Find the pressure distribution such that the velocity field given by

$v_{1}=k(x^2_{1}-x^2_{2}),\ \ v_{2}=2kx_{1}x_{2},\ \ v_{3}=0,\ \ (k=constant)$ (10.14.10)

satisfies the Navier-Stokes equation for an incompressible fluid in the absence of body force.

Question

Find the pressure distribution such that the velocity field given by

[latex]v_{1}=k(x^2_{1}-x^2_{2}),\ \ v_{2}=2kx_{1}x_{2},\ \ v_{3}=0,\ \ (k=constant)[/latex] (10.14.10)

satisfies the Navier-Stokes equation for an incompressible fluid in the absence of body force.

Accepted Answer

When written in the component form, the Navier-Stokes equation for an incompressible fluid given by (10.14.6) reads as follows, on using the identity (10.14.4) also taken in the component form:

[latex]\frac{Dv}{Dt}=\frac{\partial v}{\partial t}+(v. riangledown) extbf{v}=\frac{\partial extbf{v}}{\partial t}+ extbf{w} imes extbf{v}+\frac{1}{2} riangledown v^2[/latex] (10.14.4)

[latex]v riangledown^2 extbf{v}-\frac{1}{ ho} riangledown p+ extbf{b}=\frac{D extbf{v}}{Dt}[/latex] (10.14.6)

[latex]\begin{matrix}\frac{\partial v_{1}}{\partial t}+\left(v_{1}\frac{\partial}{\partial x_{1}}+v_{2}\frac{\partial}{\partial x_{2}} +v_{3}\frac{\partial}{\partial x_{3}} ight)v_{1}=v riangledown^2 v_{1}-\frac{1}{ ho}\frac{\partial p}{\partial x_{1}}+b_{1}\ \frac{\partial v_{2}}{\partial t}+\left(v_{1}\frac{\partial}{\partial x_{1}}+v_{2}\frac{\partial}{\partial x_{2}}+v_{3}\frac{\partial}{\partial x_{3}} ight)v_{2}=v riangledown^2v_{2}-\frac{1}{ ho}\frac{\partial p}{\partial x_{2}}+b_{2} &(10.14.11)\ \frac{\partial v_{3}}{\partial t}+\left(v_{1}\frac{\partial}{\partial x_{1}}+v_{2}\frac{\partial}{\partial x_{2}}+v_{3}\frac{\partial}{\partial x_{3}} ight)v_{3}=v riangledown^2v_{3}-\frac{1}{ ho}\frac{\partial p}{\partial x_{3}}+b_{3}\end{matrix}[/latex]

Substituting the expressions for [latex]v_i[/latex] from (10.14.10) in equations (10.14.11) and noting that [latex]b_{i}=0[/latex], we obtain

[latex]\frac{\partial p}{\partial x_{1}}=-2k^2 ho x_{1}(x^2_{1}+x^2_{2})[/latex] (10.14.12)

[latex]\frac{\partial p}{\partial x_{2}}=-2k^2 ho x_{2}(x^2_{1}+x^2_{2})[/latex] (10.14.13)

[latex]\frac{\partial p}{\partial x_{3}}=0[/latex] (10.14.14)

Thus, the pressure-gradient in the [latex]x_{3}-[/latex]direction should be 0. Accordingly, [latex]p=p(x_{1},x_{2})[/latex] so that

[latex]dp=\frac{\partial p}{\partial x_{1}}dx_{1}+\frac{\partial p}{\partial x_{2}}dx_{2}[/latex]

This yields, on using (10.14.12) and (10.14.13),

[latex]dp=-2k^2 ho(x^2_{1}+x^2_{2})(x_{1}dx_{1}+x_{2}dx_{2})=-k^2 ho d\left\{\frac{1}{2}(x^2_{1}+x^2_{2})^2 ight\}[/latex]

Hence

[latex]p=-\frac{1}{2}k^2 ho (x^2_{1}+x^2_{2})^2+C[/latex] (10.14.15)

where C is an arbitrary constant. If [latex]p^0[/latex] is the pressure at the origin, we get

[latex]p=p^0 -\frac{1}{2}k^2 ho (x^2_{1}+x^2_{2})^2[/latex] (10.14.16)

This is the required pressure distribution.

Question 10.14.1: Find the pressure distribution such that the velocity field ...

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