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FEEDING A DUAL LOAD

Some power companies use a delta–delta-connected transformer bank to feed two types of loads. These loads consist of a 240-V, three-phase industrial load and a 120/240-V, single-phase, three-wire lighting load.

One single-phase transformer supplies the single-phase, three-wire lighting load. This transformer usually is larger than the other two transformers. The 120/240-V, single-phase, three-wire service is obtained from this transformer by bringing out a tap from the mid- point of the 240-V, low-voltage secondary winding. Many transformers have two 120-V windings. As explained in Unit 13, these windings can be connected in series, with a tap brought out at the midpoint to give 120/240-V service.

Three single-phase transformers are connected as a delta–delta bank in Figure 14–5. Each transformer has two 120-V windings. When these windings are connected in series, each transformer has a total output of 240 V. The high-voltage primary windings are con- nected in closed delta. The low-voltage output windings or secondary windings are also connected in closed delta to give three-phase, 240-V service to the industrial power load. Because the middle transformer also feeds the single-phase, three-wire, 120/240-V light- ing load, a tap is made at the midpoint on the secondary output side of the transformer to give 120/240-V service. This tap feeds to the neutral wire of the single-phase, three-wire system and is grounded.

A check of the connections in Figure 14–5 shows that there is 120 V to ground on both lines A and C of the three-phase, 240-V secondary system. However, line B will have 208 V to ground. The condition can be a serious hazard and cannot be used for lighting service.

THE WYE CONNECTION

The wye connection is another standard method of connecting single-phase trans- formers to obtain three-phase voltage transformation. The wye connections must be made systematically to avoid errors. The student must be familiar with and able to use the basic voltage and current relationships for this type of connection. In Unit 10, the following information was given for the three-phase wye connection:

• The line current and the coil winding current are equal.

• The line voltage is equal to M3 times the coil winding voltage.

Figure 14–6 shows three single-phase transformers connected in wye–wye. The H2 leads of the high-voltage windings are considered to be the ends of each of the high-volt- age windings and are connected together. The H1 (beginning) lead of each transformer is

connected to one of the three line leads. When this connection is shown in a schematic diagram, it looks like the letter Y (which is written as wye). The connection may also be called a star connection.

In general, the low-voltage winding leads are marked and the polarity is shown on the transformer nameplate. However, the following procedure should be used to make the low- voltage secondary connections:

1. Energize the three-phase, wye-connected, high-voltage side of the transformer bank.

The voltage output of each of the three transformers must be the same as the name- plate rating.

2. Deenergize the primary. Connect the X2 ends of two low-voltage secondary windings, as shown in Figure 14–7A. With all three X1s open and clear, energize the primary. If the connections are made correctly, the voltage across the open ends is M3 times the secondary winding voltage. In this case, the voltage across the open ends is 208 V, as indicated in the vector diagram.

Secondaries Incorrectly Connected

The secondaries of the two transformers are shown connected incorrectly in Figure 14–7B. In this case, the voltage across the open ends is the same as the secondary voltage of each transformer. According to the vector diagram, the resultant voltage is only 120 V. The connections can be corrected by reversing the leads of transformer 2. As a result, the voltage across the open ends will be 208 V.

Secondaries Properly Connected

The correct wye connections are shown in Figure 14–8A for the secondary windings of the three transformers. The voltage across each pair of line leads is equal to M3 times the secondary coil voltage or 208 V. The vector diagram in Figure 14–8B shows the relationship between the coil voltages and the line voltages in a three-phase, wye-connected system.

The wye–wye connection can be used in those applications where the load on the secondary side is balanced. If the load consists of three-phase motor loads only, and the load currents are balanced, then the wye–wye connection can be used. This connection cannot be used if the secondary load becomes unbalanced. An unbalanced load causes a serious imbalance in the three voltages of the transformer bank.

The Neutral Wire

A fourth wire known as the neutral wire is added to eliminate unbalanced voltages. The neutral wire is connected between the source and the common point on the primary side of the transformer bank.

The diagram in Figure 14–9 shows a wye–wye-connected transformer bank having a three-phase, four-wire, 2400/4160-V input and a three-phase, four-wire, 120/208-V output. On the high-voltage input side, the neutral wire is connected to the common point. This is the point where all three high-voltage primary winding (H2 ) leads terminate. The voltage from the neutral to any one of the three line leads is 2400 V. Each high-voltage winding is connected between the neutral and one of the three line leads. This means that each high- voltage winding is connected across 2400 V. The voltage across the three line leads is M3 X 2400 V, or 4160 V. The neutral wire maintains a nearly constant voltage across each of the high-voltage windings, even when the load is unbalanced. The neutral conducts any unbalance of current between the source and the neutral point on the input side of the transformer bank. The neutral wire is grounded and helps protect the three high-voltage windings from lightning surges.

For the transformer bank shown in Figure 14–9, the three-phase, four-wire sys- tem feeds from the low-voltage side of the bank to the load. Each low-voltage wind- ing is connected between the secondary neutral and one of the three line leads. The voltage output of each secondary winding is 120 V. Thus, there is 120 V between the neutral and any one of the three secondary line leads. The voltage across the line wires is M3 X 120, or 208 V. The use of a three-phase, four-wire secondary provides two voltages that can be used for different load types:

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

2. The use of a neutral wire means that 120 V is available for lighting loads.

Neutral Wire Used on Primary and Secondary

Figure 14–10 shows the connections for a wye–wye-connected transformer bank. Note that both the primary and secondary sides contain a neutral wire. Each transformer has two 120-V, low-voltage windings connected in parallel. The voltage output for each single-phase transformer is 120 V. Because this is a three-phase, four-wire system, the

following voltages are available: (1) a three-phase, 208-V service for motor loads and (2) a 120-V, single-phase service for the lighting loads. The lighting load should be distributed evenly between the three transformers. Thus, an attempt is always made to balance the lighting circuits from the line wire to the neutral.

Nearly all wye–wye-connected transformer banks use three single-phase transformers having the same kilovolt-ampere capacity. The individual kVA ratings are added to find the capacity of the transformer bank. For example, if a bank consists of three transformers, each rated at 25 kVA, the total rating is 25 + 25 + 25 = 75 kVA.

In a wye–wye connection, a defective transformer must be replaced before the wye– wye transformer bank can be reenergized. Unlike a delta–delta bank, a wye–wye trans- former bank cannot be temporarily reconnected in an emergency using two single-phase transformers only.

Transformer Connections for Three-Phase Circuits

THE DELTA CONNECTION

A delta–delta connection is used when three single-phase transformers are used to step down as three-phase voltage of 2400 V to 240 V, three phase.

The connections for the three single-phase transformers are shown in Figure 14–1. The primary and secondary windings of the transformers are connected in delta. The high- voltage leads of the primary winding of each transformer are marked H1 and H2. The leads of the low-voltage secondary winding of each transformer are marked X1 and X2.

Primary Winding Connections

The high-voltage primary windings are connected in a closed-delta arrangement. H1 is assumed to be the beginning of each high-voltage winding and H2 is assumed to be the end. The end (H2 ) of each primary winding is connected to the beginning (H1) of the next primary winding to form a series arrangement. One three-phase line wire is connected to each junction of two windings. In other words, the primary winding of each transformer is connected directly across the line voltage. Each of the three line voltages is 2400 V and the primary winding of each transformer is correctly rated at 2400 V. Once the high-voltage primary connections are made, the three-phase 2400-V input may be energized and tested for the correct phase rotation.

Unit 13 described the high-voltage lead (H1 ) of any transformer as being on the left when the transformer is viewed from the low-voltage side. This means that the positions of H1 and H2 are fixed. Unit 13 also explained that transformers are standardized with regard to polarity. They have additive or subtractive polarity, according to their voltage ratings and capacities in kilovolt-amperes. However, there are many exceptions to these standards. To determine whether a transformer has additive or subtractive polarity, check the terminal markings or the nameplate data.

Phase Inversion

The primary side of the delta-connected transformer may be connected so as to reverse the polarity of one of its legs. Such a phase inversion of the primary side must be corrected in the secondary side. The following method can be used to correct the polarity:

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

2. Connect the end of one secondary winding to the beginning of another secondary winding, as shown in Figure 14–2A. 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. In this case, the voltage is 240 V. The resultant voltage in the vector dia- gram is the same as the two secondary winding voltages and is equal to 240 V. Note that the path from start to finish is in the direction of both voltage arrows and does not cause a reversal of the vectors. The connections of the transformer windings are shown reversed in Figure 14–2B. The path from start to finish is in the direction of the voltage arrow and opposes the other voltage arrow. The V1 vector is reversed, and the resultant voltage is equal to M3 times the secondary coil voltage. The resultant voltage is 416 V. This condition is corrected by changing the connections of the secondary leads of one of the transformers. Thus, the new connections will be the same as in Figure 14–2A.

3. The proper connections are shown in Figure 14–3A for one end of the secondary coil of the third transformer. The vector diagram shows that there is a resultant voltage opposite to each secondary voltage and having a magnitude equal to the secondary voltage. Therefore, the voltage at any corner of the delta is zero. 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 240-V, three-phase output. The three line voltages and each of the three transformer secondary voltages have the same value of 240 V. In Figure 14–3B the connections of the third secondary winding are reversed. As shown in the vector diagram, reversing the secondary voltage of the third transformer (V3) causes the resultant voltage to be twice the secondary voltage. In this case, the resultant voltage is 480 V. This condition is corrected by reversing the connections of the secondary leads of the third transformer. As a result, the voltage across the last pair of open leads is zero and the delta connection may be closed. (CAUTION: Never connect the last pair of open leads if there is a voltage difference across them. The potential difference must be zero, indicating that the winding connections are correct.)

Transformer Banks

When the primary and secondary windings of three transformers are all connected in delta, the resulting arrangement is called a delta–delta (D–D) connection. The delta connection of the primary windings is indicated by the first delta symbol. The second delta symbol indicates the connection of the secondary windings. When two or three single- phase transformers are used to transform voltages on a three-phase system, they are known as a transformer bank.

The connections shown in Figure 14–4 are for a bank of transformers connected in delta–delta. Note that it is not necessary to represent delta–delta connections as triangles.

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. Assume that the industrial load consists of three-phase motors. The current in each wire is nearly the same and is considered to be balanced. For most applications, the load is balanced and the three single-phase transformers have the same kilovolt-ampere capacity. To determine the total capacity in kilovolt-amperes of the delta–delta-connected transformer bank, the three ratings are added. For example, if each transformer is rated at 50 kVA, the total capacity is 150 kVA.

The current–voltage relationships for a delta-connected circuit were described in Unit 10. The following information applies to both delta-connected circuits and systems.

1. In any delta connection, each transformer winding is connected across two line leads.

This means that the line voltage and the transformer coil winding voltage are the same.

