SOLUTION A four-armed sprinkler is used to generate electric power. For a specified flow rate and rotational speed, the power produced is to be determined.
Assumptions 1 The flow is cyclically steady (i.e., steady from a frame of reference rotating with the sprinkler head). 2 The water is discharged to the atmosphere, and thus the gage pressure at the nozzle exit is zero. 3 Generator losses and air drag of rotating components are neglected. 4 The nozzle diameter is small compared to the moment arm, and thus we use average values of radius and velocity at the outlet.
Properties We take the density of water to be 1000 kg/m³ = 1 kg/L.
Analysis We take the disk that encloses the sprinkler arms as the control volume, which is a stationary control volume.
The conservation of mass equation for this steady-flow system is \dot{m}_{1}=\dot{m}_{2}=\dot{m}_{total}. Noting that the four nozzles are identical, we have \dot{m}_{\text {nozzle }}=\dot{m}_{\text {total }} / 4 \text { or } \dot{V}_{\text {nozzle }}=\dot{V}_{total}/4 since the density of water is constant. The average jet exit velocity relative to the rotating nozzle is
V_{\text {jet }, r}=\frac{\dot{V}_{\text {nozzle }}}{A_{\text {jet }}}=\frac{5 L / s }{\left[\pi(0.01 m )^{2} / 4\right]}\left(\frac{1 m ^{3}}{1000 L }\right)=63.66 m / s
The angular and tangential velocities of the nozzles are
\begin{gathered}\omega=2 \pi \dot{n}=2 \pi(300 rev / min )\left(\frac{1 min }{60 s }\right)=31.42 rad / s \\V_{\text {nozzle }}=r \omega=(0.6 m )(31.42 rad / s )=18.85 m / s\end{gathered}
Note that water in the nozzle is also moving at an average velocity of 18.85 m/s in the opposite direction when it is discharged. The average absolute velocity of the water jet (velocity relative to a fixed location on earth) is the vector sum of its relative velocity (jet velocity relative to the nozzle) and the absolute nozzle velocity,
\vec{V}_{\text {jet }}=\vec{V}_{\text {jet, } r}+\vec{V}_{\text {nozzle }}
All of these three velocities are in the tangential direction, and taking the direction of jet flow as positive, the vector equation can be written in scalar form using magnitudes as
V_{\text {jet }}=V_{\text {jet }, r}-V_{\text {nozzle }}=63.66-18.85=44.81 m / s
Noting that this is a cyclically steady-flow problem, and all forces and momentum flows are in the same plane, the angular momentum equation is approximated as \sum M=\sum_{\text {out }} r \dot{m} V-\sum_{\text {in }} r \dot{m} V, where r is the moment arm, all moments in the counterclockwise direction are positive, and all moments in the clockwise direction are negative.
The free-body diagram of the disk that contains the sprinkler arms is given in Fig. 6–41. Note that the moments of all forces and momentum flows passing through the axis of rotation are zero. The momentum flows via the water jets leaving the nozzles yield a moment in the clockwise direction and the effect of the generator on the control volume is a moment also in the clockwise direction (thus both are negative). Then the angular momentum equation about the axis of rotation becomes
– T _{\text {shaft }}=-4 r \dot{m}_{\text {nozzle }} V_{\text {jet }} \quad \text { or } \quad T _{\text {shaft }}=r \dot{m}_{\text {total }} V_{\text {jet }}
Substituting, the torque transmitted through the shaft is
T _{\text {shaft }}=r \dot{m}_{\text {total }} V_{\text {jet }}=(0.6 m )(20 kg / s )(44.81 m / s )\left(\frac{1 N }{1 kg \cdot m / s ^{2}}\right)=537.7 N \cdot m
since \dot{m}_{\text {total }}=\rho \dot{V}_{\text {total }}=(1 kg / L )(20 L / s )=20 kg / s
Then the power generated becomes
\dot{W}=\omega T _{\text {shaft }}=(31.42 rad / s )(537.7 N \cdot m )\left(\frac{1 kW }{1000 N \cdot m / s }\right)=16.9 kW
Therefore, this sprinkler-type turbine has the potential to produce 16.9 kW of power.
Discussion To put the result obtained in perspective, we consider two limiting cases. In the first limiting case, the sprinkler is stuck, and thus, the angular velocity is zero. The torque developed is maximum in this case, since V_{\text {nozzle }}=0 \text {. Thus } V_{\text {jet }} =V_{\text {jet, } r}=63.66 m / s \text {, giving } T _{\text {shaft, } \max }=764 N \cdot m. The power generated is zero since the generator shaft does not rotate.
In the second limiting case, the sprinkler shaft is disconnected from the generator (and thus both the useful torque and power generation are zero), and the shaft accelerates until it reaches an equilibrium velocity. Setting T _{\text {shaft }}=0 in the angular momentum equation gives the absolute water-jet velocity (jet velocity relative to an observer on earth) to be zero, V _{\text {jet }}=0. Therefore, the relative velocity V _{\text {jet, r }} and absolute velocity V _{\text {nozzle }} are equal but in opposite direction. So, the absolute tangential velocity of the jet (and thus torque) is zero, and the water mass drops straight down like a waterfall under gravity with zero angular momentum (around the axis of rotation). The angular speed of the sprinkler in this case is
\dot{n}=\frac{\omega}{2 \pi}=\frac{V_{\text {nozzle }}}{2 \pi r}=\frac{63.66 m / s }{2 \pi(0.6 m )}\left(\frac{60 s }{1 min }\right)=1013 rpm
Of course, the T _{\text {shaft }} = 0 case is possible only for an ideal, frictionless nozzle (i.e., 100 percent nozzle efficiency, as a no-load ideal turbine). Otherwise, there would be a resisting torque due to friction of the water, shaft, and surrounding air.
The variation of power produced with angular speed is plotted in Fig. 6–42. Note that the power produced increases with increasing rpm, reaches a maximum (at about 500 rpm in this case), and then decreases. The actual power produced would be less than this due to generator inefficiency (Chap. 5) and other irreversible losses such as fluid friction within the nozzle (Chap. 8), shaft friction, and aerodynamic drag (Chap. 11).
