Chapter 4, Problem 4.33P

To determine

(a)

The expression for radial flow velocity.

Answer to Problem 4.33P

The expression for radial flow velocity is $V_r=-\frac{4Q}{\pi rb}$.

Explanation of Solution

Given information:

The volume flow rate of incompressible flow towards a drain is $Q$, the area of the wedge is $A$, the effect of friction and gravity are neglected, the width of the wedge is $b$, the angle of the wedge is $45°$, the flow is assumed to be purely radial.

Write the expression for the velocity of radial flow.

$v_r=-\frac{Q}{A}$ ... (I)

Here, volume flow rate of fluid is $Q$, area of the wedge is $A$.

Write the expression for the area of the wedge

$A=\left(\frac{\theta }{2\pi }\right)\left(2\pi rb\right)$ ... (II)

Here, angle of the wedge is $\theta$, the distance from the drain where the velocity is calculated is $r$, the width of the wedge is $b$.

Calculation:

Substitute $\frac{\pi }{4}$ for $\theta$ in Equation (II)

$\begin{array}{l}A=\left(\frac{\frac{\pi }{4}}{2\pi }\right)\left(2\pi rb\right)\\ A=\left(\frac{\pi }{4}\right)\left(rb\right)\end{array}$

Substitute $\left(\frac{\pi }{4}\right)\left(rb\right)$ for $A$ in Equation (I).

$\begin{array}{c}{v}_r=-\frac{Q}{\frac{\pi }{4}\left(rb\right)}\\ =-\frac{4Q}{\pi \left(rb\right)}\end{array}$

Conclusion:

The radial velocity at distance $r$ from the wedge is $v_r=-\frac{4Q}{\pi \left(rb\right)}$.

To determine

(b)

The value of viscous term in the momentum equation in $r$ direction is zero.

Explanation of Solution

Write the expression for the velocity of radial flow in $r$ direction

$v_r=-\frac{4Q}{\pi \left(rb\right)}$

Substitute $C=\frac{4Q}{\pi b}$ in Equation (II).

$v_r=-\frac{C}{r}$

Write the expression for viscous term from momentum Equation.

$\begin{array}{c}V=v\left({\nabla }^2{v}_r-\frac{v_r}{r^2}-\frac{2}{r^2}\frac{\partial v_{\theta }}{\partial \theta }\right)\\ =\left(\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial \left(v_r\right)}{\partial r}\right)-\frac{v_r}{r^2}-\frac{2}{r^2}\frac{\partial v_{\theta }}{\partial \theta }\right)\end{array}$ ... (III)

Here, the viscous term is $V$, velocity in $r$ direction is $v_r$, the velocity in $\theta$ direction is $v_{\theta }$.

Calculation:

Substitute $-\frac{C}{r}$ for $v_r$, 0 for $\frac{2}{r^2}\frac{\partial v_{\theta }}{\partial \theta }$ in Equation (III)

$\begin{array}{c}V=\left(\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial \left(-\frac{C}{r}\right)}{\partial r}\right)+\frac{C}{r^3}-0\right)\\ =\left(\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial \left(-\frac{C}{r}\right)}{\partial r}\right)+\frac{C}{r^3}-0\right)\\ =\left(-\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{-C}{r^2}\right)+\frac{C}{r^3}-0\right)\\ =\left(\frac{1}{r}\frac{\partial }{\partial r}\left(\frac{C}{r}\right)+\frac{C}{r^3}-0\right)\end{array}$

$\begin{array}{c}V=\left(-\frac{C}{r^3}+\frac{C}{r^3}-0\right)\\ =0\end{array}$

Conclusion:

Thus, the viscous term in momentum equation in $r$ direction is $0$.

(c)

To determine

The pressure distribution $p(r)$ at boundary condition $p=p_{\text{ο}}$, $r=R$.

Answer to Problem 4.33P

The pressure distribution $p(r)$ at boundary condition $p=p_{\text{ο}}$, $r=R$ is $p=p_{\text{ο}}+\frac{\rho C^2}{2}\left(\frac{1}{R^2}-\frac{1}{r^2}\right)$.

Explanation of Solution

Given information:

The boundary condition for the pressure distribution is $p=p_{\text{ο}}$, the boundary condition for distance from the drain is $r=R$.

Write the expression for maximum normal shear stress.

$\frac{\partial v}{\partial t}+v_r\frac{\partial v}{\partial r}-\frac{1}{r}{v}_{\theta }^2=-\frac{1}{\rho }\frac{\partial p}{\partial r}+g_r+V$ (IV)

Here the viscous term is $V$, The change in radial velocity with respect to time is $\frac{\partial v}{\partial t}$, The change in radial velocity with respect to $r$ distance is $\frac{\partial v}{\partial r}$, acceleration due to gravity in r direction is $g_r$, the change in pressure with respect to distance $r$ from sink is $-\frac{1}{\rho }\frac{\partial p}{\partial r}$.

Calculation:

Substitute $0$ for $V$, $0$ for $g_r$, $0$ for $\frac{\partial v}{\partial t}$, $0$ for $\frac{1}{r}{v}_{\theta }^2$ in Equation (IV).

$v_r\frac{\partial v}{\partial r}=-\frac{1}{\rho }\frac{\partial p}{\partial r}$ ... (V)

Substitute $-\frac{C}{r}$ for $v_r$ in Equation (V).

$\begin{array}{l}\frac{C}{r}\frac{\partial \left(\frac{C}{r}\right)}{\partial r}=-\frac{1}{\rho }\frac{\partial p}{\partial r}\\ -\frac{\rho C}{r}\frac{\partial \left(\frac{C}{r}\right)}{\partial r}=\frac{\partial p}{\partial r}\\ \frac{\partial p}{\partial r}=-\frac{\rho C}{r}\times -\left(\frac{C}{r^2}\right)\\ \partial p=\frac{\rho C^2}{r^3}\partial r\end{array}$ ... (VI)

Integrate Equation (VI).

$\begin{array}{l}{\displaystyle \int dp}={\displaystyle \int \rho \frac{C^2}{r^3}\partial r}\\ {\displaystyle \int dp}=\rho C^2{\displaystyle \int \left(\frac{1}{r^3}\right)\partial r}\\ p=\rho C^2\left(-\frac{1}{2r^2}\right)\\ p=-\frac{\rho C^2}{2r^2}+K\end{array}$ ... (VII)

Here, $K$ is integrating constant

Substitute boundary condition $p_{\text{ο}}$ for $p$, $R$ for $r$ in Equation (VII)

$\begin{array}{l}{p}_{\text{ο}}=-\frac{\rho C^2}{2R^2}+K\\ K=p_{\text{ο}}+\frac{\rho C^2}{2R^2}\end{array}$

Substitute $K$ in Equation (VII)

$\begin{array}{c}p=-\frac{\rho C^2}{2r^2}+p_{\text{ο}}+\frac{\rho C^2}{2R^2}\\ =p_{\text{ο}}+\frac{\rho C^2}{2R^2}-\frac{\rho C^2}{2r^2}\\ =p_{\text{ο}}+\frac{\rho C^2}{2}\left(\frac{1}{R^2}-\frac{1}{r^2}\right)\end{array}$

Conclusion:

The pressure distribution $p(r)$ at boundary condition $p=p_{\text{ο}}$, $r=R$ is $p=p_{\text{ο}}+\frac{\rho C^2}{2}\left(\frac{1}{R^2}-\frac{1}{r^2}\right)$ ..