Chapter 6, Problem 70P

Repeal Prob. 6-69 for the case of unequal anus-the left one being 60 cm and the right one 20 cm from the axis of rotation.

To determine

(a)

The rotational speed of sprinkler.

Answer to Problem 70P

The rotational speed of sprinkler is $264.2\text{rpm}$.

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 $Q$.

The diameter of both nozzle is equal,

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

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

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

Write the expression for velocity in vertical component of each jet.

  ${\left(V_{\text{jet}}\right)}_{\text{v}}=V_{\text{jet}}\cos \theta$   ...... (V)

Here angle of nozzle with vertical is $\theta$

Write the expression for jet left absolute velocity.

  $\begin{array}{c}{\left(V_{\text{jet}}\right)}_{\text{L}}={\left(V_{\text{L}}\right)}_{\text{v}}-{\left(V_{\text{nozzel}}\right)}_{\text{L}}\\ ={\left(V_{\text{L}}\right)}_{\text{v}}-r_L\omega \end{array}$   ...... (VI)

Write the expression for right absolute velocity.

  $\begin{array}{c}{\left(V_{\text{jet}}\right)}_{\text{R}}={\left(V_{\text{R}}\right)}_{\text{v}}-{\left(V_{\text{nozzel}}\right)}_{\text{R}}\\ ={\left(V_{\text{R}}\right)}_{\text{v}}-r_{\text{R}}\omega \end{array}$   ...... (VII)

Here, the distance of left nozzle from rotation point is $r_{\text{L}}$ angular velocity of sprinkler is $\omega$

Write the expression for angular momentum equation for the sprinkler.

  $M=r_{\text{L}}{\dot{m}}_{\text{L}}{\left(V_{\text{jet}}\right)}_{\text{R}}-r_{\text{R}}{\dot{m}}_{\text{R}}{\left(V_{\text{jet}}\right)}_{\text{L}}$   ....... (VIII)

Here, distance of left nozzle from rotation centre is $r_{\text{L}}$, distance of right nozzle from rotation centre is $r_{\text{R}}$, velocity of water jet to the left nozzle is ${\left(V_{\text{jet}}\right)}_{\text{L}}$, velocity of water jet to the left nozzle is ${\left(V_{\text{jet}}\right)}_{\text{R}}$,

Substitute, $0$ for $M$,.in Equation (VIII)

  $r_{\text{L}}{\dot{m}}_{\text{L}}{\left(V_{\text{jet}}\right)}_{\text{L}}-r_{\text{R}}{\dot{m}}_{\text{R}}{\left(V_{\text{jet}}\right)}_R=0$   ....... (IX)

Write the expression for angular velocity of sprinkler in rpm.

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

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).

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

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 WEquation (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 (IV).

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

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

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

Substitute $22.104\text{m}/\text{s}$ for ${\left(V_{\text{L}}\right)}_{\text{v}}$ and $0.6\text{m}$ for $r_{\text{L}}$ in Equation (VI).

  ${\left(V_{\text{jet}}\right)}_{\text{L}}=22.104\text{m}/\text{s}-\omega \times 0.6\text{m}$

Substitute $22.104\text{m}/\text{s}$ for ${\left(V_{\text{R}}\right)}_{\text{v}}$ and $0.2\text{m}$ for $r_{\text{R}}$ in Equation (VII).

  ${\left(V_{\text{jet}}\right)}_{\text{R}}=22.104\text{m}/\text{s}-\omega \times 0.2\text{m}$

Substitute, $5\text{kg}/\text{s}$ for ${\dot{m}}_{\text{L}}$, $5\text{kg}/\text{s}$ for ${\dot{m}}_R$

  $0.6\text{m}$ for $r_{\text{L}}$, $0.2\text{m}$ for $r_{\text{R}}$, $22.104\text{m}/\text{s}-\omega \times 0.6\text{m}$ for ${\left(V_{\text{jet}}\right)}_{\text{L}}$

  $22.104\text{m}/\text{s}-\omega \times 0.2\text{m}$ for ${\left(V_{\text{jet}}\right)}_{\text{R}}$ in Equation (IX).

