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The velocity in a certain flow field is given by the equation \\[\mathbf{V}=y z \hat{\mathbf{i}}+x^{2} z \hat{\mathbf{j}}+x \hat{\mathbf{k}}\\] Determine the expressions for the three rectangular components of acceleration.

Short Answer

Expert verified
The acceleration components are: \( a_x = x^2 z^2 + xy \), \( a_y = x^3 z + yz^2 \), \( a_z = yz \).

Step by step solution

01

Know the Velocity Vector

The velocity vector is given by \( \mathbf{V} = yz \hat{\mathbf{i}} + x^2z \hat{\mathbf{j}} + x \hat{\mathbf{k}} \). We need to find the components of acceleration, which involve derivatives of these velocity components.
02

Define Acceleration Components

The acceleration vector \( \mathbf{a} \) in a flow field is given by the material derivative \( \frac{D\mathbf{V}}{Dt} = \frac{\partial \mathbf{V}}{\partial t} + (\mathbf{V} \cdot abla) \mathbf{V} \). In this problem, there is no explicit time-dependent term, so we only use \((\mathbf{V} \cdot abla) \mathbf{V}\).
03

Calculate the Gradient Operator

The gradient operator \( abla \) in rectangular coordinates is given by \( abla = \frac{\partial}{\partial x} \hat{\mathbf{i}} + \frac{\partial}{\partial y} \hat{\mathbf{j}} + \frac{\partial}{\partial z} \hat{\mathbf{k}} \). We will use this operator in calculations for \( (\mathbf{V} \cdot abla) \mathbf{V} \).
04

Calculate Each Component of Acceleration

Use the velocity vector to calculate the three components of acceleration:1. **x-component:** \( a_x = \frac{D (yz)}{Dt} = yz \frac{\partial (yz)}{\partial x} + x^2z \frac{\partial (yz)}{\partial y} + x \frac{\partial (yz)}{\partial z} \).2. **y-component:** \( a_y = \frac{D (x^2z)}{Dt} = yz \frac{\partial (x^2z)}{\partial x} + x^2z \frac{\partial (x^2z)}{\partial y} + x \frac{\partial (x^2z)}{\partial z} \).3. **z-component:** \( a_z = \frac{D x}{Dt} = yz \frac{\partial x}{\partial x} + x^2z \frac{\partial x}{\partial y} + x \frac{\partial x}{\partial z} \).Compute these manually by applying partial derivatives.
05

Compute Partial Derivatives

Calculate the needed partial derivatives:- \( \frac{\partial (yz)}{\partial x} = 0 \), \( \frac{\partial (yz)}{\partial y} = z \), \( \frac{\partial (yz)}{\partial z} = y \).- \( \frac{\partial (x^2z)}{\partial x} = 2xz \), \( \frac{\partial (x^2z)}{\partial y} = 0 \), \( \frac{\partial (x^2z)}{\partial z} = x^2 \).- \( \frac{\partial x}{\partial x} = 1 \), \( \frac{\partial x}{\partial y} = 0 \), \( \frac{\partial x}{\partial z} = 0 \).
06

Substitute and Simplify

Substitute the derivatives back into the acceleration expressions:- **x-component:** \( a_x = x^2 z^2 + xy \).- **y-component:** \( a_y = x^3 z + yz^2 \).- **z-component:** \( a_z = yz \).
07

Final Expressions for Acceleration Components

The acceleration components are:- **x-component:** \( a_x = x^2 z^2 + xy \).- **y-component:** \( a_y = x^3 z + yz^2 \).- **z-component:** \( a_z = yz \).

