Answer:
643N.m
Step-by-step explanation:
From this question we have:
Mass flow = 4kg/s
Velocity V = 400m/s
Rotation N = 1500rev/min
We get the relative velocity at exit to be:
V2 = V - r2w
400-0.5x [(2*π*1500)/60]
= 400-78.5
= 321.5m/s
Then we have to calculate the frictional torque My
Mt = Mr2 x V2
= 4x0.5x321.5
= 643Nm
From the calculations above, we get the frictional torque M on the tube to be 643Nm.