Chapter 9, Problem 89P

Consider steady, two-dimensional, incompressible flow due to a spiraling line vortex/sink flow centered on the z-axis. Streamlines and velocity components are shown in Fig. 9-87. The velocity field is $u_r=C/r$ and $u_{\theta }=K/r$ , where C and K are constants. Calculate the pressure as a function of r and $\theta$ .

FIGURE P9-87

To determine

The pressure field as a function $r$ and $\theta$.

Answer to Problem 89P

The pressure field as a function $r$ and $\theta$ is $-\frac{1}{2}\rho \left(\frac{C^2+K^2}{r^2}\right)+C_1$.

Explanation of Solution

Given information:

The flow is steady, two-dimensional and incompressible.

The velocity components are $u_r=\frac{C}{r}$ and $u_{\theta }=\frac{K}{r}$.

Here, $C$ and $K$ are constants.

Write the expression for continuity equation.

  $\frac{1}{r}\frac{\partial \left(r{u}_r\right)}{\partial r}+\frac{1}{r}\frac{\partial \left(u_{\theta }\right)}{\partial \theta }=0$........... (I)

Here, velocity components are $u_r$ and $u_{\theta }$, radius is $r$ and angle is $\theta$.

Write the expression for the $\theta$ -component of Navier-Stokes equation.

  $\left[\rho \left(\begin{array}{l}\frac{\partial u_{\theta }}{\partial t}+u_r\frac{\partial u_{\theta }}{\partial r}+\\ \frac{u_{\theta }}{r}\frac{\partial u_{\theta }}{\partial \theta }+\frac{u_r{u}_{\theta }}{r}\end{array}\right)\right]=\left[-\frac{1}{r}\frac{\partial P}{\partial \theta }+\mu \left(\begin{array}{l}\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_{\theta }}{\partial r}\right)-\frac{u_{\theta }}{r^2}\\ +\frac{1}{r^2}\frac{{\partial }^2{u}_{\theta }}{\partial {\theta }^2}-\frac{2}{r^2}\frac{\partial u_r}{\partial \theta }\end{array}\right)+\rho g_{\theta }\right]$........... (II)

Here, density is $\rho$, time is $t$, dynamic viscosity is $\mu$ and gravitational acceleration in $\theta$ -direction is $g_{\theta }$.

Write the expression for the $r$ -component of Navier-Stokes equation.

  $\left[\rho \left(\begin{array}{l}\frac{\partial u_r}{\partial t}+u_r\frac{\partial u_r}{\partial r}+\\ \frac{u_{\theta }}{r}\frac{\partial u_r}{\partial \theta }-\frac{u_{\theta }^2}{r}\end{array}\right)\right]=\left[-\frac{\partial P}{\partial r}+\mu \left(\begin{array}{l}\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_r}{\partial r}\right)-\frac{u_r}{r^2}+\\ \frac{1}{r^2}\frac{{\partial }^2{u}_r}{\partial {\theta }^2}-\frac{2}{r^2}\frac{\partial u_{\theta }}{\partial \theta }\end{array}\right)+\rho g_r\right]$........... (III)

Calculation:

Substitute $\frac{C}{r}$ for $u_r$ and $\frac{K}{r}$ for $u_{\theta }$ in Equation (I).

  $\begin{array}{l}\frac{1}{r}\frac{\partial \left(r\left(\frac{C}{r}\right)\right)}{\partial r}+\frac{1}{r}\frac{\partial \left(\frac{K}{r}\right)}{\partial \theta }=0\\ \frac{1}{r}\frac{\partial }{\partial r}\left(C\right)+\frac{1}{r}\frac{\partial }{\partial \theta }\left(\frac{K}{r}\right)=0\\ 0+0=0\\ 0=0\end{array}$........... (IV)

Since both sides of Equation (IV) are equal, therefore continuity equation is verified.

Substitute $\frac{C}{r}$ for $u_r$, $0$ for $g_{\theta }$ and $\frac{K}{r}$ for $u_{\theta }$ in Equation (II).

$\left[\rho \left(\begin{array}{l}\frac{\partial \left(\frac{K}{r}\right)}{\partial t}+\left(\frac{C}{r}\right)\frac{\partial \left(\frac{K}{r}\right)}{\partial r}+\\ \frac{\left(\frac{K}{r}\right)}{r}\frac{\partial \left(\frac{K}{r}\right)}{\partial \theta }+\frac{\left(\frac{C}{r}\right)\left(\frac{K}{r}\right)}{r}\end{array}\right)\right]=\left[-\frac{1}{r}\frac{\partial P}{\partial \theta }+\mu \left(\begin{array}{l}\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial \left(\frac{K}{r}\right)}{\partial r}\right)-\frac{\left(\frac{K}{r}\right)}{r^2}\\ +\frac{1}{r^2}\frac{{\partial }^2\left(\frac{K}{r}\right)}{\partial {\theta }^2}-\frac{2}{r^2}\frac{\partial \left(\frac{C}{r}\right)}{\partial \theta }\end{array}\right)+\rho \times 0\right]$

