Problem Statement
Known
Water enters a sprinkler through the center stem and exits tangentially through two identical nozzles. The sprinkler may be held fixed or allowed to rotate.
Use a fixed, nondeforming control volume that cuts through the shaft and surrounds the water inside the sprinkler head.
Find
Determine the shaft torque and shaft power associated with the angular momentum carried away by the exiting water.
Moment-of-Momentum Setup
The rotating-machine version of the angular-momentum balance is simplified here by using average one-dimensional velocities at each opening, a steady or steady-in-the-mean flow pattern, and only the component about the rotation axis.
The center inlet carries no axial angular-momentum flux about the shaft. The two tangential outlets carry the angular momentum that must be balanced by shaft torque.
Velocity relation
\[ \mathbf{V}=\mathbf{W}+\mathbf{U} \]For this tangential exit direction, \(U_2=r_2\omega\) and the fixed-frame tangential component is
\[ V_{\theta 2}=W_2-U_2 \]Shaft torque and power
\[ T_{\mathrm{shaft}}=-r_2V_{\theta 2}\dot{m} \] \[ \dot{W}_{\mathrm{shaft}}=T_{\mathrm{shaft}}\omega =-r_2V_{\theta 2}\dot{m}\omega \] \[ \dot{W}_{\mathrm{shaft}}=-U_2V_{\theta 2}\dot{m} \]The zero-torque rotation speed occurs when the water leaves with no tangential absolute velocity:
\[ \omega_{\mathrm{zero\ torque}}=\frac{W_2}{r_2} \]