Direct Current Motors

Transcription

1 Chapter 6 Direct Current Motors 6. INTRODUCTION People are more interested in motors than generators because motors are user oriented devices with which people are familiar. The extent to which motors are a part of today s living is dramatized by the results of a recent Federal Energy dministration study, which found that 64 percent of all the electric energy generated in this country is used for motor-driven equipment, with 49 percent being consumed by commerce and industry and 5 percent by residential users. -c motors dominate d-c motors in numbers because in most instances that is the only form in which electrical energy is available, whereas use of d-c motor necessitates a direct-current source such as a generator, battery or rectifier. Moreover, d-c motors are more expensive to manufacture than a-c motors the cost ratio is about six to one in a 5 horsepower (3.7 kilowatt) size. In view of this, the question is why do we use d-c motors at all? There are two large categories in which d-c motors are favored over a-c motors. The first is obviously in battery-operated devices such as golf carts, electric automobiles and electric fork lifts and also in the auxiliary equipment of regular automobiles and trucks such as power windows. n a-c motor won t work in any of these applications. The second category is where accurate control over speed and torque is required. This includes heavy industrial machinery such as electric locomotives, derrick hoists, gigantic electric shovels used in strip mining, steel rolling mills, textile and paper mill drives and machine tools, as well as small but extremely sensitive d-c servo systems. In most of these systems the motor is acting largely in a dynamic, constantly changing state as opposed to a steady state condition. The additional costs for generators (and their prime movers) or rectification apparatus are economically justified by the high quality performance of these d-c motors. Fortunately for our study of motors, the construction and nomenclature for parts are identical for both generators and motors, including the use of shunt and series fields. If the nameplate were missing from a machine you could not tell by physical examination of it whether it was a motor or a generator. The only difficulty one might encounter in endeavoring to use a d-c motor as a generator is that it would not build up as a self-excited generator because the shunt-field resistance is too high; otherwise they are interchangeable. It is recommended that the reader go back and review the section of Chapter 5 which covers commutation. Commutation, as it applies to the motor will be discussed below, but review of commutation in Chapter 5 is still a good idea. Figure 6. emphasizes the essential differences between motors and generators by showing a cross-sectional view of the stator and rotor for each. The rotor of a generator spins in the same direction as the torque applied on it by the prime mover. The torque developed on the armature (rotor) of the generator due to current flowing from the generator to an electrical load is opposite the applied torque and opposes the prime mover which is the cause of the motion. It takes mechanical input power to turn the crank of a generator. (Try our small demo generator in lab!) In a DC motor, the rotor spins in the same direction as the torque developed by the interaction of the current in the armature with the magnetic field from the stator. fter all, it is the developed torque which causes the rotor of a motor to spin. It makes sense that the direction of spin is the same as the direction of developed torque. nother essential difference between a motor and a generator is the direction of current through relative to the voltage drop across the armature. In a generator, current flows out of the + side of the armature. Remember that current normally flows out of the + side of a battery to a load connected to it. The same is true for the DC generator. The current on the armature of a DC motor flows into the + side of the drop across the armature coming from some external source of electrical energy. Remember, current flowing from higher

2 to lower voltage means that electrical power is transformed into some other form, which is mechanical in the case of a motor. The reader should take the time now to verify the directions for the developed torque shown in Figure 6. by using the right hand rule as presented in section 4.2 of Chapter 4. T d T a T a T d N Ω S N Ω S Generator Motor T d is the Internally Developed Torque and T a is the Externally pplied Torque Figure 6. Cross Sections of D-C Machines, Illustrating Basic Differences and Similarities. COMMUTTION IN THE DC MOTOR simple DC motor with one loop of wire on the armature and two commutator segments is shown in the diagram below. The reader should verify that the current flows in the direction shown, developing torque which spins the rotor in the clockwise direction. Every half turn, the commutator segments and brushes provide automatic mechanical switching which reverses the flow of current with respect to the rotating wires but maintains the same current direction with respect to the stator and the magnetic field. Therefore, the developed torque is always clockwise. practical motor has more loops of wire on the armature (rotor) and correspondingly more commutator segments. The diagram below implies that the magnetic field is produced by a permanent magnet on the stator. This may be the case in smaller DC motors. In larger motors, in order to produce a stronger magnetic field, stator windings are used to make an electromagnet and stronger field. Question: What change or changes could be made to reverse the direction of spin? 2

3 6.2 SEPRTELY-EXCITED MOTORS Separate excitation of a DC motor provides it with the greatest versatility. This means that the shunt and armature fields are fed from two different d-c sources. The wiring for this mode of operation is shown in Figure 6.2. I Field Source, V F I F R rmature Source, V R F Field R C E = KφΩ + - Ω T f T d Mechanical Load T a Figure 6.2 Separately-excited D-C Motor. Note that the direction of armature current is into the positive terminal of the armature. This is the reverse of its direction in the generator, which means that the direction of developed torque is also reversed. This situation was previously encountered in our study of the linear machine. Since this changes the sense of the armature resistance voltage drop, applying KVL, the voltage equation for the armature circuit becomes V = E + I R volts which differs from the generator equation only by the sign of the I R term. Because of the bilateral nature of the motor and generator, however, the equations for induced voltage (often referred to as counter emf or back emf in motors) and developed torque remain the same: E = KφΩ volts (6-2) T d = KφI newton-meters (6-3) Combining Equations (6-) and (6-2) and solving for the speed results in V Ω= I R Kφ rad / s (6-4) Which enables us to see immediately that speed varies directly with the armature voltage and inversely as the amount of air gap flux (and hence with field current). We look at this a little further, first by assuming the field current is held constant at its maximum 3

