Starting High Inertia Loads

There are a number of considerations when starting a high inerta load. It is important that the whole system is engineered correctly to prevent field problems and equipment failure.

A High inertia load, requires an extended starting time to reach full speed. The start time is a function of the load inertia, the load speed and the starting torque developed by the motor. If we reduce the start voltage or current, the start time will extend.
During start, there are two components of torque that is delvered to the load. There is the work torque which is the torque required to keep the load spinning at that speed, and there is the acceleration torque which is the torque that cause the load to increase in speed. The acceleration torque is the difference between the "work" torque and the torque developed by the motor. As we reduce the start voltage or current, the shaft torque of the motor will also reduce, but the work torque will stay the same as this is a function of the driven load. The rate of acceleration is determined by the acceleration torque. If the shaft torque is reduced to low, there will be no acceleration torque and the motor will not reach full speed.

The three key components to be considered are the motor, the starter and the supply. If any of these are not suitable for the high inertia load, the result can be disasterous.

Motor - Rotor

During start, there are high slip losses resulting in a high power disipation in the rotor of the motor. This high power dissipation causes a temperature rise in the rotor bars. The actual temperature rise is dependant on the total power dissipated and the thermal mass of the rotor bars. There is a limit to how much temperature rise the rotor can tolerate. If the temperature is high enough, the bars can actually melt, but every time the motor is started, there will be rotor bar heating and this will result in expansion and contraction of the bars. The expansion of the bars will put mechanical stresses on the bars and shorting rings and over a period of time, the interface between the bars and shorting rings can fail due to mechanical fatigue.
There is a limit to the maximum amount of energy that can be dissipated in the rotor. Exceeding maximum this will shorten the life of the rotor and result in a premature rotor failure.
Motor manufacturers usually quote the limit for motors in one of two ways. Commonly, the motor manufacturer will quote the maximum Locked Rotor Time which is the maximum time that the rotor can withstand full voltage and a stationary rotor. Loosely, we use this to indicate the maximum starting time under Direct On LIne (Full Voltage) starting conditions. The alternative rating applied to induction motors, is the maximum load inertia. This is the maximum load inertia that can be started by this motor. A higher inertia would result in to much rotor heating and premature rotor failure. The load inertia is the load inertia as seen by the shaft of the motor. If the load is spinning at a different speed than the motor, you need to compensate for the speed difference by multiplying the actual load inertia by the square of the speed ratio. i.e. if the load is spinning at half the rotor speed, the effective inertia as seen by the motor, is a quarter of the actual load inertia.

Motor - Start Characteristics

Under high start torque conditions, it is important to ensure that the motor is "efficient" during start. That means that the motor must produce a high torque for reduced current under high slip conditions. If the motor produces a very low start torque, then it is going to take much longer to start the load and this will impact on the supply and on the starter. Correct attention must be paid to the starting characteristics of the motor.
Bottom line the motor must have Good starting characteristics - high start torque for low start current, and must also be able to withstand starting the inertia of the load.

Motor - Stator

During start, there is a high start current flowing in the stator. This will result in a high temperature rise in the staor windings and if excessive will result in an insulation failure of the stator. Usually, it is the rotor that is damaged by high inertia starts rather than the stator. Stator failure usually results from continuous load conditions whereas rotor failure is generally due to starting issues.
Because the rotor bears the brunt of the start loading, is is not possible to fully protect the motor by the use of thermistors alone. Thermistor protection provides good protection for continous operation, but some form of thermal modeling is recommended for rotor protection.

Electrical Supply

Starting a high inertia load requires a lot of energy to be drawn from the supply. If you use any form of starter, (Full voltage or reduced voltage) there will be a high demand on the electrical supply during start. Reducing the voltage will reduce the instantaneous current draw, but will extend the starting time. The heating effect in the supply equipment will be the same but the voltage drop will be lower. If the supply is not strong enough to start the load, there is a problem and the supply will need to be strengthened.
A variable speed controller is one way to "start" a high inertia load without the sever overload on the motor and supply but has the disadvantage of high input harmonics, lower running efficiency and higher capital costs.

Starter

The starter must be capable of withstanding the high overload during start for the duration of the start. With a high inertia load, to get the machine to full speed in a reasonable time, a high torque must be presented to the load. Reducing the voltage too much will extend the start time to an unacceptable period.
The starter must be correctly engineered for the start current and time. This may require that the starter be increased in size by one or more frame sizes depending on the type of starter and the starting conditions.

Engineering the start.

1. Determine the load inertia, the load speed and the load speed torque curve.
2. Select a motor rated for the required continuous power for the driven load, that has a sufficiently high "maximum load inertia" or "Locked Rotor Time" to start the load, and provides a high start torque at a low start current. For comparison, take the LRT divided by the LRC as an indicator. Higher is better, and preferable use percentage numbers to keep all things equal.
3. Determine the start time under different start conditions and select a start condition that is suitable for your supply.
4. From the above, you can determine the best starter for the application.

Commonly, start currents of 450% - 500% for 30 - 60 seconds are required to start high inertia loads assuming high start torque motors are employed.

Calculations

These Calculations can be easily carried out using our Electrical Calculations software.

Step 1. Enter the load characteristics into the Edit Load Data page. The Load Torque is the work torque of the driven load. For a purely inertial load, this would be all zeros. Fan loads are usually a square law and for a freely ventillating fan would be approaching 100% at full speed.

Step 2. Enter the motor data from the motor data sheets into the Edit Motor Data page. Ensure that the effective inertia of the load is less than the "maximum Inertia" of the motor if this rating is given.

Step 3. Open the Acceleration curves and select the load and the motor and then select different starters and settings to see the effect of the reduced start current. Select the DOL start to get the maximum full voltage start time and compare this with the "maximum Locked Rotor Time" of the motor. If this is higher than the motor rating, try another motor.

Step 4. Specify a starter that can withstand the start current and time for your application.
NB this is true for the selected motor. A different motor will give different results and may need a larger starter, or may not work at all!!

You can use this software to try a number of motors and starters to determine the combination that best suits your requirements.