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Three-Phase Induction Motors

THREE-PHASE, SQUIRREL-CAGE INDUCTION MOTOR

A three-phase, squirrel-cage induction motor is shown in Figure 16–1. This motor is simple in construction and is easy to maintain. For a given horsepower rating, the physical size of this motor is small, when compared with other types of motors. It has very good speed regulation under varying load conditions. This motor is used for many industrial applications because of its low purchase price, rugged construction, and operating characteristics.

Construction

The basic structure of a three-phase, squirrel-cage induction motor consists of a stator, a rotor, and two end shields that house the bearings supporting the rotor shaft.

The stator is a three-phase winding that is placed in the slots of a laminated steel core. The winding itself is made of formed coils that are connected to give three single-phase windings spaced 120 electrical degrees apart. The three separate single-phase windings are connected in wye or delta. Three line leads from the three-phase stator windings are brought out to a terminal box mounted on the frame of the motor.

The rotor has a cylindrical core consisting of steel laminations (Figure 16–2). Aluminum bars are mounted near the surface of the rotor. These bars are brazed or welded to two aluminum end rings. Some types of squirrel-cage induction motors are smaller than others

and have aluminum end rings that are cast in one piece. The rotor shaft is supported by bearings housed in the end shields.

THE ROTATING MAGNETIC FIELD

The principle of operation for all three-phase motors is the rotating magnetic field. There are three factors that cause the magnetic field to rotate:

1. The fact that the voltages of a three-phase system are 120° out of phase with each other

2. The fact that the three voltages change polarity at regular intervals

3. The arrangement of the stator windings around the inside of the motor

Figure 16–3A shows three ac voltages 120° out of phase with each other, and the stator winding of a three-phase motor. The stator illustrates a two-pole, three-phase motor.

Two-pole means that there are two poles per phase. AC motors seldom have actual pole pieces as shown in Figure 16–3A, but they will be used here to aid in understanding how the rotating magnetic field is created in a three-phase motor. Notice that pole pieces 1A and 1B are located opposite each other. The same is true for poles 2A and 2B, and 3A and 3B. Pole pieces 1A and 1B are wound with wire that is connected to phase 1 of the three-phase system. Notice also that the pole pieces are wound in such a manner that they will always have opposite magnetic polarities. If pole piece 1A has a north magnetic polarity, pole piece 1B will have a south magnetic polarity at the same time.

The windings of pole pieces 2A and 2B are connected to line 2 of the three- phase system. The windings of pole pieces 3A and 3B are connected to line 3 of the three-phase system. These pole pieces are also wound in such a manner as to have the opposite polarity of magnetism.

To understand how the magnetic field rotates around the inside of the motor, refer to Figure 16–3B. Notice that a line labeled A has been drawn through the three voltages of the system. This line is used to illustrate the condition of the three voltages at this point in time. The arrow drawn inside the motor indicates the greatest strength of the magnetic field at the same point in time. It is to be assumed that the arrow is pointing in the direction of the north magnetic field. Notice in Figure 16–3B that phase 1 is at its maximum positive peak and that phases 2 and 3 are less than maximum. The magnetic field is, therefore, strongest between pole pieces 1A and 1B.

In Figure 16–3C, line B indicates that the voltage of line 3 is zero. The voltage of line 1 is less than maximum positive, and line 2 is less than maximum negative. The magnetic field at this point is concentrated between the pole pieces of phases 1 and 2.

In Figure 16–3D, line C indicates that line 2 is at its maximum negative peak and that lines 1 and 3 are less than maximum positive. The magnetic field at this point is concentrated between pole pieces 2A and 2B.

In Figure 16–3E, line D indicates that line 1 is zero. Lines 2 and 3 are less than maxi- mum and in opposite directions. At this point in time, the magnetic field is concentrated between the pole pieces of phase 2 and phase 3.

In Figure 16–3F, line E indicates that phase 3 is at its maximum positive peak, and lines 1 and 2 are less than maximum and in the opposite direction. The magnetic field at this point is concentrated between pole pieces 3A and 3B.

In Figure 16–3G, line F indicates that phase 2 is zero. Line 3 is less than maximum positive, and line 1 is less than maximum negative. The magnetic field at this time is concentrated between the pole pieces of phase 1 and phase 3.

In Figure 16–3H, line G indicates that phase 1 is at its maximum negative peak, and phases 2 and 3 are less than maximum and in the opposite direction. Notice that the magnetic field is again concentrated between pole pieces 1A and 1B. This time, however, the magnetic polarity is reversed because the current has reversed in the stator winding.

In Figure 16–3I, line H indicates that phase 2 is at its maximum positive peak and phases 1 and 3 are less than maximum and in the negative direction. The magnetic field is concentrated between pole pieces 2A and 2B.

In Figure 16–3J, line I indicates that phase 3 is maximum negative, and phases 1 and 2 are less than maximum in the positive direction. The magnetic field at this point is concentrated between pole pieces 3A and 3B.

In Figure 16–3K, line J indicates that phase 1 is at its positive peak, and phases 2 and 3 are less than maximum and in the opposite direction. The magnetic field is again concentrated between pole pieces 1A and 1B. Notice that in one complete cycle of the three-phase voltage, the magnetic field has rotated 360° around the inside of the stator winding.

If any two of the stator leads is connected to a different line, the relationship of the voltages will change and the magnetic field will rotate in the opposite direction. The direction of rotation of a three-phase motor can be reversed by changing any two stator leads.

INDUCTION VOLTAGE REGULATOR

A nearly constant voltage can be maintained on long distribution circuits by compensating for the voltage losses due to varying load conditions. An almost constant voltage can be maintained at the load center of the distribution circuit. In this case, an induction voltage regulator is used to compensate for the resistive and reactive voltage drops in the line wires.

The primary winding of a single-phase induction voltage regulator is wound in the slots of a laminated rotor core. The rotor can be turned through an angle of 180°. A small

motor and a worm gear mechanism drive the rotor. The stationary secondary winding is placed in the slots of a laminated stator core.

Operation of the Induction Voltage Regulator

A schematic diagram is shown in Figure 15–18 for a single-phase induction regulator. The primary winding is connected across the line voltage. The secondary winding is connected in series with one of the line wires. In one position, the flux of the primary will cut the secondary turns and induce a maximum voltage in the secondary winding. This voltage is added to the original line voltage to compensate for the line drop. When the primary winding is turned, there is less primary flux linking the secondary turns. Thus, less voltage is induced in the secondary winding. Assume that the rotor is moved 90° from the point where the maximum voltage was induced. The rotor is now placed so that none of its lines of flux link the secondary turns. This means that no voltage is induced in the secondary winding. Under these conditions, the secondary winding must not be allowed to act like a choke coil. There- fore, a short-circuited winding is placed in the rotor slots. This winding is known as a tertiary winding and acts like the short-circuited secondary of a transformer. The tertiary winding reduces the inductance of the secondary winding to a very small value that can be neglected.

The rotor is now turned until it is 180° from its original position. A maximum voltage is induced in the secondary winding again. However, at 180°, the voltage induced in the secondary is in a direction that causes this voltage to oppose the line voltage.

When the maximum flux of the primary links the secondary turns in one position, a maxi- mum voltage is induced in the secondary. In this instance, the voltage regulator acts like a voltage booster. When the primary is moved 180° from this position, the induced voltage in the secondary is also a maximum value. However, this voltage now opposes the line voltage. The amount of voltage increase or opposition can be controlled by turning the rotor to various positions.

The motor that turns the primary rotor is controlled by relays. These relays are energized by a contact-making voltmeter. When the voltage decreases, one set of contactors closes. The motor then turns in a given direction to boost the voltage back to its normal value. If the voltage becomes too high, another set of contactors is activated. The motor then turns in the opposite direction. As the primary moves to a new position, the voltage is reduced to its normal value.

The construction of the three-phase induction voltage regulator differs from that of the single-phase regulator. Three stationary primary windings are placed in the slots of a laminated stator core. The windings are connected in wye or delta to the three line wires of the distribution circuit. The three secondary windings are placed in the slots of the laminated rotor core. The secondary windings are insulated from each other. Each winding is connected in series with one of the three line wires of the three-phase circuit.

When the three-phase stationary primary windings are energized, a uniform magnetic field is set up inside the stator core. This means that the voltage induced in the three secondary windings is nearly independent of the rotor position. However, by moving the rotor with respect to the stator, there will be a change in the phase relationship between each induced voltage and its respective line voltage. This regulation gives the desired increase or decrease in the three-phase voltage of the distribution circuit.

Another common type of three-phase induction voltage regulator consists of three single-phase induction voltage regulators. These standard regulators are mounted in one enclosure. They are geared together so that they may be driven by one motor.

SUMMARY

• Instrument transformers are used with instruments and relays

1. to measure and control ac circuits.

2. to measure high voltage and current values directly.

3. to provide safety for the operator and to protect control equipment from damage due to high voltages.

4. to provide more accurate measurements and greater convenience.

• The two types of instrument transformers are the potential transformer and the current transformer.

1. The potential transformer

a. is similar in construction to a power or distribution transformer.

b. has relatively small capacity in the range of 100 to 500 VA.

c. has a primary winding designed and rated to be operated at a designated line voltage, such as 4800 V; the primary circuit voltage and the voltage rating of the primary winding of a potential transformer are the same.

d. has a low-voltage secondary rated at 120 V; thus, the ratio between the primary and secondary voltage is 4800/120 40:1 (transformer ratio).

e. normally has a subtractive polarity.

f. has one of its secondary leads grounded to minimize high-voltage hazards.

g. has a percentage error of less than 0.5%.

The actual voltage of the high-voltage circuit is determined as follows:

a. A voltmeter is used to measure the 120-V secondary.

b. The voltmeter reading is multiplied by the transformer ratio of 40:1; thus, 120 X 40 = 4800 V.

c. In some cases, a panel voltmeter is calibrated to read the actual primary value (this value includes the transformer ratio multiplier).

2. The current transformer

a. is used so that ammeters, relays, and instrument current coils need not be con- nected to high-voltage lines.

b. steps down the current by a known ratio.

c. permits the use of small and accurate instruments because they are insulated from the high-voltage line.

d. has a primary winding that is sized to carry the line current. This winding is connected in series with one of the lines. For higher currents, the line in question may be fed through a toroidal core (having no turns).

e. is always rated at 5 A at the secondary. Any standard current ratio can then be applied to the transformer.

f. generally has standard polarity markings. The H1 lead is connected to the source end of the line. The H2 lead is connected to the load end of the line. The X1 and X2 leads are not always marked by manufacturers.

g. has a grounded secondary lead to reduce the high-voltage hazard.

Also,

a. the X1 and X2 leads are connected directly to the ammeter.

b. the secondary circuit transformer must not be open when there is current in the primary. A voltage great enough to be a hazard can be induced in the secondary winding.

c. a short-circuit switch installed at the secondary terminals of the trans- former is closed when the instrument circuit is to be opened for repairs or rewiring of the metering circuit.

• When using potential transformers and current transformers with wattmeters,

1. the potential coil of the wattmeter is connected across the low-voltage output of the potential transformer.

2. the current coil of the wattmeter is connected in series with the output of the cur- rent transformer.

3. the current and voltage terminals of the wattmeter (marked ±) have the same instantaneous polarity.

4. the torque on the wattmeter movement causes the pointer to move upscale.

• Calculating power and power factor:

1. The power multiplier is the product of the current multiplier and the voltage multiplier.

Wattmeter mutiplier PT_ratio X CT_ratio

where PT is the potential transformer and CT is the current transformer.

a. Primary power in watts is equal to the wattmeter reading times the wattmeter multiplier.

b. Apparent power, in volt-amperes, is equal to the primary voltage multiplied by the line current.

2. The power factor is determined as follows:

• A three-phase, three-wire system requires two potential transformers having the same rating and two current transformers having the same rating.

• Other instruments may be added to the three-phase, three-wire circuit, including a three- phase wattmeter, a three-phase watt-hour meter, and a three-phase power factor meter.

• Portable instruments and instrument transformers are used to measure the current, the voltages, and the power in watts when testing industrial equipment.

• The secondary metering connections for a 2400/4160-V, three-phase, four-wire system are as follows:

1. The three potential transformers are connected in wye.

2. The three-phase output of these transformers consists of three secondary voltages of 120 V to neutral.

3. A 50-to-5-A current transformer is connected in series with each of the three line wires.

4. The proper phase relationship must be established when wattmeters, watt-hour meters, and other three-phase instruments are used in the circuit.

