Fig shown a lawn sprinkler with two jets each located at 30 cm from the center. The jets are 1 cm diameter. Assume no friction find the speed of rotation for discharge of 3 l/s

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written 6.2 years ago by

teamques10 ★ 68k

• modified 5.1 years ago

applied hydraulics

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written 6.2 years ago by

teamques10 ★ 68k

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d=1cm=0.01m

r=30cm=0.3m

Q=1 litre/sec

Q$\frac{\pi}{4}\times 0.01^{2}$

=7.854$\times 10^{-5}m^{2}$

Velocity of jet, $V_{1}=\frac{Q}{A}=\frac{3\times 10^{-3}}{7.854\times 10^{-5}}$=19.098m/s

Since $r_{1}=r_{2}$

Resultant Velocity $V_{A}=V_{A}=wr$

=19.098=$w\times 0.3$

w=63.66 rad/sec

Speed of rotation(N) w=$\frac{2\pi N}{60}$

63.66=$\frac{2\pi\times N}{60}$

N=608.19 rpm

The speed of rotation is 608.19 rpm

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