AUTOMATIC VOLTAGE CONTROL
An alternator will experience large changes in its terminal voltage with changes in the load current and the load power factor because of the combined effects of the armature reactance and the armature reaction. However, a relatively constant terminal voltage can be maintained under changing load conditions by the use of an automatic voltage regulator.
Automatic voltage regulators change the alternator field current to adjust for changes in the load current. As the terminal voltage decreases, a relay closes contacts in the regulator that short out a field resistor. There is a resulting increase in the field current. There are also increases in the field flux and the induced voltage. An increase in terminal voltage causes the relay to open the contacts across the field resistor. This action causes a decrease in the field current, the field flux, and the induced voltage.
A simplified schematic diagram of an automatic voltage regulator is given in Figure 12–29. Relay coil A is connected across one phase of the three-phase output of the alternator. During normal operation, relay coil A causes contacts B to open and close several times each second. The exciter generator supplies the alternator field with nearly constant values of dc excitation voltage and current. If the terminal voltage decreases, the voltage across the relay decreases. As a result, the contacts remain closed for longer time intervals. This causes the excitation current supplied to the alternator field to increase. The ac terminal voltage output of the generator rises to its original value. If the load on the alternator changes so that the terminal voltage increases, then the contacts will vibrate at a greater rate. There is a resulting decrease in the time that the resistor in series with the shunt field of the dc exciter generator is short-circuited. The terminal voltage decreases to its normal value. Alternators may be operated with many other types of automatic voltage regulators using vacuum tubes, amplidynes, magnetic amplifiers, ignition rectifiers, and silicon controlled rectifiers.
SATURATION CURVE
For a constant-speed generator, the generated voltage is a direct function of the flux value per pole. At no load, the flux of each pole is determined by the number of ampere- turns of the field pole. Because each field winding consists of a constant number of turns, the flux is proportional to the dc excitation current.
Direct Current Fundamentals explained that when iron is magnetized, the mole- cules are arranged in a definite pattern. To align more of the iron molecules in this pat- tern requires a proportional increase in the number of ampere-turns. In other words, as the number of ampere-turns increases, there is an almost proportional increase in the flux. With a decrease in the number of unaligned iron molecules, it is more difficult to increase the flux in the magnetic circuit. A point is reached at which the flux no longer increases in proportion to the increase in the magnetomotive force. This point is called the saturation point. As the magnetomotive force is increased beyond the saturation point, the flux increase becomes smaller as there are fewer unaligned molecules in the iron.
Figure 12–30 shows the connections and a typical saturation curve for an ac generator. Note in Figure 12–30A that the field winding is connected to a dc supply. The excitation current is controlled by a rheostat. An ammeter in the field circuit indicates the value of the excitation current. A voltmeter is connected across one pair of line terminals to measure the induced voltage. Because all three voltages should be of the same magnitude, three voltmeters are not required.
Plotting the Saturation Curve
The alternator is operated at the rated speed with the field circuit deenergized. The residual magnetism in the field poles causes a low voltage to appear across the terminals of the alternator. The field rheostat is adjusted to its maximum value, and the dc field circuit is energized. The field current and the induced voltage values are recorded. The field current
is then increased by fixed increments. The resulting voltage values are recorded for each setting of field current. When the maximum field current is reached, the process can be reversed to obtain a descending curve. The field current is decreased back to zero by the same stepped values.
The ascending section of the saturation curve, as shown in Figure 12–30B, starts out with a small induced voltage at a field current of zero. This voltage is due to residual magnetism. As the field current increases from zero, the induced voltage increases. The voltage increase follows an almost straight-line curve up to point A, the saturation point. As the field current continues to increase, the increase in the flux and the induced voltage becomes smaller. This means that there are few unaligned molecules in the iron of the magnetic circuit. Thus, it is more and more difficult to magnetize the circuit.
Note in Figure 12–30B that the induced voltage values are slightly higher in the descending section of the curve for given values of field current. Compare these values with the voltages in the ascending part of the curve. The slight difference in the curves is due to molecular friction. The molecules of the iron in the magnetic circuit stay aligned even after the magnetomotive force is decreased. A better name for molecular friction is hysteresis effect.
Applications of the Saturation Curve
The saturation curve can be put to a number of practical uses, including the following:
1. The curve shows the approximate value of field excitation current required to obtain the maximum ac voltage output with minimal I2 R losses in the field cir- cuit. This operating point generally occurs near the center of the knee of the curve. In Figure 12–30B, note that this area is in the region of the saturation point A.
