Chapter 4, Problem 4.30P

To determine

(a)

The validation of continuity equation.

Answer to Problem 4.30P

The continuity equation is valid for the given velocity distribution.

Explanation of Solution

Given information:

Two dimensional converging nozzle velocity distribution in $x$ direction, $y$ direction and $z$ direction is $u=U_o\left(1+\frac{x}{L}\right)$, $v=-U_o\frac{y}{L}$, $w=0$ respectively. Here, the velocity in $x$ -direction is $u$, velocity, the velocity in $y$ -direction is $v$ . the velocity in $z$ direction is $w$.

Write the expression for two dimensional incompressible continuity Equation.

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$ ... (I)

Here, the velocity of fluid along $x$ and $y$ directions are $u$ and $v$ respectively, Velocity component along $x$ direction is $\frac{\partial u}{\partial x}$, Velocity component along $y$ direction is $\frac{\partial v}{\partial y}$.

Calculation:

Substitute $U_o\left(1+\frac{x}{L}\right)$ for $u$, $-U_o\frac{y}{L}$ for $v$, in Equation (I).

$\begin{array}{c}\frac{\partial }{\partial x}\left(U_o\left(1+\frac{x}{L}\right)\right)+\frac{\partial }{\partial y}\left(-U_o\frac{y}{L}\right)=0\\ \left(U_o\left(0+\frac{1}{L}\right)\right)+\left(-U_o\frac{1}{L}\right)=0\\ \frac{U_o}{L}-\frac{U_o}{L}=0\end{array}$

Conclusion:

The given distribution satisfies the two dimensional continuity equation..

To determine

(b)

The validation of Navier stokes equation.

Answer to Problem 4.30P

Navier stokes equation is valid for the the given velocity distribution.

Explanation of Solution

Write the expression for two dimensional incompressible Navier stoke equation in $x$ direction.

$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=-\frac{1}{\rho }\frac{\partial p}{\partial x}+v\left(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}v}{\partial y^2}\right)$ ... (II)

Write the expression for two dimensional incompressible Navier stoke equation in $y$ direction.

$u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}=-\frac{1}{\rho }\frac{\partial p}{\partial y}+v\left(\frac{{\partial }^{2}v}{\partial x^2}+\frac{{\partial }^{2}v}{\partial y^2}\right)$ ... (III)

Here, the velocity of fluid along $x$ and $y$ directions are $u$ and $v$ respectively, Velocity component along $x$ direction is $\frac{\partial u}{\partial x}$, Velocity component along $y$ direction is $\frac{\partial v}{\partial y}$

Write the expression for validation for Navier stokes equation in x-direction.

$\frac{{\partial }^{2}p}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial p}{\partial y}\right)=0$ ... (IV)

Write the expression for validation for Navier stokes equation in y-direction.

$\frac{{\partial }^{2}p}{\partial y\partial x}=\frac{\partial }{\partial y}\left(\frac{\partial p}{\partial x}\right)=0$ ... (V)

Calculation:

Substitute $U_o\left(1+\frac{x}{L}\right)$ for $u$, $-U_o\frac{y}{L}$ for $v$, $0$ for $\frac{{\partial }^{2}u}{\partial x^2}$, $0$ for $\frac{{\partial }^{2}u}{\partial y^2}$ in Equation (II).

$\begin{array}{l}\left(\begin{array}{l}{U}_o\left(1+\frac{x}{L}\right)\frac{\partial \left(U_o\left(1+\frac{x}{L}\right)\right)}{\partial x}\\ +\left(-U_o\frac{y}{L}\right)\frac{\partial \left(U_o\left(1+\frac{x}{L}\right)\right)}{\partial y}\end{array}\right)=-\frac{1}{\rho }\frac{\partial p}{\partial x}+v\left(0+0\right)\\ \left(U_o\left(1+\frac{x}{L}\right)\left(\frac{U_o}{L}\right)+\left(-U_o\frac{y}{L}\right)\left(0\right)\right)=-\frac{1}{\rho }\frac{\partial p}{\partial x}+0\\ \left(\left(\frac{U_o^2}{L}\right)\left(1+\frac{x}{L}\right)+0\right)=-\frac{1}{\rho }\frac{\partial p}{\partial x}\\ -\rho \left(\frac{U_o^2}{L}\right)\left(1+\frac{x}{L}\right)=\frac{\partial p}{\partial x}\end{array}$

$\frac{\partial p}{\partial x}=\left(\frac{-\rho U_o^2}{L}\right)\left(1+\frac{x}{L}\right)$ ... (VI)

Substitute $U_o\left(1+\frac{x}{L}\right)$ for $u$, $-U_o\frac{y}{L}$ for $v$, $0$ for $\frac{{\partial }^{2}u}{\partial x^2}$, $0$ for $\frac{{\partial }^{2}u}{\partial y^2}$ in Equation (III).

