/*! 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 55 Air flows into a pipe from the r... [FREE SOLUTION] | 91Ó°ÊÓ

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Air flows into a pipe from the region between a circular disk and a cone as shown in Fig. P4.55. The fluid velocity in the gap between the disk and the cone is closely approximated by \(V=V_{0} R^{2} / r^{2},\) where \(R\) is the radius of the disk, \(r\) is the radial coordinate, and \(V_{0}\) is the fluid velocity at the edge of the disk. Determine the acceleration for \(r=0.5\) and \(2 \mathrm{ft}\) if \(V_{0}=5 \mathrm{ft} / \mathrm{s}\) and \(R=2 \mathrm{ft}\).

Short Answer

Expert verified
Acceleration at \( r = 0.5 \) ft is \(-25600 \text{ ft/s}^2\); at \( r = 2 \) ft is \(-25 \text{ ft/s}^2\).

Step by step solution

01

Identify Given Variables

We have the initial data: \( V_0 = 5 \text{ ft/s} \), \( R = 2 \text{ ft} \), and we need to find the acceleration at \( r = 0.5 \text{ ft} \) and \( r = 2 \text{ ft} \).
02

Derive Velocity Expression

The velocity, \( V(r) \), is given by the formula: \[ V(r) = V_0 \frac{R^2}{r^2} \]By substituting the known values, the velocity formula becomes: \[ V(r) = 5 \frac{2^2}{r^2} = \frac{20}{r^2} \]
03

Calculate Velocity at Given Distances

Now, calculate \( V \) at \( r = 0.5 \text{ ft} \):\[ V(0.5) = \frac{20}{(0.5)^2} = \frac{20}{0.25} = 80 \text{ ft/s} \]Next, calculate \( V \) at \( r = 2 \text{ ft} \):\[ V(2) = \frac{20}{2^2} = \frac{20}{4} = 5 \text{ ft/s} \]
04

Determine the Radial Acceleration Formula

The radial acceleration, \(a_r\), is given by the derivative of velocity with respect to time:\[ a_r = \frac{dV}{dt} \]Using the chain rule, this becomes:\[ a_r = \frac{dV}{dr}\cdot\frac{dr}{dt} = \frac{dV}{dr}\cdot V \]Given \( V = \frac{20}{r^2} \), find \( \frac{dV}{dr} \):\[ \frac{dV}{dr} = -40r^{-3} \]
05

Calculate Acceleration at r = 0.5 ft

Using \( a_r = \frac{dV}{dr}\cdot V \) with \( r = 0.5 \text{ ft} \): \[ a_r = (-40(0.5)^{-3}) \cdot 80 = (-40 \times 8) \cdot 80 = -40 \times 640 = -25600 \text{ ft/s}^2 \]
06

Calculate Acceleration at r = 2 ft

Using \( a_r = \frac{dV}{dr}\cdot V \) with \( r = 2 \text{ ft} \):\[ a_r = (-40(2)^{-3}) \cdot 5 = (-40 \times 0.125) \cdot 5 = -5 \cdot 5 = -25 \text{ ft/s}^2 \]
07

Summarize Results

The acceleration of the fluid at \( r = 0.5 \) ft is \(-25600 \text{ ft/s}^2\), and at \( r = 2 \) ft is \(-25 \text{ ft/s}^2\).

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Key Concepts

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

Radial Acceleration
In fluid mechanics, radial acceleration is a component of acceleration experienced by a fluid particle moving in a path rather than in a straight line. It is critical in scenarios where the velocity of a fluid varies with radial position, such as fluid flowing between a cone and a disk.