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Key Concepts

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

Acceleration Components
In fluid mechanics, the concept of acceleration components in a flow field is crucial for understanding how fluid particles move. Acceleration in a flow field can be thought of as the change in velocity experienced by a fluid particle. It's not just about speed, but also about direction.
In a fluid flow, the acceleration vector, denoted as \( \mathbf{a} \), can be found using the material derivative. This derivative combines two effects: the change in velocity over time and the spatial change as the fluid moves through space. Given that there's no explicit time dependence in the specified problem, we only account for spatial changes.
The acceleration components are determined by the expression \( \frac{D\mathbf{V}}{Dt} = (\mathbf{V} \cdot abla) \mathbf{V} \), which involves the velocity vector \( \mathbf{V} \) acting alongside the gradient operator. Each component x, y, and z involves different elements of this expression, capturing how velocity changes across different dimensions.
Velocity Vector
A velocity vector describes the speed and direction of a fluid particle in a flow field. Given by \( \mathbf{V} = yz \hat{\mathbf{i}} + x^2z \hat{\mathbf{j}} + x \hat{\mathbf{k}} \), it quite literally represents motion in three-dimensional space.
The velocity vector comprises three components, each linked to a spatial direction:
  • \( yz \hat{\mathbf{i}} \) – This term indicates movement along the x-axis, influenced by the variables \( y \) and \( z \).
  • \( x^2z \hat{\mathbf{j}} \) – This term governs the velocity along the y-axis, influenced by changes in \( x^2 \) and \( z \).
  • \( x \hat{\mathbf{k}} \) – This term covers velocity along the z-axis, solely dependent on \( x \).
By dissecting the velocity vector into these components, one can better visualize and compute how fluid motion occurs in each spatial direction.
Partial Derivatives
Partial derivatives are fundamental in fluid mechanics, as they help determine how a function changes as its variables slightly vary. This concept is critical when finding the acceleration components of a flow field.
For each velocity component, partial derivatives assess the rate of change concerning one variable while keeping others constant. In our problem, the velocity vector is analyzed by:
  • For \( yz \), \( \frac{\partial (yz)}{\partial x} = 0 \), implying no change concerning x as x is not part of the term.
  • For \( x^2z \), \( \frac{\partial (x^2z)}{\partial x} = 2xz \), highlighting how velocity changes concerning x.
  • For \( x \), \( \frac{\partial x}{\partial x} = 1 \), since it's a direct linear term.
Careful computation of these partial derivatives allows for precise integration into finding acceleration components, demonstrating the local behavior of the fluid throughout the field.
Gradient Operator
In fluid mechanics, the gradient operator, symbolized as \( abla \), is a vector operator essential for evaluating how a function changes in space. The gradient provides a multi-dimensional snapshot of the greatest rate of increase of the function.
The gradient operator is expressed in rectangular coordinates as:\[ \, abla = \frac{\partial}{\partial x} \hat{\mathbf{i}} + \frac{\partial}{\partial y} \hat{\mathbf{j}} + \frac{\partial}{\partial z} \hat{\mathbf{k}} \, \]
Within the context of our flow field problem, \( abla \) is used to assess how fluid velocity changes not merely in magnitude but also in direction through space. Applying the gradient operator allows you to explore how the velocity vector interacts and evolves, feeding into the larger picture of calculating the acceleration throughout the field.

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Most popular questions from this chapter

In a certain steady, two-dimensional flow field the fluid may be assumed to be ideal and the weight of the fluid (specific weight \(\left.=50 \mathrm{lb} / \mathrm{ft}^{3}\right)\) is the only body force. The \(x\) component of velocity is known to be \(u=6 x,\) which gives the velocity in \(\mathrm{ft} / \mathrm{s}\) when \(x\) is measured in feet, and the \(y\) component of velocity is known to be a function of only \(y .\) The \(y\) axis is vertical, and at the origin the velocity is zero. (a) Determine the \(y\) component of velocity so that the continuity equation is satisfied. (b) Can the difference in pressures between the points \(x=1 \mathrm{ft}, y=1 \mathrm{ft}\) and \(x=1 \mathrm{ft}, y=4 \mathrm{ft}\) be determined from the Bernoulli equation? If so, determine the value in \(\mathrm{lb} / \mathrm{ft}^{2}\). If not, explain why not.

Consider a uniform flow in the positive \(x\) direction combined with a free vortex located at the origin of the coordinate system. The streamline \(\psi=0\) passes through the point \(x=4, y=0 .\) Determine the equation of this streamline.

Determine the vorticity field for the following velocity vector: \\[\mathbf{V}=\left(x^{2}-y^{2}\right) \hat{\mathbf{i}}-2 x y \hat{\mathbf{j}}\\]

As is shown by Eq. \(6.150,\) the pressure gradient for laminar flow through a tube of constant radius is given by the expression \\[\frac{\partial p}{\partial z}=-\frac{8 \mu Q}{\pi R^{4}}\\] For a tube whose radius is changing very gradually, such as the one illustrated in Fig. \(\mathrm{P} 6.106,\) it is expected that this equation can be used to approximate the pressure change along the tube if the actual radius, \(R(z),\) is used at each cross section. The following measurements were obtained along a particular tube. $$\begin{array}{l|l|l|l|l|l|l|l|l|l|l|l} z / \ell & 0 & 0.1 & 0.2 & 0.3 & 0.4 & 0.5 & 0.6 & 0.7 & 0.8 & 0.9 & 1.0 \\ \hline R(z) / R_{o} & 1.00 & 0.73 & 0.67 & 0.65 & 0.67 & 0.80 & 0.80 & 0.71 & 0.73 & 0.77 & 1.00 \end{array}$$ Compare the pressure drop over the length \(\ell\) for this nonuniform tube with one having the constant radius \(R_{o} .\) Hint: To solve this problem you will need to numerically integrate the equation for the pressure gradient given previously.

The stream function for an incompressible, twodimensional flow field is \\[\psi=3 x^{2} y+y\\] For this flow field, plot several streamlines.

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