Chapter 4, Problem 4.29P

To determine

(a)

Whether the flow satisfy conservation of mass.

Answer to Problem 4.29P

Yes, the flow satisfy conservation of mass

Explanation of Solution

Given information:

The velocity in $x$ -direction is $-2xy$, the velocity in $y$ direction is $y^2-x^2$.

Write the expression for dot product of the gradient and the velocity vector.

$\nabla \cdot \text{V}=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}$ ... (I)

Here, velocity component along $x$ direction is $u$, velocity component along $y$ direction is $v$, the velocity component along z direction is $w$.

Calculation:

Substitute $-2xy$ for $u$, $y^2-x^2$ for $v$, $0$ for $w$ in Equation (I).

$\begin{array}{c}\nabla \cdot \text{V}=\frac{\partial }{\partial x}\left(-2xy\right)+\frac{\partial }{\partial y}\left(y^2-x^2\right)+\frac{\partial }{\partial z}\left(0\right)\\ =\frac{\partial }{\partial x}\left(-2xy\right)+\frac{\partial }{\partial y}\left(y^2-x^2\right)\\ =-2y+2y\\ =0\end{array}$

Conclusion:

Since the dot product of the gradient and velocity vector is zero therefore the field satisfies conservation of mass.

To determine

(b)

The pressure field $p(x,y)$, if the pressure at the point $\left(x=0,y=0\right)$ is equal to $p_a$.

Answer to Problem 4.29P

The pressure field $p(x,y)$, if the pressure at the point $\left(x=0,y=0\right)$ is equal to $p_a$ is $p\left(x,y\right)=-\frac{\rho }{2}\left(2x^2{y}^2+x^4+y^4\right)+p_a$.

Explanation of Solution

Given Information:

The pressure at the point $\left(x=0,y=0\right)$ is $p_a$.

Write the expression for incompressible Navier stoke Equation using $x$ relation.

$\rho \left(u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\right)=\rho g_x-\frac{\partial p}{\partial x}+\mu \left(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}+\frac{{\partial }^{2}u}{\partial z^2}\right)$ ... (II)

Here the density of the fluid is $\rho$ The acceleration due to gravity in the $x$ direction is $g_x$, Here, pressure change in $x$ direction is $\frac{\partial p}{\partial x}$, viscosity of the fluid is $\mu$.

Write the expression for incompressible Navier stoke Equation using $y$ relation.

$\rho \left(u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}\right)=\rho g_y-\frac{\partial p}{\partial y}+\mu \left(\frac{{\partial }^{2}v}{\partial x^2}+\frac{{\partial }^{2}v}{\partial y^2}+\frac{{\partial }^{2}v}{\partial z^2}\right)$ ... (III)

Here the density of the fluid is $\rho$ the acceleration due to gravity in the $y$ direction is $g_y$, Here, pressure change in $y$ direction is $\frac{\partial p}{\partial y}$.

Calculation:

Substitute $-2xy$ for $u$, $y^2-x^2$ for $v$, $0$ for $w$, 0 for $g_x$ in Equation (II).

$\begin{array}{l}\rho \left(\begin{array}{l}\left(-2xy\right)\frac{\partial \left(-2xy\right)}{\partial x}\\ +\left(y^2-x^2\right)\frac{\partial \left(-2xy\right)}{\partial y}\\ +\left(0\right)\frac{\partial \left(-2xy\right)}{\partial z}\end{array}\right)=\rho \times 0-\frac{\partial p}{\partial x}+\mu \left(\begin{array}{l}\frac{{\partial }^2\left(-2xy\right)}{\partial x^2}\\ +\frac{{\partial }^2\left(-2xy\right)}{\partial y^2}+\frac{{\partial }^2\left(-2xy\right)}{\partial z^2}\end{array}\right)\\ \rho \left(\left(-2xy\right)+\left(-2y\right)+\left(y^2-x^2\right)\left(-2x\right)+0\right)=0-\frac{\partial p}{\partial x}+\mu \left(0+0+0\right)\\ \rho \left(4x{y}^2+\left(-2x{y}^2+2x^3\right)\right)=-\frac{\partial p}{\partial x}\\ \rho \left(2x{y}^2+2x^3\right)=-\frac{\partial p}{\partial x}\end{array}$

$\begin{array}{l}2\rho \left(x{y}^2+x^3\right)=-\frac{\partial p}{\partial x}\\ \frac{\partial p}{\partial x}=-2\rho \left(x{y}^2+x^3\right)\end{array}$ ... (IV)

Substitute $-2xy$ for $u$, $y^2-x^2$ for $v$, $0$ for $w$, 0 for $g_y$ in Equation (III)

