Determine the required to hold the sprinkler stationary if the friction neglected determine the speed of sprinkler

Diameter of nozzles d=12mm =0.012m

Velocity of flow at each nozzle,

V-$V_{A}=V_{B}$=9m/s, $r_{1}=20cm$=0.2m

$r_{2}=25cm =0.25m$

Area of each nozzle, A=$\frac{\pi}{4}d^{2}=\frac{\pi}{4}\times (0.012)^{2}$

=$1.131\times10^{-4}m^{2}$

Discharger through the sprinkler

Q=V A$9\times 1.131\times 10^{-4}=10.179\times 10^{-4}m^{3}/s$

1) Torque required to hold the arm

-Torque exerted by the water flowing through nozzle A on the sprinkle

-$PQV_{A}R_{1}=100\times (10.179\times 10^{-4})\times 9\times 0.25$=2.2903 Nm

-Total torque exerted by water on sprinkler,

=2.2903+1.8322=4.1225 Nm

2) Speed of Roatation os sprinkler arm if free to rotate

Let w=Speed of rotation of sprinkler arm

Absolute Velocity of water at nozzle A,

$V_{1}=V_{A}-wr=9-2\times 0.20$

Absolute Velocity of water at nozzle B,

$V_{2}=V_{B}-wr_{2}=9-w\times 0.25$

Torque exerted by water coming out of A on Sprinkler

$PQV_{i}r_{1}$

=1000$\times (10.179\times 10^{-4})(9.w\times 0.2)0.2$

=6.203.(9.02w)

Torque exerted by water coming out of A on sprinkler

$PQV_{2}r_{2}$

=1000$\times (10.179\times 10^{-4})(9.025 w)0.25$

=0.2545(9-0.25w)

Total Torque exerted by water

=0.2036(9-0.2w)+0.2545(9-0.25w)

0.2036(9-0.2w)+0.2545(9-0.25w)=0

$0.2036\times 9-0.2036\times0.2w+0.2545\times 9-0.2545\times0.25w$

4.1229=0.104345 w

w=39.5912 rad/s