• Relevant rates
• Relevant powers

To verify that the Variable Frequency Drive/Conveyor Drive Motor design is adequate, you must verify:
Required mechanical power for all rates < Available electrical power at all rates


Figure 1. Example of decreasing conveyor belt speed with VFD

Understanding relevant rates (i.e., material handling rate, belt speed, electric motor RPM and power supply frequency) is essential in calculating all required mechanical powers and comparing them against electric motor power selected, electric power supply selected and electric motor power available.

For example, if a conveyor operator reduces conveyor speed from 330 fpm to 165 fpm to move 450 tph of stone on a 120 ft. long x 36 in. wide inclined conveyor belt using a VFD (Figure 1), will the motor fail?

Since required conveyor belt pull (Te) is 2,178 pounds to move the load at 330 fpm, then required power is 23.6 hp at that speed. If a 25-hp motor is installed to run on a 460-v/3-ph/60-hertz power supply at a belt speed of 330 fpm, then the required power is less than the installed and available power. The motor will not be overloaded at 450 tph (Figure 2).


Figure 2. Comparison of 25 hp running on 460v/3ph/60Hz power supply and 460v/3ph/30Hz power supply

If a VFD is installed to drive the 25-hp motor at 460v/3ph/30 Hz to provide a belt speed of 165 fpm, then the required belt pull is 3,760 pounds and required mechanical power is 20.3 hp. Available motor power will be 12.5 hp.

Available power = Installed power x (selected frequency/power supply frequency)
12.5 hp = 25 hp x (30Hz/60Hz)

Motor torque is nearly constant in the desired frequency spectrum, but available electric motor power is linearly proportional to rotor RPM. If rotor RPM is cut in half by reducing power supply frequency from 60 Hz to 30 Hz, then available power is 50 percent of 25 hp.


Figure 3. Relationship between pulley torque and rotor torque

Compare required pulley torque with motor design torque (Figure 3). If a 10-hp Model 320M Conveyor Drive is supplied with a 2-pole motor and appropriate gear reduction, then the 14.59 ft-pounds of rotor torque will be converted to 500 ft-pounds. of pulley torque to drive the conveyor. If the pulley diameter is 12 in., the available belt pull is 1,000 pounds.

Belt pull = pulley torque/pulley radius
1,000 pounds = 500 ft-pounds./0.5 ft.

Electric motors will fail if required mechanical power is higher than installed and available power because AC squirrel cage induction motors have no inherent ability to protect themselves from drawing more current than the stator winding is designed to carry. That is why external overcurrent protection is required.


Figure 4. Relationship between electric motor power and current draw

For example, if a model 400 Motorized Pulley is designed to run on a 460v/3ph/60Hz power supply, a 10-hpmotor will draw 12.1 amps at full load (FLA) and a 20-hp will draw 23.6 amps (FLA), (Figure 4). Each motor may run continuously at full load.

If a 10-hp motor was installed and required mechanical power was increased to 20-hp by increasing the load on the conveyor, then the 10-hp motor will attempt to supply 20-hp by drawing > 20 amps. The stator will burn because more current will flow in the windings than the motor was designed to carry. This problem could also occur if the required mechanical torque was held constant, but the power supply frequency was increased to 120 Hz.

In conclusion, conveyor operators and designers will reduce the risk of electric drive motor failure when VFDs are used to vary conveyor belt speed by confirming that required mechanical power is less than available electric power at all pertinent rates. For more information, visit rulmecacorp.com.

This newsletter was produced by North Coast Media's content marketing staff in collaboration with Rulmeca Corp.