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THE SINGLE-PHASE, THREE-WIRE SYSTEM

Nearly all residential and commercial electrical installations use a single-phase, three- wire service similar to the one shown in Figure 13–12. This type of service has a number of advantages:

1. The system provides two different voltages. The lower voltage supplies lighting and small-appliance loads, and the higher voltage supplies heavy-appliance and single- phase motor loads.

2. There are 240 V across the outside wires. Thus, the current for a given kilowatt load can be reduced by nearly half if the load is balanced between the neutral and the two outside wires. Because of the reduction in the current, the voltage drop in the circuit conductors is reduced, and the voltage at the load is more nearly constant. In addition, the following problems are minimized: dim lights, slow heating, and unsatisfactory appliance performance.

3. A single-phase, three-wire, 120/240-V system uses 37% less copper as compared to a 120-V, two-wire system having the same capacity and transmission efficiency.

Problem 6 shows why less copper is needed in such a single-phase, three-wire circuit.

PROBLEM 6

Statement of the Problem

Figure 13–13 shows a balanced single-phase, three-wire circuit. Two 10-A noninductive heater units are connected to each side of this circuit. The conductors used in this circuit are no. 12 AWG wire. The distance from the source to the load is 100 ft. Determine

1. the voltage drop in the line wires.

2. the percentage voltage drop.

3. the weight of the copper used for the three-wire system.

Solution

1. The total current taken by the two noninductive heater units connected between line 1 and the neutral wire is 10 + 10 = 20 A.

The two heater units connected between line 2 and the neutral wire also take a total current of 20 A.

The current in the neutral wire of a single-phase, three-wire noninductive circuit is the difference between the currents in the two line wires. For a balanced circuit, such as that of Figure 13–13, the current in the neutral wire is zero. This means that the actual current path for the 20-A current is the two no. 12 line wires. The actual voltage drop is

2. Because the source voltage across the two outside legs of the system is 240 V, the percentage voltage drop is 6.4/240 = 0.02666 = 2.67%.

3. The weight of the copper used in the single-phase, three-wire system is determined as follows:

Weight of 200 feet of no. 12 AWG wire = 1.98 lb

Weight of 300 feet of no. 12 AWG wire = 5.94 lb

PROBLEM 7

Statement of the Problem

A 120-V, two-wire circuit is shown in Figure 13–14. Note that the four noninductive heater units are connected in parallel. Each heater unit takes 10 A. The allowable percentage voltage drop is 3%. This value is the same as that used in problem 6 for the single-phase, three-wire system. Determine

1. the wire size in circular mils that will give the same percentage voltage drop and, as a consequence, the same transmission efficiency as specified for the circuit in problem 6.

2. the AWG wire size of the 120-V, two-wire system.

3. the weight of the copper used in the 120-V, two-wire system.

4. the amount of copper saved by using the single-phase, three-wire system.

Solution

1. Desirable voltage drop:

2. A circular mil area of 26,000 circular mils requires no. 6 AWG wire.

3. The weight of 100 ft of no. 6 AWG wire is 7.95 lb. Therefore, the total weight of the 200 ft of wire used in this circuit is 15.9 lb.

4. The actual percentage of copper used in a single-phase, three-wire system, as com- pared with an equivalent single-phase, 120-V, two-wire system, is

Therefore, the maximum amount of copper saved by using the single-phase, three- wire system is 62.65%. The actual amount of copper saved is closer to 50% because the copper used in the three-wire, single-phase system is normally sized larger than the permitted minimum size.

Circuit with Unbalanced Loads

Figure 13–15 shows a single-phase, three-wire system having unbalanced noninductive lighting and heating loads from line 1 to the neutral and from line 2 to the neutral. The function of the neutral now is to maintain nearly constant voltages across the two sides of the system in the presence of different currents. For the ac noninductive circuit in Figure 13–15, the current in the grounded neutral wire is the difference between the currents in the two outside legs. Because this is an ac circuit, the arrows show only the instantaneous directions of current. For the instant shown, X is instantaneously negative and supplies 10 A to line 1. Line 2 returns 15 A to X , which is instantaneously positive. The neutral wire conducts the difference between these two line currents. This difference is 5 A.

For the instant shown in Figure 13–15, the current in the neutral wire is in the direction from the transformer to the load. Line 1 supplies a total of 10 A, 5 A to the heater load, and 5 A to the lighting load connected between line 1 and the neutral. However, 10 A is required by the lighting load connected from the neutral to line 2. Therefore, the neutral wire supplies the difference of 5 A to meet the demands of the lighting load connected

between line 2 and the neutral wire. Kirchhoff’s current law can be applied to this circuit. Recall that the current law states that the sum of the currents at any junction point in a network system is always zero.

In a single-phase, three-wire circuit, the neutral wire is always grounded. Also, it is never fused or broken at any switch control point in the circuit. This direct path to ground by way of the neutral wire means that any instantaneous high voltage, such as that due to lightning, will be instantly discharged to ground. As a result, electrical equipment and the circuit wiring are protected from major lightning damage.

Circuit with an Open Neutral

There is an even more important reason for not breaking the neutral wire. Recall that the neutral wire helps maintain balanced voltages between each line wire and the neutral wire, even with unbalanced loads. Assume that the neutral wire in Figure 13–15 is opened accidentally. The two lighting loads are now in series across 240 V.

Figure 13–16 shows the circuit of Figure 13–15, but with an open neutral. The 48-W heater unit still takes 5 A at voltage of 240 V. However, the two lighting loads act like two resistances in series across 240 V. The total resistance of the two lighting loads in series is 24 + 12 = 36 W.

The following values can be calculated for this circuit. The current is

The lighting load connected between line 1 and the neutral wire now has 160 V through it rather than its rated voltage of 120 V. This higher voltage will cause the lamps to burn out. The lighting load connected between line 2 and the neutral wire receives only 80 V, instead of 120 V. This voltage is not sufficient for adequate operation of the load. This type of unbalanced voltage condition would be common if fuses or switch control devices were placed in the neutral wire. In practice, the neutral wire is always carried as a solid conductor through the entire circuit.

POLARITY MARKINGS

The American Standards Association (ASA) has developed a standard system of marking transformer leads. The high-voltage winding leads are marked H and H . The low-voltage winding leads are marked X and X . The H lead is always located on the left-hand side when the transformer is faced from the low-voltage side. When H is instantaneously positive, X is also instantaneously positive.

Transformers with subtractive (buck) and additive (boost) polarities are shown in Figure 13–6. In a transformer with subtractive polarity, the H and X leads are adjacent to or directly across from each other. The H and X leads of a transformer with additive polarity are diagonally across from each other. The arrows in the figure indicate the instantaneous directions of the voltage in the windings.

Standard Test Procedure

Transformer leads normally have identifying tabs or tags marked H₁, H₂ and X₁ , X₂ .

However, because it may be impossible to identify the leads because the tags are missing

or disfigured, a standard test procedure can be used.

Figure 13–7A shows a test being made on a transformer with additive (boost) polarity. In this test, a jumper lead is temporarily connected between the high-voltage lead (H ) and the low-voltage lead directly across from it. A voltmeter is connected between the other high-voltage lead (H ) and the low-voltage lead directly across from it. If the voltmeter reads the sum of the primary input voltage and the secondary voltage, the trans- former has additive (boost) polarity. The sum is 2400 V + 240 V = 2640 V. When H1 is instantaneously positive, 240 V is induced in the secondary winding. The input voltage (2400 V) is applied to X through the temporary jumper connection. This value adds to the 240 V so that the potential difference is 2640 V, as indicated on a voltmeter connected from X to H . Note that the path from X to H has the same direction as the voltage arrows. As a result, the two voltages are added.

Low-Voltage Testing

There is a hazard involved in making the previous test at high voltage values. Thus, a test using a relatively low voltage was developed to determine transformer polarity. For example, in Figure 13–7B, 240 V is used as the test voltage. This potential is usually available in the laboratory or repair shop. By impressing 240 V on the 2400-V winding, a voltage of 24 V is induced in the secondary winding of the transformer.

The voltage ratio is 10:1. The voltmeter is connected between H and the low-voltage lead X1. The reading on the voltmeter is 240 V + 24 V = 264 V. This means that an ac voltmeter, with a range of 0 to 300 V, can be used to determine the polarity of the transformer. The voltmeter is connected between the high-voltage lead (H ) and the low-voltage lead directly across from it. The meter indicates the sum of the primary and secondary voltages. A transformer with this type of polarity markings is an additive polarity type.

In Figure 13–8A, the same polarity test connections are used for a transformer with subtractive (buck) polarity. The 240 V induced in the secondary winding opposes the 2400 V entering X from the temporary jumper connection. The voltmeter is connected

between H₂ and X₂ and indicates a value of 2400 V – 240 V, or 2160 V. Figure 13–8B shows a polarity test using a low voltage of 240 V. In this case, the voltmeter indicates the difference between the primary and secondary voltages. This difference is 240 V – 24 V = 216 V.

The direction from X to H opposes the voltage arrow from X to X and is the same as the voltage arrow from H to H . Therefore, the X , X voltage is subtracted from the H , H voltage.

