Bill Colton
Baldor Electric Co.
Commerce, California
Technical Services Committee Member

When selecting a drive for an ap­plication, there are many technical and commercial issues to explore in arriving at a choice. The scope of this article is to consider some of the techni­cal issues involved.

Let us assume that the motor is already in place and the customer has asked you to help select an adjustable speed drive. You have determined that the application is “constant torque.” (Constant torque is the same amount of full load RMS [root mean square] torque over the entire speed range required by the application.) 

Determine power available
The first thing is to determine the power available for the control.

1. What is the voltage level?
2. What are the size and impedance of the transformer feeding the drive?
3. Is there a good ground in the system?
4. Does the feed have a transformer with a wye secondary and is the ground through the center tap of the wye?

The voltage level will help in select­ing the proper drive as most drives are single voltage input with a wide tolerance. The impedance level will indicate if an additional inductor or transformer is needed to protect the drive from high voltage spikes. If there is not a good ground, the customer will need to install one to avoid long term (and possibly short term) trouble. If the power is not balanced with respect to ground the customer may have control issues and possibly capacitor failures in the drive. An isolation transformer with a grounded wye secondary should be recommended.

Application
Next, you need to find additional information about the application. 
What is the expected speed range of the motor during operation?
General rules of thumb – 
    Up to 10:1 – good for a V/Hz drive
    Up to 40:1 – good for a vector drive without an encoder or speed sensor    
    Greater than 40:1– Good for a full vector drive (with feedback)

When we see the term V/Hz (Volts per Hertz) this is an indication of an inverter that has a programmed spe­cific voltage for each frequency. When a frequency is being generated by the inverter, it will be done with this voltage. There may be certain minor exceptions, but this is the intent of the V/Hz drive – which is also termed a scalar drive. 
Note that there may be other consid­erations that will drive you to a vector drive – but we will consider those later.

Overload capability
The next thing to be determined is “how much overload” is required in the application. Overload is the torque (or current) above full load (RMS) that is required in the application. This may be required for a couple of reasons. 

The first is to accelerate in a given amount of time. To accelerate a load, coupling and motor inertia in addition to the load will most likely require greater than full load torque. The faster the required acceleration, the greater the required overload. The typical for­mula used for calculating this load is:

Note: WK² is the moment of inertia expressed in a weight over a distance squared. It is typically lb-ft² or kilo­gram-centimeters². It is the product of the weight of an object and the square of the radius of gyration. The weight includes all of the moving compo­nents (motor, coupling, load, pulleys, sheaves, etc). If the application is using some type of speed reduction (pulley, reducer, etc.) it is necessary to calculate the reflected inertia to the motor.  

Reflected inertia calculation: 

Example: if the inertia is 15 lb-ft² and we have a 4:1 speed reduction, the reflected inertia to the motor would be 15(1/4)² = 0.9375 lb-ft². Since this is the torque required to “accelerate,” it needs to be added to the load torque.

The second type of overload has to do with the application. As an example, picture a conveyor with a moving belt. A large load is dropped on the moving conveyor belt. This shock load will create a momentary additional load accelerating the mass almost instanta­neously. The effect of this will be seen in the current signature of the motor.

These peaks can be measured with a meter that has a peak hold function. Measuring this will help ensure selection of a drive with adequate peak current. 

Drives have both a constant and a peak current rating. Both of these ratings must be above the application require­ment. Do not use a drive’s “horsepower” rating to size the drive; use the required current from the motor and application. This is especially true on motors with an unusual base speed or a high (e.g., 8 or more) pole count motor.

Base speed (in an AC drive) is the lowest speed (frequency) that pro­vides full voltage to the motor. This is “typically” done at either 50 or 60 Hz, although there are (unusual) instances where something other than 50 or 60 Hz is desired. This will require a specially wound motor, and the drive will need the ability to select the base speed. It is used quite often in high speed motors (8,000 – 300,000 rpm) where the maxi­mum speed is the only operational point in the profile. It may also be used to replace a very low speed motor with a more efficient 4- or 6-pole motor.

Environment
The next issue to confront is the type of environment to which the amplifier (inverter or vector drive) will be subjected. Typically, the drives come in chassis mount, enclosed chas­sis mount (IP00), NEMA 1, NEMA 12 or NEMA 4X. 

If the environment will negatively affect the life of the drive, or be unsafe due to electrical exposure, an appropri­ate enclosure (and enclosure size) for the drive should be determined and provided.

Type of control
The next item to determine is what kind of control is needed. This will include discussing with the customer the current control scheme (if any) and the desired control scheme.

This decision process should in­clude, but isn’t limited to, how the drive will start and stop (two wire control or three wire control), safety interlocks, motor protection interlocks, upstream control information, type of control (torque, speed, process, position and/or load sharing), analog inputs and outputs, digital inputs and outputs, and feedback types (speed and process).

