A lawn sprinkler discharges water upward and outward from the horizontal plane, as shown in Fig. 5.20. Initially, the angular speed of the sprinkler is zero. The total volumetric rate of discharges through the sprinkler is Q. If the area of cross section of the sprinkler arms and the discharging

Question

A lawn sprinkler discharges water upward and outward from the horizontal plane, as shown in Fig. 5.20. Initially, the angular speed of the sprinkler is zero. The total volumetric rate of discharges through the sprinkler is Q. If the area of cross section of the sprinkler arms and the discharging nozzles is A, derive an expression for the angular velocity, [latex] \omega[/latex] , of the sprinkler as a function of time.

Accepted Answer

Approach 1: Moving Reference Frame

Angular momentum conservation equation for a moving reference frame attached with the rotating sprinkler, as shown in Fig. 5.20(a), results in

[latex] \underbrace{\sum_{} \vec{M}-\underbrace{\int_{C V}\left(\vec{r} imes \vec{a}_{r e l} ight) d m}_{ ext {Term } 2}=}_{Term 1} \underbrace{\frac{\partial}{\partial t} \int_{C V} ho\left(\vec{r} imes \vec{V}_{x y z} ight) d \forall}_{ ext {Term } 3}+\underbrace{\int_{C S} ho\left(\vec{r} imes \vec{V}_{x y z} ight)\left(\vec{V}_{x y z} . \hat{n} ight) d A}_{ ext {Term } 4} [/latex] (5.52)

Term 1 [latex] =\sum \vec{M}=0[/latex] (since the sprinkler is freely rotating about its pivot)

For the term 2, we note that

[latex] \vec{a}_{r e l}=\vec{a}_{C V}+2 \vec{\omega} imes \vec{V}_{x y z}+\dot{\vec{\omega}} imes \vec{r}+\vec{\omega} imes(\vec{\omega} imes \vec{r})[/latex]

Or [latex] \vec{a}_{r e l}=0+2 \omega \hat{k} imes V_{x y z} \hat{i}+\dot{\omega} \hat{k} imes r \hat{i}+\omega \hat{k} imes(\omega \hat{k} imes r \hat{i})[/latex]

Or [latex] \vec{a}_{r e l}=2 \omega V_{x y z} \hat{j}+\dot{\omega} r \hat{j}-\omega^{2} r \hat{i}[/latex]

[latex] \vec{r} imes \vec{a}_{r e l}=r \hat{i} imes\left[2 \omega V_{x y z} \hat{j}+\dot{\omega} r \hat{j}-\omega^{2} \hat{i} ight]=\left(2 \omega V_{x y z}+\dot{\omega} r ight) \hat{k}[/latex]

Thus, Term 2 [latex] =\int_{C V}\left(\vec{r} imes \vec{a}_{r e l} ight) d m=2 \int_{0}^{R}\left(2 \omega V_{x y z}+\dot{\omega} r ight) ho A d r \hat{k}[/latex]

[latex] =2 ho A\left(\omega V_{x y z} R^{2}+\dot{\omega} \frac{R^{3}}{3} ight) \hat{k}[/latex]

Term 3 [latex] =\frac{\partial}{\partial t} \int_{C V} ho\left(\vec{r} imes \vec{V}_{x y z} ight) d \forall=\overline{0}[/latex] (Since [latex] \vec{V}_{x y z}[/latex] is time invariant)

Term 4 [latex] =\int_{C S} ho\left(\vec{r} imes \vec{V}_{x y z} ight)\left(\vec{V}_{x y z} . \hat{n} ight) d A[/latex]

[latex] =2 imes ho R \hat{i} imes V_{x y z} \hat{j} \frac{Q}{2}[/latex]

(Neglecting small length of the bent portion)

[latex] = ho R V_{x y z} Q \hat{k}[/latex]

Note here that multiplier 2 in the equation preceding the above is indicative of the fact that there are two outflow boundaries with a flow rate of [latex] \frac{Q}{2}[/latex] through each boundary.