4 setting called full field. s the armature voltage is decreased from it rated value, the difference between V and I R can vary anywhere from a maximum down to zero called a stall condition and speed consequently varies from full-field speed down to standstill. The maximum torque allowable will be the same for all settings of armature voltage because the rated armature current is the governing factor and torque depends only on I in the situation described ( Td = KφI I with a constant flux setting). Since rotational power is proportional to the product of torque and speed Pd = TdΩ6, the power capabilities are directly proportional to speed when torque is held at its maximum value. This straight-line relationship between speed and allowable power as well as the graph of speed vs. llowable torque are illustrated in the lower portions of Figure 6.3 (a) and (b). Percent of Full Field Speed Field Control rmature Control Percent of Full Field Speed Percent of llowable Torque 00 Percent of llowable Power 00 (a) (b) Figure 6.3. Control of a Separately-Excited Motor. Shaded reas Indicate Permissible Operating Regions. Next, assume armature voltage is held at the rated value while field current is gradually reduced from its full-field value. Inspection of the torque relationship Td = KφI indicates that if I is to be held at its rated value than a reduction of flux must be accompanied by a corresponding reduction of torque. In other words, as we reduce flux in order to increase speed, the torque capability of the motor is reduced. Multiplying Equations (6-3 and (6-4) together to obtain power: P = T Ω d d (6-5) 4

5 2 P = KφI V I R 6/ Kφ = VI I R d watts. (6-6) We see that if both V and I are held at their rated values, the power is constant. pplying this knowledge to Equation (6-5) it can be seen that the allowable torque must decrease as the speed is increased, to maintain constant power, forming a hyperbolic curve. Both of these phenomena are illustrated in the graphs of Figure 6.3. To summarize our information, it has been shown that by using a combination of adjustable field current (for speeds above full-field speed) and adjustable armature voltage (for speeds below full-field speed) the whole range of speeds from stall to the maximum safe speed can be covered. We have assumed the rated value of current in this development in order to establish upper operating limits for torque and power, but any current below that is obviously permissible. The watchword in applying these techniques of speed control to actual situations is to monitor the armature current at all times and to back off on the load if rated current is exceeded. n example of a constant torque load is the feed drive on a machine tool, which feeds the work into a cutting tool or vice versa. Wide speed ranges are required in order to meet the varying conditions of operation, but the torque remains constant because it consists mainly of friction in the gears. n example of a constant power load is the cutter on a milling machine or the spindle on a lathe, neither of which involves a large gear train. They require adjustable speed in order to operate most efficiently with various sizes of work or cutters. 6.3 THE WRD LEONRD SYSTEM The circuit shown in Figure 6.2 indicated a generalized symbol for the variable voltage source wired directly to the armature. Now we shall look into how this source is obtained in practice. separately-excited d-c generator driven by an a-c motor will maintain generator speed at or near a constant value, which is the distinguishing characteristic of most a-c motors. The Ward Leonard system is shown in simplified form in Figure 6.4, with a more detailed wiring diagram in Figure 6.5. The term rotary amplifier is sometimes applied to this system, the amplification arising from the fact that only a small power change in the field of the generator produces a much larger change in the mechanical power of the motor. n important feature for many applications of the Ward Leonard system is the ease of reversing motor direction. To accomplish this the slider on the generator field control resistor is moved to the center reducing terminal voltage until the motor stops, and then the slider is moved further to the left so the terminal voltage reverses polarity. This causes armature current to reverse and this reverses the torque. In some systems a rheostat and a reversing switch are used instead of a center-tapped resistor. Figure 6.4 Simplified Diagram of a Ward Leonard System. 5

6 The large initial cost of purchasing three machines to do the work of one, the additional maintenance costs and the obvious loss of overall efficiency inherent in the Ward Leonard system are offset by the following features: " wide range of speed from standstill to high speeds. " Rapid reversal of direction. " Smooth starting. " Good speed regulation. " Power in the control circuit is small. " Regenerative braking conserves energy. Figure 6.5 Wiring Diagram for Ward Leonard System, T d Is The Internally Developed Torque Of Each Of The Two D-C Machines, T a Is The Externally pplied Torque Of The -C Motor Or The Mechanical Load. The term regenerative braking refers to a braking system similar to dynamic braking, which was explained in connection with the linear motor. The difference is that in regenerative braking the energy is fed back into the electrical generating system instead of being wasted as heat in a resistor. When braking is desired, the generator field current is reduced, causing its induced emf to drop below that of the motor. This results in reversing the armature current direction, which produces a braking torque on the d-c motor, and causes the generator to motorize. This turns the a-c motor into a generator which then reverses power flow back into the a-c lines. n elevator on the way down is a good example of regenerative braking usage. The engineering choice between dynamic braking or regenerative braking depends on the duty cycle, cost of the braking resistor and relative cost of switchgear. One shipboard application of the Ward Leonard system is a diesel-electric d-c propulsion system, which provides exceptional operating flexibility at reduced speeds and under towing conditions. These systems are highly desirable for tugs, icebreakers, rescue and salvage ships, net-laying ships, fuel-oil tankers and barges, as well as survey ships. 6

