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Show that Manning's \(n\) can be expressed in terms of the Darcy friction factor, \(f,\) by the following relation: $$ n=\frac{f^{\frac{1}{2}} R^{\frac{1}{6}}}{8.86} $$ where \(R\) is the hydraulic radius of the flow. Does this relationship conclusively show that \(n\) is a function of the flow depth? Explain.

Short Answer

Expert verified
Yes, via algebra with given equations. But \(n\) isn’t solely dependent on flow depth.

Step by step solution

01

Understanding the Problem

We need to express Manning's roughness coefficient \( n \) in terms of the Darcy friction factor \( f \) using the given relation. We want to show the derivation and discuss if \( n \) is a function of the flow depth.
02

Friction Factor Formula

We start with the Darcy-Weisbach equation for head loss: \( h_f = f \frac{L}{D} \frac{V^2}{2g} \). Here, \( V \) is the velocity, \( D \) is the hydraulic diameter (4 times the hydraulic radius for open channels), and \( g \) is the acceleration due to gravity.
03

Manning's Equation

Manning’s formula for the velocity is given by \( V = \frac{1}{n}R^{\frac{2}{3}}S^{\frac{1}{2}} \), where \( R \) is the hydraulic radius, \( S \) is the slope, and \( n \) is Manning's roughness coefficient.
04

Relate Darcy and Manning

To relate these, realize that velocity \( V \) can be solved using both equations for energy loss. Equating \( \frac{1}{n}R^{\frac{2}{3}}S^{\frac{1}{2}} \) and \( \sqrt{\frac{8gR}{fL}} \), match terms involving \( f \) and \( n \) to obtain \( n \) in terms of \( f \).
05

Derive the Expression for \( n \)

Through algebraic manipulation and solving for \( n \), deduce that \( n = \frac{f^{\frac{1}{2}} R^{\frac{1}{6}}}{8.86} \), showing the mathematical relation given in the problem.
06

Discussion of Flow Dependence

Given \( R \) involves flow, this shows \( n \) depends on the hydraulic radius (related to flow depth). This doesn’t conclusively show \( n \) as a sole function of flow depth because \( R \) can change with flow area and wetted perimeter independently.

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Key Concepts

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

Darcy Friction Factor
The Darcy friction factor, often denoted as \( f \), is a dimensionless quantity used primarily in the Darcy-Weisbach equation to determine the head loss due to friction in pipes or open channels. It is a key component of flow calculations in fluid mechanics.

This factor quantifies the impact of friction along the walls of the channel on the flow of the fluid. Mathematically, it's involved in the equation:
  • \( h_f = f \frac{L}{D} \frac{V^2}{2g} \)
where:
  • \( h_f \) is the head loss,
  • \( L \) is the length of the pipe or open channel,
  • \( D \) is the hydraulic diameter, which is four times the hydraulic radius \( R \) in open channels,
  • \( V \) is the velocity of the fluid,
  • \( g \) is the acceleration due to gravity.
Determining the Darcy friction factor involves considerations such as the roughness of the channel material and the Reynolds number, an indicator of flow regime.
This factor is critical to engineers designing piping systems, as it helps calculate efficient layouts by minimizing energy losses.
Hydraulic Radius
The hydraulic radius, denoted as \( R \), is a measure used to relate the shape of a conduit to its flow characteristics. It's a crucial concept in determining flow behavior in channels and pipes. The hydraulic radius is defined as:
  • \( R = \frac{A}{P} \)
where:
  • \( A \) is the cross-sectional area of flow, and
  • \( P \) is the wetted perimeter of the channel.

    This ratio reflects how efficiently a channel can convey flow—larger hydraulic radii generally indicate more efficient flow with lower frictional resistance. The hydraulic radius is important because it provides insights into how easily fluid can move through a channel, which directly affects such things as velocity and energy loss calculations.
    The hydraulic radius is also a key variable in both Manning's and Darcy-Weisbach equations. Understanding how it relates to flow depth can help identify the dependence of certain coefficients on geometric or channel properties, which will have practical implications for water resource management and engineering projects.
Manning's Equation
Manning's Equation is an empirical formula used to estimate the velocity of flow in an open channel. It connects channel characteristics, flow properties, and the roughness of the channel. The equation looks like this:
  • \( V = \frac{1}{n} R^{\frac{2}{3}} S^{\frac{1}{2}} \)
where:
  • \( V \) is the flow velocity,
  • \( n \) is Manning's roughness coefficient,
  • \( R \) is the hydraulic radius, and
  • \( S \) is the slope of the energy line (usually the slope of the channel bed).
This equation assumes uniform, steady flow and is particularly useful in calculating flow in natural streams or man-made channels.

Manning's roughness coefficient \( n \) reflects the roughness of the channel surface. It can vary depending on the material of the channel bed, vegetation, obstructions, and the degree of meandering.
Manning’s Equation is instrumental in hydraulic design and understanding open channel flow but has limitations. It does not account for losses due to bends, contractions, or expansions in the channel, and it assumes the flow is fully turbulent. Therefore, care must be taken in its application, especially in non-practical, irregular channels.

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Most popular questions from this chapter

Water flows at \(10 \mathrm{~m}^{3} / \mathrm{s}\) in a rectangular channel of width \(5.5 \mathrm{~m}\). The slope of the channel is \(0.15 \%,\) and the Manning roughness coefficient is 0.038 . Use the following methods to estimate the flow depth \(100 \mathrm{~m}\) upstream of a section where the flow depth is \(2.2 \mathrm{~m}:\) (a) the direct integration method and (b) the standard step method. Approximately how far upstream of this section would you expect to find uniform flow?

Water discharges from a large storage reservoir via a trapezoidal channel of bottom width \(3.00 \mathrm{~m}\), side slopes 3: 1 (H:V), and longitudinal slope of \(0.5 \%\). Manning's \(n\) in the channel is estimated as \(0.025 .\) Estimate the discharge from the reservoir when the depth of the reservoir at the discharge location is \(2.00 \mathrm{~m}\).

A rectangular channel \(3 \mathrm{~m}\) wide carries \(4 \mathrm{~m}^{3} / \mathrm{s}\) of water at a depth of \(1.5 \mathrm{~m}\). If an obstruction \(15 \mathrm{~cm}\) high is placed across the channel, calculate the elevation of the water surface over the obstruction. What is the maximum height of the obstruction that will not cause a rise in the water surface upstream?

Use the Darcy-Weisbach equation to show that the head loss per unit length, \(S\), between any two sections in an open channel can be estimated by the relation $$ S=\frac{\bar{f}}{4 \bar{R}} \frac{\bar{V}^{2}}{2 g} $$ where \(\bar{f}, \bar{R},\) and \(\bar{V}\) are the average friction factor, hydraulic radius, and flow velocity, respectively, between the upstream and downstream sections.

Water flows in a trapezoidal channel that has a bottom width of \(5 \mathrm{~m}\), side slopes of \(2: 1(\mathrm{H:V}),\) and a longitudinal slope of 0.0001 . The channel has an equivalent sand roughness of \(1 \mathrm{~mm}\). Calculate the uniform flow depth in the channel when the flow rate is \(18 \mathrm{~m}^{3} / \mathrm{s}\). Is the flow hydraulically rough, smooth, or in transition? Would the Manning equation be valid in this case? Explain.

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