3. Lft-tet, Transformer Trainer Exp Manual, 9082b, Imtac

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Knowledge Engineering

TRANSFORMER TRAINER

EXPERIMENT MANUAL

MODEL : LFT-TET

Transformer Trainer Experiment Manual EFT-TET

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CONTENTS CONTENTS ............................................................................................................. i 1. OVERVIEW .......................................................................................................... 1 2. ABOUT THE TRAINER .................................................................................. 2

2.1. TECHNICAL SPECIFICATION..................................................................... 2

2.2. SETTING UP THE TRAINER ........................................................................ 3

2.3. SAFETY ON EXPERIMENT.......................................................................... 3 3. BASIC THEORY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3.1. WHY TRANSFORMERS ARE IMPORTANT

TO MODERN LIFE? ....................................................................................... 4

3.2. TYPES AND CONSTRUCTION OF TRANSFORMERS ............................. 5

3.3. THE IDEAL TRANSFORMER....................................................................... 6

3.4. THEORY OF OPERATION OF REAL SINGLE-PHASE

TRANSFORMERS ........................................................................................ 16

3.5. THE PER-UNIT SYSTEM OF MEASUREMENTS..................................... 27

3.6. TRANSFORMER VOLTAGE REGULATION AND EFFICIENCY .......... 34

3.7. TRANSFORMER TAPS AND VOLTAGE REGULATION ....................... 35

3.8 THE AUTOTRANSFORMER....................................................................... 35

3.9 THREE-PHASE TRANSFORMERS ............................................................ 43

3.10. SUMMARY ................................................................................................... 52

4. EXPERIMENT .................................................................................................. 53

4.1. POLARITY OF TRANSFORMER…………………………………….. ..... 54

4.2. SERIES CIRCUIT IN A MULTIPLE WINDING TRANSFORMER……. 57

4.3. OPEN CIRCUIT TESTING…………………………………………… ....... 61

4.4. SHORT CIRCUIT TESTING ........................................................................ 66

4.5. TURN RATIO OF PRIMARY AND SECONDARY WINDING................. 69

4.6. IMPEDANCE TRANSFORMATION........................................................... 73

4.7. VOLTAGE REGULATION .......................................................................... 76

4.8. FULL WAVE RECTIFIER............................................................................ 78

4.9. WYE – DELTA CONNECTION................................................................... 82


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4.10. WYE – WYE CONNECTION ....................................................................... 90

4.11. THREE PHASE RECTIFIER ........................................................................ 96


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1 OVERVIEW

Transformer trainer is designed for student to learn and practice about the typical application of the transformer. There are many kind of transformer applications that can be observed in this trainer. This trainer is also completed with all equipment that needed in the experiment, so makes the learning process easier.

The trainer is divided into two parts, the base station and transformer module. The base station is fixed in the special case for safety user and easy operation. In the top cover of the base station, there are two voltmeters and two ammeters that are used to measure the voltage and current. Either voltmeters or ammeters can be switch on and off in order to give protection to them when the power source is turned on. In the main base station, there are many components that support in the experiment, such as RCCB, MCB, and Digital Wattmeter. RCCB is provided in this trainer to detects current leakage and protect the user from electric shock.

The voltage and current characteristics of the transformers in the different load can be observed by change the loads of the transformer. Student can make any connections according the manual instruction using jumper cables by plug-in it into the jumper cable sockets that available on the base station.

The student can make connections from transformer module to the base station using jumper cables and absolutely have to follow the experiment manual instruction. The experiment manual is accompanied this trainer in order to give the basic theory of the transformer and how to use the trainer to the student.


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2 ABOUT THE TRAINER

2.1. TECHNICAL SPECIFICATION

A. SINGLE PHASE UNIT a. Transformer : Input : 0 – 220 – 240 Vac

(LFT-TET-P1A) Output : 0 – 12 Vac (12 A)

0 – 24 Vac (6 A)

0 – 42 Vac (3 A)

b. Transformer : Input : 0 – 220 – 240 Vac

(LFT-TET-P1B) Output : 0 – 120 Vac (5 A)

0 – 120 Vac (5 A)

c. Base Station Digital Wattmeter : 1000 Watt max.

RCCB : 40 Ampere, 30mA current leakage protection

MCB : 6 Ampere

AC Voltmeter : 0 – 500 V

0 – 250 V

AC Ammeter : 0 – 1 A

0 – 10 A

Bridge diode : 10 A, 250 Vac max

B. THREE PHASE UNIT a. Transformer : Input : Three phase 0 – 380 – 415 Vac

(LFT-TET-P3) Output : Three phase 0 – 41.5 – 415 Vac

b. Base Station Digital Wattmeter : 1000 Watt max.

RCCB : 40 Ampere, 30mA current leakage protection

MCB : 6 Ampere

AC Voltmeter : 0 – 500 V

0 – 500 V

AC Ammeter : 0 – 1 A

0 – 5 A

3 Phase Rectifier : 10 A, 450 Vac max


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C. OPTIONAL ITEMS (in separated unit)

1. Single Phase Load Unit (LFT-TET-01)

2. Single Phase Variable Power Supply (LFT-TET-02)

3. Three Phase Load Unit (LFT-TET-06)

4. Variable Three Phase Power Supply (LFT-TET-07)

2.2. SETTING UP THE TRAINER a. Prepare the equipment required on your experiment.

b. Switch off all of the switches (RCCB, MCB, and meter switches).

c. Make connections between the base station to transformer unit and other equipment following the experiment. Use different cable colors for different connections.

d. Plug the power cord to three phase power source.

e. The trainer is ready to be used.

2.4. SAFETY ON EXPERIMENT a. Ensure the power cable and jumper cables are in best condition. No crack or ripped.

b. Ensure the RCCB and MCB are OFF when making or changing the connection leads.

c. Ensure the power is OFF when removing the connection leads.

d. Ask your instructor to recheck your wiring before applying power to the circuit.

e. Always use the different cable color for different connections.


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3 BASIC THEORY

A transformer is a device that changes ac electric power at one voltage level to ac electric power at another voltage level through the action of a magnetic field. It consists of two or more coils of wire wrapped around a common ferromagnetic core. These coils are (usually) not directly connected. The only connection between the coils is the common magnetic flux present within the core. One of the transformer windings is connected to a source of ac electric power, and the second (and perhaps third) transformer winding supplies electric power to loads. The transformer winding connected to the power source is called the primary winding or input winding, and the winding connected to the loads is called the secondary winding or output winding. If there is a third winding on the transformer, it is called the tertiary winding.

Figure 3.1 The first practical modem transformer, built by William Stanley in 1885. Note that the core is made up of individual sheets of metal (laminations).

3.1. WHY TRANSFORMERS ARE IMPORTANT TO MODERN LIFE? The first power distribution system in the United States was a 120-V dc system invented by Thomas A. Edison to supply power for incandescent light bulbs. Edison’s s first central power station went into operation in New York City in September 1882. Unfortunately, his power system generated and transmitted power at such low voltages that very large currents were necessary to supply significant amounts of power. These high currents caused huge voltage drops and power losses in the transmission lines, severely restricting the service area of a generating station. In the 1880s, central power stations were located every few city blocks to overcome this problem. The fact that power could not be transmitted far with low-voltage dc power systems meant that generating stations had to be small and localized and so were relatively inefficient.