2. The line current in a delta-connected transformer bank is equal to 1.73 times the coil winding current. In a closed-delta transformer connection, each line wire is fed by two transformer coil currents that are out of phase. Because the two coil currents are out of phase, they do not add together directly, but must be added vectorially to obtain the line current.

PROBLEM 1

Statement of the Problem

Three single-phase transformers are connected in delta–delta, as shown in Figure 14–4. Each transformer is rated at 50 kVA, 2400/240 V. The current in each of the three secondary line leads is 300 A. The secondary line voltage is 240 V across all three phases. The three- phase load connected to the transformer has a lagging power factor of 0.75. Determine

1. the kilovolt-ampere load on the transformer bank.

2. the kilowatt load on the transformer bank.

3. the percentage load on the transformer bank, in terms of its rated kVA capacity.

4. the current in each secondary winding.

5. the voltage across each secondary winding.

6. the kVA load on each transformer.

7. the line current on the primary side (assuming that the losses are negligible).

8. the kVA input to the transformer bank, if the losses are neglected.

Solution

Transformer Connections for Three-Phase Circuits

THE DELTA CONNECTION

A delta–delta connection is used when three single-phase transformers are used to step down as three-phase voltage of 2400 V to 240 V, three phase.

The connections for the three single-phase transformers are shown in Figure 14–1. The primary and secondary windings of the transformers are connected in delta. The high- voltage leads of the primary winding of each transformer are marked H1 and H2. The leads of the low-voltage secondary winding of each transformer are marked X1 and X2.

Primary Winding Connections

The high-voltage primary windings are connected in a closed-delta arrangement. H1 is assumed to be the beginning of each high-voltage winding and H2 is assumed to be the end. The end (H2 ) of each primary winding is connected to the beginning (H1) of the next primary winding to form a series arrangement. One three-phase line wire is connected to each junction of two windings. In other words, the primary winding of each transformer is connected directly across the line voltage. Each of the three line voltages is 2400 V and the primary winding of each transformer is correctly rated at 2400 V. Once the high-voltage primary connections are made, the three-phase 2400-V input may be energized and tested for the correct phase rotation.

Unit 13 described the high-voltage lead (H1 ) of any transformer as being on the left when the transformer is viewed from the low-voltage side. This means that the positions of H1 and H2 are fixed. Unit 13 also explained that transformers are standardized with regard to polarity. They have additive or subtractive polarity, according to their voltage ratings and capacities in kilovolt-amperes. However, there are many exceptions to these standards. To determine whether a transformer has additive or subtractive polarity, check the terminal markings or the nameplate data.

Phase Inversion

The primary side of the delta-connected transformer may be connected so as to reverse the polarity of one of its legs. Such a phase inversion of the primary side must be corrected in the secondary side. The following method can be used to correct the polarity:

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

2. Connect the end of one secondary winding to the beginning of another secondary winding, as shown in Figure 14–2A. 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. In this case, the voltage is 240 V. The resultant voltage in the vector dia- gram is the same as the two secondary winding voltages and is equal to 240 V. Note that the path from start to finish is in the direction of both voltage arrows and does not cause a reversal of the vectors. The connections of the transformer windings are shown reversed in Figure 14–2B. The path from start to finish is in the direction of the voltage arrow and opposes the other voltage arrow. The V1 vector is reversed, and the resultant voltage is equal to M3 times the secondary coil voltage. The resultant voltage is 416 V. This condition is corrected by changing the connections of the secondary leads of one of the transformers. Thus, the new connections will be the same as in Figure 14–2A.

3. The proper connections are shown in Figure 14–3A for one end of the secondary coil of the third transformer. The vector diagram shows that there is a resultant voltage opposite to each secondary voltage and having a magnitude equal to the secondary voltage. Therefore, the voltage at any corner of the delta is zero. 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 240-V, three-phase output. The three line voltages and each of the three transformer secondary voltages have the same value of 240 V. In Figure 14–3B the connections of the third secondary winding are reversed. As shown in the vector diagram, reversing the secondary voltage of the third transformer (V3) causes the resultant voltage to be twice the secondary voltage. In this case, the resultant voltage is 480 V. This condition is corrected by reversing the connections of the secondary leads of the third transformer. As a result, the voltage across the last pair of open leads is zero and the delta connection may be closed. (CAUTION: Never connect the last pair of open leads if there is a voltage difference across them. The potential difference must be zero, indicating that the winding connections are correct.)

Transformer Banks

When the primary and secondary windings of three transformers are all connected in delta, the resulting arrangement is called a delta–delta (D–D) connection. The delta connection of the primary windings is indicated by the first delta symbol. The second delta symbol indicates the connection of the secondary windings. When two or three single- phase transformers are used to transform voltages on a three-phase system, they are known as a transformer bank.

The connections shown in Figure 14–4 are for a bank of transformers connected in delta–delta. Note that it is not necessary to represent delta–delta connections as triangles.

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. Assume that the industrial load consists of three-phase motors. The current in each wire is nearly the same and is considered to be balanced. For most applications, the load is balanced and the three single-phase transformers have the same kilovolt-ampere capacity. To determine the total capacity in kilovolt-amperes of the delta–delta-connected transformer bank, the three ratings are added. For example, if each transformer is rated at 50 kVA, the total capacity is 150 kVA.

The current–voltage relationships for a delta-connected circuit were described in Unit 10. The following information applies to both delta-connected circuits and systems.

1. In any delta connection, each transformer winding is connected across two line leads.

This means that the line voltage and the transformer coil winding voltage are the same.

2. The line current in a delta-connected transformer bank is equal to 1.73 times the coil winding current. In a closed-delta transformer connection, each line wire is fed by two transformer coil currents that are out of phase. Because the two coil currents are out of phase, they do not add together directly, but must be added vectorially to obtain the line current.

PROBLEM 1

Statement of the Problem

Three single-phase transformers are connected in delta–delta, as shown in Figure 14–4. Each transformer is rated at 50 kVA, 2400/240 V. The current in each of the three secondary line leads is 300 A. The secondary line voltage is 240 V across all three phases. The three- phase load connected to the transformer has a lagging power factor of 0.75. Determine

1. the kilovolt-ampere load on the transformer bank.

2. the kilowatt load on the transformer bank.

3. the percentage load on the transformer bank, in terms of its rated kVA capacity.

4. the current in each secondary winding.

5. the voltage across each secondary winding.

6. the kVA load on each transformer.

7. the line current on the primary side (assuming that the losses are negligible).

8. the kVA input to the transformer bank, if the losses are neglected.

Solution

TAP-CHANGING TRANSFORMERS

There is some voltage loss in high-voltage distribution circuits. This means that the impressed voltage on the primary winding of a stepdown transformer may be lower than the primary voltage indicated on the nameplate. As a result, the secondary terminal voltage is also lower than the nameplate value.

Because this condition is undesirable, provisions can be made for changing taps to obtain the voltage expected. In this way, minor adjustments can be made in the trans- former ratio to compensate for the voltage loss in the high-voltage primary line. A simple method of doing this is to bring out the tap points of the high-voltage winding to a terminal block inside the transformer. The tap connections can be changed on the terminal board to vary the number of turns of the primary winding.

Tap-Changing Switch

Figure 13–33 shows the connections for a typical tap-changing switch used with the high-voltage windings of a step down transformer. Each full coil winding on the high-volt- age side is rated at 2400 V. The voltage rating of the two coil windings in series is 4800 V. The transformer steps the voltage down to 120/240 V. The tap-changing switch contacts are shown in position 1. This means that both full coil windings are in series. If the switch is moved to position 2, section A is removed from the primary coil winding on the left, resulting in a decrease in the turns ratio between the primary and secondary windings. Thus, the secondary voltage increases. When the tap-changing switch is moved in sequence to

positions 3, 4, and 5, sections B, C, and D of the primary windings are removed in the same order. As each section of the primary winding is removed from the circuit, the turns ratio decreases and the secondary voltage increases.

Special tap switches, immersed in oil, are used on some large transformers. This type of switch permits the tap connections to be changed under load. In general, the tap- changing mechanism is placed in a compartment separated from the transformer tank. As a result, oil is confined near the arc-interrupting area of the tap-changing mechanism and there is no reduction in the high dielectric strength of the oil in the transformer case. A small motor drives the tap-changing mechanism. The motor is operated from a control panel.

FREQUENCY AND VOLTAGE

A transformer must be operated on an ac circuit at the frequency for which it is designed. If a lower frequency is used, the reactance of the primary winding will decrease. As a result, there will be a marked increase in the exciting current. This increase causes the flux density in the core to increase, with the result that the saturation of the core is above normal. This greater saturation causes a greater heat loss and the efficiency of the transformer is lowered. If the frequency is greater than the nameplate frequency value, the reactance will increase and the exciting current will decrease. There will be a lower flux density, but the core loss will remain nearly constant.

If the voltage is increased to a high enough value above the nameplate rating, excessive heating will take place at the windings. The flux density of the core will increase and the saturation of the core will be above normal. Transformers are designed so that they can be operated, without overheating, at a voltage that is 5% above the nameplate rating. If a transformer is operated at a voltage lower than the nameplate rating, the power output will be reduced in proportion to the reduction in voltage.

TRANSFORMER IMPEDANCE

Transformer impedance is determined by the physical construction of the trans- former. Factors such as the amount and type of core material, wire size used to con- struct the windings, the number of turns, and the degree of magnetic coupling between the windings greatly affect the transformer’s impedance. Impedance is expressed as a percent and is measured by connecting a short circuit across the low-voltage winding of the transformer and then connecting a variable voltage source to the high-voltage winding (Figure 13–34). The variable voltage is then increased until rated current flows in the low-voltage winding. The transformer impedance is determined by calculating the percentage of variable voltage as compared to the rated voltage of the high-voltage winding.

Example

Assume that the transformer shown in Figure 13–34 is a 2400/480-V, 15-kVa trans- former. To determine the impedance of the transformer, first compute the full-load current rating of the secondary winding:

Next, increase the source voltage connected to the high-voltage winding until a current of 31.25 A flows in the low voltage winding. For the purpose of this example, assume that voltage value to be 138 V. Finally, determine the percentage of applied voltage as compared to the rated voltage:

Discussion

Transformer impedance is a major factor in determining the amount of voltage drop that a transformer will exhibit between no-load and full-load conditions and in determining the amount of current flow in a short-circuit condition. Short-circuit current can be computed using the following formula:

TRANSFORMER NAMEPLATE DATA

The information listed on the nameplate for a typical transformer is shown in Table 13–1. The transformer capacity is rated in kilovolt-amperes, not in kilowatts. The kilo- volt-ampere capacity is used because the load connected to the secondary determines the power factor of the output circuit of the transformer. The kVA rating represents the full-load output of the transformer, rather than the input. The nameplate also lists the voltage ratings of the high- and low-voltage windings, the frequency, the phase, the percent impedance, the polarity, the maximum temperature rise, and the gallons of trans- former oil required.

