AMAZON multi-meters discounts AMAZON oscilloscope discounts
<< cont. from part 1
IGBTs (Insulated Gate Bipolar Transistors)
IGBTs have fast become the power semiconductor of choice with drive manufacturers from fractional to well over 1000 HP. The IGBT carries all of the advantages of the bipolar and Darlington transistors, but offers one additional feature-increased switching time.
As you will learn in later sections, the switching time of the power semi conductors will have a major role in shaping the output waveform. The faster the switching time, the more exact, electrically, the drive output, compared with pure sine wave power from the utility. An additional benefit of faster switching times is the reduction of audible noise generated by the motor. IGBTs can switch in the 12-16 kHz range, compared with standard transistors of 1-3 kHz.
Higher switching frequencies mean the audible noise is moved out of the normal hearing range. The motor appears to be quieter, compared with a motor operated on a transistor drive. FIG. 33 traces power device switch times and the number of circuit cards required to operate the devices.
As seen in FIG. 33, switching times of the SCR or GTO are in the 3- to10-µs range (500 Hz-1 kHz). On the other hand, switching times of the bipolar transistor are in the 1- to 1.5-µs range (1-4 kHz). The IGBT has the fastest switching time of all: 0.1 to 0.8 µs (5-16 kHz).
The SCR and GTO gained popularity in the early 1960s, and superseded thermionic power devices such as vacuum tubes. AC drives of that era required 12-20 circuit cards to gate on the power devices, turn them off, and verify they were off.
In the early 1980s, bipolar and power Darlington transistors gained popularity because of the switching times as well as the need for only 3-7 circuit cards (the base driver circuitry that caused the transistor to conduct was smaller, therefore requiring less circuit card space).
The currently used IGBT gained acceptance in the early 1990s and caused a further decrease in circuit card requirements. In many cases, hybrid circuitry is included on a power control card, which means the total circuit card count can be down to 2.
The circuitry needed for power semiconductor switching will be examined in greater detail in a later section.
FIG. 33.
Power device switching times and circuit cards required (ABB Inc.)
Mechanical Principles
Load Types
There are literally hundreds of applications for AC and DC drives. Though it would be almost impossible to cover all possible types, we will review some of the more common applications.
When we look at applications, it is important to consider the load connected to the motor. The load is the determining factor in what type, style, size, and power producing device the motor should be. The drive is the device that will control the motor. In simple terms, it is best to start with the application (load), work backward to the motor (prime mover), then to the drive (controller), and then to the automated or manual control (speed control or PLC).
There are three major categories of mechanical loads -- Constant Torque, Constant Horsepower and Variable Torque. Any drive application will fall into one of these three categories. When reviewing the following application types, keep in mind that the drive's primary responsibility it to control the application, protect the motor, and protect itself.
Constant Torque Applications
These types of applications will make up the majority of what we would consider to be industrial applications. FIG. 34 shows a graph of the characteristics of a constant torque application.
FIG. 34. Constant torque application
As shown in FIG. 34, we can see that torque required in this application is 100% and remains constant from zero to base speed (100%). In other words, the rated torque (100%) is required to operate this application, no matter what the speed. In most cases, this application will be operated from only zero to base speed.
It is also evident that this type of application possesses a variable horsepower characteristic. In a constant torque application, horsepower is directly proportional to the speed (i.e., at 50% speed, 50% HP is required, etc.). While the horsepower consumed by the motor varies directly with speed, the torque required to power the load remains constant.
In reality, the load may vary from above rated torque (100%) to below rated level and anywhere in between, until the motor attains full speed (in many cases, base speed). In many cases, as much as 150% torque is required to break the load away from a stand-still. The 150% dashed line indicates a possible overload requirement from a motor and drive, if connected to this type of load.
Many applications fall into the constant torque category. The standard belt conveyor shown in Section 1 (Figure 1) is a prime example. A ball mill in a cement plant is another example. FIG. 35 shows this process of grinding aggregate into powder for cement.
FIG. 35. Ball mill in a cement plant
In FIG. 35, the degree of grinding is controlled by the speed of the AC drive. Speeds of less than 120 rpm cause the ball mill cylinder to internally grind the large raw material into smaller particles and eventually into fine granulated powder. The torque required from the motor can be constant from zero to the maximum speed. Overload torque is needed to initially start the mill rotating-with the amount of overload dictated by the amount of material inside the mill at start-up.
Extruders also fall into the constant torque category. A typical extruder, such as a screw extruder for plastic, often requires overload capability, especially when the raw plastic has cooled somewhat and is less than fluid.
FIG. 36 shows a plastic screw extruder. In FIG. 36, the AC drive provides variable speed motor operation. It drives the screw vanes at speeds regulated by the temperature of the plastic or the rate of production and die size (output extrusion hole). AC drives provide a high degree of efficiency, while maintaining a constant rate of production accuracy.
