Given the following velocity field, determine if the Eulerian acceleration is zero:ν=Axi-Axj,where A is a number.

Question

Given the following velocity field, determine if the Eulerian acceleration is zero:[latex] u =Ax\hat{i}-Ay\hat{j},[/latex] where A is a number.

Accepted Answer

Although [latex] u[/latex] is independent of time and thus does not have a local component of acceleration (i.e., it is a steady flow), we must check the convec-tive part. Note, therefore, that the Cartesian components of [latex] u[/latex] are

[latex] u _{x}=Ax, u _{y}=-Ay[/latex]

and the velocity gradients are

[latex]\frac{\partial u _{x}}{\partial x}=A, \frac{\partial u _{x}}{\partial y}=0, \frac{\partial u _{y}}{\partial x}=0, \frac{\partial u _{y}}{\partial y}=-A. [/latex]

Hence, from Eq. (7.22),

[latex]a_{x}=\frac{\partial u _{x}}{\partial t}+ u _{x}\frac{\partial u _{x}}{\partial x}+ u _{y}\frac{\partial u _{x}}{\partial y}+ u _{z}\frac{\partial u _{x}}{\partial z},[/latex]

[latex]a_{y}=\frac{\partial u _{y}}{\partial t}+ u _{x}\frac{\partial u _{y}}{\partial x}+ u _{y}\frac{\partial u _{y}}{\partial y}+ u _{z}\frac{\partial u _{y}}{\partial z},[/latex]

[latex]a_{z}=\frac{\partial u _{z}}{\partial t}+ u _{x}\frac{\partial u _{z}}{\partial x}+ u _{y}\frac{\partial u _{z}}{\partial y}+ u _{z}\frac{\partial u _{z}}{\partial z}.[/latex] (7.22)

[latex]a=\left[0+Ax(A)+(-Ay)(0)+0 ight]\hat{i}+\left[0+Ax(0)+(-Ay)(- A)+0 ight]\hat{j}+0\hat{k},[/latex]
or
[latex] a=A^{2}\left(x\hat{i}+y\hat{j} ight)[/latex]

and, consequently, the acceleration is not zero.

Question 7.1: Given the following velocity field, determine if the Euleria...

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