Question

A cylindrical shaft 90 mm in diameter rotates about a vertical axis inside a cylindrical tube of length 50 cm and 95 mm internal diameter. if the space between them is filled with oil of viscosity 2 poise, find the power lost in friction for a shaft speed of 240 rpm.

Answer 1

Answer:

14.5 watt

Explanation:

shear stress = (.2*(2πN/60)*.09/2)/.0025. = 90.478N/m2

Drag force = shear stress*Area(π*.09*.5)= 12.79N

Torque = Drag force * Radius= .576 N-m

power = Torque*2πN/60. = 14.5 watt

Answer 2

Answer:7.64 watt

Explanation:

dynamic viscosity ( ,u)=0.2 pa sec

dia of cyl shaft d =0.09m

length of fixed cyl tube L=0.5m

int dia of fixed cyl. tube D=0.095m

space btw tube and shaft dy=D-d=0.005m

rotating speed of shaft N =240rpm

tangential velocity du=πdN/60=1.13m/sec

shear stress{,u(du/dy)}=45.2N/m²

shear force F=shear stress×surface area

F=45.2×πdL=6.39N

Torque=F×(D/2)=0.304Nm

power loss P=2πNT/60=7.64watt