FIG. 36. Plastic screw extruder
FIG. 37. Rotating feed table
Another application combines the use of a rotating feed turntable and a conveyor. FIG. 37 shows a Rotating Feed Table, which combines both liner motion (conveyor) and rotating motion of a turntable.
As seen in FIG. 37, the flow rate of the granules can be controlled by the motor speed that is dictated by the AC drive. This type of AC-drive application is highly efficient because of the ability of the drive to accurately regulate motor speed. With accurate motor-speed control, an optimum amount of granules can be distributed to the conveyor, which may then be fed to another process, such as heating or coating.
Constant Horsepower Applications
In this type of application, the horsepower required remains constant, while the torque drops off as a ratio of 1/speed^2. FIG. 38 shows the constant horsepower type of load.
Applications of this type operate in the above base speed and below base speed area. The fact that torque drops off as 1/speed^2 is an advantage in applications of this type. FIG. 39 shows a typical constant horsepower application-machine tools.
FIG. 39. Machine tool application (cross section of work piece)
As seen in FIG. 39, a slow speed is required when the tool bit is taking off large amounts of material. In doing so, a greater amount of torque is required (operating in the "Below Base Speed" section of FIG. 38). Once the work piece is at the approximate diameter, the speed is increased, and the tool bit is required to take off minimal amounts of material. This process produces a fine, smooth surface (operating in the "Above Base Speed" section of FIG. 38).
Another example of a constant horsepower application is a center driven winder. FIG. 40 shows another use of above and below base speed operation.
FIG. 40. Center driven winder (cross section of web roll)
As seen in FIG. 40, basically the same scenario exits as with machine tool applications (above and below base speed operation). Very low inertia exists when the take-up roll is first started. This translates to little torque required to start the winding process, thereby allowing a high speed. Once the winding process is started, increasing amounts of inertia are seen at the take-up roll, translating into more torque required, thereby causing the need for a slower speed.
The key to a proper winding process is to have adequate control of torque and speed on the web at all times. If little torque is available at the take-up roll, "telescoping" or bunching of the material is a result. This would be like that of a window shade that is let-go from only one corner. If too much torque is available on the take-up roll, stretching of the material could result, or even worse, a complete break in the web.
Since speed and torque are interactive in this application, it is not necessarily an application for a simple speed-controlled drive. DC drives have found their use in these types of winding applications because of the ease of controlling speed and torque simultaneously. Newer types of AC drives (vector or torque controllers) have also found their way into the control of this type of application in recent years. A more complete discussion of drives specifics in winder applications will be covered in later sections.
Variable Torque Applications
In this type of application, below base speed, torque required varies as speed. It should also be noted that the horsepower required varies as speed. Because of this relationship, this type of application is sometimes referred to as the cubed exponential application. FIG. 41 shows the characteristics of this type of application.
This type of application is a prime candidate for energy savings using AC drives. As seen in the example, to obtain 50% flow rate, only 1/8 or 12.5% horsepower, is required (½ × ½ × ½). It should also be noted that the torque and horsepower curves normally end at the 100% spot (100% speed and 100% torque, flow, and horsepower). Because it is easy to over speed a motor using a drive, users may have a tendency to increase speed above base, to obtain more CFM output. This is not normally recommended because of the mechanical limitations of the motor and the characteristics of the application.
FIG. 41. Variable torque application
To obtain increased speed above base, a higher output frequency (Hz) must be supplied by the drive. With several Hz output above base, the torque and horsepower curves would follow the natural path indicated by the dashed lines. As shown in the figure, just a few hertz increase in out put frequency would cause a much greater demand for motor torque and horsepower. However, flow would continue to increase in a somewhat "linear" manner.
This tremendous increase in torque and horsepower from the motor is now coupled with the fact that decreased motor torque is a natural result of above base speed operation. (Details on this characteristic will be discussed in Section 3 - General Principles of Operation.) The end result is a slight increase in drive output frequency, causing a very large requirement for torque and horsepower, at the same time motor torque is dropping rapidly. Nearly all of the fan systems engineered today take this phenomenon into account. It is rare to need to over-speed a fan, unless the above characteristics are addressed (duct work changes or over sizing the motor, or both).
Because HVAC fan systems are engineered for below base speed operation, typical overload requirements are set at 10%. AC drives that are rated to NEMA standards, will allow 110% current for 1 minute as an overload capability.
As shown in the discussion above, this type of application has two main examples: centrifugal fans and pumps. FIG. 42 shows a standard HVAC fan application.
