Final answer:
To reduce the speed of a 575V - 60Hz motor by 30%, the new frequency required from the VFD is 42Hz, and the new required output voltage is approximately 402.5V, maintaining the original volts/hertz ratio.
Step-by-step explanation:
The subject of this question is Engineering, particularly related to electric motors and Variable Frequency Drives (VFD). When it comes to reducing the speed of a 575V - 60Hz 3-phase motor by 30%, the new frequency can be calculated as:
New Frequency (f NEW) = Original Frequency × (1 - Reduction Percentage)
f NEW = 60Hz × (1 - 0.30) = 60Hz × 0.70 = 42Hz
To maintain the V/Hz ratio constant, which is necessary to keep the magnetic flux in the motor within operational limits, we calculate the initial V/Hz ratio and apply it to the new frequency:
Original V/Hz ratio = Original Voltage / Original Frequency
V/Hz = 575V / 60Hz = 9.583 V/Hz
We use this V/Hz ratio to find the new required output voltage (E L-Lnew):
New Voltage (E L-Lnew) = V/Hz ratio × New Frequency
E L-Lnew = 9.583 V/Hz × 42Hz = 402.486 V
Therefore, to reduce the motor speed by 30%, a VFD should output approximately 402.5V at 42Hz.