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The invention of the transformer and the concurrent development of ac power sources eliminated forever these restrictions on the range and power level of power systems. A transformer ideally changes one ac voltage level to another voltage level without affecting the actual power supplied. If a transformer steps up the voltage level of a circuit, it must decrease the current to keep the power into the device equal to the power out of it. Therefore, ac electric power can be generated at one central location, its voltage stepped up for transmission over long distances at very low losses, and its voltage stepped down again for final use. Since the transmission losses in the lines of a power system are proportional to the square of the current in the lines, raising the transmission voltage and reducing the resulting transmission currents by a factor of 10 with transformers reduces power transmission losses by a factor of 100. Without the transformer, it would simply not be possible to use electric power in many of the ways it is used today. In a modem power system, electric power is generated at voltages of 12 to 25 kV. Transformers step up the voltage to between 110 kV and nearly 1000 kV for transmission over long distances at very low losses. Transformers then step down the voltage to the 12- to 34.5-kv range for local distribution and finally permit the power to be used safely in homes offices and factories at voltages as low as 120 V. 3.2. TYPES AND CONSTRUCTION OF TRANSFORMERS The principal purpose of a transformer is to convert ac power at one voltage level to ac power of the same frequency at another voltage level. Transformers are also used for a variety of other purposes (e.g., voltage sampling, current sampling and impedance transformation), but this chapter is primarily devoted to the power transformer. Power transformers are constructed on one of two types of cores. One type of construction consists of a simple rectangular laminated piece of steel with the transformer windings wrapped around two sides of the rectangle. This type of construction is known as core form and is illustrated in Figure 3.2. The other type consists of a three-legged laminated core with the windings wrapped around the center leg. This type of construction is known as shell form and is illustrated in Figure 3.3. In either case, the core is constructed of thin laminations electrically isolated from each other in order to minimize eddy currents.

The primary and secondary windings in a physical transformer are wrapped one on top of the other with the low-voltage winding innermost. Such an arrangement serves two purposes: 1. It simplifies the problem of insulating the high-voltage winding from the core. 2. It results in much less leakage flux than would be the case if the two windings were

separated by a distance on the core. Power transformers are given a variety of different names, depending on their use in power systems. A transformer connected to the output of a generator and used to step its voltage up to transmission levels (110+ kV) is sometimes called a unit transformer The transformer at the other end of the transmission line, which steps the voltage down from transmission levels to distribution levels (from 2.3 to 34.5 kV), is called a substation transformer. Finally, the transformer that takes the distribution voltage and steps it down to the final voltage at which the power is actually used (110, 208, 220 V, etc.) is called a distribution transformer All these devices are essentially the same, the only difference among them is their intended use.


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In addition to the various power transformers, two special-purpose transformers are used with electric machinery and power systems. The first of these special transformers is a device specially designed to sample a high voltage and produce a low secondary voltage directly proportional to it. Such a transformer is called a potential transformer. A power transformer also produces a secondary voltage directly proportional to its primary voltage; the difference between a potential transformer and a power transformer is that the potential transformer is designed to handle only a very small current. The second type of special transformer is a device designed to provide a secondary current much smaller than but directly proportional to its primary current. This device is called a current transformer. Both special-purpose transformers are discussed in a later section of this chapter. 3.3. THE IDEAL TRANSFORMER An ideal transformer is a loss less device with an input winding and an output winding. The relationships between the input voltage and the output voltage, and between the input current and the output current, are given by two simple equations. Figure 3.4 shows an ideal transformer. The transformer shown in Figure 3.4 has NP turns of wire on its primary side and NS turns of wire on its secondary side. The relationship between the voltage vP(t) applied to the primary side of the transformer and the voltage vS(t) produced on the secondary side is

)()(

tVtV

S

P = S

P

NN = a

Figure 3.2.

Core-form transformer construction.


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Figure. 3.3

(a) Shell-form transformer construction. (b) A typical shell-form transformer

Figure 3.4 (a) Sketch of an ideal transformer. (b) Schematic symbols of a transformer.


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where a is defined to be the turns ratio of the transformer:

a =S

p

NN

The relationship between the current iP(t) flowing into the primary side of the transformer and the current iS(t) flowing out of the secondary side of the transformer is NPiP(t) = NSiS(t)

or )()(

titi

S

P = a1

In terms of phasor quantities, these equations are

S

P

VV

= a

and

S

P

II

= a1

Notice that the phase angle of VP is the same as the angle of VS and the phase angle of IP is the same as the phase angle of IS. The turns ratio of the ideal transformer affects the magnitudes of the voltages and currents, but not their angles. The equations above describe the relationships between the magnitudes and angles of the voltages and currents on the primary and secondary sides of the transformer, but they leave one question unanswered: Given that the primary circuit’s voltage is positive at a specific end of the coil, what would the polarity of the secondary circuit’s voltage be? In real transformers, it would be possible to tell the secondary’s polarity only if the transformer were opened and its windings examined. To avoid this necessity, transformers utilize the dot convention. The dots appearing at one end of each winding in Figure 3.4 tell the polarity of the voltage and current on the secondary side of the transformer. The relationship is as follows: 1. If the primary voltage is positive at the dotted end of the winding with respect to the

undotted end, then the secondary voltage will be positive at the dotted end also. Voltage polarities are the same with respect to the dots on each side of the core.

2. If the primary current of the transformer flows into the dotted end of the primary winding, the secondary current will flow out of the dotted end of the secondary winding.


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Power in an Ideal Transformer The power supplied to the transformer by the primary circuit is given by the equation

Pin = VPIP cos θP

where θP is the angle between the primary voltage and the primary current. The power supplied by the transformer secondary circuit to its loads is given by the equation. Pout = VSIS cos θS where θS is the angle between the secondary voltage and the secondary current. Since voltage and current angles are unaffected by an ideal transformer, θP-θS = θ. The primary and secondary windings of an ideal transformer have the same power factor.

How does the power going into the primary circuit of the ideal transformer compare to the power coming out of the other side? It is possible to find out through a simple application of the voltage and current equations. The power out of a transformer is Pout = VSIS cos θS Applying the turns-ratio equations gives VS = VS /a and IS = a IP , so

Pout = a

VP (aIP) cos θ

Pout = VPIP cos θ = Pin Thus, the output power of an ideal transformer is equal to its input power.

The same relationship applies to reactive power Q and apparent power S:

Qin = VPIP sin θ = VSIS sin θ = Qout and Sin = VPIP = VSIS = Sout Impedance Transformation through a Transformer The impedance of a device or an element is defined as the ratio of the phasor voltage across it to the phasor current flowing through it:


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ZL = L

L

IV

One of the interesting properties of a transformer is that, since it changes voltage and current levels, it changes the ratio between voltage and current and hence the apparent impedance of an element. To understand this idea, refer to Figure 3.21. If the secondary current is called IS and the secondary voltage VS, then the impedance of the load is given by

ZL = S

S

IV

The apparent impedance of the primary circuit of the transformer is

Z′L = P

P

IV

Since the primary voltage can be expressed as

VP = aVS

and the primary current can be expressed as

IP = a

VS

the apparent impedance of the primary is

Z′L = P

P

IV =

aIaV

S

S

/=

S

S

IVa2

Z′L = a2ZL

With a transformer, it is possible to match the magnitude of load impedance to source impedance simply by picking the proper turns ratio.


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Figure 3.5 (a) Definition of impedance. (b) Impedance scaling through a transformer

Analysis of Circuits Containing Ideal Transformers If a circuit contains an ideal transformer, then the easiest way to analyze the circuit for its voltages and currents is to replace the portion of the circuit on one side of the transformer by an equivalent circuit with the same terminal characteristics. After the equivalent circuit has been substituted for one side, then the new circuit (without a transformer present) can be solved for its voltages and currents. In the portion of the circuit that was not replaced, the solutions obtained will be the correct values of voltage and current for the original circuit. Then the turns ratio of the transformer can be used to determine the voltages and currents on the other side of the transformer. The process of replacing one side of a transformer by its equivalent at the other side’s voltage level is known as referring the first side of the transformer to the second side. How is the equivalent circuit formed? Its shape is exactly the same as the shape of the original circuit. The polarities of voltage sources in the equivalent circuit will be reversed from their direction in the original circuit if the dots on one side of the transformer windings are reversed compared to the dots on the other side of the transformer windings. The solution for circuits containing ideal transformers is illustrated in the following example. Example 3.1 A single-phase power system consists of a 480-V 60-Hz generator supplying a load Zload = 4 + j3 Ω through a transmission line of impedance Zline = 0.18 + j0.24 Ω. Answer the following questions about this system. (a) If the power system is exactly as described above (Figure 3.6a), what will the voltage at

the load be? What will the transmission line losses be? (b) Suppose a 1:10 step-up transformer is placed at the generator end of the transmission line

and a 10:1 step-down transformer is placed at the load end of the line (Figure 3.6b). What will the load voltage be now? What will the transmission line losses be now?


