Question

A motor equipped with a flywheel has to supply a load torque of 600n-m for 10 sec followed by a no load period long enough for the flywheel to regain its full speed, it is desired to limit the motor torque to 450 n-m. What should be the moment of inertia of the flywheel? The no load speed of the motor is 600 rpm and it has a slip of 8% at torque of 400 n-m. Assume the motor speed torque characteristic to be a straight line in the range of operation. Motor has inertia of 10 kg-m2​

Answer 1

Answer:

Torque = 14.13 N-M

Explanation:

Given that,

a flywheel acquired on angular speed of 50 revolution in 10.5 sec

10.5 seconds = 50 revolutions

1 second = 50/10.5

= 4.76 revolution

so,

here,

Frequency of the revolution(f) = 4.76

so,

angular velocity = 2πf

= 9.52π rad/s

Here,

given the initial angular velocity(W•)

= 0 rad/s

and final angular velocity(W) =

9.52π tad/s

time taken(t) = 10.5 s

Now,

W = W• + αt

where,

α is the angular acceleration

putting the values,

9.52 = 0 + 10.5α

α = 9.52/10.5

α = 0.9 rad/s²

Now,

we have,

Torque = Iα

where,

I is the moment of inerta

that is

5 kg m²

putting the values,

Torque = 5 × 0.9

Torque = 4.5 π NM

π = 3.14

so,

torque = 4.5 × 3.14

Torque = 14.13 N-M

Explanation:

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