• The primary and secondary windings of an autotransformer are in the form of one con- tinuous winding.

1. This winding is tapped at certain points to obtain the desired voltages.

2. The winding is wound on a laminated silicon steel core. As a result, the primary and secondary sections of the winding are in the same magnetic circuit.

3. The calculations necessary for autotransformer circuits involve two new quantities.

a. Conductive power is the product of the current conducted to the secondary side from the primary side and the voltage on the secondary side:

Pconductive IP X VS

b. Transformed power is the product of the difference between the primary and secondary voltages and the secondary current:

Ptransformed (VP – VS) X IS

4. Total power, watts conductive power + transformer power.

5. The autotransformer can be used to step up a voltage.

a. The load is connected across the full winding.

b. The source voltage is impressed across only part of the winding.

c. With a noninductive load,

d. The current input to the autotransformer at a known input voltage can be found knowing the power input:

6. The core and the copper losses in an autotransformer are smaller than those in a constant-voltage transformer.

7. The efficiency of the autotransformer is slightly better than that of the conven- tional transformer.

8. The exciting current and the leakage reactance in the autotransformer are small.

As a result, it is an accepted practice to assume that the input and the output are the same.

• Autotransformers are used when a small increase or decrease in voltage is required

1. on long-distance distribution lines, both single-phase and three-phase, to compensate for the line voltage drop by increasing the line voltage.

2. in electronic circuits where several voltages are required.

3. in manual and automatic motor starters for three-phase systems to reduce the startup voltage applied to three-phase motors; the decreased voltage reduces the starting current surge.

4. to provide a multivoltage source as needed in electronics laboratories and for some types of single-phase variable speed motors, such as the repulsion motor.

• Autotransformers are not used between a high-voltage source and a low-voltage circuit such as a lighting circuit. If the low-voltage section is open circuited, the entire high- voltage input is applied across the low-voltage output.

• A constant-current transformer takes power into its primary winding at a constant volt- age and a variable current. It delivers power from the secondary winding at a constant current and a variable voltage.

• A series street lamp circuit is one application of the constant-current transformer. If a street lamp burns out, the total impedance of the series circuit decreases. If a constant voltage were maintained, the current would increase and cause other lamps to burn out. To prevent this, a small cutout device is placed in the socket of each lamp. When a lamp filament does burn out, the entire voltage of the circuit is applied across the cutout device, which short-circuits the socket and closes the series circuit.

• When used in a series street lighting circuit, a typical constant-current transformer functions as follows:

1. The primary is stationary. Its input comes from a constant voltage source.

2. The secondary winding moves on the center leg of the core. Its movement is balanced by a counterweight–dashpot arrangement. The output of the secondary is connected to the series street lighting circuit.

3. If the total impedance of the series circuit decreases because lamps burn out, there is an increase of current in the series circuit, the movable secondary coil, and the primary coil.

4. The force of repulsion between the primary and secondary coils will increase and cause the secondary coil to move away from the primary coil. As a result, there is less primary flux linking the secondary, and the induced voltage in the secondary will decrease. Thus, the secondary current also decreases.

5. When the counterweight system and the repulsion effect between the coils are equal and opposite, the secondary coil maintains a constant current.

• The primary of the street lighting system transformer (constant-current transformer) is generally energized from an oil switch. The oil switch is operated by a relay that is controlled by a phototube and an amplifier circuit.

• An induction voltage regulator is used to compensate for the resistive and reactive volt- age drops in long-distance distribution circuits.

1. The primary winding of the regulator is wound in the slots of a laminated rotor core. The winding is connected across the line voltage.

2. The secondary winding is stationary and is placed in the slots of a laminated stator core. The secondary winding is connected in series with one of the line wires.

3. The rotor can be turned through an angle of 180°. A small motor and worm gear mechanism drive the rotor.

4. When the maximum flux of the primary links the secondary turns in one position, a maximum voltage is induced in the secondary. This voltage is added to the incoming line voltage.

5. At the center position of the rotor (90°), none of the flux from the primary coil cuts the secondary winding. Thus, no voltage is induced in the secondary.

a. A short-circuited winding known as a tertiary winding is placed in the rotor slot.

b. This winding acts like the short-circuited secondary of a transformer. It pre- vents the secondary from acting as a choke coil when there is no induced voltage in it. The tertiary winding reduces the inductance of the secondary winding to a very small value that can be neglected.

6. As the rotor is moved from the 90° position to the 180° position, the voltage increases until it reaches a maximum at 180°. This induced voltage is in a direction that causes the secondary voltage to oppose the line voltage. Thus, this volt- age is subtracted from the line voltage.

7. The amount of voltage increase or opposition can be controlled by turning the rotor to various positions.

a. The motor that turns the primary rotor is controlled by relays. The relays are energized by a contact-making voltmeter.

b. One set of contacts closes when the line voltage decreases.

c. The motor then turns in a given direction to boost the voltage back to its nor- mal value.

d. If the voltage becomes too high, another set of contactors closes and the motor turns in the opposite direction.

e. As the primary rotor moves to a new position, the voltage is reduced to its normal value.

• Construction of the three-phase induction voltage regulator:

1. The regulator has three primary windings connected in wye or delta to the three line wires of the distribution circuit.

2. Each of the three secondary windings is connected in series with one of the three line wires of the three-phase circuit.

3. The primary windings are stationary and the secondary windings are movable.

4. The motor control system is the same as for a single-phase voltage regulator.

Achievement Review

1. a. What are the two functions of an instrument potential transformer?

b. What are the two functions of an instrument current transformer?

2. In what ways does a current transformer differ from a constant-potential transformer?

3. Explain why the secondary circuit of an instrument current transformer must be closed when there is current in the primary circuit.

4. A 2300/115-V potential transformer and a 100/5-A current transformer are installed and connected on a single-phase line. A voltmeter, an ammeter, and a wattmeter are connected to the secondaries of the instrument transformers. The voltmeter indicates 110 V, the ammeter reads 4 A, and the wattmeter indicates 352

V. Draw a schematic wiring diagram for this circuit. Indicate the polarities of the terminals of the instrument transformers. Also indicate which current and voltage terminals of the wattmeter should be marked with the polarity.

5. Using the circuit in question 4, determine

a. the primary voltage.

b. the primary current, in amperes.

c. the primary power, in watts.

d. the circuit power factor.

6. Instrument potential and current transformers are installed to meter the voltage, current, and power of a 2400-V, three-phase, three-wire distribution circuit. Two 2400/120-V instrument potential transformers and two 100/5-A instrument cur- rent transformers are used. All three secondary voltages are 116 V. Two ammeters are connected to the secondaries of the current transformers. Each meter indicates 3.5 A. Two single-phase wattmeters are used. Wattmeter 1 indicates 400 W and wattmeter 2 reads 160 W. Draw a schematic wiring diagram of this circuit. Indi- cate the polarities on the instrument transformers. Also show which current and potential terminals on each of the wattmeters are to have polarity markings.

7. Using the circuit in question 6, determine

a. the primary voltage.

b. the primary current.

c. the total three-phase primary power, in kVA.

d. the total primary apparent power, in kVA.

e. the circuit power factor.

8. Determine the power factor of the circuit in question 6. Use the power factor curve for the two-wattmeter method given in Unit 10. Compare this value with the calculated power factor obtained in question 7.

9. a. What is an autotransformer?

b. What is one advantage to the use of an autotransformer?

c. What is one limitation of an autotransformer?

10. Explain what is meant by the following terms as they relate to autotransformers:

a. Transformed power

b. Conductive power

11. An autotransformer is shown in Figure 15–19. It is used to step down 600 V to 480

V. The 480-V output is to supply a 24-kW noninductive load. Neglect the losses of the transformer. Determine

a. the load current.

b. the input current.

c. the current between terminals 1 and 2.

d. the current between terminals 3 and 2.

e. what part of the power, in watts, supplied to the load is the transformed power.

f. what part of the power supplied to the load is the conductive power.

12. The autotransformer shown in Figure 15–20 is used to boost the voltage on a long, single-phase line. The voltage is to be changed from 2000 V to 2400 V. The 2400-V output supplies a load of 72 kW. The load has a power factor of unity. Neglect the losses of the transformer and determine

a. the current output to the load.

b. the current input to the transformer.

c. the current in coil section 2–1.

d. the current in coil section 2–3.

e. the part of the power, in watts, supplied to the load due to

(1) transformed power.

(2) conductive power.

13. a. What is a constant-current transformer?

b. Explain the reactions that take place in a constant-current transformer so that a constant current is maintained when there are changes in the load.

14. A constant-current transformer is located in an outdoor substation. The transformer delivers an output of 29.7 kW at a potential of 4500 V to a street lighting series circuit. The circuit consists of 50 lamps having the same rating. The power factor of this load is unity.

a. What current flows in this street lighting series circuit?

b. If 20 lamps are cut out of this series circuit, what output voltage is required from the constant-current transformer so that the current is the same as that for the 50-lamp circuit?

c. Explain why a series street lighting circuit does not go out when the filament of one lamp burns out.

15. Explain the operations of a single-phase induction voltage regulator and a three- phase induction voltage regulator.

INDUCTION VOLTAGE REGULATOR

A nearly constant voltage can be maintained on long distribution circuits by compensating for the voltage losses due to varying load conditions. An almost constant voltage can be maintained at the load center of the distribution circuit. In this case, an induction voltage regulator is used to compensate for the resistive and reactive voltage drops in the line wires.

The primary winding of a single-phase induction voltage regulator is wound in the slots of a laminated rotor core. The rotor can be turned through an angle of 180°. A small

motor and a worm gear mechanism drive the rotor. The stationary secondary winding is placed in the slots of a laminated stator core.

Operation of the Induction Voltage Regulator

A schematic diagram is shown in Figure 15–18 for a single-phase induction regulator. The primary winding is connected across the line voltage. The secondary winding is connected in series with one of the line wires. In one position, the flux of the primary will cut the secondary turns and induce a maximum voltage in the secondary winding. This voltage is added to the original line voltage to compensate for the line drop. When the primary winding is turned, there is less primary flux linking the secondary turns. Thus, less voltage is induced in the secondary winding. Assume that the rotor is moved 90° from the point where the maximum voltage was induced. The rotor is now placed so that none of its lines of flux link the secondary turns. This means that no voltage is induced in the secondary winding. Under these conditions, the secondary winding must not be allowed to act like a choke coil. There- fore, a short-circuited winding is placed in the rotor slots. This winding is known as a tertiary winding and acts like the short-circuited secondary of a transformer. The tertiary winding reduces the inductance of the secondary winding to a very small value that can be neglected.

The rotor is now turned until it is 180° from its original position. A maximum voltage is induced in the secondary winding again. However, at 180°, the voltage induced in the secondary is in a direction that causes this voltage to oppose the line voltage.

When the maximum flux of the primary links the secondary turns in one position, a maxi- mum voltage is induced in the secondary. In this instance, the voltage regulator acts like a voltage booster. When the primary is moved 180° from this position, the induced voltage in the secondary is also a maximum value. However, this voltage now opposes the line voltage. The amount of voltage increase or opposition can be controlled by turning the rotor to various positions.

The motor that turns the primary rotor is controlled by relays. These relays are energized by a contact-making voltmeter. When the voltage decreases, one set of contactors closes. The motor then turns in a given direction to boost the voltage back to its normal value. If the voltage becomes too high, another set of contactors is activated. The motor then turns in the opposite direction. As the primary moves to a new position, the voltage is reduced to its normal value.

The construction of the three-phase induction voltage regulator differs from that of the single-phase regulator. Three stationary primary windings are placed in the slots of a laminated stator core. The windings are connected in wye or delta to the three line wires of the distribution circuit. The three secondary windings are placed in the slots of the laminated rotor core. The secondary windings are insulated from each other. Each winding is connected in series with one of the three line wires of the three-phase circuit.

When the three-phase stationary primary windings are energized, a uniform magnetic field is set up inside the stator core. This means that the voltage induced in the three secondary windings is nearly independent of the rotor position. However, by moving the rotor with respect to the stator, there will be a change in the phase relationship between each induced voltage and its respective line voltage. This regulation gives the desired increase or decrease in the three-phase voltage of the distribution circuit.

Another common type of three-phase induction voltage regulator consists of three single-phase induction voltage regulators. These standard regulators are mounted in one enclosure. They are geared together so that they may be driven by one motor.

SUMMARY

• Instrument transformers are used with instruments and relays

1. to measure and control ac circuits.