2. For small differences between the ascending and descending portions of the curve for given values of the dc field current, the hysteresis effect in the magnetic circuit is small. However, a large spread between the ascending and descending portions of the curve means that the hysteresis effect in the magnetic curve is significant.
ALTERNATOR NAMEPLATE DATA
The capacity of an alternator is given in kilovolt-amperes rather than kilowatts. This practice is followed because the machine may be required to supply a load with a power factor other than unity. Once the kVA output and the power factor are known, the kilowatt output can be determined from the following equation:
The alternator nameplate also contains the following data: the full-load terminal voltage, the rated full-load current per terminal, the number of phases, the frequency, the speed in r/min, the operating power factor, the dc field current and voltage, and the maximum temperature rise. Table 12–1 summarizes the data found on a typical nameplate.
ALTERNATOR EFFICIENCY
DC generators and alternators have nearly the same losses. The fixed or stray power losses include the bearing and brush friction losses, windage loses, and iron losses. (The iron losses include eddy current losses and hysteresis losses.) The copper losses include the I2 R losses in the armature windings and the power expended in the separately excited field circuit.
To determine the efficiency of an alternator, it can be loaded to its rated capacity and the values of the input and output power can then be measured. These values are substituted in the following equation to find the efficiency:
However, the kVA rating of the alternator can be very high. In this case, it is very difficult to find a suitable loading device having the proper voltage, current, and power factor ratings for a desired load condition. As a result, the efficiency of such alternators is determined using their losses, where
In the following problems, alternator losses and percent efficiency are to be determined.
PROBLEM 3
Statement of the Problem
A 480-V, 60-Hz, single-phase alternator delivers 18 kW to a load with a 75% lagging power factor. The generator has an efficiency of 80%. The stray power losses are 2000 W. The separately excited field takes 8 A from a 125-V dc source. Determine
1. the load current.
2. the copper losses in the alternator armature.
3. the effective resistance of the alternator armature.
4. the horsepower delivered by the prime mover.
Solution
Statement of the Problem
A 25-kVA, 250-V, 60-Hz, 1800-r/min, single-phase alternator has an effective resistance of 0.1 D and an inductive reactance of 0.5 D in its armature windings. The generator is delivering the rated load output at a unity power factor to a noninductive heating load. At the rated load and unity power factor, determine
1. the load current.
2. the copper losses in the armature windings.
3. the efficiency of the alternator if the input is 38 hp.
Solution
An alternator should be operated at or near the rated full-load output to obtain the maximum efficiency. At points where the load is light, the fixed losses are a large part of the input. Thus, the efficiency at such points is low. As the load output of an alternator increases, the fixed losses become a much smaller part of the input. This results in a marked increase in the efficiency. Alternators with capacities in the order of 200,000 kVA may have efficiency ratings as high as 96% at the full-load output.
PARALLELING ALTERNATORS
Power-generating stations operate several alternators in parallel. This practice is preferred to the use of a single large generator for the following reasons:
1. The use of several ac generators means that periodic maintenance and repairs can be made to one alternator. Because the other machines are operating in parallel, they can supply the load and prevent a power interruption due to a generator failure.
2. The load requirements for any central generating station change continually.
During light-load periods, the load demands can be met by one or two alternators
operating at a high efficiency. This means that these alternators will be operating at a high efficiency. As the load demands increase during certain periods of the day, other alternators can be connected in parallel to meet the peak demands. This procedure is more economical than the use of one huge machine operating, at certain periods of each day, at only a small fraction of its full-load capacity. Such operation results in a low efficiency.
3. Electrical power requirements are continually increasing. To meet these increased demands, utility companies can increase the physical size of their generating plants so that more alternators can be installed. These machines can be operated in parallel with the existing generating equipment. This procedure is a convenient and economi- cal way of increasing the generating capacity of a utility.
Paralleling AC Generators
To parallel dc generators, the voltages and the polarities of the generators must be the same. For ac generators, however, it must be remembered that the output voltages continuously change in both magnitude and polarity at a definite frequency. Therefore, to parallel alternators, the following conditions must be observed:
• The output voltages of the alternators must be equal.
• The frequencies of the alternators must be the same.
• The output voltages of the alternators must be in phase.
