Chapter 6, Problem 56P

To determine

(a)

The rotational speed of the sprinkler.

Answer to Problem 56P

The rotational speed of the sprinkler is $22.17\text{rpm}$.

Explanation of Solution

Given information:

The volume flow rate of the water is $35\text{L}/\text{s}$, the discharge area for the smaller jet is $3{\text{cm}}^2$, the distance of the smaller jet from the axis of rotation is $50\text{cm}$, the discharge area for the larger jet is $5{\text{cm}}^2$, the distance of the larger jet from the axis of rotation is $35\text{cm}$.

Write the expression for the velocity at the smaller section of the sprinkler.

  $V_s=\frac{Q_s}{A_s}$   ....... (I)

Here, the cross sectional area at the smaller section of the sprinkler is $A_s$ and the flow rate is $Q_s$.

Write the expression for the velocity at the larger section of the sprinkler.

  $V_L=\frac{Q_L}{A_L}$   ....... (II)

Here, the cross sectional area at the larger section of the sprinkler is $A_L$ and the flow rate is $Q_L$.

Write the expression for the velocity of the jet with respect to smaller jet.

  $V_1=r_1\omega$   ....... (III)

Here, the distance of the smaller jet from the axis of rotation is $r_1$.

Write the expression for the velocity of the jet with respect to larger jet.

  $V_2=r_2\omega$   ....... (IV)

Here, the distance of the larger jet from the axis of rotation is $r_2$.

Write the expression for the absolute velocity of the water with respect to larger section of sprinkler.

  $V_{j2}=V_L-V_2$   ....... (V)

Write the expression for the absolute velocity of the water with respect to smaller section of sprinkler.

  $V_{j1}=V_s-V_1$   ....... (VI)

Write the expression for the mass flow rate of the water at larger section of sprinkler.

  ${\dot{m}}_L=\rho A_L{V}_L$   ....... (VII)

Here, the density of the water is $\rho$, the cross sectional area at the larger section is $A_L,$ and the velocity at the larger section is $V_L$.

Write the expression for the mass flow rate of the water at smaller section of sprinkler.

  ${\dot{m}}_s=\rho A_s{V}_s$   ....... (VIII)

Here, the cross sectional area at the smaller section is $A_L$ and the velocity at the smaller section is $V_L$.

Write the expression for the mass flow rate of the water.

  $\dot{m}=\rho Q$   ....... (IX)

Here, the volume flow rate of the water is $Q$.

Write the expression for the total mass flow rate.

  $\dot{m}={\dot{m}}_s+{\dot{m}}_L$   ....... (X)

Write the expression for the discharge for the smaller section.

  $Q_s=\frac{Q}{2}$   ....... (XI)

Write the expression for the discharge for the larger section.

  $Q_L=\frac{Q}{2}$   ....... (XII)

Write the expression for the angular speed using the angular momentum equation.

  $r_2{\dot{m}}_L\left(V_{j2}\right)-r_1{\dot{m}}_s\left(V_{j1}\right)=0$   ....... (XIII)

Write the expression for the rotation speed of the sprinkler.

  $n=\frac{\omega \times 60}{2\pi }$   ....... (XIV)

Calculation:

Substitute $3{\text{cm}}^2$ for $A_s$ in Equation (I).

  $\begin{array}{c}{V}_s=\frac{Q_s}{3{\text{cm}}^2}\\ =\frac{Q_s}{3{\text{cm}}^2\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)}\\ =3333.33{\text{m}}^{-2}{Q}_s\end{array}$

Substitute $5{\text{cm}}^2$ for $A_s$ in Equation (II).

  $\begin{array}{c}{V}_L=\frac{Q_L}{5{\text{cm}}^2}\\ =\frac{Q_L}{5{\text{cm}}^2\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)}\\ =2000{\text{m}}^{-2}{Q}_L\end{array}$

Substitute $50\text{cm}$ for $r_1$ in Equation (III).

  $\begin{array}{c}{V}_1=\left(50\text{cm}\right)\omega \\ =\left(50\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\omega \\ =\left(0.50\text{m}\right)\omega \end{array}$

Substitute $35\text{cm}$ for $r_2$ in Equation (IV).

  $\begin{array}{c}{V}_2=\left(35\text{cm}\right)\omega \\ =\left(35\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\omega \\ =\left(0.35\text{m}\right)\omega \end{array}$

Substitute $\left(0.35\text{m}\right)\omega$ for $V_2$ and $2000{\text{m}}^{-2}{Q}_L$ for $V_L$ in Equation (V).

  $V_{j2}=\left(2000{\text{m}}^{-2}{Q}_L\right)-\left(\left(0.35\text{m}\right)\omega \right)$   ....... (XV)

Substitute $\left(0.5\text{m}\right)\omega$ for $V_1$ and $3333.33{\text{m}}^{-2}{Q}_s$ for $V_s$ in Equation (VI).

