Respuesta :

Answer:

712.5 rpm

Explanation:

Given data:

number of poles = 4

percentage of full load torque = 200%

Calculate how fast in RPMs a typical NEMA class B motor shaft will rotate

assuming a frequency of ; 50 Hz

number of poles = 4

[tex]N_{s} = \frac{120 *50}{4}[/tex] =  1500 rpm

For an Induction  Class B motor it has a maximum slippage of 5%

therefore the operating speed = 0.95 * 1500 = 1425 rpm

Note ; An induction motor load torque is inversely proportional to its speed hence a t 200% of full load torque the speed in RPM at which the class B motor shaft will rotate = 1425 / 2 = 712.5

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