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Chapter 6: Problem 1

The velocity in a certain two-dimensional flow field is given by the equation $$\mathbf{v}=2 x t \hat{\mathbf{i}}-2 y t \hat{\mathbf{j}}$$ where the velocity is in ft'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 at x=y=2, t=0 is (0,0) ft/s. The local components of the acceleration are (4,4) ft/s^2. The convective components of the acceleration are (0,0) ft/s^2. Therefore, the total components of the acceleration are also (4,4) ft/s^2. The magnitude of the acceleration is 4\sqrt{2} ft/s^2 and its direction is 45 degrees.

Step by step solution

01

Find The Derivative Of The Velocity function

The velocity function is given by \( \mathbf{v}=2 x t \hat{\mathbf{i}}-2 y t \hat{\mathbf{j}} \). To find the acceleration, you need the derivative of the velocity function. According to the chain rule, the derivative of a product of two functions is given by the derivative of the first times the second plus the first times the derivative of the second. This leads to the following derivations: \( \frac{dv_x}{dt}= 2x + 2xt\), \( \frac{dv_y}{dt}= -2y - 2yt\).
02

Local and convective components of acceleration

In fluid dynamics, the local acceleration refers to the rate at which velocity changes at a particular point in time. Using the derivatives calculated in the previous step, we can state the components of local acceleration in the \(x\) and \(y\) directions as: \( a_{Lx} = \frac{dv_x}{dt}= 2x + 2xt\) and \( a_{Ly} = \frac{dv_y}{dt}= -2y - 2yt \). The convective acceleration refers to the rate at which velocity changes due to the movement of fluid particles from one point to another in field-oriented applications. for convective acceleration \( A_{Cx} = v \frac{dv_x}{dx} = 2 x t *2t = 4xt^2\) and \( A_{Cy} = -v \frac{dv_x}{dy} = -2 x t *2t = -4xt^2\)
03

Evaluate at given coordinates

Substituting the given coordinates x=y=2, and time t=0 into the expressions for velocity and acceleration, we get: \( v_x = v_y = 2*2*0 = 0 ft/s \), \( a_{Lx} = a_{Ly} = 2*2 + 2*2*0 = 4 ft/s^2 \), \( a_{Cx} = a_{Cy} = 2*2*0*4*0^2 = 0 ft/s^2\).
04

Calculate The Total Acceleration

The total acceleration is the vector sum of the local and convective components. Therefore, the total acceleration in the x and y directions can be calculated: \( a_x = a_{Lx} + a_{Cx} = 4 + 0 = 4 ft/s^2 \), \( a_y = a_{Ly} + a_{Cy} = 4 + 0 = 4 ft/s^2\). The magnitude of the acceleration \( a = \sqrt{a_x^2 + a_y^2} = \sqrt{4^2+4^2} = 4\sqrt{2} ft/s^2\). The direction can be found using \( arctan(a_y/a_x) = arctan(1) = 45^\circ \).

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Most popular questions from this chapter

The three components of velocity in a flow field are given by $$\begin{aligned}u &=x^{2}+y^{2}+z^{2} \\\v &=x y+y z+z^{2} \\\w &=-3 x z-z^{2} / 2+4\end{aligned}$$ (a) Determine the volumetric dilatation rate and interpret the results. (b) Determine an expression for the rotation vector. Is this an irrotational flow field?

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