SUMMARY

• A transformer transfers energy from one ac circuit to another. This transfer is made with a change in the voltage, but with no change in the frequency.

• A transformer is a stationary device.

1. In its simplest form, it consists of a laminated iron core. The input and output windings are wound on this common core.

2. A transformer does not require any moving parts to transfer energy. There are no friction or windage losses and other losses are slight. Maintenance and repair costs are relatively low because of the lack of rotating parts.

• At full load, the efficiency of a transformer is between 96% and 97%. For a transformer with a very high capacity, the efficiency may be as high as 99%.

• Exciting current:

1. The exciting current is present when the primary winding of a transformer is connected to an alternating voltage source.

2. This current exists even when there is no load connected to the secondary.

3. It sets up an alternating flux in the core. This flux links the turns of both windings and induces a voltage in both windings. The total induced voltage in each winding is directly proportional to the number of turns in that winding. The alternating voltage induced in the secondary has the same frequency, but is opposite in direction to the primary winding voltage. The induced voltage in the primary winding opposes the source voltage and limits the excitation current of the primary.

• The induced voltages in the primary and secondary windings are directly proportional to the number of turns:

• Leakage flux:

1. This is a flux that does not link all of the turns of all of the coil windings.

2. This flux has its magnetic path in air and not in the core.

3. Leakage flux from the primary voltage does not link the turns of the secondary winding.

4. Leakage flux from the secondary links the secondary windings, but not the primary winding.

5. The leakage flux uses part of the impressed primary voltage to cause a reactance voltage drop; thus, both the secondary flux linkages and the secondary induced voltage are reduced.

6. Leakage flux from the secondary voltage is proportional to the secondary cur- rent; thus, there is a reactance voltage drop in the secondary winding. (Both the primary and the secondary leakage fluxes reduce the secondary terminal voltage of the transformer as the load increases.)

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

• The primary current at no load is called the exciting current.

1. This current is usually 2% to 5% of the full-load current.

2. It supplies the alternating flux and the losses in the transformer core. These losses are known as core losses and consist of eddy current losses and hysteresis losses. The losses cause I2 R losses.

3. Exciting current consists of a large quadrature (magnetizing) component and a small in-phase component to supply the core losses.

4. The power factor for the exciting current is from 5% to 10% lagging.

• Voltages are induced in the metal core by the large quadrature (magnetizing) component of the exciting current.

1. Eddy currents result and circulate through the core.

a. Eddy currents can be reduced by assembling the core structure from lamination coated with insulating varnish.

b. If the laminations are not varnished, the oxide coating on each lamination still reduces the eddy current losses.

2. The core structure also has a hysteresis loss.

a. This loss is due to molecular friction or the opposition offered by the mol- ecules to changes in their direction.

b. To reduce the amount of power needed to overcome these losses, a special silicon steel is used in the core construction.

• Core losses can be measured by the core loss open-circuit test.

1. The losses of a transformer at no load are small.

2. In the open-circuit test, the voltmeter measures the primary voltage.

3. The voltmeter must be disconnected before a reading is taken with the wattmeter. This is done so that the wattmeter will not indicate the power taken by the voltmeter.

4. The wattmeter indicates the core loss in watts, and the ammeter indicates the exciting current.

5. The effective resistance of the windings of a transformer is approximately 10% more than the dc or ohmic resistance.

6. Generally, the ac or effective resistance of a transformer is obtained by measuring the dc resistance of the winding and multiplying it by 1.1.

• The efficiency of a transformer is a ratio of the output in watts to the input in watts.

• The preferred method of determining the efficiency of a transformer is to measure the losses and add this value to the nameplate output to obtain the input.

• The impedance voltage is that voltage required to cause the rated current to flow through the impedance of the primary and secondary windings.

• The ratio of the impedance voltage to the rated terminal voltage yields the percentage impedance voltage, which is in the range of 3% to 5%.

• The voltage regulation of a transformer is the percentage change in the secondary terminal voltage from the full-load condition to the no-load condition, whereas the impressed primary voltage is held constant.

• The polarity of the transformer lead-in wires is marked according to a standard developed by the American Standards Association (ASA).

1. The high-voltage winding leads are marked H₁ and H₂ . H₁ is on the left-hand side when the transformer is faced from the low-voltage side.

2. The low-voltage winding leads are marked X₁ and X₂ .

3. When H is instantaneously positive, X₁ is also instantaneously positive.

4.In a transformer with subtractive polarity, the H₁ and X₁ leads are adjacent or directly across from each other.

5. In a transformer with additive polarity, the H and X leads are diagonally across from each other.

• If paralleled transformers are to deliver an output that is proportional to the kVA ratings

1. the percentage impedance values of each transformer must be the same.

2. the secondary terminal voltages must be the same.

3. the instantaneous polarity of the secondary leads of each transformer must be correct.

• If it is impossible to identify the transformer leads because the identifying tags are missing or disfigured, a standard test procedure can be used to determine the polarity markings.

1. In this test, a jumper lead is temporarily connected between the high-voltage lead and the low-voltage lead directly across from it.

2. A voltmeter is connected from the other high-voltage lead to the second low- voltage lead.

3. If the voltmeter reads the sum of the primary input voltage and the secondary volt- age, the transformer has additive polarity.

4. If the voltmeter reads the primary input voltage minus the secondary voltage, the transformer has subtractive polarity.

5. There is a hazard involved in making the test at high-voltage values. Thus, a test using a relatively low voltage was developed to determine transformer polarity.

• NEMA and ASA standards for transformers:

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

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

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

• When stepdown transformers are operated in parallel, it may be necessary to remove one from service for repairs.

1. The low-voltage side of the transformer is always disconnected from the line wires before the primary fuses are opened.

2. A sign reading “DANGER—FEEDBACK” must be placed at each primary fuse to minimize the hazard of feedback voltage from one secondary winding to the other primary winding.

• Distribution transformer:

1. A distribution transformer has two high-voltage windings, rated at 2400 V each.

2. Small metal links are used to connect the two high-voltage windings either in series, for 4800-V primary service, or in parallel, for a 2400-V input. (Note that there are only two external high-voltage leads, which are marked with the standard designations H and H .)

3. The connections are made from the two windings at a terminal block slightly below the level of the insulating oil in the transformer case.

4. There are two secondary windings. These windings may be connected in parallel for 120-V service, or in series for 240-V service.

5. To obtain 120/240-V service, the two 120-V secondary windings are connected in series and a grounded neutral wire is connected between leads X₂ and X₃ .

• Advantages of a single-phase, three-wire service:

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

2. There are 240 V across the outside wires. Thus, the current for a given kilowatt load can be reduced by nearly half if the load is balanced between the neutral and the two outside wires.

a. Because of the reduction in the current, the voltage drop in the circuit conductors is reduced. Thus, the size of the conductor may be reduced and the voltage at the load is more nearly constant.

b. In addition, the following problems are minimized: dim lights, slow heating, and unsatisfactory appliance performance.

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

• For the ac noninductive circuit, the current in the grounded neutral wire is the difference between the currents in the two outside legs.

1. For a balanced load on a three-wire, single-load service, the grounded neutral wire carries no current.

2. In a single-phase, three-wire circuit, the neutral wire is always grounded. The neutral wire is never fused or broken at any switch control point in the circuit.

3. With a direct path to ground by way of the neutral wire, any instantaneous high-voltage surge, such as that due to lightning, is instantly discharged to ground. In this way, the electrical equipment and the circuit wiring are protected from major lightning damage.

• Danger of an open neutral:

1. On a three-wire, single-phase system, the neutral wire helps maintain balanced voltages between each line wire and the neutral, even with unbalanced loads.

2. If the neutral wire is opened accidentally, there is an open circuit in the return path for the current from each device connected to the two outside wires back to the transformer.

3. Because of the open circuit, the devices are connected in series across the outside wires, or 240 V.

4. Some devices, such as lamps and motors, may have too great a voltage drop and will burn out. Other devices may receive a reduced voltage that does not permit proper operation of the load.

• A characteristic curve can be plotted showing the percent efficiency of a transformer.

1. The load–voltage characteristic curve usually shows that there is little change in the terminal voltage from a full-load condition to no-load condition on a trans- former whose efficiency is very high.

2. With a lagging power factor load, the secondary terminal voltage change is slightly greater than it is for a unity power factor load.

3. With a leading power factor load, the terminal voltage increases with the load, giving a negative voltage regulation.

• Cooling of transformers

1. The core-and-coil assembly is placed in a pressed steel tank and is completely covered with an insulating oil.

2. The insulating oil removes heat from the core and the coil windings. It also insu- lates the windings from the core and the transformer case.

3. The core-and-coil assembly contains channels or ducts to permit the circulation of the oil to remove the heat.

4. As the oil gains heat, its density decreases and it rises. As it rises, it contacts the tank walls and transfers the heat to the tank. The walls are cooled by air circula- tion. As the oil loses heat, its specific gravity increases and it returns to the bottom of the tank. This action is repeated over and over.

5. To obtain better cooling, the surface area of a transformer can be increased by adding tubes or fins to the steel tank assembly.

6. The insulating oil used in transformers is a high-grade oil that must be kept clean and moisture free.

7. Pyranol (manufactured by the General Electric Company) is a synthetic dielectric that is an effective, nonflammable, nonexplosive liquid cooling and insulating agent.

8. Other methods of cooling transformers include forced air circulation, natural air circulation, natural circulation of oil with water cooling, and forced oil circulation (for large transformers).

• Some voltage is lost in high-voltage distribution circuits.

1. The impressed voltage on the primary winding may be lower than indicated on the nameplate of the transformer.

2. As a result, the secondary terminal voltage is also lower than the nameplate value.

This condition is undesirable, and provisions can be made for changing taps to obtain the desired voltage.

3. The turns ratio of the transformer is adjusted by changing the taps.

a. The change in the tap connections can be made at the terminal board inside the transformer.

b. A tap-changing switch can be used to vary the number of turns of the primary winding, resulting in a change in the turns ratio.

c. A special tap switch, immersed in oil, may be used on some large transformers. This type of switch permits the tap connections to be changed under load. A small motor drives the tap-changing mechanism and is operated from a control panel.

• A transformer must be operated on an ac circuit at the frequency for which it is designed.

1. If a lower frequency is used, the reactance of the primary will decrease. The exciting current will increase, causing an increase in the flux density. The saturation of the core is above normal and there is a greater heat loss. Thus, the efficiency of the transformer is decreased.

2. If the frequency is greater than the nameplate value, the reactance will increase and the exciting current will decrease. There will be a lower flux density, but the core loss will remain nearly constant.

• A transformer must be operated at its rated primary voltage, as indicated on the nameplate.

1. If the voltage is increased to a high enough value above the nameplate rating, excessive heating will take place at the windings. The flux density of the core will increase, and saturation of the core will be above normal. Transformers are designed to operate without overheating at a voltage that is 5% above the rated voltage.

