Torque to hold two-armed sprinkler

Given: total flow $Q = 10$ L/s $= 0.01$ m³/s entering vertically at the axis; two equal nozzles of $d = 12$ mm; $r_1 = 0.6$ m, $r_2 = 0.2$ m; the jets leave horizontally at $\theta = 60^\circ$ to the tangential direction (assumed from the figure). The arm is held stationary ($\omega = 0$), and both jets give torque in the same sense.

Jet velocity

The nozzles are equal, so each nozzle carries $q = Q/2 = 0.005$ m³/s:

$$\begin{aligned} a &= \frac{\pi}{4}(0.012)^2 = 1.131\times 10^{-4}\ \text{m}^2\\ V &= \frac{q}{a} = \frac{0.005}{1.131\times 10^{-4}} = 44.21\ \text{m/s} \end{aligned}$$

Torque

Water enters along the axis with zero angular momentum. The angular momentum leaving per second is $\rho q\,(r_1V\cos\theta + r_2V\cos\theta)$. The torque needed to hold the arm equals this rate of change of angular momentum:

$$\begin{aligned} T &= \rho\,q\,V\cos\theta\,(r_1 + r_2)\\ &= 1000\times 0.005\times 44.21\times\cos60^\circ\times(0.6 + 0.2)\\ &= 1000\times 0.005\times 44.21\times 0.5\times 0.8 = 88.4\ \text{N·m} \end{aligned}$$

Individually, jet 1 gives 66.3 N·m and jet 2 gives 22.1 N·m.

Answer: torque required to hold the sprinkler stationary $\approx 88.4$ N·m. (It is equal and opposite to the torque the jets exert on the arm. If $\theta$ were measured from the arm instead of from the tangential direction, replace $\cos60^\circ$ by $\sin60^\circ$.)