Chapter 5, Problem 114P

A circular thin plate is placed on the top of a tube. as shown in the figure. (a) Find the exit velocity from the gap. (b) Find the velocity and pressure distributions in the gap between plate and tube’s tip for d/2 ρ air = 1.2 k g / m 3 and $P_{\text{atm}}=101,325\text{Pa}$ .

To determine

(a)

The exit velocity from the gap.

Answer to Problem 114P

The exit velocity from the gap is $0.045\text{m}/\text{s}$

Explanation of Solution

Given information:

The diameter of tube is $3\text{mm}$, the gap between plate is $1.5\text{mm}$, the diameter of disc is $10\text{cm}$.

Atmospheric pressure is $101325\text{Pa}$, velocity of air entering the tube is $3\text{m}/\text{s}$ the density of remains constant.

Write the expression for area of tube.

  $A_{\text{t}}=\frac{\pi }{4}{d}^2$   ...... (I)

Here, diameter of tube is $d$.

Write the expression for area of exit.

  $A_{\text{e}}=\pi Ds$   ...... (II)

Here, diameter of dic is $D$ and the gap is $s$.

Write the expression for continuity equation of flow.

  ${\rho }_{\text{t}}{A}_{\text{t}}{V}_{\text{t}}={\rho }_{\text{e}}{A}_{\text{e}}{V}_{\text{e}}$   ...... (III)

Here, density of air at tube entry is ${\rho }_{\text{t}}$, area of tube entry is $A_{\text{t}}$ and velocity of air at entry is $V_{\text{t}}$ density of air at tube exit is ${\rho }_{\text{e}}$, area of exit is $A_{\text{e}}$ and velocity of air at exit is $V_{\text{e}}$.

Calculation:

Substitute $3\text{mm}$ for $d$ in equation (I).

  $\begin{array}{c}{A}_{\text{t}}=\frac{\pi }{4}{\left(3\text{mm}\right)}^2\\ =\frac{\pi }{4}{\left(3\text{mm}\times \frac{\text{1}\text{m}}{1000\text{mm}}\right)}^2\\ =\frac{\pi }{4}{\left(0.003\text{m}\right)}^2\\ =7.068\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Substitute $1.5\text{mm}$ for $s$ and $10\text{cm}$ for $D$ in Equation (II).

  $\begin{array}{c}{A}_{\text{e}}=\pi \left(10\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(1.5\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\\ =\pi \left(0.1\text{m}\right)\left(0.0015\text{m}\right)\\ =4.712\times {10}^{-4}{\text{m}}^2\end{array}$

Substitute $7.068\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_{\text{t}}$, $4.712\times {10}^{-4}{\text{m}}^2$ for $A_{\text{e}}$, $\rho$ for ${\rho }_{\text{t}}$, $\rho$ for ${\rho }_{\text{e}}$

  $3\text{m}/\text{s}$ for $V_{\text{t}}$ in Equation (III).

  $\begin{array}{l}\rho \left(7.068\times {10}^{-6}{\text{m}}^{\text{2}}\right)\left(3\text{m}/\text{s}\right)=\rho \left(4.712\times {10}^{-4}{\text{m}}^2\right)V_{\text{e}}\\ \left(7.068\times {10}^{-6}{\text{m}}^{\text{2}}\right)\left(3\text{m}/\text{s}\right)=\left(4.712\times {10}^{-4}{\text{m}}^2\right)V_{\text{e}}\\ V_{\text{e}}=\frac{\left(7.068\times {10}^{-6}{\text{m}}^{\text{2}}\right)\left(3\text{m}/\text{s}\right)}{\left(4.712\times {10}^{-4}{\text{m}}^2\right)}\\ V_{\text{e}}=0.045\text{m}/\text{s}\end{array}$

Conclusion:

The exit velocity from the gap is $0.045\text{m}/\text{s}$.

To determine

(b)

The velocity distribution at distance $r$.

The pressure distribution at distance $r$.

Answer to Problem 114P

The velocity distribution at distance $r$ is $\frac{2.249\times {10}^{-3}}{r}\text{m}/\text{s}$.

The pressure distribution at distance $r$.is $\frac{3.0348\times {10}^{-6}}{r^2}\text{Pa}$.

Explanation of Solution

Given information:

The diameter of tube is $3\text{mm}$, the gap between plate is $1.5\text{mm}$.

Write the expression for area of exit.

  $A_{\text{e}}=2\pi rs$   ...... (IV)

Here, radius is $r$ and the gap is $s$.

Write the expression for continuity equation of flow.

  ${\rho }_{\text{t}}{A}_{\text{t}}{V}_{\text{t}}={\rho }_{\text{r}}{A}_{\text{r}}{V}_{\text{r}}$   ...... (V)

Here, density of air at tube entry is ${\rho }_{\text{t}}$, area of tube entry is $A_{\text{t}}$ and velocity of air at entry is $V_{\text{t}}$ density of air at tube exit is ${\rho }_r$, area of exit is $A_{\text{r}}$ and velocity of air at exit is $V_{\text{r}}$.

Write the expression for Bernoulli equation.