$\left[\rho \left(\begin{array}{l}\frac{\partial }{\partial t}\left(\frac{K}{r}\right)+\left(\frac{C}{r}\right)\frac{\partial }{\partial r}\left(\frac{K}{r}\right)\\ +\frac{K}{r^2}\frac{\partial }{\partial \theta }\left(\frac{K}{r}\right)+\frac{CK}{r^3}\end{array}\right)\right]=\left[-\frac{1}{r}\frac{\partial P}{\partial \theta }+\mu \left(\begin{array}{l}\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\left(\frac{K}{r}\right)\right)-\frac{K}{r^3}+\\ \frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}\left(\frac{K}{r}\right)-\frac{2}{r^2}\frac{\partial }{\partial \theta }\left(\frac{C}{r}\right)\end{array}\right)\right]$

$\rho \left(0-\frac{CK}{r^3}+0+\frac{CK}{r^3}\right)=-\frac{1}{r}\frac{\partial P}{\partial \theta }+\mu \left(\frac{1}{r}\frac{\partial }{\partial r}\left(-\frac{K}{r}\right)-\frac{K}{r^3}+0-0\right)$

$0=-\frac{1}{r}\frac{\partial P}{\partial \theta }+\mu \left(\frac{K}{r^3}-\frac{K}{r^3}+0-0\right)$

  $\begin{array}{l}-\frac{1}{r}\frac{\partial P}{\partial \theta }=0\\ \frac{\partial P}{\partial \theta }=0\end{array}$........... (V)

Integrate Equation (V) with respect to $\theta$.

  $P\left(r,\theta \right)=0+g\left(r\right)$........... (VI)

Here, arbitrary function is $g\left(r\right)$.

Differentiate Equation (VI) with respect to $r$.

  $\frac{\partial P}{\partial r}=g^{\prime }\left(r\right)$........... (VII)

Substitute $\frac{C}{r}$ for $u_r$, $0$ for $g_r$ and $\frac{K}{r}$ for $u_{\theta }$ in Equation (III).

$\left[\rho \left(\begin{array}{l}\frac{\partial \left(\frac{C}{r}\right)}{\partial t}+\left(\frac{C}{r}\right)\frac{\partial \left(\frac{C}{r}\right)}{\partial r}+\\ \frac{\left(\frac{K}{r}\right)}{r}\frac{\partial \left(\frac{C}{r}\right)}{\partial \theta }-\frac{{\left(\frac{K}{r}\right)}^2}{r}\end{array}\right)\right]=\left[-\frac{\partial P}{\partial r}+\mu \left(\begin{array}{l}\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial \left(\frac{C}{r}\right)}{\partial r}\right)-\frac{\left(\frac{C}{r}\right)}{r^2}\\ +\frac{1}{r^2}\frac{{\partial }^2\left(\frac{C}{r}\right)}{\partial {\theta }^2}-\frac{2}{r^2}\frac{\partial \left(\frac{K}{r}\right)}{\partial \theta }\end{array}\right)+\rho \times 0\right]$

$\left[\rho \left(\begin{array}{l}\frac{\partial }{\partial t}\left(\frac{C}{r}\right)+\left(\frac{C}{r}\right)\frac{\partial }{\partial r}\left(\frac{C}{r}\right)\\ +\frac{K}{r^2}\frac{\partial }{\partial \theta }\left(\frac{C}{r}\right)-\frac{K^2}{r^3}\end{array}\right)\right]=\left[-\frac{\partial P}{\partial r}+\mu \left(\begin{array}{l}\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\left(\frac{C}{r}\right)\right)-\frac{C}{r^3}+\\ \frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}\left(\frac{C}{r}\right)-\frac{2}{r^2}\frac{\partial }{\partial \theta }\left(\frac{K}{r}\right)\end{array}\right)\right]$

$\rho \left(0-\frac{C^2}{r^3}+0-\frac{K^2}{r^3}\right)=-\frac{\partial P}{\partial r}+\mu \left(\frac{1}{r}\frac{\partial }{\partial r}\left(-\frac{C}{r}\right)-\frac{C}{r^3}+0-0\right)$

$\rho \left(-\frac{C^2}{r^3}-\frac{K^2}{r^3}\right)=-\frac{\partial P}{\partial r}+\mu \left(\frac{C}{r^3}-\frac{C}{r^3}\right)$

  $\begin{array}{l}\rho \left(-\frac{C^2}{r^3}-\frac{K^2}{r^3}\right)=-\frac{\partial P}{\partial r}\\ \frac{\partial P}{\partial r}=\rho \left(\frac{C^2+K^2}{r^3}\right)\end{array}$

Substitute $\rho \left(\frac{C^2+K^2}{r^3}\right)$ for $\frac{\partial P}{\partial r}$ in Equation (VII).

  $\rho \left(\frac{C^2+K^2}{r^3}\right)=g^{\prime }\left(r\right)$........... (VIII)

Integrate Equation (VIII) with respect to $r$.

  $g\left(r\right)=-\frac{1}{2}\rho \left(\frac{C^2+K^2}{r^2}\right)+C_1$

Here, arbitrary constant is $C_1$.

Substitute $-\frac{1}{2}\rho \left(\frac{C^2+K^2}{r^2}\right)+C_1$ for $g\left(r\right)$ in Equation (VI).

  $P\left(r,\theta \right)=-\frac{1}{2}\rho \left(\frac{C^2+K^2}{r^2}\right)+C_1$........... (IX)

Conclusion:

The pressure field as a function $r$ and $\theta$ is $-\frac{1}{2}\rho \left(\frac{C^2+K^2}{r^2}\right)+C_1$.