Using small motors for highcycle applications

The traditional method for handling high cycle applications incorporates a clutch/brake module or a servomotor. However, both methods can be expensive. For example, the clutch/brake module requires periodic maintenance. These methods may be more than what is needed for typical highcycle applications.
This article provides general guidelines on applying small ac induction motors (1/2 hp to 3 hp) with an ac inverter to handle applications requiring approximately 10 to 30 starts and stops per minute.
Most applications for NEMA frame motors involve starting at the beginning of the day and running continuously until the end of the shift or production run. So the NEMA motor may start and stop one time per hour. This is normal duty. A distinction must be made between high cycle and very high cycle applications: a highcycle application may involve up to 30 motor starts and stops per minute (Fig. 1), whereas a very highcycle application may consist of 50 to 60 starts and stops per minute or more (Fig. 2). Typical highcycle applications include pickandplace equipment, palletizers, positioning applications, and packaging applications (see “Capabilities of small NEMA ac induction motors”).
Inertia and motor design
In order to get the maximum benefit from a small ac motor, it must be properly selected for a given application and load inertia. When considering applications that require multiple starts and stops per minute, the first question that should be asked is: “What is the load inertia (Wk2) that is transferred to the motor shaft?”
Inertia is a function of kinetic energy. Every moving object — whether it is a motor rotor, flywheel, or load traveling down a conveyor — has stored kinetic energy, which is its mass times the square of its velocity. An object that is at rest or one that is moving at constant speed must have an external force to accelerate or decelerate the object. The inertia of an object affects directly the amount of force necessary for a change in speed or direction. Therefore, in highcycle applications, the inertia of a motor rotor is very important. The “inertia transferred to the motor shaft” equation is the conversion of all the external load inertia to its equivalent inertia if rotating at motor speed.
Inertia transferred to the motor = Load Inertia *(Load rpm/Motor rpm) 2
Generally, a motor with a smalldiameter longlength rotor will perform better than one with a largediameter shortlength rotor. Typically, motor inertia is defined (Wk2) in the units lbft2or lbin2. Multiply lbft2by 144 to get lbin2. This is true because the “k2” or radius of gyration of a cylinder rotating about its axis is:
k 2 = r 2 /2
with “r” being the radius of the cylinder (or in this case, the radius of the motor rotor).
A motor with a higher slip, high resistance rotor, and a high torqueperamp design will be more forgiving in these applications.
Duty cycle
Duty cycle is defined as all the different loads (torque) to which a motor will be subjected at a given speed over a defined period of time. Most repetitivecycling applications have four basic elements. The first is the acceleration step. Generally, in the acceleration step, the load is being accelerated from rest, which is where the motor must deliver the peak torque of the cycle. The second step is the steady speed, or period where the machine is moving at a constant rate. Typically, this step has the lowest motor torque requirement.
The third step is the deceleration speed. In this period, the motor makes a transition from a high speed to a low or zero speed. The torque requirements for this step are similar to the acceleration step and may require more or less torque, depending on the time required for stopping. The last step is the dwell time. This is the period of time the motor “rests” before the next cycle begins. The total time required for all four steps is the cycle time. This duty cycle is represented by a trapezoidal profile (Fig. 3).
Calculating motor load requirements
The first step in calculating motor requirements is to calculate the total system inertia, including the motor inertia. Typically, there are only two types of driven loads — rotary and linear. Rotating drivenloads include fans, flywheels, and turntables. Conveyors and indexing equipment are examples of lineardriven loads. Generally, the driven load turns at a slower speed than the motor speed. Because of this, the drivenload inertia must be converted to the inertia as “seen” at the motor shaft — in other words, the transferred to motor shaft speed. Use the following equation when converting rotating driven inertia to transferred to the motor shaft inertia:
Rotating driven inertia = Load inertia * (Load rpm/Motor rpm) 2
When converting linear driven inertia:
Linear driven inertia = W * {(V/(6.28*N 2
W= weight in lbs.
V= Linear velocity (ft./min.)