  $\frac{P_1}{\rho g}+\frac{V_1{}^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{V_2{}^2}{2g}+z_2$   ...... (VI)

Here, pressure at entry is $P_1$, velocity of air at entry is $V_1$, height at entry is $z_1$, pressure at exit is $P_2$, velocity of air at exit is $V_2$ and height at exit is $z_2$.

Calculation:

Substitute $1.5\text{mm}$ for $s$ in Equation (IV).

  $\begin{array}{c}{A}_{\text{e}}=2\pi \left(1.5\text{mm}\right)r\\ =2\pi \left(1.5\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)r\\ =9.42477\times {10}^{-3}r{\text{m}}^{\text{2}}\end{array}$

Substitute $7.068\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_{\text{t}}$, $9.42477\times {10}^{-3}r{\text{m}}^{\text{2}}$ for $A_r$, $\rho$ for ${\rho }_{\text{t}}$, $\rho$ for ${\rho }_e$

  $3\text{m}/\text{s}$ for $V_{\text{t}}$ in Equation (V).

  $\begin{array}{l}\rho \left(7.068\times {10}^{-6}{\text{m}}^{\text{2}}\right)\left(3\text{m}/\text{s}\right)=\rho \left(9.42477\times {10}^{-3}r{\text{m}}^{\text{2}}\right)V_r\\ \left(7.068\times {10}^{-6}{\text{m}}^{\text{2}}\right)\left(3\text{m}/\text{s}\right)=\left(9.42477\times {10}^{-3}r{\text{m}}^{\text{2}}\right)V_r\\ V_r=\frac{\left(7.068\times {10}^{-6}{\text{m}}^{\text{2}}\right)\left(3\text{m}/\text{s}\right)}{\left(9.42477\times {10}^{-3}r{\text{m}}^{\text{2}}\right)}\\ V_r=\frac{2.249\times {10}^{-3}}{r}\text{m}/\text{s}\end{array}$

Substitute $3\text{m}/\text{s}$ for $V_{\text{t}}$, $0$ for $z_1$

  $1.2\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0.045\text{m}/\text{s}$ for $V_r$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0$ for $z_2$, $0$ for $P_2$ in Equation (VI).

  $\begin{array}{l}\frac{P_1}{\rho g}+\frac{{\left(3\text{m}/\text{s}\right)}^2}{2g}+0=\frac{0}{\rho g}+\frac{{\left(0.045\text{m}/\text{s}\right)}^2}{2g}+0\\ \frac{P_1}{\rho g}+\frac{{\left(3\text{m}/\text{s}\right)}^2}{2g}=\frac{{\left(0.045\text{m}/\text{s}\right)}^2}{2g}\\ \frac{P_1}{g}=\left(1.2\text{kg}/{\text{m}}^{\text{3}}\right)\frac{{\left(0.045\text{m}/\text{s}\right)}^2-{\left(3\text{m}/\text{s}\right)}^2}{2g}\\ P_1=-5.3987\text{kg}/\text{m}\cdot {\text{s}}^2\end{array}$

Substitute $3\text{m}/\text{s}$ for $V_{\text{t}}$, $0$ for $z_1$

  $1.2\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$

  $\frac{2.249\times {10}^{-3}}{r}\text{m}/\text{s}$ for $V_r$

  $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0$ for $z_2$

  $-5.3987\text{kg}/\text{m}\cdot \text{s}$ for $P_1$ in Equation (VI).

  $\begin{array}{l}\frac{-5.3987\text{kg}/\text{m}\cdot \text{s}}{\rho g}+\frac{{\left(3\text{m}/\text{s}\right)}^2}{2g}+0=\frac{P_{\text{r}}}{\rho g}+\frac{{\left(\frac{2.249\times {10}^{-3}}{r}{\text{m}/\text{s}}^2\right)}^2}{2g}+0\\ P_{\text{r}}=\left(1.2\text{kg}/{\text{m}}^{\text{3}}\right)\frac{{\left(\frac{2.249\times {10}^{-3}}{r}{\text{m}/\text{s}}^2\right)}^2}{2}+5.3987\text{kg}/\text{m}\cdot \text{s}-\left(1.2\text{kg}/{\text{m}}^{\text{3}}\right)\frac{{\left(3\text{m}/\text{s}\right)}^2}{2}\\ P_{\text{r}}=\frac{3.0348\times {10}^{-6}}{r^2}\text{kg}/\text{m}\cdot {\text{s}}^2+5.3987\text{kg}/\text{m}\cdot {\text{s}}^2-5.3987\text{kg}/\text{m}\cdot {\text{s}}^2\\ P_r=\frac{3.0348\times {10}^{-6}}{r^2}\text{kg}/\text{m}\cdot {\text{s}}^2\times \frac{1\text{Pa}}{1\text{kg}/\text{m}\cdot {\text{s}}^2}\end{array}$

  $P_r=\frac{3.0348\times {10}^{-6}}{r^2}\text{Pa}$

Conclusion:

The velocity distribution at distance $r$ is $\frac{2.249\times {10}^{-3}}{r}\text{m}/\text{s}$.

The pressure distribution at distance $r$.is $\frac{3.0348\times {10}^{-6}}{r^2}\text{Pa}$.