2. to measure high voltage and current values directly.

3. to provide safety for the operator and to protect control equipment from damage due to high voltages.

4. to provide more accurate measurements and greater convenience.

• The two types of instrument transformers are the potential transformer and the current transformer.

1. The potential transformer

a. is similar in construction to a power or distribution transformer.

b. has relatively small capacity in the range of 100 to 500 VA.

c. has a primary winding designed and rated to be operated at a designated line voltage, such as 4800 V; the primary circuit voltage and the voltage rating of the primary winding of a potential transformer are the same.

d. has a low-voltage secondary rated at 120 V; thus, the ratio between the primary and secondary voltage is 4800/120 40:1 (transformer ratio).

e. normally has a subtractive polarity.

f. has one of its secondary leads grounded to minimize high-voltage hazards.

g. has a percentage error of less than 0.5%.

The actual voltage of the high-voltage circuit is determined as follows:

a. A voltmeter is used to measure the 120-V secondary.

b. The voltmeter reading is multiplied by the transformer ratio of 40:1; thus, 120 X 40 = 4800 V.

c. In some cases, a panel voltmeter is calibrated to read the actual primary value (this value includes the transformer ratio multiplier).

2. The current transformer

a. is used so that ammeters, relays, and instrument current coils need not be con- nected to high-voltage lines.

b. steps down the current by a known ratio.

c. permits the use of small and accurate instruments because they are insulated from the high-voltage line.

d. has a primary winding that is sized to carry the line current. This winding is connected in series with one of the lines. For higher currents, the line in question may be fed through a toroidal core (having no turns).

e. is always rated at 5 A at the secondary. Any standard current ratio can then be applied to the transformer.

f. generally has standard polarity markings. The H1 lead is connected to the source end of the line. The H2 lead is connected to the load end of the line. The X1 and X2 leads are not always marked by manufacturers.

g. has a grounded secondary lead to reduce the high-voltage hazard.

Also,

a. the X1 and X2 leads are connected directly to the ammeter.

b. the secondary circuit transformer must not be open when there is current in the primary. A voltage great enough to be a hazard can be induced in the secondary winding.

c. a short-circuit switch installed at the secondary terminals of the trans- former is closed when the instrument circuit is to be opened for repairs or rewiring of the metering circuit.

• When using potential transformers and current transformers with wattmeters,

1. the potential coil of the wattmeter is connected across the low-voltage output of the potential transformer.

2. the current coil of the wattmeter is connected in series with the output of the cur- rent transformer.

3. the current and voltage terminals of the wattmeter (marked ±) have the same instantaneous polarity.

4. the torque on the wattmeter movement causes the pointer to move upscale.

• Calculating power and power factor:

1. The power multiplier is the product of the current multiplier and the voltage multiplier.

Wattmeter mutiplier PT_ratio X CT_ratio

where PT is the potential transformer and CT is the current transformer.

a. Primary power in watts is equal to the wattmeter reading times the wattmeter multiplier.

b. Apparent power, in volt-amperes, is equal to the primary voltage multiplied by the line current.

2. The power factor is determined as follows:

• A three-phase, three-wire system requires two potential transformers having the same rating and two current transformers having the same rating.

• Other instruments may be added to the three-phase, three-wire circuit, including a three- phase wattmeter, a three-phase watt-hour meter, and a three-phase power factor meter.

• Portable instruments and instrument transformers are used to measure the current, the voltages, and the power in watts when testing industrial equipment.

• The secondary metering connections for a 2400/4160-V, three-phase, four-wire system are as follows:

1. The three potential transformers are connected in wye.

2. The three-phase output of these transformers consists of three secondary voltages of 120 V to neutral.

3. A 50-to-5-A current transformer is connected in series with each of the three line wires.

4. The proper phase relationship must be established when wattmeters, watt-hour meters, and other three-phase instruments are used in the circuit.

• The primary and secondary windings of an autotransformer are in the form of one con- tinuous winding.

1. This winding is tapped at certain points to obtain the desired voltages.

2. The winding is wound on a laminated silicon steel core. As a result, the primary and secondary sections of the winding are in the same magnetic circuit.

3. The calculations necessary for autotransformer circuits involve two new quantities.

a. Conductive power is the product of the current conducted to the secondary side from the primary side and the voltage on the secondary side:

Pconductive IP X VS

b. Transformed power is the product of the difference between the primary and secondary voltages and the secondary current:

Ptransformed (VP – VS) X IS

4. Total power, watts conductive power + transformer power.

5. The autotransformer can be used to step up a voltage.

a. The load is connected across the full winding.

b. The source voltage is impressed across only part of the winding.

c. With a noninductive load,

d. The current input to the autotransformer at a known input voltage can be found knowing the power input:

6. The core and the copper losses in an autotransformer are smaller than those in a constant-voltage transformer.

7. The efficiency of the autotransformer is slightly better than that of the conven- tional transformer.

8. The exciting current and the leakage reactance in the autotransformer are small.

As a result, it is an accepted practice to assume that the input and the output are the same.

• Autotransformers are used when a small increase or decrease in voltage is required

1. on long-distance distribution lines, both single-phase and three-phase, to compensate for the line voltage drop by increasing the line voltage.

2. in electronic circuits where several voltages are required.

3. in manual and automatic motor starters for three-phase systems to reduce the startup voltage applied to three-phase motors; the decreased voltage reduces the starting current surge.

4. to provide a multivoltage source as needed in electronics laboratories and for some types of single-phase variable speed motors, such as the repulsion motor.

• Autotransformers are not used between a high-voltage source and a low-voltage circuit such as a lighting circuit. If the low-voltage section is open circuited, the entire high- voltage input is applied across the low-voltage output.

• A constant-current transformer takes power into its primary winding at a constant volt- age and a variable current. It delivers power from the secondary winding at a constant current and a variable voltage.

• A series street lamp circuit is one application of the constant-current transformer. If a street lamp burns out, the total impedance of the series circuit decreases. If a constant voltage were maintained, the current would increase and cause other lamps to burn out. To prevent this, a small cutout device is placed in the socket of each lamp. When a lamp filament does burn out, the entire voltage of the circuit is applied across the cutout device, which short-circuits the socket and closes the series circuit.

• When used in a series street lighting circuit, a typical constant-current transformer functions as follows:

1. The primary is stationary. Its input comes from a constant voltage source.

2. The secondary winding moves on the center leg of the core. Its movement is balanced by a counterweight–dashpot arrangement. The output of the secondary is connected to the series street lighting circuit.

3. If the total impedance of the series circuit decreases because lamps burn out, there is an increase of current in the series circuit, the movable secondary coil, and the primary coil.

4. The force of repulsion between the primary and secondary coils will increase and cause the secondary coil to move away from the primary coil. As a result, there is less primary flux linking the secondary, and the induced voltage in the secondary will decrease. Thus, the secondary current also decreases.

5. When the counterweight system and the repulsion effect between the coils are equal and opposite, the secondary coil maintains a constant current.

• The primary of the street lighting system transformer (constant-current transformer) is generally energized from an oil switch. The oil switch is operated by a relay that is controlled by a phototube and an amplifier circuit.

• An induction voltage regulator is used to compensate for the resistive and reactive volt- age drops in long-distance distribution circuits.

1. The primary winding of the regulator is wound in the slots of a laminated rotor core. The winding is connected across the line voltage.

2. The secondary winding is stationary and is placed in the slots of a laminated stator core. The secondary winding is connected in series with one of the line wires.

3. The rotor can be turned through an angle of 180°. A small motor and worm gear mechanism drive the rotor.

4. When the maximum flux of the primary links the secondary turns in one position, a maximum voltage is induced in the secondary. This voltage is added to the incoming line voltage.

5. At the center position of the rotor (90°), none of the flux from the primary coil cuts the secondary winding. Thus, no voltage is induced in the secondary.

a. A short-circuited winding known as a tertiary winding is placed in the rotor slot.

b. This winding acts like the short-circuited secondary of a transformer. It pre- vents the secondary from acting as a choke coil when there is no induced voltage in it. The tertiary winding reduces the inductance of the secondary winding to a very small value that can be neglected.

6. As the rotor is moved from the 90° position to the 180° position, the voltage increases until it reaches a maximum at 180°. This induced voltage is in a direction that causes the secondary voltage to oppose the line voltage. Thus, this volt- age is subtracted from the line voltage.

7. The amount of voltage increase or opposition can be controlled by turning the rotor to various positions.

a. The motor that turns the primary rotor is controlled by relays. The relays are energized by a contact-making voltmeter.

b. One set of contacts closes when the line voltage decreases.

c. The motor then turns in a given direction to boost the voltage back to its nor- mal value.

d. If the voltage becomes too high, another set of contactors closes and the motor turns in the opposite direction.

e. As the primary rotor moves to a new position, the voltage is reduced to its normal value.

• Construction of the three-phase induction voltage regulator:

1. The regulator has three primary windings connected in wye or delta to the three line wires of the distribution circuit.

2. Each of the three secondary windings is connected in series with one of the three line wires of the three-phase circuit.

3. The primary windings are stationary and the secondary windings are movable.

4. The motor control system is the same as for a single-phase voltage regulator.

Achievement Review

1. a. What are the two functions of an instrument potential transformer?

b. What are the two functions of an instrument current transformer?

2. In what ways does a current transformer differ from a constant-potential transformer?

3. Explain why the secondary circuit of an instrument current transformer must be closed when there is current in the primary circuit.

4. A 2300/115-V potential transformer and a 100/5-A current transformer are installed and connected on a single-phase line. A voltmeter, an ammeter, and a wattmeter are connected to the secondaries of the instrument transformers. The voltmeter indicates 110 V, the ammeter reads 4 A, and the wattmeter indicates 352

V. Draw a schematic wiring diagram for this circuit. Indicate the polarities of the terminals of the instrument transformers. Also indicate which current and voltage terminals of the wattmeter should be marked with the polarity.

5. Using the circuit in question 4, determine

a. the primary voltage.

b. the primary current, in amperes.

c. the primary power, in watts.

d. the circuit power factor.

6. Instrument potential and current transformers are installed to meter the voltage, current, and power of a 2400-V, three-phase, three-wire distribution circuit. Two 2400/120-V instrument potential transformers and two 100/5-A instrument cur- rent transformers are used. All three secondary voltages are 116 V. Two ammeters are connected to the secondaries of the current transformers. Each meter indicates 3.5 A. Two single-phase wattmeters are used. Wattmeter 1 indicates 400 W and wattmeter 2 reads 160 W. Draw a schematic wiring diagram of this circuit. Indi- cate the polarities on the instrument transformers. Also show which current and potential terminals on each of the wattmeters are to have polarity markings.

7. Using the circuit in question 6, determine

a. the primary voltage.

b. the primary current.

c. the total three-phase primary power, in kVA.

d. the total primary apparent power, in kVA.

e. the circuit power factor.

8. Determine the power factor of the circuit in question 6. Use the power factor curve for the two-wattmeter method given in Unit 10. Compare this value with the calculated power factor obtained in question 7.

9. a. What is an autotransformer?

b. What is one advantage to the use of an autotransformer?

c. What is one limitation of an autotransformer?

10. Explain what is meant by the following terms as they relate to autotransformers:

a. Transformed power

b. Conductive power

11. An autotransformer is shown in Figure 15–19. It is used to step down 600 V to 480

V. The 480-V output is to supply a 24-kW noninductive load. Neglect the losses of the transformer. Determine

a. the load current.

b. the input current.

c. the current between terminals 1 and 2.

d. the current between terminals 3 and 2.

e. what part of the power, in watts, supplied to the load is the transformed power.

f. what part of the power supplied to the load is the conductive power.

12. The autotransformer shown in Figure 15–20 is used to boost the voltage on a long, single-phase line. The voltage is to be changed from 2000 V to 2400 V. The 2400-V output supplies a load of 72 kW. The load has a power factor of unity. Neglect the losses of the transformer and determine

a. the current output to the load.

b. the current input to the transformer.

c. the current in coil section 2–1.

d. the current in coil section 2–3.

e. the part of the power, in watts, supplied to the load due to

(1) transformed power.

(2) conductive power.

13. a. What is a constant-current transformer?

b. Explain the reactions that take place in a constant-current transformer so that a constant current is maintained when there are changes in the load.

