Lawn sprinkler has two nozzles of diameters 3 mm each is connected across a tap of water. The nozzles are at distance of 40 cm and 30cm from the centre of tap. The rate of water through the tap is $100 cm^3/s$. The nozzle discharge water in the downword directions. Determine the angular speed at which the sprinkler will rotate free.

Mumbai University > Civil Engineering > Sem 5 > Applied Hydraulics 1

Marks: 10M

Year: May 2016

Given:

Dia of nozzles A and B

$D = D_A = D_B = 3 mm = 0.003m \\ \therefore Area A = \frac{\pi}{4} \times (0.003)^2 = 7.068 \times 10^{-6} = 0.000007068m^2 \\ \text{Discharge Q} = 100cm^3/s$

Assuming the discharge to be equally divided between the two nozzles, we have

$Q_A = Q_B = \frac{Q}{2} = \frac{100}{2} = 50 cm^3/s = 50 \times 10^{-6} m^3/s$

velocity of water at the outlet of each nozzle

$V_A = V_B = \frac{Q_A}{A} = \frac{50 \times 10^{-6}}{0.000007068} = 7.074 m/s$

The jet of water coming out from nozzles A and b is having velocity 7.074 m/s. These jets of water will exerts force in the opposite direction i.e force exerted by the jets will be in the upward direction. the torque exerted will also be in the opposite direction. Hence torque at B will be in the anti-clockwise direction and A in the clockwise direction. But torque at B is more than the torque at A and hence sprinkle, if free, will rotate in the anticlockwise direction.

Let $\omega$ = angular velocity of sprinkler

Then absolute velocity of water at A

$V_1 = V_A + \omega \times x_A$

Where $x_A$ = distance of nozzle A from the centre of tap

Where $x_A$ = distance of nozzle A from the centre of tap = 30cm = 0.3m $\hspace{2cm}$ {$\omega \times x_A$ = tangential velocity due to rotation}

$V_1 = (7.074 + \omega \times 0.3) m/s$

Here $\omega \times x_A$ is added to $V_A$ as $V_A$ and tangential velocity due rotation $(\omega \times x_A)$ are ion the same direction.

Similarly absolute velocity of water at B

$V_2 = V_B - \text{tangential velocity due to rotation}\\ \hspace{0.5cm} = 7.074 - \omega \times x_B \hspace{2cm} \text{where} x_B = 40cm = 0.4m \\ \hspace{0.5cm} = (7.074 - \omega \times 0.4)$

Now applying equation we get

$T = \rho Q [V_2x_2 - V_1x_1]\\ \hspace{0.5cm} = \rho Q [V_2x_B - V_1 x_A]\\ \hspace{0.5cm} = 1000 \times 50 \times 10^{-6} [(7.074 - 0.4\omega) \times 0.4 - (7.074 + 0.3\omega) \times 0.3]$

The moment of momentum of the fluid entering sprinkler is given zero and also there is no external torque applied on the sprinkler. Hence resultant external torque is zero.

i.e T = 0

$\therefore$ 1000 x 50 x $10^-6$ [(7.074 - 0.4 w) x 0.4 - (7.074 + 0.3 w) x 0.3]= 0

$\therefore$ {(7.074 x 0.4w) x 0.4 - (7.074 + 0.3 w) x 0.3]= 0

$\therefore$ 7.074 x 0.4 - 0.16 w - 7.074 x 0.3 - 0.09w = 0

$\therefore$ 2.8296 - 0.16w - 2.1222 - 0.09w = 0

$\therefore$ 0.7074 - 0.25 w = 0

$\therefore$ 0.25 w = 0.7074

$\therefore$ w = 0.7074 / 0.25 = 2.8296 rad/sec