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Chapter 4: Problem 27

A nozzle is designed to accelerate the fluid from \(V_{1}\) to \(V_{2}\) in a linear fashion, That is, \(V=a x+b,\) where \(a\) and \(b\) are constants. If the flow is constant with \(V_{t}=10 \mathrm{m} / \mathrm{s}\) at \(x_{1}=0\) and \(V_{2}=25 \mathrm{m} / \mathrm{s}\) at \(x_{2}=1 \mathrm{m},\) determine the local acceleration, the convective acceleration, and the acceleration of the fluid at points (1) and (2).

Short Answer

Expert verified
Local acceleration is 0; convective is 150 m/s虏 at x鈧�=0 and 375 m/s虏 at x鈧�=1.

Step by step solution

01

Understand the Equation

The velocity of the fluid is given by the linear equation \( V=ax+b \). We need to find \( a \) and \( b \) using the given data points: \( V_1 = 10 \, \text{m/s} \) at \( x_1 = 0 \) and \( V_2 = 25 \, \text{m/s} \) at \( x_2 = 1 \, \text{m} \).
02

Solve for Constants a and b

Using the first condition \( V_1 = a \cdot 0 + b = 10 \), we find \( b = 10 \). Using the second condition \( V_2 = a \cdot 1 + 10 = 25 \), we solve for \( a \):\\[ a = 25 - 10 = 15 \]\Thus, \( a = 15 \) and \( b = 10 \).
03

Local Acceleration

Local acceleration is the time derivative of velocity for the changes in time but here only spatial change is provided and no temporal term, so local acceleration is zero since \(\frac{\partial V}{\partial t} = 0.\)
04

Convective Acceleration

Convective acceleration is given by \( V \frac{dV}{dx} \). First, we find \( \frac{dV}{dx} = a = 15 \). Thus, convective acceleration is \( V \cdot 15 \).
05

Calculate Acceleration at Point x_1

At \( x_1 = 0 \), \( V_1 = 10 \, \text{m/s} \). Convective acceleration is \( 10 \cdot 15 = 150 \, \text{m/s}^2 \). The total acceleration is therefore \( 150 \, \text{m/s}^2 \) since local acceleration is zero.
06

Calculate Acceleration at Point x_2

At \( x_2 = 1 \), \( V_2 = 25 \, \text{m/s} \). The convective acceleration is \( 25 \cdot 15 = 375 \, \text{m/s}^2 \). The total acceleration at this point is \( 375 \, \text{m/s}^2 \).
07

Summary of Results

The local acceleration is 0 at both points. The convective acceleration is \( 150 \, \text{m/s}^2 \) at \( x_1 = 0 \) and \( 375 \, \text{m/s}^2 \) at \( x_2 = 1 \). Total acceleration equals the convective acceleration at both points.

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Key Concepts

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

Local Acceleration
Local acceleration is a key concept in fluid dynamics. It represents changes in flow velocity due to variation over time. In scenarios where the velocity changes purely with respect to position, like in our nozzle problem, the local acceleration becomes zero.
This is because local acceleration is calculated by the time derivative of velocity, expressed as \( \frac{\partial V}{\partial t} \). When there鈥檚 no time-dependent change, such as the nozzle鈥檚 design to alter velocity along its length only, this derivative is zero. This means no local acceleration occurs as fluid flows through the nozzle, since every part of the fluid experiences the same velocity change at any given moment.
Therefore, in fluid dynamics, when considering problems set in steady, position-only dependent scenarios, you can often expect local acceleration to be zero as is shown. This simplifies analysis by allowing focus on other acceleration forms.
Convective Acceleration
Convective acceleration is integral to understanding fluid behavior, particularly in flows where velocity is dependent on spatial variables. Unlike local acceleration, convective acceleration arises because of spatial changes in velocity, necessitating fluid speed differentials as the flow progresses.
It is calculated using the expression \( V \frac{dV}{dx} \), where \( \frac{dV}{dx} \) is the spatial derivative of velocity. This derivative signifies the change rate of velocity with respect to position along the nozzle. In the problem provided, the derivative \( \frac{dV}{dx} \) is constant at 15, obtained from the linear velocity profile \( V = 15x + 10 \). Thus, convective acceleration depends on the local speed \( V \) at each point:
  • At \( x_1 = 0 \), with \( V_1 = 10 \, \text{m/s} \), the convective acceleration is \( 10 \times 15 = 150 \, \text{m/s}^2 \).
  • At \( x_2 = 1 \), with \( V_2 = 25 \, \text{m/s} \), it increases to \( 25 \times 15 = 375 \, \text{m/s}^2 \).
Therefore, understanding convective acceleration is crucial for predicting how fluid speeds change spatially, affecting overall flow dynamics.
Velocity Profile
A velocity profile describes how velocity varies with respect to position within a flow field. In the context of the exercise, the velocity profile is expressed as a linear equation \( V = ax + b \). Understanding this linear relation is essential for analyzing and predicting fluid flow behavior across the nozzle.
The constants \( a \) and \( b \) are derived from known velocity conditions at designated points, ensuring the equation reflects actual flow behavior. Given that:
  • \( V_1 = 10 \, \text{m/s} \) at \( x_1 = 0 \) provides that \( b = 10 \).
  • \( V_2 = 25 \, \text{m/s} \) at \( x_2 = 1 \) allows us to solve for \( a \), making \( a = 15 \).
Thus, the profile \( V = 15x + 10 \) effectively depicts how velocity transitions from one point in the nozzle to another. Analyzing the velocity profile reveals how and where velocity increases occur, which is crucial for applications ranging from engineering designs to scientific predictions of fluid behavior under different conditions.

