Find the discharge through a trapezoidal channel of width 6 m and side slope of 1 horizontal to 3 vertical. The depth of flow of water is 3 m and Chezy’s constant, C = 60. The slope of the bed of the channel is given 1 in 5000.

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Trapezoidal Channel Flow Calculation

Problem Statement

Find the discharge through a trapezoidal channel of width 6 m and side slope of 1 horizontal to 3 vertical. The depth of flow of water is 3 m and Chezy's constant, C = 60. The slope of the bed of the channel is given 1 in 5000.

Given Data & Constants

  • Bottom width of channel, \(B = 6 \, \text{m}\)
  • Side slope = 1 Horizontal to 3 Vertical
  • Depth of flow, \(d = 3 \, \text{m}\)
  • Chezy's constant, \(C = 60\)
  • Bed slope, \(i = 1 \text{ in } 5000 = \frac{1}{5000}\)

Solution

1. Calculate Geometric Properties

First, we determine the top width of the water surface. With a slope of 1H to 3V, the horizontal distance for a vertical depth \(d\) is \(d/3\).

$$ \text{Top Width, } T = B + 2 \times \left(\frac{d}{3}\right) = 6 + 2 \times \left(\frac{3}{3}\right) = 6 + 2 = 8 \, \text{m} $$ $$ \text{Area of flow, } A = \frac{B+T}{2} \times d = \frac{6+8}{2} \times 3 = 21 \, \text{m}^2 $$

Next, we find the length of the wetted side slopes to calculate the wetted perimeter.

$$ \text{Length of one slope} = \sqrt{d^2 + \left(\frac{d}{3}\right)^2} = \sqrt{3^2 + 1^2} = \sqrt{10} \approx 3.162 \, \text{m} $$ $$ \text{Wetted Perimeter, } P = B + 2 \times (\text{Length of slope}) = 6 + 2 \times 3.162 = 12.324 \, \text{m} $$ $$ \text{Hydraulic Mean Depth, } m = \frac{A}{P} = \frac{21}{12.324} \approx 1.704 \, \text{m} $$

2. Calculate Velocity and Discharge

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

$$ V = C \sqrt{m \cdot i} $$ $$ V = 60 \times \sqrt{1.704 \times \frac{1}{5000}} = 60 \times \sqrt{0.0003408} $$ $$ V = 60 \times 0.01846 \approx 1.108 \, \text{m/s} $$ $$ Q = A \times V = 21 \, \text{m}^2 \times 1.108 \, \text{m/s} \approx 23.26 \, \text{m}^3/\text{s} $$
Final Result:

The discharge through the trapezoidal channel is approximately \(23.26 \, \text{m}^3/\text{s}\).

Explanation of Chezy's Formula

Chezy's formula is an empirical equation used in fluid dynamics to calculate the mean velocity of uniform flow in an open channel. The formula, \(V = C \sqrt{m \cdot i}\), relates the velocity to three key factors:

  • Chezy's Constant (C): An empirical value that accounts for the roughness of the channel bed and sides. A higher value of C means a smoother channel and thus a higher velocity.
  • Hydraulic Mean Depth (m): Also known as the hydraulic radius, this is the ratio of the cross-sectional area of the flow to the wetted perimeter. It represents the efficiency of the channel's shape in conveying water.
  • Bed Slope (i): The gradient or slope of the channel bed. This is the primary driving force for the flow, as it is the component of gravity that acts in the direction of motion.

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A main pipe of diameter 500mm branches in two pipes of diameter 300mm each in the horizontal plane. Angle between the branches is 600, which is symmetrical with respect to the main pipe. Flow discharge through the main pipe is 1.0 m3/s, which is equally divided into the branch pipes. If the pressure intensity at the main pipe is 400KPa, find the magnitude and direction of resultant force in the bend. Assume no loss of energy due to branch junction and in pipe sections.

A main pipe of diameter 500mm branches in two pipes of diameter 300mm each in the horizontal plane. Angle between the branches is 60°, which is symmetrical with respect to the main pipe. Flow discharge through the main pipe is 1.0 m³/s, which is equally divided into the branch pipes. If the pressure intensity at the main pipe is 400 kPa, find the magnitude and direction of resultant force in the bend. Assume no loss of energy due to branch junction and in pipe sections.

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