A lawn sprinkle shown in the figure has 12 mm diameter nozzle at the end of a rotating arm and discharges water with velocity of 15 m/s. Determine:

• Torque required to hold the rotating arm stationary
• Constant speed of rotation of arm if free to rotate

SOLUTION:

Given,

Diameter of each nozzle = 12 mm =0.012 m

Area of each nozzle,`A=(pi*d^2)/4=0.000113 m^2`

Velocity relative to the nozzle,`V_A=V_B=15 `m/s

Let `omega` be the angular velocity of the sprinkle.

Now,

Absolute velocity,`V_1=V_A+omega*r_A`

For stationary arm,`omega=0`

Thus,`V_1=V_A` 

Similarly, `V_2=V_B`

Now,

`Q=A*V=0.000113**5=0.001695``m^3`/s.

Now,

The torque exerte.... Show More

SOLUTION:

Given,

Diameter of each nozzle = 12 mm =0.012 m

Area of each nozzle,`A=(pi*d^2)/4=0.000113 m^2`

Velocity relative to the nozzle,`V_A=V_B=15 `m/s

Let `omega` be the angular velocity of the sprinkle.

Now,

Absolute velocity,`V_1=V_A+omega*r_A`

For stationary arm,`omega=0`

Thus,`V_1=V_A` 

Similarly, `V_2=V_B`

Now,

`Q=A*V=0.000113**5=0.001695``m^3`/s.

Now,

The torque exerted by water coming through nozzle A on the sprinkler is,

`=rho*Q*V_A**r_A`

`=1000**0.001695**15**0.3`

`=7.627 Nm`

Similarly, Torque exerted by the water coming through nozzle B on the sprinkler is,

`=rho*Q*V_B**r_B`

`=1000**0.001695**15**0.375`

`=9.534 Nm`

Thus the total torque exerted by the wayer on sprinkler is,

`=7.627+9.534`

`=17.161 Nm`

This is the total required torque required to hold the rotating arm stationary.

Let `omega` be the angular speed of rotation of the sprinkler

Now, the absolute velocity of flow of water at nozzle A and B becomes,

`V_1=15-0.3omega` 

`V_2=15-0.375omega`

Now the torque exerted by water coming out at sprinkler A is,

`=rho*Q*V_1*r_1`

`=1000**0.001695 (15-0.3omega)**0.3`

`=0.5085 (15-0.3omega)` 

Thia acts in clockwise direction.

Similarly the torque exerted by water coming out at sprinkler B is,

`=rho*Q*V_1*r_1`

`=1000**0.001695 (15-0.375omega)**0.375`

`=0.6356 (15-0.375omega)` 

This acts in clockwise direction.

Now,

Since the moment of the momentum of flow entering is zero and no external torque is applied on the sprinkle, the resultant torque must be zero.i.e,

`T_r=0`

or,`0.5085 (15-0.3omega) + 0.6356 (15-0.375omega)=0`

or,`omega=43.93`rad/sec.

`=43.93**60/(2pi)`

`omega=419.5 RPM`