7 Solid State Control Replace of the d-c generator and a-c motor combination with an electronic rectifier system is entirely feasible and such systems are in widespread use today, sharing about half of the market. Whether or not it can still be called a Ward Leonard system is a moot question. The most important single development which make this possible has been the availability of silicon controlled rectifiers (SCR s), also called thyristors, with large current capacities (hundreds of amperes). single-phase system is shown in Figure 6.6 to illustrate the basic circuit. By controlling the duration in each half cycle, and thus the average currents in field and armature, the SCR s allow a wide variety of speed and torque settings. In large systems three-phase sources are used, but the method of control is essentially the same as depicted here. For many large motor requirements such as paper mills, steel mills and in strip-mining the basic Ward Leonard system is still preferred because of the high current requirements and also because the high rotational inertia of the generator allows it to smooth out peak load demands better than an electronic system can. Figure 6.6 Separately-Excited D-C Motor Operated From -C Source By Means Of Bridge Rectifies. The SCR s Control verage Currents In Field nd rmature. " EXMPLE 6- " In the Ward Leonard system of Figure 6.5 the generator is driven at a constant speed of 800 rpm and has an armature resistance of 0.25 ohm. The d-c motor also has an armature resistance of 0.25 ohm. Neglect mechanical losses in all three machines. Initially the terminal voltage is 24 volts and armature current 32 amperes, with a motor speed of 900 rpm. 7

8 a. Calculate the generated emf s in both generator and motor. b. Calculate the developed torque and converted power in the motor. c. If the load is uncoupled from the motor, the armature current drops to zero (why?). Calculate the no-load speed of the motor. d. The original load is recoupled to the motor and the generator field current is reduced until the terminal voltage is 62 volts. ssuming a constant torque load, calculate the new motor current, speed, torque and converted power. e. ssume the same conditions as in (d) except it is not a constant torque load and armature current is observed to be 20 amperes. Solution a. E = V + I R = x025. = 28 volts g E = V I R = 24 32x025. = 20 volts m b. Converted power = EmI = 20x32 = 3840 watts T = P / Ω = 3840 / 2π900 / 60 = 9. 3 N-m d d 6 c. E = kφn Motor condition : E = kφn Motor condition 2: E2 = kφn2 n = n Ek 2 φ 2 900x 28 x rpm Ek φ = 20 = 2030 d. Td = KφI Motor condition : T = KΦI Motor condition 2: T 3 = KΦI 3 T3 = T = 9. 3 N-m (from (b) - - constant torque load ) I KT φ 3 = I = 32xx = 32 KT φ 3 amperes E3 = V I3R = 62 32x025. = 58 n = n Ek 3 φ 3 900x 58 x rpm Ek φ = 20 = 920 P3 = E3I3 = 58x32 = 860 watts Check: P3 = T3Ω 3 = 9. 3x2π 920 / 60 = 860 watts e. E 4 = V I 4 = 62 20( 025. ) = 59. 5watts P4 = E4I4 = 59. 5x20 = 90 watts 8

9 n = n Ek 4 φ x x rpm Ek =. φ 20 = 940 T P N-m 4 = 4 / Ω = 90 / 2π 940 / 606= 2. Check: T = Tx K φ N-m K x I xx 20 I =. φ 32 =2. " EXMPLE 6-2 " Solution ssume the same Ward Leonard system described in Example 6-, including the initial set of operating conditions. a. The motor field current is reduced from its full-field value, reducing flux per pole by 5 percent. ssuming a constant torque load, calculate the new armature current. b. Calculate the new speed. c. Calculate the new converted power. a. Td = KφI T5 = T = 9. 3 N-m I = Kφ T5 I x x amperes 5 32 Kφ T = 085. = b. E = E I R volts 5 g 5 2 = = 9 n n Ek 5 φ x x rpm = = = 2220 Ekφ c. P5 = E5I5 = 9x37. 7 = 4490 watts Check: P5 = T5Ω 5 = 9. 3x2πx2220 / 60 = 4490 watts 6.4 SHUNT MOTORS When the high cost of a separately excited motor is not warranted, but a considerable degree of speed control is needed, the shunt motor offers an excellent solution. single constant voltage source supplies current to both the armature and the field. The wiring of the basic shunt motor is shown in Figure

10 I L I R C I F R Voltage Source, V R F E = KφΩ + Field φ - Ω T f T d Mechanical Load T a Figure 6.7. Shunt Motor With Field Control Of Speed. By adjusting the field rheostat, R C, speeds ranging from full-field speed up to the maximum safe speed (about 50 percent above rated) can be obtained. With any given field current the equilibrium speed drops as the load torque increases, since E Ω, I = V E / R, T I. Which is to say that an increased load demand 6 d causes a decrease in speed which reduces the counter emf, which simultaneously causes an increase in I causing more torque to be developed. The motor continues slowing down until a new equilibrium point is reached where the developed torque is precisely the amount needed. s with voltage regulation, we can describe the speed drop quantitatively with 6 / S.R. = Speed Regulation = n n n, (6-7) NL FL FL where full-load speed is the speed at rated torque. The speed-torque curves for several settings of the field rheostat are shown in Figure 6.8; note that the speed regulations are not the same for all the curves, although the speed drops are the same. The decrease in speed from no-load to full-load is small for shunt motors (about 5 to 0 percent of rated speed), which is tolerable for most applications. Under normal operating conditions, the DC shunt motor is very nearly a constant speed machine. 0

11 Speed Typical Load Characteristic I F4 < I F3 I F3 < I F2 I F2 < I F I F Full-Field Setting Rated Torque Torque Figure 6.8 Speed-Torque Curves For Several Field Rheostat Settings For Shunt Motor, lso Shown Is The Speed-Torque Curve For Typical Load. load characteristic is shown superimposed on the same graph, i.e. the speed-torque relationship for a particular load. Intersections of the load curve with the motor curves indicate possible steady state operating points. Noting that speed control by means of varying the field current has a bottom speed limit which corresponds to full-field conditions, we are interested in finding a method of extending speed control below this limit. To see how this can be done when restricted to a fixed voltage source (The speed can easily be varied up and down if the voltage source can be increased and decreased) let us look again at the speed relationship of Equation (6-4), repeated here for convenience. V I R Ω= rad /sec. (6-8) kφ We have already exploited the denominator by using field rheostat adjustment, and the terminal voltage, V, is a fixed quantity; this leaves the I R term as a possibility. If we supplement the armature resistance with an external rheostat we would modify the equation to read V I R + RS6 Ω= rad /sec. (6-9) kφ which indicates that for constant V,I and, the speed would decrease as R s is increased. Implementation of this scheme would modify the wiring as shown in Figure 6.9. The effectiveness of this system of speed control can be demonstrated by means of an example problem.

