A four-armed water sprinkler has nozzles at right angles to the 30-cm-long arms and at 45° angles with the ground. If the outlet diameters are 8 mm and 4 kg/s of water exits the four nozzles, find the rotational speed.

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[latex]V_e = \frac{\dot{m} }{ρA_e}=\frac{4}{1000 imes 4 imes π imes 0.004^2} =19.89 m/s. Velocity in arm = V . [/latex]

[latex]\pmb M_I = ∫_{c.v.}\pmb r×(2\pmb Ω×\pmb V)ρd[/latex] V[latex] = 4 ∫_0^{0.3} r\pmb {\widehat{i}} ×(−2Ω\pmb {\widehat{k}} ×V\pmb {\widehat{i}} )ρ Adr [/latex]

[latex] = 8ρ AVΩ\pmb {\widehat{k}} ∫_0^{0.3} rdr = −0.36ρ AVΩ\pmb {\widehat{k}} [/latex]

[latex] \pmb {ΣM} = \pmb 0 and \frac{d}{dt} \int_{ c.v.} (\pmb r ×\pmb V)ρ d [/latex] V[latex] =0[/latex]

[latex] ∫_{c.s.} (\pmb r ×\pmb V) \pmb V⋅\pmb {\widehat{n} }ρdA = 0.3\pmb{\widehat{i}} ×( 0.707V_e\pmb{\widehat{j}}+ 0.707V_e\pmb{\widehat{K}})V_e ho A_e[/latex]

The z-component of [latex] ∫_{c.s.}\pmb r ×\pmb V (\pmb V⋅\pmb {\widehat{n} })ρdA =0.3×0.707 V_e^2 A_e ho[/latex]

Finally, [latex]−(M_I )_z = 0.36ρ AVΩ = 4×0.3×0.707V_e^2 A_e ho. Using AV = A_e V_e[/latex] ,

[latex] 0.36Ω = 4×0.3×0.707×19.89. ∴ Ω = \underline{46.9 rad / s} [/latex].

Question 4.164: A four-armed water sprinkler has nozzles at right angles to ...

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