2. If the transformer is operated at a voltage lower than the nameplate rating, the power output will be reduced in proportion to the voltage decrease.

• Inrush current for a transformer can be very high compared to a choke.

1. The magnetic domains in the core material of a transformer “remember” where they were set when the power is turned off.

2. Three factors that determine the amount of inrush current for a transformer are the amount of applied voltage, the resistance of the wire in the primary winding, and the flux change of the magnetic field in the core.

3. Magnetic domains or molecules can be made to reset to a neutral position by inserting an airgap in the core material.

• Transformer impedance is determined by the physical construction of the transformer.

1. Transformer impedance is expressed as a percent.

2. Transformer impedance is used to calculate the short-circuit current of a transformer.

Achievement Review

1. Define the term static transformer.

2. Explain what is meant by the terms primary and secondary, as applied to transformers.

3.	a. b.	What is a stepdown transformer? What is a stepup transformer?
4.	a. b.	What is the relationship between the induced voltages on the primary and secondary windings and the number of turns of the two windings? What is the difference between the impressed voltage and the induced voltage in the primary winding?

5. The primary winding of a transformer is rated at 115 V and the secondary winding is rated at 300 V. The primary winding has 500 turns. How many turns does the secondary winding have?

6. The transformer in question 5 has a full-load secondary output of 300 VA at 300 V.

a. What is the full-load secondary current?

b. Determine the full-load primary current (neglect all losses).

c. What is the relationship between the primary current and the secondary current and the number of turns on the two windings?

7. a. What is an exciting current?

b. What two functions does it perform?

8. a. Explain what is meant by core losses.

b. How can the core losses be kept to a minimum?

c. Draw a circuit diagram to show how the core loss, in watts, is measured for a 5-kVA transformer rated at 2400 to 240 V. The low-voltage side is used as the primary winding.

d. What safety precautions should be observed in measuring the core loss on the low-voltage side?

e. Using accurate instruments, should the core loss be the same for the high- voltage winding as for the primary?

9. The core loss of the transformer in question 8 is measured using the low-voltage side as the primary. The following data are obtained:

Voltmeter reading = 240 V

Exciting current = 1.0 A Wattmeter reading = 24 W

Determine

a. the core loss, in watts.

b. the power factor and the phase angle.

c. the in-phase component of the exciting current required for the core losses.

d. the lagging quadrature (magnetizing) component of the current.

10. Explain why the secondary ampere-turns oppose the magnetizing flux of the pri- mary winding.

11. List, in sequence, the reactions that take place in a transformer that result in an increase in the primary current input with an increase in the secondary current output to the load.

12. a. What is the primary leakage flux?

b. What is the secondary leakage flux?

c. What effect does leakage flux have on the operation of a transformer?

d. How can the leakage flux of a transformer be kept to a minimum?

13. Using diagrams, describe the three types of core construction used in transformers.

14. a. List the losses that occur in a transformer.

b. Which losses remain nearly constant at all load points? Why?

c. Which losses change when there is a change in the load? Why?

15. A 5-kVA, 240/120-V, 60-Hz transformer has a core loss of 32 W. The transformer has an effective resistance of 0.05 W in the high-voltage winding and 0.0125 W in the low-voltage winding. Determine the efficiency of the transformer at the rated load and unity power factor.

16. A 25-kVA, 2400/240-V, 60-Hz stepdown transformer has a core loss of 120 W. The effective resistance of the primary winding is 1.9 W. The effective resistance of the secondary winding is 0.02 W. Determine

a. the full-load current rating of the high-voltage and low-voltage windings. (Neglect any losses and assume that the input and the output are the same.)

b. the total copper losses of the transformer at the rated load.

c. the efficiency of the transformer at the rated load and a unity power factor.

17. Determine the efficiency of the transformer in question 15 at 50% of the rated load output and 0.l75 lagging power factor.

18. A 50-kVA, 4600/230-V, 60-Hz, single-phase, stepdown transformer is designed so that at the condition of full load and a unity power factor, the core losses equal the copper losses. The copper losses are equally divided between the two wind- ings. At full load the efficiency is 96.5%. Determine

a. the core losses.

b. the total copper losses.

c. the rated current of the primary and secondary windings. (Neglect the losses and assume that the input and the output are the same.)

d. the effective resistance of the primary and secondary windings.

19. Explain why it is hard to arrive at an accurate value of the efficiency of a trans- former using measurements of the power input and output.

20. a. Draw a diagram of the connections for the impedance short-circuit test for a single-phase transformer, rated at 20 kVA, 60 Hz, 4800/240 V. The high- voltage winding is the input side and is supplied from a 240-V ac source with a variable resistor in series.

b. Why should the measurements for the short-circuit test (part a) be made on the high-voltage side of the transformer?

21. The following data were obtained during the short-circuit test for the 20-kVA, 4800/240-V transformer in question 20:

Impedance voltage = 160 V

Ammeter reading = 4.2 A Wattmeter reading = 280 W

Another test shows the core loss to be 120 W. Determine

a. the efficiency at the rated load and a unity power factor.

b. the efficiency at 50% of the rated load and 0.80 lagging power factor.

22. a How are transformer leads marked?

b. Explain what is meant by additive polarity and subtractive polarity.

23. A transformer is used to step down a primary voltage of 4800 V to 240 V to sup- ply a single-phase load. Additional equipment is installed, making the kVA load greater than the full-load rating of the transformer. A second transformer, having the same kVA capacity, is to be operated in parallel with the first transformer.

a. List three requirements that are to be observed if the transformers are to share the load properly.

b. Explain the procedure used in connecting the second transformer in parallel.

24. a. There is a question about the accuracy of the turns ratio of a transformer rated at 2400/240 V. A 240-V test source is available in the laboratory. Assuming that voltmeters are available with the proper scales, explain how the voltage ratio can be determined. Draw a schematic diagram to clarify the answer.

b. Explain how the polarity of this transformer may be obtained using the 240-V test source.

25. A 30-kVA, 2400/240-V, 60-Hz transformer is used as a stepdown transformer.

The core loss is 150 W, the rated primary current is 12.5 A, the resistance of the primary winding is 1.5 W, the rated secondary current is 125 A, and the resistance of the secondary winding is 0.015 W. At the rated kVA capacity and a 0.75 lagging power factor, determine

a. the total copper losses.

b. the efficiency of the transformer.

26. A 20-kVA standard distribution transformer is rated at 2400/4800 V on the high- voltage side, and 120/240 V on the low-voltage side. The transformer has an efficiency of 97% when it is operating at the rated load as a stepdown transformer supplying a noninductive lighting load.

a. Draw a diagram to show the internal and external transformer connections required to step down the voltage from 2400 V to a 120/240-V, single- phase, three-wire service. Show the polarity markings for the transformer terminals for additive polarity.

b. What is the full-load secondary current if the load is balanced?

c. What is the power input to the primary side at the rated load?

d. If the no-load voltage of the 240-V secondary is 4 V more than the full- load voltage, what is the percentage voltage regulation?

27. Explain why the single-phase, 120/240-V, three-wire system is preferred to the single-phase, 120-V, two-wire system.

28.	a. b.	Why is oil used in transformers? What advantage is there in using Pyranol, instead of oil, in a transformer?
29.	a. b.	Explain the process of cooling a transformer by the natural circulation of oil and air. How can the heat-dissipating surface be increased for transformers with very large kVA ratings?

30. a. What is the purpose of a tap-changing switch as used on a transformer?

b. What are some of the applications of a transformer with tap-changing facilities?

31. Explain why a transformer has a high efficiency from 10% of the rated load to full load.

32. List the data commonly found on a transformer nameplate.

33. A 50-kVA single-phase transformer has a secondary voltage of 240 V. The name- plate indicates that the transformer has an impedance of 2.35%. What is the short- circuit current for this transformer?

34. A three-phase transformer is rated at 75 kVA and has a secondary voltage of 208

V. The nameplate indicates a transformer impedance of 3.5%. What is the short- circuit current for this transformer?

PRACTICE PROBLEMS FOR UNIT 13

Find the missing values in the following problems. Refer to Figure 13–35 and the formulas under the Transformers section of Appendix 15.

Find the missing values in the following problems. Refer to Figure 13–36 and the formulas listed under the Transformers section of Appendix 15.

[Note: Total primary current is equal to the sum of the primary currents needed to supply power to each secondary.]

TAP-CHANGING TRANSFORMERS

There is some voltage loss in high-voltage distribution circuits. This means that the impressed voltage on the primary winding of a stepdown transformer may be lower than the primary voltage indicated on the nameplate. As a result, the secondary terminal voltage is also lower than the nameplate value.

Because this condition is undesirable, provisions can be made for changing taps to obtain the voltage expected. In this way, minor adjustments can be made in the trans- former ratio to compensate for the voltage loss in the high-voltage primary line. A simple method of doing this is to bring out the tap points of the high-voltage winding to a terminal block inside the transformer. The tap connections can be changed on the terminal board to vary the number of turns of the primary winding.

Tap-Changing Switch

Figure 13–33 shows the connections for a typical tap-changing switch used with the high-voltage windings of a step down transformer. Each full coil winding on the high-volt- age side is rated at 2400 V. The voltage rating of the two coil windings in series is 4800 V. The transformer steps the voltage down to 120/240 V. The tap-changing switch contacts are shown in position 1. This means that both full coil windings are in series. If the switch is moved to position 2, section A is removed from the primary coil winding on the left, resulting in a decrease in the turns ratio between the primary and secondary windings. Thus, the secondary voltage increases. When the tap-changing switch is moved in sequence to

positions 3, 4, and 5, sections B, C, and D of the primary windings are removed in the same order. As each section of the primary winding is removed from the circuit, the turns ratio decreases and the secondary voltage increases.

Special tap switches, immersed in oil, are used on some large transformers. This type of switch permits the tap connections to be changed under load. In general, the tap- changing mechanism is placed in a compartment separated from the transformer tank. As a result, oil is confined near the arc-interrupting area of the tap-changing mechanism and there is no reduction in the high dielectric strength of the oil in the transformer case. A small motor drives the tap-changing mechanism. The motor is operated from a control panel.

FREQUENCY AND VOLTAGE

A transformer must be operated on an ac circuit at the frequency for which it is designed. If a lower frequency is used, the reactance of the primary winding will decrease. As a result, there will be a marked increase in the exciting current. This increase causes the flux density in the core to increase, with the result that the saturation of the core is above normal. This greater saturation causes a greater heat loss and the efficiency of the transformer is lowered. If the frequency is greater than the nameplate frequency value, the reactance will increase and the exciting current will decrease. There will be a lower flux density, but the core loss will remain nearly constant.

If the voltage is increased to a high enough value above the nameplate rating, excessive heating will take place at the windings. The flux density of the core will increase and the saturation of the core will be above normal. Transformers are designed so that they can be operated, without overheating, at a voltage that is 5% above the nameplate rating. If a transformer is operated at a voltage lower than the nameplate rating, the power output will be reduced in proportion to the reduction in voltage.