FIG. 42. HVAC fan application
In this example, we see that the centrifugal fan is coupled to the motor by means of a belt. The air inlet allows air to enter the fan assembly. Depending on the status of the outlet damper (open, part open, closed), the fan will blow the amount of air dictated by some control system, typically an actuator. The area in the dashed box indicates a three-phase start contactor and motor overloads to protect the motor, in case of an over-current condition. Also shown is the fact that what's inside the dashed box can be replaced by an AC drive.
As can be imagined, the fan continues to run at full speed, 24 hours a day.
This is required, regardless of the status of the outlet damper, because there is no device controlling the speed of the motor. The device that controls airflow has no effect on the speed of the motor and fan.
This method of variable flow control is quite inefficient. FIG. 43 shows a graph of how the fan actually operates on the basis of the system shown in FIG. 42. The fan curve indicates the fixed speed fan. The system curve indicates the status of the outlet damper and also the size and shape of the sheet metal duct work.
To reduce the airflow in this fixed-speed system, the actuator must close down the outlet damper, thereby restricting air output. If 50% airflow is desired, the position of the outlet damper must be modified by the actuator, which changes the system curve (indicated by the dashed curve). The air output is reduced; however, the pressure in the system duct work has increased, placing additional load on the motor. This additional load translates to more horsepower required to operate with less airflow.
If the start contactor assembly were replaced by a variable-frequency AC drive, the outlet damper would be locked in full open position (indicated by the existing system curve). For every output frequency of the AC drive, a new fan curve occurs (indicated by the dashed curves).
FIG. 43. Operation of a centrifugal fan (Courtesy of ABB Inc.)
As seen, where the new fan curves cross the fixed system curve, a different percentage of flow output is achieved. It can also be seen that the amount of pressure in the duct work never exceeds 100%. In essence, if there is less pressure, the motor doesn't work as hard and energy is saved, com pared with full-speed operation.
This same principle works for centrifugal pumps. Reduced speed is a definite advantage when operating a pump, which leads to reduced energy use. However, when speed reduction is contemplated, the optimum efficiency of the pump must be taken into consideration. Pumps have "efficiency islands" around the crossing of the pump and system curves. Any energy savings gained can be lost if operation is done outside these efficiency islands because of greater pump inefficiency.
Speed, Torque, and Horsepower
Understanding speed, torque, and horsepower is essential to understanding mechanical systems. The following discussion will help you gain a basic insight into these relationships, and will help prepare you for a discussion of inertia.
Torque and horsepower are two very important characteristics that deter mine the size of the motor required for a particular drive application. The difference between the two can be explained using the illustration in the next section.
Torque
Torque is basically a turning effort. Suppose we have a worker that is required to raise a construction elevator by hand. FIG. 44 shows this example and demonstrates the process of applying torque.
This worker must supply the required torque to raise the elevator from the first floor to the second floor (FIG. 44).
FIG. 44. Using torque to turn a hand crank
In this example, the worker cannot raise the elevator even though he applies 1 lb-ft of torque.
Note: Torque = Force × Distance. In this case, the worker applies 1 lb of force on the crank, which is 1 foot in length. We say the torque is one pound times one foot or one lb-ft.
If the worker wants to raise the elevator, more torque must be applied to the hand crank. The worker has two choices: either place more force on the crank or increase the length of the crank. The worker chooses to increase the length of the extension on the hand crank. The elevator now moves, even though the worker does not apply any more force. FIG. 45 shows this procedure.
In this case, the worker has developed 5 lb-ft of torque, more than enough torque to turn the crank and raise the elevator.
If the worker turns the crank twice as fast, the torque remains the same.
Regardless of how fast the worker turns the crank, as long as the worker is turning it at a steady rate, the torque is unchanged.
Horsepower (HP)
Horsepower takes into account the speed at which the worker turns the crank. Turning the crank faster takes more horsepower than turning it slowly. Horsepower is a rate of doing work.
FIG. 45. Increased length of the hand crank
Note: By definition, 1 horsepower equals 33,000 ft-lb per minute. In other words, to lift a 33,000-lb weight 1 ft in 1 minute would take 1 HP.
By using the formula below, we can determine the horsepower developed by the worker. We know the system in FIG. 44 is developing 5 lb-ft of torque.
Formula:
HP =T× N/5252
T = torque in lb-ft
N = speed in rpm
The worker turns the crank at 5 rpm as shown in FIG. 44. By inserting the known information into the formula, we find that the worker is developing approximately .005 HP (5 × 5 ÷ 5252 = .0047). As seen by the formula, horsepower is directly related to the speed of turning the crank. If the worker turns the crank twice as fast (10 rpm), he will develop almost .010 HP.
Inertia
Inertia is the measurement of an object's resistance to change in velocity (speed). The measurement holds true whether the object is at a stop or moving at a constant velocity. By definition, inertia deals with the presence of mechanical "resistance." Inertia is resistant to changes in speed or direction.
Note: In electrical terms, an inductor opposes changes in current, due to resistance.