12 I L I I F R C R S R Voltage Source, V R F E = KφΩ + Field φ - Ω T f T d Mechanical Load T a " EXMPLE 6-3 " Solution Figure 6.9 Shunt Motor With Field nd rmature Control Of Speed shunt motor is wired as shown in Figure 6.9, with V = 25 volts; internal motor resistances are 0.25 ohm in the armature and 65 ohms in the shunt field. With both rheostats set for zero resistance the speed is 450 rpm and the line current is 32 amperes. a. Calculate field and armature currents. b. Calculate the developed torque and converted power. With the armature rheostat readjusted to.5 ohm, and assuming a constant torque load: c. Calculate the new speed and converted power. d. Calculate the voltage across the armature terminals. e. Calculate heat loss in rheostat R s as a percent of total power from the source. f. Calculate the no-load speed. 6 6 a. I = V / R + R = 25 / = 9. amperes F F C I = IL IF = = 30. amperes b. E = V I R + R = = 22. volts 6 6 S P = T Ω = EI = 22. x30. = 3650 watts T d d d = P / Ω = 3650 / 2π450 / 60 = N-m d 6 2

13 c. T2 = T (given) T I I amperes T x K 2 φ 2 = = 30. xx = 30. Kφ 6 6 E2 = V I R + RS = = 76. volts n = n Ek 2 φ 450x 76 x RPM Ek =. φ 22. = 90 2 P2 = E2I = 76. x30. = 2290 watts d. V = V I R = x5. = volts S 6 Power loss in rheostat x.5 e. Power input to motor = I R = 30. S VI 25 x 32 = 360 = 34% L 4000 f. t no load, I a = 0 E3 = V IR + RS6 = 25 0 = 25 volts n n E 3 kφ 25 3 = x = 450 x = rpm E kφ From this example it can be concluded that the armature resistance method of reducing speed below the full field value is effective and simple. There are three drawbacks to this type of control, however, which detract from its worth. First, it is seen to be an inefficient way to use energy, since a large proportion of input power is wasted as heat. simple calculation will reveal that this power is equal to the difference in the converted powers; i.e., the power has simply shifted from the motor to the rheostat. The second drawback is that the armature rheostat has to have a large power rating, which makes it expensive. Finally, for any given setting of the armature rheostat, speed variation with changing load is high, which is undesirable for most applications. This can be seen by referring to the speed-torque curves in Figure 6.0, where the curve for R s = 0 has been added for reference. third curve has also been added, based on the hypothetical assumption that the armature voltage had been reduced to 80 volts by using a separate source. Curve (a) shows a moderate drop in speed for a shunt motor of 45 rpm. Curve (b) on the other hand shows a drop of 585 rpm, while curve (c) shows the same drop as curve (a), 45 rpm (the reader should be able to confirm this). gain we see that the separately excited motor is superior for speed control, but the question arises: is armature resistance control worthwhile? Yes indeed! In spite of its drawbacks, speed control with an armature rheostat offers a fairly simple way of achieving a given speed below the full-field value. This is especially true in an experimental situation where quick results are wanted with equipment at hand, and it is not to be used over an extended time period. For example, a bank of parallel-connected light bulbs makes a fairly inexpensive but high wattage resistor. variation on the method, which provides better speed regulation but reduced torque capability, is to add a second rheostat in parallel with the armature, however, space requirements preclude treating it in detail here. 3

14 Speed in rpm rpm (a) R S = rpm (c) (b) R S =.5Ω 0 0 Torque in N-m 24 Figure 6.0 Speed-Torque Curves Relating to Example 6-3. Curve (c) Indicates How The Motor Would Have Reacted If rmature Voltage Had Been Reduced By Separate Excitation To 79.9 volts. 6.5 D-C MOTOR STRTING major problem can arise when a shunt motor is started by suddenly closing a switch and thus applying the full source voltage to both field and armature circuits. If we consider the basic armature current equation I = V E R (6-0) 6 / volts for zero speed (standstill), when E = KφΩ = 0, then I = V / R, which may be sufficiently large to activate the circuit breaker or open a fuse. In other words, a shunt motor should not be started across the line because it may be automatically disconnected exceptions are small motors up to 2 horsepower (.5 kilowatt), which are able to accelerate rapidly up to full speed before the circuit breakers can react. The remedy for this problem is to add a starting resistor in series with the armature, designed to limit the initial surge of current to about.5 to 2 times the rated value. Moreover, this resistor is sectionalized so that it is cut out in two or more steps instead of all at once. The more steps, the smoother the acceleration but also the more costly it is. There are two main types of starting resistors, the manual starting box which has a lever-type handle, and the automatic starter. 4

