Question

A large lawn sprinkler with four identical arms is to be converted into a turbine to generate electric power by attaching a generator to its rotating head. Water enters the sprinkler from the base along the axis ofrotation at a rate of 15 kg/s and leaves the nozzles in the tangential direction at a velocity of 50 m/s relative to the rotating nozzle. The sprinkler rotates at a rate of 400 rpm in a horizontal plane. The normal distance between the axis of rotation and the center of each nozzle is 30 cm. Estimate the electric power produced.

Answer 1

28.22 kW

Step-by-step explanation:

Angular velocity of nozzles

`\omega=2n\pi` where
`\omega` is angular velocity, n is rotations per second

`\omega=2*400/60*\pi`= 41.8879 rad/s

Tangential velocity of nozzle

`V_(nozzle)=r\omega`=0.3m* 41.8879 rad/s =12.56637 m/s

Where r is normal distance between the axis of rotation and the centre of each nozzle,
`V_(nozzle) is the tangential velocity of nozzle and \omega is the angular velocity </p><p>The average absolute velocity of water jet </p><p>[tex]V_(jet)=V_(jet, r)-V_(nozzle)`=50-12.56637=37.43363 m/s

Where
`V_(jet)` is average absolute velocity, [/ex]V_{jet, r}[/tex] is velocity relative to the rotating nozzle

Angular momentum equation is expressed as

`M=rmV_(out)-rmV_{in`}

`T_(shaft)=-4rm_(noz)V_(jet)`=4*0.3*15*37.43363=673.8053 Nm

Energy generated, Q

`Q=\omega *T_(shaft)=(41.8879 rad/s*673.8053 Nm)*\frac {1kW}{1000Nm/s}`=28.22429kW

Therefore, electric power produced is 28.22kW