N= Motor rpm
Once all the load inertia components are calculated and reflected to the motor shaft, the values are added together to get the total Wk2. This information can then be used in the following formula:
Inertia Torque (T) = (Wk 2 * Change in rpm) / (308 * t)
T = Torque in lbft
Wk 2 = total inertia in lbft 2
Change in rpm = change in motor rpm (during acceleration from rest the motor may change from 0 to 1800 rpm)
t = time in seconds
308 is a constant.
This formula can be used to calculate the inertia torque required for each step of the cycle.
Calculating motor torque
The motor torque required to drive the application is different at each step of the duty cycle. For example, the friction and losses in the belt on a horizontal roller conveyor generate the load torque. In this scenario, the acceleration torque usually is the highest torque required (peak torque) during the duty cycle because it includes the inertia torque plus the load torque. The load torque is generated during steadystate running. The deceleration torque is the torque required to decelerate the load minus the load torque. Therefore:
Acceleration torque = T + load torque
Deceleration torque = T – load torque
Index torque = load torque
Dwell = zero torque
After the torque for each segment is calculated, the RMS torque can be calculated. Because the motor in this example sees peak torque only during acceleration, a smaller and less expensive motor can be used. Motors have the ability to provide power above their load rating for a short period of time. The RMS torque takes into account each load segment and the corresponding duration, and compares it to the motor’s full load torque rating:
The RMS torque is then compared to the full load torque rating of the motor. If the RMS torque is less than the full load torque, the motor will work provided the motor and the ac inverter have enough capacity to overcome the calculated peak torque. This formula works for short duration duty cycles of 5 minutes or less. C is a factor that accounts for motor frame construction where C = 2 for TEFC motors. Consult the motor manufacturer when applying this formula.
VFD considerations
The variable frequency drive (VFD) is an excellent piece of equipment to use in conjunction with a small ac motor for cycling applications. One of the best features of a VFD is that it allows a controlled start and stop of a process. Some considerations include sizing the drive for the output current (torque), including peak torque. Also, dynamic braking may be a required option to stop overhauling loads. Overhauling loads occur when the load inertia tries to overspeed the motor during the deceleration phase of the duty cycle.
When this happens, the motor becomes a generator and “pushes” power back into the drive causing the dc bus voltage to rise, which could cause a drive fault. To avoid this, there is a device called a snubber resistor kit that can be added to most drives. It will absorb the regeneration power created by the motor.
Gear reducer considerations
Many times, highcycle applications require a gear reducer. The main reason for using a gear reducer is to increase output torque and decrease output speed. Also, the reducer decreases the reflected load inertia by the square of its gear ratio. So a 2:1 gearreduction ratio will decrease the load inertia by 4. Gear reducers that are subjected to cycling and reversing loads must also have higher service factors. Check with your gearing manufacture for sizing information.
Conclusion
Often, it becomes necessary for an engineer to properly size industrial motors to perform a duty cycles that are cyclical in nature with many starts and stops. This type of general motion control application requires special consideration. However, following some general guidelines, it may be possible to apply VFDrated motors to perform tasks normally reserved for higher performance ac motors or servomotors. Properly applied, NEMA induction motors can be used in general motion control. They can be an economical solution normally filled by more costly traditional methods.
More Info:
The author is available to answer questions about this article. Mr. Spees can be reached at 8642812486 or cjspees@powersystems.rockwell.com . Find more information on highcycle motor applications at reliance.com or plantengineering.com . Article edited by Jack Smith, Senior Editor, Plant Engineering magazine, 6302888783, jsmith@reedbusiness.com .
Capabilities of small NEMA AC induction motors
Most small ac induction motors are limited to about 10 to 15 starts per minute (across the ac line start) under very lightly loaded conditions as outlined in this table. Inrush current causes heating that must be dissipated by running the motor with an adequate amount of time between starts. Failure to do so will cause motor insulation failure. The number of starts of which a motor is capable depends on the inertia of the externally connected load, the friction load, and duty cycle. There may be some variance from the parameters shown in the table.
Frame Type  HP  Speed (rpm)  Number of starts per minute 
TEFC  1800  15  
TEFC  1  1800  15 
TEFC  2  1800  12 
TEFC  3  1800  10 