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Chapter 5: Problem 20

Calculate the velocity of propagation relative to the fluid of a small surface wave along a very wide channel in which the water is \(1.6 \mathrm{~m}\) deep. If the velocity of the stream is \(2 \mathrm{~m} \mathrm{~s}^{-1}\) what will be the Froude number?

Short Answer

Expert verified
The Froude number is approximately 0.505.

Step by step solution

01

Identify the Relevant Formula for Wave Velocity

To calculate the wave velocity relative to the fluid, we need to use the wave velocity formula for small surface waves in shallow water. The formula is given by:\[v = \sqrt{g \cdot d}\]where:- \(v\) is the wave velocity,- \(g\) is the acceleration due to gravity \(9.81 \ m/s^2\), and- \(d\) is the depth of the water \(1.6 \ m\).
02

Calculate the Wave Velocity

Substitute the known values into the formula:\[v = \sqrt{9.81 \ m/s^2 \cdot 1.6 \ m}\]Calculate the expression under the square root first:\[9.81 \ m/s^2 \times 1.6 \ m = 15.696 \ m^2/s^2\]Then, find the square root:\[v = \sqrt{15.696 \ m^2/s^2} \approx 3.96 \ m/s\]
03

Define and Calculate the Froude Number

The Froude Number \(Fr\) is given by:\[Fr = \frac{V}{v}\]where:- \(V\) is the velocity of the stream \(2 \ m/s\),- \(v\) is the wave velocity we calculated \(3.96 \ m/s\).
04

Perform Froude Number Calculation

Substitute the values into the Froude Number formula:\[Fr = \frac{2 \ m/s}{3.96 \ m/s} \approx 0.505\]Therefore, the Froude Number for the given conditions is approximately \(0.505\).

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Key Concepts

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

Wave Velocity
Wave velocity in fluid dynamics refers to the speed at which waves travel through a medium. To determine this speed in the context of shallow water waves, we use the formula \(v = \sqrt{g \cdot d}\). This equation highlights two main factors that dictate wave velocity:
  • The acceleration due to gravity \(g\), which is approximately \(9.81\, m/s^2\) on Earth.
  • The depth of the water \(d\), which in the exercise is \(1.6\, m\).
Combining these, we find the wave velocity is approximately \(3.96\, m/s\). It's crucial to understand that this formula is specific to small surface waves in shallow water conditions.
In situations where water is deeper or where wave heights are large, the equation can change, showing the contextual dependency of wave velocity calculations.
Shallow Water
Shallow water waves behave differently than waves in deeper waters. In fluid dynamics, 'shallow' refers to conditions where the water depth \(d\) is much less than the wavelength of the wave. Such environments require specific consideration because:
  • The water depth directly influences the speed and behavior of the wave.
  • Shallow waters tend to modify how energy is dispersed within the wave.
  • The interaction with the bottom is more significant, impacting wave speed.
The property of shallow water significantly reduces wave velocity when compared to deeper water scenarios.
This is why, for our situation, the wave velocity formula includes the water depth as a key component, emphasizing that wave motion in shallow water is more closely related to the depth than to other characteristics.
Fluid Dynamics
Fluid dynamics is the study of how fluids (liquids and gases) move and interact with forces. It covers numerous real-world applications ranging from engineering, meteorology, to oceanography. In our exercise, understanding the fluid dynamics principle helps connect concepts such as:
  • Wave velocity: Influenced by fluid motion and forces acting upon the fluid.
  • Froude number: A dimensionless number providing insight into flow dynamics.
  • Pressure and density variations: Affecting how waves propagate through the medium.
By calculating the Froude number, we assess the flow regime of the water, which indicates whether it is subcritical, critical, or supercritical. Knowing the flow type helps predict how waves will interact with structures or the riverbed, essential for engineering applications.

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Most popular questions from this chapter

A flat plate is struck normally by a jet of water \(50 \mathrm{~mm}\) in diameter with a velocity of \(18 \mathrm{~m} \mathrm{~s}^{-1}\). Calculate \((a)\) the force on the plate when it is stationary, \((b)\) the force on the plate when it moves in the same direction as the jet with a velocity of \(6 \mathrm{~m} \mathrm{~s}^{-1},(c)\) the work done per second and the efficiency in case \((b)\).

A ram-jet engine consumes \(20 \mathrm{~kg}\) of air per second and \(0.6 \mathrm{~kg}\) of fuel per second. The exit velocity of the gases is \(520 \mathrm{~m} \mathrm{~s}^{-1}\) relative to the engine and the flight velocity is \(200 \mathrm{~m} \mathrm{~s}^{-1}\) absolute. What is the power developed?

A jet of water delivers \(85 \mathrm{dm}^{3} \mathrm{~s}^{-1}\) at \(36 \mathrm{~m} \mathrm{~s}^{-1}\) onto a series of vanes moving in the same direction as the jet at \(18 \mathrm{~m} \mathrm{~s}^{-1}\). If stationary, the water which enters tangentially would be diverted through an angle of \(135^{\circ}\). Friction reduces the relative velocity at exit from the vanes to \(0.80\) of that at entrance. Determine the magnitude of the resultant force on the vanes and the efficiency of the arrangement. Assume no shock at entry.

A jet of water \(50 \mathrm{~mm}\) in diameter with a velocity of \(18 \mathrm{~m} \mathrm{~s}^{-1}\) strikes a flat plate inclined at an angle of \(25^{\circ}\) to the axis of the jet. Determine the normal force exerted on the plate \((a)\) when the plate is stationary, \((b)\) when the plate is moving at \(4.5 \mathrm{~m} \mathrm{~s}^{-1}\) in the direction of the jet, and \((c)\) determine the work done and the efficiency for case \((b)\).

Oil flows through a pipeline \(0.4 \mathrm{~m}\) in diameter. The flow is laminar and the velocity at any radius \(r\) is given by \(u=\left(0.6-15 r^{2}\right) \mathrm{m} \mathrm{s}^{-1}\). Calculate \((a)\) the volume rate of flow, (b) the mean velocity, \((c)\) the momentum correction factor.
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