ASA AND NEMA STANDARDS

The American Standards Association (ASA) and the National Electrical Manufacturers Association (NEMA) developed the following standards that relate to the polarity of transformers:

1. Additive polarity shall be standard for single-phrase transformers up to 200 kVA, and having voltage ratings not in excess of 9000 V.

2. Subtractive polarity shall be standard for all single-phase transformers larger than 200 kVA, regardless of the voltage rating.

3. Subtractive polarity shall be standard for all single-phase transformers in sizes of 200 kVA and below, having high voltage ratings above 9000 V.

The polarity of a single-phase transformer must be known before it can be connected in parallel with other single-phase transformers or in a three-phase bank. This information normally is provided on the transformer lead tags on the nameplate of the machine. However, when such information is not available, the standard polarity test just explained should be used to determine the polarity.

TRANSFORMERS IN PARALLEL

Single-phase transformers often must be operated in parallel. Several conditions must be satisfied to ensure that the current outputs of the transformers will be in proportion to the kVA capacity of the transformers. These conditions are as follows: (1) the transformers must have the same secondary terminal voltages; (2) the transformer polarities must be correct; and (3) each transformer must have the same percent impedance.

Two stepdown transformers are shown in Figure 13–9. If these transformers have the same voltage ratings, percent impedance values, and additive polarity, they can be connected in parallel. The following steps are used to connect the transformers:

1. The high-voltage leads (H ) of both transformers are connected to one line wire. The other two primary high-voltage leads (H ) are connected to the other line wire.

2. The low-voltage leads (X ) of both transformers are connected to one secondary line wire. The other two low-voltage leads are connected to the other low-voltage line wire.

These two transformers satisfy the three conditions listed previously. As a result, they will both deliver secondary currents to the load in proportion to their kVA ratings.

Transformers of Unknown Polarities

When transformers are supplied by different manufacturers and it is not known whether they have additive or subtractive polarity, the following test may be used. It is assumed that one transformer operates as a stepdown transformer to supply energy to the 120-V bus bars. This transformer is called transformer 1. Transformer 2 is to be paralleled with the first transformer. Transformer 2 has the same voltage ratings and percent impedance as transformer 1, but its polarity is not known.

Figure 13–10 shows that transformer 1 has additive polarity. Regardless of the polarity of transformer 2, its H lead is always on the left-hand side when viewed from the low-voltage side of the transformer. This means that the H lead is connected to the same high-voltage line wire as the H lead of transformer 1. Thus, the H lead of transformer is connected to the other side of the high-voltage line. One of the low-voltage leads of transformer 2 is connected to one side of the 120-V secondary. A voltmeter is connected between the other side of the 120-V secondary and that unconnected secondary lead of transformer 2. If transformer 2 has subtractive polarity, the voltmeter reading is twice the secondary coil voltage. In this case, the voltmeter indicates 240 V.

The instantaneous voltage directions are shown in Figure 13–10. The reason why the voltmeter indicates 240 V is evident by reviewing these instantaneous voltage directions. Assume that the X lead of transformer 2 is connected to the secondary line wire

where the voltmeter is already connected. There will be a potential difference of 240 V at the connection point, resulting in a short circuit.

In Figure 13–11, the low-voltage lead X of transformer 2 is reconnected to the other secondary line wire. The voltmeter now shows a zero potential because the secondary leads of both transformers have the same instantaneous polarity. The voltmeter can be removed and the final connections made without fear of a short circuit.

Stepdown Transformers in Parallel

When stepdown transformers are operated in parallel, as in Figure 13–9, it may be necessary to remove one of the transformers from service for repairs. To do this, the low-voltage side of the transformer is always disconnected from the 120-V line wires before the primary fuses are opened. Remember that the 120-V line wires are still ener- gized by the other transformer. If the primary fuses are opened but the low-voltage transformer leads are still connected to the 120-V line, there will be a serious safety hazard. The low-voltage winding will become a high-voltage secondary. A worker may be electrocuted if it is assumed that the high-voltage winding is deenergized because the primary fuses are open. Although the primary fuses are open, there is still 4800 V across the terminals of the high-voltage winding. As a result, a sign reading “DANGER— FEEDBACK” must be placed at each primary fuse to minimize this hazard.

THE DISTRIBUTION TRANSFORMER

A typical distribution transformer is shown in Figure 13–12. The transformer has two high-voltage windings that are rated at 2400 V each. These windings are connected to a terminal block. The block is located slightly below the level of the insulating oil in the transformer case. Small metal links are used to connect the two high-voltage windings either in series, for a 4800-V primary service, or in parallel, for a 2400-V input.

The two 2400-V primary coils shown in Figure 13–12 are connected in series. A metal link connects terminals B and C for 4800-V operation. Operation at 2400 V is obtained by connecting a metal link between terminals A and B. A second link is used to connect terminals C and D to place the two 2400-V coils in parallel. Although the distribution transformer has two high-voltage coils, note that there are only two external high-voltage

leads. These leads are marked with the standard designations H and H . These leads are permanently connected to the terminal block with lead H attached to terminal A and lead attached to terminal D.

There are four low-voltage leads. When lead H is instantaneously positive, leads X and are also instantaneously positive. At the same time, leads X and X are instantaneously negative. If leads X₂ and X₃ are commoned together, and leads X and X are commoned together, then the low-voltage coils are connected in parallel to supply an output of 120 V.

If leads X and X are connected together, the two low-voltage coils are connected in series. The resulting output is 240 V across leads X₁ and X₄ .

If there is a requirement for a 120/240-V, single-phase, three-wire service, then the following connections must be made: The two 120-V secondary windings are connected in series and a grounded neutral wire is connected between leads X₂ and X₃ . These connections are shown in Figure 13–12. The resulting service provides 120 V for lighting and small-appliance loads and 240 V for heavy-appliance and single-phase, 240-V motor loads.

POLARITY MARKINGS

The American Standards Association (ASA) has developed a standard system of marking transformer leads. The high-voltage winding leads are marked H and H . The low-voltage winding leads are marked X and X . The H lead is always located on the left-hand side when the transformer is faced from the low-voltage side. When H is instantaneously positive, X is also instantaneously positive.

Transformers with subtractive (buck) and additive (boost) polarities are shown in Figure 13–6. In a transformer with subtractive polarity, the H and X leads are adjacent to or directly across from each other. The H and X leads of a transformer with additive polarity are diagonally across from each other. The arrows in the figure indicate the instantaneous directions of the voltage in the windings.

Standard Test Procedure

Transformer leads normally have identifying tabs or tags marked H₁, H₂ and X₁ , X₂ .

However, because it may be impossible to identify the leads because the tags are missing

or disfigured, a standard test procedure can be used.

Figure 13–7A shows a test being made on a transformer with additive (boost) polarity. In this test, a jumper lead is temporarily connected between the high-voltage lead (H ) and the low-voltage lead directly across from it. A voltmeter is connected between the other high-voltage lead (H ) and the low-voltage lead directly across from it. If the voltmeter reads the sum of the primary input voltage and the secondary voltage, the trans- former has additive (boost) polarity. The sum is 2400 V + 240 V = 2640 V. When H1 is instantaneously positive, 240 V is induced in the secondary winding. The input voltage (2400 V) is applied to X through the temporary jumper connection. This value adds to the 240 V so that the potential difference is 2640 V, as indicated on a voltmeter connected from X to H . Note that the path from X to H has the same direction as the voltage arrows. As a result, the two voltages are added.

Low-Voltage Testing

There is a hazard involved in making the previous test at high voltage values. Thus, a test using a relatively low voltage was developed to determine transformer polarity. For example, in Figure 13–7B, 240 V is used as the test voltage. This potential is usually available in the laboratory or repair shop. By impressing 240 V on the 2400-V winding, a voltage of 24 V is induced in the secondary winding of the transformer.

The voltage ratio is 10:1. The voltmeter is connected between H and the low-voltage lead X1. The reading on the voltmeter is 240 V + 24 V = 264 V. This means that an ac voltmeter, with a range of 0 to 300 V, can be used to determine the polarity of the transformer. The voltmeter is connected between the high-voltage lead (H ) and the low-voltage lead directly across from it. The meter indicates the sum of the primary and secondary voltages. A transformer with this type of polarity markings is an additive polarity type.

In Figure 13–8A, the same polarity test connections are used for a transformer with subtractive (buck) polarity. The 240 V induced in the secondary winding opposes the 2400 V entering X from the temporary jumper connection. The voltmeter is connected

between H₂ and X₂ and indicates a value of 2400 V – 240 V, or 2160 V. Figure 13–8B shows a polarity test using a low voltage of 240 V. In this case, the voltmeter indicates the difference between the primary and secondary voltages. This difference is 240 V – 24 V = 216 V.

The direction from X to H opposes the voltage arrow from X to X and is the same as the voltage arrow from H to H . Therefore, the X , X voltage is subtracted from the H , H voltage.

