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emmainna [20.7K]
3 years ago
5

Air enters into the hollow propeller tube at A with a mass flow of 4 kg/s and exits at the ends B and C with a velocity of 400 m

/s, measured relative to the tube. If the tube rotates at 1500 rev/min, determine the frictional torque M on the tube.
Physics
1 answer:
mash [69]3 years ago
6 0

Answer:

643N.m

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.

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V_{2-2.01}=\frac{Y(2.01)-Y(2)}{2.01-2}=-24.16ft/s

Now, for the instantaneous velocity, we derive the expression for y:

V(t) = 40 - 32t    For t=2s   V(2) = -24ft/s

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