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Solution (a) Figure 3.6a shows the power system without transformers. Here IG = Iline = Iload. The line

current in this system is given by

Figure. 3.6 The power system of Example 3-1 (a) without and (b) with transformers

at the ends of the transmission line.

Iline = loadline ZZ

V+

= )34()24.018.0(

º0480Ω+Ω+Ω+Ω

∠jj

V

= 24.318.4º0480

j+∠

= º8.3729.5

º0480∠∠

= 90.8∠ – 37.8° A Therefore the load voltage is Vload = Iline Zload = (90.8 ∠ – 37.8° A)(4 Ω + j3 Ω) = (90.8 ∠ – 37.8° A)(5∠ 36.9° Ω)

= 454∠ – 0.9° V


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and the line losses are Ploss = (Iline)2 Rline = (90.8 A)2 (0.18 Ω) = 1484 W

(b) Figure 3.7b shows the power system with the transformers. To analyze this system, it is

necessary to convert it to a common voltage level. This is done in two steps:

1. Eliminate transformer T2 by referring the load over to the transmission line’s voltage level.

2. Eliminate transformer T1 by referring the transmission line’s elements and the equivalent load at the transmission line’s voltage over to the source side.

The value of the load’s impedance when reflected to the transmission system’s voltage is Z′load = a2 Zload

= 2

110

⎟⎠⎞

⎜⎝⎛ (4 Ω + j3 Ω)

= 400 Ω + j300 Ω The total impedance at the transmission line level is now Zeq = Zline + Z′load = 400.18 + j300.24 Ω = 500.3∠ 36.88° Ω This equivalent circuit is shown in Figure 3.8a. The total impedance at the transmission line level (Zline + Z′load) is now reflected across T1 to the source’s voltage level:

Z′eq = a2Zeq = a2(Zline + Z′load)

=

2

101⎟⎠⎞

⎜⎝⎛

(0.18 Ω + j0.24 Ω + 400 Ω + j300 Ω

= (0.0018 Ω + j0.0024 Ω + 4 Ω + j3 Ω) = 5.003 ∠ 36.88° Ω


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Figure 3.7 (a) System with the load referred to the transmission system voltage level. (b) System with the load and

transmission line referred to the generator’s voltage level. Notice that Z″load = 4 + j3 Ω and Z′line = 0.0018 + j0.0024. The resulting equivalent circuit is shown in Figure 3.7b. The generator’s current is

IG = Ω∠

∠º88.36003.5

º0480 V

= 95.94 ∠ −36.88° A Knowing the current IG, we can now work back and find Iline and Iload. Working back through T1, we get NP1IG = NSlIline Iline = NP1 IG

= 101 (95.94 ∠ −36.88° A)

= 9.594∠ −36.88° A Working back through T2 gives

NP2Iline = NS2Iload


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Iload = NP2 Iline NS2

= 101 (9.594∠ −36.88° A)

= 95.94 ∠−36.88° A

It is now possible to answer the questions originally asked. The load voltage is given by Vload = Iload Zload = (95.94∠ −36.88° A)(5 ∠ 36.87° Ω = 479.7 ∠ 0.01° V and the line losses are given by Ploss = (Iline)2 Rline = (9.594 A)2 (0.18 Ω) = 16.7 W Notice that raising the transmission voltage of the power system reduced transmission losses by a factor of nearly 90. Also, the voltage at the load dropped much less in the system with transformers compared to the system without transformers. This simple example dramatically illustrates the advantages of using higher-voltage transmission lines as well as the extreme importance of transformers in modem power systems.


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3.4. THEORY OF OPERATION OF REAL SINGLE-PHASE TRANSFORMERS The ideal transformers described in section 3.3 can of course never actually be made. What can be produced are real transformers, two or more coils of wire physically wrapped around a ferromagnetic core. The characteristics of a real transformer approximate the characteristics of an ideal transformer, but only to a degree. This section deals with the behavior of real transformers. To understand the operation of a real transformer, refer to Figure 3.8. It shows a transformer consisting of two coils of wire wrapped around a transformer core. The primary of the transformer is connected to an ac power source, and the secondary winding is open-circuited. The hysteresis curve of the transformer is shown in Figure 3.9. The basis of transformer operation can be derived from Faraday’s law:

eind = dtdλ

where λ is the flux linkage in the coil across which the voltage is being induced. The flux linkage λ is the sum of the flux passing through each turn in the coil added over all the turns of the coil:

λ = ∑=

N

ii

1

φ

Figure 3.8 Sketch of real transformer with no load attached to its secondary.


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Figure 3.9

The hysteresis curve of the transformer. The total flux linkage through a coil is not just Nφ, where N is the number of turns in the coil, because the flux passing through each turn of a coil is slightly different from the flux in the other turns, depending on the position of the turn within the coil. However, it is possible to define an average flux per turn in a coil. If the total flux linkage in all the turns of the coils is λ and if there are N turns, then the average flux per turn is given by

φ = Nλ

and Faraday’s law can be written as

eind = N dtdφ

The Voltage Ratio across a Transformer If the voltage of the source in Figure 3.8 is vp(t), then that voltage is placed directly across the coils of the primary winding of the transformer. How will the transformer react to this applied voltage? Faraday’s law explains what will happen. When equation above is solved for the average flux present in the primary winding of the transformer, the result is

φ = PN

1∫ dttVP )(

This equation states that the average flux in the winding is proportional to the integral of the voltage applied to the winding, and the constant of proportionality is the reciprocal of the number of turns in the primary winding 1/NP.

This flux is present in the primary coil of the transformer. What effect does it have on the secondary coil of the transformer? The effect depends on how much of the flux reaches the secondary coil. Not all the flux produced in the primary coil also passes through the


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secondary coil—some of the flux lines leave the iron core and pass through the air instead (see Figure 3.10). The portion of the flux that goes through one of the transformer coils but not the other one is called leakage flux. The flux in the primary coil of the transformer can thus be divided into two components: a mutual flux, which remains in the core and links both windings, and a small leakage flux, which passes through the primary winding but returns through the air, bypassing the secondary winding:

Figure 3.10 Mutual and leakage fluxes in a transformer core.

Pφ = φM + φLP where φP = total average primary flux

φM = flux component linking both primary and secondary coils φLP = primary leakage flux

There is a similar division of flux in the secondary winding between mutual flux and leakage flux which passes through the secondary winding but returns through the air, by passing the primary winding: Sφ = φM + φLS where φS = total average secondary flux φM = flux component linking both primary and secondary coils φLS = secondary leakage flux

With the division of the average primary flux into mutual and leakage components, Faraday’s law for the primary circuit can be reexpressed as

vP(t) = NP dt

d Pφ


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= NP dtMdφ + NP

dtd LPφ

The first term of this expression can be called eP(t), and the second term can be called eLP(t). If this is done, then equation can be rewritten as vP(t) = eP(t) + eLP(t) The voltage on the secondary coil of the transformer can also be expressed in terms of Faraday’s law as

vS(t) = NS dt

d Sφ

= NS dtMdφ + NS

dtLSdφ

= eS(t) + eLS(t) The primary voltage due to the mutual flux is given by

eP(t) = NP dt

d Mφ

and the secondary voltage due to the mutual flux is given by

eS(t) = NS dt

d Mφ

Notice from these two relationships that

P

P

Nte )(

= dt

d Mφ = S

S

Nte )(

Therefore,

)()(

tete

S

P = S

P

NN

This equation means that the ratio of the primary voltage caused by the mutual flux to the secondary voltage caused by the mutual flux is equal to the turns ratio of the transformer. Since in a well-designed transformer φM >> φLP and φM >> φLS, the ratio of the total voltage on the primary of a transformer to the total voltage on the secondary of a transformer is approximately


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)()(

tvtv

S

P = S

P

NN = a

The smaller the leakage fluxes of the transformer are, the closer the total transformer voltage ratio approximates that of the ideal transformer discussed in section 3.3. The Magnetization Current in a Real Transformer When an ac power source is connected to a transformer as shown in Figure 3.8, a current flows in its primary circuit, even when the secondary circuit is open-circuited. This current is the current required to produce flux in a real ferromagnetic core. It consists of two components: 1. The magnetization current iM, which is the current required to produce the flux in the transformer core. 2. The core-loss current ih+e, which is the current required to make up for hysteresis and

eddy current losses. Figure 3.11 shows the magnetization curve of a typical transformer core. If the flux in the transformer core is known, then the magnitude of the magnetization current can be found directly from Figure 3.11. Ignoring for the moment the effects of leakage flux, we see that the average flux in the core is given by

φ = ∫ dttvN P

P

)(1

If the primary voltage is given by the expression vP(t) = vM cos ωt V, then the resulting flux must be

φ = ∫ dttCosVN M

P

..1 ω

= WbtSinN

V

P

M ..ωω

If the values of current required to produce a given flux (Figure 3.11a) are compared to the flux in the core at different times, it is possible to construct a sketch of the magnetization current in the winding on the core. Such a sketch is shown in Figure 3.11b. Notice the following points about the magnetization current:


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Figure 3.11 (a) The magnetization curve of the transformer core.