Substituting Terms 1 through 4 into Eq. (5.52), we obtain

[latex] -2 ho A\left(\omega V_{x y z} R^{2}+\dot{\omega} \frac{R^{3}}{3} ight)= ho R V_{x y z} Q[/latex]

Where [latex] V_{x y z}=\frac{Q}{2 A}[/latex] (Relative speed of flow through each limb)

[latex] -2 ho A\left(\omega \frac{Q}{2 A} R^{2}+\frac{d \omega}{d t} \frac{R^{3}}{3} ight)= ho R \frac{Q}{2 A} Q[/latex] (5.53)

Equation (5.53) may be cast in the form,

[latex] \frac{d \omega}{d t}=a-b \omega[/latex] (5.53a)

Where [latex] a=-\frac{3 Q^{2}}{4 A R^{2}}, b=\frac{3 Q}{2 A R}[/latex]

[latex] \int_{0}^{\omega} \frac{d \omega}{a-b \omega}=\int_{0}^{t} d t[/latex]

Integrating and rearranging the terms, we get

[latex] \omega=\frac{a}{b}\left[1-\frac{1}{\exp (b t)} ight][/latex] (5.54)

Approach 2: Stationary Reference Frame

Writing the angular momentum conservation equation relative to the stationary reference frame X Y Z, as shown in Fig. 5.20(b), results in

[latex] \underbrace{\sum_{} \vec{M}=\underbrace{\frac{\partial}{\partial t} \int_{C V} ho(\vec{r} imes \vec{V}) d \forall}_{ ext {Term } 2}}_{Term 1}+\underbrace{\int_{C S} ho(\vec{r} imes \vec{V})\left(\vec{V}_{r} . \hat{n} ight) d A}_{ ext {Term } 3}[/latex] (5.55)

Term 1:

[latex] \sum \vec{M}=0[/latex]

For Term 2, we note that:

[latex] \vec{V}=\vec{V}_{ ext {sprinkler }}+\vec{V}_{r}[/latex]

Where [latex] \vec{V}_{r}[/latex] is the velocity of water relative to sprinkler.

Thus, considering the right arm (horizontal portion of the sprinkler),

[latex] \vec{V}=(\omega \hat{k} imes r \hat{i})+V_{r} \hat{i}=\omega r \hat{j}+V_{r} \hat{i}[/latex]

[latex] \vec{r} imes \vec{V}=r \hat{i} imes\left(\omega r \hat{j}+V_{r} \hat{i} ight)=\omega r^{2} \hat{k}[/latex]

Term 2 [latex] =\frac{\partial}{\partial t} \int_{C V} ho(\vec{r} imes \vec{V}) d \forall=2 \frac{\partial}{\partial t} \int_{0}^{R} ho(\vec{r} imes \vec{V}) d \forall[/latex]

[latex] =2 \frac{\partial}{\partial t} \int_{0}^{R} ho \omega r^{2} \hat{k} A d r=2 ho A \frac{R^{3}}{3} \frac{d \omega}{d t} \hat{k}[/latex]

Term 3 [latex] =\int_{C S} ho(\vec{r} imes \vec{V})\left(\vec{V}_{r} . \hat{n} ight) d A[/latex]

[latex] =2\left[ ho R \hat{i} imes\left\{\omega R \hat{j}+V_{r} \hat{j} ight\} \frac{Q}{2} ight][/latex]

[latex] =\left(2 ho \omega R^{2} Q+ ho R V_{r} Q ight) \hat{k}[/latex]

Substituting terms 1 through 3 into Eq. (5.55), we obtain

[latex] 0=2 ho A \frac{R^{3}}{3} \frac{d \omega}{d t}+2 ho \omega R^{2} Q+ ho R V_{r} Q[/latex]

The solution of this differential equation may be obtained in a manner similar to the approach adapted in the context of moving-reference-frame-based method adapted for this problem.

Question 5.15: A lawn sprinkler discharges water upward and outward from th...

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