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A fluid flow has velocity components of \(u=\left(x^{2}-y^{2}\right.\) \(+3 y) \mathrm{m} / \mathrm{s}\) and \(v=(y+2 x y) \mathrm{m} / \mathrm{s}\), where \(x\) and \(y\) are in meters. Determine the magnitude of the velocity and acceleration of a particle at point \((1 \mathrm{~m}, 2 \mathrm{~m})\).

Short Answer

Expert verified
The velocity magnitude at point (1m, 2m) is 6.71 m/s, and the acceleration is 0 m/s^2.

Step by step solution

01

Substitute the points into velocity components

Firstly, substitute \(x = 1m\) and \(y = 2m\) into \(u = x^2 - y^2 + 3y\) and \(v = y + 2xy\). So, \(u = 1^2 - 2^2 + 3*2 = 3 m/s\) and \(v = 2 + 2*1*2 = 6 m/s\).
02

Calculate the Magnitude of the Velocity

The formula to calculate the magnitude of a 2D velocity vector is \(|V| = \sqrt{u^2 + v^2}\). So, \(|V| = \sqrt{3^2 + 6^2} = \sqrt{45} = 6.71 m/s.\)
03

Calculate the time derivatives of velocity components

The acceleration vector components are the time derivatives of the velocity components. However, in this case, we were not given any dependency on time, so the time derivatives of both components are 0. Meaning, \(u'(t) = 0\) and \(v'(t) = 0\). Therefore, the acceleration components are zero.
04

Calculate the Magnitude of the Acceleration

The formula to calculate the magnitude of a 2D acceleration vector is \(|A| = \sqrt{u'^2 + v'^2}\). But as \(u' = 0\) and \(v' = 0\), the magnitude also equals to \(|A| = \sqrt{0^2 + 0^2} = 0 m/s^2\).

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Key Concepts

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

Velocity Components in Fluid Mechanics
In fluid mechanics, understanding the velocity of a fluid particle is crucial as it describes the particle's speed and direction. Velocity components, typically represented by u and v in two dimensions, break this velocity into parts aligned with a coordinate system. In our exercise, we are given the velocity components as u = (x^2 - y^2 + 3y) m/s and v = (y + 2xy) m/s, with x and y indicating the particle's position in meters along the respective axes.

These are scalar fields defining the velocity in the x-direction (u) and y-direction (v) at any point in the flow. The flow's total velocity at any point is thus a vector field combining these two scalar fields. Recognizing the multidimensional nature of fluid flow is essential for predicting and understanding complex behaviors in fluid dynamics.
Calculating Velocity Magnitude
Calculating the magnitude of velocity gives insight into how fast a fluid particle is moving, regardless of its direction. Based on Pythagorean theorem, the magnitude of velocity in two dimensions is found using the formula |V| = sqrt(u^2 + v^2), where u and v are the velocity components. In our problem, after substituting x = 1 m and y = 2 m into the given equations, we calculate the velocity components as u = 3 m/s and v = 6 m/s. Thus, the magnitude of velocity is |V| = sqrt(3^2 + 6^2) = sqrt(45) = 6.71 m/s.

This step is pivotal as it transitions from a vectorial description of fluid flow, which includes direction, to a scalar quantity that solely reflects speed. Understanding how to compute this magnitude is a fundamental skill in fluid mechanics.
Time Derivatives of Velocity
The concept of acceleration in fluid mechanics is tied to the change of velocity over time. Acceleration components are thus time derivatives of velocity components, designated as u'(t) and v'(t) for u and v, respectively. In the absence of explicit time-dependence in the velocity components in our exercise, the time derivatives are zero. This implies a steady flow, where the speed and direction of a fluid particle at a particular point do not change with time. Consequently, the acceleration components are u'(t) = 0 and v'(t) = 0, leading to a zero acceleration magnitude.

Recognizing situations where acceleration is present or absent is key for analyzing fluid flow. It allows predictions on how forces interact within the fluid, a concept that underpins much of fluid dynamics.

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Most popular questions from this chapter

The water flow is defined by a two-dimensional fluid flow field as \(\mathbf{V}=\left[\left(3 y^{2}-x\right) \mathbf{i}+(2 x+y) \mathbf{j}\right] \mathrm{m} / \mathrm{s},\) where \(x\) and \(y\) are in meters. Determine the magnitude of the velocity of a water particle located at \((3 \mathrm{~m},-2 \mathrm{~m})\), and its direction measured counterclockwise from the \(x\) axis.

In a fluid flow, the velocity components of a fluid particle are defined by \(u=\left(x^{2}+3\right) \mathrm{m} / \mathrm{s}\) and \(v=(-4 x y) \mathrm{m} / \mathrm{s}\) where \(x\) and \(y\) are in meters. Determine the equation of the streamline that passes through point \((3 \mathrm{~m}, 1 \mathrm{~m})\), and find the velocity of a particle located at this point. Draw this streamline.

A fluid flow has velocity components of \(u=\left(3 y^{2}\right) \mathrm{m} / \mathrm{s}\) and \(v=(6 x y) \mathrm{m} / \mathrm{s},\) where \(x\) and \(y\) are in meters. Determine the equation of the streamline passing through point \((1 \mathrm{~m}, 2 \mathrm{~m})\). Also, what is the acceleration of a particle at this point? Is the flow steady or unsteady?

A velocity field is defined by \(u=(4 x) \mathrm{m} / \mathrm{s}\) and \(v=(2 t) \mathrm{m} / \mathrm{s},\) where \(t\) is in seconds and \(x\) is in meters. Determine the pathline that passes through point \((2 \mathrm{~m}, 6 \mathrm{~m})\) when \(t=1 \mathrm{~s}\). Plot this pathline for \(0.25 \mathrm{~m} \leq x \leq 4 \mathrm{~m}\).

The velocity field for a fluid is defined by \(u=\left[y /\left(x^{2}+y^{2}\right)\right] \mathrm{m} / \mathrm{s}\) and \(v=\left[4 x /\left(x^{2}+y^{2}\right)\right] \mathrm{m} / \mathrm{s},\) where \(x\) and \(y\) are in meters. Determine the acceleration of a particle located at point \((2 \mathrm{~m}, 0)\) and a particle located at point \((4 \mathrm{~m}, 0)\). Sketch the equations that define the streamlines that pass through these points.

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