A jet of water 100 mm in diameter leaves a nozzle with a mean velocity of 36 m/s as shown in Fig. 2.23. It is deflected by a series of vanes moving with a velocity of 15 m/s in the direction at 30° to the direction of the jet so that it leaves the vane with an absolute velocity which is at right

Question

A jet of water 100 mm in diameter leaves a nozzle with a mean velocity of $36 {m}/{s}$ as shown in Fig. 2.23. It is deflected by a series of vanes moving with a velocity of $15 {m}/{s}$ in the direction at $30^{\circ}$ to the direction of the jet so that it leaves the vane with an absolute velocity which is at right angles to the direction of the vane. Owing to friction the velocity of the fluid relative to the vanes at exit is equal to 0.85 of relative velocity at inlet.

Calculate

(a) inlet angle and outlet angle of the vane for no shock entry and exit.
(b) force exerted on the series of vanes in the direction of motion of vanes.

Calculate

(a) inlet angle and outlet angle of the vane for no shock entry and exit.
(b) force exerted on the series of vanes in the direction of motion of vanes.

Accepted Answer

(a) With reference to the Fig. 2.23 given quantities are:

[latex]V_{1}=36 {m}/{s}; u=15 {m}/{s}; \alpha_{1}=30^{\circ}; \alpha _{2}=90^{\circ}[/latex]

[latex]V_{r2}=0.85 V_{r1} ; d=100 mm[/latex]

To determine the inlet angle [latex]\beta_{1}[/latex], consider the inlet velocity triangle ABD. The velocity of the fluid relative to vane at inlet in [latex]V_{r1}[/latex] must be tangential to the vane and make an angle [latex]\beta_{1}[/latex] with the direction of motion

[latex] an \beta _{1} = \frac{CD}{BC} = \frac{V_{1}\sin \alpha _{1}}{\left(V_{1}\cos \alpha _{1}-u ight) }[/latex]

Substituting [latex]V_{1}=36 {m}/{s}; u=15 {m}/{s}; \alpha_{1}=30^{\circ}[/latex]

[latex] an \beta _{1} = \frac{36 imes 0.5}{\left(36 imes 0.866-15 ight) } = 1.13 herefore \beta _{1} = 48^{\circ}[/latex]

To determine outlet angle [latex]\beta_{2}[/latex], as [latex]V_{2}[/latex] has no component in the direction of motion, outlet angle [latex]\beta_{2}[/latex] in given by

[latex]\cos \beta _{2}=\frac{u}{V_{r2}}[/latex], but [latex] V_{r2}=0.85 V_{r1}[/latex]

and from the inlet triangle

[latex]V_{r1} = \frac{CD}{\sin \beta _{1}} = \frac{V_{1}\sin \alpha _{1}}{\sin \beta _{1}} = \frac{36 imes \sin 30}{\sin 48}[/latex]

[latex]V_{r1} = 24.2 {m}/{s}[/latex]

[latex]V_{r2} = 0.85 imes V_{r1} = 0.85 imes 24.2 = 20.57[/latex]

[latex]\cos \beta _{2} = \frac{u}{V_{r2}} =\frac{15}{20.57} = 0.729[/latex]

[latex] herefore \beta_{2} = 43^{\circ}[/latex].

(b) Since the jet strikes series of vanes perhaps mounted on the periphery of a wheel so that each vane moves on its place is taken by the next in the series, the average length of the jet does not alter and whole flow from the nozzle of diameter d is deflected by the vanes.
Mass flow rate of the fluid from the nozzle

[latex]\dot{m} = ho imes \frac{\pi }{4}d^{2} imes V_{1} = 1000 imes \frac{\pi }{4}\left(0.1 ight)^{2} imes 36 = 282.8 {kg}/{s}[/latex]

Velocity component of [latex]V_{1}[/latex] in direction of [latex]X = V_{1} \cos \alpha _{1}[/latex]

[latex]V_{1x} = 36 imes 0.866 = 31.17 {m}/{s}[/latex]

[latex]V_{2x} = V_{2}\cos \alpha _{2} = V_{2}\cos 90^{\circ} = 0[/latex]

Force on the direction of motion

[latex]F = \dot{m}\left(V_{1x}-V_{2x} ight) = 282.8\left(31.17 ight) = 8814 N[/latex]

[latex]F = 8814 N[/latex]

Question 2.12: A jet of water 100 mm in diameter leaves a nozzle with a mea...

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