  $V_{j1}=\left(3333.33{\text{m}}^{-2}{Q}_s\right)-\left(\left(0.5\text{m}\right)\omega \right)$   ....... (XVI)

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $5{\text{cm}}^2$ for $A_L$ and $2000Q_L$ for $V_L$ in Equation (VII).

  $\begin{array}{c}{\dot{m}}_L=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(5{\text{cm}}^2\right)\left(2000{\text{m}}^{-2}{Q}_L\right)\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(5{\text{cm}}^2\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)\right)\left(2000{\text{m}}^{-2}{Q}_L\right)\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.0005{\text{m}}^2\right)\left(2000{\text{m}}^{-2}{Q}_L\right)\\ =1000\text{kg}/{\text{m}}^3{Q}_L\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $3{\text{cm}}^2$ for $A_s$ and $3333.33Q_s$ for $V_s$ in Equation (VIII).

  $\begin{array}{c}{\dot{m}}_s=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(3{\text{cm}}^2\right)\left(3333.33{\text{m}}^{-2}{Q}_s\right)\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(3{\text{cm}}^2\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)\right)\left(3333.33{\text{m}}^{-2}{Q}_s\right)\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.0003{\text{m}}^2\right)\left(3333.33{\text{m}}^{-2}{Q}_s\right)\\ \approx 1000\text{kg}/{\text{m}}^3{Q}_s\end{array}$

Substitute $35\text{L}/\text{s}$ for $Q$ and $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in Equation (IX).

  $\begin{array}{c}\dot{m}=\left(35\text{L}/\text{s}\right)\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\\ =\left(35\text{L}/\text{s}\right)\left(\frac{{10}^{-3}{\text{m}}^{\text{3}}/\text{s}}{1\text{L}/\text{s}}\right)\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\\ =35\text{kg}/\text{s}\end{array}$

Substitute $50\text{kg}/\text{s}$ for $\dot{m}$, $1000\text{kg}/\text{m }{Q}_s$ for ${\dot{m}}_s$ and $1000\text{kg}/\text{m }{Q}_L$ for ${\dot{m}}_L$ in Equation (X).

  $50\text{kg}/\text{s}=1000\text{kg}/\text{m }{Q}_s+1000\text{kg}/\text{m }{Q}_L$

Substitute $35\text{L}/\text{s}$ for $Q$ in Equation (XI).

  $\begin{array}{c}{Q}_s=\frac{35\text{L}/\text{s}}{2}\\ =\frac{35\text{L}/\text{s}}{2}\left(\frac{{10}^{-3}{\text{m}}^{\text{3}}/\text{s}}{1\text{L}/\text{s}}\right)\\ =17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\end{array}$

Substitute $35\text{L}/\text{s}$ for $Q$ in Equation (XII).

  $\begin{array}{c}{Q}_L=\frac{35\text{L}/\text{s}}{2}\\ =\frac{37\text{L}/\text{s}}{2}\left(\frac{{10}^{-3}{\text{m}}^{\text{3}}/\text{s}}{1\text{L}/\text{s}}\right)\\ =17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\end{array}$

Substitute $17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$ for $Q_s$, $17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$ for $Q_L$, $35\text{cm}$ for $r_2$, $1000\text{kg}/\text{m }{Q}_s$ for ${\dot{m}}_s$ and $1000\text{kg}/\text{m }{Q}_L$ for ${\dot{m}}_L$, $50\text{cm}$ for $r_1$, $\left(2000Q_L-0.35\omega \right)$ for $V_{j2}$ and $\left(3333.33Q_L-0.5\omega \right)$ for $V_{j1}$ in Equation (XIII).

  $\begin{array}{l}\left[\begin{array}{l}\left(35\text{cm}\right)\left(1000\text{kg}/\text{m }\right)\left(17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)\left(\left(\begin{array}{l}2000\left(17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)-\\ 0.35\omega \end{array}\right)\right)-\\ \left(50\text{cm}\right)\left(1000\text{kg}/\text{m }\left(17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)\right)\left(\begin{array}{l}3333.33\left(17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)\\ -0.5\omega \end{array}\right)\end{array}\right]=0\\ \left[\begin{array}{l}250\omega -166666500\text{kg}\cdot \text{cm}/\text{m }\left(\frac{1\text{m}}{100\text{cm}}\right){\left(17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)}^2-\\ 122.5\omega +70000000\text{kg}\cdot \text{cm}/\text{m }\left(\frac{1\text{m}}{100\text{cm}}\right){\left(17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)}^2\end{array}\right]=0\\ 127.5\omega =1666665{\left(17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)}^2-700000{\left(17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)}^2\\ \omega =2.32\text{rad}/\text{s}\end{array}$

Substitute $2.32\text{rad}/\text{s}$ for $\omega$ in Equation (XIV).

