Question 14.SP.8: A nozzle discharges a stream of water of cross-sectional are......
A nozzle discharges a stream of water of cross-sectional area A with a velocity v_{A}. The stream is deflected by a single blade that moves to the right with a constant velocity V. Assuming that the water moves along the blade at constant speed, determine (a) the components of the force F exerted by the blade on the stream, (b) the velocity V for which maximum power is developed.
STRATEGY: Because you have a steady stream of particles, apply the principle of impulse and momentum.
MODELING: Choose the system to be the particles in contact with the blade and the particles striking the blade in the time Δt, and use a coordinate system that moves with the blade at a constant velocity V. The particles of water strike the blade with a relative velocity u_{A} = v_{A} − V and leave the blade with a relative velocity u_{B}, as shown in Fig. 1. Because the particles move along the blade at a constant speed, the relative velocities u_{A}~and~u_{B} have the same magnitude u. Denoting the density of water by ρ, the mass of the particles striking the blade during the time interval Δt is Δm = Aρ(v_{A} − V)Δt; an equal mass of particles leaves the blade during Δt. The impulse–momentum diagram for this system is shown in Fig. 2.
ANALYSIS:
a. Components of Force Exerted on Stream. Recalling that u_{A}~and~u_{B} have the same magnitude u and omitting the momentum Σm_{i}v_{i} that appears on both sides, applying the principle of impulse and momentum gives you
\underrightarrow{+} x components:\qquad(\Delta m)u-F_{x}\Delta t=(\Delta m)u\cos\theta
+ ↑ y components: \qquad\qquad+{F}_{y}\Delta t=(\Delta m)u\,\sin\theta
Substituting Δm = Aρ(v_{A} – V)Δt and u = v_{A} − V, you obtain
\mathbf{F}_{x}=A\rho(\nu_{A}-V)^{2}(1-\cos\theta)\leftarrow\quad\quad \mathbf{F}_{y}=A\rho(\nu_{A}-V)^{2}\sin\theta\uparrow◂
b. Velocity of Blade for Maximum Power. You can obtain the power by multiplying the velocity V of the blade by the component F_{x} of the force exerted by the stream on the blade.
Power = F_{x}V=A\rho(\nu_{A}-V)^{2}(1-\cos\theta)V
Differentiating the power with respect to V and setting the derivative equal to zero, you have
{\frac{d(\mathrm{power})}{d\nu}}=A\rho(\nu_{A}^{2}-4\,\nu_{A}\,V+3\,V^{2})(1-\cos\theta)=0
V=\nu_{A}\qquad V=\textstyle{\frac{1}{3}}\nu_{A} For maximum power V = {\frac{1}{3}}v_{A}→ ◂
REFLECT and THINK: These results are valid only when a single blade deflects the stream. Different results appear when a series of blades deflects the stream, as in a Pelton-wheel turbine (see Prob. 14.81).
