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The velocity in a certain two-dimensional flow field is given by the equation \\[\mathbf{V}=2 x \hat{t} \hat{\mathbf{i}}-2 y t \hat{\mathbf{j}}\\] where the velocity is in \(\mathrm{ft} / \mathrm{s}\) when \(x, y,\) and \(t\) are in feet and seconds, respectively. Determine expressions for the local and convective components of acceleration in the \(x\) and \(y\) directions. What is the magnitude and direction of the velocity and the acceleration at the point \(x=y=2 \mathrm{ft}\) at the time \(t=0 ?\)

Short Answer

Expert verified
The velocity is zero at the given point and time. The acceleration has a magnitude of \(4\sqrt{2}\) and is directed at \(-45^\circ\).

Step by step solution

01

Determine Velocity Components

Given the velocity field \( \mathbf{V}=2 x \hat{t} \hat{\mathbf{i}}-2 y t \hat{\mathbf{j}} \), the components of velocity are \( u = 2x \hat{t} \) for the \( x \)-direction and \( v = -2yt \) for the \( y \)-direction.
02

Write down Local Acceleration Expressions

The local acceleration is the time derivative of the velocity components: \( \frac{\partial u}{\partial t} = 2x \) and \( \frac{\partial v}{\partial t} = -2y \).
03

Compute Convective Acceleration Components

Convective acceleration in the \( x \)-direction is \( u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} \), and for the \( y \)-direction, it's \( u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} \). First, calculate the partial derivatives: \( \frac{\partial u}{\partial x} = 2t \), \( \frac{\partial u}{\partial y} = 0 \), \( \frac{\partial v}{\partial x} = 0 \), and \( \frac{\partial v}{\partial y} = -2t \). Thus, convective acceleration terms become \( a_c^x = 2x \cdot 2t = 4xt \) and \( a_c^y = -2yt \cdot (-2t) = 4y t^2 \).
04

Calculate Total Acceleration Components

Combine local and convective components: \( a^x = \frac{\partial u}{\partial t} + a_c^x = 2x + 4xt = 2x(1 + 2t) \), and \( a^y = \frac{\partial v}{\partial t} + a_c^y = -2y + 4yt^2 \).
05

Evaluate Velocity and Acceleration at Given Point

For \( x = 2 \), \( y = 2 \), and \( t = 0 \), velocity components are \( u = 4 \times 0 = 0 \) and \( v = 0 \times (-4) = 0 \). Thus, the velocity is zero. For acceleration, \( a^x = 2 \times 2 \times 1 = 4 \) and \( a^y = -4 \).
06

Calculate Magnitude and Direction of Velocity and Acceleration

The magnitude of velocity \( |\mathbf{V}| = \sqrt{0^2 + 0^2} = 0 \). The magnitude of acceleration \( |\mathbf{a}| = \sqrt{4^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2} \). The direction of acceleration is given by \( \theta = \tan^{-1}\left(\frac{-4}{4}\right) = -45^\circ \).

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Key Concepts

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

Velocity Components
In fluid dynamics, understanding the velocity components of a flow is crucial, as they describe how the fluid particles are moving in space. In a two-dimensional flow field, the velocity vector \( \mathbf{V} \) can be broken down into its individual components. Given the velocity equation \( \mathbf{V}=2 x \hat{t} \hat{\mathbf{i}}-2 y t \hat{\mathbf{j}} \), you'll find the velocity in the \( x \)-direction, often denoted as \( u \), and the velocity in the \( y \)-direction, known as \( v \). Playing the role of components, these expressions allow us to investigate specific movements:\[ u = 2x \hat{t} \] and \[ v = -2yt \].
These components indicate that as time \( t \) or spatial variables \( x \) and \( y \) change, the velocities in each direction change too. This relationship helps in analyzing how a fluid flows through a space or around an object over time. Evaluating these relationships helps predict fluid behavior at any given point in the flow field.
Acceleration
Acceleration in fluid dynamics gives insights into how fast a fluid particle speeds up or slows down. Overall acceleration consists of two parts:
  • Local Acceleration
  • Convective Acceleration
Each component plays a role in the fluid's motion. We'll delve into these to see how they contribute to the total acceleration the fluid experiences.
This helps reveal whether a fluid speeds up uniformly or if it varies due to spatial factors, aiding engineers in designing more efficient fluid systems.
Convective Acceleration
Convective acceleration arises due to spatial changes in velocity. This is particularly prominent in non-uniform flow fields where fluid velocity varies from one point to another. It's a product of the fluid's velocity itself moving through a gradient. To calculate it, we consider the effect of moving through space at varying speeds. In the \( x \)-direction, convective acceleration is determined by:
  • \( a_c^x = u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} \)
In the \( y \)-direction, it is:
  • \( a_c^y = u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} \)
For the given velocity components, the convective acceleration components work out to be:
  • \( a_c^x = 4xt \)
  • \( a_c^y = 4y t^2 \)
This mechanism helps highlight how changes in velocity will alter fluid motion over spatial regions within a flow field.
Local Acceleration
Local acceleration in fluid dynamics is tied to temporal changes, describing how velocity at a specific point changes with time irrespective of its position. So, it's essentially about how a point in a flow speeds up or slows down over time.
The expressions for local acceleration derive from the time derivative of the velocity components. Here, they appear as:
  • \( \frac{\partial u}{\partial t} = 2x \)
  • \( \frac{\partial v}{\partial t} = -2y \)
This tells us that in this flow field, as time changes, the velocity at any point \( (x, y) \) changes according to these derivatives. This component is crucial to understand phenomena like heating or cooling effects in a fluid flow over time, enabling engineers to predict time-dependent changes in fluid movement.

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Most popular questions from this chapter

The velocity potential for a given two-dimensional flow field is \\[\phi=\left(\frac{5}{3}\right) x^{3}-5 x y^{2}\\] Show that the continuity equation is satisfied and determine the corresponding stream function.

Consider two sources having equal strengths located along the \(x\) axis at \(x=0\) and \(x=2 \mathrm{m},\) and a sink located on the \(y\) axis at \(y=2 \mathrm{m} .\) Determine the magnitude and direction of the fluid velocity at \(x=5 \mathrm{m}\) and \(y=0\) due to this combination if the flowrate from each of the sources is \(0.5 \mathrm{m}^{3} / \mathrm{s}\) per \(\mathrm{m}\) and the flowrate into the sink is \(1.0 \mathrm{m}^{3} / \mathrm{s}\) per \(\mathrm{m}\).

A viscous fluid is contained between two infinitely long, vertical, concentric cylinders. The outer cylinder has a radius \(r_{o}\) and rotates with an angular velocity \(\omega .\) The inner cylinder is fixed and has a radius \(r_{i}\). Make use of the Navier-Stokes equations to obtain an exact solution for the velocity distribution in the gap. Assume that the flow in the gap is axisymmetric (neither velocity nor pressure are functions of angular position \(\theta\) within the gap) and that there are no velocity components other than the tangential component. The only body force is the weight.

The velocity potential for a certain inviscid, incompressible flow field is given by the equation \\[\phi=2 x^{2} y-\left(\frac{2}{3}\right) y^{3}\\] where \(\phi\) has the units of \(\mathrm{m}^{2} / \mathrm{s}\) when \(x\) and \(y\) are in meters. Determine the pressure at the point \(x=2 \mathrm{m}, y=2 \mathrm{m}\) if the pressure at \(x=1 \mathrm{m}\) \(y=1 \mathrm{m}\) is \(200 \mathrm{kPa}\). Elevation changes can be neglected, and the fluid is water.

It is proposed that a two-dimensional, incompressible flow field be described by the velocity components \\[\begin{array}{l}u=A y \\\v=B x\end{array}\\] where \(A\) and \(B\) are both positive constants. (a) Will the continuity equation be satisfied? (b) Is the flow irrotational? (c) Determine the equation for the streamlines and show a sketch of the streamline that passes through the origin. Indicate the direction of flow along this streamline.

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