A control with the customer’s required inputs should be selected, which should include a discussion with the customer about available in­put selections. The discussion may also include ancillary devices that can make up for something not included on your control choice (such as isolated input cards, encoder splitters, and pulse train to analog devices).

Type of communication
Requirements for any bus com­munication need to be determined. This can be for programming, control or just process monitoring. There are many different bus languages pres­ent in today’s drive world. There are some standards, but there are a lot of manufacturer specific languages as well. Some of the more common bus languages are:
BACnet         Modbus
Device Net    Modbus Plus
EtherCat       Modbus RTU
Ethernet        Modbus TCP/IP
Ethernet IP    Power link
Lonworks       ProfiBus
Metasys N2   Profibus DP

Braking
Application requirements for brak­ing also need to be determined. Typi­cally braking in a control comes in three varieties: DC injection braking, regenerative braking (to a shunt or resistor, and line regenerative braking).

Requirements for braking:

1. Anti-windmilling – the load may need to be brought to a stop before beginning control.
2. Emergency Stop (E stop) – bring­ing down the motor and load to a quick stop in an emergency. 
   Caution: Make sure you dis­cuss with your customer what Emergency Stop means. In some locations, they want the output current to the motor interrupted. This is typically called “disen­  abling” the drive. Make sure the term is defined with the custom­er and that you are both comfort­able with the definition.
3. Controlled Stop – slowing down the motor and load to a controlled stop during normal operation.
4. Bringing the motor from a higher speed to a lower speed in a rela­tively quick time (determined by the inertia to the motor).
5. Pulsating loads – these are often forgotten; but a high inertia load that is quickly removed will allow the rotor to accelerate (spring ahead) and regenerate power back into the control, often  times causing nuisance tripping unless there is a means to absorb the energy.
6. Overhauling Loads – the load is driving the motor in the direc­tion the motor is trying to rotate, such that the load is pulling the motor “ahead” and the motor has to hold back to keep from  running away. (Think of cruise control in your car; even though you are going forward, you may be on such a downgrade that you have to brake to maintain speed, or you would run away.)
7. Reverse Driving – this is where the motor is used as a load or a brake such as a dynamometer or a valve control, or hoist floating the load.

DC Injection Braking is used for anti-windmilling and the end of an E-Stop or Controlled stop on inverters and some vector drives without encod­ers or speed sensors.

Regenerative Braking (to a shunt or resistor) is most common and used for E-stop, Controlled Stops, Reducing speeds, Pulsating loads and some Overhauling loads. While it can handle continual overhauling loads, and continual reverse driving, it may become too large, expensive or inefficient.

Line Regeneration is used for con­tinuous overhauling loads, reverse driving loads and for recovering energy for slowing down or stopping high inertia loads.

Speed regulation
Another issue in the selection of the amplifier (inverter or vector) is speed regulation. If it is a speed regulated system, how close does it need to be to the commanded speed?  If it is on an inverter it has the nor­mal slip of the motor for creating the torque. So, if it is a 1750 rpm motor, it takes approximately 50 rpm of slip to create full torque. Those 50 rpm are a constant (not the percentage), so when trying to run at 300 rpm (synchronous speed), the motor will run approximately 250 rpm (300 – 50) under full load. If more precise speed is needed a sensorless vector drive may need to be investigated. The regulation is actually dependent on the manufacturer that is selected; but typical specifications are within 1% of set speed (the desired run speed). With a full vector drive the motor should be within a couple of rpm of the desired speed.

Other considerations
If the customer wants a motor selected as well as the amplifier there are a few other application consider­ations to incorporate. 

The speed range needs to be consid­ered. Remember that the cooling fan be­comes less effective at lower speeds and requires more power (windage) and becomes extremely noisy at elevated speeds. An alternate cooling system may be used with an external air source such as TEBC (Totally Enclosed Blower Cooled) or Pipe in and Pipe out. In some more exotic high speed applications, liquid cooling is employed. The motor needs to be protected from dirt and grime as in any application. In addition, the cooling must be effective for the speed range. Some motors are simply oversized (physically), employing a TENV (totally enclosed, non-ventilated) cooling scheme.

To improve motor reliability some type of bearing current mitigating system can be employed, such as isolated bearings, or a shaft ground­ing system. 

Operation at other than 50 or 60 Hz base speed
For some applications a customer may want something other than a traditional 50 or 60 Hz base speed. For example, consider the case of an application requiring constant torque up to 2500 rpm.   

There are several methods of handling this. 

One is to employ a speed increaser on a 4-pole motor, or a speed reduc­tion on a 2-pole motor. These speed changers might be gearboxes, or sheaves, chain, etc. The motor would run up to the base speed, but the load would run to 2500 rpm.

