VFD Overload

The impedance of the motor is proportional to frequency.

Below the rated frequency, the VFD voltage is proportional to frequency and current (torque) is constant. At the rated frequency, voltage x current = rated power of the motor. Above the rated frequency, the VFD voltage is held constant and the current (torque) decreases as the motor impedance increases with frequency. Above the rated frequency motor power (torque x rpm) is constant.

Z_motor = 2*pi*f*L

http://lhp.co.in/index_without_right.php?file=vfd_school

"Overload" in an AC induction motor is about exceeding the motor's designed thermal damage limits. Heat in the motor is a function of the current flow (I²R losses or "copper losses"), magnetic hysterisis (eddy current or "magnetic") losses, bearing friction and windage. But the vast majority (>70%) of the heat is coming from the copper losses based on current flow. So if your motor is designed to handle 50A, but it is only pulling 10A, there is no "overload" taking place.

As to "saturation", that is why you use a VFD. It is controlling BOTH the frequency AND the voltage, so that the motor always gets the proper ratio of V and hz that it was designed for. So in your example the motor was designed for 380V 50Hz, a ratio of 7.6 V/hz. At 25Hz then, it will reduce the voltage to 190V*. No saturation.

*In reality, the required torque of a centrifugal fan is much lower than 50% at 50% speed, it will be 50% cubed, so 12.5%. So that means at that speed, the motor torque could be much lower and the VFD will, if told to, reduce the voltage even more, essentially UNDER fluxing the motor as a way to reduce those magnetic losses. It's not much, but over time it does add up.

"In reality, the required torque of a centrifugal fan is much lower than 50% at 50% speed, it will be 50% cubed, so 12.5%."

No, the torque varies as speed squared, the power as speed cubed.