contestada

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.

Respuesta :

opudodennis

Answer:

28.22 kW

Explanation:

Angular velocity of nozzles

[tex]\omega=2n\pi[/tex] where [tex]\omega[/tex] is angular velocity, n is rotations per second

[tex]\omega=2*400/60*\pi[/tex]= 41.8879 rad/s

Tangential velocity of nozzle

[tex]V_{nozzle}=r\omega[/tex]=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, [tex]V_{nozzle} is the tangential velocity of nozzle and \omega is the angular velocity

The average absolute velocity of water jet

[tex]V_{jet}=V_{jet, r}-V_{nozzle}[/tex]=50-12.56637=37.43363 m/s

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

Angular momentum equation is expressed as

[tex]M=rmV_{out}-rmV_{in[/tex]}

[tex]T_{shaft}=-4rm_{noz}V_{jet}[/tex]=4*0.3*15*37.43363=673.8053 Nm

Energy generated, Q

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

Therefore, electric power produced is 28.22kW

Otras preguntas