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The Froude number, Fr, at any cross section of an open channel is defined by the relation $$\mathrm{Fr}=\frac{\bar{V}}{\sqrt{g D_{\mathrm{h}}}}$$ where \(\bar{V}\) is the average velocity, \(g\) is the acceleration due to gravity, and \(D_{\mathrm{h}}\) is the hydraulic depth. The hydraulic depth is defined as \(A / T,\) where \(A\) is the flow area and \(T\) is the top width of the flow area. (a) Show that Fr is dimensionless. (b) Determine the value of Fr in a trapezoidal channel that has a bottom width of \(3 \mathrm{~m}\), side slopes \(2.5: 1(\mathrm{H}: \mathrm{V}),\) an average velocity of \(0.4 \mathrm{~m} / \mathrm{s},\) and a flow depth of \(1.5 \mathrm{~m} .\)

Short Answer

Expert verified
(a) The Froude number is dimensionless. (b) Fr is approximately 0.1301 for the trapezoidal channel.

Step by step solution

01

Show Froude Number is Dimensionless

The Froude number is given as \( \mathrm{Fr} = \frac{\bar{V}}{\sqrt{g D_{\mathrm{h}}}} \). To show \( \mathrm{Fr} \) is dimensionless, we analyze each component's dimensions. The average velocity, \( \bar{V} \), has the dimension of \([L T^{-1}]\). The acceleration due to gravity, \( g \), has the dimension of \([L T^{-2}]\). The hydraulic depth \( D_h = \frac{A}{T} \) has the dimension of length \([L]\). Therefore, \( \sqrt{g D_{\mathrm{h}}} \) has the dimension \(\sqrt{[L T^{-2}] [L]} = [L T^{-1}]\). Thus, \( \mathrm{Fr} = \frac{[L T^{-1}]}{[L T^{-1}]} = [1]\), indicating it is dimensionless.
02

Calculate Flow Area and Top Width for Trapezoidal Channel

First, calculate the flow area \( A \) and top width \( T \) for a trapezoidal cross-section. The base width \( b = 3 \mathrm{~m} \), the flow depth \( y = 1.5 \mathrm{~m} \), and side slopes are \(2.5:1\). **Flow Area**: The total top width of the channel is \( T = b + 2 \cdot (2.5y) = 3 + 2 \cdot (2.5 \times 1.5) = 10.5 \mathrm{~m} \). **Flow Area**: Use the formula \( A = (b + \text{top width}) \cdot y / 2 = (3 + 10.5) \cdot 1.5 / 2 = 10.125 \mathrm{~m}^2 \).
03

Compute Hydraulic Depth

The hydraulic depth \( D_h \) is given by \( D_h = \frac{A}{T} = \frac{10.125}{10.5} \approx 0.9643 \mathrm{~m} \).
04

Calculate Froude Number for Trapezoidal Channel

Substitute the values into the Froude number equation. Use \( \bar{V} = 0.4 \mathrm{~m/s} \), \( g = 9.81 \mathrm{~m/s}^2 \), and \( D_h = 0.9643 \mathrm{~m} \). Compute \( \sqrt{g D_h} = \sqrt{9.81 \times 0.9643} \approx 3.072 \mathrm{~m/s} \), so \( \mathrm{Fr} = \frac{0.4}{3.072} \approx 0.1301 \).

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Key Concepts

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

Dimensionless Quantity
The concept of a dimensionless quantity is pivotal in the field of fluid dynamics, and the Froude number is an excellent example of such a quantity. A dimensionless number, like the Froude number, does not have any physical dimensions such as length, mass, or time attached to it. This means they remain constant irrespective of the measurement units applied. This property is useful for comparing different situations or setups using non-dimensional scaling methods.

In the case of the Froude number, it is expressed mathematically as \( \mathrm{Fr} = \frac{\bar{V}}{\sqrt{g D_{\mathrm{h}}}} \). Here, all components—velocity \( \bar{V} \) and the square root of the product of gravity \( g \) and hydraulic depth \( D_{\mathrm{h}} \)—maintain consistency by canceling out each other's dimensions. As demonstrated in the solution, when these dimensions are analyzed, they result in \([1]\), which defines the number as dimensionless. Understanding this concept helps in interpreting and applying mathematical models to real-world fluid flow scenarios.
Hydraulic Depth
Hydraulic depth is a critical concept in understanding the behavior of water flow through open channels. It is defined as the flow area \( A \) divided by the top width \( T \) of the channel, expressed as \( D_{\mathrm{h}} = \frac{A}{T} \). It offers insights into the efficiency and capacity of the channel to convey water.