14. A constant-current transformer is located in an outdoor substation. The transformer delivers an output of 29.7 kW at a potential of 4500 V to a street lighting series circuit. The circuit consists of 50 lamps having the same rating. The power factor of this load is unity.

a. What current flows in this street lighting series circuit?

b. If 20 lamps are cut out of this series circuit, what output voltage is required from the constant-current transformer so that the current is the same as that for the 50-lamp circuit?

c. Explain why a series street lighting circuit does not go out when the filament of one lamp burns out.

15. Explain the operations of a single-phase induction voltage regulator and a three- phase induction voltage regulator.

CONSTANT-CURRENT TRANSFORMERS

Street lamp circuits may be connected in series or in parallel. The method used depends on the policies of the local utility company. If the lamps are connected in series, some method must be provided to short out any defective lamps. Remember that if one lamp in a series circuit burns out, all of the lamps in the circuit will go out. Shunting in a series circuit is accomplished by placing a small cutout device in the lamp socket. The device consists of two metal contacts that are separated by a film of insulating material. If one lamp filament burns out,

the entire voltage of the series circuit is applied across the cutout device. This momentary voltage punctures the insulation between the metal contacts and shorts out the lamp.

Figure 15–16 is a drawing of a typical lamp socket used on a series street lighting circuit. As the lamps burn out, the total impedance of the series circuit decreases. If a constant voltage were maintained, the current would increase and cause other lamps to burn out.

Most series street lighting circuits require a current of 6.6 A. The voltage of the circuit will depend on the number of lamps in series. A constant 6.6 A is maintained in the circuit by a constant-current transformer. This type of transformer takes power into its primary winding at a constant voltage and a variable current. It delivers power from the secondary winding at a constant current and a variable voltage.

Functioning of the Constant-Current Transformer

The primary and secondary coils of the constant-current transformer are both placed on a silicon steel laminated core. The primary winding is stationary. Its input comes from a constant-voltage source. The secondary winding moves on the center leg of the core. Its movement is balanced by a counterweight–dashpot arrangement. The output of the secondary winding is connected to the series street lighting circuit.

A constant-current transformer is shown in Figure 15–17. The primary winding is energized from a constant-voltage source. The position of the movable secondary winding is adjusted by a counterweight until the desired value of secondary current is obtained. If there are a number of lamp failures in the series circuit, the total impedance of the series circuit decreases. At the same time, the current in the series circuit, the current in the movable secondary coil, and the current in the stationary primary coil will increase.

The force of repulsion between the coils will increase (as described by Lenz’s law). This force causes the secondary coil to move away from the primary coil. As a result, there is less primary flux linking the turns of the secondary winding, and the induced voltage in the secondary will decrease. The secondary coil continues to move away from the primary coil until the secondary current returns to its normal value of 6.6 A. When the current reaches this value, the repulsion force between the coils is balanced by the counterweight system.

Energizing the Primary

The primary of the constant-current transformer may be energized from an oil switch. The transformer is operated by a phototube and an amplifier circuit combined with a relay system. For street lighting applications, this type of control operates as follows. When the normal light intensity decreases below a set level, the phototube causes the amplifier circuit to operate the relay system and energize the transformer primary. The secondary coil moves so that the required 6.6 A is maintained in the circuit. When the light intensity increases above the set level, the phototube and the amplifier circuit operate the relays to open the coil switch. Thus, both the transformer and the street lighting circuit are deenergized.

CONSTANT-CURRENT TRANSFORMERS

Street lamp circuits may be connected in series or in parallel. The method used depends on the policies of the local utility company. If the lamps are connected in series, some method must be provided to short out any defective lamps. Remember that if one lamp in a series circuit burns out, all of the lamps in the circuit will go out. Shunting in a series circuit is accomplished by placing a small cutout device in the lamp socket. The device consists of two metal contacts that are separated by a film of insulating material. If one lamp filament burns out,

the entire voltage of the series circuit is applied across the cutout device. This momentary voltage punctures the insulation between the metal contacts and shorts out the lamp.

Figure 15–16 is a drawing of a typical lamp socket used on a series street lighting circuit. As the lamps burn out, the total impedance of the series circuit decreases. If a constant voltage were maintained, the current would increase and cause other lamps to burn out.

Most series street lighting circuits require a current of 6.6 A. The voltage of the circuit will depend on the number of lamps in series. A constant 6.6 A is maintained in the circuit by a constant-current transformer. This type of transformer takes power into its primary winding at a constant voltage and a variable current. It delivers power from the secondary winding at a constant current and a variable voltage.

Functioning of the Constant-Current Transformer

The primary and secondary coils of the constant-current transformer are both placed on a silicon steel laminated core. The primary winding is stationary. Its input comes from a constant-voltage source. The secondary winding moves on the center leg of the core. Its movement is balanced by a counterweight–dashpot arrangement. The output of the secondary winding is connected to the series street lighting circuit.

A constant-current transformer is shown in Figure 15–17. The primary winding is energized from a constant-voltage source. The position of the movable secondary winding is adjusted by a counterweight until the desired value of secondary current is obtained. If there are a number of lamp failures in the series circuit, the total impedance of the series circuit decreases. At the same time, the current in the series circuit, the current in the movable secondary coil, and the current in the stationary primary coil will increase.

The force of repulsion between the coils will increase (as described by Lenz’s law). This force causes the secondary coil to move away from the primary coil. As a result, there is less primary flux linking the turns of the secondary winding, and the induced voltage in the secondary will decrease. The secondary coil continues to move away from the primary coil until the secondary current returns to its normal value of 6.6 A. When the current reaches this value, the repulsion force between the coils is balanced by the counterweight system.

Energizing the Primary

The primary of the constant-current transformer may be energized from an oil switch. The transformer is operated by a phototube and an amplifier circuit combined with a relay system. For street lighting applications, this type of control operates as follows. When the normal light intensity decreases below a set level, the phototube causes the amplifier circuit to operate the relay system and energize the transformer primary. The secondary coil moves so that the required 6.6 A is maintained in the circuit. When the light intensity increases above the set level, the phototube and the amplifier circuit operate the relays to open the coil switch. Thus, both the transformer and the street lighting circuit are deenergized.

AUTOTRANSFORMERS

The conventional transformer has separate primary and secondary windings that are electrically insulated from each other. The autotransformer, on the other hand, has a primary winding and a secondary winding in the form of one continuous winding. This winding is tapped at certain points to obtain the desired voltages. The winding is wound on a laminated silicon steel core. Therefore, the primary and secondary sections of the winding are in the same magnetic circuit.

Figure 15–8 shows how the autotransformer is connected into a typical circuit. The autotransformer can be viewed as having a primary side of terminals 1–3 and a secondary side of terminals 2–3. The calculations necessary for autotransformer circuits involve two new quantities: conductive power and transformed power.

Conductive power is the product of the current conducted to the secondary side from the primary side and the voltage on the secondary side.

Transformed power is the product of the difference between the primary and secondary voltages and the secondary current.

As an example of the calculation of these power values, consider the circuit shown in Figure 15–8. The 240-V input is connected across the entire winding at terminals 1 and 3.

Figure 15–8 indicates that 18 A is supplied from the source to terminal 1. This cur- rent passes from terminal 1 down through the winding to terminal 2. Note that the load requires 24 A. As a result, there must be a current of 6 A up from terminal 3 to terminal 2 at this same instant: 6 A + 18 A 24 A, as required by the load. The current entering terminal 1 is 18 A at a potential of 240 V. The current at terminal 2 is at a potential of only 180 V. There is a potential drop of 60 V between terminals 1 and 2. This change represents a power value of 60 X 18 = 1080 W. This power is the transformed power. The product of 180 V and 6 A also equals 1080 W. The section of the winding between terminals 1 and 2 represents the primary winding with an input of 1080 W. The section of the winding between terminals 3 and 2 represents the secondary winding with an output of 1080 W.

The 1080-W output provides only a part of the power required by the load. The load requires a total of 4320 W. The 18-A current from terminal 1 leaves the winding at terminal 2 at a potential of 180 volts. The product of 180 V and 18 A is a part of the power, in watts, that is conducted directly through a section of the windings to supply the load:

This value of power is the conductive power. The sum of the transformed power and the conductive power equals the total power supplied to the load, or 4320 W:

Output, watts transformed power + conductive power

3240 W + 1080 W 4320 W

Autotransformer Used to Step Up Voltage

The autotransformer can also be used to step up a voltage. If, in Figure 15–8, a voltage of 180 V is impressed across the section of the winding from terminals 2 to 3, the output voltage across terminals 1 and 3 will be 240 V.

An autotransformer used to step up a voltage is shown in Figure 15–9. A heating load is connected across the full winding, which has an output of 250 V. The current supplied to the load is

This means that if losses are neglected, the input to the transformer will be 5000 W. The current input to the autotransformers at an input voltage of 200 V can be found knowing the power input:

It is assumed that the line wire connected to terminal 2 is negative. A cur- rent of 25 A enters the winding at terminal 2. The load connected between terminals 1 and 3 requires only 20 A. Therefore, 20 A appears in the winding from terminal 2 to terminal 1. The remaining 5 A at terminal 2 appears in the winding between terminal 2 and terminal 3. Between terminal 2 and terminal 3, there is a loss of 200 V at 5 A. This change represents a transfer of 1000 W of power from the winding between terminals 2 and 3 to the flux of the core. Therefore, this section of the coil is the primary winding with a power input of 1000 W. The 20-A current between terminals 2 and 1 increases in potential by 50 V. The product of 50 V and 20 A is 1000 W. The winding between terminals 2 and 1 is the secondary winding and has a power output of 1000 W. This value is the part of the total power output that is transferred by transformer action from one section of the winding to a second section of the winding. The conductive power is considered to be the actual 20-A current at 200 V between terminals 2 and 1:

As compared to a constant-voltage transformer, the core and the copper losses in an autotransformer are smaller. The autotransformer has a slightly better efficiency than does the conventional transformer. The exciting current and the leakage reactance in the autotransformer are small. Therefore, it is an accepted practice to assume that the input and output are the same.

Applications of Autotransformers

Autotransformers are used in applications requiring small increases or decreases in voltage. Several typical uses of autotransformers are as follows:

• On long single-phase and three-phase distribution lines where the line voltage must be increased to compensate for voltage drops in the line wires

• In electronic circuits where several voltages are required

• In three-phase manual and automatic motor starters to reduce the startup voltage applied to three-phase motors; the decreased voltage reduces the starting current surge

• To provide a multivoltage source, as required by electronics laboratories

Multivoltages from an Autotransformer

Figure 15–10 shows one method of providing a number of voltages from one source. Taps are connected to the single winding of an autotransformer to provide a range of volt- ages, from very low values to values above the input voltage. For example, voltages ranging from 15 to 260 V may be obtained from an autotransformer with a 240-V input. This type of multivoltage supply can be used on some types of single-phase variable-speed motors, such as the repulsion motor. When the tap points are connected to a switching mechanism, a range of above- and below-normal speeds is available from the motor.

Another common autotransformer is often referred to as a variable transformer. This type of transformer has a sliding brush that can provide any voltage from 0 to full output voltage. The sliding brush permits the turns ratio to be changed to any setting. Many of these transformers can be connected to provide a higher output voltage than the input volt- age. Autotransformers that are connected to 120 Vac can often provide output voltages of 140 Vac. The connection diagram of a variable autotransformer is shown in Figure 15–11. In the circuit shown in Figure 15–12, 120 volts is connected to terminals 2 and 4. The load is connected between the sliding brush connected to terminal 3 and terminal 4. This connection will permit the transformer to supply any voltage between 0 and 120 Vac to the load. If the 120-volt supply is changed to terminals 1 and 4, Figure 15–13, the autotrans- former becomes a stepup transformer.

The autotransformer can now supply 0 to 140 Vac to the load. The part of the wind- ing connected to power is the primary, and the part of the winding connected to the load is the secondary. In Figure 15–13, the primary winding is connected across terminals 1 and 4. Therefore, the winding between these two terminals is the primary. The sliding brush, however, can access all the windings between terminals 4 and 2. The number of turns between terminals 2 and 4 is greater than between terminals 1 and 4, permitting an increase in voltage between terminals 3 and 4. A variable transformer is shown in Figure 15–14.

Autotransformer Limitations

There are certain applications for which autotransformers should not be used. For example, an autotransformer should not be used to step down 2400 to 120 V for a lighting load. Recall that the primary and secondary windings of the autotransformer are connected electrically through the same winding. Thus, if the low-voltage section is open circuited, the entire high-voltage input will be applied across the low-voltage output.