15 Manual Starters Constructional details of a manual starter are illustrated in Figure 6.. These starters are used for motors up to 25 horsepower. The handle is moved gradually from left to right. When it reaches the right end, there is an electromagnet (not shown) which holds the spring-loaded handle in that position. If power fails, the electromagnet de-energizes and the handle is pulled to the starting position. Otherwise, the return of power would find the motor without a starting resistor, which would cause the breakers to trip. Figure 6. Manual Starting Box, Wired To Shunt Motor. utomatic Starter utomatic starters short out sections of the starting resistor by means of electromagnetic contactors (relays). ll the operator does is push a button which starts a timing sequence which actuates the contactors one by one until they are all closed. Figure 6.2 illustrates the wiring of a typical automatic starter. Momentarily pushing the start button down activates relay coil M, contact M closes and timing motor TM energizes. One by one in preset intervals contacts TM, TM2 and TM3 close, energizing relay coils, 2 and 3 respectively which in turn close contacts, 2 and 3. Thus, the series resistance begins at R +R2 +R3 then is R2 + R3, then R3 and finally zero. ll contacts shown stay closed until the stop button is pushed down, which de-energizes all relays. Manual starters are not used for large motors because as a manual starter becomes larger it requires more effort to move the handle and also the proximity of the operator to the location of heavy current switching is dangerous with automatic starters. The control buttons can be safely located some distance away from the actual switching equipment. On the other hand, manual starters for small motors have the advantages of being simpler in construction and maintenance and allowing the operator to vary the timing of the starting sequence. Both types of starters are used in small motor installations. 5

16 " EXMPLE 6-4" Figure 6.2 Ladder Diagram For Time-Driven utomatic Starter. Solution multi-section starting resistor is to be used with a shunt motor which has an armature resistance of 0.25 ohm. The supply voltage is 25 volts and the load is such that the armature draws 30 amperes at a speed of 500 rpm under normal conditions. a. Calculate the total resistance of the starting resistor if the initial armature current is to be limited to 55 amperes. b. Calculate the resistance of the first resistor to be shorted out if this will coincide with a speed of 400 rpm and the current surge is again to be limited to 55 amperes. c. Calculate the initial surge current if a starting resistor were not used, neglecting the inductance. 6 a. V = E+ I R+ RS circuit resistance: R + R = V E / I = 25 0 / 55 = ohms 6 6 S 6

17 starter resistance: R S = = 25. ohms b. Under normal conditions: E = V I R = x 0.25= 23. volts t a speed of 400 rpm: E E nk 2 φ = =23. x x 32.4 volts nkφ 500 Total resistance needed to limit current surge: RS2 + R = V E26/ I = / 55 = 68. ohms R S 2 = = 56. ohms Difference between the resistance values: R = = = ohms c. I = V / R = 25 / 025. = 000 amperes! 6.6 SERIES MOTORS If the shunt field coils (many turns of fine wire) were removed from the pole structures and replaced with series field coils (few turns of heavy wire) which were then connected in series with the armature, the motor would have a drastically different speed-torque characteristic. The big difference is that now the magnetic flux is proportional to the armature current instead of being fixed at a constant amount. lthough the relationship between flux and armature current is not entirely a straight line because of saturation, recall that a considerable portion of the magnetization curve, below the knee is fairly straight. Our analysis will be simplified if we assume operation in this region, where φ = KI Webers (6-) Which in turn means that T = KφI = K I 2 Newton - meters d S (6-2) and that E = KφΩ = K I Ω Volts S (6-3) Now, solving Equation (6-3) for I and substituting it in Equation (6-2), we have 2 T K I K = = d S S E K Ω S 2 = 2 E KSΩ6 2, (6-4) which indicates that speed is inversely proportional to the square root of the torque Ω T d (6-5) 7

18 This accounts for the shape of the speed-torque curve in Figure 6.3. more tangible way to demonstrate the shape of the series motor curve relative to the shunt motor curve is to solve Example Problem 6-5. I Series Motor V S Series Field φ E + (a) Wiring Diagram - R Ω T f T d T a Mechanical Load Percent of Rated Speed 00 Shunt Motor Rated Torque (b) Speed vs Torque Comparison Torque " EXMPLE 6-5 " Figure 6.3 Series Motor nd Characteristics. Solution Two identical mixers in the galley of a missile cruiser are driven by a shunt motor and a series motor respectively. Both are rated at 25 volts, 30 amperes (armature current) and 000 rpm, and have armature resistances of 0.25 ohm (assume negligible series field resistance). Both have been operating under rated conditions, mixing batter for half an hour, when a cook pours three gallons of milk into each batter, which lightens the torque load to 65 percent of its former value. Calculate the new speed for each motor assuming operation on the linear portion of the magnetization curve. a. Solve for the counter emf under rated conditions: E = V I R = x 0.25 = 2.3 volts b. Solve for the new shunt motor speed: T2 Kφ I 2= I =30 x 0.65 x = 9.5 mperes T Kφ E 2 = V I R = x 0.25 = 22.6 volts n n E 2 kφ 2 = =000 x 22.6 x = 00 rpm E kφ 2.3 c. Solve for the new series motor speed: 8

19 From Equation (6-2): I 2 T2 K S 2 = I = 30 x x = mperes T K 2 S " EXMPLE 6-6 " Solution E 2 = V I R = x 0.25 = 22.0 volts = 2 From Equation (6-3): n n E E x I 2 2 I n 2 = 000 x x x K S K x x = 250 rpm The series motor s speed has leveled off at 240 rpm higher than the speed for the shunt motor, about 24 percent higher, thus verifying the graphical comparison in Figure 6.3(b). By carrying the situation to an extreme, it is possible to make a very import point concerning the operation of series motors the danger of over speeding. Suppose the motors of Example 6-6 were coupled to the mixing mechanisms by vee-belts or chains and the belts or chains broke, leaving the motors with no load torque. Calculate the no-load speeds of the motors. a. For the shunt motor: I T2 Kφ = I = 30 x 0 x = 0 amperes T Kφ 2 E 2 = V I R = 25 0 = 25 volts n n E 2 Kφ = =000 x E Kφ 2 b. For the series motor: I T2 KS 2 = I = 30 x 0 x = zero T K 2 S E = V I R = 25 0 = 25 volts x = 030 rpm S 2π π 9

