Calculating load % of AC motors when operated at a frequency less than rated value
When operating an induction motor at a lower frequency, the most important change in performance is the reduction in speed. The synchronous speed is given by RPM = 120 X f / P where f is frequency (Hz) and P is the number of motor poles (an even integer number). The rated speed at 50 Hz for the given motor is 1475 RPM. The synchronous speed is a little higher. The nearest speed that satisfies the synchronous speed formula at 50 Hz is 1500 (1500 = 120 X 50 / 4), so this is a 4-pole motor. Since the rated speed is given as 1475 RPM, we know that the slip at rated load is 1500 - 1475 = 25 RPM. Horsepower = Torque (Lbs - Ft) X RPM / 5252, so the rated torque is 214 Lbs. - Ft. Since the motor's rated power is given in horsepower rather than watts, I am assuming that all of the ratings should follow the USA units of measurement. Units can be converted as necessary.

At 40 Hz, the synchronous speed will be 1200 RPM. The slip at rated load can be held to 25 RPM at reduced frequency by reducing the voltage the proper amount, approximately maintaining a constant V/Hz ratio. That means the rated speed will be 1175 RPM at 40 Hz. The rated load torque can be maintained at the 50 Hz value. Rated power at 40 Hz will then be 216 X 1175 / 5252 = 48.3 Hp.

If the motor voltage can be controlled in a way that will maintain the slip RPM, the motor's rated torque can be maintained all the way to zero speed. In order to operate at that torque continuously, the motor will need to be adequately cooled. At 40 Hz, the motor may not need any special cooling provisions. Below that, adequate cooling will become progressively more difficult. Determining and maintaining the optimum voltage also becomes progressively more difficult as the frequency is reduced.

With the optimum voltage applied, the motor current will be approximately proportional to torque and will remain constant as the speed is reduced with constant load torque.