Possible Hazards with Autotransformers

The normal operation of an autotransformer is shown in Figure 15–15A. Figure 15–15B shows a break in the low-voltage section of the winding. The entire 2400-V input is applied across the break in the winding and across the lighting load. Another voltage hazard will exist if one of the high-voltage leads becomes grounded. In this case, there will be a high voltage between ground and one of the low-voltage wires. Because of these hazards, an autotrans- former is not used in applications requiring a high voltage to be transformed to 120 or 240 V.

AUTOTRANSFORMERS

The conventional transformer has separate primary and secondary windings that are electrically insulated from each other. The autotransformer, on the other hand, has a primary winding and a secondary winding in the form of one continuous winding. This winding is tapped at certain points to obtain the desired voltages. The winding is wound on a laminated silicon steel core. Therefore, the primary and secondary sections of the winding are in the same magnetic circuit.

Figure 15–8 shows how the autotransformer is connected into a typical circuit. The autotransformer can be viewed as having a primary side of terminals 1–3 and a secondary side of terminals 2–3. The calculations necessary for autotransformer circuits involve two new quantities: conductive power and transformed power.

Conductive power is the product of the current conducted to the secondary side from the primary side and the voltage on the secondary side.

Transformed power is the product of the difference between the primary and secondary voltages and the secondary current.

As an example of the calculation of these power values, consider the circuit shown in Figure 15–8. The 240-V input is connected across the entire winding at terminals 1 and 3.

Figure 15–8 indicates that 18 A is supplied from the source to terminal 1. This cur- rent passes from terminal 1 down through the winding to terminal 2. Note that the load requires 24 A. As a result, there must be a current of 6 A up from terminal 3 to terminal 2 at this same instant: 6 A + 18 A 24 A, as required by the load. The current entering terminal 1 is 18 A at a potential of 240 V. The current at terminal 2 is at a potential of only 180 V. There is a potential drop of 60 V between terminals 1 and 2. This change represents a power value of 60 X 18 = 1080 W. This power is the transformed power. The product of 180 V and 6 A also equals 1080 W. The section of the winding between terminals 1 and 2 represents the primary winding with an input of 1080 W. The section of the winding between terminals 3 and 2 represents the secondary winding with an output of 1080 W.

The 1080-W output provides only a part of the power required by the load. The load requires a total of 4320 W. The 18-A current from terminal 1 leaves the winding at terminal 2 at a potential of 180 volts. The product of 180 V and 18 A is a part of the power, in watts, that is conducted directly through a section of the windings to supply the load:

This value of power is the conductive power. The sum of the transformed power and the conductive power equals the total power supplied to the load, or 4320 W:

Output, watts transformed power + conductive power

3240 W + 1080 W 4320 W

Autotransformer Used to Step Up Voltage

The autotransformer can also be used to step up a voltage. If, in Figure 15–8, a voltage of 180 V is impressed across the section of the winding from terminals 2 to 3, the output voltage across terminals 1 and 3 will be 240 V.

An autotransformer used to step up a voltage is shown in Figure 15–9. A heating load is connected across the full winding, which has an output of 250 V. The current supplied to the load is

This means that if losses are neglected, the input to the transformer will be 5000 W. The current input to the autotransformers at an input voltage of 200 V can be found knowing the power input:

It is assumed that the line wire connected to terminal 2 is negative. A cur- rent of 25 A enters the winding at terminal 2. The load connected between terminals 1 and 3 requires only 20 A. Therefore, 20 A appears in the winding from terminal 2 to terminal 1. The remaining 5 A at terminal 2 appears in the winding between terminal 2 and terminal 3. Between terminal 2 and terminal 3, there is a loss of 200 V at 5 A. This change represents a transfer of 1000 W of power from the winding between terminals 2 and 3 to the flux of the core. Therefore, this section of the coil is the primary winding with a power input of 1000 W. The 20-A current between terminals 2 and 1 increases in potential by 50 V. The product of 50 V and 20 A is 1000 W. The winding between terminals 2 and 1 is the secondary winding and has a power output of 1000 W. This value is the part of the total power output that is transferred by transformer action from one section of the winding to a second section of the winding. The conductive power is considered to be the actual 20-A current at 200 V between terminals 2 and 1:

As compared to a constant-voltage transformer, the core and the copper losses in an autotransformer are smaller. The autotransformer has a slightly better efficiency than does the conventional transformer. The exciting current and the leakage reactance in the autotransformer are small. Therefore, it is an accepted practice to assume that the input and output are the same.

Applications of Autotransformers

Autotransformers are used in applications requiring small increases or decreases in voltage. Several typical uses of autotransformers are as follows:

• On long single-phase and three-phase distribution lines where the line voltage must be increased to compensate for voltage drops in the line wires

• In electronic circuits where several voltages are required

• In three-phase manual and automatic motor starters to reduce the startup voltage applied to three-phase motors; the decreased voltage reduces the starting current surge

• To provide a multivoltage source, as required by electronics laboratories

Multivoltages from an Autotransformer

Figure 15–10 shows one method of providing a number of voltages from one source. Taps are connected to the single winding of an autotransformer to provide a range of volt- ages, from very low values to values above the input voltage. For example, voltages ranging from 15 to 260 V may be obtained from an autotransformer with a 240-V input. This type of multivoltage supply can be used on some types of single-phase variable-speed motors, such as the repulsion motor. When the tap points are connected to a switching mechanism, a range of above- and below-normal speeds is available from the motor.

Another common autotransformer is often referred to as a variable transformer. This type of transformer has a sliding brush that can provide any voltage from 0 to full output voltage. The sliding brush permits the turns ratio to be changed to any setting. Many of these transformers can be connected to provide a higher output voltage than the input volt- age. Autotransformers that are connected to 120 Vac can often provide output voltages of 140 Vac. The connection diagram of a variable autotransformer is shown in Figure 15–11. In the circuit shown in Figure 15–12, 120 volts is connected to terminals 2 and 4. The load is connected between the sliding brush connected to terminal 3 and terminal 4. This connection will permit the transformer to supply any voltage between 0 and 120 Vac to the load. If the 120-volt supply is changed to terminals 1 and 4, Figure 15–13, the autotrans- former becomes a stepup transformer.

The autotransformer can now supply 0 to 140 Vac to the load. The part of the wind- ing connected to power is the primary, and the part of the winding connected to the load is the secondary. In Figure 15–13, the primary winding is connected across terminals 1 and 4. Therefore, the winding between these two terminals is the primary. The sliding brush, however, can access all the windings between terminals 4 and 2. The number of turns between terminals 2 and 4 is greater than between terminals 1 and 4, permitting an increase in voltage between terminals 3 and 4. A variable transformer is shown in Figure 15–14.

Autotransformer Limitations

There are certain applications for which autotransformers should not be used. For example, an autotransformer should not be used to step down 2400 to 120 V for a lighting load. Recall that the primary and secondary windings of the autotransformer are connected electrically through the same winding. Thus, if the low-voltage section is open circuited, the entire high-voltage input will be applied across the low-voltage output.

Possible Hazards with Autotransformers

The normal operation of an autotransformer is shown in Figure 15–15A. Figure 15–15B shows a break in the low-voltage section of the winding. The entire 2400-V input is applied across the break in the winding and across the lighting load. Another voltage hazard will exist if one of the high-voltage leads becomes grounded. In this case, there will be a high voltage between ground and one of the low-voltage wires. Because of these hazards, an autotrans- former is not used in applications requiring a high voltage to be transformed to 120 or 240 V.

Special Transformer Applications

INSTRUMENT TRANSFORMERS

Instrument transformers are used with instruments and relays to measure and control ac circuits. Large and expensive devices are required to measure high voltages and currents directly. However, instrument transformers used with small standard devices provide a way of measuring high voltage and current values safely. The use of instrument transformers also means increased safety for the operator, less chance of control equipment damage due to high voltages, more accurate measurements, and greater convenience.

There are two types of instrument transformers: the instrument potential transformer and the instrument current transformer.

The Potential Transformer

The potential transformer is similar to a power transformer or a distribution transformer. However, its capacity is relatively small when compared to that of power transformers. Typical ratings of potential transformers range from 100 to 500 VA. The low-voltage side of the trans- former generally is wound for 120 V. The load on the secondary (low-voltage) side of the potential transformer consists of the potential coils of various instruments. In some cases, the potential coils of relays and other control equipment are also connected to the secondary. In almost all cases, the load is light. Thus, potential transformers require a capacity no greater than 100 to 500 VA.

The primary circuit voltage and the voltage rating of the primary winding of a potential transformer are the same. As an example of the use of a potential transformer, assume that the voltage of a 4800-V, single-phase line must be measured. In this case, the primary of the potential transformer is rated at 4800 V. The low-voltage secondary is rated at 120 V. The ratio between the primary and secondary voltages is 4800/120 40/1.

A voltmeter can be connected across the secondary of the potential transformer to mea- sure the primary voltage, or 120 V. To determine the actual voltage of the high-voltage circuit, the voltmeter reading is multiplied by the transformer ratio. Thus, 120 X 40 4800 V. In some cases, the voltmeter is calibrated to indicate the actual value of the voltage on the primary side. As a result, it is not necessary to apply the multiplier to the instrument read- ing. This means that errors are minimized.

Figure 15–1 shows the connections for a potential transformer with a 4800-V pri- mary input and a 120-V output to a voltmeter. Note that the transformer has subtractive polarity. All instrument potential transformers now manufactured have subtractive polar- ity. One of the secondary leads is grounded to minimize high-voltage hazards.

The ratio between the primary and secondary voltages of a potential transformer is very accurate. Normally, the percentage error is less than 0.5%.

The Current Transformer

The second type of instrument transformer is the current transformer. Current trans- formers are used to avoid the need to connect ammeters, instrument current coils, and relays directly to high-voltage lines. Current transformers step down the current by a known ratio. Thus, small and accurate instruments and control devices can be used because they are insulated from the high-voltage line.

The primary winding of a current transformer is connected in series with one of the line wires. The primary winding consists of a few turns of heavy wire wound on a laminated iron core. The secondary coil has more turns of smaller wire and is wound on the same core as the primary coil. The current rating of the primary winding is the maximum current that the winding will be required to conduct. For higher currents, the line in question may be fitted through a toroidal core with no turns at all. The core than acts as the primary winding and has no connection to the line. The secondary winding is always rated at 5 A.

To illustrate the operation of a current transformer, assume that the current rating of the primary winding is 100 A. The secondary winding has the standard rating of 5 A. The primary winding consists of three turns of wire, and the secondary winding has 60 turns. The ratio between the primary and the secondary currents is 100 A/5 A, or 20:1. In other words, the primary current is 20 times greater than the secondary current. Note that the number of turns and the current in the primary and secondary windings are related by an inverse proportion.

Stepping Down Current. The current transformer shown in Figure 15–2 is used to step down the current in a 4800-V, single-phase circuit. The primary is rated at 100 A. Because the secondary has the standard 5-A rating, the transformer ratio is 20:1. In other words, there is 20 A in the primary winding for each ampere in the secondary winding. If the ammeter indicates 4 A, the actual current in the primary is 20 times this secondary current, or 80 A.

Polarity Markings. The polarity markings of the current transformer are also shown in Figure 15–2. The high-voltage primary terminals are marked H1 and H2. The secondary terminals are marked X1 and X2. Electrons enter H1 and leave X1 at the same instant. Some manufacturers mark only the H1 and X1 leads. The H1 lead connects to the line wire feeding from the source. The H2 lead connects to the line wire feeding to the load. The secondary

leads connect directly to the ammeter. One of the secondary leads is grounded to reduce the high-voltage hazard. A common current transformer with a ratio of 200:5 is shown in Figure 15–3.

Current Transformer Precaution. The secondary circuit of a current transformer must not be opened when there is current in the primary winding. An absence of current in the sec- ondary winding means that there is no secondary magnetomotive force opposing the primary mmf. The primary current becomes an exciting current. The flux of this current is unopposed by a secondary mmf. The primary flux causes a high voltage to be induced in the secondary

winding. This voltage is great enough to be a hazard. If the instrument circuit must be opened when there is current in the primary winding, a short-circuiting switch is installed at the secondary terminals of the current transformer. This switch is closed before the instrument circuit is opened to make repairs or rewire the metering circuit.

Primary-to-Secondary Ratios. Current transformers have very accurate ratios between the primary and secondary currents. The percent error generally is less than 0.5.

