Monday, January 12, 2015

Special Transformer Applications : 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.

image

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 PTratio X CTratio

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:

image• 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,image

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

image

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.

image

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.

image

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.