TRANSFORMER IMPEDANCE

Transformer impedance is determined by the physical construction of the trans- former. Factors such as the amount and type of core material, wire size used to con- struct the windings, the number of turns, and the degree of magnetic coupling between the windings greatly affect the transformer’s impedance. Impedance is expressed as a percent and is measured by connecting a short circuit across the low-voltage winding of the transformer and then connecting a variable voltage source to the high-voltage winding (Figure 13–34). The variable voltage is then increased until rated current flows in the low-voltage winding. The transformer impedance is determined by calculating the percentage of variable voltage as compared to the rated voltage of the high-voltage winding.

Example

Assume that the transformer shown in Figure 13–34 is a 2400/480-V, 15-kVa trans- former. To determine the impedance of the transformer, first compute the full-load current rating of the secondary winding:

Next, increase the source voltage connected to the high-voltage winding until a current of 31.25 A flows in the low voltage winding. For the purpose of this example, assume that voltage value to be 138 V. Finally, determine the percentage of applied voltage as compared to the rated voltage:

Discussion

Transformer impedance is a major factor in determining the amount of voltage drop that a transformer will exhibit between no-load and full-load conditions and in determining the amount of current flow in a short-circuit condition. Short-circuit current can be computed using the following formula:

TRANSFORMER NAMEPLATE DATA

The information listed on the nameplate for a typical transformer is shown in Table 13–1. The transformer capacity is rated in kilovolt-amperes, not in kilowatts. The kilo- volt-ampere capacity is used because the load connected to the secondary determines the power factor of the output circuit of the transformer. The kVA rating represents the full-load output of the transformer, rather than the input. The nameplate also lists the voltage ratings of the high- and low-voltage windings, the frequency, the phase, the percent impedance, the polarity, the maximum temperature rise, and the gallons of trans- former oil required.

SUMMARY

• A transformer transfers energy from one ac circuit to another. This transfer is made with a change in the voltage, but with no change in the frequency.

• A transformer is a stationary device.

1. In its simplest form, it consists of a laminated iron core. The input and output windings are wound on this common core.

2. A transformer does not require any moving parts to transfer energy. There are no friction or windage losses and other losses are slight. Maintenance and repair costs are relatively low because of the lack of rotating parts.

• At full load, the efficiency of a transformer is between 96% and 97%. For a transformer with a very high capacity, the efficiency may be as high as 99%.

• Exciting current:

1. The exciting current is present when the primary winding of a transformer is connected to an alternating voltage source.

2. This current exists even when there is no load connected to the secondary.

3. It sets up an alternating flux in the core. This flux links the turns of both windings and induces a voltage in both windings. The total induced voltage in each winding is directly proportional to the number of turns in that winding. The alternating voltage induced in the secondary has the same frequency, but is opposite in direction to the primary winding voltage. The induced voltage in the primary winding opposes the source voltage and limits the excitation current of the primary.

• The induced voltages in the primary and secondary windings are directly proportional to the number of turns:

• Leakage flux:

1. This is a flux that does not link all of the turns of all of the coil windings.

2. This flux has its magnetic path in air and not in the core.

3. Leakage flux from the primary voltage does not link the turns of the secondary winding.

4. Leakage flux from the secondary links the secondary windings, but not the primary winding.

5. The leakage flux uses part of the impressed primary voltage to cause a reactance voltage drop; thus, both the secondary flux linkages and the secondary induced voltage are reduced.

6. Leakage flux from the secondary voltage is proportional to the secondary cur- rent; thus, there is a reactance voltage drop in the secondary winding. (Both the primary and the secondary leakage fluxes reduce the secondary terminal voltage of the transformer as the load increases.)

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

• The primary current at no load is called the exciting current.

1. This current is usually 2% to 5% of the full-load current.

2. It supplies the alternating flux and the losses in the transformer core. These losses are known as core losses and consist of eddy current losses and hysteresis losses. The losses cause I2 R losses.

3. Exciting current consists of a large quadrature (magnetizing) component and a small in-phase component to supply the core losses.

4. The power factor for the exciting current is from 5% to 10% lagging.

• Voltages are induced in the metal core by the large quadrature (magnetizing) component of the exciting current.

1. Eddy currents result and circulate through the core.

a. Eddy currents can be reduced by assembling the core structure from lamination coated with insulating varnish.

b. If the laminations are not varnished, the oxide coating on each lamination still reduces the eddy current losses.

2. The core structure also has a hysteresis loss.

a. This loss is due to molecular friction or the opposition offered by the mol- ecules to changes in their direction.

b. To reduce the amount of power needed to overcome these losses, a special silicon steel is used in the core construction.

• Core losses can be measured by the core loss open-circuit test.

1. The losses of a transformer at no load are small.

2. In the open-circuit test, the voltmeter measures the primary voltage.

3. The voltmeter must be disconnected before a reading is taken with the wattmeter. This is done so that the wattmeter will not indicate the power taken by the voltmeter.

4. The wattmeter indicates the core loss in watts, and the ammeter indicates the exciting current.

5. The effective resistance of the windings of a transformer is approximately 10% more than the dc or ohmic resistance.

6. Generally, the ac or effective resistance of a transformer is obtained by measuring the dc resistance of the winding and multiplying it by 1.1.

• The efficiency of a transformer is a ratio of the output in watts to the input in watts.

• The preferred method of determining the efficiency of a transformer is to measure the losses and add this value to the nameplate output to obtain the input.

• The impedance voltage is that voltage required to cause the rated current to flow through the impedance of the primary and secondary windings.

• The ratio of the impedance voltage to the rated terminal voltage yields the percentage impedance voltage, which is in the range of 3% to 5%.

• The voltage regulation of a transformer is the percentage change in the secondary terminal voltage from the full-load condition to the no-load condition, whereas the impressed primary voltage is held constant.

• The polarity of the transformer lead-in wires is marked according to a standard developed by the American Standards Association (ASA).

1. The high-voltage winding leads are marked H₁ and H₂ . H₁ is on the left-hand side when the transformer is faced from the low-voltage side.

2. The low-voltage winding leads are marked X₁ and X₂ .

3. When H is instantaneously positive, X₁ is also instantaneously positive.

4.In a transformer with subtractive polarity, the H₁ and X₁ leads are adjacent or directly across from each other.

5. In a transformer with additive polarity, the H and X leads are diagonally across from each other.

• If paralleled transformers are to deliver an output that is proportional to the kVA ratings

1. the percentage impedance values of each transformer must be the same.

2. the secondary terminal voltages must be the same.

3. the instantaneous polarity of the secondary leads of each transformer must be correct.

• If it is impossible to identify the transformer leads because the identifying tags are missing or disfigured, a standard test procedure can be used to determine the polarity markings.

1. In this test, a jumper lead is temporarily connected between the high-voltage lead and the low-voltage lead directly across from it.

2. A voltmeter is connected from the other high-voltage lead to the second low- voltage lead.

3. If the voltmeter reads the sum of the primary input voltage and the secondary volt- age, the transformer has additive polarity.

4. If the voltmeter reads the primary input voltage minus the secondary voltage, the transformer has subtractive polarity.

5. There is a hazard involved in making the test at high-voltage values. Thus, a test using a relatively low voltage was developed to determine transformer polarity.

• NEMA and ASA standards for transformers:

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

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

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

• When stepdown transformers are operated in parallel, it may be necessary to remove one from service for repairs.

1. The low-voltage side of the transformer is always disconnected from the line wires before the primary fuses are opened.

2. A sign reading “DANGER—FEEDBACK” must be placed at each primary fuse to minimize the hazard of feedback voltage from one secondary winding to the other primary winding.

• Distribution transformer:

1. A distribution transformer has two high-voltage windings, rated at 2400 V each.

2. Small metal links are used to connect the two high-voltage windings either in series, for 4800-V primary service, or in parallel, for a 2400-V input. (Note that there are only two external high-voltage leads, which are marked with the standard designations H and H .)

3. The connections are made from the two windings at a terminal block slightly below the level of the insulating oil in the transformer case.

4. There are two secondary windings. These windings may be connected in parallel for 120-V service, or in series for 240-V service.

5. To obtain 120/240-V service, the two 120-V secondary windings are connected in series and a grounded neutral wire is connected between leads X₂ and X₃ .

• Advantages of a single-phase, three-wire service:

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

2. There are 240 V across the outside wires. Thus, the current for a given kilowatt load can be reduced by nearly half if the load is balanced between the neutral and the two outside wires.

a. Because of the reduction in the current, the voltage drop in the circuit conductors is reduced. Thus, the size of the conductor may be reduced and the voltage at the load is more nearly constant.

b. In addition, the following problems are minimized: dim lights, slow heating, and unsatisfactory appliance performance.

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

• For the ac noninductive circuit, the current in the grounded neutral wire is the difference between the currents in the two outside legs.

1. For a balanced load on a three-wire, single-load service, the grounded neutral wire carries no current.

2. In a single-phase, three-wire circuit, the neutral wire is always grounded. The neutral wire is never fused or broken at any switch control point in the circuit.

3. With a direct path to ground by way of the neutral wire, any instantaneous high-voltage surge, such as that due to lightning, is instantly discharged to ground. In this way, the electrical equipment and the circuit wiring are protected from major lightning damage.

• Danger of an open neutral:

1. On a three-wire, single-phase system, the neutral wire helps maintain balanced voltages between each line wire and the neutral, even with unbalanced loads.

2. If the neutral wire is opened accidentally, there is an open circuit in the return path for the current from each device connected to the two outside wires back to the transformer.

3. Because of the open circuit, the devices are connected in series across the outside wires, or 240 V.

4. Some devices, such as lamps and motors, may have too great a voltage drop and will burn out. Other devices may receive a reduced voltage that does not permit proper operation of the load.

• A characteristic curve can be plotted showing the percent efficiency of a transformer.

1. The load–voltage characteristic curve usually shows that there is little change in the terminal voltage from a full-load condition to no-load condition on a trans- former whose efficiency is very high.

2. With a lagging power factor load, the secondary terminal voltage change is slightly greater than it is for a unity power factor load.

3. With a leading power factor load, the terminal voltage increases with the load, giving a negative voltage regulation.

• Cooling of transformers

1. The core-and-coil assembly is placed in a pressed steel tank and is completely covered with an insulating oil.

2. The insulating oil removes heat from the core and the coil windings. It also insu- lates the windings from the core and the transformer case.

3. The core-and-coil assembly contains channels or ducts to permit the circulation of the oil to remove the heat.

4. As the oil gains heat, its density decreases and it rises. As it rises, it contacts the tank walls and transfers the heat to the tank. The walls are cooled by air circula- tion. As the oil loses heat, its specific gravity increases and it returns to the bottom of the tank. This action is repeated over and over.

5. To obtain better cooling, the surface area of a transformer can be increased by adding tubes or fins to the steel tank assembly.

