(a) Referring to Fig. 13-5, at what angle should a jet of water moving at 50.0 ft/sec impinge upon a series of blades moving at 20.0 ft/sec in order that the water will strike tangent to the blades (no shock)?
(b) What power is developed if the flow is 2.86 cfs?
(c) What is the efficiency of the blades?

Question

Accepted Answer

(a) [latex] ext { velocity of water }= ext { velocity of water/blade } rightarrow ext { velocity of blade }[/latex].

or [latex]50.0 ext { at } \angle heta_{x}= ext { ? at } 40^{\circ} ightarrow 20.0 ightarrow[/latex]

[latex] ext { From the vector diagram of Fig. 13-5(b), } 50 \cos heta_{x}=20.0+x, 50 \sin heta_{x}=y ext {, and } an 40^{\circ}=[/latex]

[latex]y / x=0.8391 . ext { Solving these equations, } heta_{x}=25^{\circ} 05^{\prime}[/latex].

(b) Solving Fig. l3-5(b) for the velocity of the water with respect to the blades,

[latex]y=50 \sin heta_{x}=50 \sin 25^{\circ} 05^{\prime}=21.20 ft / sec \quad ext { and } \quad V_{w / b}=y /\left(\sin 40^{\circ} ight)=33.0 ft / sec[/latex].

[latex] ext { Also, absolute } V_{x_{2}}=3.3 ft / sec ext { to the left, from Fig. 13-5(c). Then }[/latex]

[latex] ext { force } F_{x}=\frac{62.4 imes 2.86}{32.2}[50 imes 0.906-(-3.3)]=269 lb \quad ext { and } \quad ext { power } E_{x}=269 imes 20=5380 ft - lb / sec[/latex].

[latex] ext { (c) } ext { Efficiency }=\frac{5380}{\frac{1}{2} M(50)^{2}}=\frac{5380}{6928}=77.7 \% [/latex].

Question 13.11: (a) Referring to Fig. 13-5, at what angle should a jet of wa...

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