(b) The magnetization current caused by the flux in the transformer core.

1. The magnetization current in the transformer is not sinusoidal. The higher-frequency components in the magnetization current are due to magnetic saturation in the transformer core.

2. Once the peak flux reaches the saturation point in the core, a small increase in peak flux requires a very large increase in the peak magnetization current.

3. The fundamental component of the magnetization current lags the voltage applied to the core by 90°.

4. The higher-frequency components in the magnetization current can be quite large compared to the fundamental component. In general, the further a transformer core is driven into saturation, the larger the harmonic components will become.


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The other component of the no-load current in the transformer is the current required to supply power to make up the hysteresis and eddy current losses in the core. This is the core-loss current. Assume that the flux in the core is sinusoidal. Since the eddy currents in the core are proportional to dφ/dt, the eddy currents are largest when the flux in the core is passing through 0 Wb. Therefore, the core-loss current is greatest as the flux passes through zero. The total current required to make up for core losses is shown in Figure 3.12. Notice the following points about the core-loss current: 1. The core-loss current is nonlinear because of the nonlinear effects of hysteresis. 2. The fundamental component of the core-loss current is in phase with the voltage applied

to the core. The total no-load current in the core is called the excitation current of the transformer. It is just the sum of the magnetization current and the core-loss current in the core: iex = im + ih+e The total excitation current in a typical transformer core is shown in Figure 3.13

Figure 3.12 The core-loss current in a transformer.

Figure 3.13 The total excitation current in a transformer.


LABTECH 23

Determining the Values of Components in the Transformer Model It is possible to experimentally determine the values of the inductances and resistances in the transformer model. An adequate approximation of these values can be obtained with only two tests, the open-circuit test and the short-circuit test. In the open-circuit test, a transformer’s secondary winding is open-circuited, and its primary winding is connected to a full-rated line voltage. The open-circuit test connections are shown in Figure 3.14. Full line voltage is applied to the primary of the transformer, and the input voltage, input current, and input power to the transformer are measured. From this information, it is possible to determine the power factor of the input current and therefore both the magnitude and the angle of the excitation impedance.

The easiest way to calculate the values of RC and XM is to look first at the admittance of the excitation branch. The conductance of the core-loss resistor is given by

Gc = CR

1

and the susceptance of the magnetizing inductor is given by

BM = MX

1

Since these two elements are in parallel, their admittances add, and the total excitation admittance is YE = Gc - jBM

= CR

1 – MX

j 1

Figure 3.14 Connection for transformer open-circuit test.

The magnitude of the excitation admittance (referred to the primary circuit) can be found from the open-circuit test voltage and current:

EY = OC

OC

VI


LABTECH 24

The angle of the admittance can be found from a knowledge of the circuit power factor. The open-circuit power factor (PF) is given by

PF = cos θ

=OCOC

OC

IVP

and the power-factor angle θ is given by

θ = cos –1

OC

OC

VP

The power factor is always lagging for a real transformer, so the angle of the current always lags the angle of the voltage by θ degrees. Therefore, the admittance YE is

YE = OC

OC

VI ∠ -θ

= OC

OC

VI ∠ -θ cos –1 PF

By comparing equations above, it is possible to determine the values of RC and XM directly from the open-circuit test data. In the short-circuit test, the secondary terminals of the transformer are short-circuited, and the primary terminals are connected to a fairly low-voltage source, as shown in Figure 3.15. The input voltage is adjusted until the current in the short-circuited windings is equal to its rated value. (Be sure to keep the primary voltage at a safe level. It would not be a good idea to burn out the transformer’s windings while trying to test it.) The input voltage, current, and power are again measured. Since the input voltage is so low during the short-circuit test, negligible current flows through the excitation branch. If the excitation current is ignored, then all the voltage drop in the transformer can be attributed to the series elements in the circuit. The magnitude of the series impedances referred to the primary side of the transformer is

SEZ = SC

SC

IV

The power factor of the current is given by

PF = cos θ

=SCSC

SC

IVP

and is lagging. The current angle is thus negative, and the overall impedance angle θ is positive:


LABTECH 25

θ = cos-1 SCSC

SC

IVP

Therefore,

ZSE = °−∠°∠θSC

SC

IV 0

= °∠0SC

SC

IV

The series impedance ZSE is equal to ZSE = Req + jXeq =(Rp + a2Rs) + j(Xp + a2Xs)

IIIIIII

Figure 3.15 Connection for transformer short-circuit test.

It is possible to determine the total series impedance referred to the primary side by using this technique, but there is no easy way to split the series impedance into primary and secondary components. Fortunately, such separation is not necessary to solve normal problems. Also these same tests may be performed on the secondary side of the transformer if it is more convenient to do so because of voltage levels or other reasons. If the tests are performed on the secondary side, the results will naturally yield the equivalent circuit impedances referred to the secondary side of the transformer instead of to the primary side. Example 3.2 The equivalent circuit impedances of a 20-kVA, 8000V 240-V, 60-Hz transformers are to be determined. The open-circuit test and the Short-circuit test were performed on the primary side of the transformer, and the following data were taken:

Open-circuit Test (on primary)

Short-circuit Test (on primary)

VOC = 8000V VSC = 489 V IOC = 0.214 A ISC = 2.5 A POC = 400 W PSC = 240 W


LABTECH 26

Find the impedances of the approximate equivalent circuit referred to the primary side, and sketch that circuit. Solution. The power factor during the open-circuit test is:

PF = cos θ = OCOC

OC

IVP

= cos θ = )214.0)(8000(

400AV

W

= 0.234 lagging The excitation admittance is given by

YE = OC

OC

VI ∠ – Cos-1 PF

= VA

8000214.0 ∠ – Cos-1 0.234

= 0.0000268∠ – 76.5º Ω

= 0.0000063 – j 0.0000261

= CR

1 – j MX

1

Therefore, RC = 1 . 0.0000063 = 159kΩ

XM = 0000261.0

1

0.000026 1 = 38.4kΩ The power factor during the short-circuit test is:

PF = Cos θ = SCSC

SC

IVP

=

Cos θ = )5.2)(489(

240AV

W = 0.196 lagging

The series impedance is given by

ZSE = SC

SC

IV ∠ - Cos-1 PF


LABTECH 27

= AV

5.2489 ∠ 78.7º

= 195.6 ∠ 78.7º = 38.4 + j l92 Ω Therefore, the equivalent resistance and reactance are Req =38.4 Ω X eq =192 Ω The resulting simplified equivalent circuit is shown in Figure 3.16. 3.5. THE PER-UNIT SYSTEM OF MEASUREMENTS As the relatively simple Example 3.2 showed, solving circuits containing transformers can be quite a tedious operation because of the need to refer all the different voltage levels on different sides of the transformers in the system to a common level. Only after this step has been taken can the system be solved for its voltages and currents.

Figure 3.16 The equivalent circuit of example 3.2

There is another approach to solving circuits containing transformers which eliminates the need for explicit voltage-level conversions at every transformer in the system. Instead, the required conversions are handled automatically by the method itself, without ever requiring the user to worry about impedance transformations. Because such impedance transformations can be avoided, circuits containing many transformers can be solved easily with less chance of error. This method of calculation is known as the per-unit (pu) system of measurements. In the per-unit system, the voltages, currents, powers, impedances, and other electrical quantities are not measured in their usual SI units (volts, amperes, watts, ohms, etc.). Instead, each electrical quantity is measured as a decimal fraction of some base level. Any quantity can be expressed on a per-unit basis by the equation

Quantity per unit = Actual value . base value of quantity where “actual value” is a value in volts, amperes, ohms, etc.