ASA AND NEMA STANDARDS

The American Standards Association (ASA) and the National Electrical Manufacturers Association (NEMA) developed the following standards that relate to the polarity of transformers:

1. Additive polarity shall be standard for single-phrase transformers up to 200 kVA, and having voltage ratings not in excess of 9000 V.

2. Subtractive polarity shall be standard for all single-phase transformers larger than 200 kVA, regardless of the voltage rating.

3. Subtractive polarity shall be standard for all single-phase transformers in sizes of 200 kVA and below, having high voltage ratings above 9000 V.

The polarity of a single-phase transformer must be known before it can be connected in parallel with other single-phase transformers or in a three-phase bank. This information normally is provided on the transformer lead tags on the nameplate of the machine. However, when such information is not available, the standard polarity test just explained should be used to determine the polarity.

TRANSFORMERS IN PARALLEL

Single-phase transformers often must be operated in parallel. Several conditions must be satisfied to ensure that the current outputs of the transformers will be in proportion to the kVA capacity of the transformers. These conditions are as follows: (1) the transformers must have the same secondary terminal voltages; (2) the transformer polarities must be correct; and (3) each transformer must have the same percent impedance.

Two stepdown transformers are shown in Figure 13–9. If these transformers have the same voltage ratings, percent impedance values, and additive polarity, they can be connected in parallel. The following steps are used to connect the transformers:

1. The high-voltage leads (H ) of both transformers are connected to one line wire. The other two primary high-voltage leads (H ) are connected to the other line wire.

2. The low-voltage leads (X ) of both transformers are connected to one secondary line wire. The other two low-voltage leads are connected to the other low-voltage line wire.

These two transformers satisfy the three conditions listed previously. As a result, they will both deliver secondary currents to the load in proportion to their kVA ratings.

Transformers of Unknown Polarities

When transformers are supplied by different manufacturers and it is not known whether they have additive or subtractive polarity, the following test may be used. It is assumed that one transformer operates as a stepdown transformer to supply energy to the 120-V bus bars. This transformer is called transformer 1. Transformer 2 is to be paralleled with the first transformer. Transformer 2 has the same voltage ratings and percent impedance as transformer 1, but its polarity is not known.

Figure 13–10 shows that transformer 1 has additive polarity. Regardless of the polarity of transformer 2, its H lead is always on the left-hand side when viewed from the low-voltage side of the transformer. This means that the H lead is connected to the same high-voltage line wire as the H lead of transformer 1. Thus, the H lead of transformer is connected to the other side of the high-voltage line. One of the low-voltage leads of transformer 2 is connected to one side of the 120-V secondary. A voltmeter is connected between the other side of the 120-V secondary and that unconnected secondary lead of transformer 2. If transformer 2 has subtractive polarity, the voltmeter reading is twice the secondary coil voltage. In this case, the voltmeter indicates 240 V.

The instantaneous voltage directions are shown in Figure 13–10. The reason why the voltmeter indicates 240 V is evident by reviewing these instantaneous voltage directions. Assume that the X lead of transformer 2 is connected to the secondary line wire

where the voltmeter is already connected. There will be a potential difference of 240 V at the connection point, resulting in a short circuit.

In Figure 13–11, the low-voltage lead X of transformer 2 is reconnected to the other secondary line wire. The voltmeter now shows a zero potential because the secondary leads of both transformers have the same instantaneous polarity. The voltmeter can be removed and the final connections made without fear of a short circuit.

Stepdown Transformers in Parallel

When stepdown transformers are operated in parallel, as in Figure 13–9, it may be necessary to remove one of the transformers from service for repairs. To do this, the low-voltage side of the transformer is always disconnected from the 120-V line wires before the primary fuses are opened. Remember that the 120-V line wires are still ener- gized by the other transformer. If the primary fuses are opened but the low-voltage transformer leads are still connected to the 120-V line, there will be a serious safety hazard. The low-voltage winding will become a high-voltage secondary. A worker may be electrocuted if it is assumed that the high-voltage winding is deenergized because the primary fuses are open. Although the primary fuses are open, there is still 4800 V across the terminals of the high-voltage winding. As a result, a sign reading “DANGER— FEEDBACK” must be placed at each primary fuse to minimize this hazard.

THE DISTRIBUTION TRANSFORMER

A typical distribution transformer is shown in Figure 13–12. The transformer has two high-voltage windings that are rated at 2400 V each. These windings are connected to a terminal block. The block is located slightly below the level of the insulating oil in the transformer case. Small metal links are used to connect the two high-voltage windings either in series, for a 4800-V primary service, or in parallel, for a 2400-V input.

The two 2400-V primary coils shown in Figure 13–12 are connected in series. A metal link connects terminals B and C for 4800-V operation. Operation at 2400 V is obtained by connecting a metal link between terminals A and B. A second link is used to connect terminals C and D to place the two 2400-V coils in parallel. Although the distribution transformer has two high-voltage coils, note that there are only two external high-voltage

leads. These leads are marked with the standard designations H and H . These leads are permanently connected to the terminal block with lead H attached to terminal A and lead attached to terminal D.

There are four low-voltage leads. When lead H is instantaneously positive, leads X and are also instantaneously positive. At the same time, leads X and X are instantaneously negative. If leads X₂ and X₃ are commoned together, and leads X and X are commoned together, then the low-voltage coils are connected in parallel to supply an output of 120 V.

If leads X and X are connected together, the two low-voltage coils are connected in series. The resulting output is 240 V across leads X₁ and X₄ .

If there is a requirement for a 120/240-V, single-phase, three-wire service, then the following connections must be made: The two 120-V secondary windings are connected in series and a grounded neutral wire is connected between leads X₂ and X₃ . These connections are shown in Figure 13–12. The resulting service provides 120 V for lighting and small-appliance loads and 240 V for heavy-appliance and single-phase, 240-V motor loads.

Transformers

TRANSFORMER EFFICIENCY

A transformer does not require any moving parts to transfer energy. This means that there are no friction or windage losses, and the other losses are slight. The resulting efficiency of a transformer is high. At full load, the efficiency of a transformer is between 96% and 97%. For a transformer with a very high capacity, the efficiency may be as high as 99%. Transformers can be used for very high voltages because there are no rotating windings and the stationary coils can be submerged in insulating oil. Transformer maintenance and repair costs are relatively low because of the lack of rotating parts.

THE EXCITING CURRENT

When the primary winding of a transformer is connected to an alternating voltage, there will be a small current in the input winding. This current is called the exciting current and exists even when there is no load connected to the secondary.

The exciting current sets up an alternating flux in the core. This flux links the turns of both windings as it increases and decreases in opposite directions. As the flux links the turns of the secondary winding, an alternating voltage is induced in the secondary. This voltage has the same frequency as, but its direction is opposite that of, the primary winding voltage. The same voltage is induced in each turn of both windings because the same flux links the turns of both windings. As a result, the total induced voltage in each winding is directly proportional to the number of turns in that winding.

PRIMARY AND SECONDARY VOLTAGE RELATIONSHIPS

The relationship between the induced voltage and the number of turns in a winding is given in the following expression

In this expression, V is the voltage induced in the primary according to Lenz’s law.

This induced voltage is only 1% or 2% less than the applied primary voltage in a typical transformer. Thus, V and V , respectively, are used to represent the input and output voltages of the transformer.

PROBLEM 1

Statement of the Problem

A transformer has 300 turns on its high-voltage winding and 150 turns on its low-voltage winding. It is used as a stepdown transformer. With 240 V applied to the high- voltage primary winding, determine the induced voltage on the secondary winding.

Solution

PRIMARY AND SECONDARY CURRENT RELATIONSHIPS

When a load is connected across the terminals of the secondary winding, the instantaneous direction of the current will tend to oppose the effect that is producing the current.

As an example of this effect, consider the simple transformer diagram of Figure 13–2. A noninductive load is connected to the terminals of the secondary winding. The secondary current sets up a magnetomotive force that opposes the flux (f) of the primary winding. As a result, both the primary flux and the counterelectromotive force in the primary winding are reduced. The primary current increases because the impressed primary voltage has less opposi- tion from the counterelectromotive force (induced voltage). The increase in the primary current supplies the energy required by the load connected to the secondary winding.

The ampere-turns of the primary winding increase the magnetizing flux.

It was stated at the beginning of this unit that the exciting current is small when compared to the rated current. Most transformer calculations neglect the exciting current. In addition, it is assumed that the primary and secondary ampere-turns are equal, as deter- mined by the following equation

PROBLEM 2

Statement of the Problem

The transformer shown in Figure 13–2 delivers 25 A at 120 V to a load with a unity power factor. Neglect the exciting current and determine

1. the primary current.

2. the secondary ampere-turns.

3. the power in watts taken by the load.

Solution

Transformers

TRANSFORMER EFFICIENCY

A transformer does not require any moving parts to transfer energy. This means that there are no friction or windage losses, and the other losses are slight. The resulting efficiency of a transformer is high. At full load, the efficiency of a transformer is between 96% and 97%. For a transformer with a very high capacity, the efficiency may be as high as 99%. Transformers can be used for very high voltages because there are no rotating windings and the stationary coils can be submerged in insulating oil. Transformer maintenance and repair costs are relatively low because of the lack of rotating parts.