6. The insulating oil used in transformers is a high-grade oil that must be kept clean and moisture free.

7. Pyranol (manufactured by the General Electric Company) is a synthetic dielectric that is an effective, nonflammable, nonexplosive liquid cooling and insulating agent.

8. Other methods of cooling transformers include forced air circulation, natural air circulation, natural circulation of oil with water cooling, and forced oil circulation (for large transformers).

• Some voltage is lost in high-voltage distribution circuits.

1. The impressed voltage on the primary winding may be lower than indicated on the nameplate of the transformer.

2. As a result, the secondary terminal voltage is also lower than the nameplate value.

This condition is undesirable, and provisions can be made for changing taps to obtain the desired voltage.

3. The turns ratio of the transformer is adjusted by changing the taps.

a. The change in the tap connections can be made at the terminal board inside the transformer.

b. A tap-changing switch can be used to vary the number of turns of the primary winding, resulting in a change in the turns ratio.

c. A special tap switch, immersed in oil, may be used on some large transformers. This type of switch permits the tap connections to be changed under load. A small motor drives the tap-changing mechanism and is operated from a control panel.

• A transformer must be operated on an ac circuit at the frequency for which it is designed.

1. If a lower frequency is used, the reactance of the primary will decrease. The exciting current will increase, causing an increase in the flux density. The saturation of the core is above normal and there is a greater heat loss. Thus, the efficiency of the transformer is decreased.

2. If the frequency is greater than the nameplate value, the reactance will increase and the exciting current will decrease. There will be a lower flux density, but the core loss will remain nearly constant.

• A transformer must be operated at its rated primary voltage, as indicated on the nameplate.

1. If the voltage is increased to a high enough value above the nameplate rating, excessive heating will take place at the windings. The flux density of the core will increase, and saturation of the core will be above normal. Transformers are designed to operate without overheating at a voltage that is 5% above the rated voltage.

2. If the transformer is operated at a voltage lower than the nameplate rating, the power output will be reduced in proportion to the voltage decrease.

• Inrush current for a transformer can be very high compared to a choke.

1. The magnetic domains in the core material of a transformer “remember” where they were set when the power is turned off.

2. Three factors that determine the amount of inrush current for a transformer are the amount of applied voltage, the resistance of the wire in the primary winding, and the flux change of the magnetic field in the core.

3. Magnetic domains or molecules can be made to reset to a neutral position by inserting an airgap in the core material.

• Transformer impedance is determined by the physical construction of the transformer.

1. Transformer impedance is expressed as a percent.

2. Transformer impedance is used to calculate the short-circuit current of a transformer.

Achievement Review

1. Define the term static transformer.

2. Explain what is meant by the terms primary and secondary, as applied to transformers.

3.	a. b.	What is a stepdown transformer? What is a stepup transformer?
4.	a. b.	What is the relationship between the induced voltages on the primary and secondary windings and the number of turns of the two windings? What is the difference between the impressed voltage and the induced voltage in the primary winding?

5. The primary winding of a transformer is rated at 115 V and the secondary winding is rated at 300 V. The primary winding has 500 turns. How many turns does the secondary winding have?

6. The transformer in question 5 has a full-load secondary output of 300 VA at 300 V.

a. What is the full-load secondary current?

b. Determine the full-load primary current (neglect all losses).

c. What is the relationship between the primary current and the secondary current and the number of turns on the two windings?

7. a. What is an exciting current?

b. What two functions does it perform?

8. a. Explain what is meant by core losses.

b. How can the core losses be kept to a minimum?

c. Draw a circuit diagram to show how the core loss, in watts, is measured for a 5-kVA transformer rated at 2400 to 240 V. The low-voltage side is used as the primary winding.

d. What safety precautions should be observed in measuring the core loss on the low-voltage side?

e. Using accurate instruments, should the core loss be the same for the high- voltage winding as for the primary?

9. The core loss of the transformer in question 8 is measured using the low-voltage side as the primary. The following data are obtained:

Voltmeter reading = 240 V

Exciting current = 1.0 A Wattmeter reading = 24 W

Determine

a. the core loss, in watts.

b. the power factor and the phase angle.

c. the in-phase component of the exciting current required for the core losses.

d. the lagging quadrature (magnetizing) component of the current.

10. Explain why the secondary ampere-turns oppose the magnetizing flux of the pri- mary winding.

11. List, in sequence, the reactions that take place in a transformer that result in an increase in the primary current input with an increase in the secondary current output to the load.

12. a. What is the primary leakage flux?

b. What is the secondary leakage flux?

c. What effect does leakage flux have on the operation of a transformer?

d. How can the leakage flux of a transformer be kept to a minimum?

13. Using diagrams, describe the three types of core construction used in transformers.

14. a. List the losses that occur in a transformer.

b. Which losses remain nearly constant at all load points? Why?

c. Which losses change when there is a change in the load? Why?

15. A 5-kVA, 240/120-V, 60-Hz transformer has a core loss of 32 W. The transformer has an effective resistance of 0.05 W in the high-voltage winding and 0.0125 W in the low-voltage winding. Determine the efficiency of the transformer at the rated load and unity power factor.

16. A 25-kVA, 2400/240-V, 60-Hz stepdown transformer has a core loss of 120 W. The effective resistance of the primary winding is 1.9 W. The effective resistance of the secondary winding is 0.02 W. Determine

a. the full-load current rating of the high-voltage and low-voltage windings. (Neglect any losses and assume that the input and the output are the same.)

b. the total copper losses of the transformer at the rated load.

c. the efficiency of the transformer at the rated load and a unity power factor.

17. Determine the efficiency of the transformer in question 15 at 50% of the rated load output and 0.l75 lagging power factor.

18. A 50-kVA, 4600/230-V, 60-Hz, single-phase, stepdown transformer is designed so that at the condition of full load and a unity power factor, the core losses equal the copper losses. The copper losses are equally divided between the two wind- ings. At full load the efficiency is 96.5%. Determine

a. the core losses.

b. the total copper losses.

c. the rated current of the primary and secondary windings. (Neglect the losses and assume that the input and the output are the same.)

d. the effective resistance of the primary and secondary windings.

19. Explain why it is hard to arrive at an accurate value of the efficiency of a trans- former using measurements of the power input and output.

20. a. Draw a diagram of the connections for the impedance short-circuit test for a single-phase transformer, rated at 20 kVA, 60 Hz, 4800/240 V. The high- voltage winding is the input side and is supplied from a 240-V ac source with a variable resistor in series.

b. Why should the measurements for the short-circuit test (part a) be made on the high-voltage side of the transformer?

21. The following data were obtained during the short-circuit test for the 20-kVA, 4800/240-V transformer in question 20:

Impedance voltage = 160 V

Ammeter reading = 4.2 A Wattmeter reading = 280 W

Another test shows the core loss to be 120 W. Determine

a. the efficiency at the rated load and a unity power factor.

b. the efficiency at 50% of the rated load and 0.80 lagging power factor.

22. a How are transformer leads marked?

b. Explain what is meant by additive polarity and subtractive polarity.

23. A transformer is used to step down a primary voltage of 4800 V to 240 V to sup- ply a single-phase load. Additional equipment is installed, making the kVA load greater than the full-load rating of the transformer. A second transformer, having the same kVA capacity, is to be operated in parallel with the first transformer.

a. List three requirements that are to be observed if the transformers are to share the load properly.

b. Explain the procedure used in connecting the second transformer in parallel.

24. a. There is a question about the accuracy of the turns ratio of a transformer rated at 2400/240 V. A 240-V test source is available in the laboratory. Assuming that voltmeters are available with the proper scales, explain how the voltage ratio can be determined. Draw a schematic diagram to clarify the answer.

b. Explain how the polarity of this transformer may be obtained using the 240-V test source.

25. A 30-kVA, 2400/240-V, 60-Hz transformer is used as a stepdown transformer.

The core loss is 150 W, the rated primary current is 12.5 A, the resistance of the primary winding is 1.5 W, the rated secondary current is 125 A, and the resistance of the secondary winding is 0.015 W. At the rated kVA capacity and a 0.75 lagging power factor, determine

a. the total copper losses.

b. the efficiency of the transformer.

26. A 20-kVA standard distribution transformer is rated at 2400/4800 V on the high- voltage side, and 120/240 V on the low-voltage side. The transformer has an efficiency of 97% when it is operating at the rated load as a stepdown transformer supplying a noninductive lighting load.

a. Draw a diagram to show the internal and external transformer connections required to step down the voltage from 2400 V to a 120/240-V, single- phase, three-wire service. Show the polarity markings for the transformer terminals for additive polarity.

b. What is the full-load secondary current if the load is balanced?

c. What is the power input to the primary side at the rated load?

d. If the no-load voltage of the 240-V secondary is 4 V more than the full- load voltage, what is the percentage voltage regulation?

27. Explain why the single-phase, 120/240-V, three-wire system is preferred to the single-phase, 120-V, two-wire system.

28.	a. b.	Why is oil used in transformers? What advantage is there in using Pyranol, instead of oil, in a transformer?
29.	a. b.	Explain the process of cooling a transformer by the natural circulation of oil and air. How can the heat-dissipating surface be increased for transformers with very large kVA ratings?

30. a. What is the purpose of a tap-changing switch as used on a transformer?

b. What are some of the applications of a transformer with tap-changing facilities?

31. Explain why a transformer has a high efficiency from 10% of the rated load to full load.

32. List the data commonly found on a transformer nameplate.

33. A 50-kVA single-phase transformer has a secondary voltage of 240 V. The name- plate indicates that the transformer has an impedance of 2.35%. What is the short- circuit current for this transformer?

34. A three-phase transformer is rated at 75 kVA and has a secondary voltage of 208

V. The nameplate indicates a transformer impedance of 3.5%. What is the short- circuit current for this transformer?

PRACTICE PROBLEMS FOR UNIT 13

Find the missing values in the following problems. Refer to Figure 13–35 and the formulas under the Transformers section of Appendix 15.

Find the missing values in the following problems. Refer to Figure 13–36 and the formulas listed under the Transformers section of Appendix 15.

[Note: Total primary current is equal to the sum of the primary currents needed to supply power to each secondary.]

MAGNETIC DOMAINS

Magnetic materials contain tiny magnetic structures in their molecular material, known as domains. These domains can be affected by outside sources of magnetism. Figure 13–22 illustrates a magnetic domain that has not been polarized by an outside magnetic source.

Now assume that the north pole of a magnet is placed toward the top of the material that contains the magnetic domains (Figure 13–23). Notice the structure of the domain has changed to realign the molecules in the direction of the outside magnetic field.

If the polarity of the magnetic pole is changed (Figure 13–24), the molecular structure of the domain will change to realign itself with the new magnetic lines of flux. This external influence can be produced by an electromagnet as well as a permanent magnet.

