Question

Extra Credit: Determine the velocity of the water leaving the nozzle Nozzle exit area 30 mm2 r-600 mm a) If the sprinkler head is stationary b) If the sprinkler head rotates at 600 rpm Q 1000 mlls

I want the right answer plzzzz

Answer 1

(a)

$Q=AV\Rightarrow V=Q/A$

here two nozzles are used, so total area = 2x(area of one nozzle) = 2x30mm² =60mm² = 60x10^-6 m²

Q= 1000ml/s = 1000x10^-6 m³/s

$V_{W}=\frac{1000\times 10^{-6}}{60\times 10^{-6}}=16.667m/s$

(b)

angular velocity of nozzle,

$\omega=\frac{2\pi N}{60} = \frac{2\pi \times 600}{60} =62.53 rad/s$

so, tangential velocity of nozzle,

$V_{N}=r\omega=0.6\times 62.83=37.70m/s$

so velocity of water = -37.70m/s

so actual velocity of water due to discharge rate and rotaation of sprinkler head = 16.667-37.7 = -21.033m/s

so exiting water will have velocity of 21.033 m/s in the opposite direction of the rotation.