the jets is 4cm. Find the speed of rotation and the torque required to hold the sprinkler stationary

Diameter of jet d=1cm =0.01 m

Q=3l/s=$\frac{3}{1000}m^{3}/s=0.003m^{3}/s$

l=64cm =0.64 m

$r_{1}=r_{2}\frac{l}{2}=\frac{0.64}{2}$=0.32m

Area of jet

A=$\frac{\pi}{4}\times d^{2}=\frac{\pi}{4}\times(0.01)^{2}$

=$0.7854\times 10^{-4}m^{2}$

Quantity of discharge from each nozzle

$Q_{1}=\frac{Q}{2}=\frac{0.003}{2}=0.0015m^{3}/s$

Velocity of jet,

V=$V_{A}=V_{B}=\frac{Q_{1}}{A}=\frac{0.0015}{0.7854\times 10^{-4}}$=19.1m/s

Absolute Velocity if water leaving the jets

$V_{1}=V_{A}-w.r_{2} \ \ \ \ \ V_{2}=V_{a}-w.r_{2}$

$V_{1}=19.1-w\times 0.82 \ \ \ \ \ \ V_{2}=19.1-w\times 0.32$

1. Speed of rotation, w

T=$PQV_{1}r_{1}+(PQV_{2}r_{2})$

o=$PQ(V_{1}r_{1}+V_{2}r_{2})$

o=$V_{1}r_{1}+V_{2}r_{2}$

=$(19.1-0.32w)0.32+(19.1-0.32w)\times 0.32$=0

w=59.69 rad/s

2.Torque required to hold the sprinkler T

$T=PQ_{1}V_{A}r_{1}+PQ_{1}V_{B}r_{2}$

=$PQ(V_{A}+V_{B})r$

=$1000\times 0.0015(19.1+19.1)$0.32

=18.336 kN m