In certain types of cores, the molecular structure of the domain will snap back to its neutral position when the magnetizing force is removed. This type of core is used in the construction of reactors or chokes (Figure 13–25). A core of this type is constructed by

separated sections of the steel laminations with an airgap. This airgap breaks the path of the magnet through the core material and is responsible for the domains returning to their neutral position once the magnetizing force is removed.

The core construction of a transformer, however, does not contain an airgap. The steel laminations are connected together in such a manner as to produce a very low reluctance path for the magnetic lines of flux. In this type of core, the domains remain in their set posi- tion once the magnetizing force has been removed. This type of core “remembers” where it was last set. This was the principle of operation of the core memory of early computers. It is also the reason why transformers can have extremely high inrush currents when they are first connected to the power line.

The amount of inrush current in the primary of a transformer is limited by the following three factors:

1. The amount of applied voltage.

2. The resistance of the wire in the primary winding.

3. The flux change of the magnetic field in the core. The amount of flux change deter- mines the amount of inductive reactance produced in the primary winding when power is applied.

Figure 13–26 illustrates a simple isolation-type transformer. The alternating cur- rent applied to the primary winding produces a magnetic field around the winding. As the current changes in magnitude and direction, the magnetic lines of flux change also. Because the lines of flux in the core are continually changing polarity, the magnetic domains in the core material are changing also. As stated previously, the magnetic domains in the core of a transformer remember their last set position. For this reason, the point on the waveform where current is disconnected from the primary winding can have a great bearing on the amount of inrush current when the transformer is reconnected to power. For example, assume that the power supplying the primary winding is disconnected at the zero crossing point (Figure 13–27). In this instance, the magnetic domains would be set at the neutral point. When power is restored to the primary winding, the core material can be magnetized by either magnetic polarity. This permits a change of

flux, which is the dominant current-limiting factor. In this instance, the amount of inrush current would be relatively low.

If the power supplying current to the primary winding is interrupted at the peak point of the positive or negative half-cycle, however, the domains in the core material will be set at that position. Figure 13–28 illustrates this condition. It is assumed that the current was stopped as it reached its peak positive point. If the power should be reconnected to the primary winding during the positive half-cycle, only a very small amount of flux change can take place. Because the core material is saturated in the positive direction, the primary winding of the transformer is essentially an air core inductor, which greatly decreases the inductive characteristics of the winding. The inrush current in this situation would be limited by the resistance of the winding and a very small amount of inductive reactance.

This characteristic of transformers can be demonstrated with a clamp-on ammeter that has a “peak hold” capability. If the ammeter is connected to one of the primary leads, and power is switched on and off several times, it can be seen that the amount of inrush current will vary over a wide range.

MULTIPLE TAPPED WINDINGS

It is not uncommon for transformers to be designed with windings that have more than one set of lead wires connected to the primary or secondary. The trans- former shown in Figure 13–29 contains a secondary winding rated at 24 V. The primary winding contains several taps, however. One of the primary lead wires is labeled C and is the common for the other leads. The other leads are labeled 120, 208, and 240, respectively. This transformer is designed in such a manner that it can be connected to different primary voltages without changing the value of the secondary voltage. In this example, it is assumed that the secondary winding has a total of 120 turns of wire. To maintain the proper turns ratio, the primary would have 600 turns of wire between C and 120, 1040 turns between C and 208, and 1200 turns between C and 240.

The transformer shown in Figure 13–30 contains a single primary winding. The secondary winding, however, has been tapped at several points. One of the secondary lead wires is labeled C and is common to the other lead wires. When rated voltage is applied to

the primary, voltages of 12, 24, and 48 V can be obtained at the secondary. It should also be noted that this arrangement of taps permits the transformer to be used as a center-tapped transformer for two of the voltages. If a load is placed across the lead wires labeled C and 24, the lead wire labeled 12 becomes a center tap. If a load is placed across the C and 48 lead wires, the 24 lead wire becomes a center tap.

In this example, it is assumed the primary winding has 300 turns of wire. To produce the proper turns ratio, it would require 30 turns of wire between C and 12, 60 turns of wire between C and 24, and 120 turns of wire between C and 48.

The transformer shown in Figure 13–31 is similar to the transformer in Figure 13–30. The transformer in Figure 13–31, however, has multiple secondary windings instead of a single secondary winding with multiple taps. The advantage of the transformer in Figure 13–31 is that the secondary windings are electrically isolated from each other. These secondary windings can be either stepup or stepdown depending on the application of the transformer.

COOLING TRANSFORMERS

The core-and-coil assembly is placed in a pressed steel tank and is completely covered with an insulating oil. This insulating oil removes heat from the core and the coil windings and insulates the windings from the core and the transformer case. The core-and-coil assembly for a typical transformer contains channels or ducts. These ducts permit the oil to circulate and remove the heat. As the oil gains heat, its density decreases and it rises. As it rises and circulates in the transformer, it contacts the tank walls and the heat is transferred from the oil to the tank walls. The walls are cooled by air circulation. The specific gravity of the oil increases as it loses heat. As a result, the oil flows down to the bottom of the tank and again circulates up through the coil ducts to repeat the cooling process. Transformers having a very large

kVA capacity require more than the available surface area of the transformer case for cooling.

The surface area of a transformer can be increased by the use of tubes or fins added to the steel tank assembly (Figure 13–32). This increased surface area will take more heat from the oil and will radiate it faster to the surrounding air in a given time.

Transformer Oil

The insulating oil used in transformers is a high-grade oil that must be kept clean and moisture-free. This type of oil should be checked periodically to determine whether there is any change in its insulating ability. If traces of moisture or foreign materials are found, the oil must be filtered or replaced. The insulating fluid in some transformers is a nonflammable, nonexplosive liquid. One fluid of this type is Pyranol, manufactured by the General Electric Company. This liquid is a synthetic dielectric that is an effective cooling and insulating agent.

Oil-cooled transformers are considered a fire hazard in some locations. Air-cooled transformers are used under these conditions. Such transformers permit the natural circulation of air to remove the heat from the coils and the core. A perforated metal enclosure pro- vides mechanical protection to the coil and windings and permits air to circulate through the windings.

In addition to the method described, other ways of cooling transformers include forced air circulation, natural air circulation, natural circulation of oil with water cooling, and forced oil circulation (for large transformers).

MAGNETIC DOMAINS

Magnetic materials contain tiny magnetic structures in their molecular material, known as domains. These domains can be affected by outside sources of magnetism. Figure 13–22 illustrates a magnetic domain that has not been polarized by an outside magnetic source.

Now assume that the north pole of a magnet is placed toward the top of the material that contains the magnetic domains (Figure 13–23). Notice the structure of the domain has changed to realign the molecules in the direction of the outside magnetic field.

If the polarity of the magnetic pole is changed (Figure 13–24), the molecular structure of the domain will change to realign itself with the new magnetic lines of flux. This external influence can be produced by an electromagnet as well as a permanent magnet.

In certain types of cores, the molecular structure of the domain will snap back to its neutral position when the magnetizing force is removed. This type of core is used in the construction of reactors or chokes (Figure 13–25). A core of this type is constructed by

separated sections of the steel laminations with an airgap. This airgap breaks the path of the magnet through the core material and is responsible for the domains returning to their neutral position once the magnetizing force is removed.

The core construction of a transformer, however, does not contain an airgap. The steel laminations are connected together in such a manner as to produce a very low reluctance path for the magnetic lines of flux. In this type of core, the domains remain in their set posi- tion once the magnetizing force has been removed. This type of core “remembers” where it was last set. This was the principle of operation of the core memory of early computers. It is also the reason why transformers can have extremely high inrush currents when they are first connected to the power line.

The amount of inrush current in the primary of a transformer is limited by the following three factors:

1. The amount of applied voltage.

2. The resistance of the wire in the primary winding.

3. The flux change of the magnetic field in the core. The amount of flux change deter- mines the amount of inductive reactance produced in the primary winding when power is applied.

Figure 13–26 illustrates a simple isolation-type transformer. The alternating cur- rent applied to the primary winding produces a magnetic field around the winding. As the current changes in magnitude and direction, the magnetic lines of flux change also. Because the lines of flux in the core are continually changing polarity, the magnetic domains in the core material are changing also. As stated previously, the magnetic domains in the core of a transformer remember their last set position. For this reason, the point on the waveform where current is disconnected from the primary winding can have a great bearing on the amount of inrush current when the transformer is reconnected to power. For example, assume that the power supplying the primary winding is disconnected at the zero crossing point (Figure 13–27). In this instance, the magnetic domains would be set at the neutral point. When power is restored to the primary winding, the core material can be magnetized by either magnetic polarity. This permits a change of

flux, which is the dominant current-limiting factor. In this instance, the amount of inrush current would be relatively low.

If the power supplying current to the primary winding is interrupted at the peak point of the positive or negative half-cycle, however, the domains in the core material will be set at that position. Figure 13–28 illustrates this condition. It is assumed that the current was stopped as it reached its peak positive point. If the power should be reconnected to the primary winding during the positive half-cycle, only a very small amount of flux change can take place. Because the core material is saturated in the positive direction, the primary winding of the transformer is essentially an air core inductor, which greatly decreases the inductive characteristics of the winding. The inrush current in this situation would be limited by the resistance of the winding and a very small amount of inductive reactance.

This characteristic of transformers can be demonstrated with a clamp-on ammeter that has a “peak hold” capability. If the ammeter is connected to one of the primary leads, and power is switched on and off several times, it can be seen that the amount of inrush current will vary over a wide range.

MULTIPLE TAPPED WINDINGS

It is not uncommon for transformers to be designed with windings that have more than one set of lead wires connected to the primary or secondary. The trans- former shown in Figure 13–29 contains a secondary winding rated at 24 V. The primary winding contains several taps, however. One of the primary lead wires is labeled C and is the common for the other leads. The other leads are labeled 120, 208, and 240, respectively. This transformer is designed in such a manner that it can be connected to different primary voltages without changing the value of the secondary voltage. In this example, it is assumed that the secondary winding has a total of 120 turns of wire. To maintain the proper turns ratio, the primary would have 600 turns of wire between C and 120, 1040 turns between C and 208, and 1200 turns between C and 240.

The transformer shown in Figure 13–30 contains a single primary winding. The secondary winding, however, has been tapped at several points. One of the secondary lead wires is labeled C and is common to the other lead wires. When rated voltage is applied to

the primary, voltages of 12, 24, and 48 V can be obtained at the secondary. It should also be noted that this arrangement of taps permits the transformer to be used as a center-tapped transformer for two of the voltages. If a load is placed across the lead wires labeled C and 24, the lead wire labeled 12 becomes a center tap. If a load is placed across the C and 48 lead wires, the 24 lead wire becomes a center tap.

In this example, it is assumed the primary winding has 300 turns of wire. To produce the proper turns ratio, it would require 30 turns of wire between C and 12, 60 turns of wire between C and 24, and 120 turns of wire between C and 48.