  $\begin{array}{l}\left(0.6\text{m}\right)\left(5\text{kg}/\text{s}\right)\left(22.104\text{m}/\text{s}-\omega \times 0.6\text{m}\right)-\left(0.2\text{m}\right)\left(5\text{kg}/\text{s}\right)\left(22.104\text{m}/\text{s}-\omega \times 0.2\text{m}\right)=0\\ \left(6\text{m}\right)\left(22.104\text{m}/\text{s}-\omega \times 0.6\text{m}\right)=\left(2\text{m}\right)\left(22.104\text{m}/\text{s}-\omega \times 0.2\text{m}\right)\\ \left(3\right)\left(22.104\text{m}/\text{s}-\omega \times 0.6\text{m}\right)=\left(22.104\text{m}/\text{s}-\omega \times 0.2\text{m}\right)\\ \left(66.312\text{m}/\text{s}-\omega \times 1.8\text{m}\right)=22.104\text{m}/\text{s}-\omega \times 0.2\text{m}\end{array}$

  $\begin{array}{l}\omega \times 1.6\text{m}=66.312\text{m}/\text{s}-22.104\text{m}/\text{s}\\ \omega =27.65\frac{\text{m}/\text{s}}{\text{m}}\times \text{rad}\\ \omega \text{=}27.65\text{rad}/\text{s}\end{array}$

Substitute $27.65\text{rad}/\text{s}$ for $\omega$ in Equation (X).

  $\begin{array}{c}N=\frac{60\times 27.65\text{rad}/\text{s}}{2\pi }\\ =264.2\text{rad}/\text{s}\times \frac{\text{rpm}}{\text{rad}/\text{s}}\\ =264.2\text{rpm}\end{array}$

Conclusion:

The rotational speed of sprinkler is $264.2\text{rpm}$.

To determine

(b)

The net torque on sprinkler.

Answer to Problem 70P

The net torque on sprinkler is $33.2\text{N}\cdot \text{m}$.

Explanation of Solution

Given information:

The distance of left nozzle from rotation centre is $0.6\text{m,}$ the distance of right nozzle from rotation centre is $0.2\text{m}$.

Write the expression for torque due to left nozzle.

  $T_{\text{L}}=r_{\text{L}}{\dot{m}}_{\text{L}}{\left(V_{\text{jet}}\right)}_{\text{L}}$   ....... (XI)

Here, distance from rotation centre to left nozzle is $r_{\text{L}}$

Write the expression for torque due to left nozzle.

  $T_{\text{R}}=r_{\text{R}}{\dot{m}}_{\text{R}}{\left(V_{\text{jet}}\right)}_{\text{R}}$   ....... (XII)

Here, distance from rotation centre to left nozzle is $r_{\text{R}}$

Write expression for net torque.

  $T=T_{\text{L}}+T_{\text{R}}$   ...... (XIII)

Calculation:

Substitute $0.6\text{m}$ for $r_{\text{L}}$, $4.99\text{kg}/\text{s}$ for ${\dot{m}}_{\text{L}}$, $27.65\text{rad}/\text{s}$ for $\omega$, $r_{\text{L}}{\dot{m}}_{\text{L}}\left(22.104\text{m}/\text{s}-\omega \times 0.6\text{m}\right)$ for ${\left(V_{\text{jet}}\right)}_{\text{L}}$ in Equation (XI).

  $\begin{array}{c}{T}_{\text{L}}=\left(0.6\text{m}\right)\left(4.99\text{kg}/\text{s}\right)\left(22.123\text{m}/\text{s}-0.6\text{m}\times \text{27}\text{.65}\text{rad}/\text{s}\right)\\ =\left(0.6\text{m}\right)\left(4.99\text{kg}/\text{s}\right)\left(5.533\text{m}/\text{s}\right)\\ =16.6\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\times \frac{\text{1}\text{N}\cdot \text{m}}{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =16.6\text{1}\text{N}\cdot \text{m}\end{array}$

Substitute $0.2\text{m}$ for $r_{\text{R}}$, $4.99\text{kg}/\text{s}$ for ${\dot{m}}_R$, $27.63\text{rad}/\text{s}$ for $\omega$ and $r_{\text{R}}{\dot{m}}_{\text{R}}\left(22.104\text{m}/\text{s}-\omega \times 0.2\text{m}\right)$ for ${\left(V_{\text{jet}}\right)}_{\text{R}}$ in Equation (XII).

  $\begin{array}{c}{T}_{\text{R}}=\left(0.2\text{m}\right)\left(4.99\text{kg}/\text{s}\right)\left(22.123\text{m}/\text{s}-0.2\text{m}\times \text{27}\text{.63}\text{rad}/\text{s}\right)\\ =\left(0.2\text{m}\right)\left(4.99\text{kg}/\text{s}\right)\left(16.593\text{m}/\text{s}\right)\\ =16.6\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\times \frac{\text{1}\text{N}\cdot \text{m}}{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =16.6\text{1}\text{N}\cdot \text{m}\end{array}$

Substitute $16.6\text{1}\text{N}\cdot \text{m}$ for $T_{\text{R}}$ and $16.6\text{1}\text{N}\cdot \text{m}$ for $T_{\text{L}}$ in Equation (XIII).

  $\begin{array}{c}T=16.6\text{kN}+16.6\text{kN}\\ =33.2\text{kN}\end{array}$

Conclusion:

The torque required to prevent the rotation of sprinkler is $33.2\text{N}\cdot \text{m}$.