Construction. A bar-type construction is often used when the primary winding of a cur- rent transformer has a large rating. The primary winding consists of a straight copper bus bar that passes through the center of a hollow metal core. The secondary winding is wound on the core. All standard current transformers rated at 1000 A or larger have this type of structure. In some cases, current transformers with ratings smaller than 1000 A also have the bar-type structure.

Measuring Current, Voltage, and Power

A potential transformer and a current transformer are shown in Figure 15–4. These transformers are being used with standard instrument movements to measure the voltage,

current, and power for a 4600-V single-phase circuit. The potential transformer is rated at 4800/120 V, and the current transformer is rated at 50/5 A.

The voltmeter and the potential coil of the wattmeter are connected in parallel across the low-voltage output of the potential transformer. The voltage across the potential coils of both instruments is the same.

The ammeter and the current coil of the wattmeter are connected in series across the secondary output of the current transformer. This means that the same current appears in the current coils of both instruments.

When lead H1 of the potential transformer and lead H1 of the current transformer are instantaneously positive, the X1 leads of both transformer secondaries are also instan- taneously positive. The current and voltage terminals of the wattmeter (marked ±) have the same instantaneous polarity. Thus, the torque on the wattmeter movement causes the pointer to move upscale. The secondary side of each instrument transformer is grounded to minimize high-voltage hazards.

Calculating the Primary Voltage and Current. In Figure 15–4, the voltmeter reading is 112.5 V, the ammeter reading is 4 A, and the wattmeter reading is 450 W. The primary voltage can be found as follows:

Calculating Power and Power Factor. The primary power must be determined using a power multiplier. Recall that power is the product of the voltage. the current, and the power factor. For this example, the power multiplier is the product of the current multiplier and the voltage multiplier:

Metering Connections

Three-phase circuits are used for most high-voltage and current transmission and distribution lines. This means that the three-phase, three-wire system requires two potential transformers having the same rating, and two current transformers having the same rating.

The connections shown in Figure 15–5 are for a three-phase, three-wire system using instrument transformers and measuring devices. The two potential transformers are connected in open delta to the 4800-V, three-phase line. There are three secondary voltages of 120 V each. Two current transformers are used. The primary winding of one current transformer is connected in series with line A. The primary winding of the other current transformer is connected in series with line C.

It can be seen by checking the respective primary and secondary circuit paths for the ammeters that each ammeter is connected correctly. Other instruments that may be added to the circuit include a three-phase wattmeter, a three-phase watt-hour meter, or a three-phase power factor meter. The proper phase relationships must be maintained when three-phase instruments are connected into a secondary circuit. If the instruments are not connected properly, their readings will be incorrect. For the three-phase, three-wire metering system

of Figure 15–5, both the interconnected secondary potential circuit and the interconnected secondary current circuit are grounded. Grounding serves as protection from high-voltage hazards.

Testing Metering Connections

When a test is to be made on industrial equipment such as a three-phase motor, temporary metering connections are commonly made using portable instruments and instrument transformers. Figure 15–6 shows the connections required to obtain measurements of the cur- rents, the voltage, and the power in watts for a 20-hp, 480-V, three-phase motor. Two potential transformers, rated at 480/120 V, are connected in open delta. Two current transformers, each rated at 50/5 A, are connected in line wires A and C.

The circuit in Figure 15–6 shows one method of measuring the three-phase power using the two-wattmeter method. The current coil of wattmeter 2 is in the secondary circuit of the current transformer and is in series with line wire A. The potential coil of wattmeter 2 is connected across the secondary of the potential transformer whose primary winding is

connected to line wire A and B. The current coil of wattmeter 1 in the secondary circuit of the current transformer is in series with line wire C. The potential coil of wattmeter 1 is connected across the secondary of the potential transformer whose primary winding is connected to line wires B and C. The polarity terminals of both wattmeters are marked and are connected into the circuit correctly.

To check the connections, note that when primary line wires A and C are instantaneously positive, line wire B is instantaneously negative. The H1 terminals on the current transformers are positive. The X1 terminals on both current transformers are positive at the same instant. This means that the marked polarity current terminal on each wattmeter is instantaneously positive. Leads A and C from the low-voltage side of the potential trans- formers feed to the marked polarity potential terminals of the two wattmeters. These leads are also positive at this same instant. Therefore, both wattmeters indicate upscale if the power factor of the load is greater than 0.50 lag.

PROBLEM 1

Statement of the Problem

Figure 15–6 shows that the three secondary voltages are 120 V each. Both ammeters indicate 2.7 A. Wattmeter 1 reads 317 W, and wattmeter 2 indicates 121 W. The motor is delivering its rated capacity of 20 hp. Determine

1. the input, in watts.

2. the power factor.

3. the motor efficiency.

Solution

1. The total true power input is determined as follows:

The ratio of the two power values may be called the power factor ratio for the two- wattmeter method given in Unit 10. This ratio can be determined using the two- wattmeter readings or the primary power values:

If the ratio of 0.38 is projected on the power factor ratio curve, it will be 0.78. This value agrees with the calculated power factor of 0.78 lag.

3. When the motor is delivering its rated horsepower output, its efficiency is

Secondary Metering Connections

Figure 15–7 shows the secondary metering connections for a 2400/4160-V, three- phase, four-wire system. The three potential transformers are connected in wye. The three-phase output of these transformers consists of three secondary voltages of 120 V to neutral. A 50-to-5-A current transformer is connected in series with each of the three line wires.

When wattmeters, watt-hour meters, and other three-phase instruments are used in a circuit, their readings will be correct only if they are connected with the proper phase relationship. Three single-phase wattmeters are shown in the circuit in Figure 15–7. These instruments make it easier to check the instantaneous directions of current and voltage. In an actual installation, a three-element indicating wattmeter and possibly a three-element watt-hour meter would be used. However, the connections are the same. Note that both the low-voltage potential network and the secondary current network are grounded to minimize high-voltage hazards.

Special Transformer Applications

INSTRUMENT TRANSFORMERS

Instrument transformers are used with instruments and relays to measure and control ac circuits. Large and expensive devices are required to measure high voltages and currents directly. However, instrument transformers used with small standard devices provide a way of measuring high voltage and current values safely. The use of instrument transformers also means increased safety for the operator, less chance of control equipment damage due to high voltages, more accurate measurements, and greater convenience.

There are two types of instrument transformers: the instrument potential transformer and the instrument current transformer.

The Potential Transformer

The potential transformer is similar to a power transformer or a distribution transformer. However, its capacity is relatively small when compared to that of power transformers. Typical ratings of potential transformers range from 100 to 500 VA. The low-voltage side of the trans- former generally is wound for 120 V. The load on the secondary (low-voltage) side of the potential transformer consists of the potential coils of various instruments. In some cases, the potential coils of relays and other control equipment are also connected to the secondary. In almost all cases, the load is light. Thus, potential transformers require a capacity no greater than 100 to 500 VA.

The primary circuit voltage and the voltage rating of the primary winding of a potential transformer are the same. As an example of the use of a potential transformer, assume that the voltage of a 4800-V, single-phase line must be measured. In this case, the primary of the potential transformer is rated at 4800 V. The low-voltage secondary is rated at 120 V. The ratio between the primary and secondary voltages is 4800/120 40/1.

A voltmeter can be connected across the secondary of the potential transformer to mea- sure the primary voltage, or 120 V. To determine the actual voltage of the high-voltage circuit, the voltmeter reading is multiplied by the transformer ratio. Thus, 120 X 40 4800 V. In some cases, the voltmeter is calibrated to indicate the actual value of the voltage on the primary side. As a result, it is not necessary to apply the multiplier to the instrument read- ing. This means that errors are minimized.

Figure 15–1 shows the connections for a potential transformer with a 4800-V pri- mary input and a 120-V output to a voltmeter. Note that the transformer has subtractive polarity. All instrument potential transformers now manufactured have subtractive polar- ity. One of the secondary leads is grounded to minimize high-voltage hazards.

The ratio between the primary and secondary voltages of a potential transformer is very accurate. Normally, the percentage error is less than 0.5%.

The Current Transformer

The second type of instrument transformer is the current transformer. Current trans- formers are used to avoid the need to connect ammeters, instrument current coils, and relays directly to high-voltage lines. Current transformers step down the current by a known ratio. Thus, small and accurate instruments and control devices can be used because they are insulated from the high-voltage line.

The primary winding of a current transformer is connected in series with one of the line wires. The primary winding consists of a few turns of heavy wire wound on a laminated iron core. The secondary coil has more turns of smaller wire and is wound on the same core as the primary coil. The current rating of the primary winding is the maximum current that the winding will be required to conduct. For higher currents, the line in question may be fitted through a toroidal core with no turns at all. The core than acts as the primary winding and has no connection to the line. The secondary winding is always rated at 5 A.

To illustrate the operation of a current transformer, assume that the current rating of the primary winding is 100 A. The secondary winding has the standard rating of 5 A. The primary winding consists of three turns of wire, and the secondary winding has 60 turns. The ratio between the primary and the secondary currents is 100 A/5 A, or 20:1. In other words, the primary current is 20 times greater than the secondary current. Note that the number of turns and the current in the primary and secondary windings are related by an inverse proportion.

Stepping Down Current. The current transformer shown in Figure 15–2 is used to step down the current in a 4800-V, single-phase circuit. The primary is rated at 100 A. Because the secondary has the standard 5-A rating, the transformer ratio is 20:1. In other words, there is 20 A in the primary winding for each ampere in the secondary winding. If the ammeter indicates 4 A, the actual current in the primary is 20 times this secondary current, or 80 A.

Polarity Markings. The polarity markings of the current transformer are also shown in Figure 15–2. The high-voltage primary terminals are marked H1 and H2. The secondary terminals are marked X1 and X2. Electrons enter H1 and leave X1 at the same instant. Some manufacturers mark only the H1 and X1 leads. The H1 lead connects to the line wire feeding from the source. The H2 lead connects to the line wire feeding to the load. The secondary

leads connect directly to the ammeter. One of the secondary leads is grounded to reduce the high-voltage hazard. A common current transformer with a ratio of 200:5 is shown in Figure 15–3.

Current Transformer Precaution. The secondary circuit of a current transformer must not be opened when there is current in the primary winding. An absence of current in the sec- ondary winding means that there is no secondary magnetomotive force opposing the primary mmf. The primary current becomes an exciting current. The flux of this current is unopposed by a secondary mmf. The primary flux causes a high voltage to be induced in the secondary

winding. This voltage is great enough to be a hazard. If the instrument circuit must be opened when there is current in the primary winding, a short-circuiting switch is installed at the secondary terminals of the current transformer. This switch is closed before the instrument circuit is opened to make repairs or rewire the metering circuit.

Primary-to-Secondary Ratios. Current transformers have very accurate ratios between the primary and secondary currents. The percent error generally is less than 0.5.

Construction. A bar-type construction is often used when the primary winding of a cur- rent transformer has a large rating. The primary winding consists of a straight copper bus bar that passes through the center of a hollow metal core. The secondary winding is wound on the core. All standard current transformers rated at 1000 A or larger have this type of structure. In some cases, current transformers with ratings smaller than 1000 A also have the bar-type structure.

Measuring Current, Voltage, and Power

A potential transformer and a current transformer are shown in Figure 15–4. These transformers are being used with standard instrument movements to measure the voltage,

current, and power for a 4600-V single-phase circuit. The potential transformer is rated at 4800/120 V, and the current transformer is rated at 50/5 A.

The voltmeter and the potential coil of the wattmeter are connected in parallel across the low-voltage output of the potential transformer. The voltage across the potential coils of both instruments is the same.

The ammeter and the current coil of the wattmeter are connected in series across the secondary output of the current transformer. This means that the same current appears in the current coils of both instruments.

When lead H1 of the potential transformer and lead H1 of the current transformer are instantaneously positive, the X1 leads of both transformer secondaries are also instan- taneously positive. The current and voltage terminals of the wattmeter (marked ±) have the same instantaneous polarity. Thus, the torque on the wattmeter movement causes the pointer to move upscale. The secondary side of each instrument transformer is grounded to minimize high-voltage hazards.

Calculating the Primary Voltage and Current. In Figure 15–4, the voltmeter reading is 112.5 V, the ammeter reading is 4 A, and the wattmeter reading is 450 W. The primary voltage can be found as follows:

Calculating Power and Power Factor. The primary power must be determined using a power multiplier. Recall that power is the product of the voltage. the current, and the power factor. For this example, the power multiplier is the product of the current multiplier and the voltage multiplier:

Metering Connections

Three-phase circuits are used for most high-voltage and current transmission and distribution lines. This means that the three-phase, three-wire system requires two potential transformers having the same rating, and two current transformers having the same rating.

