Chapter 6, Problem 69P

Water enters vertically and steadily at a rate of 10 Ls into the sprinkler shown in Fig. P6-69. Both water jets have a diameter of 1.2 cm. Disregarding any frictional effects, determine (a) the rotational speed of the sprinkler in rpm and (b) the torque required to prevent the sprinkler from rotating.

To determine

(a)

The rotational speed of sprinkler.

Answer to Problem 69P

The rotational speed of sprinkler is $527.71\text{rpm}$.

Explanation of Solution

Given information:

The volume flow rate is $10\text{L}/\text{s}$, diameter of both jets is $1.2\text{cm}$,angle of nozzle with vertical is $60{}^{\text{ο}}$.

Write the expression for mass flow rate entering the sprinkler.

  $\dot{m}=\rho Q$   ....... (I)

Here, density of water is $\rho$,volume flow rate is $Q$.

Write the expression for volume flow rate from each nozzle.

  $\dot{Q}=\frac{Q}{2}$   ....... (II)

Write the expression for jet area.

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

Here, diameter is $d$

Write expression for velocity of water jet,the diameter of both nozzle are equal.

  $V_{\text{jet}}=\frac{\dot{Q}}{A_{\text{jet}}}$........ (IV)

Here, volume flow rate from each jet is $\dot{Q}$.

Write the expression for angular momentum equation for the sprinkler.

  $M=r\dot{m}{V}_{\text{out}}-r\dot{m}{V}_{\text{in}}$   ....... (V)

Here, distance of nozzle from sprinkler centre is $r$,inlet velocity of water to the sprinkler is $V_{\text{in}}$, exit velocity of water from sprinkler is $V_{\text{out}}$.

Substitute, $0$ for ${\displaystyle \sum M}$, $V_{\text{r}}\cos \theta$ for $V_{\text{in}}$ and $-V_{\text{r}}\cos \theta$ for $V_{\text{out}}$.in Equation (V)

  $\begin{array}{l}0=-r\dot{m}{V}_{\text{r}}\cos \theta -r\dot{m}{V}_{\text{r}}\cos \theta \\ r\dot{m}\left(-2V_{\text{r}}\cos \theta \right)=0\\ -2r\dot{m}{V}_{\text{r}}\cos \theta =0\\ V_{\text{r}}=0\end{array}$   ....... (VI)

Here, angle of sprinkler with vertical is $\theta$.

Write the expression for relative velocity of jet with respect to nozzle.

  $V_r={\left(V_{\text{jet}}\right)}_{\text{v}}-V_{{}_{\text{nozzle}}}$   ....... (VII)

Here, vertical component of jet velocity is ${\left(V_{\text{jet}}\right)}_{\text{v}}$, velocity of nozzle is $V_{\text{nozzle}}$

Substitute $V_{\text{jet}}\cos \theta$ for ${\left(V_{\text{jet}}\right)}_{\text{v}}$ in Equation (VII)

  $V_r=V_{\text{jet}}\cos \theta -V_{{}_{\text{nozzle}}}$   ....... (VIII)

Substitute 0 for $V_{\text{r}}$ in Equation (VIII)

  $\begin{array}{l}0=V_{\text{jet}}\cos \theta -V_{{}_{\text{nozzle}}}\\ V_{{}_{\text{nozzle}}}=V_{\text{jet}}\cos \theta \end{array}$   ....... (IX)

Write the expression for angular velocity of sprinkler.

  $\omega =\frac{2V_{\text{nozzle}}}{L}$   ....... (X)

Here, length of rod is $L$.

Write the expression for angular velocity of sprinkler in rpm.

  $N=\frac{60\omega }{2\pi }$   ....... (XI)

Calculation:

Refer to table "Properties of saturated water" to obtain the density of water as $998\text{kg}/{\text{m}}^{\text{3}}$.

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $10\text{L}/\text{s}$ for $Q$ in Equation (I).