The transformer shown in Figure 13–31 is similar to the transformer in Figure 13–30. The transformer in Figure 13–31, however, has multiple secondary windings instead of a single secondary winding with multiple taps. The advantage of the transformer in Figure 13–31 is that the secondary windings are electrically isolated from each other. These secondary windings can be either stepup or stepdown depending on the application of the transformer.

COOLING TRANSFORMERS

The core-and-coil assembly is placed in a pressed steel tank and is completely covered with an insulating oil. This insulating oil removes heat from the core and the coil windings and insulates the windings from the core and the transformer case. The core-and-coil assembly for a typical transformer contains channels or ducts. These ducts permit the oil to circulate and remove the heat. As the oil gains heat, its density decreases and it rises. As it rises and circulates in the transformer, it contacts the tank walls and the heat is transferred from the oil to the tank walls. The walls are cooled by air circulation. The specific gravity of the oil increases as it loses heat. As a result, the oil flows down to the bottom of the tank and again circulates up through the coil ducts to repeat the cooling process. Transformers having a very large

kVA capacity require more than the available surface area of the transformer case for cooling.

The surface area of a transformer can be increased by the use of tubes or fins added to the steel tank assembly (Figure 13–32). This increased surface area will take more heat from the oil and will radiate it faster to the surrounding air in a given time.

Transformer Oil

The insulating oil used in transformers is a high-grade oil that must be kept clean and moisture-free. This type of oil should be checked periodically to determine whether there is any change in its insulating ability. If traces of moisture or foreign materials are found, the oil must be filtered or replaced. The insulating fluid in some transformers is a nonflammable, nonexplosive liquid. One fluid of this type is Pyranol, manufactured by the General Electric Company. This liquid is a synthetic dielectric that is an effective cooling and insulating agent.

Oil-cooled transformers are considered a fire hazard in some locations. Air-cooled transformers are used under these conditions. Such transformers permit the natural circulation of air to remove the heat from the coils and the core. A perforated metal enclosure pro- vides mechanical protection to the coil and windings and permits air to circulate through the windings.

In addition to the method described, other ways of cooling transformers include forced air circulation, natural air circulation, natural circulation of oil with water cooling, and forced oil circulation (for large transformers).

TRANSFORMER CHARACTERISTICS

The characteristic curves of a 50-kVA transformer are shown in Figure 13–17. The efficiency of this transformer is very high even when the load is as low as 10% of the rated load. Note that the efficiency curve is nearly flat between 20% of the rated load to about 20% overload. The load–voltage characteristics show that there is little change in

the terminal voltage from a full-load condition to a no-load condition. This condition was shown in the previous voltage regulation calculations. With a lagging power factor load, the secondary terminal voltage change is slightly greater than it is for a unity power factor load. However, the percentage of voltage regulation is satisfactory. With the leading power factor load, the terminal voltage increases with the load, giving a negative voltage regulation.

TRANSFORMER CONSTRUCTION

The single-phase transformer is a simple alternating-current device. The core consists of thin annealed sheets (laminations) of silicon steel. These laminations may be in the form of rectangular strips or L-shaped stampings. When separate sheets are used to make the core, the sheets are placed with butt joints in each layer. The core can also be made by winding a long strip of silicon steel to the desired size. In the laminated core, the sheets are stacked so that the butt joints are staggered in successive layers to ensure a low reluctance. Because the spiral-wound core has no joints, it has a very low reluctance.

Three basic designs of cores are available for use in transformers. These cores are known as the core type, the shell type, and the combined core-and-shell type. Figure 13–18 shows the core-type structure for a transformer.

In the core-type structure, the windings surround the laminated silicon steel core. The low-voltage winding generally is placed next to the core with the high-voltage winding wound over the low-voltage winding. These windings are carefully insulated from each other. The advantage of placing the low-voltage coils next to the core is that a reduction can be made in the material required to insulate the high-voltage windings. Many simplified transformer drawings show the primary winding on one leg of the core and the secondary winding on the other leg. However, such an arrangement gives rise to excessive leakage flux. In practice, both the primary and secondary windings are placed on the same leg of the core to minimize the leakage.

Shell-Type Core

A shell-type core is shown in Figure 13–19. In this structure, the silicon steel core surrounds the windings. The entire flux passes through the center leg of the core and then divides. One-half of the flux passes through each of the two outside legs.

The coil arrangement and the flux path for a shell-type core are shown in Figure 13–20. The low-voltage coil windings are placed next to the laminated core. The high- voltage windings are placed between the low-voltage windings. This coil arrangement provides adequate insulation between the coils. Because the high-voltage coils are not adjacent to the core structure, less insulation is required. Pancake coils wound with rectangular copper wire may be used with this type of core. The fact that the coil windings surround the core and the core surrounds the coils means that the leakage flux is minimized.

Type H Core

A type H core with coil windings is shown in Figure 13–21. The core is shaped like a cross when viewed from above. The coils are constructed so that the high-voltage wind- ings are located between the low-voltage windings. This arrangement of coils minimizes the insulation required. As a result, the only high-voltage insulation required is placed between the high-voltage windings and the low-voltage windings. The structure of this core-and-coil assembly keeps the leakage flux to a minimum because the coil wind- ings are placed on a center leg and are surrounded by the four outside legs of the core structure. The H core is often used for distribution transformers with two high-voltage windings, each rated at 2400 V, and two low-voltage windings, each rated at 120 V. Such a transformer can be used to step down either 2400 V or 4800 V to 120 V, 240 V, or 120/240 V.

The GE 220 Class Transformer

A 220 class transformer developed by the General Electric Company has a wound core. That is, a long strip of silicon steel is wound in a tight spiral around the insulated windings. This type of core has the following advantages:

1. It can be manufactured more easily than the conventional core, consisting of lamina- tions stacked and clamped together.

2. The magnetic circuit path is relatively short and has a large cross section.

3. The construction of the core helps reduce the flux leakage.

4. The flux path direction is always along the grain of the silicon steel, thus reducing the iron losses.

TRANSFORMER INRUSH CURRENT

Although transformers and reactors are both inductive devices, there is a great difference in their operating characteristics. Reactors are often used to prevent inrush current from becoming excessive when a circuit is first turned on. Transformers, how- ever, can produce extremely high inrush currents when power is first applied to the primary winding. The type of core used is primarily responsible for this difference in characteristics.

TRANSFORMER CHARACTERISTICS

The characteristic curves of a 50-kVA transformer are shown in Figure 13–17. The efficiency of this transformer is very high even when the load is as low as 10% of the rated load. Note that the efficiency curve is nearly flat between 20% of the rated load to about 20% overload. The load–voltage characteristics show that there is little change in

the terminal voltage from a full-load condition to a no-load condition. This condition was shown in the previous voltage regulation calculations. With a lagging power factor load, the secondary terminal voltage change is slightly greater than it is for a unity power factor load. However, the percentage of voltage regulation is satisfactory. With the leading power factor load, the terminal voltage increases with the load, giving a negative voltage regulation.

TRANSFORMER CONSTRUCTION

The single-phase transformer is a simple alternating-current device. The core consists of thin annealed sheets (laminations) of silicon steel. These laminations may be in the form of rectangular strips or L-shaped stampings. When separate sheets are used to make the core, the sheets are placed with butt joints in each layer. The core can also be made by winding a long strip of silicon steel to the desired size. In the laminated core, the sheets are stacked so that the butt joints are staggered in successive layers to ensure a low reluctance. Because the spiral-wound core has no joints, it has a very low reluctance.

Three basic designs of cores are available for use in transformers. These cores are known as the core type, the shell type, and the combined core-and-shell type. Figure 13–18 shows the core-type structure for a transformer.

In the core-type structure, the windings surround the laminated silicon steel core. The low-voltage winding generally is placed next to the core with the high-voltage winding wound over the low-voltage winding. These windings are carefully insulated from each other. The advantage of placing the low-voltage coils next to the core is that a reduction can be made in the material required to insulate the high-voltage windings. Many simplified transformer drawings show the primary winding on one leg of the core and the secondary winding on the other leg. However, such an arrangement gives rise to excessive leakage flux. In practice, both the primary and secondary windings are placed on the same leg of the core to minimize the leakage.

Shell-Type Core

A shell-type core is shown in Figure 13–19. In this structure, the silicon steel core surrounds the windings. The entire flux passes through the center leg of the core and then divides. One-half of the flux passes through each of the two outside legs.

The coil arrangement and the flux path for a shell-type core are shown in Figure 13–20. The low-voltage coil windings are placed next to the laminated core. The high- voltage windings are placed between the low-voltage windings. This coil arrangement provides adequate insulation between the coils. Because the high-voltage coils are not adjacent to the core structure, less insulation is required. Pancake coils wound with rectangular copper wire may be used with this type of core. The fact that the coil windings surround the core and the core surrounds the coils means that the leakage flux is minimized.

Type H Core

A type H core with coil windings is shown in Figure 13–21. The core is shaped like a cross when viewed from above. The coils are constructed so that the high-voltage wind- ings are located between the low-voltage windings. This arrangement of coils minimizes the insulation required. As a result, the only high-voltage insulation required is placed between the high-voltage windings and the low-voltage windings. The structure of this core-and-coil assembly keeps the leakage flux to a minimum because the coil wind- ings are placed on a center leg and are surrounded by the four outside legs of the core structure. The H core is often used for distribution transformers with two high-voltage windings, each rated at 2400 V, and two low-voltage windings, each rated at 120 V. Such a transformer can be used to step down either 2400 V or 4800 V to 120 V, 240 V, or 120/240 V.

The GE 220 Class Transformer

A 220 class transformer developed by the General Electric Company has a wound core. That is, a long strip of silicon steel is wound in a tight spiral around the insulated windings. This type of core has the following advantages:

1. It can be manufactured more easily than the conventional core, consisting of lamina- tions stacked and clamped together.

2. The magnetic circuit path is relatively short and has a large cross section.

3. The construction of the core helps reduce the flux leakage.

4. The flux path direction is always along the grain of the silicon steel, thus reducing the iron losses.

TRANSFORMER INRUSH CURRENT

Although transformers and reactors are both inductive devices, there is a great difference in their operating characteristics. Reactors are often used to prevent inrush current from becoming excessive when a circuit is first turned on. Transformers, how- ever, can produce extremely high inrush currents when power is first applied to the primary winding. The type of core used is primarily responsible for this difference in characteristics.

THE SINGLE-PHASE, THREE-WIRE SYSTEM

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

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

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

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

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

PROBLEM 6

Statement of the Problem

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

1. the voltage drop in the line wires.

2. the percentage voltage drop.

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

Solution

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

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

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

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

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

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

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

PROBLEM 7

Statement of the Problem

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

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

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

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

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

Solution

1. Desirable voltage drop:

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

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

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

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

Circuit with Unbalanced Loads

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

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

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

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

Circuit with an Open Neutral

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

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

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

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