In mechanical terms, inertia opposes changes in speed, also due to resistance.
Inertia is the reason why your auto can't immediately accelerate from 0- 65 mph. As logic would dictate, a 4.5-L engine would be needed to power a large luxury auto, and only a 2.5-L engine would be needed for a small auto. If we apply previously learned concepts of speed, torque, and horse power, the larger the auto, more horsepower is needed to accelerate from 0-65 mph in 15 seconds. FIG. 46 shows a comparison of horsepower and speed, in relation to inertia.
FIG. 46. Relationship of inertia with speed and horsepower
As seen in FIG. 46, a higher amount of horsepower is needed to over come inertia and accelerate the auto, to achieve a speed of 65 mph in 15 s (example A of the figure). The previous would be true if the engines were sized properly in both the large and small autos. However, if the small car needed quicker acceleration capability, we would use the same 4.5-liter engine as in the large auto. The small auto would achieve quicker acceleration because of its oversized engine (example B of the figure).
It should be noted here that the principles of inertia, torque, horsepower, and speed apply to an electromechanical motor. With that in mind, it should also be noted that the extra horsepower of the 4.5-L engine would be needed only on acceleration. A large amount of horsepower is needed for quick acceleration, but only a small amount is needed to keep the auto moving at a constant speed.
This principle is similar to pushing someone on a swing. A large amount of effort is needed to get the person started in a swinging motion, but only a small tap every cycle is needed to keep the person going. The amount of inertia is greater when at a standstill, than when the swing is already in motion.
In relation to rotating machinery, a flywheel would be termed a high-inertia load. It would take a large amount of energy to quickly bring the fly wheel up and back down in speed. The main reason is that high inertia loads have high resistance to change (in this case, change in speed).
Inertia has a significant impact on which size of drive is chosen to accelerate a high inertia load. We will take a brief look at some commonly used formulas to calculate acceleration and inertia.
The following information will assist you in your general understanding of inertia-related calculations. The logical place to start is to label terms and units of measure. When looking at inertia, two different ways of labeling are commonly used: WK^2 and WR^2. Both terms relate the formula to inertia.
WR^2 refers to the inertia of a rotating object that was calculated by assuming the weight of the object was concentrated around the rim-at a certain distance. That distance is termed R and is the distance from the center of the object. (By strict definition, WR^2 is the weight of an object [W] times its radius [R] squared.) In this definition, inertia is that property by which an object in motion will stay in motion, until acted upon by another force.
An example of an inertia calculation of this type is a flywheel.
WK^2 refers to the inertia of a rotating object that was calculated by assuming the weight of the object is concentrated at some smaller distance. The smaller distance is termed K. (By strict definition, WK^2 is the weight of an object times its radius of gyration value squared.) Examples of this type of inertia calculation would be cylinders, pulleys, or gears.
The units of measurement for inertia is the lb-ft^2 (pronounced pound-feet squared). Another unit sometimes found in inertia calculations is in-lb sec^2 , used in calculations for the moment of inertia (motion control applications using servos). Initially, calculations using lb-ft^2 will be used to demonstrate calculations of inertia.
The next section will present some commonly used formulas using torque, horsepower, inertia, and time. These formulas will be used as examples for calculations based on rotating machinery, which need to be considered before drive selection can begin.
Accelerating Torque
The following formula calculates the torque required for rotating motion to begin:
where:
T = acceleration torque in lb-ft
WK^2 = total reflected inertia (total of motor, gear reducer, and load) in lb-ft^2
? N = change in speed required in seconds t = time to accelerate the total system load
A term above that may not be familiar, is the term reflected inertia. Basically, this term is used to describe the inertia found at the motor shaft, and is a standard term found throughout the rotating machinery industry. A speed reducer (gear box or belt coupler) changes the inertia that the motor shaft actually sees. More on this subject will be discussed later in this section.
Acceleration Time
The next formula is a rearranged version of the previous one, only it allows for the calculation of the time needed to accelerate, given a specified amount of torque, inertia, and change in speed.
where:
t = time to accelerate the total system load WK^2
= total reflected inertia (total of motor, gear reducer, load) in lb-ft^2
delta N = change in speed required in seconds
T = acceleration torque in lb-ft
The following are examples of how inertia is calculated and how much time is needed to accelerate a machine with the specified requirements.
One note regarding calculations should be added here. When performing inertia calculations, two measurement units are commonly used: lb-ft^2 and in-lb-sec^2. For the most part, many calculations are defined in lb-ft^2 , which is the units of measure for WK^2 or WR^2. However, in many motion control (servo) applications, inertia is defined in terms of in-lb-sec^2 , which is the units of measure for the moment of inertia (J). When performing inertia calculations, be consistent with the formulas and units of measure used. For most practical motion-control applications involving inertia given as J, the following conversions can be used to convert lb-ft^2 to in-lb sec^2 and vice versa.
To convert a calculated answer of lb-ft^2 to in-lb-sec2 , divide the answer by 2.68.
To convert a calculated answer of in-lb-sec^2 to lb-ft 2 , multiply the answer by 2.68.
Solid Cylinder Inertia Calculations
To calculate the inertia of a solid cylinder, the following formula is used.
WK^2 = .000681 × ? × L × D4 where:
WK^2 = inertia of a cylinder in lb-ft^2
rho = density of cylinder material in lb/in^3
L = length of the cylinder in inches
D = diameter of the cylinder in inches
Note: The units of measure for WK^2 are in lb-ft^2. Refer to FIG. 47.
FIG. 47. Inertia of solid cylinders
For an example of a solid cylinder, let's consider a solid aluminum roll that has a length of 72 inches and a diameter of 18 inches. Its inertia would be:
WK^2 = .000681 × 0.0977 × 72 × 184
WK^2 = .000681 × 0.0977 × 72 × 104976
WK^2 = 502.62 lb-ft^2
Now that the inertia has been calculated (WK^2 ), the time it would take to accelerate the system can be determined as follows:
From 0 to 1200 rpm, with an acceleration torque available of 30 lb-ft, the formula for acceleration time will be used:
Therefore:
If the time calculated is too long, then the easiest item to control would be the amount of available acceleration torque-meaning the motor. The motor would have to be upsized to have increased available acceleration torque. Motor torque will again be discussed in the section on DC and AC motors.
Hollow Cylinder Inertia Calculations
To calculate the inertia of a hollow cylinder, basically the same formula can be used, but without the inertia of the hollow section. The formula would be:
WK^2 = .000681 × ? × L × (D2 4 - D1 4 )
FIG. 48 shows a hollow cylinder with the formula parts.
where:
WK^2 = inertia of a cylinder in lb-ft^2
rho = density of cylinder material in lb/in^3
L = length of the cylinder in inches
D2 = outside diameter of the cylinder in inches
D1 = inside diameter of the cylinder in inches.
If we take the previous example and only solve inertia for the hollow cylinder above, we would go through the following calculations. We will assume the same information regarding the cylinder:
FIG. 48. Inertia of hollow cylinders
Solid aluminum roll has a length of 72 inches, an outside diameter of 18 inches, and an inside hollow diameter of 12 inches. Therefore, we would go through the following calculations:
WK^2 = .000681 × 0.0977 × 72 × (184 - 124 )
WK^2 = .000681 × 0.0977 × 72 × (104976 - 20736)
WK^2 = .000681 × 0.0977 × 72 × 84240
WK^2 = 403.34 lb-ft^2
If we use the above inertia, we can again find the time it would take to accelerate the system. We will use the same example:
From 0 to 1200 rpm, with an acceleration torque available of 30 lb-ft, the formula for acceleration time will be used:
Therefore:
We have determined that the amount of time to accelerate the hollow-cylinder system is definitely less, compared with a solid cylinder. However, it should be noted that the amount of time reduction is not as much as expected. With 66% of the cylinder removed, there was only a reduction of about 13 s of acceleration time. This indicates a characteristic of inertia-the largest amount of inertia is concentrated around the rim, as opposed to the center part of the object.
The same formulas would be used to determine the inertia of pulleys, sheaves, and other rotating objects. Simply break up the objects into solid and hollow cylinders and apply the formulas previously discussed. Then add all the inertia components calculated and determine the time required for acceleration of the system.
Reflected Inertia
One additional note should be stated about inertia. The term reflected inertia is the inertia actually found at the motor shaft. This term is standard throughout the industry. A speed reducer (gear box or belt type of coupling device) changes the inertia that the motor actually sees. FIG. 49 shows this phenomenon.
FIG. 49. Reflected inertia of a system
As seen in FIG. 49, if an adjustable speed drive was sized for an inertia of 12 lb-ft^2 , it may be much larger than actually necessary. Reflected inertia also includes inertia of the actual motor.
Power transmission devices (gear boxes, belts, and pulleys) serve to make the motor's job easier. In our next section, we will look at gear boxes and speed reducers in more detail and determine what relationship exists between speed, torque, and horsepower.
Mechanical Devices
Introduction
As you may recall in an earlier section on horsepower, a worker had to apply a certain amount of torque and horsepower to raise the elevator.
The job would have been much easier if some sort of gearing device was applied to the crank, thereby changing the ratio of one revolution of the pulley. Mechanical devices such as belts, sheaves, pulleys, and gear boxes allow this change in ratio.
An easy place to start the discussion of ratios is the common 15-speed bicycle (FIG. 50).
FIG. 50. Gear ratios-first gear
To get enough torque to start the bicycle in motion, the rider would place the gears into first gear. Notice that the rear wheel sprocket (first gear sprocket) is smaller than the pedal sprocket.
This means that the rear gear sprocket would make five revolutions com-pared with one revolution of the pedal sprocket. As seen in FIG. 50, first gear indicates a 1:5 ratio (one revolution of the driving shaft, in this case the pedal) compared with five revolutions of the output shaft (rear wheel sprocket).
FIG. 51. Gear ratios-fifteenth gear
If we assume that the rider did not place the bicycle into first gear and tried to start out in fifteenth gear, we would have a scenario seen in FIG. 51.
As seen in FIG. 51, the ratio of the rear sprocket to the pedal sprocket has changed to 1:1. Basically there is no mechanical advantage for the rider and therefore no reason to own a 15-speed bicycle. Essentially, the gear arrangement on a bicycle reduces the amount of reflected inertia to the pedal sprocket. This makes the job of starting the bicycle in motion much easier, with less torque applied by the rider. We could consider this action that of breakaway torque required to start the system in motion.
The only item that changed in this example is the amount of inertia reflected back to the pedals, by way of the gears. The weight and inertia of the rider remained constant, no matter what gear was selected. We will use these same principles and apply them to the following power trans mission devices.
Belts and Pulleys
Flat and V-Belts
A common arrangement for belt-pulley type transmission systems is the flat belt shown in FIG. 52. Belt-pulley arrangements are sometimes called belt-sheave, meaning that there is more than one pulley attached together, giving several selections of gear ratios.
FIG. 52. Belt-pulley arrangements
For smaller pulley diameters, the flat belt provides low weight and inertia to the power system. A disadvantage of the flat belt is that they rely on maximum friction with minimum slippage to function properly.
The v-belt arrangement allows more surface area with the inside of the pulley, which increases the friction capability (FIG. 51). V-belts pro vide shock absorption between the drive and driven shafts due to its inherent v-shaped design. According to some machinery designers, a speed ratio of 7:1 is satisfactory for a v-belt system. Most flat and v-belts are constructed of nylon or durable fibers imbedded in rubber.
Both these types of systems require a certain amount of friction to operate properly. In many cases, some type of tensioning device (spring arrangement) is used to keep the belts tight and efficiency at maximum.
In some applications, a precise belt system is needed. A specific revolution in the driven shaft would produce a specific revolution in the output shaft, with little to no slippage to reduce efficiency. This type of belt system would be the synchronous belt system.
Synchronous Belts
The synchronous belt system is sometimes known as the timing belt system and is shown in FIG. 53.
FIG. 53. Synchronous belts system
The synchronous belt system is basically a flat belt with a series of evenly spaced teeth on the inside of the belt. This system provides the advantages of a belt, with positive grip features and little slip.
Chains and Sprockets
Chain Drive
Chain drives are often used in the same manner as synchronous belts and provide positive synchronization between transmission shafts, especially when large amounts of torque are required (FIG. 53).
The stretching action of belt systems is not a problem for a chain drive.
However, depending on the number of teeth in the drive gear sprockets, slight pulsations (whipping action) could occur in the drive gear sprockets.
This can cause a noise or vibration problem if the pulsations are great enough.
Couplings, Gearboxes, and Speed Reducers
In many cases, some positive method of connecting two shafts together is required when the shafts operate along the same centerline. The two most common types of couplings are rigid and flexible couplings.
The two most common types of rigid coupling are the flange and the sleeve type.
Flange Coupling
As the name would imply, the flange type uses two metallic flanges slipped over the ends of the two shafts and bolted together. Each individual flange is either pressure fit over the shaft keyway, or locked onto the shaft by some other method such as a set screw (FIG. 54).
FIG. 54. Flange and sleeve-type couplings: Flange Type; Sleeve Type;
Set Screws Fastening Bolts
Flexible Couplings
The function of the flexible coupling is the same as the rigid coupling. The difference between the two is the flexing or twisting capability of the coupling parts (within limits).
A certain amount of flexing is expected with this type of coupling. Flexing may be caused by: misalignment of the shafts, shaft end movement due to insufficient motor and machine mounting, or simple vibration between the connected devices.
The flexible part of the coupling could be considered a "mechanical fuse." If the flexible part fails, it is a sign of mechanical drive trouble in the system. The mechanical system should be thoroughly checked out before the flexible part is replaced, which is the same for an electrical system replacement. We will take a brief look at the two types of flexible couplings: mechanically flexible and elastically flexible.
Mechanically Flexible
These types of couplings obtain their flexibility from the rolling or sliding of the mating parts. As expected, these parts do require some type of periodic lubrication. Typical examples are gear, chain, and sprocket and disc types, shown in FIG. 55.
Elastically Flexible
This type of coupling (sometimes referred to as elastomeric) obtains its flexibility from the stretching or compressing of a material such as rubber or plastic. The sliding or flexing that takes place is minimal, and lubrication is not required. These types of couplings come in designs such as jaw, clamped, or unclamped donut, and the tire (FIG. 56).
FIG. 55. Mechanically flexible couplings
FIG. 56. Elastically flexible couplings
FIG. 57. Speed-reducer characteristics
Gearboxes and Speed Reducers
Gearboxes and speed reducers transmit mechanical power from the motor shaft to the driven load. A simple speed reducer contains two different sized gears, called spur gears. Spur gears are straight teeth cut parallel to the axis of rotation. Speed reducers may have more than two gears, which provide several fixed speed outputs. If that is the case, some type of clutch arrangement will exist to change gears. FIG. 57 shows the gear relationship within a speed-reducer case.
As seen in FIG. 57, this type of device provides an efficient means of transmitting positive speed, direction, and torque. In many cases, this device will change speed, with a corresponding change in torque or output direction. The gear reducer acts as a torque amplifier, increasing the torque output by a factor proportional to the ratio, less an efficiency factor.
As seen in FIG. 57, if a 1150-rpm motor delivers 4.5 lb-ft of torque to the input shaft, then 153 lb-ft of torque is present at the output shaft (given the efficiency of 85%). The formula used is lb-ft × reducer ratio × efficiency of the speed reducer. Therefore, 4.5 × 40 × 0.85 = 153 lb-ft of output torque.
To change the output speed by more than one fixed ratio, more than two gears are required. The same is true if the direction of rotation of the out put shaft needs to be identical to that of the input shaft. The type of gear used depends on the application (i.e., horizontal axis of rotation, vertical to horizontal axis of rotation, etc.). FIG. 58 shows several types of gears used in common speed reducers.
As shown in FIG. 58, the speed reducer is operating at 85% efficiency.
A brief look at mechanical efficiency would be helpful in understanding the actual output of the device. As previously stated, efficiency is the ratio of output power to input power and is expressed in a percent. Therefore the formula for efficiency would be:
(%) Efficiency = output × 100 /input
When considering the efficiency of a belt type of device, the more friction available, the less slippage that occurs, and the higher the efficiency. The output shaft turns simultaneously with the input shaft.
FIG. 58. Speed reducer gear types
In belt systems, several factors aid in decreasing total system efficiency: (1) Losses that occur due to friction of the rubber and cords as the belt stretches and flexes. (2) Friction of the belt as it enters and leaves the pulley or sheave. (3) Bearing friction caused by excessive tension on the pulley (or sheave) causing a high level of drag in the bearings.
Efficiencies of belt transmission systems can be as high as 90-98%. When proper maintenance and belt tension is maintained, these efficiencies can be sustained throughout much of the system's usable life.
Section Review
Electrical principles play a major role in understanding how electrical power is modified and controlled in a drive unit. All electrical circuits have three main factors: current, voltage, and resistance. Current is actually the flow of electrons that cause the work to happen in a circuit. Resistance is the opposition to current flow, which is present in any circuit (all matter in almost all modern electrical circuits have some resistance). Voltage is the electrical pressure that tries to overcome resistance.
Two types of voltage are available: direct current and alternating current.
Direct current is typically found in battery circuits and its output value does not fluctuate, until the battery goes dead. Alternating current is typically found as a household and industrial power source. The output value fluctuates depending on the time base of transmission (positive to negative to positive output). Frequency is the number of times complete cycles are seen in a power system (e.g., 60 Hz). Two types of AC are available: single phase and three phase.
Electromagnetism is the ability to produce a magnetic field through the use of a voltage and a coil. Electromagnetism plays an important role in the transmission of voltage through inductors, coils, and motor windings.
Devices using this principle are: inductors, relays, contactors, and trans formers. Inductance is the ability to block AC voltage and allow DC to flow. Capacitance is the ability to block DC and allow AC to flow. Both inductance and capacitance play a role in the filter circuit in an AC drive.
Capacitance in an inductive circuit (motors) tend to improve the power factor of the entire system. The power factor is the measure of the efficient use of the current waveform and is stated as a ratio between the utility generated voltage and current waveform.
Semiconductors are a mix between conductors and insulators; they need an electrical push to drive the device to conduct current. Typical semiconductor devices include diodes (which allow current flow in one direction), thyristors (SCRs that conduct current only when triggered on), GTOs (which can be latched on or off depending on the polarity of the gate signal), transistors (which control the amount of current flow determined by the amount of trigger signal), and IGBTs (specialized transistors that have extremely high speed switch on and off times).
There are three basic types of mechanical loads that are encountered by any AC or DC drive-system: constant torque (e.g., conveyors), variable torque (e.g., centrifugal fans/pumps), and constant horsepower (e.g., machine tools).
Speed, torque, and horsepower play a major role in the operation of any application. Speed affects how much horsepower is required to perform the function-faster speed requires more horsepower. Torque is a turning effort and defines the ability of a system to start and keep moving at a specified rate. Inertia (WK^2 or WR^2 ) is the measurement of an object's resistance to change in speed. This measurement is needed to determine the acceleration time available from a drive system.
Gears, belts, pulleys (sheaves), chains, and sprockets all work to allow a smooth transmission of mechanical power, and in some cases, change speed and direction. Types of belts and pulleys include flat and v-belts, synchronous belts, and chains and sprockets.
Couplings, gearboxes, and speed reducers offer a positive connection point between the motor and application. Couplings are available in the following designs: flange, sleeve, and flexible (mechanically and elastically).
Speed reducers offer an effective means of changing speed delivered to the application, as well as the torque developed by the motor. If the speed is reduced at the output of the reducer, torque is increased by approximately the same amount. Speed reducers are available in various types such as helical gears, spur gears, worm gears, and bevel gears.
QUIZ
1. For current to flow, what must be added to overcome resistance?
2. What devices are used to measure current and voltage, respectively?
3. What is DC?
4. What is magnetic flux?
5. What is electromagnetism?
6. What does 60 Hz refer to?
7. Capacitance has the ability to block __ and let __ pass.
8. Inductance has the ability to block _ and let __ pass.
9. Give an example as to where inductance may be helpful.
10. What is power factor?
11. What is one benefit of placing an inductor ahead of the drive?
12. What two circuits make up a relay circuit?
13. How does a transformer conduct output voltage without any physical connection?
14. Indicate a use for capacitors in the DC bus circuit of an AC drive.
15. Explain the operation of a diode.
16. How does an SCR operate?
17. How does a GTO operate?
18. What is a Darlington bipolar transistor?
19. How do IGBTs differ from the common transistor?
20. What are the characteristics of a constant torque load?
21. Indicate two applications that fall into the constant horsepower category.
22. How does an AC drive save the operator money when connected to a centrifugal fan?
23. What is the definition of torque?
24. What is the definition of horsepower?
25. What is the definition of inertia?
26. If a speed reducer has a 10:1 ratio, with 3 lb-ft and 1200-rpm input shaft, what speed and torque would the output shaft be (assume 85% efficiency)?
ANSWERS-- Section 2
1. Voltage
2. Ammeter and voltmeter
3. Direct current (current flowing in one direction only; its value does fluctuate until the power supply is removed).
4. Magnetic flux is the relationship between the north and south poles of a magnet, with flux being the magnetic field itself.
5. The ability to produce a magnetic field through the use of a voltage and a coil.
6. 60 Hz refers to the number of complete cycles the AC waveform goes through during normal operations.
7. DC, AC
8. AC, DC
9. Inductance is useful in the DC bus circuit of an AC drive (filters out unwanted AC ripple) so that the voltage delivered to the next stage will be as clean.
10. Power factor, given in %, is the measurement of the phase difference between the utility-generated voltage, and current waveform.
11. It is used to reduce any surge current that may enter the input section of the drive due to a lightning strike or other high-voltage disturbance.
12. Control circuit and the power circuit.
13. With the alternating current, the coil is set-up where voltage is induced into the secondary part of the coil.
14. The alternating sequence of the AC allows the transfer of voltage through magnetism properties.
15. A diode conducts current in one direction only. The device continues to operate until the AC waveform goes into the negative direction, at which time the conducting of current stops.
16. An SCR conducts current in one direction, and only when the trigger circuit (gate) is powered on. The device shuts off when it is subjected to an opposite polarity between the anode and the cathode.
17. A GTO is similar to an SCR, in that it can be turned on and off at will. The GTO is turned on or off, depending on the polarity of the gate signal.
18. A device that basically has two or more power transistors, internally connected in one package. It can be turned on and off by the use of a command signal (base).
19. The difference is in the ability to switch on and off rapidly. The common transistor can switch between 1 kHz and 4 kHz while the IGBT can switch at 16 kHz.
20. Maximum torque may be required between zero and base speed. In addition, many drive manufacturers allow for 150% overload for 1 minute current before the drive automatically shuts down.
21. Machine tools and center driven winder applications.
22. The drive has the ability, through software, to simulate the torque application curve for a centrifugal fan. The drive develops a V/Hz curve, allowing the motor to develop the torque that matches the customer's torque curves. The drive can also allow the motor to develop the braking horse power used by the application for only fractions of a time 23. Torque is a turning motion, twisting of a device. It is also defined as Force X Distance.
24. Horsepower is the rate at which work gets done.
25. Inertia is the measurement of an object's resistance to change. Measurements are taken or calculated in WK^2 or WR^2 .
26. 102 rpm, with 30 lb-ft of torque.