  $\begin{array}{c}\dot{m}=\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(10\text{L}/\text{s}\right)\\ =\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(10\text{L}/\text{s}\times \frac{1{\text{m}}^{\text{3}}/\text{s}}{1000\text{L}/\text{s}}\right)\\ =\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.01m^3/\text{s}\right)\\ =9.98\text{kg}/\text{s}\end{array}$

Substitute $10\text{L}/\text{s}$ for $Q$ in Equation (II).

  $\begin{array}{c}\dot{Q}=\frac{10\text{L}/\text{s}}{2}\\ =5\text{L}/\text{s}\times \frac{1{\text{m}}^{\text{3}}/\text{s}}{1000\text{L}/\text{s}}\\ =0.005{\text{m}}^{\text{3}}/\text{s}\end{array}$

Substitute $1.2\text{cm}$ for $d$ in equation (III).

  $\begin{array}{c}{A}_{\text{jet}}=\frac{\pi }{4}{\left(1.2\text{cm}\right)}^2\\ =\frac{\pi }{4}{\left(1.2\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ =\frac{\pi }{4}{\left(0.012\text{m}\right)}^2\\ =0.0001130{\text{m}}^{\text{2}}\end{array}$

Substitute $0.005{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$ and $0.0001130{\text{m}}^{\text{2}}$ for $A_{\text{jet}}$ in Equation (III).

  $\begin{array}{c}{V}_{\text{jet}}=\frac{0.005{\text{m}}^{\text{3}}/\text{s}}{0.0001130{\text{m}}^{\text{2}}}\\ =44.209\text{m}/\text{s}\end{array}$

Substitute $60{}^{\text{ο}}$ for $\theta$ and $44.209\text{m}/\text{s}$ for $V_{\text{jet}}$ in Equation (IX).

  $\begin{array}{c}{V}_{{}_{\text{nozzle}}}=\left(44.209\text{m}/\text{s}\right)\cos 60{}^{\text{ο}}\\ =22.104\text{m}/\text{s}\end{array}$

Substitute $22.104\text{m}/\text{s}$ for $V_{{}_{\text{nozzle}}}$ and $0.8\text{m}$ for $L$ in Equation (X).

  $\begin{array}{c}\omega =\frac{2\times 22.104\text{m}/\text{s}}{0.8\text{m}}\\ =55.262\frac{\text{m}/\text{s}}{\text{m}}\times \text{rad}\\ \text{=55}\text{.262}\text{rad}/\text{s}\end{array}$

Substitute $\text{55}\text{.262}\text{rad}/\text{s}$ for $\omega$ in Equation (XI).

  $\begin{array}{c}N=\frac{60\times \left(55.262\text{rad}/\text{s}\right)}{2\pi }\\ =527.71\text{rad}/\text{s}\times \frac{\text{rpm}}{\text{rad}/\text{s}}\\ =527.71\text{rpm}\end{array}$

Conclusion:

The rotational speed of sprinkler is $527.71\text{rpm}$.

To determine

(b)

The torque required to prevent the rotation of sprinkler.

Answer to Problem 69P

The torque required to prevent the rotation of sprinkler is $88.418\text{N}\cdot \text{m}$.

Explanation of Solution

Given information:

The distance of nozzle from sprinkler centre is $0.4\text{m}$

Write the expression for torque to prevent rotation.

  $T_{\text{shaft}}=r\dot{m}{V}_{\text{jet}}\cos \theta$   ....... (XII)

Here, distance from sprinkler centre to nozzle is $r$

Calculation:

Substitute $0.4\text{m}$ for $r$, $9.98\text{kg}/\text{s}$ for $\dot{m}$, $44.209\text{m}/\text{s}$ for $V_{\text{jet}}$ and $60{}^{\text{ο}}$ for $\theta$ in Equation (XII).

  $\begin{array}{c}{T}_{\text{shaft}}=\left(0.4\text{m}\right)\left(9.98\text{kg}/\text{s}\right)\left(44.209\text{m}/\text{s}\right)\cos 60{}^{\text{ο}}\\ =88.418\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\times \frac{1\text{N}\cdot \text{m}}{\text{1}{\text{kg×m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =88.418\text{N}\cdot \text{m}\end{array}$

Conclusion:

The torque required to prevent the rotation of sprinkler is $88.418\text{N}\cdot \text{m}$.