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The nozzle shown in Figure 5.50 is designed such that the velocity, \(\mathbf{v}\), at any point near the centerline of the nozzle under steady- state conditions can be estimated by $$ \mathbf{v}=\frac{4}{0.8-x} \mathbf{i} \mathrm{m} / \mathrm{s} $$ where \(x\) is the distance in meters from the center of the nozzle entrance, measured along the centerline of the nozzle. The centerline velocities at the entrance and exit of the nozzle are \(4 \mathrm{~m} / \mathrm{s}\) and \(20 \mathrm{~m} / \mathrm{s}\), respectively, and the length of the nozzle is \(0.6 \mathrm{~m}\). What is the acceleration of a fluid element at the center of the midsection of the nozzle, where \(x=0.2 \mathrm{~m} ?\)

Short Answer

Expert verified
The acceleration at the midsection of the nozzle is approximately 74.07 m/s².

Step by step solution

01

Understand the Velocity Function

The velocity function for the nozzle is given by \( \mathbf{v} = \frac{4}{0.8-x} \mathbf{i} \mathrm{~m/s} \). This equation describes how the velocity of fluid changes along the nozzle's centerline, where \( x \) is the distance from the entrance. The velocity is influenced by the position \( x \).
02

Determine the Derivative of the Velocity Function

We need to find the acceleration, which is the derivative of velocity with respect to time. Using the given velocity function: \( \mathbf{v} = \frac{4}{0.8-x} \), first find the derivative of \( \mathbf{v} \) with respect to \( x \), using the chain rule: \[ \frac{dv}{dx} = \frac{d}{dx} \left( \frac{4}{0.8-x} \right) = \frac{4}{(0.8-x)^2} \].
03

Relate Position and Velocity

We need \( \frac{dv}{dt} \), the acceleration, but we currently have \( \frac{dv}{dx} \). Use the chain rule: \( \frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt} \). Since \( \frac{dx}{dt} \) is the velocity, \( \mathbf{v} = \frac{4}{0.8-x} \), \( \frac{dv}{dt} = \frac{4}{(0.8-x)^2} \times \frac{4}{0.8-x} \).
04

Substitute Given Values for x to Find Acceleration

Substitute \( x = 0.2 \) into the expression for acceleration: \[ \frac{dv}{dt} = \frac{4}{(0.8 - 0.2)^2} \times \frac{4}{0.8 - 0.2} \]. Simplifying further: \( \frac{dv}{dt} = \frac{4}{0.36} \times \frac{4}{0.6} = \frac{16}{0.36 \times 0.6} \). Calculate this to find \( \frac{dv}{dt} = \frac{16}{0.216} \).
05

Calculate Acceleration Value

Perform the division: \( \frac{16}{0.216} \approx 74.07 \mathrm{~m/s^2} \). This is the acceleration at the middle of the nozzle when \( x = 0.2 \mathrm{~m} \).

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Key Concepts

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

Nozzle Design
Nozzle design is essential in fluid mechanics to control the way fluid flows through various stages and outputs in a system. A nozzle is commonly used to accelerate the fluid by constricting it and converting pressure energy into kinetic energy.
In the example exercise, the nozzle is structured along a centerline, through which fluid dynamics such as velocity and acceleration are analyzed. The nozzle influences the velocity of fluid based on its position, creating different accelerations at different points along the length of the nozzle. Nozzle design considerations often include things like pressure drop, energy conversion efficiency, and achieving desired flow patterns. To understand the effects of the nozzle shape, engineers look at how variables like velocity and pressure change with distance along the nozzle's length.
Velocity Function
The velocity function is crucial in fluid dynamics as it describes how the speed of a fluid element changes at different points in space or time. In the given exercise, the velocity function is defined as:
\[ \mathbf{v} = \frac{4}{0.8-x} \mathbf{i} \; \text{m/s} \]
where \(x\) represents the distance from the nozzle entrance along the centerline. The function illustrates that as \(x\) increases, the denominator of the fraction becomes smaller, resulting in a higher velocity. Such a function helps in visualizing how the fluid is accelerating through the nozzle due to changes in the nozzle's cross-sectional area. This type of information is typically used to determine how efficiently a nozzle performs and can be useful when adjusting design parameters to meet specific performance goals or constraints.
Acceleration Calculation
Acceleration in the context of the fluid flow through a nozzle is defined mathematically as the rate of change of velocity with time. To find the acceleration along the nozzle's centerline, the first step involves determining the derivative of the velocity function concerning the position \(x\), resulting in:
\[ \frac{dv}{dx} = \frac{4}{(0.8-x)^2} \]
The next step is using the chain rule to relate this with time, as \( \frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt} \), where the expression for \( \frac{dx}{dt} \) is the velocity itself. By evaluating these expressions at a given point (like \(x = 0.2\) meters), you obtain the acceleration.
This computed acceleration value reflects how quickly the fluid velocity is changing at that specific location within the nozzle. Such calculations are vital in designing efficient fluid systems where controlling the rate of speed change is necessary to maintain stable and controllable performance across different operating conditions.
Steady-State Conditions
Steady-state conditions in fluid dynamics mean that variables such as velocity, pressure, and flow rate are constant over time, even if they vary across different locations within the system. This implies that there are no temporal changes to the flow characteristics, simplifying analysis and allowing predictions over long periods.
In the nozzle exercise, the use of steady-state conditions suggests that, at any given point along the nozzle, properties affecting the fluid flow (like velocity) remain unchanged over time. This makes it easier to predict behavior and calculate aspects like velocity and acceleration without accounting for dynamic changes.
Understanding steady-state conditions helps streamline calculations, as it allows for simplifications in predicting system behavior. By focusing on spatial rather than temporal variations, engineers can effectively design efficient flow systems that reliably manage fluid behavior.