In practical terms, hydraulic depth indicates how deep the water would be if the flow area was uniformly rectangular across its width. This parameter aids in predicting the channel's capability to handle varying water volumes, potentially assisting with flood management and irrigation planning. By simplifying complex river or stream cross-sections into a single value, hydraulic depth allows engineers to apply theoretical models and compute key parameters like the Froude number accurately.
Open Channel Flow
Open channel flow refers to fluid flow with a free surface open to atmospheric pressure, unlike closed conduits like pipes. This often includes flows in rivers, streams, and human-made channels. The governing principles of open channel flow differ from those of closed conduits because of the presence of a free surface. Changes in the surface level can occur due to various factors, including flow rate fluctuations and channel shape.

Understanding open channel flow is crucial for efficient water resource management, predicting natural event impacts like floods, and designing infrastructure like canals. Open channel flow can be further categorized based on flow uniformity (steady or unsteady) and flow regime (laminar or turbulent), each having distinct characteristics affecting flow measurement and management strategies.
Trapezoidal Channel
A trapezoidal channel is a type of open channel flow design with a distinct trapezoidal cross-section. Its base width is accompanied by sloped sides which help in managing debris, ease construction, and provide stability. Key features of a trapezoidal channel include:
  • **Base Width**: The straight bottom part of the channel. In our exercise, this is 3 meters.
  • **Side Slopes**: These extend upwards from the base. For instance, a 2.5:1 (horizontal: vertical) slope means for every 2.5-meter horizontal distance, there is a 1-meter rise.
  • **Flow Area**: This is calculated considering both the base and the top resulting from the sloped sides, contributing to the total water conveyance capacity.

Trapezoidal channels are widely used in applications such as irrigation and flood management and are preferred for their practical benefits, including lower chances of erosion and ease of maintenance. Understanding their geometry and flow characteristics helps in accurately computing hydraulic elements like depth and velocity, which are essential for efficient water flow management.

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Most popular questions from this chapter

A 1: 60 scale model of a ship is used in a water tank to simulate a ship speed of 10 \(\mathrm{m} / \mathrm{s}\). What should be the model speed? If a towing force of \(10 \mathrm{~N}\) is measured in the model, what force is expected on the prototype? Neglect viscous effects.

Consider the energy equation for flow of a fluid in a closed conduit, which can be expressed in the form $$\frac{p_{1}}{\gamma}+\frac{V_{1}^{2}}{2 g}+z_{1}=\frac{p_{2}}{\gamma}+\frac{V_{2}^{2}}{2 g}+z_{2}+h_{\mathrm{f}}$$ where \(p_{1}, V_{1}\), and \(z_{1}\) are the pressure, average velocity, and centerline elevation at an upstream section of the conduit, \(p_{2}, V_{2},\) and \(z_{2}\) are the corresponding variables at a downstream section, \(h_{f}\) is the head loss between the two sections, \(\gamma\) is the specific weigh of the fluid, and \(g\) is the gravity constant. State the SI unit of energy and state the SI unit(s) of the terms (not the individual variables) in Equation 6.27. Explain why each of the terms in Equation 6.27 represents energy and express Equation 6.27 as a nondimensional equation.

A boat of length \(L\) is designed to move through water at a velocity \(V\) while generating bow waves of height \(H\). To study the range of wave heights that will be generated by the boat, a scale model of the boat is constructed and tested in a hydraulics laboratory. Viscous effects might be important at the laboratory scale. (a) What nondimensional relationship will you use to design your laboratory study? (b) How will you use this relationship? (c) If your model scale is \(1: 10,\) the model length is \(60 \mathrm{~cm},\) and you measure a wave height of \(5 \mathrm{~cm}\) when the model boat is moving at \(30 \mathrm{~cm} / \mathrm{s}\), what are the corresponding conditions in the prototype? (d) How does the ratio of wave height to boat length in the model compare with the corresponding ratio in the prototype?

When an object of a given shape falls very slowly in a fluid, the steady-state velocity of the object is observed to depend only on the size of the object and the density and viscosity of the fluid. Use dimensional analysis to determine the functional relationship between the steady-state velocity and the influencing variables.

The momentum equation that describes the motion of a nonviscous incompressible fluid in a three-dimensional \((x y z)\) flow field can be expressed in the form $$\rho\left[\frac{\partial \mathbf{V}}{\partial t}+u \frac{\partial \mathbf{V}}{\partial x}+v \frac{\partial \mathbf{V}}{\partial y}+w \frac{\partial \mathbf{V}}{\partial z}\right]=-\nabla p-\rho g \mathbf{k}$$ where \(\rho\) is the density of the fluid, \(\mathbf{V}\) is the velocity vector with components \(u, v,\) and \(w, p\) is the pressure, and \(g\) is the gravity constant. Consider the case in which the only relevant scales are the length scale, \(L,\) and the velocity scale, \(V\). Note that time can be normalized by \(L / V\) and pressure can be normalized by \(\rho V^{2}\). Express Equation 6.29 in normalized form, using the Froude number, Fr, defined as \(\mathrm{Fr}=V / \sqrt{g L},\) in the final expression. What happens to the effect of gravity on the flow as the Froude number becomes large?

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