THE EXCITING CURRENT

When the primary winding of a transformer is connected to an alternating voltage, there will be a small current in the input winding. This current is called the exciting current and exists even when there is no load connected to the secondary.

The exciting current sets up an alternating flux in the core. This flux links the turns of both windings as it increases and decreases in opposite directions. As the flux links the turns of the secondary winding, an alternating voltage is induced in the secondary. This voltage has the same frequency as, but its direction is opposite that of, the primary winding voltage. The same voltage is induced in each turn of both windings because the same flux links the turns of both windings. As a result, the total induced voltage in each winding is directly proportional to the number of turns in that winding.

PRIMARY AND SECONDARY VOLTAGE RELATIONSHIPS

The relationship between the induced voltage and the number of turns in a winding is given in the following expression

In this expression, V is the voltage induced in the primary according to Lenz’s law.

This induced voltage is only 1% or 2% less than the applied primary voltage in a typical transformer. Thus, V and V , respectively, are used to represent the input and output voltages of the transformer.

PROBLEM 1

Statement of the Problem

A transformer has 300 turns on its high-voltage winding and 150 turns on its low-voltage winding. It is used as a stepdown transformer. With 240 V applied to the high- voltage primary winding, determine the induced voltage on the secondary winding.

Solution

PRIMARY AND SECONDARY CURRENT RELATIONSHIPS

When a load is connected across the terminals of the secondary winding, the instantaneous direction of the current will tend to oppose the effect that is producing the current.

As an example of this effect, consider the simple transformer diagram of Figure 13–2. A noninductive load is connected to the terminals of the secondary winding. The secondary current sets up a magnetomotive force that opposes the flux (f) of the primary winding. As a result, both the primary flux and the counterelectromotive force in the primary winding are reduced. The primary current increases because the impressed primary voltage has less opposi- tion from the counterelectromotive force (induced voltage). The increase in the primary current supplies the energy required by the load connected to the secondary winding.

The ampere-turns of the primary winding increase the magnetizing flux.

It was stated at the beginning of this unit that the exciting current is small when compared to the rated current. Most transformer calculations neglect the exciting current. In addition, it is assumed that the primary and secondary ampere-turns are equal, as deter- mined by the following equation

PROBLEM 2

Statement of the Problem

The transformer shown in Figure 13–2 delivers 25 A at 120 V to a load with a unity power factor. Neglect the exciting current and determine

1. the primary current.

2. the secondary ampere-turns.

3. the power in watts taken by the load.

Solution

LEAKAGE FLUX

Some of the lines of flux produced by the primary winding do not link the turns of the secondary winding in most transformers. The magnetic circuit for this primary leakage flux is in air. In other words, the leakage flux does not follow the circuit path through the core. This flux links the turns of the primary winding, but it does not link the turns of the secondary winding. Because the leakage flux uses part of the impressed primary voltage, there is a reactance voltage drop in the primary winding. The result is that both the secondary flux linkages and the secondary induced voltage are reduced.

A second leakage flux links the secondary turns but not the primary turns. This flux also has its magnetic path in air and not in the core. The secondary leakage flux is proportional to the secondary current. There is a resulting reactance voltage drop in the secondary winding. Both the primary and the secondary leakage fluxes reduce the secondary terminal voltage of the transformer as the load increases. The leakage flux of

a transformer can be controlled by the type of core used. The placement of the primary and secondary coils on the legs of the core is also a controlling factor in the amount of leakage flux produced.

EXCITING CURRENT AND CORE LOSSES

When there is no load attached to the secondary winding, the current input to the primary winding usually ranges from 2% to 5% of the full-load current. The primary current at no load is called the exciting current. This current supplies the alternating flux and the losses in the transformer core. These losses are known as core losses and consist of eddy current losses and hysteresis losses. As the magnitude of the alternating flux increases and decreases, the metal core is cut by the flux, as are the turns of the primary and secondary coils. Voltages are thus induced in the metal core and give rise to eddy currents. These eddy currents circulate through the core and cause I2 R loses, which must by supplied by the exciting current. Eddy current losses can be reduced by laminating the core structure. Each lamination normally is coated with a film of insulating varnish. When a protective film of varnish is not used, the oxide coating on each lamination still reduces the eddy cur- rent losses to some extent.

The core structure also experiences a hysteresis loss. In each second, the millions of molecules in the core structure are reversed many times by the alternating flux. Power is required to overcome this molecular friction in the core. This power is supplied by the primary winding. To decrease the amount of power used, a special steel such as silicon steel is used. Eddy current and hysteresis losses are called the core losses. In a typical transformer, these losses are relatively small.

To supply the magnetizing flux, the exciting current has a relatively large component of quadrature or magnetizing current. A smaller in-phase component of current supplies the core losses. For an actual transformer, the no-load power factor ranges between 0.05 and 0.10 lagging. This means that the phase angle between the exciting current and the impressed primary voltage is between 84° and 87° lagging. The exciting current can be measured by a method known as the core loss open-circuit test.

Measuring Core Losses

The connections for the core loss test are shown in Figure 13–3A. The high-voltage winding of the transformer is rated at 2400 V. The low-voltage winding is rated at 240 V and is used as the primary side of the transformer. This arrangement makes it convenient to use 240 V for the potential circuits of the wattmeter and voltmeter. The test circuit can also use 240 V safely.

CAUTION: The high-voltage winding leads and terminal connections must be well insulated and barricaded so that no one can contact this high-voltage circuit. This test circuit can be considered hazardous because the transformer operates as a stepup transformer with a 2400-V potential across the leads of the high-voltage secondary winding.

The losses of a transformer at no load are small. This means that instrument errors must be checked. The dashed-line voltmeter connections in Figure 13–3A mean that the

voltmeter is to be disconnected when a wattmeter reading is to be taken. The voltmeter must be disconnected so that the wattmeter will not indicate the power taken by the voltmeter.

If the rated voltage and frequency are used in this circuit, then the rated alternating flux exists in the core. The resulting core loss is normal. The wattmeter indicates the core loss in watts, and the ammeter indicates the exciting current.

The error introduced by the copper loss in the primary can be neglected. The following example shows why this is so. The primary resistance (R ) equals 0.007 W. The primary copper loss is 102 times 0.007 equals 0.7 W. This value of 0.7 W is small enough to be neglected, when compared to the 140 W indicated by the wattmeter. This assumption still holds for smaller transformers because they have smaller values of exciting current and core losses.

The core loss can also be measured when the high-voltage side of the transformer is used as the primary winding. In this case, a 2400-V source is required. After compensating for instrument losses, it is found that the core loss is the same as for the case when the

low-voltage side of the transformer is used as the primary. This result is to be expected because both windings are wound on the same core. Because the same number of ampere- turns will produce the same alternating flux, when either winding is used as the primary, the core loss in watts will be the same for both cases.

Figure 13–3B shows the vector relationship between the exciting current (10 A), its in-phase component (1 A), and its quadrature (magnetizing) lagging component. The phase angle between the line voltage and the exciting current is 84.3°. The core loss is 240 W.

PROBLEM 3

Statement of the Problem

Figure 13–3A shows the connections for a core loss test on a 50-kVA, 60-Hz, single-phase transformer. The high-voltage winding of the transformer is rated at 2400 V and the low-voltage winding is rated at 240 V. The low-voltage winding is used as the primary for the core loss test. With 240 V applied to the primary winding, the wattmeter indicates a core loss of 240 W. The ammeter indicates an exciting current of 10 A. Determine

1. the power factor and the phase angle.

2. the in-phase component of the current.

3. the quadrature lagging or magnetizing component of the current.

Solution

COPPER LOSSES USING DIRECT CURRENT

The copper losses of a transformer consist of the I2 R losses in both the primary and secondary windings. If the effective resistance of each winding is known, the copper losses of a transformer can be found readily. Recall that the approximate ac resistance of an alternator is found by multiplying the dc resistance by 1.4 or 1.5. For transformers, however, the windings are not embedded in the slots of a stator, but consist of coils wound on a core. This means that the difference between the ohmic resistance and the effective resistance is small. Generally, the ac or effective resistance of a transformer is obtained by measuring the dc resistance of the winding and multiplying it by 1.1.

Figure 13–4 shows the connections required to measure the dc, or ohmic resistance, of a winding. A current-limiting resistance is used in this circuit. It is important that small dc currents be used. The voltmeter should be disconnected before the circuit is deenergized because the windings are highly self-inductive. The large value of the induced voltage could damage the voltmeter movement and pointer.

PROBLEM 4

Statement of the Problem

The resistance for each winding of a 50-kVA, 2400/240-V, 60-Hz, single-phase, step- down transformer is measured with direct current. The dc resistance of the high-voltage winding is 0.68 W. The low-voltage winding has a dc resistance of 0.0065 W. Determine

1. the effective resistance of each winding.

2. the total copper losses at full load.