When these conditions are met, the alternators are said to be in synchronism. The following steps describe the actual process of synchronizing two three-phase alternators:
1. Assume that alternator 1 is supplying energy to the bus bars of the station at the rated voltage and frequency.
2. An incoming machine, alternator 2, is to be synchronized with alternator 1 for the first time. The speed of alternator 2 is increased until it turns at the value required to give the desired frequency. The voltage of generator 2 is adjusted by means of its field rheostat until it is equal to that of generator 1.
3. The three voltages of the incoming generator must be in phase with the respective voltages of generator 1. To accomplish this, the phase sequence of the two alternators and their frequencies must be the same. The use of synchronizing lamps is a simple way to check these relationships.
Synchronizing Two Alternators
Three-Lamp Method. The circuit shown in Figure 12–31 is used to synchronize two three phase alternators. Alternator 1 supplies energy to the load. Alternator 2 is to be paralleled with alternator 1. Three lamps are connected across the switches as shown in the figure. Each lamp is rated at the terminal voltage of the alternator. When both machines are operating, either of two actions will occur:
1. The three lamps will go on and off in unison. The rate at which the lamps go on and off depends on the frequency difference between alternator 2 and alternator 1.
2. The three lamps will light and go off, but not in unison. The rate at which this occurs depends on the frequency difference between the two generators. In this case, the phase rotation, or phase sequence, of alternator 2 is not the same as that of alternator 1. The phase sequence of alternator 2 must be corrected so that it will be the same as that of alternator 1. By interchanging the connections of any two leads of alternator 2, the phase sequence can be changed. The three synchronizing lamps should now go on and off in unison, indicating that the phase sequence is correct. A slight adjustment in the speed of the prime mover for alternator 2 makes the frequency of alternator 2 the same as that of alternator 1. As the frequency difference between the alternators decreases, the rate at which the synchronizing lamps increase and decrease in light intensity also decreases. Thus, the rate at which the lamps change in light intensity represents the difference in frequency between the two alternators.
For example, assume that the frequency of alternator 1 is 60 Hz and the frequency of the incoming generator (alternator 2) is 59 Hz. The frequency difference between the alternators is 1 Hz. This means that the synchronizing lamps will come on and go off once each second. When the lamps are off, the instantaneous electrical polarity of alternator 2 is the same as that of alternator 1. The switch for alternator 2 can be closed at this point and the ac generators will be in parallel.
Three Lamps Dark Method. One way of connecting the synchronizing lamps is called the three lamps dark method. In this method, the synchronizing lamps are connected directly across the switch from the blade to the jaw. The three lamps dark method can always be used to determine the phase sequence of an alternator. Once the phase sequence is known, permanent connections can be made between the stator windings, the switching equipment, and the station bus bars. From this point on, it is not necessary to repeat the process of determining the phase sequence each time the alternator is paralleled. Figure 12–32A shows the connections for the three lamps dark method to determine the phase sequence. This method is also used to indicate when alternators are in synchronism.
Two Bright, One Dark Method. The connections are shown in Figure 12–32B for another method of synchronizing lamps. This method is called the two bright, one dark method. This method is never used to determine the phase sequence. It is used only to indicate the synchronism of two alternators. When the incoming alternator is in synchronism (1) the two lamps in line wires 1 and 2 will have a maximum brightness, and (2) the lamp in line wire 3 will be dark.
There is one disadvantage to both lamp connections shown in Figure 12–32. Using the three dark method, there may be a large voltage difference across the synchronizing lamps (even when they are dark). Thus, a large voltage and phase difference may be present when an attempt is made to bring the incoming alternator into the bus bar circuit system with other machines. A large disturbance in the electrical system may result and damage the alternator windings.
Use of the Synchroscope
Once the phase sequence is known to be correct and permanent connections are made, a synchroscope can be used. This single-phase instrument indicates synchronism accurately. Unit 11 gives information on the construction and operation of synchroscopes.
A synchroscope gives an accurate indication of the differences in the frequency–phase (en dash between words, no space between dash and "phase" relationship between two voltages. The voltage from one phase of the three-phase bus bar system is connected to one set of synchroscope coils. The voltage from the same phase of the incoming alternator is connected to another set of synchroscope coils. A pointer is attached to the synchroscope rotor and rotates over a dial face. When the pointer stops, the frequencies of the two alter- nators are the same. When the pointer stops in a vertical upward position, the frequencies are equal and the voltages are in phase. This means that the alternators are in synchronism and the alternator switch can be closed to parallel both machines.
Figure 12–33 shows the synchroscope connections required to obtain an indication of the synchronism of two alternators. The voltage of the same phase from each alternator is applied to the coils of the synchroscope through special synchroscope switches. Each switch has one position marked “run” and another position marked “incoming.” This flex- ibility in making connections allows the synchroscope to be used when either alternator is
being synchronized with the station bus bars. The use of two voltmeter switches means that one voltmeter can be used to measure the voltages of all three phases of either alternator. Generally, Figure 12–33 shows only the panelboard connections for the synchroscope and the voltmeter. In an actual installation, the alternator panelboard will also contain a three- phase wattmeter, a three-phase power factor meter, ammeters, and control switches.
Load Sharing
Once the two alternators are operating in parallel, the load is shared between them. The load taken by each machine is proportional to the kVA rating of the machine. The division of the load between dc generators is obtained by changing the field excitation of each generator until the load is shared. However, the same method cannot be used to divide the kilowatt load between two alternators in parallel. Keep in mind that alternators in parallel must turn at a fixed speed to maintain a constant frequency. The input to steam turbines, waterwheels, and diesel units is controlled by sensitive governors. Because the governor control holds the input to these prime movers at a fixed value, the input to the alternator
will also be a fixed value. Therefore, for machines in parallel, the true power output (in kilowatts) will show very little change even when the field excitation is changed.
A different method must be used to adjust the kilowatt load between alternators in par- allel. The prime movers for such alternators should have drooping speed–load characteris- tic curves. For the alternators shown in Figure 12–33, assume that alternator 1 operates at 60 Hz. Alternator 2 also operates at 60 Hz, once it is paralleled with alternator 1. Alternator 2 should deliver very little load because it cuts the system frequency line (60 Hz) at a point close to zero. On the other hand, alternator 1 is heavily loaded. Its drooping speed–load curve cuts the system frequency line a large distance from zero.
Figure 12–34 illustrates this condition. Point A indicates that alternator 1 supplies most of the kilowatt load. At point B, alternator 2 delivers very little power to the bus bars. To divide the load equally between the two machines, the input to the prime mover must be increased. This is done by slightly opening the governor on the prime mover of alternator 2. As a result, the horsepower input to alternator 2 increases, causing the power output of alter- nator 2 to increase as well. At the same time, the governor on the prime mover of alternator 1 is closed very slightly. This action decreases the input to the prime mover of generator 1. This decrease in the horsepower input to generator 1 causes a decrease in the power output (in kilowatts) of generator 1. Careful adjustments of the governors of both prime movers can result in both speed–load characteristic curves cutting the system frequency line at the same load point (Figure 12–35).
The governors used on the prime movers usually have electrical controls to ensure that both alternators feed the same amount of power (in kilowatts) into the station bus bars. These controls are operated from the instrument panelboard. Accurate adjustments of the
governors can be made using these controls to obtain a satisfactory load division between machines. For two machines operating in parallel, the load division between the machines must be obtained without causing a change in the frequency. For example, if the governor of alternator 2 is opened slightly, the frequency of the two alternators will increase. To maintain the original frequency, the speed–load curve of alternator 1 must be lowered at the same time that the speed–load curve of alternator 2 is raised. Refer to Figure 12–35 and note that the drooping speed–load curve of alternator 2 is raised and that of alternator 1 is lowered. As a result, both curves cut the original system frequency line (60 Hz) at the same load point.
Alternators operating in parallel in various central stations often will operate in a network power system. A faulty governor will cause the speed of an alternator to increase so that the machine is pulled out of synchronism. However, certain reactions within the generator will prevent this condition. Also, if the governor malfunctions so that it cuts out completely and shuts off the input to the prime mover, the alternator will operate as a synchronous motor until the governor fault is corrected.
Effect of Field Excitation
It was stated earlier in this unit that any change in the division of the kilowatt load between the alternators is due to changing the input to the prime movers of the machines. The field excitation to the alternators is not changed. What is the effect of changing the field excitation on alternators operating in parallel?
Following the previous explanation of the paralleling of alternators, assume that even after alternator 2 is connected in parallel with alternator 1, alternator 2 delivers only a
small kilowatt input to the bus bars. It is also assumed that both generators operate at unity power factor with a unity power factor load. The vector diagram for this circuit is shown in Figure 12–36. The diagram shows that the currents delivered by the respective alternators (I and I ) are in phase with the line voltage. The arithmetic sum of the two alternator currents equals the total current supplied to the load (I ). This total current is also in phase with the line voltage (because the load has a unity power factor).