LABTECH 28

It is customary to select two base quantities to define a given per-unit system. The ones usually selected are voltage and power (or apparent power). Once these base quantities have been selected, all the other base values are related to them by the usual electrical laws. In a single-phase system, these relationships are: Pbase’ Qbase’ or Sbase = Vbase Ibase

Zbase = base

base

IV

Y base = base

base

VI

and Zbase = base

base

SV 2)(

Once the base values of S (or P) and V have been selected, all other base values can be computed easily. In a power system, a base apparent power and voltage are selected at a specific point in the system. A transformer has no effect on the base apparent power of the system, since the apparent power into a transformer equals the apparent power out of the transformer. On the other hand, voltage changes when it goes through a transformer, so the value of Vbase changes at every transformer in the system according to its turns ratio. Because the base quantities change in passing through a transformer, the process of referring quantities to a common voltage level is automatically taken care of during per-unit conversion. Example 3.3 A simple power system is shown in Figure 3.17. This system contains a 480-V generator connected to an ideal 1:10 step-up transformer, a transmission line, an ideal 20:1 step-down transformer, and a load. The impedance of the transmission line is 20 + j60 fl, and the impedance of the load is l0<30o. The base values for this system are Chosen to be 480 V and 10 kVA at the generator. (a) Find the base voltage, current, impedance, and apparent power at every point in the

power system. (b) Convert this system to its per-unit equivalent circuit. (c) Find the power supplied to the load in this system. (d) Find the power lost in the transmission line.

Figure 3.17 The power system of Example 3.3.


LABTECH 29

Solution (a) In the generator region, Vbase = 480 V and Sbase = 10 kVA, so

I base 1 = 1base

base

VS

= VVA

480000.10

= 20.83A

Zbase 1 = 1

1

base

base

IV

= A

V83.80

480

= 23.04Ω

The turns ratio of transformer T1 is a = 1/10 = 0.1, so the base voltage in the transmission line region is

Vbase 2 = a

Vbase1

= 1.0

480V

= 4800 V The other base quantities are: S base 2 = 10 kVA

I base 2 = VVA

4800000.10

= 2.083 A

Z base 2 = A

V083.2

4800

= 2304 Ω The turns ratio of transformer T2 is a = 20/1 = 20, so the base voltage in the load region is:

V base 3 = a

Vbase2

= 20

4800V

= 240 V


LABTECH 30

The other base quantities are: S base 3 = 10 kVA

I base 3 = VVA

240000.10

= 41.67 A

Z base 3 = A

V67.41

240

= 5.76 Ω (b) To convert a power system to a per-unit system, each component must be divided by its

base value in its region of the system. The generator’s per-unit voltage is its actual value divided by its base value:

V G,pu = V

V480

0480 °∠

= 1.0∠ 0º pu The transmission line’s per-unit impedance is its actual value divided by its base value:

Z line,pu = ΩΩ+

23046020 j

= 0.0087 + j 0.0260 pu The load’s per-unit impedance is also given by actual value divided by base value:

Z load,pu = ΩΩ°∠

76.53010

= 1.736∠ 30º pu The per-unit equivalent circuit of the power system is shown in Figure 3.18.

Figure 3.18 The per-unit equivalent circuit for Example 3.3.


LABTECH 31

(c) The current flowing in this per-unit power system is:

Ipu = PUtot

PU

ZV

,

= )30736.1()0260.00087.0(

01°∠++

°∠j

= )868.0503.1()0260.00087.0(

01jj +++

°∠

= 894.0512.1

01j+°∠

= °∠

°∠6.30757.1

01

= 0.569∠ -30.6º pu Therefore, the per-unit power of the load is: Pload, pu = I2

puRpu = (0.569)2(1.503) = 0.487 and the actual power supplied to the load is: Pload = Pload,pu Sbase = (0.487)(10,000 VA) = 4870 W (d) The per-unit power lost in the transmission line is: Pline,pu =I2

puRline,pu = (0.569)2(0.0087) = 0.00282 and the actual power lost in the transmission line is: Pline = Pline,pu Sbase = (0.00282)(10,000 VA) = 28.2W When only one device (transformer or motor) is being analyzed, its own ratings are usually used as the base for the per-unit system. If a per-unit system based on the transformer’s own ratings is used, a power or distribution transformer’s characteristics will not vary much over a wide range of voltage and power ratings. For example, the series resistance of a transformer is usually about 0.01 per unit, and the series reactance is usually between 0.02 and 0.10 per unit. In general, the larger the transformer is the smaller the series impedances. The magnetizing reactance is usually between about 10 and 40 per unit, while the core-loss resistance is usually between about 50 and 200 per unit. Because per unit values provide a convenient and meaningful way to compare transformer characteristics when they are of different sizes, transformer impedances are normally given in per-unit or as a percentage on


LABTECH 32

the transformer’s nameplate. The same idea applies to synchronous and induction machines as well: Their per-unit impedances fall within relatively narrow ranges over quite large size ranges. If more than one machine and one transformer are included in a single power system, the system base voltage and power maybe chosen arbitrarily, but the entire system must have the same base. One common procedure is to choose the system base quantities to be equal to the base of the largest component in the system. Per-unit values given to another base can be converted to the new base by converting them to their actual values (volts, amperes, ohms, etc.) as an in-between step. Alternatively, they can be converted directly by the equations.

(P,Q,S)PU on base 2 = (P,Q,S) PU on base 1 2

1

base

base

SS

Figure 3.19 (a) Atypical 13.2 kV to 120/240V distribution transformer. (b) A cutaway view of the distribution transformer

showing the shell-form transformer inside it.

VPU on base 2 = VPU on base 1 2

1

base

base

VV

(R, X, Z) PU on base 2 = (R, X, Z) PU on base1 )()()()(

12

2

22

1

basebase

basebase

SVSV


LABTECH 33

Example 3.4 Sketch the approximate per-unit equivalent circuit for the transformer in Example 3.2. Use the transformer’s ratings as the system base.

Solution. The transformer in Example 3.2 is rated at 20 kVA, 8000/240 V. The approximate equivalent circuit (Figure 3.16) developed in the example was referred to the high-voltage side of the transformer, so to convert it to per-unit, the primary circuit base impedance must be found. On the primary, Vbasel = 8000 V Sbase 1 = 20,000 VA

Zbase 1= 1

21)(

base

base

SV

= VA

V000.20

2)8000(

= 3200 Ω

Therefore,

ZSE,PU = Ω

Ω+3200

1924.38 j

= 0.012 + j0.06PU

RC,PU = ΩΩ

3200159k

= 49.7 PU

ZM,PU = ΩΩ

32004.38 k

= 12PU The per-unit approximate equivalent circuit, expressed to the transformer’s own base, is shown in Figure 3.20.


LABTECH 34

Figure 3.20

The per-unit equivalent circuit of Example 3.4. 3.6. TRANSFORMER VOLTAGE REGULATION AND EFFICIENCY Because a real transformer has series impedances within it, the output voltage of a transformer varies with the load even if the input voltage remains constant. To conveniently compare transformers in this respect, it is customary to define a quantity called voltage regulation (VR). Full-load voltage regulation is a quantity that compares the output voltage of the transformer at no load with the output voltage at full load. It is defined by the equation

%100 x VRflS,

flS,ln,S,

VVV −

=

Since at no load, VS = VP/a, the voltage regulation can also be expressed as

%100 x /VRflS,

flS,P

VVaV −

=

If the transformer equivalent circuit is in the per-unit system, then voltage regulation can be expressed as

%100VR,f,

,l,, xV

VVPUlS

PfSPUP U−=

Usually it is a good practice to have as small a voltage regulation as possible. For an ideal transformer, VR = 0 percent. It is not always a good idea to have a low-voltage regulation, though—sometimes high-impedance and high-voltage regulation transformers are deliberately used to reduce the fault currents in a circuit.