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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 lineariy with time, going from \(60^{\circ} \mathrm{F}\) at 9 A,M. to \(80^{\circ} \mathrm{F}\) by 1 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 eats lunch at a rest stop 30 mi from her home at noon; (c) as she arrives enthusiastically at the beach at 1 P.M., pedaling 20 mph.

"A. 9 Consider a ball thrown with initial speed \(V_{0}\) at an angle of \(\theta\) as shown in Fig. P4.9 \(a\). As discussed in beginning physics, if friction is negligible the path that the ball takes is given by \\[ y=(\tan \theta) x-\left[g /\left(2 V_{0}^{2} \cos ^{2} \theta\right)\right] x^{2} \\] That is, \(y=c_{1} x+c_{2} x^{2},\) where \(c_{1}\) and \(c_{2}\) are constants. The path is a parabola. The pathline for a stream of water leaving a smalt for this water stream are given in the following table. (a) Use the given data to determine appropriate values for \(c_{1}\) and \(c_{2}\) in the above equation and, thas, show that these water particles also follow a parabolic pathline, (b) Use your values of \(c_{1}\) and \(c_{2}\) to determine the speed of the water, \(V_{0}\), leaving the nozzle. $$\begin{array}{ll} x(\text { in. }) & y \text { (in.) } \\ \hline 0 & 0 \\ 0.25 & 0.13 \\ 0.50 & 0.16 \\ 0.75 & 0.13 \\ 1.0 & 0.00 \\ 1.25 & -0.20 \\ 1.50 & -0.53 \\ 1.75 & -0.90 \\ 2.00 & -1.43 \\ \hline \end{array}$$

(See, "Toliow those particles," Section 4.1 .) Two photographs of four particles in a flow past a sphere are superposed as shown in Fig, P4.62. The time interval between the photos is \(\Delta t=0.002 \mathrm{s} .\) The locations of the particles, as determined from the photos, are shown in the table. (a) Determine the fluid velocity for these particles. (b) Plot a graph to compare the results of part (a) with the theoretical velocity which is given by \(V=V_{0}\left(1+a^{3} / x^{3}\right),\) where \(a\) is the sphere radius and \(V_{0}\) is the fluid speed far from the sphere, $$\begin{array}{ccc} \hline \text { Particle } & x \text { at } t=0 \mathrm{s}(\mathrm{ft}) & x \text { at } t=0.002 \mathrm{s}(\mathrm{ft}) \\ \hline 1 & -0.500 & -0.480 \\ 2 & -0.250 & -0.232 \\ 3 & -0.140 & -0.128 \\ 4 & -0.120 & -0.112 \\ \hline \end{array}$$

Show that the streamlines for a flow whose velocity components are \(u=c\left(x^{2}-y^{2}\right)\) and \(v=-2 c x y,\) where \(c\) is a constant, are given by the equation \(x^{2} y-y^{3} / 3=\) constant. At which point (points) is the flow parallel to the \(y\) axis? At which point (points) is the fluid stationary?

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