Chapter 9, Problem 66EP

To determine

(a)

The axial speeds $\left(\text{ft}/\text{s}\right)$ at the nozzle entrance and at the nozzle exit.

Answer to Problem 66EP

The axial speed $\text{ft}/\text{s}$ at the nozzle entrance is $3.2688\text{ft}/\text{s}$.

The axial speed $\text{ft}/\text{s}$ at the nozzle exit is $41.7404\text{ft}/\text{s}$.

Explanation of Solution

Given information:

The diameter at the entrance of the nozzle is $0.50\text{in}$, the diameter at exit of the nozzle is $0.14\text{in}$. The length of the nozzle is $0.20\text{in}$. The volume flow rate through nozzle is $2\text{gal}/\text{min}$.

Write the expression for axial speeds at the nozzle entrance.

  $u_{z,entrance}=-\frac{4\dot{V}}{\pi D_{entrance}^2}$   ....... (I)

Here, the volume flow rate is $\dot{V}$, the axial speed at the entrance of the nozzle is $u_{z,entrance}$, and the diameter at the entrance of the nozzle is $D_{entrance}$.

Write the expression for axial speeds at the nozzle exit.

  $u_{z,exit}=-\frac{4\dot{V}}{\pi D_{exit}^2}$   ....... (II)

Here, the axial speed at the exit of the nozzle is $u_{z,exit}$, and the diameter at the exit of the nozzle is $D_{exit}$.

Calculation:

Substitute $0.50\text{in}$ for $D_{entrance}$ and $2\text{gal}/\text{min}$ for $\dot{V}$ in Equation (I).

  $\begin{array}{c}{u}_{z,entrance}=-\frac{4\left(2\text{gal}/\text{min}\right)}{\pi {\left(0.50\text{in}\right)}^2}\\ =-\frac{8\text{gal}/\text{min}\left(\frac{0.1337{\text{ft}}^3}{\text{gal}}\right)}{0.7853{\text{in}}^2}\\ =-\frac{1.0696{\text{ft}}^3/\min \left(\frac{144{\text{in}}^2}{{\text{ft}}^2}\right)}{0.7853{\text{in}}^2}\\ =-\frac{154.022\text{ft}\cdot {\text{in}}^2/\min \left(\frac{1\text{min}}{60\text{s}}\right)}{0.7853{\text{in}}^2}\end{array}$

  $\begin{array}{c}{u}_{z,entrance}=-\frac{2.56704\text{ft}\cdot {\text{in}}^2/\text{s}}{0.7853{\text{in}}^2}\\ =-\frac{2.56704\text{ft}/\text{s}}{0.7853}\\ =3.2688\text{ft}/\text{s}\end{array}$

Substitute $0.14\text{in}$ for $D_{exit}$ and $2\text{gal}/\text{min}$ for $\dot{V}$ in Equation (II).

  $\begin{array}{c}{u}_{z,exit}=-\frac{4\left(2\text{gal}/\text{min}\right)}{\pi {\left(0.14\text{in}\right)}^2}\\ =-\frac{8\text{gal}/\text{min}\left(\frac{0.1337{\text{ft}}^3}{\text{gal}}\right)}{\mathrm{0.0.0615}{\text{in}}^2}\\ =-\frac{1.0696{\text{ft}}^3/\min \left(\frac{144{\text{in}}^2}{{\text{ft}}^2}\right)}{\mathrm{0.0.0615}{\text{in}}^2}\\ =-\frac{154.022\text{ft}\cdot {\text{in}}^2/\min \left(\frac{1\text{min}}{60\text{s}}\right)}{\mathrm{0.0.0615}{\text{in}}^2}\end{array}$

  $\begin{array}{c}{u}_{z,exit}=-\frac{2.56704\text{ft}\cdot {\text{in}}^2/\text{s}}{0.0615{\text{in}}^2}\\ =-\frac{2.56704\text{ft}/\text{s}}{0.0615}\\ =41.7404\text{ft}/\text{s}\end{array}$

Conclusion:

The axial speeds $\text{ft}/\text{s}$ at the nozzle entrance is $3.2688\text{ft}/\text{s}$.

