Chapter 4, Problem 4.38P

To determine

(a)

The validation of Navier stokes Equation for two values of $n$ in cylindrical coordinate system.

Answer to Problem 4.38P

The Navier stokes Equation is satisfied for two values of $n$ equal to $\pm 1$.

Explanation of Solution

Given information:

The flow is incompressible in nature, the cylindrical coordinate are $v_r=0$, $v_{\theta }=C{r}^n$, $v_z=0$ . Here, C is constant.

Write the expression for momentum equation in $\theta$ direction.

$\frac{\partial v}{\partial t}+\left(\text{V}\cdot \nabla \right)v_{\theta }+\frac{1}{r}{v}_r{v}_{\theta }=-\frac{1}{\rho r}\frac{\partial p}{\partial \theta }+g_{\theta }+v\left({\nabla }^2{v}_{\theta }-\frac{v_{\theta }}{r^2}+\frac{2}{r^2}\frac{\partial v_r}{\partial \theta }\right)$        …… (I)

Here, the cylindrical coordinates are $(r,\theta ,z)$, the velocity components are $v_r$, $v_{\theta }$ and $v_z$, gravitation effect in $\theta$ direction is $g_{\theta }$, the pressure is $p$, the density of the fluid is $\rho$, the change in velocity with respect to time in $\theta$ direction is $\frac{\partial v}{\partial t}$, the pressure difference in the $\theta$ direction is $\frac{\partial p}{\partial \theta }$.

Since the flow is steady state so $\frac{\partial v}{\partial t}$ = 0.

Since the flow is irrotational so $\left(\text{V}\cdot \nabla \right)v_{\theta }=0$

Since there is no pressure difference so $-\frac{1}{\rho r}\frac{\partial p}{\partial \theta }=0$.

Since $v_r=0$ so $\frac{1}{r}{v}_r{v}_{\theta }=0$.

Calculation:

Substitute $0$ for $\frac{\partial v}{\partial t}$, $0$ for $g_{\theta }$, $0$ for $\frac{\partial v}{\partial t}$, $0$ for $\left(\text{V}\cdot \nabla \right)v_{\theta }$, 0 for $-\frac{1}{\rho r}\frac{\partial p}{\partial \theta }$, $0$ for $\frac{1}{r}{v}_r{v}_{\theta }$, $0$ for $v_r$ in Equation (I)

$\begin{array}{l}0+0+0=0+0+\mu \left({\nabla }^2{v}_{\theta }-\frac{v_{\theta }}{r^2}+\frac{2}{r^2}\frac{\partial \left(0\right)}{\partial \theta }\right)\\ 0=\mu \left({\nabla }^2{v}_{\theta }-\frac{v_{\theta }}{r^2}\right)\end{array}$        …… (II)

Substitute $C{r}^n$ for $v$ in Equation (II).

$\mu \left({\nabla }^2\left(C{r}^n\right)-\frac{C{r}^n}{r^2}\right)=0$        …… (III)

Substitute $\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)$ for ${\nabla }^2$ in Equation (III).

$\begin{array}{l}\mu \left(\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial \left(C{r}^n\right)}{\partial r}\right)-\frac{C{r}^n}{r^2}\right)=0\\ \mu \left(\frac{1}{r}\frac{\partial }{\partial r}\left(rnC{r}^{n-1}\right)-\left(C{r}^n\cdot r^{-2}\right)\right)=0\\ \mu \left(\frac{1}{r}\left(rn(n-1)C{r}^{n-2}+nC{r}^{n-1}\right)-\left(C{r}^{n-2}\right)\right)=0\\ \mu \left(\frac{1}{r}\left(rn(n-1)C{r}^{n-2}+nC{r}^{n-1}\right)-\left(C{r}^{n-2}\right)\right)=0\end{array}$

$\begin{array}{l}\mu \left(\left(n(n-1)C{r}^{n-2}+\left(\frac{nC{r}^{n-1}}{r}\right)\right)-\left(C{r}^{n-2}\right)\right)=0\\ \mu \left(\left(n^2-n\right)C{r}^{n-2}+nC{r}^{n-2}-C{r}^{n-2}\right)=0\\ \mu \left(n^{2}C{r}^{n-2}-nC{r}^{n-2}+nC{r}^{n-2}-C{r}^{n-2}\right)=0\\ \mu \left(n^{2}C{r}^{n-2}-C{r}^{n-2}\right)=0\\ \left(n^{2}C{r}^{n-2}-C{r}^{n-2}\right)=0\end{array}$

$\begin{array}{l}C{r}^{n-2}\left(n^2-1\right)=0\\ \left(n^2-1\right)=0\\ n^2=1\\ n=\sqrt{1}\end{array}$

$n=\pm 1$

Conclusion:

The Navier stokes Equation is satisfied for two values of $n$ equal to $\pm 1$.

To determine

(b)

The pressure distribution for $n=1$.

The pressure distribution for $n=-1$.

Answer to Problem 4.38P

The pressure distribution for $n=1$ is $p_o+\frac{\rho C^2}{2}\left(r^2-R^2\right)$.

