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Chapter 10

Q. 10.14.1

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.

Step-by-Step

Verified Solution

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:

\frac{Dv}{Dt}=\frac{\partial v}{\partial t}+(v.\triangledown)\textbf{v}=\frac{\partial \textbf{v}}{\partial t}+\textbf{w}\times \textbf{v}+\frac{1}{2}\triangledown v^2               (10.14.4)

v\triangledown^2\textbf{v}-\frac{1}{\rho}\triangledown p+\textbf{b}=\frac{D\textbf{v}}{Dt}               (10.14.6)

\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}}\right)v_{1}=v\triangledown^2 v_{1}-\frac{1}{\rho}\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}}\right)v_{2}=v\triangledown^2v_{2}-\frac{1}{\rho}\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}}\right)v_{3}=v\triangledown^2v_{3}-\frac{1}{\rho}\frac{\partial p}{\partial x_{3}}+b_{3}\end{matrix}

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

\frac{\partial p}{\partial x_{1}}=-2k^2\rho x_{1}(x^2_{1}+x^2_{2})                (10.14.12)

\frac{\partial p}{\partial x_{2}}=-2k^2 \rho x_{2}(x^2_{1}+x^2_{2})                (10.14.13)

\frac{\partial p}{\partial x_{3}}=0                            (10.14.14)

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

dp=\frac{\partial p}{\partial x_{1}}dx_{1}+\frac{\partial p}{\partial x_{2}}dx_{2}

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

dp=-2k^2\rho(x^2_{1}+x^2_{2})(x_{1}dx_{1}+x_{2}dx_{2})=-k^2 \rho d\left\{\frac{1}{2}(x^2_{1}+x^2_{2})^2\right\}

Hence

p=-\frac{1}{2}k^2 \rho (x^2_{1}+x^2_{2})^2+C                    (10.14.15)

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

p=p^0 -\frac{1}{2}k^2 \rho (x^2_{1}+x^2_{2})^2                        (10.14.16)

This is the required pressure distribution.