20 n n E 2 I 25 = = 000 x Infinity E I 2.3 x gain we see that the trend in speed as shown by the speed-torque curve is verified, but more than that, the consequences of allowing a series motor to be uncoupled are seen to be disastrous. The high-pitched whine of a runaway motor is enough to send all personnel running, with no hope of automatic disconnection by a circuit breaker or fuse since the armature current is small and getting smaller. Destruction by centrifugal force is inevitable, with deadly flying missiles a realistic possibility. What can be done to prevent this from happening? series motor should not be coupled to its load by a belt or chain, which could break. Instead, it should be used only where it can be directly coupled to the load or to gearing, unless an operator is in constant attendance (as with a crane). typical application for series motors is in traction systems, such as rapid-transit trains used in urban transportation. The choice of a series motor is based on the characteristics of this kind of load, which match the series motor characteristics well. For example if the vehicle is traveling up an incline, which demands more torque, the series motor slows down more than a shunt motor would. On the other hand, when running along a level track, which demands less torque, the speed is naturally higher with the lighter load. In other words, the series motor slows down or speeds up, where appropriate, automatically, whereas a shunt motor would require direct control to effect the wide speed range needed. In automotive terms, the series motor and the shunt motor correspond to the automatic and stick-shift transmissions respectively. Other applications where the series motor curve fits the job requirements are in portable electric drills and in cranes and hoists (raise a three-ton load slowly but raise an empty hook fast). In traction and hoist work another advantage of the series motor is its large starting torque. The additional flux created by armature current in the series field under stall conditions produces more starting torque for a given amount of current than in the shunt motor. 6.7 COMPOUND MOTORS Compound motors are equipped with both shunt and series fields. Combining the two kinds of field windings in one machine results in a speed-torque characteristic which lies between the series and shunt curves, as seen in Figure 6.4. The compound motor has the drooping characteristic of the series motor, although not as severe, but it has a singular advantage over the series motor in that the danger of running away has been avoided, since the shunt field supplies a magnetic field even at no load. However, a danger does exist in wiring the motor in such a way that the series field mmf opposes the shunt field mmf. No one should deliberately wire a compound motor in this differential mode because of its strange behavior. This includes starting in the wrong direction and later reversing its direction of rotation, as well as being unstable with heavy loads a sudden rise in speed and tripping of the breakers. The wiring diagram of Figure 6.4 designates the correct field wiring by placement of polarity dots on the windings. There should be no wiring problems with new motors as they are supplied with correct wiring diagrams and the wires are tagged for identification. In the absence of this information, start the motor (but don t run it) with only the series field connected and then start it with only the shunt field connected, making note of how connections were made and whether the motor rotated in the proper direction or not. This information should make it possible to tag the wires and prepare a proper wiring diagram. The starting of compound or series motors requires the use of starting resistors, as does the starting of shunt motors. Most jobs formerly done by compound motors have been taken over by induction motors, which are a-c motors of simpler construction and lower cost. 20

21 Compound Series V S Shunt Field I F Series Field φ I R + E - Percent of Rated Speed 00 Shunt Rated Torque (a) Wiring Diagram Torque (b) Speed vs Torque Comparison Figure 6.4 Compound Motor nd Characteristics. 6.8 COMPENSTING ND COMMUTTING WINDINGS Compensating Windings Motors subject to rapidly changing load demands such as occur in steel mills must be equipped with special windings to counteract the mmf of the armature winding. The principle which we are concerned with here is the voltage of self induction which is developed when a rapid change in current is made, the familiar e = Ldi / dt = dλ / dt voltage. For example, a sudden decrease in torque demand results in a decrease in armature current, and since the armature winding has the property of inductance, an induced voltage results. Since the ends of each armature coil are connected to adjacent commutator segments, the induced voltage appears across the gap between segments. Now if this voltage exceeds about 35 volts, arcing occurs between the segments. When this takes place simultaneously across all the commutator gaps at once, the commutator becomes a ring of fire which amounts to a short circuit across the armature terminals, rendering the motor inoperative. The only way to prevent this from happening on motors subject to this severe type of duty is to neutralize the inductance of the armature circuit. Since the unit of inductance is defined as the number of flux linkages per ampere of current, the flux linkages are reduced by introducing another winding physically located next to the armature winding but having an mmf equal and opposite to the armature mmf. This extra winding, called a compensating winding, gets its name from its function of compensating for the armature mmf. The total effect of the armature mmf on the air gap flux is called armature reaction. The compensating winding is connected in series with the armature so it always has the same current as in the armature, and thus automatically compensates for all values of armature current; connections are made internally at the factory. Figure 6.6(b) shows the physical location of the windings, which are located in slots cut into the faces of the pole shoes. ddition of compensating windings adds greatly to the cost of a machine and so are only installed in motors and generators destined for the severe types of duty described above or in motors made for a wide range of field control. 2

