Chapter 6, Problem 89P

Water enters a two-armed lawn sprinkler along the vertical axis at a rate of 75 L s. and leaves the sprinkler nozzles as 2-cm diameter jets at an angle of $\theta$front the tangential direction, as shown in Fig. P6-89. The length of cacti sprinkler arm is 052 in. Disregarding any frictional effects, determine the rate of rotation n of the sprinkler in rev/min for (a) $\theta =0°$. (b) $\theta =30°$. and (c) $\theta =60°$ .

To determine

(a)

The rate of rotation of sprinkler.

Answer to Problem 89P

The rate of rotation of the sprinkler is $2192\text{rpm}$.

Explanation of Solution

Given information:

Length of sprinkler arm is $0.52\text{m}$, the vane angle is $0°$ and the flow rate is $75\text{L}/\sec$.

Write the expression for the area of outlet of jet.

  $A=\frac{\pi }{4}{D}^2$  ...... (I)

Here, the diameter of jet is $D$

Write the expression for outlet velocity.

  $V=\frac{\dot{V}}{A}$  ...... (II)

Here, the volume flow rate is $\dot{V}$ and the area of jet is $A$.

Write the expression for the radial velocity

  $V_r=V_{jet}-V_{nozzle}$  ...... (III)

Here, the jet velocity is $V_{jet}$ and nozzle velocity is $V_{nozzle}$

Substitute $0$ for $V_r$ in Equation (III).

  $\begin{array}{l}0=V_{jet}-V_{nozzle}\\ V_{jet}=V_{nozzle}\end{array}$

Write the expression for jet velocity.

  $V_{jet}=V_t\cos \theta$

Write the expression for the angular momentum about axis of rotation.

  $\omega =\frac{V_{nozzle}}{r}$  ...... (IV)

Here, the length of arm of sprinkler is $r$ and the velocity of water through nozzle is $V_{nozzle}$.

Write the expression for rate of rotation of impellor.

  $\dot{n}=\frac{\omega }{2\pi }$....... (V)

Here, the angular velocity is $\omega$.

Substitute $2\text{cm}$ for $D$ in Equation (I).

  $\begin{array}{c}A=\frac{\pi }{4}\times {\left(2\text{cm}\right)}^2\\ =0.785\times {\left(2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\\ =0.000314{\text{m}}^2\end{array}$

Substitute $75\text{L}/\sec$ for $\dot{V}$ and $0.000314{\text{m}}^2$ for $A$ in Equation (II).

  $\begin{array}{c}V=\frac{75\text{L}/\sec }{2\times 0.000314{\text{m}}^2}\\ =119366\text{L}/\sec \left(\frac{0.001{\text{m}}^3/\sec }{1\text{L}/\sec }\right)\\ =119.366{\text{m}}^3/\sec \end{array}$

Substitute $0°$ for $\theta$, and $119.366\text{m}/\sec$ for $V_t$ in Equation (III).

  $\begin{array}{c}{V}_{nozzle}=119.366\text{m}/\sec \times \cos 0°\\ =119.366\text{m}\cdot \text{rad}/\sec \end{array}$

Substitute $119.366\text{m}\cdot \text{rad}/\sec$ for $V_{nozzle}$ and $0.52\text{m}$ for $r$ in Equation (IV).

  $\begin{array}{c}\omega =\frac{119.366\text{m}\cdot \text{rad}/\sec }{0.52\text{m}}\\ =229.55\text{rad}/\sec \end{array}$

Substitute $229.55\text{rad}/\sec$ for $\omega$ in Equation (V).

  $\begin{array}{c}\dot{n}=\frac{229.55\text{rad}/\sec }{2\pi }\left(\frac{60\sec }{1\min }\right)\\ =2192\text{rpm}\end{array}$

Conclusion:

The rate of rotation of the sprinkler is $2192\text{rpm}$.

To determine

(b)

The rate of rotation of sprinkler.

Answer to Problem 89P

The rate of rotation of the sprinkler is $1898\text{rpm}$.

