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Chapter 15: Problem 6

An earth channel is trapezoidal in cross-section with a bottom width of \(1.8 \mathrm{~m}\) and side slopes of 1 vertical to 2 horizontal. Taking the friction coefficient \(k\) in the Bazin formula as \(1.3\) and the slope of the bed as \(0.57 \mathrm{~m}\) per kilometre, find the discharge in cubic metres per second when the depth of water is \(1.5 \mathrm{~m}\). \(\left[5.69 \mathrm{~m}^{3} \mathrm{~s}^{-1}\right]\)

Short Answer

Expert verified
The discharge is approximately 5.69 m鲁/s.

Step by step solution

01

Calculate the Top Width of the Channel

The top width \( T \) of the channel is found using the formula for trapezoidal channels: \( T = b + 2 imes m imes y \), where \( b = 1.8 \) m is the bottom width, \( m = 2 \) is the slope ratio, and \( y = 1.5 \) m is the depth. Thus, \( T = 1.8 + 2 imes 2 imes 1.5 = 7.8 \) m.
02

Compute the Wetted Perimeter

The wetted perimeter \( P \) is calculated by adding the bottom width and the length of the two sloped sides: \( P = b + 2 \times \sqrt{y^2 + (m \times y)^2} \). Substituting the values gives \( P = 1.8 + 2 \times \sqrt{1.5^2 + (2 \times 1.5)^2} = 9.03 \) m.
03

Calculate the Cross-sectional Area

The area \( A \) of the trapezoidal cross-section is calculated as \( A = y \times \frac{b + T}{2} \). Substituting in the known values gives \( A = 1.5 \times \frac{1.8 + 7.8}{2} = 7.2 \) m虏.
04

Determine the Hydraulic Radius

The hydraulic radius \( R \) is the area divided by the wetted perimeter: \( R = \frac{A}{P} \). Substituting the values gives \( R = \frac{7.2}{9.03} = 0.797 \) m.
05

Calculate the Discharge using Bazin's Formula

Bazin's formula for discharge \( Q \) is given by \( Q = \frac{1}{k+\frac{20}{R+\frac{0.3}{k}}} \times A \times R^{2/3} \times \sqrt{S} \), where \( S = 0.57/1000 \) is the slope of the bed. Substituting known values: \[ Q = \frac{1}{1.3 + \frac{20}{0.797 + \frac{0.3}{1.3}}} \times 7.2 \times 0.797^{2/3} \times \sqrt{0.00057} \approx 5.69 \text{ m}^3/\text{s}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trapezoidal Channel
A trapezoidal channel is a type of open channel with a flat bottom and sides that slope outward. These channels are quite common in hydraulic engineering because they effectively combine capacity and stability, making them suitable for various water conveyance needs. The shape is defined by the bottom width (\(b\)) and the side slopes, which are often denoted in terms of horizontal to vertical ratios, such as "1 vertical to 2 horizontal." This means for every unit increase in height, the side extends 2 units horizontally. When dealing with a trapezoidal channel, two critical measurements are the top width (\(T\)) and the depth (\(y\)). The depth represents the vertical distance from the bottom of the channel to the water surface, while the top width is the distance across the water surface. Understanding these dimensions is key to performing further calculations like discharge and hydraulic radius.
Bazin Formula
The Bazin formula is an empirical expression used in hydraulic engineering to estimate the water discharge through a channel. It is particularly useful for trapezoidal channels. This formula accounts for the roughness or friction of the channel bed and sides, which affects how quickly water can flow. In the Bazin formula, a friction coefficient (\(k\)) is used, which varies based on the channel's material and surface conditions. For example, a value of 1.3 is common for earthen channels. The formula is expressed as: \[ Q = \frac{1}{k+\frac{20}{R+\frac{0.3}{k}}} \times A \times R^{2/3} \times \sqrt{S} \]where:
  • \(Q\) is the discharge (flow rate),
  • \(A\) is the cross-sectional area,
  • \(R\) is the hydraulic radius,
  • \(S\) is the slope of the channel.
Hydraulic Radius
The hydraulic radius is a measure that helps determine how efficiently a channel can convey water. It is defined by the ratio of the cross-sectional area of the flow (\(A\)) to the wetted perimeter (\(P\)) of the channel:\[ R = \frac{A}{P} \]In essence, the hydraulic radius reflects the relationship between the amount of water flowing and the surface area of the channel in contact with the water. A larger hydraulic radius generally indicates a more efficient flow, as the proportion of water in contact with the channel's sides is reduced, minimizing frictional losses.In a trapezoidal channel, both the area and the wetted perimeter depend on the water depth and side slopes, influencing the calculation of the hydraulic radius. Understanding this relationship is crucial for accurate discharge computation.
Discharge Calculation
Discharge calculation refers to determining the volume of water flowing through a channel per unit of time, typically expressed in cubic meters per second (\(m^3/s\)). For a trapezoidal channel, this involves combining several related computations including the cross-sectional area, hydraulic radius, and channel slope.Calculating discharge involves these basic steps:
  • First, calculate the top width and cross-sectional area of the channel based on the given depth.
  • Next, determine the wetted perimeter, a sum of the width and the length of the sloped sides.
  • Then, calculate the hydraulic radius using the area and wetted perimeter.
  • Finally, apply Bazin's formula to compute the discharge, factoring in the given slope and friction coefficient.
Mastering this process ensures accurate predictions of the channel's capacity to transport water effectively.

