Answer:
712.5 rpm
Step-by-step 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
= 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