Three-Arm Lawn Sprinkler: 22-Agric-A4 Fluid Flow, December 2016, Q3

Problem 3: Three-Arm Lawn Sprinkler (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

The numbers below reproduce that verified solution.

Given. A three-arm sprinkler with frictionless (negligible collar) bearing, total flow $Q=2.7\ \text{m}^3/\text{hr}$ split equally among the three arms, arm radius $R=15\ \text{cm}$, nozzle diameter $d=7\ \text{mm}$, nozzle jet angled at $\theta$ from the tangent to the arm's circular path.

Quantity	Value
Total flow, $Q$	2.7 m³/hr $=7.5\times10^{-4}$ m³/s
Number of arms	3 (equal split)
Arm radius, $R$	0.15 m
Nozzle diameter, $d$	7 mm

Find. The steady (free-wheeling) rotation rate $n$, in rev/min, for (a) $\theta=0^\circ$ and (b) $\theta=40^\circ$.

Figure 3 — Top view of the three-arm sprinkler: each nozzle exits at angle $\theta$ from the tangent to the $R=15$ cm circle; the tangential component of the relative jet drives the rotation.

Approach. With collar friction negligible, the sprinkler spins up to the torque-free steady state, i.e. until the exiting jet's absolute tangential velocity is zero. Get the nozzle's relative exit speed from per-arm continuity, then set the tangential component of that relative velocity equal to the tip speed $\omega R$.

Relative jet speed at each nozzle. Each of the 3 arms carries one-third of the total flow through a nozzle of area $a=\tfrac{\pi}{4}d^2$: $$\begin{aligned} q_{\text{arm}} &= \frac{Q}{3} = 2.5\times10^{-4}\ \text{m}^3/\text{s} \\ a &= \frac{\pi}{4}(0.007)^2 = 3.849\times10^{-5}\ \text{m}^2 \end{aligned}$$ $$V_{\text{rel}} = \frac{q_{\text{arm}}}{a} = \boxed{V_{\text{rel}} \approx 6.50\ \text{m/s}}.$$
Torque-free (free-wheeling) condition. The angular-momentum theorem gives driving torque $T=\rho Q R(V_{\text{rel}}\cos\theta-\omega R)$; zero collar friction requires $T=0$ at steady state, i.e. zero absolute tangential jet velocity: $$\omega = \frac{V_{\text{rel}}\cos\theta}{R}.$$
(a) $\theta=0^\circ$. $$\omega = \frac{6.50(\cos 0^\circ)}{0.15} = 43.31\ \text{rad/s} \;\Rightarrow\; \boxed{n = \omega\cdot\frac{60}{2\pi} \approx 413.6\ \text{rev/min}}.$$
(b) $\theta=40^\circ$. $$\omega = \frac{6.50(\cos 40^\circ)}{0.15} = 33.18\ \text{rad/s} \;\Rightarrow\; \boxed{n \approx 316.8\ \text{rev/min}}.$$ Angling the jets toward the tangent reduces the tangential thrust component and so lowers the free-spin speed.

Quantity	Result
Relative jet velocity, $V_{\text{rel}}$	6.50 m/s
(a) $n$ at $\theta=0^\circ$	413.6 rev/min
(b) $n$ at $\theta=40^\circ$	316.8 rev/min

Concept & Theory

Topic. Angular momentum in a rotating control volume — the reaction-driven lawn-sprinkler problem.

Key relations. Nozzle continuity $V_{\text{rel}}=q_{\text{arm}}/a$; angular-momentum torque $T=\rho Q R(V_{\text{rel}}\cos\theta-\omega R)$; torque-free spin speed $\omega=V_{\text{rel}}\cos\theta/R$.

Why this works. A frictionless sprinkler is a zero-torque device, not a maximum-power one — nothing external absorbs the reaction torque, so steady state occurs exactly when the fluid leaves each nozzle with zero tangential velocity in the fixed (lab) frame.

Common pitfall. Using the full nozzle velocity $V_{\text{rel}}$ rather than its tangential component $V_{\text{rel}}\cos\theta$ once $\theta\neq0$ — only the tangential component does work against rotation.

Source. White, Fluid Mechanics, Ch. 3 (Integral Relations) — Angular-Momentum Theorem, Rotating Sprinkler example.

Question 3 of 4: Three-Arm Lawn Sprinkler

Problem 3: Three-Arm Lawn Sprinkler (equal value)