LABTECH 35

3.7. TRANSFORMER TAPS AND VOLTAGE REGULATION In previous sections of this chapter, transformers were described by their turns ratios or by their primary-to-secondary-voltage ratios. Throughout those sections, the turns ratio of a given transformer was treated as though it were completely fixed. In almost all real distribution transformers, this is not quite true. Distribution transformers have a series of taps in the windings to permit small changes in the turns ratio of the transformer after it has left the factory. A typical installation might have four taps in addition to the nominal setting with spacings of 2.5 percent of full-load voltage between them. Such an arrangement provides for adjustments up to 5 percent above or below the nominal voltage rating of the transformer. Example 3.5 A 500-kVA, 13,200/480-V distribution transformer has four 2.5 percent taps on its primary winding. What are the voltage ratios of this transformer at each tap setting? Solution The five possible voltage ratings of this transformer are +5.0% tap 13,860/480V +2.5% tap 13,530/480V Nominal rating 13,200/480V -2.5% tap 12,870/480V -5.0% tap 12,540/480V The taps on a transformer permit the transformer to be adjusted in the field to accommodate variations in local voltages. However, these taps normally cannot be changed while power is being applied to the transformer. They must be set once and left alone. Sometimes a transformer is used on a power line whose voltage varies widely with the load. Such voltage variations might be due to a high line impedance between the generators on the power system and that particular load (perhaps it is located far out in the country). Normal loads need to be supplied an essentially constant voltage. How can a power company supply a controlled voltage through high-impedance lines to loads which are constantly changing? One solution to this problem is to use a special transformer called a tap changing under load (TCUL) transformer or voltage regulator. Basically, a TCUL transformer is a transformer with the ability to change taps while power is connected to it. A voltage regulator is a TCUL transformer with built-in voltage sensing circuitry that automatically changes taps to keep the system voltage constant. Such special transformers are very common in modern power systems. 3.8. THE AUTOTRANSFORMER On some occasions it is desirable to change voltage levels by only a small amount. For example, it may be necessary to increase a voltage from 110 to 120 V or from 13.2 to 13.8 kV. These small rises may be made necessary by voltage drops that occur in power systems a long way from the generators. In such circumstances, it is wasteful and excessively expensive to wind a transformer with two full windings, each rated at about the same voltage. A special-purpose transformer, called an autotransformer, is used instead.


LABTECH 36

A diagram of a step-up autotransformer is shown in Figure 3.21. In Figure 3.21a, the two coils of the transformer are shown in the conventional manner. In Figure 3.21b, the first winding is shown connected in an additive manner to the second winding. Now, the relationship between the voltage on the first winding and the voltage on the second winding is given by the turns ratio of the transformer. However, the voltage at the output of the whole transformer is the sum of the voltage on the first winding and the voltage on the second winding. The first winding here is called the common winding, because its voltage appears on both sides of the transformer. The smaller winding is called the series winding, because it is connected in series with the common winding. A diagram of a step-down autotransformer is shown in Figure 3.22. Here the voltage at the input is the sum of the voltages on the series winding and the common winding, while the voltage at the output is just the voltage on the common winding.

Figure 3.21 A transformer with its windings (a) connected in the conventional manner and

(b) reconnected as an auto-transformer.

Because the transformer coils are physically connected, a different terminology is used for the autotransformer than for other types of transformers. The voltage on the common coil is called the common voltage VC and the current in that coil is called the common current IC The voltage on the series coil is called the series voltage VSE, and the current in that coil is called the series current ISE. The voltage and current on the low-voltage side of the transformer are called VL and IL respectively, while the corresponding quantities on the high-voltage side of the transformer are called VH and IH. The primary side of the autotransformer (the side with power into it) can be either the high-voltage side or the low-voltage side, depending on whether the autotransformer is acting as a step-down or a step-up transformer. From Figure 3.21b the voltages and currents in the coils are related by the equations

SE

C

SE

C

NN

VV

=

The voltages in the coils are related to the voltages at the terminals by the equations VL =VC VH = VC + VSE and the currents in the coils are related to the currents at the terminals by the equations IL = IC + ISE


LABTECH 37

IH = ISE Voltage and Current Relationships in an Autotransformer What is the voltage relationship between the two sides of an autotransformer? It is quite easy to determine the relationship between VH and VL The voltage on the high side of the autotransformer is given by

Figure 3.22 A step-down autotransformer connection.

VH = VC + VSE But

SE

C

SE

C

NN

VV

= , so

VH = VC + CC

SE VNN

Finally, noting that VL = VC, we get

VH = VL + LC

SE VNN

= LC

CSE VN

NN + , or

CSE

C

H

L

NNN

VV

+=

The current relationship between the two sides of the transformer can be found by noting that IL = IC + ISE

IC = SEC

SE INN x , so


LABTECH 38

IL = SEC

SE INN x +ISE

Finally, noting that IH= ISE , we find

IL = HC

SE INN x + IH

= HC

CSE xIN

NN + , or

C

CSE

H

L

NNN

II +=

The Apparent Power Rating Advantage of Autotransformers It is interesting to note that not all the power traveling from the primary to the secondary in the autotransformer goes through the windings. As a result, if a conventional transformer is reconnected as an autotransformer, it can handle much more power than it was originally rated for. To understand this idea, refer again to Figure 3.21b. Notice that the input apparent power to the autotransformer is given by Sin = VLIL and the output apparent power is given by Sout = VHIH It is easy to show, by using the voltage and current equations, that the input apparent power is again equal to the output apparent power: Sin = Sout = SIO where SIO is defined to be the input and output apparent powers of the transformer. However, the apparent power in the transformer windings is SW = VCIC = VSEISE The relationship between the power going into the primary (and out the secondary) of the transformer and the power in the transformer’s actual windings can be found as follows: SW = VCIC = VL (IL - IH) = VLIL - VLIH

SW = VLIL - VLIL CSE

C

NNN+


LABTECH 39

= VLIL CSE

CCSE

NNNNN

+−+ )(

= SIO CSE

SE

NNN+

Therefore, the ratio of the apparent power in the primary and secondary of the autotransformer to the apparent power actually traveling through its windings is

SE

CSE

W

IO

NNN

SS +

=

Equation above describes the apparent power rating advantage of an autotransformer over a conventional transformer. Here SIO is the apparent power entering the primary and leaving the secondary of the transformer, while SW is the apparent power actually traveling through the transformer’s windings (the rest passes from primary to secondary without being coupled through the transformer’s windings). Note that the smaller the series winding, the greater the advantage. For example, a 5000-kVA autotransformer connecting a 110-kV system to a 138-kV system would have an NC/NSE turns ratio of 110:28. Such an autotransformer would actually have windings rated at

SW = CSE

SEIO

NNNS+

= (5000 kVA) 11028

28+

= 1015 kVA

Figure 3.23 The autotransformer of Example 3.6


LABTECH 40

The autotransformer would have windings rated at only about 1015 kVA, while a con-ventional transformer doing the same job would need windings rated at 5000 kVA. The autotransformer could be 5 times smaller than the conventional transformer and also would be much less expensive. For this reason, it is very advantageous to build transformers between two nearly equal voltages as autotransformers. The following example illustrates autotransformer analysis and the rating advantage of autotransformers. Example 3.6. A 100-VA 120/12-V transformer is to be connected so as to form a step-up autotransformer (see Figure 3.23). A primary voltage of 120 V is applied to the transformer. (a) What is the secondary voltage of the transformer? (b) What is its maximum voltampere rating in this mode of operation? (c) Calculate the rating advantage of this autotransformer connection over the transformer’s

rating in conventional 120/12-V operation. Solution. To accomplish a step-up transformation with a 120-V primary, the ratio of the turns on the common winding NC to the turns on the series winding NSE in this transformer must be 120:12 (or 10:1). (a) This transformer is being used as a step-up transformer. The secondary voltage is VH :

VH = LC

CSE VN

NN +

= V120120

12012 +

= 132 V (b) The maximum volt-ampere rating in either winding of this transformer is 100 VA. How

much input or output apparent power can this provide? To find out, examine the series winding. The voltage VSE on the winding is 12 V, and the volt-ampere rating of the winding is 100 VA. Therefore, the maximum series winding current is

ISE,max = SEV

S max

= VVA

12100

= 8.33 A Since ISE is equal to the secondary current IS (or IH) and since the secondary voltage VS = VH = 132 V, the secondary apparent power is Sout = VSIS = VHIH = (132 V)(8.33 A) = 1100 VA = Sin


LABTECH 41

(c) The rating advantage can be calculated from part (b) (d) From part b,

W

IO

SS =

VAVA

1001100

= 11

W

IO

SS =

SE

CSE

NNN +

= 12

12012 +

= 11 By either equation, the apparent power rating is increased by a factor of 11. It is not normally possible to just reconnect an ordinary transformer as an auto-transformer and use it in the manner of Example 3.6, because the insulation on the low-voltage side of the ordinary transformer may not be strong enough to withstand the full output voltage of the autotransformer connection. In transformers built specifically as autotransformers, the insulation on the smaller coil (the series winding) is made just as strong as the insulation on the larger coil. It is common practice in power systems to use autotransformers whenever two voltages fairly close to each other in level need to be transformed, because the closer the two voltages are, the greater the autotransformer power advantage becomes. They are also used as variable transformers, where the low-voltage tap moves up and down the winding. This is a very convenient way to get a variable ac voltage. Such a variable autotransformer is shown in Figure 3.24. The principal disadvantage of autotransformers is that, unlike ordinary transformers, there is a direct physical connection between the primary and the secondary circuits, so the electrical isolation of the two sides is lost. If a particular application does not require electrical isolation, then the autotransformer is a convenient and inexpensive way to tie nearly equal voltages together. The Internal Impedance of an Autotransformer

Autotransformers have one additional disadvantage compared to conventional transformers. It turns out that, compared to a given transformer connected in the conventional manner, the effective per-unit impedance of an autotransformer is smaller by a factor equal to the reciprocal of the power advantage of the autotransformer connection. The proof of this statement is left as a problem at the end of the chapter.


LABTECH 42

Figure 3.24 A variable-voltage autotransformer. (b) Cutaway view of the autotransformer.

The reduced internal impedance of an autotransformer compared to a conventional two-winding transformer can be a serious problem in some applications where the series impedance is needed to limit current flows during power system faults (short circuits). The effect of the smaller internal impedance provided by an autotransformer must be taken into account in practical applications before autotransformers are selected. Example 3.7 A transformer is rated at 1000 kVA, 12/1.2 kV, 60 Hz when it is operated as a conventional two-winding transformer. Under these conditions, its series resistance and reactance are given as 1 and 8 percent per unit, respectively. This transformer is to be used as a 13.2/12—kV step-down autotransformer in a power distribution system. In the autotransformer connection, (a) what is the transformer’s rating when used in this manner and (b) what is the transformer’s series impedance in per-unit?

Solution (a) The NC/NSE turns ratio must be 12:1.2 or 10:1. The voltage rating of this transformer will

be 13.2/12 kV, and the apparent power (voltampere) rating will be

SIO = WSE

CSE SN

NN +

= kVA10001101+

= 11.000kVA (b) The transformer’s impedance in a per-unit system when connected in the conventional

manner is Zeq = 0.01 + j0.08 PU separate windings The apparent power advantage of this autotransformer is 11, so the per-unit impedance of the autotransformer connected as described is

Zeq = 11

08.001.0 j+

= 0.00091 + j 0.00727 PU autotransformer


LABTECH 43

3.9. THREE-PHASE TRANSFORMERS Almost all the major power generation and distribution systems in the world today are three-phase ac systems. Since three-phase systems play such an important role in modem life, it is necessary to understand how transformers are used in them. Transformers for three-phase circuits can be constructed in one of two ways. One approach is simply to take three single-phase transformers and connect them in a three-phase bank. An alternative approach is to make a three-phase transformer consisting of three sets of windings wrapped on a common core. These two possible types of transformer construction are shown in Figures 3.25 and 3.26. The construction of a single three-phase transformer is the preferred practice today, since it is lighter, smaller, cheaper, and slightly more efficient. The older construction approach was to use three separate transformers. That approach had the advantage that each unit in the bank could be replaced individually in the event of trouble, but that does not outweigh the advantages of a combined three-phase unit for most applications. However, there are still a great many installations consisting of three single-phase units in service. Three-Phase Transformer Connections A three-phase transformer consists of three transformers, either separate or combined on one core. The primaries and secondaries of any three-phase transformer can be independently connected in either a wye (Y) or a delta (A). This gives a total of four possible connections for a three-phase transformer bank: 1. Wye—wye (Y - Y) 2. Wye—delta (Y - Δ) 3. Delta—wye (Δ - Y) 4. Delta—delta (Δ - Δ) These connections are shown in Figure 3.27 The key to analyzing any three-phase transformer bank is to look at a single transformer in the bank. Any single transformer in the bank behaves exactly like the single-phase transformers already studied. The impedance, voltage regulation, efficiency, and similar calculations for three-phase transformers are done on a per-phase basis, using exactly the same techniques already developed for single-phase transformers. The advantages and disadvantages of each type of three-phase transformer connection are discussed below.


LABTECH 44

Figure 3.25 A three-phase transformer bank composed of independent transformers.

Figure 3.26 A three-phase transformer wound on a single three-legged core.

WYE-WYE CONNECTION. The Y-Y connection of three-phase transformers is shown in Figure 3.27a. In a Y-Y connection, the primary voltage on each phase of the transformer is given by VøP = VLP / √3. The primary-phase voltage is related to the secondary-phase voltage by the turns ratio of the transformer. The phase voltage on the secondary is then related to the line voltage on the secondary by VLS = √3VØs. Therefore, overall the voltage ratio on the transformer is

SLS

LP

VPV

VV

θ

θ

33

= = a Y - Y


LABTECH 45

Figure 3.27 Three-phase transformer connections and wiring diagrams: (a) Y-Y

The Y-Y connection has two very serious problems: 1. If loads on the transformer circuit are unbalanced, then the voltages on the phases of the

transformer can become severely unbalanced. 2. Third-harmonic voltages can be large. If a three-phase set of voltages is applied to a Y—Y transformer, the voltages in any phase will be 120º apart from the voltages in any other phase. However, the third-harmonic components of each of the three phases will be in phase with each other, since there are three cycles in the third harmonic for each cycle of the fundamental frequency. There are always some third-harmonic components in a transformer because of the nonlinearity of the core, and these components add up. The result is a very large third-harmonic component of voltage on top of the 50- or 60-Hz fundamental voltage. This third-harmonic voltage can be larger than the fundamental voltage itself. Both the unbalance problem and the third-harmonic problem can be solved using one of two techniques: 1. Solidly ground the neutrals of the transformers, especially the primary winding’s neutral.

This connection permits the additive third-harmonic components to cause a current flow in the neutral instead of building up large voltages. The neutral also provides a return path for any current imbalances in the load.

2. Add a third (tertiary) winding connected in Δ to the transformer bank. If a third Δ -connected winding is added to the transformer, then the third-harmonic components of voltage in the A will add up, causing a circulating current flow within the winding. This


LABTECH 46

suppresses the third-harmonic components of voltage in the same manner as grounding the transformer neutrals.

The Δ -connected tertiary windings need not even be brought out of the transformer case, but they often are used to supply lights and auxiliary power within the substation where it is located. The tertiary windings must be large enough to handle the circulating currents, so they are usually made about one-third the power rating of the two main windings. One or the other of these correction techniques must be used any time a Y—Y transformer is installed. In practice, very few Y—Y transformers are used, since the same jobs can be done by one of the other types of three-phase transformers. WYE-DELTA CONNECTION. The Y-Δ connection of three-phase transformers is shown in Figure 3-27b. In this connection, the primary line voltage is related to the primary phase voltage by VLP = √3VθP, while the secondary line voltage is equal to the secondary phase voltage VLS = VθS. The voltage ratio of each phase is

S

P

VVθ

θ = a

so the overall relationship between the line voltage on the primary side of the bank and the line voltage on the secondary side of the bank is

LS

LP

VV =

S

P

VVθ

θ3

S

P

VVθ

θ = a3 Y - Δ

The Y-Δ connection has no problem with third-harmonic components in its voltages, since they are consumed in a circulating current on the A side. This connection is also more stable with respect to unbalanced loads, since the A partially redistributes any imbalance that occurs. This arrangement does have one problem, though. Because of the connection, the secondary voltage is shifted 30º relative to the primary voltage of the transformer. The fact that a phase shift has occurred can cause problems in paralleling the secondaries of two transformer banks together. The phase angles of transformer secondaries must be equal if they are to be paralleled, which means that attention must be paid to the direction of the 30º phase shift occurring in each transformer bank to be paralleled together.