The field excitation of alternator 2 can be increased and the field of alternator 1 can be weakened in an attempt to divide the kilowatt load equally between the two alternators. However, the power output of each generator remains nearly the same. (It is assumed that the input to the prime mover of each alternator is unchanged.) There is an increase in the current output of each alternator. Also, both machines no longer operate at a unity power factor. It is apparent that when the field excitation current of alternator 2 is increased, the internal induced voltage of this machine also increases.
For this case, the power factor decreases from unity to a value in the lagging quadrant. Thus, a greater induced voltage is required to maintain the same terminal voltage. If the field of alternator 1 is weakened, the internal induced voltage of this machine decreases. In addition, the power factor of alternator 1 will decrease from unity to a value in the lead- ing quadrant. A leading power factor means that the same terminal voltage can be main- tained with a lower internal induced voltage. (This unit has already explained the effects of armature reactance and armature reaction on the terminal voltage of an alternator for different power factor loads.) The armature reactance and armature reaction give rise to internal operating conditions that adjust themselves. As a result, the terminal voltage and the power output of each alternator in parallel remain nearly constant with changes in the field excitation.
Figure 12–37 shows the circuit conditions after the field excitation of both alternators is changed. When the field excitation of alternator 2 is strengthened, the current value of this machine increases and lags the line voltage. However, the in-phase component of cur- rent for this machine is unchanged. When the field of alternator 1 is weakened, the current of this machine increases and leads the line voltage. The in-phase current of alternator 1 also is unchanged by a change in the field excitation. The quadrature current surges back and forth between the alternators. This current causes greater I2R copper losses in both generators. Because the in-phase current supplied by each alternator remains the same, the power (in kilowatts) supplied by each generator to the bus bars is unchanged. Of course, the current and power will be determined by the load.
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PROBLEM 5
Statement of the Problem
Two single-phase alternators are each rated at 30 kVA, 240 V, 60 Hz. The alternators are operated in parallel to supply power to a load having a unity power factor. The load requires 48 kW at 240 V. Both alternators operate at a unity power factor. The output of alternator 1 is 30 kW and the output of alternator 2 is 18 kW. Determine
1. the current supplied by each alternator.
2. the total current supplied to the load.
The vector diagram is the same as the one shown in Figure 12–36. Both alternators operate at a unity power factor. Thus, the current of each alternator is in phase with the line voltage. The total line current is also in phase with the line voltage because the load operates at a unity power factor.
Solution
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The arithmetic sum of the two alternator currents equals the total current supplied to the load. This is true because each current value is in phase with the line voltage.
PROBLEM 6
Statement of the Problem
An attempt is made to redistribute the load between the alternators in problem 5 by weakening the field excitation of alternator 1 and strengthening the field excitation of alternator 2. The resulting power output of both machines remains nearly the same, as in the previous case. Refer to Figure 12–37. (The input to the prime movers remains the same.) After the field of alternator 1 is weakened and the field of alternator 2 is strengthened, the power factor of alternator 1 is 0.90 leading. Determine
1. the current delivered by alternator 1.
2. the quadrature (quad.) current that circulates between the alternators.
3. the current delivered by alternator 2.
4. the value of lagging power factor for alternator 2.
5. the current required by the load.
Solution
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4. cos 02 = 0.779 lagging power factor
5. The power factor of the load is determined by the electrical characteristics of the load alone. Therefore, the power factor remains at unity. If the terminal voltage is kept to 240 V, the load current will remain the same:
Problem 6 shows that the power output of alternators is not increased by changing the field excitation of the alternators. Only the VARs load on each alternator is increased when the field excitation is changed. To distribute the load properly, the input to the prime mover of alternator 1 must be decreased and the input to the prime mover of alternator 2 must be increased. If the input to the prime mover for alternator 1 is decreased, the drooping speed–load curve of this unit will be lowered. If the input to the prime mover of alternator 2 is increased at the same time, the drooping speed–load curve of this unit will rise. Adjusting the governor controls of both units causes the two drooping speed–load curves to cut the system frequency line (60 Hz) at the same load point. This means that the kilowatt outputs of the generators are equal and the power factor for both units will be unity, if there is no change in the field excitation of either alternator after the machines are in parallel.
The vector diagram for the two alternators is shown in Figure 12–38. An equal distribution of load is obtained by changing the input to the prime movers of the genera- tors. The power factor of each alternator is unity. The currents for both alternators are 100
A. The kilowatt load on each unit is
Labels: Alternating current fundamentals
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