Solution

2. The transformer losses are small. Thus, when determining the full-load current rating of each winding, it can be assumed that the volt-ampere input and output are the same:

TRANSFORMER LOSSES AND EFFICIENCY

Transformer losses consist of copper losses and core losses. Copper losses are the I2 R losses in the primary and secondary windings. These losses increase as the load current in the primary and secondary windings increases. The copper losses can be calculated from the current and the effective resistance for each winding.

Transformer efficiency is the ratio of the output in watts to the input in watts. The load connected to the transformer secondary often has a power factor other than unity. In such cases, the output is the product of the secondary voltage and current plus the power factor of the load. The input equals the output plus the total losses. These losses include the cop- per losses and the core losses.

The core losses can be measured as shown in Figure 13–3A. These losses consist of eddy current and hysteresis losses. The core losses remain nearly constant at all load points if the frequency and primary voltage remain constant.

If the losses are known or can be calculated, for any given load point, the transformer efficiency can be determined. The basic efficiency formula is as follows:

The efficiency of a transformer can be found by loading it at various percentages of the load from no load to full load and measuring the input and output power values. However, the losses are quite small. Unless extremely accurate instruments are used, the

results will be of little value. For example, the efficiency of a transformer at full load is usually in the range from 96% to 98%. Typically, an indicating wattmeter can have an error of one percent. This means that the error in the calculations can be as much as 50%. In addition, this method requires various loading devices having the correct cur- rent, voltage, and power factor ratings. It is both inconvenient and costly to provide such loading devices. This is particularly true for transformers having extremely large kVA capacities.

The preferred method of determining the efficiency of a transformer is to measure the losses and add this value to the nameplate output to obtain the input. The following example shows how the efficiency is obtained by measuring the losses.

PROBLEM 5

Statement of the Problem

The 50-kVA, 2400/240-V transformer described in problems 3 and 4 delivers the rated load output at a unity power factor. The core loss was found to be 240 W. The primary cop- per loss was 325.4 W and the secondary copper loss was 312.4 W. Find the efficiency at the rated output and unity power factor.

Solution

The Short-Circuit Test

Another method of measuring the copper losses of a transformer is the short-circuit test. In this test, the high-voltage winding is used as the primary and the low-voltage wind- ing is short-circuited.

The connections for the short-circuit test are shown in Figure 13–5. A variable resistor is in series with the primary winding. In this way, the current input can be con- trolled. The series resistor is adjusted until the full-load current circulated in both the primary and secondary windings. When the low-side winding of the transformer is short- circuited, only 3% to 5% of the rated primary voltage is required to obtain the full-load current in both windings. This voltage is called the impedance voltage. In other words, the impedance voltage is that voltage required to cause the rated current to flow through the impedances of the primary and secondary windings. The ratio of the impedance voltage to the rated terminal voltage yields the percentage impedance voltage, which is in the range of 3% to 5%.

The wattmeter in Figure 13–5 indicates the total copper losses of the transformer referred to the primary side. The wattmeter reading includes the core losses, which are so small that they can be neglected. The core losses are small because the voltage impressed on the primary winding in this test is very low.

To make the short-circuit test, the core loss of the transformer must be determined using the connections of Figure 13–3A. Using the 50-kVA transformer described in the previous problems, the core loss is 240 W. The readings on the instruments shown in Figure 13–5 will be as follows: a current of 21 A, a power value of 640 W, and a volt- age of 80 V. The value of 640 W represents the total copper losses of the transformer. If the dc ohmic values are used to compute the copper losses in the high- and low-voltage windings, then the total losses are

This value is nearly the same as the constant of 1.10 used to convert the ohmic resistance to the effective resistance for transformers.

The entire equivalent effective resistance of the transformer referred to the primary side can be found for a wattmeter reading of 640 W and a current of 21 A (at almost full load). When referred to the primary side, this equivalent resistance, R , is

The secondary is the output side of the transformer. Therefore, the entire resistance of the transformer must be referred to the secondary side as an equivalent effective  resistance, R_{OS :}

At the rated load, the total copper losses of this transformer are determined as follows. The full-load current is

This value of efficiency, as determined by the short-circuit test, is the same as the efficiency found in problem 5.

Assume that the same transformer is operated at 50% of the rated current capacity to supply a load with a unity power factor. The core loss is the same because the magnetizing flux is the same. However, the copper loss will decrease to one-fourth of its full- load value. Note that this decrease occurs because of the squaring operation in the I2 R formula. The I2 multiplier is only one-fourth of the original value. Use the previous formula and calculate the actual efficiency for this load condition:

If the transformer supplies a load having a power factor other than unity, the output (in watts) will decrease. However, the losses will be the same as those for a unity power factor load. For example, if the 50-kVA transformer supplies the rated output to a load with a power factor of 60% lagging, the efficiency is decreased:

LEAKAGE FLUX

Some of the lines of flux produced by the primary winding do not link the turns of the secondary winding in most transformers. The magnetic circuit for this primary leakage flux is in air. In other words, the leakage flux does not follow the circuit path through the core. This flux links the turns of the primary winding, but it does not link the turns of the secondary winding. Because the leakage flux uses part of the impressed primary voltage, there is a reactance voltage drop in the primary winding. The result is that both the secondary flux linkages and the secondary induced voltage are reduced.

A second leakage flux links the secondary turns but not the primary turns. This flux also has its magnetic path in air and not in the core. The secondary leakage flux is proportional to the secondary current. There is a resulting reactance voltage drop in the secondary winding. Both the primary and the secondary leakage fluxes reduce the secondary terminal voltage of the transformer as the load increases. The leakage flux of

a transformer can be controlled by the type of core used. The placement of the primary and secondary coils on the legs of the core is also a controlling factor in the amount of leakage flux produced.

EXCITING CURRENT AND CORE LOSSES

When there is no load attached to the secondary winding, the current input to the primary winding usually ranges from 2% to 5% of the full-load current. The primary current at no load is called the exciting current. This current supplies the alternating flux and the losses in the transformer core. These losses are known as core losses and consist of eddy current losses and hysteresis losses. As the magnitude of the alternating flux increases and decreases, the metal core is cut by the flux, as are the turns of the primary and secondary coils. Voltages are thus induced in the metal core and give rise to eddy currents. These eddy currents circulate through the core and cause I2 R loses, which must by supplied by the exciting current. Eddy current losses can be reduced by laminating the core structure. Each lamination normally is coated with a film of insulating varnish. When a protective film of varnish is not used, the oxide coating on each lamination still reduces the eddy cur- rent losses to some extent.

The core structure also experiences a hysteresis loss. In each second, the millions of molecules in the core structure are reversed many times by the alternating flux. Power is required to overcome this molecular friction in the core. This power is supplied by the primary winding. To decrease the amount of power used, a special steel such as silicon steel is used. Eddy current and hysteresis losses are called the core losses. In a typical transformer, these losses are relatively small.

To supply the magnetizing flux, the exciting current has a relatively large component of quadrature or magnetizing current. A smaller in-phase component of current supplies the core losses. For an actual transformer, the no-load power factor ranges between 0.05 and 0.10 lagging. This means that the phase angle between the exciting current and the impressed primary voltage is between 84° and 87° lagging. The exciting current can be measured by a method known as the core loss open-circuit test.

Measuring Core Losses

The connections for the core loss test are shown in Figure 13–3A. The high-voltage winding of the transformer is rated at 2400 V. The low-voltage winding is rated at 240 V and is used as the primary side of the transformer. This arrangement makes it convenient to use 240 V for the potential circuits of the wattmeter and voltmeter. The test circuit can also use 240 V safely.

CAUTION: The high-voltage winding leads and terminal connections must be well insulated and barricaded so that no one can contact this high-voltage circuit. This test circuit can be considered hazardous because the transformer operates as a stepup transformer with a 2400-V potential across the leads of the high-voltage secondary winding.

The losses of a transformer at no load are small. This means that instrument errors must be checked. The dashed-line voltmeter connections in Figure 13–3A mean that the

voltmeter is to be disconnected when a wattmeter reading is to be taken. The voltmeter must be disconnected so that the wattmeter will not indicate the power taken by the voltmeter.

If the rated voltage and frequency are used in this circuit, then the rated alternating flux exists in the core. The resulting core loss is normal. The wattmeter indicates the core loss in watts, and the ammeter indicates the exciting current.

The error introduced by the copper loss in the primary can be neglected. The following example shows why this is so. The primary resistance (R ) equals 0.007 W. The primary copper loss is 102 times 0.007 equals 0.7 W. This value of 0.7 W is small enough to be neglected, when compared to the 140 W indicated by the wattmeter. This assumption still holds for smaller transformers because they have smaller values of exciting current and core losses.

The core loss can also be measured when the high-voltage side of the transformer is used as the primary winding. In this case, a 2400-V source is required. After compensating for instrument losses, it is found that the core loss is the same as for the case when the

low-voltage side of the transformer is used as the primary. This result is to be expected because both windings are wound on the same core. Because the same number of ampere- turns will produce the same alternating flux, when either winding is used as the primary, the core loss in watts will be the same for both cases.

Figure 13–3B shows the vector relationship between the exciting current (10 A), its in-phase component (1 A), and its quadrature (magnetizing) lagging component. The phase angle between the line voltage and the exciting current is 84.3°. The core loss is 240 W.

