Three-Arm Lawn Sprinkler: 22-Agric-A4 Fluid Flow, December 2013, Q3

Question 2: 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.

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 $\theta$ measured 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
Water	20°C, negligible collar friction

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

Figure 2 — 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 until the net torque is zero, i.e. until the absolute tangential velocity of the exiting jet is zero. Get the nozzle's relative exit speed from continuity per arm, 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 $a=\tfrac{\pi}{4}d^2$: $$q_{\text{arm}} = \frac{Q}{3} = \frac{7.5\times10^{-4}}{3} = 2.5\times10^{-4}\ \text{m}^3/\text{s}, \qquad a = \frac{\pi}{4}(0.007)^2 = 3.849\times10^{-5}\ \text{m}^2,$$ $$V_{\text{rel}} = \frac{q_{\text{arm}}}{a} = \frac{2.5\times10^{-4}}{3.849\times10^{-5}} = \boxed{V_{\text{rel}} \approx 6.50\ \text{m/s}}.$$
Torque-free (free-wheeling) condition. The angular-momentum flux gives a driving torque $T=\rho Q R\big(V_{\text{rel}}\cos\theta-\omega R\big)$ (per White's rotating-sprinkler control volume). With zero collar friction, steady state requires $T=0$, i.e. the jet's absolute tangential velocity vanishes: $$V_{\text{rel}}\cos\theta = \omega R \;\Rightarrow\; \omega = \frac{V_{\text{rel}}\cos\theta}{R}.$$
(a) $\theta=0^\circ$. $$\omega = \frac{6.50(\cos 0^\circ)}{0.15} = \frac{6.50}{0.15} = 43.31\ \text{rad/s} \;\Rightarrow\; \boxed{n = \omega\cdot\frac{60}{2\pi} \approx 413.6\ \text{rev/min}}.$$
(b) $\theta=40^\circ$. $$\omega = \frac{6.50(\cos 40^\circ)}{0.15} = \frac{6.50(0.766)}{0.15} = 33.17\ \text{rad/s} \;\Rightarrow\; \boxed{n \approx 316.8\ \text{rev/min}}.$$ Angling the jets toward the tangent ($\theta$ up from $0^\circ$) 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 (Euler turbine-type) 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 not a maximum-power device; it is a zero-torque device, since nothing external absorbs the reaction torque. That happens exactly when the fluid leaves each nozzle with zero tangential velocity in the fixed (lab) frame — any tangential "kick" left in the jet, forward or backward, would still be accelerating or decelerating the arm. Setting the absolute tangential jet speed to zero is therefore the correct steady-state condition, distinct from the (higher-torque, lower-speed) condition that maximizes shaft power in a bearing-loaded sprinkler.

Common pitfall. Using the full nozzle velocity $V_{\text{rel}}$ (rather than its tangential component $V_{\text{rel}}\cos\theta$) in the torque balance once $\theta\neq0$ — only the tangential component does work against rotation; the radial component contributes no torque about the hub.

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

Question 3 of 6: Three-Arm Lawn Sprinkler

Question 2: Three-Arm Lawn Sprinkler (equal value)