \text{m}^2 \times 1.335 \, \text{m/s} \approx 4.878 \, \text{m}^3/\text{s} $$
Final Result:
The discharge through the circular pipe is approximately \(4.88 \, \text{m}^3/\text{s}\).