Another is to use a 4-pole motor and increase the frequency. Rpm = 120 x frequency/poles. 2500 = 120x/4 = 83 Hz. This will provide constant horsepower above base speed for ap­proximately 1.5 – 2.0 times base speed (depending on the breakdown torque of the motor). This means your base horsepower will still be available at 2500 rpm.

The following is an example of applying the method for a 100 hp, 60 Hz motor application.
100 hp = T x 2500 rpm / 5252 – Therefore, torque will be reduced to 72% of the 60 Hz base speed torque. Since this is a constant torque appli­cation, if 100 hp was required at 60 Hz, it will also be required at 83 Hz. Therefore hp = (300 lb-ft x 2500)/5252 = 142 hp. So, it would be appropriate to select a 150 hp, 4-pole motor. 

A third approach is to use a 2-pole motor operating at 42 hertz (2500 rpm on a 2-pole motor). Once again hp = T x rpm / 5252 – meaning that horse­power at 60 Hz will be 44 % higher than at 42 Hz. This isn’t a real good solution either.

The best design approach is to set the base speed at 2500 rpm. This can be done in either a 2- or a 4-pole de­sign. If rapid acceleration/decelera­tion is required, a 2-pole rating may be chosen due to the lower inertia rotor. If a lower speed is needed in the application, a 4-pole motor will be better controlled at the very low speeds. 

If you choose a 4-pole motor, the base speed becomes 83 Hz. Assuming a 460 VAC system, the base speed will be 460 VAC at 83 Hz (2500 rpm) and volts/hertz = 5.54. A motor wound for this volts/hertz base speed must be selected and the inverter or vector drive must be set for this base speed to ensure that the proper power is applied to the motor terminals.

This same concept is crucial in high speed test stand motors. For example, consider an application requiring 50 hp at 12,000 rpm with a 460 VAC power source.

A 2-pole motor is selected. At this speed a bipolar shaft driven fan should be avoided and a sepa­rately powered blower would be preferred; or a liquid cooled motor could be provided. 

RPM = line frequency (Hz) x 120 /Poles  Therefore:  
RPM x Poles/120 = line frequency (Hz) 
and 
(12,000 rpm) x (2 Poles) / 120 = 200 Hz

An inverter that can output a mini­mum of 200 Hz is required. The motor should be wound for 460 VAC/200 Hz = 2.3 volts / hertz. Special consider­ation will need to be made to ensure the rotor can withstand the centrifugal force, and that the bearings (and lubri­cant) can survive the operating speed.

This motor will be smaller than you might expect. The torque at 12,000 rpm for 50 hp = (50 hp) x (5252)/12,000 = 21.88 lb-ft. The required frame size would be approximately a 215 T (132 M). 

This motor will draw approxi­mately 200 amps – so the amplifier would have to be capable of a con­tinuous current of at least 200 amps. Note that a normal 50 hp motor would only draw about 60 amps. The special volts-per-hertz brings the current way up (much like a lower voltage motor needs higher current). 

At this point the question is raised: Why go to all that work? Why not just oversize a 2- pole motor by 1 rat­ing and bring the frequency up? The calculations indicate that the motor will be much too weak at the high speed. To illustrate that, reverse the calculations.

In order to get 50 hp at 12,000 rpm with a base speed of 3600 rpm assume a breakdown torque high enough to achieve a 2:1 constant horsepower range (all the way up to 7,200 rpm). Cal­culate down by the square of the speed increase from that point, meaning the horsepower at 7200 rpm needs to be (12,000 / 7,200)2 x 50 = 138 hp. Thus the motor would be sized up to 150 hp.This motor would be in a 445T (280M) frame which would be extremely hard to keep together at 12,000 rpm.  Note the current for a 150 hp motor is going to be approaching the 200-amp mark just like the special motor! And now there would be a very large machine moving at a high velocity. The special base speed approach is much better.

Conclusion
By helping your customer become acquainted with the options for con­stant torque applications you will show your expertise and knowledge in both products and an understand­ing of the customer’s needs. In do­ing so, you should be able to help the customer select, install, set up and apply the products needed for a successful application. You will also show your ability to apply new tech­nology to help the customer become more productive and successful. This can also help transition you from being a supplier to being a partner with the customer in automating and transitioning their business for tomor­row’s competitive environment. If you don’t help your customers keep up with new technology applied at their facilities, they will at some point cease to remain in business. Or they may transition to another firm that is able to help them stay competitive. It is better to be in the driver’s seat and help your customers remain suc­cessful.

Categories: Design/theory/application, Drives/controls

Tags: Variable frequency drives Adjustable speed drives

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