The connections shown in Figure 15–5 are for a three-phase, three-wire system using instrument transformers and measuring devices. The two potential transformers are connected in open delta to the 4800-V, three-phase line. There are three secondary voltages of 120 V each. Two current transformers are used. The primary winding of one current transformer is connected in series with line A. The primary winding of the other current transformer is connected in series with line C.

It can be seen by checking the respective primary and secondary circuit paths for the ammeters that each ammeter is connected correctly. Other instruments that may be added to the circuit include a three-phase wattmeter, a three-phase watt-hour meter, or a three-phase power factor meter. The proper phase relationships must be maintained when three-phase instruments are connected into a secondary circuit. If the instruments are not connected properly, their readings will be incorrect. For the three-phase, three-wire metering system

of Figure 15–5, both the interconnected secondary potential circuit and the interconnected secondary current circuit are grounded. Grounding serves as protection from high-voltage hazards.

Testing Metering Connections

When a test is to be made on industrial equipment such as a three-phase motor, temporary metering connections are commonly made using portable instruments and instrument transformers. Figure 15–6 shows the connections required to obtain measurements of the cur- rents, the voltage, and the power in watts for a 20-hp, 480-V, three-phase motor. Two potential transformers, rated at 480/120 V, are connected in open delta. Two current transformers, each rated at 50/5 A, are connected in line wires A and C.

The circuit in Figure 15–6 shows one method of measuring the three-phase power using the two-wattmeter method. The current coil of wattmeter 2 is in the secondary circuit of the current transformer and is in series with line wire A. The potential coil of wattmeter 2 is connected across the secondary of the potential transformer whose primary winding is

connected to line wire A and B. The current coil of wattmeter 1 in the secondary circuit of the current transformer is in series with line wire C. The potential coil of wattmeter 1 is connected across the secondary of the potential transformer whose primary winding is connected to line wires B and C. The polarity terminals of both wattmeters are marked and are connected into the circuit correctly.

To check the connections, note that when primary line wires A and C are instantaneously positive, line wire B is instantaneously negative. The H1 terminals on the current transformers are positive. The X1 terminals on both current transformers are positive at the same instant. This means that the marked polarity current terminal on each wattmeter is instantaneously positive. Leads A and C from the low-voltage side of the potential trans- formers feed to the marked polarity potential terminals of the two wattmeters. These leads are also positive at this same instant. Therefore, both wattmeters indicate upscale if the power factor of the load is greater than 0.50 lag.

PROBLEM 1

Statement of the Problem

Figure 15–6 shows that the three secondary voltages are 120 V each. Both ammeters indicate 2.7 A. Wattmeter 1 reads 317 W, and wattmeter 2 indicates 121 W. The motor is delivering its rated capacity of 20 hp. Determine

1. the input, in watts.

2. the power factor.

3. the motor efficiency.

Solution

1. The total true power input is determined as follows:

The ratio of the two power values may be called the power factor ratio for the two- wattmeter method given in Unit 10. This ratio can be determined using the two- wattmeter readings or the primary power values:

If the ratio of 0.38 is projected on the power factor ratio curve, it will be 0.78. This value agrees with the calculated power factor of 0.78 lag.

3. When the motor is delivering its rated horsepower output, its efficiency is

Secondary Metering Connections

Figure 15–7 shows the secondary metering connections for a 2400/4160-V, three- phase, four-wire system. The three potential transformers are connected in wye. The three-phase output of these transformers consists of three secondary voltages of 120 V to neutral. A 50-to-5-A current transformer is connected in series with each of the three line wires.

When wattmeters, watt-hour meters, and other three-phase instruments are used in a circuit, their readings will be correct only if they are connected with the proper phase relationship. Three single-phase wattmeters are shown in the circuit in Figure 15–7. These instruments make it easier to check the instantaneous directions of current and voltage. In an actual installation, a three-element indicating wattmeter and possibly a three-element watt-hour meter would be used. However, the connections are the same. Note that both the low-voltage potential network and the secondary current network are grounded to minimize high-voltage hazards.

BUS DUCTS AND PANEL PROBLEMS

Triplen harmonic currents can also cause problems with neutral bus ducts and connecting lugs. A neutral bus is sized to carry the rated phase current. Because triplen harmonics can cause the neutral current to be higher than the phase current, it is possible for the neutral bus to become overloaded.

Electrical panels and bus ducts are designed to carry currents that operate at 60 Hz. Harmonic currents produce magnetic fields that operate at higher frequencies. If these fields should become mechanically resonant with the panel or bus duct enclosures, the panels and bus ducts can vibrate and produce buzzing sounds at the harmonic frequency.

Telecommunications equipment is often affected by harmonic currents. Tele- communication cable is often run close to power lines. To minimize interference, communication cables are run as far from phase conductors as possible and as close to the neutral conductor as possible. Harmonic currents in the neutral conductor induce high- frequency currents into the communication cable. These high-frequency currents can be heard as a high-pitched buzzing sound on telephone lines.

Determining Harmonic Problems on Single-Phase Systems

There are several steps that can be followed in determining whether there is a problem with harmonics. One step is to do a survey of the equipment. This is especially important in determining whether there is a problem with harmonics in a single-phase system.

1. Make an equipment check. Equipment such as personal computers, printers, and fluorescent lights with electronic ballast are known to produce harmonics. Any piece of equipment that draws current in pulses can produce harmonics.

2. Review maintenance records to see whether there have been problems with circuit breakers tripping for no apparent reason.

3. Check transformers for overheating. If the cooling vents are unobstructed and the transformer is operating excessively hot, harmonics could be the problem. Check transformer currents with an ammeter capable of indicating a true RMS current value. Make sure that the voltage and current ratings of the transformer have not been exceeded.

It is necessary to use an ammeter that responds to true RMS current when making this check. Some ammeters respond to the average value, not the RMS value. Meters that respond to the true RMS value generally state this on the meter. Meters that respond to the average value are generally less expensive and do not state that they are RMS meters.

Meters that respond to the average value use a rectifier to convert the alternating current into direct current. This value must be increased by a factor of 1.111 to change the average reading into the RMS value for a sine-wave current. True RMS responding meters calculate the heating effect of the current. The chart in Figure 14–34 shows some of the differences between average indicating meters and true RMS meters. In a distorted waveform, the true RMS value of current will no longer be average × 1.111 (Figure 14–35). The distorted wave- form generally causes the average value to be as much as 50% less than the RMS value.

value of 36.8 A, and the average ammeter indicates a value of 24.8 A. Determine the ratio of the two measurements by dividing the average value by the true RMS value:

A ratio of 1 would indicate no harmonic distortion. A ratio of 0.5 would indicate extreme harmonic distortion. This method does not reveal the name or sequence of the harmonic distortion, but it does indicate whether there is a problem with harmonics.

The most accurate method for determining whether there is a harmonics problem is to use a harmonic analyzer. The harmonic analyzer will determine the name, sequence, and amount of harmonic distortion present in the system.

Determining Harmonic Problems on Three-Phase Systems

Determining whether a problem with harmonics exists in a three-phase system is similar to determining the problem in a single-phase system. Because harmonic problems in a three-phase system generally occur in a wye-connected, four-wire system, this example will assume a delta-connected primary and wye-connected secondary with a center-tapped neutral as shown in Figure 14–32. To test for harmonic distortion in a three- phase, four-wire system, measure all phase currents and the neutral current with both a true RMS indicating ammeter and an average indicating ammeter. It will be assumed that the three-phase system being tested is supplied by a 200-kVA transformer, and the current values shown in Figure 14–37 were recorded. The current values indicate that a problem with harmonics does exist in the system. Note the higher current measurements made with the true RMS indicating ammeter, and also the fact that the neutral current is higher than any phase current.

Dealing with Harmonic Problems

After it has been determined that harmonic problems exist, something must be done to deal with them. It is generally not practical to remove the equipment causing the harmonic distortion, so other methods must be employed. It is a good idea to consult a power quality expert to determine the exact nature and amount of harmonic distortion present. Some general procedures for dealing with harmonics follow:

1. In a three-phase, four-wire system, the 60-Hz part of the neutral current can be reduced by balancing the current on the phase conductors. If all phases have equal current flow, the neutral current would be zero.

2. If triplen harmonics are present on the neutral conductor, harmonic filters can be added at the load. These filters can help reduce the amount of harmonics on the line.

3. Pull extra neutral conductors. The ideal situation would be to use a separate neutral for each phase, instead of using a shared neutral.

4. Install a larger neutral conductor. If it is impractical to supply a separate neutral conductor for each phase, increase the size of the common neutral.

5. Derate or reduce the amount of load on the transformer. Harmonic problems generally involve overheating of the transformer. In many instances it is necessary to derate the transformer to a point that it can handle the extra current caused by the harmonic distortion. When this is done, it is generally necessary to add a second transformer and divide the load between the two.

Determining Transformer Harmonic Derating Factor

Probably the most practical and straightforward method for determining the derating factor for a transformer is recommended by the Computer & Business Equipment Manufacturers Association. To use this method, two ampere measurements must be made. One is the true RMS current of the phases and the second is the instantaneous peak phase current. The instantaneous peak current can be determined with an oscilloscope connected to a cur- rent probe or with an ammeter capable of measuring the peak value. Many of the digital clamp-on ammeters are capable of measuring the average, true RMS, and peak values of current. For this example, it will be assumed that peak current values are measured for the 200-kVA transformer discussed previously. These values are added to the previous data obtained with the true RMS and average indicating ammeters (Figure 14–38). The formula for determining the transformer harmonic derating factor (THDF) is

This formula will produce a derating factor somewhere between 0 and 1.0. Because the instantaneous peak value of current is equal to the RMS value × 1.414, if the current wave- forms are sinusoidal (no harmonic distortion), the formula will produce a derating factor of 1.0. Once the derating factor is determined, multiply the derating factor by the kVA capacity of the transformer. The product will be the maximum load that should be placed on the transformer.

SUMMARY

• Most electrical energy is generated by three-phase alternating-current generators.

1. Three-phase systems are used to transmit and distribute this electrical energy.

2. For economy, the generated three-phase voltage is stepped up to an extremely high voltage by three-phase transformers for long-distance transmission over the three-phase system.

3. Three-phase transformers are used at distribution points to step down the high voltage to a safe and usable level.

• A large three-phase transformer is completely enclosed in one container. Three single- phase transformers, connected in a bank, can be used to transform the voltages of three- phase systems at distribution points.

1. These transformers may be connected in any one of four standard ways:

a. Delta–delta

b. Wye–wye

c. Delta–wye

d. Wye–delta

2. Another method is the open-delta (or V) connection. This connection uses only two single-phase transformers to transform the voltages on a three-phase system.

• The delta–delta connection:

1. The delta–delta connection is used to supply an industrial load by stepping down a 2400-V, three-phase, three-wire service to a 240-V, three-phase, three-wire service.

2. For most applications, the load is balanced and the three single-phase transformers have the same kilovolt-ampere rating.

3. The primary and the secondary windings are connected in delta.

4. To determine the total capacity in kilovolt-amperes of a delta–delta-connected transformer bank, the three kVA ratings are added.

• For any delta-connected circuits and systems:

1. Each transformer winding is connected across the two line leads; thus, the line voltage and the transformer coil winding voltage are the same.

2. The line current is equal to M3 or 1.73 times the coil winding current.

3. In a closed-delta transformer connection, each line wire is fed by two trans- former coil currents that are out of phase. These coil currents do not add directly, but must be added vectorially to obtain the line current.

• Feeding a dual load using a delta–delta connection—some power companies use a delta–delta-connected transformer bank to feed two types of loads:

1. A 240-V, three-phase industrial load.

2. A 120/240-V, single-phase, three-wire lighting load.

a. The transformer supplying the lighting load is larger than the other two trans- formers of the three-phase system.

b. The neutral wire is tied to the midpoint of the 240-V, low-voltage secondary winding.

c. Many transformers have two 120-V windings. These windings are connected in series and the neutral is brought out at the midpoint to give a 120/240-V service.

3. Both lines A and C have 120 V to the grounded neutral. Line B has approximately 208 V to ground. This connection can be a serious hazard and cannot be used for lighting service.

• Procedure for making a closed-delta connection:

1. Two primary winding leads are brought out of the transformer and are marked H1 and H2. It is assumed that H1 is the beginning and H2 is the ending of each high-voltage winding. The H2 end of each primary winding is connected to the beginning (H1) of the next primary winding to form a series arrangement.