  $\begin{array}{c}n=\frac{\left(2.32\text{rad}/\text{s}\right)\times 60}{2\pi }\\ =\frac{139.31\text{rad}/\text{s}}{2\pi }\\ =22.17\text{rpm}\end{array}$

Conclusion:

The rotational speed of the sprinkler is $22.17\text{rpm}$.

To determine

(b)

The total torque required to prevent the sprinkler.

Answer to Problem 56P

The total torque required to prevent the sprinkler is $709.58\text{N}\cdot \text{m}$.

Explanation of Solution

Write the expression for the torque on the sprinkler due to the larger jet.

  $T_2=r_2{\dot{m}}_L\left(V_{j2}\right)$   ....... (XVII)

Write the expression for the torque on the sprinkler due to the smaller jet.

  $T_1=r_1{\dot{m}}_s\left(V_{j1}\right)$   ....... (XVIII)

Write the expression for the total torque.

  $T=T_1+T_2$   ....... (XIX)

Write the expression for the mass flow rate for the smaller section.

  ${\dot{m}}_s=1000\text{kg}/{\text{m}}^3{Q}_s$   ....... (XX)

Write the expression for the mass flow rate for the larger section.

  ${\dot{m}}_L=1000\text{kg}/{\text{m}}^3{Q}_L$   ....... (XXI)

Calculation:

Substitute $2.32\text{rad}/\text{s}$ for $\omega$ and $17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$ for $Q_L$ in Equation (XV).

  $\begin{array}{c}{V}_{j2}=\left(2000{\text{m}}^2\left(17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)\right)-\left(\left(0.35\text{m}\right)\left(2.32\text{rad}/\text{s}\right)\right)\\ =\left(35{\text{m}}^2\left({\text{m}}^{\text{3}}/\text{s}\right)\right)-\left(\left(\text{m}\right)\left(0.812\text{rad}/\text{s}\right)\right)\\ =34.18\text{m}/\text{s}\end{array}$

Substitute $2.32\text{rad}/\text{s}$ for $\omega$ and $17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$ for $Q_s$ in Equation (XVI).

  $\begin{array}{c}{V}_{j1}=\left(3333.33{\text{m}}^2\left(17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)\right)-\left(\left(0.5\text{m}\right)\left(2.32\text{rad}/\text{s}\right)\right)\\ =\left(58.33{\text{m}}^2\left({\text{m}}^{\text{3}}/\text{s}\right)\right)-\left(\left(1.16\text{rad}/\text{s}\right)\right)\\ =57.17\text{m}/\text{s}\end{array}$

Substitute $17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$ for $Q_s$ in Equation (XX).

  $\begin{array}{c}{\dot{m}}_s=1000\text{kg}/{\text{m}}^3\left(17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)\\ =17.5\text{kg}/\text{s}\end{array}$

Substitute $17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}$ for $Q_s$ in Equation (XXI).

  $\begin{array}{c}{\dot{m}}_L=1000\text{kg}/{\text{m}}^3\left(17.5\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)\\ =17.5\text{kg}/\text{s}\end{array}$

Substitute $35\text{cm}$ for $r_2$, $17.5\text{kg}/\text{s}$ for ${\dot{m}}_L$ and $34.18\text{m}/\text{s}$ for $V_{j2}$ in Equation (XVII).

  $\begin{array}{c}{T}_2=\left(35\text{cm}\right)\left(17.5\text{kg}/\text{s}\right)\left(34.18\text{m}/\text{s}\right)\\ =\left(35\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\left(17.5\text{kg}/\text{s}\right)\left(34.18\text{m}/\text{s}\right)\\ =209.35\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\left(\frac{\text{1}\text{N}\cdot \text{m}}{\text{1}\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =209.35\text{N}\cdot \text{m}\end{array}$

Substitute $50\text{cm}$ for $r_1$, $17.5\text{kg}/\text{s}$ for ${\dot{m}}_s$ and $57.17\text{m}/\text{s}$ for $V_{j1}$ in Equation (XVIII).

  $\begin{array}{c}{T}_1=\left(50\text{cm}\right)\left(17.5\text{kg}/\text{s}\right)\left(57.17\text{m}/\text{s}\right)\\ =\left(50\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\left(17.5\text{kg}/\text{s}\right)\left(57.17\text{m}/\text{s}\right)\\ =500.23\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\left(\frac{\text{1}\text{N}\cdot \text{m}}{\text{1}\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =500.23\text{N}\cdot \text{m}\end{array}$

Substitute $500.23\text{N}\cdot \text{m}$ for $T_1$ and $209.35\text{N}\cdot \text{m}$ for $T_2$ in Equation (XIX).

  $\begin{array}{c}T=500.23\text{N}\cdot \text{m}+209.35\text{N}\cdot \text{m}\\ =709.58\text{N}\cdot \text{m}\end{array}$

Conclusion:

The total torque required to prevent the sprinkler is $709.58\text{N}\cdot \text{m}$.