The pressure distribution for $n=-1$ is $p_o+\frac{\rho C^2}{2}\left(\frac{1}{R^2}-\frac{1}{r^2}\right)$

Explanation of Solution

Given information:

The flow is incompressible in nature, the cylindrical coordinate are $v_r=0$, $v_{\theta }=C{r}^n$, $v_z=0$ . Here C is constant, $p=p\left(r\right)$ and at $r=R$ pressure is $p_o$

Write the expression for momentum equation in $r$ direction.

$\frac{\partial v}{\partial t}+\left(\text{V}\cdot \nabla \right)v_r-\frac{1}{r}{v}_{\theta }^2=-\frac{1}{\rho }\frac{\partial p}{\partial r}+g_r+v\left({\nabla }^2{v}_r-\frac{v_r}{r^2}-\frac{2}{r^2}\frac{\partial v_{\theta }}{\partial \theta }\right)$        …… (IV)

Here the viscous term is $v\left({\nabla }^2{v}_r-\frac{v_r}{r^2}-\frac{2}{r^2}\frac{\partial v_{\theta }}{\partial \theta }\right)$, The change in radial velocity with respect to time is $\frac{\partial v}{\partial t}$, acceleration due to gravity in r direction is $g_r$, the change in pressure with respect to distance $r$ is $-\frac{1}{\rho }\frac{\partial p}{\partial r}$.

Calculation:

Substitute $0$ for $v\left({\nabla }^2{v}_r-\frac{v_r}{r^2}-\frac{2}{r^2}\frac{\partial v_{\theta }}{\partial \theta }\right)$, $0$ for $g_r$, $0$ for $\frac{\partial v}{\partial t}$, $0$ for $\frac{1}{r}{v}_{\theta }^2$, consider steady flow equation in $r$ direction with no viscous force and with no gravity in Equation (IV).

$0+0-\frac{1}{r}{v}_{\theta }^2=-\frac{1}{\rho }\frac{\partial p}{\partial r}+0+0$

$\begin{array}{c}-\frac{1}{r}{v}_{\theta }^2=-\frac{1}{\rho }\frac{\partial p}{\partial r}\\ \frac{\rho }{r}{v}_{\theta }^2=\frac{\partial p}{\partial r}\end{array}$        …… (V)

Substitute $C{r}^n$ for $v_{\theta }$ in Equation (V).

$\begin{array}{l}\rho \frac{{\left(C{r}^n\right)}^2}{r}=\frac{\partial p}{\partial r}\\ \end{array}$        …… (VI)

Case-1:

Integrate Equation (VI)

${\displaystyle \underset{R}{\overset{r}{\int }}\rho \frac{{\left(C{r}^n\right)}^2}{r}\partial r}={\displaystyle \underset{p_o}{\overset{p}{\int }}\partial p}$        …… (VII)

Substitute $1$ for $n$ in Equation (VII)

$\begin{array}{l}{\displaystyle \underset{R}{\overset{r}{\int }}\rho \frac{{\left(C{r}^1\right)}^2}{r}\partial r}={\displaystyle \underset{p_o}{\overset{p}{\int }}\partial p}\\ {\displaystyle \underset{R}{\overset{r}{\int }}\rho C^{2}r\partial r}={\displaystyle \underset{p_o}{\overset{p}{\int }}\partial p}\\ \rho C^2{\left[\frac{r^2}{2}\right]}_R^r={\left[p\right]}_{p_o}^p\\ \frac{\rho C^2}{2}\left(r^2-R^2\right)=p-p_o\end{array}$

$p=\frac{\rho C^2}{2}\left(r^2-R^2\right)+p_o$

Case-2:

Substitute $-1$ for $n$ in Equation (VII)

$\begin{array}{c}{\displaystyle \underset{R}{\overset{r}{\int }}\rho \frac{{\left(C{r}^{-1}\right)}^2}{r}\partial r}={\displaystyle \underset{p_o}{\overset{p}{\int }}\partial p}\\ {\displaystyle \underset{R}{\overset{r}{\int }}\rho C^2{r}^{-3}\partial r}={\displaystyle \underset{p_o}{\overset{p}{\int }}\partial p}\\ \rho C^2{\left[\frac{r^{-2}}{-2}\right]}_R^r={\left[p\right]}_{p_o}^p\\ \rho C^2{\left[-\frac{1}{2r^2}\right]}_R^r={\left[p\right]}_{p_o}^p\end{array}$

$\begin{array}{l}\frac{\rho C^2}{2}{\left[-\frac{1}{r^2}+\frac{1}{R^2}\right]}_R^r=p-p_o\\ \frac{\rho C^2}{2}\left(\frac{1}{R^2}-\frac{1}{r^2}\right)=p-p_o\\ p=p_o+p\frac{\rho C^2}{2}\left(\frac{1}{R^2}-\frac{1}{r^2}\right)\end{array}$

Conclusion:

The pressure distribution for $n=1$ is $p_o+\frac{\rho C^2}{2}\left(r^2-R^2\right)$.

The pressure distribution for $n=-1$ is $p_o+\frac{\rho C^2}{2}\left(\frac{1}{R^2}-\frac{1}{r^2}\right)$.

For $n=1$, the flow represents the case of solid body rotation

For $n=-1$, the flow represents the case of irrotational potential vortex.