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Most popular questions from this chapter

The velocity field in a hurricane can be approximated by a forced vortex between the center and a radial distance of \(R\) from the center and as a free vortex beyond the radial distance \(R\). In such an approximation, the \(\theta\) -component of the velocity can be approximated by $$ v_{\theta}=\left\\{\begin{array}{ll} r \omega, & 0 \leq r \leq R \\ \frac{\Gamma}{2 \pi r}, & r \geq R \end{array}\right. $$ where \(\omega\) is the rotational speed of the forced vortex, \(\Gamma\) is the circulation of the free vortex, and \(R\) is the match point of the two velocity distributions. The radial component of the velocity, \(v_{r},\) is equal to zero. (a) Explain why \(R\) is where the wind speed is a maximum. (b) If the undisturbed pressure outside the hurricane is \(p_{0}\) and the maximum wind speed is \(V_{\max },\) determine an expression for the pressure at the match point in terms of the given variables and the relevant properties of the air. (c) For an intense Category 4 hurricane, the maximum velocity is \(251 \mathrm{~km} / \mathrm{h}\), the matchpoint radius is \(15 \mathrm{~km}\), and conditions outside the hurricane are standard sea-level conditions. Determine the angular speed, \(\omega\), and the circulation, \(\Gamma\), of the hurricane.

A two-dimensional velocity field in the \(r \theta\) plane is described by the velocity components \(v_{r}=-6 / r \mathrm{~m} / \mathrm{s}\) and \(v_{\theta}=3 / r \mathrm{~m} / \mathrm{s}\), where \(r\) and \(\theta\) are polar coordinates in meters and radians, respectively. The gravity force acts in the negative \(z\) -direction, and the fluid has a density of \(1.20 \mathrm{~kg} / \mathrm{m}^{3}\). Calculate the pressure gradients in the \(r\) -, \(\theta-,\) and \(z\) -directions at \(r=2 \mathrm{~m}\) and \(\theta=\pi / 4 \mathrm{rad}\).

The \(x\) and \(z\) components of the velocity in a three-dimensional flow field are given by: \(u=a y z-b x y^{2}\) and \(w=2 a z+b x z+c x^{2}\). If the fluid is incompressible, determine the \(y\) component of the velocity, \(v\), as a function of \(x, y,\) and \(z\)

Consider the velocity field, \(\mathbf{V},\) given by $$ \mathbf{V}=x z \mathbf{i}-2 y z \mathbf{j}+3 x y \mathbf{k} \mathbf{m} / \mathbf{s} $$ where \(x, y\), and \(z\), are the coordinate locations in meters. Find the dilatation rate of the fluid at \((x, y, z)=(2 \mathrm{~m}, 3 \mathrm{~m}, 1 \mathrm{~m})\) and assess whether the fluid is being compressed or expanded.

SAE 30 oil at \(20^{\circ} \mathrm{C}\) flows between two horizontal \(0.50-\mathrm{m}\) -wide parallel plates separated by \(30 \mathrm{~mm}\). The length of the top and bottom plates in the direction of flow is \(2 \mathrm{~m}\), the bottom plate is stationary, the top plate moves at \(3 \mathrm{~cm} / \mathrm{s}\), an adverse pressure gradient of \(700 \mathrm{~Pa} / \mathrm{m}\) is applied between the plates, and the flow is laminar between the plates. Determine the flow rate between the plates and the force that must be applied to move the top plate.

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