PROBLEM 3

Statement of the Problem

Figure 13–3A shows the connections for a core loss test on a 50-kVA, 60-Hz, single-phase transformer. The high-voltage winding of the transformer is rated at 2400 V and the low-voltage winding is rated at 240 V. The low-voltage winding is used as the primary for the core loss test. With 240 V applied to the primary winding, the wattmeter indicates a core loss of 240 W. The ammeter indicates an exciting current of 10 A. Determine

1. the power factor and the phase angle.

2. the in-phase component of the current.

3. the quadrature lagging or magnetizing component of the current.

Solution

COPPER LOSSES USING DIRECT CURRENT

The copper losses of a transformer consist of the I2 R losses in both the primary and secondary windings. If the effective resistance of each winding is known, the copper losses of a transformer can be found readily. Recall that the approximate ac resistance of an alternator is found by multiplying the dc resistance by 1.4 or 1.5. For transformers, however, the windings are not embedded in the slots of a stator, but consist of coils wound on a core. This means that the difference between the ohmic resistance and the effective resistance is small. Generally, the ac or effective resistance of a transformer is obtained by measuring the dc resistance of the winding and multiplying it by 1.1.

Figure 13–4 shows the connections required to measure the dc, or ohmic resistance, of a winding. A current-limiting resistance is used in this circuit. It is important that small dc currents be used. The voltmeter should be disconnected before the circuit is deenergized because the windings are highly self-inductive. The large value of the induced voltage could damage the voltmeter movement and pointer.

PROBLEM 4

Statement of the Problem

The resistance for each winding of a 50-kVA, 2400/240-V, 60-Hz, single-phase, step- down transformer is measured with direct current. The dc resistance of the high-voltage winding is 0.68 W. The low-voltage winding has a dc resistance of 0.0065 W. Determine

1. the effective resistance of each winding.

2. the total copper losses at full load.

Solution

2. The transformer losses are small. Thus, when determining the full-load current rating of each winding, it can be assumed that the volt-ampere input and output are the same:

TRANSFORMER LOSSES AND EFFICIENCY

Transformer losses consist of copper losses and core losses. Copper losses are the I2 R losses in the primary and secondary windings. These losses increase as the load current in the primary and secondary windings increases. The copper losses can be calculated from the current and the effective resistance for each winding.

Transformer efficiency is the ratio of the output in watts to the input in watts. The load connected to the transformer secondary often has a power factor other than unity. In such cases, the output is the product of the secondary voltage and current plus the power factor of the load. The input equals the output plus the total losses. These losses include the cop- per losses and the core losses.

The core losses can be measured as shown in Figure 13–3A. These losses consist of eddy current and hysteresis losses. The core losses remain nearly constant at all load points if the frequency and primary voltage remain constant.

If the losses are known or can be calculated, for any given load point, the transformer efficiency can be determined. The basic efficiency formula is as follows:

The efficiency of a transformer can be found by loading it at various percentages of the load from no load to full load and measuring the input and output power values. However, the losses are quite small. Unless extremely accurate instruments are used, the

results will be of little value. For example, the efficiency of a transformer at full load is usually in the range from 96% to 98%. Typically, an indicating wattmeter can have an error of one percent. This means that the error in the calculations can be as much as 50%. In addition, this method requires various loading devices having the correct cur- rent, voltage, and power factor ratings. It is both inconvenient and costly to provide such loading devices. This is particularly true for transformers having extremely large kVA capacities.

The preferred method of determining the efficiency of a transformer is to measure the losses and add this value to the nameplate output to obtain the input. The following example shows how the efficiency is obtained by measuring the losses.

PROBLEM 5

Statement of the Problem

The 50-kVA, 2400/240-V transformer described in problems 3 and 4 delivers the rated load output at a unity power factor. The core loss was found to be 240 W. The primary cop- per loss was 325.4 W and the secondary copper loss was 312.4 W. Find the efficiency at the rated output and unity power factor.

Solution

The Short-Circuit Test

Another method of measuring the copper losses of a transformer is the short-circuit test. In this test, the high-voltage winding is used as the primary and the low-voltage wind- ing is short-circuited.

The connections for the short-circuit test are shown in Figure 13–5. A variable resistor is in series with the primary winding. In this way, the current input can be con- trolled. The series resistor is adjusted until the full-load current circulated in both the primary and secondary windings. When the low-side winding of the transformer is short- circuited, only 3% to 5% of the rated primary voltage is required to obtain the full-load current in both windings. This voltage is called the impedance voltage. In other words, the impedance voltage is that voltage required to cause the rated current to flow through the impedances of the primary and secondary windings. The ratio of the impedance voltage to the rated terminal voltage yields the percentage impedance voltage, which is in the range of 3% to 5%.

The wattmeter in Figure 13–5 indicates the total copper losses of the transformer referred to the primary side. The wattmeter reading includes the core losses, which are so small that they can be neglected. The core losses are small because the voltage impressed on the primary winding in this test is very low.

To make the short-circuit test, the core loss of the transformer must be determined using the connections of Figure 13–3A. Using the 50-kVA transformer described in the previous problems, the core loss is 240 W. The readings on the instruments shown in Figure 13–5 will be as follows: a current of 21 A, a power value of 640 W, and a volt- age of 80 V. The value of 640 W represents the total copper losses of the transformer. If the dc ohmic values are used to compute the copper losses in the high- and low-voltage windings, then the total losses are

This value is nearly the same as the constant of 1.10 used to convert the ohmic resistance to the effective resistance for transformers.

The entire equivalent effective resistance of the transformer referred to the primary side can be found for a wattmeter reading of 640 W and a current of 21 A (at almost full load). When referred to the primary side, this equivalent resistance, R , is

The secondary is the output side of the transformer. Therefore, the entire resistance of the transformer must be referred to the secondary side as an equivalent effective  resistance, R_{OS :}

At the rated load, the total copper losses of this transformer are determined as follows. The full-load current is

This value of efficiency, as determined by the short-circuit test, is the same as the efficiency found in problem 5.

Assume that the same transformer is operated at 50% of the rated current capacity to supply a load with a unity power factor. The core loss is the same because the magnetizing flux is the same. However, the copper loss will decrease to one-fourth of its full- load value. Note that this decrease occurs because of the squaring operation in the I2 R formula. The I2 multiplier is only one-fourth of the original value. Use the previous formula and calculate the actual efficiency for this load condition:

If the transformer supplies a load having a power factor other than unity, the output (in watts) will decrease. However, the losses will be the same as those for a unity power factor load. For example, if the 50-kVA transformer supplies the rated output to a load with a power factor of 60% lagging, the efficiency is decreased:

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

PROBLEM 4

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.

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

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

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

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

PROBLEM 4

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.

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

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

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

HUNTING

If the torque output of a prime mover pulsates, it may cause the rotor of the alternator to be pulled ahead of, and then behind, its normal running position. A diesel engine is one example of a prime mover that has a pulsating torque output. If an alternator is used with a diesel engine, the alternator rotor periodically will move slightly faster and then slower. This pulsating or oscillating effect is called hunting. It causes the current to surge back and forth between alternators operating in parallel. This condition is unsatisfactory and may become cumulative, resulting in such a large increase in the current between the alternators that the overload relays open the circuit.

Hunting can be corrected by the use of a heavy flywheel. A damper winding is often used in the rotating field structure to minimize the pulsating torque. Figure 12–2 showed a damper winding, which is often called an amortisseur winding. This winding is embedded in slots in each of the pole faces of the rotating field. The amortisseur winding consists of heavy conductors that are brazed or welded to two end rings. At the instant that hunting develops, the path of the armature flux changes so that it cuts the short-circuited conductors of the amortissuer winding. This change in the flux path produces induced currents in the damper winding. The currents oppose the force producing them (by Lenz’s law). The proper design of the damper winding ensures that the effects of hunting are canceled by the induced currents in the short-circuited conductors.

SUMMARY

• An emf is generated

1. when there is relative motion between the armature conductors and the field.

2. in armature conductors when rotating in a magnetic field with stationary field poles.

3. in stationary armature conductors when the field poles rotate past the conductors.

• A dc generator has stationary field poles and rotating armature conductors. The alternating voltage induced in the rotating armature conductors is changed to a direct voltage at the brushes by means of the commutator.

• AC generators (alternators) do not use commutators.

• Alternators are classified into two groups:

1. Revolving armature machines with stationary fields; the kilovolt-ampere capacity and the low-voltage rating of these machines are usually rather small.

2. Revolving field machines with stationary armature (stator) conductors.

a. The field poles rotate inside the stator.

b. Higher voltages can be generated without insulation failure.

c. Higher current values can be obtained without arcing or heat production at the brushes and slip rings.

• There are two types of revolving fields:

1. The salient field rotor (used with slow-speed alternators up to 1800 r/min).

2. The cylindrical field rotor (used for speeds from 1800 to 3600 r/min; most steam turbine-driven alternators have cylindrical rotors).

