Question

-ircuit using OTL and OTU. Name your 1/O correspondingly. ON 2 Question 1. (a) A steel tube having inner and outer dimeters of 1.5 in. and 2 in., respectively, is turning at 45 rev/min. If the allowable shear stress (Tallow) of the tube material is 10 ksi, determine the horse power (hp) transmitted by the tube. (20 points) $4,714ad w 27 60 P=TW

Answer 1

Title - Horse Power Transmission

Solution : Given: Inner. Diameter, Di= 1.5 in = 38.1 mm Outer Diameter, Do = 2 in 50.8 mm 45 rpm > speed Allowable shear stress, Callow lo ksi 2. Callow 68.9476 MPa To find: The Horse power Chp) trasmitted by the tube We know, Power, P = Torque x Angular Velocity .. P = Tow (i) . Lets calculate w, 2 X TT X 45 4.7123 rad/s o = 2IN T, Now lets calculate Torque, formula, Using Torsion ? allow J R where T Torque J = polar moment of Inertia Callow Allow able shear stress R = Radius of extreme fibre

substituting values in above equation to find Torque Ti 68.9476 T I [D."-D:-] Do/2 32 T gy 68.9476 50.8/2. I [50.84-38.14] 32 T 1213.2163 N-m Now, we know, Power, Р T. w p . 1213.2163 X 4.7123 .. P = 5717.0391 Watts we know, 745.7 watts 1 hp 5717.0391 watts 7.66 hp The horse power (hp) transmitted by the tube is 7.66 hp.

Summary - To find the horse power, we require the torque and angular velocity. Angular velocity is obtained by the given speed of rotation in rpm. Torque is calculated using the torsion formula as the allowable shear stress and dimensions of the tube were given.