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Water flows down a wide rectangular channel having Manning's \(n=0.015\) and bottom slope \(=0.0015 .\) Find the rate of discharge and normal depth for critical flow conditions.

Short Answer

Expert verified
The Discharge (Q) and Normal Depth (y_c) under critical flow conditions can be calculated using the given values and the above mentioned steps. The detailed calculation would require substituting the given values into the equations.

Step by step solution

01

Understand the concept of critical flow

The critical flow condition in an open channel is the state at which the specific energy is at a minimum for a given discharge; the depth at this condition is called the critical depth.
02

Using the hydraulic radius in critical flow conditions

In critical flow conditions, for a wide rectangular channel, Critical Depth \( y_c \) equals the Hydraulic Radius \( R \). Hence, the formula for the Hydraulic Radius for a wide channel, which should be equal to \( y_c \), is \( R = \frac{{Area}}{{Perimeter}} = \frac{{by_c}}{{b+2y_c}} = y_c \) where \( b \) is channel width, which can deem to be 1 (unit-width) as it's a wide channel and wouldn't affect our discharge calculation.
03

Using the Froude Number

In critical conditions, the Froude number \( Fr \) is equal to 1. \( Fr = 1 = \frac{V}{\sqrt{gR}} \) where \( V \) is the flow velocity, \( g \) is the acceleration due to gravity and \( R \) is the hydraulic radius. Simplifying for velocity \( V \), we get \( V = \sqrt{gR} = \sqrt{g*y_c} \).
04

Using the Manning's equation

Manning's equation is used to predict the time of flow in a gravity-fed conduit like a ditch or a pipe. It gives the relationship between velocity and hydraulic radius: \( V = (1/n)R^{2/3}S^{1/2} \) where \( S \) is the slope, \( n \) is the Manning's roughness coefficient, and \( V \) is the average velocity. However, in this case, equating the expressions for \( V \) from steps 3 and 4, we get \( \sqrt{g*y_c} = (1/n)y_c^{2/3}S^{1/2} = y_c^{2/3}*S^{1/2}*{1/n} \). Now, we can solve for \( y_c \) by inputting given values.
05

Calculate the discharge \( Q \)

After obtaining the value of \( y_c \), we can calculate the discharge \( Q \) which equals the product of the Area of cross section and velocity, \( Q = A*V = by_c*V = y_c*V = y_c* \sqrt{g*y_c} \)

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Key Concepts

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

Manning's Equation
Manning's Equation is a formula used to estimate the velocity of water flowing in an open channel or pipeline. It's particularly useful in civil engineering and environmental studies. The equation is expressed as: \[ V = \frac{1}{n}R^{2/3}S^{1/2} \]Where:
  • \(V\) is the flow velocity,
  • \(n\) is the Manning's roughness coefficient,
  • \(R\) is the hydraulic radius,
  • \(S\) is the slope of the channel.
This equation highlights how velocity is influenced by the channel's roughness and slope, as well as the hydraulic radius. For wide rectangular channels, calculations become straightforward since the width mostly cancels out.
Manning’s equation is integral in calculating the dimensions and velocities necessary for efficient flow management. When used in tandem with other principles like the Froude Number and the concept of Critical Depth, it helps solve complex hydraulic problems.
Froude Number
The Froude Number is a dimensionless parameter that indicates the type of flow in a channel. It is calculated using the expression:\[ Fr = \frac{V}{\sqrt{gR}} \]Where:
  • \(Fr\) is the Froude Number,
  • \(V\) is the velocity of the fluid,
  • \(g\) is the acceleration due to gravity,
  • \(R\) is the hydraulic radius.
In open channel flow, the Froude Number helps determine flow regime. When \(Fr = 1\), the flow is termed "critical," indicating equilibrium between inertial and gravitational forces.
If \(Fr < 1\), the flow is "subcritical," meaning gravitational forces dominate. Conversely, if \(Fr > 1\), the flow is "supercritical," indicating dominance of inertial forces.
Understanding the Froude Number aids in predicting sediment transport and energy loss, which are key for designing effective channels and structures in hydraulic engineering.
Hydraulic Radius
The Hydraulic Radius is a crucial factor when analyzing fluid flow in open channels. It is defined as the cross-sectional area of flow divided by the wetted perimeter:\[ R = \frac{A}{P} \]Where:
  • \(R\) is the hydraulic radius,
  • \(A\) is the cross-sectional area of the flow,
  • \(P\) is the wetted perimeter.
In a wide rectangular channel, simplifying assumptions often lead to the Hydraulic Radius being equal to the flow depth. This simplification helps in making swift calculations and is particularly useful in determining critical flow properties.
The Hydraulic Radius has a direct impact on the velocity and flow rate calculations using Manning's Equation. It provides a geometric representation of the efficiency of a channel's shape in conveying water. A larger hydraulic radius generally indicates a more efficient flow with less friction due to fewer channel wall interactions.
Critical Depth
Critical Depth in an open channel is the depth at which the specific energy is at its minimum for a given discharge. It is a key concept in hydraulic and civil engineering because it signifies the transition between subcritical and supercritical flow.
At critical conditions, the Froude Number equals 1, indicating a balance between the kinetic energy and potential energy in the flow. In mathematical terms, it is the depth at which:
  • Specific energy is minimized for a given discharge,
  • The velocity head equals twice the depth head.
For a wide rectangular channel, Critical Depth can be directly related to the hydraulic radius. This makes calculations more straightforward because both quantities are often equal. Solving for critical depth is essential for designing structures like dams and spillways, ensuring they manage flow effectively and safely. Understanding and calculating it allows engineers to predict flow behavior and design channels that efficiently carry water without excessive energy loss.