22 Figure 6.5. Cross-section of commutator, illustrating the phenomenon of ring-fire. Commutating Windings and Interpoles If you look inside any d-c machine except a small one (less than one kilowatt) you will observe narrow poles placed between the main poles. These are called interpoles, and the windings placed on them are called commutating windings. Their purpose is to improve the commutation process by reducing sparking between brushes and the commutator to a minimum. The reason for the sparking is that as a given commutator segment rotates away from the brush, a flow of charge jumps the gap. The commutation process takes place while the coil sides are located between poles, and is limited to the finite time interval during which the commutator segments are shorted by the brush. In addition to the reversal of induced emf which takes place, the current also reverses its direction. Unfortunately, the commutation process does not go as smoothly as it should, because of two extraneous voltages induced in the coil. The first of these is the voltage of self-induction (Ldi/dt) and the second is the speed voltage (BLu) induced in the coil as it moves in the magnetic field produced by the armature mmf (armature reaction) sketched in Figure 6.6 (a). Both of these voltages delay the commutation process, which leads to the sparking. The remedy for this problem is the introduction of the commutating windings on the interpoles. They produce localized mmfs between poles, where commutation takes place of sufficient strength to cancel the effect of the armature mmf there*, and produce additional flux to induce a speed voltage which will cancel the voltage of self-induction in the coil being commutated. Like the compensating winding, the commutating winding is internally wired at the factory, in series with the armature. ll larger machines have interpoles, some machines have compensating windings also. *this is strictly a local cancellation, i.e. only in the immediate vicinity of the interpole, and hence does not preclude the need for a compensating winding. 22

23 Figure 6.6 Effect On The Magnetic Circuit Of Introducing Compensating Winding. 6.9 LOSSES ND EFFICIENCY Machine Ratings Every rotating machine has a metal nameplate secured to the frame. long with the manufacturer s name and identification numbers are the maximum ambient temperature for which the machine was designed (typically 40( C) and the class of insulation (,B.F or H), the classes of insulation differing in the amount of allowable temperature rise within the machine. Older machines state only the allowable temperatures rise, instead of the ambient temperature and insulation class. The reason for all this concern with heat is that it is the limiting factor which determines the amount of current that can be allowed in the armature. Modern wire insulation used in machinery is made of mica, glass fiber, asbestos and silicone, bonded together with synthetic resins. Prolonged operation under high temperatures reduces its insulting qualities and therefore temperature limits have to be set (by the National Electrical Manufacturers ssociation NEM) which will assure a reasonably long life for the machine. If a machine is to be operated continuously in surroundings which have a higher ambient temperature than stated on the nameplate, the user should derate the machine accordingly, i.e. run it at less than its rating; this may mean buying a 7.5 horsepower motor to do a 5 horsepower job. It is best to consult the manufacturer, if possible, in situations like this. Sometimes a motor has a short-time rating, such as 5, 5, 30 or 60 minutes, which is higher than the continuous rating, for example, a motor with a 30- minute rating of 0 horsepower on a continuous basis. Many applications require intermittent operation of a motor, the duty cycle consisting of periods when it is operating at different power levels and some rest periods. Choice of the proper motor size in situations like this is aided by the use of the rms power method. To use this method, first determine the duty cycle required of the motor and then use a formula basically the same as for finding rms current or voltage: rms power = 2 (power x time) running time + standstill time / 2 (6-6) Conversion Factor: horsepower = 746 watts (6-7) 23

24 "EXMPLE 6-7 " grab-bucket hoist for unloading coal from a barge is subject to the following duty cycle. Solution Calculate the minimum motor rating required for the job. Status Time Period Output Power Close bucket 6 sec 30 KW Hoist 0 sec 60 KW Open bucket 3 sec 20 KW Lower bucket 0 sec 35 KW Rest 6 sec 0 rms power = x6+ 60 x0+ 20 x3+ 35 x0+ 0 x6 ( ) + 6 / 2) rms power = 38.5 KW = 5.6 horsepower The nearest standard size would be a 60 horsepower motor (44.8 KW). Other nameplate ratings, in addition to current, are speed, voltage and output power. The power rating on a machine refers to it output power capability. t present generators are in kilowatts and motors in horsepower, but as the conversion to the metric system progresses the unit for motor power will be changed to kilowatts also. Voltages should not exceed the rated values by more than 0 percent, and speeds should be kept below 50 percent of rated, in general. Types of Losses in D-C Machines The losses which occur in d-c machines are divided into two main categories; rotational and electrical, and are listed in Table 6.. Friction and windage losses vary directly with speed only but the core losses depend on both speed and flux density. However, at the level we are treating the subject here, it will suffice to assume the rotational losses remain constant over the full range of operation. The diagrams in Figures 6.7 and 6.8 aid in visualizing the distribution of power with motors and generators. 24

25 Table 6.. LOSSES IN D.C. MCHINES. Rotational Losses. Windage: Friction of air on moving parts. 2. Friction: In bearings and between brushes and commutator. 3. Core or Iron: Hysteresis and eddy current losses B. Electrical Losses. Shunt Field: Heat loss in shunt field coils and field rheostat - V L I F 2. rmature Resistance: Heat loss in armature coils and also including all fields in series with 2 armature (series, compensating, commutating) = I R. Electrical Mechanical Electrical Input Power = V L I L P Developed = EI = T d Ω Mechanical Output Power to the Shaft = T a Ω Shunt Field Resistance Loss = V L I F rmature Resistance Loss = I 2 R Rotational Losses (Windage, Friction, etc.) Figure 6.7 Power Distribution In D-C Shunt Motor. Mechanical Electrical Mechanical Input Power From the Prime Mover = T a Ω P Developed = T d Ω = EI Electrical Output Power = V L I L Rotational Losses rmature Resistance Loss = I 2 R Shunt Field Resistance Loss = V L I F Figure 6.8 Power Distribution In D-C Shunt Generator 25