The axial speeds $\text{ft}/\text{s}$ at the nozzle exist is $41.7404\text{ft}/\text{s}$.

To determine

(b)

The several streamlines in the $rz$ -plane inside the plane and design the appropriate nozzle shape.

Answer to Problem 66EP

The following figure represents the stream lines.

  

Explanation of Solution

Given information:

The diameter at the entrance of the nozzle is $0.50\text{in}$, the diameter at exist of the nozzle is $0.14\text{in}$. The length of the nozzle is $0.20\text{in}$. The volume flow rate through nozzle is $2\text{gal}/\text{min}$.

Write the expression for axial speeds at the nozzle entrance.

  $u_{z,entrance}=-\frac{4\dot{V}}{\pi D_{entrance}^2}$   ....... (I)

Here, the volume flow rate is $\dot{V}$, the axial speed at the entrance of the nozzle is $u_{z,entrance}$, and the diameter at the entrance of the nozzle is $D_{entrance}$.

Write the expression for axial speeds at the nozzle exit.

  $u_{z,exit}=-\frac{4\dot{V}}{\pi D_{exit}^2}$   ....... (II)

Here, the axial speed at the exit of the nozzle is $u_{z,exit}$, and the diameter at the exit of the nozzle is $D_{exit}$.

Write the expression for stream function of the flow filed.

  $\psi =\frac{r^2}{2}\left(u_{z,entrance}+-\frac{u_{z,exit}-u_{z,entrance}}{L}z\right)+C$   ....... (III)

Here, the stream function of the flow field is $\psi$, the radius is $r$, the distance in $z$ direction is $z$ and the length of nozzle is $L$.

Calculation:

Substitute $0.50\text{in}$ for $D_{entrance}$ and $2\text{gal}/\text{min}$ for $\dot{V}$ in Equation (I).

  $\begin{array}{c}{u}_{z,entrance}=-\frac{4\left(2\text{gal}/\text{min}\right)}{\pi {\left(0.50\text{in}\right)}^2}\\ =-\frac{8\text{gal}/\text{min}\left(\frac{0.1337{\text{ft}}^3}{\text{gal}}\right)}{0.7853{\text{in}}^2}\\ =-\frac{1.0696{\text{ft}}^3/\min \left(\frac{144{\text{in}}^2}{{\text{ft}}^2}\right)}{0.7853{\text{in}}^2}\\ =-\frac{154.022\text{ft}\cdot {\text{in}}^2/\min \left(\frac{1\text{min}}{60\text{s}}\right)}{0.7853{\text{in}}^2}\end{array}$

  $\begin{array}{c}{u}_{z,entrance}=-\frac{2.56704\text{ft}\cdot {\text{in}}^2/\text{s}}{0.7853{\text{in}}^2}\\ =-\frac{2.56704\text{ft}/\text{s}}{0.7853}\\ =3.2688\text{ft}/\text{s}\end{array}$

Substitute $0.14\text{in}$ for $D_{exit}$ and $2\text{gal}/\text{min}$ for $\dot{V}$ in Equation (II).