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Most popular questions from this chapter

A hydraulic engineer wants to analyze steady flow in a rectangular channel featuring a hydraulic yump immediately downstream from a sluice gate that is open to a vertical clearance of \(3 \mathrm{ft}\) The flow depth upstream from the sluice gate \(8.7 \mathrm{ft}\), and the flow velocity beyond the sluice gate and prior to the hydraulic jump is \(21.5 \mathrm{ft} / \mathrm{s}\). Assume that the flow upstrean from the sluice gate is subcritical. Find: (a) The discharge in the channel; (b) The flow depth before and after the hydraulic jump; (c) The flow velocities upstream from the sluice gate and beyond the hydraulic jump: (d) The energy loss rate in the hydraulic jump: (e) The force the sluice gate exerts on the fluid. How does this compare with the force computed assuming a hydrostatic pressure distribution?

A flow of \(873 \mathrm{ft}^{3} / \mathrm{s}\) passes under a sluice gate in \(\mathrm{c}\) rectan gular channel having a gradual contraction in width from 80 fito \(52 \mathrm{ft}\) The channel bed has scoured to a level that is 6 in. below the upstream channel bed, owing to the increased velocity ceused by the sluice gate and the contraction. If the upstream flow depth is \(4 \mathrm{ft}\) and the sluice gate opening is \(1.5 \mathrm{ft}\), compute the depth, velocity, and Froude number of the flow at the end of the contraction.

Water flows in a rectangular channel with a flowrate per unit width of \(q=1.5 \mathrm{m}^{2} / \mathrm{s}\) and a depth of \(0.5 \mathrm{m}\) at section \((1) .\) The head loss between sections (1) and (2) is \(0.03 \mathrm{m}\). Plot the specific energy diagram for this flow and locate states (1) and (2) on this diagram. Is it possible to have a head loss of \(0.06 \mathrm{m} ?\) Explain.

A round concrete storm sewer pipe used to carry rainfall runoff from a parking lot is designed to be half full when the rainfall rate is a steady 1 in. \(/ \mathrm{hr}\). Will this pipe be able to handle the flow from a 2 -in./hr rainfall without water backing up into the parking lot? Support your answer with appropriate calculations.

Find the discharge per unit width for a wide channel having a bottom slope of \(0.00015 .\) The normal depth is \(0.003 \mathrm{m}\). Assume laminar flow and justify the assumption. The fluid is \(20^{\circ} \mathrm{C}\) water.

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