26 " EXMPLE 6-8 " Calculate the efficiency of a shunt motor which has an armature resistance of 0.25 ohms and a field resistance, including rheostat, of 90 ohms. The motor is wired to a 25 volt source and draws 32 amperes, at a speed of 800 rpm. separate test is run with the motor uncoupled from the load and adjusted for 800 rpm, to determine rotational losses, at which time the armature current is found to be.5 amperes. Solution F C amperes a. Field current = V / R + R = / =. Field Loss = VI F = 25x4. = 75 watts b. rmature current = I I = = amperes L F 2 2 rmature resistance loss = IR = x025. = 7 watts c. Rotational loss = EI 0at no load5= ( ). 5 = 87watts d. Total losses = = 479 watts e. Efficiency = Input - Losses Input 25 x = x00% = 88% 25 x REVERSL OF DIRECTION ccording to the basic torque equation Td = KφI,it is possible to reverse the direction of a motor by reversing the direction of the magnetic field or the armature current, but not both. In practice it can be done by reversing either the armature or the field connections on a shunt motor. With a compound motor, both series and shunt fields must be reversed to avoid a differentially compounded motor. series motor requires changing either the series field or the armature connections. In no case will simple reversing the wires to the source produce reversal. 6. SUMMRY " Separately-excited motors receive power from two different sources. This allows for the most flexibility in speed control by varying the field current and/or armature voltage " Speed control by field current adjustment has a lower limit of speed where field current is a maximum, called the full-field speed. The upper limit is governed only by safety considerations. " Lowering the armature voltage extends speed control below the full-field speed down to standstill. " The Ward Leonard system, consisting of a motor-driven generator directly connected to the armature of the motor to be controlled, is an effective but expensive means of achieving complete control. 26

27 " Solid state rectifiers using SCR s provide an alternative to the motor-driven generator of the Ward Leonard system, especially popular in smaller systems. " The shunt motor offers an economical way of achieving speed control since it uses the same source for field and armature. Speed control is effected by means of field and armature rheostats. " The starting of a d-c motor requires a starting resistor be placed in the armature circuit to limit starting current until the rotor accelerates to a moderate speed. Exceptions are small motors of less than 2 horsepower. " series motor has the field winding connected in series with the armature instead of across the line. The low speed-high torque and high speed-low torque characteristic make it desirable in traction applications and many others. " Compound motors have both shunt and series fields, which provides a motor with the higher starting torque of a series motor without excessive speed at light loads. " Compensating and commutating windings improve the operation of motors. ll except small machines have commutating windings to prevent sparking at the brushes. few special machines have compensating windings for severe duty applications. " Losses can be classified in three major categories, field resistance losses, armature resistance losses and rotational losses (friction, windage, hysteresis and eddy current). " The preferred way to reverse a motor is to reverse the armature connections only. 6.2 QUESTIONS Q6. Explain why DC motors are a better choice than C motors for some applications. Q6.2 Describe what compensating windings do and what problems they remedy. Q6.3 Explain why a series motor is the best choice for some applications. Q6.4 Why should a series motor always be directly coupled to the mechanical load or to the gearing? Q6.5 Why must special circuitry containing resistors and switches by used to start large DC motors? Q6.6 Describe the advantages of a Ward Leonard or similar type system. 27

28 6.3 PROBLEMS P6. Repeat Example 6- of the text if the speed at which the generator is driven is constant at 800 rpm., the armature resistance of the generator is 0.08 Ohms, the armature resistance of the motor is 0. Ohms, initially the terminal voltage equals 30 Volts, the armature current is 28 mps and the speed of the motor is 2000 rpm. P6.2 Repeat Example 6-2 of the text for the system described in P6. above. P6.3 shunt motor is wired as shown in Figure 6.9 of the text, with V T = 30 Volts, armature resistance equal to 0. Ohms and shunt field resistance equal to 70 Ohms. With both rheostats set for zero resistance the speed is 550 rpm and the line current is 29 mps. Repeat Example 6-3 for the system described above. P6.4 multi-section starting resistor is to be used with a shunt motor which has an armature resistance of 0.5 Ohms. The supply voltage is 35 Volts and the load is such that the armature draws 25 mps at a speed of 400 rpm under normal conditions. Repeat Example 6-4 for these conditions. P6.5 Calculate the efficiency of a shunt motor which has an armature resistance of 0.09 Ohms and a field resistance, including rheostat, of 80 Ohms. The motor is wired to a 30 Volt source and draws 34 mps at a speed of 800 rpm. separate test is run with the motor uncoupled from the load and adjusted for 800 rpm for which the armature current is measured to be.8. P6.6 separately-excited DC motor has a rated output of Hp at 800 rpm. It operates on 55 Volts, the armature resistance is 0.36 Ohms and the counter emf (E) generated is 53 Volts for rated load. The field current is supplied by a separate 50 volts power supply. The following resistances apply to the field circuit: R F = 00 Ohms and R C =40 Ohms. Sketch the equivalent circuit for the motor and determine: a. the value of armature current for rated load, b. the output torque at rated load, c. the power converted from electrical to mechanical form, d. The mechanical power losses, e. The torque developed, f. The armature copper loss, g. nd the efficiency of the motor. h. If a.5 Ohm resistance is placed in series with the motor s armature, the speed of the motor slows to 694 rpm. ssume that the output of the motor remains at Hp and determine I a and E. i. If the field circuit is operating in the linear range of the magnetization curve, what value of R C is needed to increase the speed back to 800 rpm from that of part (h)? j. Sketch the power flow diagram for this motor. P6.7 DC shunt motor is rated to deliver an output of 0 Hp at 800 rpm. This machine is energized by 220 Volt mains which supply an input line current of 44 mps. The total resistance in the shunt field circuit is 25.7 Ohms (R C = 25.7 Ohms and R F = 00 Ohms). The armature resistance, R is 0.85 Ohms. For these rated conditions, determine: a. the shunt field current, I F, 28