  $\begin{array}{c}{u}_{z,exit}=-\frac{4\left(2\text{gal}/\text{min}\right)}{\pi {\left(0.14\text{in}\right)}^2}\\ =-\frac{8\text{gal}/\text{min}\left(\frac{0.1337{\text{ft}}^3}{\text{gal}}\right)}{\mathrm{0.0.0615}{\text{in}}^2}\\ =-\frac{1.0696{\text{ft}}^3/\min \left(\frac{144{\text{in}}^2}{{\text{ft}}^2}\right)}{\mathrm{0.0.0615}{\text{in}}^2}\\ =-\frac{154.022\text{ft}\cdot {\text{in}}^2/\min \left(\frac{1\text{min}}{60\text{s}}\right)}{\mathrm{0.0.0615}{\text{in}}^2}\end{array}$

  $\begin{array}{c}{u}_{z,exit}=-\frac{2.56704\text{ft}\cdot {\text{in}}^2/\text{s}}{0.0615{\text{in}}^2}\\ =-\frac{2.56704\text{ft}/\text{s}}{0.0615}\\ =41.7404\text{ft}/\text{s}\end{array}$

Substitute $41.7404\text{ft}/\text{s}$ for $u_{z,exit}$, $3.2688\text{ft}/\text{s}$ for $u_{z,entrance}$, $0.20\text{in}$ for $L$ and $0$ for $C$ in Equation (III).

  $\begin{array}{c}\psi =\frac{r^2}{2}\left(3.2688\text{ft}/\text{s}+\left(-\frac{41.7404\text{ft}/\text{s}-3.2688\text{ft}/\text{s}}{0.20\text{in}}\right)z\right)\\ \psi =\frac{r^2}{2}\left(3.2688\text{ft}/\text{s}-\frac{41.7404\text{ft}/\text{s}-3.2688\text{ft}/\text{s}}{0.20\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)}z\right)\\ \psi =\frac{r^2}{2}\left(3.2688\text{ft}/\text{s}-230.514z{\text{s}}^{-1}\right)\\ r=\mp \sqrt{\frac{2\psi }{\left(3.2688\text{ft}/\text{s}-230.514z{\text{s}}^{-1}\right)}}\end{array}$   ...... (IV)

Substitute $\text{0}$ for $z$ and $\text{0}\text{.25}\text{in}$ for $r$ in Equation (IV).

  $\begin{array}{c}\text{0}\text{.25}\text{in}=\mp \sqrt{\frac{2\psi }{\left(3.2688\text{ft}/\text{s}-230.514\left(0\right){\text{s}}^{-1}\right)}}\\ \text{0}\text{.25}\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)=\mp \sqrt{\frac{2\psi }{3.2688\text{ft}/\text{s}}}\\ \psi =\frac{{\left(0.0208\text{ft}\right)}^2}{2}\left(3.2688\text{ft}/\text{s}\right)\\ \psi =0.000707{\text{ft}}^3/\text{s}\end{array}$

Substitute $\text{0}$ for $z$ and $\text{0}\text{.07}\text{in}$ for $r$ in Equation (IV).

  $\begin{array}{c}\text{0}\text{.07}\text{in}=\mp \sqrt{\frac{2\psi }{\left(3.2688\text{ft}/\text{s}-230.514\left(0\right){\text{s}}^{-1}\right)}}\\ \text{0}\text{.07}\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)=\mp \sqrt{\frac{2\psi }{3.2688\text{ft}/\text{s}}}\\ \psi =\frac{{\left(0.0058334\text{ft}\right)}^2}{2}\left(3.2688\text{ft}/\text{s}\right)\\ \psi =0.0000463{\text{ft}}^3/\text{s}\end{array}$

The different values of stream function and radius are shown below the table.

$\psi \left({\text{ft}}^3/\text{s}\right)$	$z$	$r\left(\text{in}\right)$	$r^{\prime }$
$0.00071$	$0$	$\text{0}\text{.25}$	$-\text{0}\text{.25}$
$0.00071$	$0.02$	$0.0014$	$-0.0014$
$0.0000463{\text{ft}}^3/\text{s}$	$0$	$0.07$	$-0.01796$
$0.0000463{\text{ft}}^3/\text{s}$	$0.02$	$0.011377$	$-0.011377$

The following figure represents the stream line with varying of length.

  

  Figure-(1)

For design, the shape of the nozzle is like the graph of the boundary layers.

Conclusion:

The following figure represents the stream lines.