To determine radial acceleration, we use the derivative of velocity with respect to time. This process often involves the chain rule of calculus. Specifically, the formula involves multiplying the derivative of velocity with respect to the radial coordinate and the velocity itself:
  • Given velocity as a function of radius, \( V(r) = \frac{20}{r^2} \), differentiate with respect to \( r \).
  • The derivative, \( \frac{dV}{dr} \), is \( -40r^{-3} \).
  • Radial acceleration becomes \( a_r = \frac{dV}{dr} \cdot V \).
At \( r = 0.5 \) ft, the acceleration is found to be quite intense at \(-25600 \text{ ft/s}^2\). This high value indicates rapid changes in speed within a small radial distance, showing the sensitivity of the fluid flow to changes in radius.
Velocity Profile
The velocity profile describes how the speed of a fluid varies across different positions in its flow path. It is essential for predicting how quickly fluid particles move at different points, especially when factors like geometric constraints are involved. Using the formula for velocity in our problem:
  • The given function is \( V(r) = \frac{20}{r^2} \), showing inversely proportional relationship with the square of the radial coordinate \( r \).
  • As \( r \) decreases, the velocity \( V \) increases significantly, such as at \( r = 0.5 \) ft with \( V(0.5) = 80 \text{ ft/s} \).
  • Conversely, at larger \( r = 2 \) ft, the velocity reduces back to its minimum at \( V(2) = 5 \text{ ft/s} \).
The velocity profile is critical in designing systems where controlled flow is necessary, like in chemical reactors or HVAC systems. It allows us to predict how fluid will behave and adjust system geometry and flow rates accordingly.
Differential Calculus
Differential calculus is indispensable for analyzing changes in fluid properties, as it allows us to compute rates of change like velocity and acceleration. In fluid flow problems, it enables us to perform the following:
  • Determine the velocity function \( V(r) \) and its derivative \( \frac{dV}{dr} \).
  • By using the power rule of differentiation, we found \( \frac{dV}{dr} = -40r^{-3} \), illustrating a rapid rate of change.
  • Calculate acceleration through advanced calculus techniques such as the chain rule, which links radial and temporal changes.
In this context, differential calculus provides a robust framework for predicting fluid behavior under varying conditions. It supports the design and optimization of numerous engineering systems where fluid dynamics play a key role.

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Most popular questions from this chapter

A bicyclist leaves from her home at 9 A.M. and rides to a beach \(40 \mathrm{mi}\) away. Because of a breeze off the ocean, the temperature at the beach remains \(60^{\circ} \mathrm{F}\) throughout the day. At the cyclist's home the temperature increases linearly with time, going from \(60^{\circ} \mathrm{F}\) at \(9 \mathrm{A} . \mathrm{M}\). to \(80^{\circ} \mathrm{F}\) by \(1 \mathrm{P.M.}\) The temperature is assumed to vary linearly as a function of position between the cyclist's home and the beach. Determine the rate of change of temperature observed by the cyclist for the following conditions: (a) as she pedals 10 mph through a town 10 mi from her home at 10 A.M.; (b) as she cats lunch at a rest stop \(30 \mathrm{mi}\) from her home at noon; (c) as she arrives enthusiastically at the beach at 1 P.M., pedaling 20 mph.

The velocity field of a flow is given by \(\mathbf{V}=\) \((5 z-3) i+(x+4) \hat{\mathbf{j}}+4 y \hat{\mathbf{k}} \mathrm{ft} / \mathrm{s},\) where \(x, y,\) and \(z\) are in feet. Determine the fluid speed at the origin \((x=y=z=0)\) and on the \(x\) axis \((y=z=0)\)

A 10 -ft-diameter dust devil that rotates one revolution per second travels across the Martian surface (in the \(x\) -direction) with a speed of \(5 \mathrm{ft} / \mathrm{s}\), Plot the pathline etched on the surface by a fluid particle \(10 \mathrm{ft}\) from the center of the dust devil for time \(0 \leq t \leq 3\) s. The particle position is given by the sum of that for a stationary swirl \([x=10 \cos (2 \pi t), y=10 \sin (2 \pi t)]\) and that for a uniform velocity \((x=5 t, y=\text { constant }),\) where \(x\) and \(y\) are in feet and \(t\) is in seconds.

\(A\) fluid flows along the \(x\) axis with a velocity given by \(\mathbf{V}=(x / t) \hat{\mathbf{i}},\) where \(x\) is in feet and \(t\) in seconds. (a) Plot the speed for \(0 \leq x \leq 10\) ft and \(t=3 \mathrm{s}\). (b) Plot the speed for \(x=7\) ft and \(2 \leq t \leq 4\) (c) Determine the local and convective acceleration. (d) Show that the acceleration of any fluid particle in the flow is zero. (e) Explain physically how the velocity of a particle in this unsteady flow remains constant throughout its motion.

The velocity field of a flow is given by \(\mathbf{V}=2 x^{2}+\hat{\mathbf{i}}+\) \(\left[4 y(t-1)+2 x^{2} t\right] \hat{j} \mathrm{m} / \mathrm{s},\) where \(x\) and \(y\) are in meters and \(t\) is in seconds. For fluid particles on the \(x\) axis, determine the speed and direction of flow.

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