Explanation of Solution

Given information:

The vane angle is $30°$.

Substitute $2\text{cm}$ for $D$ in Equation (I).

  $\begin{array}{c}A=\frac{\pi }{4}\times {\left(2\text{cm}\right)}^2\\ =0.785\times {\left(2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\\ =0.000314{\text{m}}^2\end{array}$

Substitute $75\text{L}/\sec$ for $\dot{V}$ and $0.000314{\text{m}}^2$ for $A$ in Equation (II).

  $\begin{array}{c}V=\frac{75\text{L}/\sec }{2\times 0.000314{\text{m}}^2}\\ =119366\text{L}/\sec \left(\frac{0.001{\text{m}}^3/\sec }{1\text{L}/\sec }\right)\\ =119.366{\text{m}}^3/\sec \end{array}$

Substitute $30°$ for $\theta$, and $119.366\text{m}/\sec$ for $V_t$ in Equation (III).

  $\begin{array}{c}{V}_{nozzle}=119.366\text{m}/\sec \times \cos 30°\\ =103.368\text{m}\cdot \text{rad}/\sec \end{array}$

Substitute $119.366\text{m}\cdot \text{rad}/\sec$ for $V_{nozzle}$ and $0.52\text{m}$ for $r$ in Equation (IV).

  $\begin{array}{c}\omega =\frac{103.368\text{m}\cdot \text{rad}/\sec }{0.52\text{m}}\\ =198.786\text{rad}/\sec \end{array}$

Substitute $198.786\text{rad}/\sec$ for $\omega$ in Equation (V).

  $\begin{array}{c}\dot{n}=\frac{198.786\text{rad}/\sec }{2\pi }\\ =\frac{198.786\text{rad}/\sec }{2\pi }\left(\frac{60\sec }{1\min }\right)\\ =1898\text{rpm}\end{array}$

Conclusion:

The rate of rotation of the sprinkler is $1898\text{rpm}$.

To determine

(c)

The rate of rotation of sprinkler.

Answer to Problem 89P

The rate of rotation of the sprinkler is $1096\text{rpm}$.

Explanation of Solution

Given information:

The vane angle is $60°$.

Substitute $2\text{cm}$ for $D$ in Equation (I).

  $\begin{array}{c}A=\frac{\pi }{4}\times {\left(2\text{cm}\right)}^2\\ =0.785\times {\left(2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\\ =0.000314{\text{m}}^2\end{array}$

Substitute $75\text{L}/\sec$ for $\dot{V}$ and $0.000314{\text{m}}^2$ for $A$ in Equation (II).

  $\begin{array}{c}V=\frac{75\text{L}/\sec }{2\times 0.000314{\text{m}}^2}\\ =119366\text{L}/\sec \left(\frac{0.001{\text{m}}^3/\sec }{1\text{L}/\sec }\right)\\ =119.366{\text{m}}^3/\sec \end{array}$

Substitute $60°$ for $\theta$, and $119.366\text{m}/\sec$ for $V_t$ in Equation (III).

  $\begin{array}{c}{V}_{nozzle}=119.366\text{m}/\sec \times \cos 60°\\ =59.683\text{m}\cdot \text{rad}/\sec \end{array}$

Substitute $59.683\text{m}\cdot \text{rad}/\sec$ for $V_{nozzle}$ and $0.52\text{m}$ for $r$ in Equation (IV).

  $\begin{array}{c}\omega =\frac{59.683\text{m}\cdot \text{rad}/\sec }{0.52\text{m}}\\ =114.775\text{rad}/\sec \end{array}$

Substitute $114.775\text{rad}/\sec$ for $\omega$ in Equation (V).

  $\begin{array}{c}\dot{n}=\frac{114.775\text{rad}/\sec }{2\pi }\\ =\frac{114.775\text{rad}/\sec }{2\pi }\left(\frac{60\sec }{1\min }\right)\\ =1096\text{rpm}\end{array}$

Conclusion:

The rate of rotation of the sprinkler is $1096\text{rpm}$.