2. One three-phase primary source line is connected to each junction of two windings. With a 2400-V source, each of the primary windings has a line voltage of 2400 V impressed across it.

3. Once the high-voltage primary connections are made, the three-phase, 2400-V input may be energized and tested for the correct phase rotation. If the phase rotation is incorrect, the circuit is deenergized and any two line wires are interchanged.

4. To determine whether a transformer has additive or subtractive polarity, the terminal markings or the nameplate data can be checked.

5. The secondary closed-delta connection is the same as the primary connection, with X1 substituted for H1 and X2 substituted for H2. The primary side of the delta-connected transformer may be connected so as to cause a polarity reversal in one of its legs. A phase inversion on the primary side must be corrected on the secondary side.

a. Determine whether the voltage output of each of the three transformers is the same as the voltage rating on the nameplate.

b. Connect the end of one secondary winding to the beginning of another secondary winding. If the connection is correct, the voltage across the open ends of the two transformers should be the same as the output of each transformer. If the connections are not correct, the resultant voltage is M3 times the secondary rated voltage. The phase reversal is corrected by interchanging the connections to one of the coils.

c. With the correct voltage on the two coils, add the end of the third secondary winding coil to the beginning of the second coil. The resultant voltage is opposite to each secondary voltage and has a magnitude equal to the secondary voltage. If the voltage is zero across the last pair of open leads, they can be connected together. A line wire is attached at each of the connection points. These three line wires form the three-phase output. If the voltage reading is not zero, reverse the connections of the last transformer coil before the delta connection is closed and the output lines are added.

• The percentage load on a transformer bank connected in closed delta, in terms of its rated capacity, is

• The total output of a closed-delta connection is

• The wye–wye connection:

1. This connection can be used when the load on the secondary side is balanced. If the secondary load consists of three-phase motors only and the load currents are balanced, the wye–wye connection can be used.

2. This connection cannot be used if the secondary load becomes unbalanced. An unbalanced load causes a serious imbalance in the three voltages of the trans- former bank.

3. A fourth wire, known as the neutral wire, is added to eliminate unbalanced volt- ages. It is connected between the source and the common point on the primary side of the transformer bank.

a. Each high-voltage winding is connected between the neutral and one of the three line leads. If the source is 2400/4160 V, the voltage across each high-voltage winding is 2400 V. The voltage across the three line leads is M3 X 2400, or 4160 V.

b. The neutral wire maintains a constant voltage across each of the high-volt- age windings, even when the load is unbalanced.

c. The neutral wire is grounded and helps protect the three high-voltage windings from lightning surges.

4. With the secondary connected in wye, each low-voltage winding is connected between the secondary neutral and one of the three line leads.

a. For a four-wire, three-phase system rated at 2400/4160 V and 120/280 V, the voltage output of each secondary winding is 120 V. There are 120 V between the neutral wire and any one of the three secondary line leads.

b. The voltage across the line wires is M3 X 120, or 208 V.

(1) 208-V, three-phase service is available for industrial power loads such as three-phase motors.

(2) 120 V is available for lighting loads.

5. Nearly all wye–wye-connected transformers use three single-phase transformers having the same kVA capacity:

6. In a wye–wye connection, a defective transformer must be replaced before the bank can be reenergized.

• The delta–wye connection:

1. This connection is also used for transformation to step up or step down a voltage.

2. The primary windings are connected in delta and the secondary windings are connected in wye.

3. This connection is used at a generator station to step up the voltage. The input voltage is stepped up by the transformer ratio. This voltage is further increased by M3, or 1.73 times the secondary coil voltage. The high voltage is then connected directly to three-phase transmission lines.

4. The insulation requirements are reduced for the secondary windings because the coil voltage is only 58% of the line voltage. (This fact is very important when the secondary side has a very high voltage.)

5. This connection can also be used at the distribution point to step down the voltage.

6. The delta–wye connection can be used for a three-phase, four-wire system for power and lighting.

a. In a 208/120-V system, 208 V supplies the three-phase power load and 120 V supplies the lighting load.

b. In a 480/277-V system, 480 V supplies the three-phase power load and 277 V supplies the lighting load.

(1) Standard 120-V fluorescent lighting fixtures are used with special ballasts for operation on 277-V circuits.

(2) A 277/480-V, three-phase, four-wire system has the following advantages:

(a) The I2 R drop is reduced in feeders and branch circuits, resulting in an increase in the operating efficiency.

(b) Smaller sizes of copper conductors, conduits, and equipment can be used to save up to 25% of the installation cost.

(c) The load demands on the 277/480-V system can be increased with a minimum of changes and expense.

c. In a building using a 277/480-V, three-phase system, the additional require- ment of a 120-V service uses little power. A transformer is added to step down the voltage from 480 V to 120/240 V. A small air-cooled transformer can be used to supply the 120-V service and is centrally located.

• The wye–delta connection:

1. This connection is used to step down relatively high transmission line voltages at the load center. A transformer bank of this type is commonly used to step down voltages of 60,000 V or more.

2. Advantages of the delta–wye connection are

a. the three-phase voltage is reduced by the transformer ratio times 1.73.

b. the insulation requirements are reduced for the high-voltage windings because the wye primary coil voltage is only 58% of the primary line voltage.

• For the delta–wye and wye–delta connections, the three single-phase transformers generally have the same kVA capacity, and the total capacity of the transformer bank, in kVA, is obtained by adding the kVA ratings of the three transformers.

• The open-delta connection (V connection):

1. This connection uses two transformers only.

2. It can be used in emergency situations when one of three transformers is damaged in a three-phase service. The damaged transformer is cut out of the system, and the configuration of the open-delta connection is used.

3. With two transformers supplying the service, the total capacity is only 58% of the capacity of the closed-delta connection.

4. On the secondary side, the line voltage and the line current are the same as the secondary coil current and voltage.

5. The original transformer installation may consist of an open-delta bank. As the industrial power load requirements increase, a third transformer may be added. When the third transformer is added, a delta–delta bank (closed-delta bank) is formed.

6. The total kVA capacity for an open-delta bank is Total kVA = kVA1 + kVA2 X 0.866

• Advantages of an enclosed three-phase transformer:

1. The operating efficiency of a three-phase transformer is slightly higher than the overall efficiency of three separate single-phase transformers.

2. The three-phase transformer weighs less and requires less space than do three separate single-phase transformers.

3. The three-phase transformer supplies the same kVA output and costs less than three single-phase transformers.

4. The installation equipment required (such as bus bars, switchgear, and wiring) is easier to install and is less complex than that required by a transformer bank consisting of three single-phase transformers.

• If the three-phase transformer develops a problem, such as a defective phase winding, then the entire three-phase unit must be taken out of service.

• T-connected transformers are similar to an open-delta connection in that only two trans- formers are required to make the connection.

1. Before two transformers can be connected T, one transformer must have a 50% tap on both the primary and secondary windings. It is preferable for the second transformer to have an 86.6% tap.

2. T-connected transformers are generally wound specially for the purpose and are contained in the same case.

3. The phase balance of the T-connected transformer is better than that of an open- delta bank.

4. The T-connected transformer can be connected to supply three-phase, four-wire service in a manner similar to that for a four-wire wye connection.

• Scott-connected transformers are similar to T-connected ones in that the main trans- former must have a 50% tap on both the primary and secondary, and the teaser trans- former must have an 86.6% tap.

1. The Scott connection is used to produce two-phase power from a three-phase connection.

2. A two-phase system has voltages 90° out of phase with each other.

• The zigzag connection is generally used for grounding purposes.

• Harmonics can cause heating problems in ac circuits.

• Harmonics are generally caused by devices that cause pulsations on the line such as switching power supplies and variable-frequency drives.

• The third harmonic generally causes overload conditions on neutral conductors.

• It is sometimes necessary to derate transformers because of harmonic problems.

Achievement Review

1. An industrial plant has a 2400-V, three-phase, three-wire service. It uses three 100-kVA, single-phase transformers. Each transformer is rated at 2400/240 V. The transformers supply a 240-V, three-phase, three-wire system. Each trans- former has additive polarity.

a. Draw a connection diagram showing how the transformer bank is connected.

Mark the polarity of all transformer leads.

b. A balanced load of 200 kW at a lagging power factor of 0.70 is supplied by the transformer bank. Determine

1. the secondary line current.

2. the secondary coil current.

3. the primary coil current.

4. the primary line current.

2. What is the percentage of load in kVA on the transformer bank described in question 1?

3. Explain a procedure that may be used to connect the secondary windings of single-phase transformers in closed delta.

4. a. Assume that one of the transformers in the delta–delta transformer bank in question 1 is cut out of service because it is damaged by lightning. Assuming that plant operations are cut to a minimum so that the peak load does not exceed the capacity of the two remaining transformers, show how they could be connected during this emergency.

b. What is the maximum balanced load, in kilowatts, at a lagging power factor of 0.70, that may be connected to the transformer bank during the emergency described in part a of this question?

5. Explain a procedure that may be used to connect the secondary windings of three single-phase transformers in wye.

6. Three single-phase, 20-kVA, 2400/120-V transformers are connected in a three- phase transformer bank. The transformers have additive polarity. The bank steps down the 2400/4160-V, three-phase, four-wire primary service to a 120/208-V, three-phase, four-wire secondary service. This service supplies both a 120-V, single-phase lighting load and a 208-V, three-phase motor load. Draw a schematic diagram of the connections for this transformer bank.

7. a. What is the maximum balanced three-phase load, in kVA, that can be connected to the transformer bank in question 6?

b. If the load has a lagging power factor of 0.80, what is the maximum output in kW that can be obtained from the transformer bank?

8. Give two practical applications for a delta–wye transformer bank. Give one practical application for a wye–delta transformer bank.

9. A three-phase, four-wire, wye-connected transmission system has a voltage of approximately 60,550 V across the three line wires. In addition, there is 35,000 V from each line wire to the neutral. This voltage is to be stepped down at a substation to supply energy to a 5000-V, three-phase, three-wire distribution sys- tem. Each transformer is rated at 35,000/5000 V, 2000 kVA. Each transformer has subtractive polarity. Draw a schematic wiring diagram for this circuit and determine

a. the full-load capacity of the bank, in kVA.

b. the line current on the secondary side when the transformer bank is loaded to its rated capacity.

c. the line current on the primary side when the transformer bank is loaded to its rated capacity, neglecting any losses.

10. What are typical applications of a 277/480-V, three-phase, four-wire, wye- connected system? What are the advantages of such a system?

11. a. List several advantages of a three-phase transformer, as compared with a three-phase bank consisting of three single-phase transformers.

b. Give one disadvantage of the use of a three-phase transformer.

12. The high-voltage windings of a three-phase transformer are connected in wye. The low-voltage windings are connected in delta. The actual ratio between the high- voltage windings and their respective low-voltage windings is 10:1. The primary side of the transformer is supplied from a three-phase, four-wire, 2400/4160-V circuit. The secondary output is a three-phase, three-wire service for a three-phase industrial motor load. Determine

a. the secondary voltage.

b. the kVA output of the bank, when the current supplied to the motor load on the secondary is 60 A in each line wire.

c. the line current in each line wire on the high-voltage primary side of the trans- former, neglecting any losses.

13. What is the frequency of the second harmonic?

14. Of the following, identify those that are considered triplen harmonics: third, sixth, ninth, twelfth, fifteenth, and eighteenth.

15. Would a positive rotating harmonic or a negative rotating harmonic be more harmful to an induction motor? Explain your answer.

16. What instrument should be used to determine what harmonics are present in a power system?

17. A 22.5-kVA single-phase transformer is tested with a true RMS ammeter and an ammeter that indicates the peak value. The true RMS reading is 94 A. The peak reading is 204 A. Should this transformer be derated, and if so, by how much?

PRACTICE PROBLEMS FOR UNIT 14

In the following problems, three single-phase control transformers have been con- nected to operate with a primary voltage of 240 V and a secondary voltage of 120 V. This gives each transformer a turns ratio of 2:1. It will be assumed that a line-to-line voltage of 208 V is connected to the primary winding, and that the load connected to the secondary has an impedance of 4 !1 in each phase. Find the unknown values in each of the following problems. It will be necessary to use the formulas in the Transformers and Three-Phase Connections sections of Appendix 15.

[Note: When computing values of voltage and current between the primary and secondary windings, use phase values and not line values.] In the following problems: