/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 When the floodgates in a channel... [FREE SOLUTION] | 91Ó°ÊÓ

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When the floodgates in a channel are opened, water flows along the channel downstream of the gates with an increasing speed given by \(V=4(1+0.1 t) \mathrm{ft} / \mathrm{s},\) for \(0 \leq t \leq 20 \mathrm{s}\) where \(t\) is in seconds. For \(t \geq 20\) s the speed is a constant \(V=12 \mathrm{ft} / \mathrm{s}\). Consider a location in the curved channel where the radius of curvature of the streamlines is \(50 \mathrm{ft}\). For \(t=10 \mathrm{s}\) determine (a) the component of acceleration along the streamline, (b) the component of acceleration normal to the streamline, and (c) the net acceleration (magnitude and direction). Repeat for \(t=30 \mathrm{s}\)

Short Answer

Expert verified
For \(t = 10\) s: \(a_s = 0.4\) ft/s\(^2\), \(a_n = 1.28\) ft/s\(^2\), net acceleration \(= 1.34\) ft/s\(^2\). For \(t = 30\) s: \(a_s = 0\), \(a_n = 2.88\) ft/s\(^2\), net acceleration \(= 2.88\) ft/s\(^2\).

Step by step solution

01

Understanding the Velocity Function

The velocity of the water flow is given by two equations. When \(0 \leq t \leq 20\) seconds, the velocity \(V\) is given by \(V = 4(1 + 0.1t)\) ft/s. For \(t \geq 20\) seconds, the velocity is constant at \(12\) ft/s.
02

Finding Acceleration Along the Streamline for \(t = 10\) s

For \(t = 10\) seconds, we first calculate the streamline velocity:\[ V = 4(1 + 0.1 \times 10) = 4 \times 2 = 8 \text{ ft/s} \]The component of acceleration along the streamline can be found by differentiating \(V\) with respect to \(t\):\[ a_s = \frac{dV}{dt} = \frac{d}{dt}[4(1+0.1t)] = 4 \times 0.1 = 0.4 \text{ ft/s}^2 \]
03

Finding Acceleration Normal to the Streamline for \(t = 10\) s

The normal acceleration is given by:\[ a_n = \frac{V^2}{R} \text{ where } R = 50 \text{ ft} \]Using \(V = 8 \text{ ft/s}\):\[ a_n = \frac{8^2}{50} = \frac{64}{50} = 1.28 \text{ ft/s}^2 \]
04

Finding Net Acceleration for \(t = 10\) s

The net acceleration is the vector sum of the tangential and normal components:\[a = \sqrt{a_s^2 + a_n^2} = \sqrt{0.4^2 + 1.28^2} = \sqrt{0.16 + 1.6384} = \sqrt{1.7984} \approx 1.34 \text{ ft/s}^2 \]
05

Finding Acceleration for \(t = 30\) s Along the Streamline

For \(t \geq 20\) seconds, \(V = 12\) ft/s. The velocity is constant, thus the derivative \(a_s = \frac{dV}{dt} = 0\) ft/s\(^2\).
06

Finding Normal Acceleration for \(t = 30\) s

With \(V = 12\) ft/s, calculate the normal acceleration:\[ a_n = \frac{V^2}{R} = \frac{12^2}{50} = \frac{144}{50} = 2.88 \text{ ft/s}^2 \]
07

Finding Net Acceleration for \(t = 30\) s

Since \(a_s = 0\), the net acceleration equals the normal acceleration:\[ a = \sqrt{a_s^2 + a_n^2} = \sqrt{0 + 2.88^2} = 2.88 \text{ ft/s}^2 \]

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Key Concepts

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

Acceleration Along the Streamline
When examining fluid dynamics, it is important to understand how acceleration behaves along a streamline, which is the path followed by a fluid particle. The acceleration along the streamline, also known as tangential acceleration, reflects the change in speed of the fluid along its path.

In our problem, the velocity of water in the channel is initially described by the function \( V = 4(1 + 0.1t) \) ft/s for \( 0 \leq t \leq 20 \). Acceleration along the streamline can be identified by taking the derivative of this velocity with respect to time \( t \). This derivative tells us how the speed changes over time. For example, at \( t = 10 \) seconds, by differentiating the velocity equation, we find an acceleration of \( a_s = 0.4 \text{ ft/s}^2 \).

For times \( t \geq 20 \) seconds, the velocity becomes constant at \( V = 12 \text{ ft/s} \). Here, the acceleration along the streamline reduces to zero since the speed no longer changes. This aspect highlights the important concept that acceleration is linked to changing velocities.
Normal Acceleration
Normal acceleration, or radial acceleration, occurs perpendicular to the flow direction along a curved path. It indicates how quickly the direction of the fluid particle's velocity vector is changing, which is crucial for understanding the behavior in curved channels.

The formula for calculating normal acceleration is \( a_n = \frac{V^2}{R} \), where \( V \) is the velocity and \( R \) is the radius of curvature of the path. For instance, at \( t = 10 \text{ seconds} \) with a computed velocity of \( V = 8 \text{ ft/s} \) and a given radius \( R \) of 50 ft, normal acceleration comes out to be \( 1.28 \text{ ft/s}^2 \).

When \( t = 30 \) seconds, the velocity is 12 ft/s, and using the same formula, we derive \( a_n = 2.88 \text{ ft/s}^2 \). This acceleration remains even when the speed is constant, illustrating how curvature affects the motion of the fluid.
Net Acceleration
Net acceleration is a comprehensive measure that combines both the tangential and normal components of acceleration. It provides the total rate of change of velocity of the fluid particle, including both changes in speed and changes in direction.

Net acceleration is calculated using the Pythagorean theorem: \( a = \sqrt{a_s^2 + a_n^2} \). At \( t = 10 \) seconds, using our calculated values, the net acceleration is approximately \( 1.34 \text{ ft/s}^2 \). This value is the magnitude of the vector sum of both acceleration components.

For \( t = 30 \) seconds, since the velocity is constant and thus \( a_s = 0 \), the net acceleration equals the normal component, \( 2.88 \text{ ft/s}^2 \). Understanding net acceleration is key to predicting the overall motion and behavior of the fluid in the channel.

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Most popular questions from this chapter

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