$\begin{array}{l}\rho \left(\begin{array}{l}\left(-2xy\right)\frac{\partial \left(y^2-x^2\right)}{\partial x}\\ +\left(y^2-x^2\right)\frac{\partial \left(y^2-x^2\right)}{\partial y}\\ +0\times \frac{\partial \left(y^2-x^2\right)}{\partial z}\end{array}\right)=\rho \times 0-\frac{\partial p}{\partial y}+\mu \left(\begin{array}{c}\frac{{\partial }^2\left(y^2-x^2\right)}{\partial x^2}\\ +\frac{{\partial }^2\left(y^2-x^2\right)}{\partial y^2}+\frac{{\partial }^2\left(y^2-x^2\right)}{\partial z^2}\end{array}\right)\\ \rho \left(\left(-2xy\right)+\left(-2x\right)+\left(y^2-x^2\right)\left(-2y\right)+0\right)=0-\frac{\partial p}{\partial y}+\mu \left(\left(-2\right)+\left(2\right)+0\right)\\ \rho \left(4x{y}^2+\left(2y^3-2x^{2}y\right)\right)=-\frac{\partial p}{\partial y}\end{array}$

$\begin{array}{l}-2\rho \left(x^{2}y+y^3\right)\partial y=\partial p\\ \partial p=-2\rho \left(x^{2}y+y^3\right)\partial y\end{array}$

Integrate the equation on both sides

$\begin{array}{l}{\displaystyle \int \partial p}=-2\rho {\displaystyle \int \left(x^{2}y+y^3\right)\partial y}\\ p=-2\rho \left(x^2\frac{y^2}{2}+\frac{y^4}{4}\right)+f\left(x\right)+\text{constant}\end{array}$ ... (V)

Differentiate the Equation with respect to $x$.

$\begin{array}{c}\frac{\partial p}{\partial x}=\frac{\partial }{\partial x}\left(-2\rho \left(x^2\frac{y^2}{2}+\frac{y^4}{4}\right)+f\left(x\right)+\text{constant}\right)\\ =\left(-2\rho \times \left(2x\frac{y^2}{2}+0\right)+\frac{\partial \left(f\left(x\right)\right)}{\partial x}+0\right)\\ =\left(-2\rho \times \left(x{y}^2+0\right)+\frac{\partial \left(f\left(x\right)\right)}{\partial x}\right)\\ =-2\rho x{y}^2+\frac{\partial \left(f\left(x\right)\right)}{\partial x}\end{array}$

Substitute $-2\rho x{y}^2+\frac{\partial \left(f\left(x\right)\right)}{\partial x}$ for $\frac{\partial p}{\partial x}$ in Equation (IV).

$\begin{array}{l}-2\rho x{y}^2+\frac{\partial \left(f\left(x\right)\right)}{\partial x}=-2\rho \left(x{y}^2+x^3\right)\\ \frac{\partial \left(f\left(x\right)\right)}{\partial x}=-2\rho \left(x{y}^2+x^3\right)+2\rho x{y}^2\\ \frac{\partial \left(f\left(x\right)\right)}{\partial x}=-2\rho x{y}^2-2\rho x^3+2\rho x{y}^2\\ \frac{\partial \left(f\left(x\right)\right)}{\partial x}=-2\rho x^3\end{array}$

$\partial \left(f\left(x\right)\right)=-2\rho x^3\partial x$

Integrating the above equation.

$\begin{array}{l}{\displaystyle \int \partial \left(f\left(x\right)\right)}=-2\rho {\displaystyle \int \left(x^3\right)\partial x}\\ f\left(x\right)=-2\rho \left[\frac{x^4}{4}\right]\end{array}$

Substitute $-2\rho \left[\frac{x^4}{4}\right]$ for $f\left(x\right)$ in Equation (V).

$\begin{array}{c}p=-2\rho \left(x^2\frac{y^2}{2}+\frac{y^4}{4}\right)+\left(-2\rho \frac{x^4}{4}\right)+\text{constant}\\ \text{=}-2\rho \left(x^2\frac{y^2}{2}+\frac{y^4}{4}+\frac{x^4}{4}\right)+\text{constant}\\ \text{=}-2\rho \times \frac{1}{4}\left(2x^2{y}^2+y^4+x^4\right)+\text{constant}\\ \text{=}-\frac{\rho }{2}\left(2x^2{y}^2+y^4+x^4\right)+\text{constant}\end{array}$ ... (VI).

Substitute $0$ for $x$, $0$ for $y$, $p_a$ for $p$ in Equation (VI).

$\begin{array}{l}{p}_a=-\frac{\rho }{2}\left(2\times 0+0+0\right)+\text{constant}\\ p_a=\text{constant}\end{array}$

Substitute $p_a$ for $\text{constant}$ in Equation (VI).

$p=-\frac{\rho }{2}\left(2x^2{y}^2+y^4+x^4\right)+p_a$

Conclusion:

The pressure field for $p\left(x,y\right)$ is $-\frac{\rho }{2}\left(2x^2{y}^2+x^4+y^4\right)+p_a$ ..