• An ac generator cannot supply its own field current:

1. The field excitation is direct current and is supplied from an external direct-current source.

2. Slip rings and brushes are used to take the excitation current from the external source to the field windings.

3. The field voltage is usually in the range between 100 and 250 V.

4. The amount of power delivered to the field circuit is relatively small.

• The induced field voltage is large enough to damage equipment. To eliminate this dan- ger, a special field discharge switch is used.

1. In the closed position, the field discharge switch acts like a normal double-pole, single-throw switch.

2. When the switch is opened, an auxiliary blade closes just before the main switch blades open.

3. When the main switch blades are fully open, a circuit path still exists through the auxiliary switch blade.

4. This path goes through the field discharge resistor, bypassing the field rheostat and the ammeter.

5. The voltage induced in the field coils by the collapsing magnetic field dissipates quickly as a current through the field discharge resistor.

6. A large machine may use a field contactor or field circuit breaker for the same purpose; each of these devices has two normally open main poles and one over- lapping normally closed discharge pole.

• Large three-phase, revolving field-type ac generators can use one of several types of stationary armature (stator) windings; these windings generally consist of an even number of coils spaced around the perimeter of the stator core.

1. The formed coil is machine wound and insulated before it is installed in the slots of the stator core.

2. A full-pitch coil spans the distance between poles of opposite polarity.

3. A fractional-pitch coil is smaller than the distance between poles of opposite polarity. All of the coils of one single-phase winding are known as a phase belt.

a. Connecting the formed coils of a phase belt (single-phase winding) in series yields a maximum voltage.

b. Reconnecting the same coils in parallel yields the maximum current possible at a lower voltage value.

4. A three-phase alternator has three separate windings that are placed in the slots of the stator core.

a. The windings are arranged so that three voltages are produced and are 120 electrical degrees apart.

b. The three single-phase windings (phase belts) may be connected in either delta or wye.

c. The wye connection is more commonly used because it gives a higher terminal voltage: 1.73 times the phase winding voltage.

• Eddy currents in the stator core due to the flux of the revolving field can be reduced by the use of a laminated core. Such a core consists of thin strips of steel clamped together.

• Ventilating ducts in the core of the stator and ventilating passages in the steel frame prevent the temperature of the stator windings from becoming too high.

1. When the alternator has a salient field rotor, it acts as a fan to aid cooling.

2. High-speed alternators with cylindrical rotors normally have a cooling system that completely encloses the alternator. Either air or hydrogen is used in the system. Hydrogen is more efficient than air because it has almost seven times the heat conductivity of air. However, hydrogen is explosive and is costly to replace.

• When an armature conductor of a generator is cut by a rotating magnetic field, the induced voltage is

• Controlling the field current:

1. The voltage induced in any alternator depends on the field strength and the speed of the rotor.

2. To maintain a fixed frequency, an alternator must operate at a constant speed.

3. The magnitude of the generated voltage depends on the dc field excitation.

4. The following method can be used to change or control the terminal voltage:

a. A rheostat is connected in series with the separately excited field circuit.

b. If an alternator is operated at a constant speed with a fixed field excitation cur- rent, the terminal voltage will change with an increase in the load current.

c. The actual change in voltage is influenced by the power factor of the load circuit and the impedance of the armature windings.

• Percent voltage regulation is the change in the terminal voltage from a full-load to a no-load condition at a constant speed and a fixed field excitation current:

• Vector diagrams can be used to determine the induced voltage of an alternator for different load power factors:

• At a lagging power factor, the induced voltage is greater than when the load power factor was unity.

• At a leading power factor, the induced voltage is less than when the load power factor was unity. In this case, the induced voltage is less than the terminal voltage.

• Synchronous reactance:

1. The voltage drop due to the inductive reactance and the armature reaction have the same effect on the terminal voltage.

2. Both of these effects are proportional to the armature current.

3. These two effects are known as the synchronous reactance, X .

The synchronous impedance test is used to determine values of R and XLS

1. In this test, the field excitation current and the alternator speed are kept at constant values.

2. The line terminals of the alternator are shorted through an ammeter when the test switch is closed.

3. The field excitation current increases until the current in the armature is nearly 150% of the rated full-load current. This value of current is recorded.

4. The test switch is opened and the reading of a voltmeter connected across the generator terminal is recorded.

5. The synchronous impedance is

• A relatively constant alternator terminal voltage can be maintained under changing load conditions by the use of an automatic voltage regulator:

1. As the terminal voltage decreases, a relay closes contacts in the regulator to short out a field resistor. There is a resulting increase in the field current, field flux, and induced voltage.

2. An increase in the terminal voltage causes the relay to open the contacts across the field resistor. This action causes a decrease in the field current, field flux, and induced voltage.

3. Many other types of automatic voltage regulators may be used with alternators.

These regulators may use vacuum tubes, amplidynes, magnetic amplifiers, ignition rectifiers, controlled silicon rectifiers, or solid-state control devices.

• Saturation point:

1. For a constant-speed generator, the generated voltage is a direct function of the flux value per pole.

2. The flux is determined by the number of ampere-turns of the field pole. Because the number of turns on each field winding is constant, the flux is proportional to the dc excitation.

3. Increasing the dc excitation increases the flux; therefore, the induced voltage also increases.

4. A point is reached at which the flux no longer increases in proportion to the increase in dc excitation. This point is called the saturation point.

• Reading the saturation curve:

1. The saturation curve shows an increase in induced voltage with an increase in the field current in an alternator.

2. The ascending curve shows an increase in induced voltage as the field current is increased; the descending curve represents a decrease in induced voltage as the field current is decreased.

3. The first part of the ascending curve is nearly vertical because the induced voltage is almost directly proportional to the increase in field current.

4. When the alternator is being driven and the field current is zero, there is a small induced voltage. This voltage is due to the effects of residual magnetism after the magnetomotive force is removed from the field.

5. As the ascending curve reaches the saturation point, it flattens out. From this point to the maximum induced voltage on the curve, the increase in induced voltage is not proportional to the increase in field current.

6. The knee of the ascending curve is located immediately before and after the saturation point.

7. After reaching the maximum induced voltage, the field current is decreased to zero. The resulting plot of these events is the descending curve.

8. Note that the descending curve has a slightly higher induced voltage than does the ascending curve. This reaction is due to molecular friction or hysteresis effects in which the molecules of the iron in the magnetic circuit stay aligned even after the magnetomotive force is decreased.

9. The operating point of a particular generator occurs near the center of the knee of the ascending curve. At this point, there is a maximum ac voltage output with minimal I2 R losses in the field circuit.

• The alternator nameplate contains the following data:

1. The capacity of the alternator, in kVA

2. The full-load terminal voltage

3. The rated full-load current per terminal

4. The number of phases

5. The frequency

6. The speed in r/min

7. The operating power factor

8. The dc field current and voltage

9. The maximum temperature rise

• Power losses in an alternator consist of fixed or stray losses such as

1. the bearing and brush friction losses.

2. windage losses.

3. iron losses, including eddy current and hysteresis losses.

4. copper losses, including the I2 R losses in the armature windings and the power expended in the separately excited field circuit.

• Alternator efficiency:

2. An alternator should be operated at or near the rated full-load output to obtain the maximum efficiency.

3. Alternators with capacities in the order of 200,000 kVA may have efficiency ratings as high as 96% at the full-load output.

• Advantages of paralleling alternators:

1. They aid in scheduling of maintenance and emergency repairs on alternators.

2. They allow the alternators on the line to be operated near their full-load rating (high efficiency range). Another alternator may be paralleled with the first one to meet peak demands.

3. The generating plant capacity may be expanded to meet increased power demands by installing more alternators. These machines then operate in parallel with the existing generating equipment.

• To parallel ac generators:

1. Observe the following conditions:

a. The output voltages of the alternators must be equal.

b. The frequencies of the alternators must be the same.

c. The output voltages of the alternators must be in phase.

2. If these three conditions are met, the alternators are said to be in synchronism.

• Synchronizing two alternators:

1. Three-lamp method

a. The three lamps go on and off in unison depending on the frequency difference between the two alternators.

(1) For this case, a slight adjustment in the speed of the prime mover for the alternator coming on line will make the frequency of this machine the same as the alternator presently on the line.

(2) When all three lights go out, the instantaneous electrical polarity of the second machine will equal that of the alternator on the line. The second machine can be brought on line, and the generators will be paralleled.

b. The three lamps will light and go off, but not in unison. In this case, the phase sequence or phase rotation of the second generator is not the same as the alternator on the line. By interchanging the connections of any two leads of the second alternator, the phase sequence can be changed. The steps in part 1 can be followed to adjust the frequency of the second machine so that it can be paralleled.

2. Three lamps dark method

a. This method is used to determine the phase sequence of an alternator.

b. Once the phase sequence is known, permanent connections can be made between the stator windings, the switching equipment, and the station bus bars. It is not necessary to determine the phase sequence each time the alter- nator is paralleled once the equipment is marked correctly.

c. This method is also used to indicate when alternators are in synchronism.

3. Two bright, one dark method:

a. This method is never used to determine the phase sequence.

b. It is used to indicate the synchronism of two alternators.

c. When the incoming alternator is in synchronism (ready to be paralleled), two lamps in lines 1 and 2 will have a maximum brightness and the lamp in line 3 will be dark.

