Question 14.7: 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 which 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.
a. Components of Force Exerted on Stream. We choose a coordinate system which moves with the blade at a constant velocity V. The particles of water strike the blade with a relative velocity \text u _A=\text v _A- V and leave the blade with a relative velocity \text u _B. Since the particles move along the blade at a constant speed, the relative velocities \text u _A and \text u _B have the same magnitude u. Denoting the density of water by r, the mass of the particles striking the blade during the time interval Δt is Δm = Ar\left(v_A-V\right) Δt; an equal mass of particles leaves the blade during Δt. We apply the principle of impulse and momentum to the system formed by the particles in contact with the blade and the particles striking the blade in the time Δt.
Recalling that \text u _A and \text u _B have the same magnitude u, and omitting the momentum \sum m_i\text v _i which appears on both sides, we write
\overset{+}{\text y}~x~ components: ~~~~~~~(\Delta m) u-F_x \Delta t=(\Delta m) u~\cos u+x y components: +F_y \Delta t=(\Delta m) u~ \sin u
Substituting \Delta m=A r\left(v_A-V\right) \Delta t \text { and } u=v_A-V, we obtain
F _x=\operatorname{Ar}\left(v_A-V\right)^2(1-\cos u ) ~\text z \quad\quad\quad F _y=\operatorname{Ar}\left(v_A-V\right)^2 \sin \text u ~ \text xb. Velocity of Blade for Maximum Power. The power is obtained by multiplying the velocity V of the blade by the component F_x of the force exerted by the stream on the blade.
\text { Power }=F_x V=\operatorname{Ar}\left(v_A-V\right)^2(1-\cos \text u ) VDifferentiating the power with respect to V and setting the derivative equal to zero, we obtain
\frac{d(\text { power })}{d V}=\operatorname{Ar}\left(v_A^2-4 v_A V+3 V^2\right)(1-\cos\text u )=0 \\ \\ V=v_A \quad\quad V=\frac{1}{3} v_A \quad\quad \text { For maximum power } V =\frac{1}{3} v_A ~ \text yNote. These results are valid only when a single blade deflects the stream. Different results are obtained when a series of blades deflects the stream, as in a Pelton-wheel turbine. (See Prob. 14.81.)
