Figure shows an unsymmetrical sprinkle. It has a frictionless shaft and equal flow through each nozzle with a velocity of 8 m/s relative to nozzle. Find speed of rotation in r.p.m.

SOLUTION:

Given,

`r_A=0.4m`

`r_B=0.6 m`

Velocity relative to the nozzle,`V_A=V_B=8`m/s

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

Now,

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

`V_1=8+0.4omega` (positive because the tangential velocity is acting in the same direction as flow velocity )

Similarly, absolute velocity at point B is,

`V_2=V_B-omegar_B`

`V_2=8-0.6omega` (negative because the tangential velocity is acting in opposite direction to the flow velocity )

Now the torque exerted by w....Show More

SOLUTION:

Given,

`r_A=0.4m`

`r_B=0.6 m`

Velocity relative to the nozzle,`V_A=V_B=8`m/s

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

Now,

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

`V_1=8+0.4omega` (positive because the tangential velocity is acting in the same direction as flow velocity )

Similarly, absolute velocity at point B is,

`V_2=V_B-omegar_B`

`V_2=8-0.6omega` (negative because the tangential velocity is acting in opposite direction to the flow velocity )

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

`=rho*Q*V_1*r_1`

This acts in clockwise direction.

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

`=rho*Q*V_2*r_2`

This acts in anticlockwise direction.

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,`rho*Q (V_2r_2-V_1r_1)=0`

or,`V_2*r_2=V_1*r_1`

or,`(8+0.4omega)**0.4=(8-0.6omega)**0.6`

or,`omega=3.077`rad/s

or,`omega=3.077**60/(2pi)`

or`omega=29.4 r p m`