4. There are disadvantages to both methods of testing:

a. There may be a large voltage difference across the synchronizing lamps (even when they are dark); thus, a large voltage difference and phase difference may be present.

b. When an attempt is made to bring the incoming alternator into the bus bar circuit system with the other machines, a large disturbance in the electrical system may result in damage to the alternator windings.

• Use of the synchroscope:

1. Once the phase sequence is known to be correct and permanent connections are made, a synchroscope can be used.

2. The single-phase synchroscope indicates synchronism accurately.

3. A synchroscope gives an accurate indication of the differences in the frequency– phase relationship between two voltages.

a. A pointer rotates over a dial face. When the pointer stops, the frequencies of the two alternators are the same.

b. 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 syn- chronism and the alternator switch can be closed to parallel both machines.

• Once the two alternators are operating in parallel, the load is shared between them.

1. The load taken by each machine is proportional to the kVA rating of the machine.

2. Changing the field excitation of each dc generator will divide the load between the generators connected in parallel. The same method cannot be used to divide the kilowatt load between the two alternators in parallel.

3. A different method is used to divide the load between two alternators.

a. The prime movers for such alternators should have drooping speed–load characteristic curves.

b. To divide the load equally between the two machines, the input to the prime mover must be increased. The governor on the prime mover of the light-load alternator is opened slightly. As a result, there is an increase in both the horse- power of the light-load alternator and the power output of the alternator. At the same time, the governor on the prime mover of the heavy-load alternator is closed very slightly. This action causes decreases in the input to the prime mover, the horsepower input to the heavy-load alternator, and the power out- put (in kW) of the same alternator.

c. Careful adjustment of the governors of both prime movers can result in the speed–load characteristic curve of each cutting the system frequency line at the same load point. Thus, there is a satisfactory load division between the machines.

d. This adjustment of the prime movers must be obtained without causing a change in the frequency.

• When alternators are operating in parallel, a faulty governor may cause the speed of an alternator to increase; thus, the machine may be pulled out of synchronism.

1. Certain reactions within the alternator will prevent this condition.

2. 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.

• Hunting

1. is a pulsating or oscillating effect.

2. of the prime mover causes a torque that may cause the rotor of the alternator to be pulled ahead of, and then behind, its normal running position.

3. causes the current to surge back and forth between alternators operating in parallel; this condition is unsatisfactory and may become cumulative, resulting in the circuit being opened by the overload relays.

4. can be corrected by using

a. a heavy flywheel.

b. a damper winding in the rotating field structure. Such a winding is often called an amortisseur winding. The proper design of this winding ensures that the effects of hunting are canceled by the induced currents in the short- circuited conductors.

Achievement Review

1. List two advantages of the rotating field alternator, as compared with the rotating armature ac generator.

2.	a. b.	Where is the salient pole rotor used? Where is the cylindrical rotor used?
3.	a.	Explain why a field discharge resistor is used with the separately excited field circuit.
	b.	Draw a schematic wiring diagram of a separately excited field circuit of an alternator. Include a field discharge switch, a field discharge resistor, an ammeter, and a field rheostat. Connect the circuit so that the field rheostat and the ammeter are not in the field discharge circuit path.
4.	a.	What methods are used to cool the windings of a high-speed turbine-driven alternator having a large kVA capacity?
	b.	There are fewer problems in cooling the windings of slow-speed alternators using salient pole rotors. Why?
5.	A	25-kVA, 250-V, 60-Hz, 1800-r/min, single-phase alternator has an

armature resistance of 0.12 W and an armature reactance of 0.5 W. The generator delivers the rated load output at a power factor of 1.00 (unity) to a non- inductive load. Determine the induced voltage. (Neglect any armature reaction.)

6. a. Determine the induced voltage of the alternator in question 5 when it delivers the rated output to a load with a lagging power factor of 0.8660.

b. Determine the induced voltage of the alternator in question 5 when it delivers the rated output to a load with a leading power factor of 0.8660.

7. a. Define voltage regulation as it is used with alternators.

b. The full-load terminal voltage of an alternator is 240 V. The load is removed.

The no-load terminal voltage increases to 265 V at the same speed and field excitation. What is the percent voltage regulation of the alternator?

8. A three-phase, wye-connected alternator is rated at 2000 kVA, 4800 V, 60 Hz. In a short-circuit synchronous impedance test, the field excitation current is increased until the three line ammeters indicate nearly 150% of 360 A, the rated line cur- rent. The field excitation current and the speed are kept constant and the three-pole switch is opened. The voltmeter indicates 2250 V. The dc resistance between the line terminals is 0.4 W, and the ratio of effective to ohmic resistance is 1.5. Determine

a. the synchronous impedance.

b. the effective resistance.

c. the synchronous reactance.

9. Determine the no-load voltage and the percent voltage regulation for the alterna- tor in question 8. Assume that the alternator is delivering the rated current to a noninductive, unity power factor load.

10. a. Determine the no-load voltage and the percent voltage regulation for the alter- nator in question 8 assuming that it is delivering the rated current to a load with a lagging power factor of 0.8.

b. Find the no-load voltage and the percent voltage regulation for the same alter- nator when it delivers the rated current to a load with a leading power factor of 0.8.

11. Explain what is meant by the terms

a. synchronous reactance.

b. synchronous impedance.

12. a. Draw a typical saturation curve for an alternator.

b. Give two reasons why saturation curves are used.

13. a. What are the fixed losses of an alternator?

b. What are the copper losses of an alternator?

c. How is the full-load efficiency of an alternator determined?

14. A three-phase, wye-connected alternator is rated at 500 kVA, 2400 V, 60 Hz.

Determine

a. the full-load kilowatt output of the generator at 80% lagging power factor.

b. the full-load current per line terminal for the alternator.

c. the full current rating of each of the phase windings.

d. the voltage rating of each of the phase windings.

15. A three-phase, delta-connected, diesel-driven alternator is rated at 50 kVA, 240 V, 60 Hz.

a. Determine

(1) the full-load current rating per line terminal for the alternator.

(2) the full-load current rating of each of the phase windings.

(3) the voltage rating of each of the phase windings.

b. If the alternator has a rated speed of 240 r/min, how many poles are required for the rotating field?

16. A 5-kVA, 208-V, three-phase alternator is connected in wye.

a. Determine

(1) the line current per terminal at full load.

(2) the coil current at full load.

(3) the voltage of each phase winding.

b. Assuming that this alternator is reconnected in delta, compute the new terminal voltage and the current at full load.

17. A three-phase, 60-Hz, wye-connected turbine-driven alternator has three single- phase windings. Each winding is at 8000 V and 625 A. The alternator has four poles. Determine

a. the kilovolt-ampere rating of the alternator.

b. the kilowatt output of the alternator when it delivers the rated current to a load with an 80% lagging power factor.

c. the line voltage.

d. the rated full-load line current.

e. the speed in r/min of the revolving field of the alternator.

18. For the turbine-driven, high-speed alternator described in question 17, answer the following questions:

a. Why is the field, rather than the armature, the rotating member?

b. Why is a cylindrical rotating field used, rather than one with salient poles?

c. How is the dc excitation current supplied to the rotor of the rotating field alternator?

19. List three reasons why ac generators are operated in parallel.

20. A three-phase, wye-connected alternator is rated at 10,000 kVA, 11,000 V, and 60 Hz. Determine

a. the full-load kilowatt output of the ac generator at a lagging power factor of 80%.

b. the full-load line current of the alternator.

c. the voltage rating of each of the three windings.

d. the horsepower input to the alternator when it delivers the rated load output at an efficiency of 92% and a lagging power factor of 80%.

21. A 240-V, single-phase, 60-Hz, revolving field alternator delivers 30 kW to a non- inductive load. The generator efficiency is 86%, and the stray power losses are 2000 W. The separately excited field requires 6 A at 240 V, dc. Determine

a. the full-load current.

b. the copper losses in the stator winding.

22. a. List the steps, in chronological order, required to place a three-phase alterna- tor in parallel with another ac generator. Assume that this is the first time that the alternator is placed in service.

b. After paralleling the alternators, what means are used to redistribute the kilo- watt load between the two alternators? Assume that the frequency is held constant.

23. a. Diagram the “three lamps dark” method of synchronizing an alternator to the bus bars.

b. How is this method used to determine whether the phase sequence of an alternator is correct with reference to the station bus bars?

24. a. Diagram the “two lamps bright, one lamp dark” method.

b. How is this method used to determine when an incoming alternator is in phase with the bus bars?

25. a. What does the term hunting mean as applied to slow-speed alternators?

b. How does the amortisseur or damping winding on the field rotor decrease the effects of hunting?

c. What other way is used to minimize the effects of hunting?

26. Explain how equal load distribution is obtained between alternators in parallel.

27. Explain what happens when an attempt is made to shift the kilowatt load between alternators by changing the field excitation. Assume that the input to the prime movers of the alternators is not changed.

List the data commonly found on the nameplate of an alternator.