$\begin{array}{l}\left[U_o\left(1+\frac{x}{L}\right)\right]\frac{\partial \left(-U_o\frac{y}{L}\right)}{\partial x}+\left(-U_o\frac{y}{L}\right)\frac{\partial \left(-U_o\frac{y}{L}\right)}{\partial y}=-\frac{1}{\rho }\frac{\partial p}{\partial y}+v\left(0+0\right)\\ U_o\left(1+\frac{x}{L}\right)\left(0\right)+\left(-U_o\frac{y}{L}\right)\left(-U_o\frac{1}{L}\right)=-\frac{1}{\rho }\frac{\partial p}{\partial y}+0\\ 0+\left(\frac{U_o^{2}y}{L^2}\right)=-\frac{1}{\rho }\frac{\partial p}{\partial y}\\ -\rho \left(\frac{U_o^{2}y}{L^2}\right)=\frac{\partial p}{\partial y}\end{array}$

$\frac{\partial p}{\partial y}=-\rho \left(U_o^2\frac{y}{L^2}\right)$ ... (VII)

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

$\begin{array}{c}\frac{\partial }{\partial x}\left(-\rho \left(\frac{U_o^{2}y}{L^2}\right)\right)=0\\ 0=0\end{array}$

Hence the equation $v=-U_o\frac{y}{L}$ has an exact solution to Navier stokes

Substitute $\left(\frac{-\rho U_o^2}{L}\right)\left(1+\frac{x}{L}\right)$ for $\frac{\partial p}{\partial x}$ in Equation (V).

$\begin{array}{c}\frac{\partial }{\partial y}\left(\left(\frac{-\rho U_o^2}{L}\right)\left(1+\frac{x}{L}\right)\right)=0\\ 0=0\end{array}$

Hence the equation $u=U_o\left(1+\frac{x}{L}\right)$ has an exact solution to Navier stokes

Conclusion:

Navier stokes equation is valid for the given velocity distribution.

To determine

(c)

The pressure distribution $p(x,y)$.

Answer to Problem 4.30P

The pressure distribution $p(x,y)$ is $-\frac{\rho U_o^2}{L}\left(x+\frac{x^2}{2L}+\frac{y^2}{2L}\right)+p_o$

Explanation of Solution

Given information:

The pressure at the origin is $p_a$.

Write the expression for pressure distribution in $x$ direction.

$p_x={\displaystyle \int \frac{\partial p}{\partial x}}dx$ ... (VIII)

Write the expression for pressure distribution in $y$ direction.

$p_y={\displaystyle \int \frac{\partial p}{\partial y}}dy$ ... (IX)

Write the expression for pressure distribution

$p=p_x+p_y$ ... (X)

Calculation:

Substitute $\left(\frac{-\rho U_o^2}{L}\right)\left(1+\frac{x}{L}\right)$ for $\frac{\partial p}{\partial x}$ in Equation (VIII).

$\begin{array}{c}{p}_x={\displaystyle \int \left(\frac{-\rho U_o^2}{L}\right)\left(1+\frac{x}{L}\right)}dx\\ =\left(\frac{-\rho U_o^2}{L}\right)\left(x+\frac{x^2}{2L}\right)\end{array}$ ... (XI)

Substitute $-\rho \left(U_o^2\frac{y}{L^2}\right)$ for $\frac{\partial p}{\partial y}$ in Equation (IX).

$\begin{array}{c}{p}_y={\displaystyle \int \left(-\rho \left(U_o^2\frac{y}{L^2}\right)\right)}dy\\ =-\rho \left(U_o^2\frac{y^2}{2L^2}\right)+C\end{array}$ ... (XII)

Substitute $p_a$ for $C$ in Equation (XII).

$-\rho \left(U_o^2\frac{y^2}{2L^2}\right)+P_a$

Substitute $\left(\frac{-\rho U_o^2}{L}\right)\left(x+\frac{x^2}{2L}\right)$ for $p_x$ and $-\rho \left(U_o^2\frac{y^2}{2L^2}\right)+P_a$ for $p_y$ in Equation (X)

$\begin{array}{c}p=\left(\frac{-\rho U_o^2}{L}\right)\left(x+\frac{x^2}{2L}\right)-\rho \left(U_o^2\frac{y^2}{2L^2}\right)+P_a\\ =\frac{-\rho U_o^2}{L}\left(x+\frac{x^2}{2L}+\frac{y^2}{2L^2}\right)+P_a\end{array}$

Conclusion:

The pressure field $p(x,y)$ is $\frac{-\rho U_o^2}{L}\left(x+\frac{x^2}{2L}+\frac{y^2}{2L^2}\right)+P_a$.