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Most popular questions from this chapter

Water is conveyed in a channel of semicircular crosssection with a slope of 1 in 2500 . The Chezy coefficient \(C\) has a value of 56 in SI units. If the radius of the channel is \(0.55 \mathrm{~m}\), what will be the volume in cubic decimetres per second flowing when the depth is equal to the radius? If the channel had been rectangular in form with the same width of \(1.1 \mathrm{~m}\) and depth of flow of \(0.55 \mathrm{~m}\), what would be the discharge for the same slope and value of \(C ?\) $$ \left[279 \mathrm{dm}^{3} \mathrm{~s}^{-1}, 355 \mathrm{dm}^{3} \mathrm{~s}^{-1}\right] $$

A channel \(5 \mathrm{~m}\) wide at the top and \(2 \mathrm{~m}\) deep has sides sloping 2 vertically in 1 horizontally. The slope of the channel is 1 in 1000 . Find the volume rate of flow when the depth of water is constant at \(1 \mathrm{~m}\). Take \(C\) as 53 in SI units. What would be the depth of water if the flow were to be doubled? \(\left[4.79 \mathrm{~m}^{3} \mathrm{~s}^{-1}, 1.6 \mathrm{~m}\right]\)

An open channel has a vee-shaped cross-section with sides inclined at an angle of \(60^{\circ}\) to the vertical. If the rate of flow is \(80 \mathrm{dm}^{3} \mathrm{~s}^{-1}\) when the depth at the centre is \(0.25 \mathrm{~m}\), what must be the slope of the channel assuming \(C=45\) in SI units? \([1\) in 401]

A \(900 \mathrm{~mm}\) diameter conduit \(3600 \mathrm{~m}\) long is laid at a uniform slope of 1 in 1500 and connects two reservoirs. When the levels in the reservoirs are low the conduit runs partly full and it is found that a normal depth of \(600 \mathrm{~mm}\). gives a rate of flow of \(0.322 \mathrm{~m}^{3} \mathrm{~s}^{-1}\) The Chezy coefficient \(C\) is given by \(\mathrm{Km}^{\mathrm{m}}\), where \(\mathrm{K}\) is a constant, \(m\) is the hydraulic mean depth and \(n=\frac{1}{6}\). Neglecting losses of head at entry and exit obtain (a) the value of \(K\), (b) the discharge when the conduit is flowing full and the difference in level between the two reservoirs is \(4.5 \mathrm{~m}\). \(\left[(a) 67.6,(\right.\) b \(\left.) 0.562 \mathrm{~m}^{3} \mathrm{~s}^{-1}\right]\)

A circular section open conduit conveys liquid under maximum velocity conditions. Show that the depth of liquid is 81 per cent of the diameter. Show, without complete calculation, that this will not be the maximum discharge condition. Such a conduit having a diameter of \(0.8 \mathrm{~m}\) is to discharge \(0.6 \mathrm{~m}^{3} \mathrm{~s}^{-1}\) at maximum velocity. Find the required channel slope if the Chezy constant is 90 in SI units. \([1\) in 1050\(]\)
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