LABTECH 47

Figure 3.27 (continued) (b) Y-Δ.

The connection shown in Figure 3.27b will cause the secondary voltage to be lagging if the system phase sequence is abc. If the system phase sequence is acb, then the connection shown in Figure 3.54b will cause the secondary voltage to be leading the primary voltage by 30º.

Figure 3.27 (continued) (c) Δ.-Y


LABTECH 48

DELTA-WYE CONNECTION. A Δ-Y connection of three-phase transformers is shown in Figure 3.27c. In a Δ-Y connection, the primary line voltage is equal to the primary-phase voltage VLP = VθP while the secondary voltages are related by VLS = 3 VθS. Therefore, the line-to-line voltage ratio of this transformer connection is

S

P

LS

LP

VV

VV

θ

θ

3=

aV

VLS

LP 3= Δ-Y

This connection has the same advantages and the same phase shift as the Y—A transformer. The connection shown in Figure 3.27c makes the secondary voltage lag the primary voltage by 30º, as before.

Figure 3.27 (continued) (d) Δ-Δ

DELTA-DELTA CONNECTION. The Δ-Δ connection is shown in Figure 3.27d. In a Δ-Δ connection, VLP = VθP and VLS = VθS,so the relationship between primary and secondary line voltages is

aVV

VLSVLP

S

P==

φ

φ Δ- Δ

This transformer has no phase shift associated with it and no problems with unbalanced loads or harmonics.


LABTECH 49

The Per-Unit System for Three-Phase Transformers The per-unit system of measurements applies just as well to three-phase transformers as to single-phase transformers. The single-phase base equations apply to three-phase systems on a per-phase basis. If the total base voltampere value of the transformer bank is called Sbase, then the base voltampere value of one of the transformers SIØ base

SIØ,base = 3baseS

and the base phase current and impedance of the transformer are

IØ,base = base

baseI

VS

,,

φ

φ

IØ,base = base

base

VS

,3 φ

Z,base = ( )

baseI

base

SV

,

2,

φ

φ

Z,base = ( )

base

base

SV 2

,3 φ

Line quantities on three-phase transformer banks can also be represented in the per-unit system. The relationship between the base line voltage and the base phase voltage of the transformer depends on the connection of windings. If the windings are connected in delta, VL,base = VØ,base, while if the windings are connected in wye, VL,,base = √3VØ,base. The base line current in a three-phase transformer bank is given by

IL,base = baseL

base

VS

,3

The application of the per-unit system to three-phase transformer problems is similar to its application in the single-phase examples already given. Example 3.8. A 50-kVA 13,800/208-V Δ-Y distribution transformer has a resistance of 1 Percent and a reactance of 7 percent per unit. (a) What is the transformer’s phase impedance referred to the high-voltage side? (b) Calculate this transformer’s voltage regulation at full load and 0.8 PF lagging, using the

calculated high-side impedance. (c) Calculate this transformer’s voltage regulation under the same conditions, using the per-

unit system.


LABTECH 50

Solution (a) The high-voltage side of this transformer has a base line voltage of 13,800 V and a base apparent power of 50 kVA. Since the primary is A-connected, its phase voltage is equal to its line voltage. Therefore, its base impedance is

Zbase = ( )

base

base

SV 2

,3 φ

= ( )000.50800.133 2

= 11.426Ω The per-unit impedance of the transformer is Zeq = 0.01 + jO.07 pu so the high-side impedance in ohms is Zeq = Zeq,PUZbase = (0.01 + j0.07PU)(11.426Ω) = 114.2 + j800Ω (b) To calculate the voltage regulation of a three-phase transformer bank, determine the

voltage regulation of any single transformer in the bank. The voltages on a single transformer are phase voltages, so

VR = %100xaV

aVV

S

SP

φ

φφ −

The rated transformer phase voltage on the primary is 13.800 V, so the rated phase current on the primary is given by

IØ = φV

S3

The rated apparent power S = 50 kVA, so IØ =

)800.13(3000.50

VVA

= 1.208 A


LABTECH 51

The rated phase voltage on the secondary of the transformer is 208 V/√3 = 120V. When referred to the high-voltage side of the transformer, this voltage becomes VØS = aVØS = 13.800 V. Assume that the transformer secondary is operating at the rated voltage and current, and find the resulting primary phase voltage: T

ØP = aVØS +ReqIØ + jXeqIØ =13,800∠ 0ºV+(114.2Ω)(1.208∠ -6.87ºA)+(j800 Ω)(1.208∠ -36.87ºA) =13,800 + 138 ∠ -36.87º + 966.4 ∠ 53.13º =13,800 + 110.4 -j82.8 + 579.8 +j773.1 =14,490 +j690.3 =14,506∠ 2.73º V Therefore,

VR = %100xaV

aVV

S

SP

φ

φφ −

= %100800.13

800.13506.14 x−

= 5.1 % (c) In the per-unit system, the output voltage is 1∠ 0º, and the current is 1 ∠ -36.87º.

Therefore, the input voltage is VP = 1∠ 0º+ (0.01)(1∠ -36.87º) + (j0.07)(1∠ -36.87º) = 1 + 0.008 - j0.006 + 0.042 + j0.056 = 1.05 +j0.05 = 1.051∠ 2.730 The voltage regulation is

VR = %1000.1

0.1051.1 x−

= 5.1% Of course, the voltage regulation of the transformer bank is the same whether the calculations are done in actual ohms or in the per-unit system.


LABTECH 52

3.10. SUMMARY A transformer is a device for converting electric energy at one voltage level to electric energy at another voltage level through the action of a magnetic field. It plays an extremely important role in modem life by making possible the economical long-distance transmission of electric power. When a voltage is applied to the primary of a transformer, a flux is produced in the core as given by Faraday’s law. The changing flux in the core then induces a voltage in the secondary winding of the transformer. Because transformer cores have very high permeability, the net magnetomotive force required in the core to produce its flux is very small. Since the net magnetomotive force is very small, the primary circuit’s magnetomotive force must be approximately equal and opposite to the secondary circuit’s magnetomotive force. This fact yields the transformer current ratio. A real transformer has leakage fluxes that pass through either the primary or the secondary winding, but not both. In addition there are hysteresis, eddy current, and copper losses. These effects are accounted for in the equivalent circuit of the transformer. Transformer imperfections are measured in a real transformer by its voltage regulation and its efficiency. The per-unit system of measurement is a convenient way to study systems containing transformers, because in this system the different system voltage levels disappear. In addition, the per-unit impedances of a transformer expressed to its own ratings base fall within a relatively narrow range, providing a convenient check for reasonableness in problem solutions. An autotransformer differs from a regular transformer in that the two windings of the autotransformer are connected. The voltage on one side of the transformer is the voltage across a single winding, while the voltage on the other side of the transformer is the sum of the voltages across both windings. Because only a portion of the power in an autotransformer actually passes through the windings, an autotransformer has a power rating advantage compared to a regular transformer of equal size. However, the connection destroys the electrical isolation between a transformer’s primary and secondary sides. The voltage levels of three-phase circuits can be transformed by a proper combination of two or three transformers. Potential transformers and current transformers can sample the voltages and currents present in a circuit. Both devices are very common in large power distribution systems.


LABTECH 53

4

EXPERIMENT


LABTECH 54

SINGLE PHASE TRANSFORMER

4.1. POLARITY OF TRANSFORMER

OBJECTIVE After completing this experiment, you will be able to determine the polarity of the transformer.

EQUIPMENT 1. Single Phase Transformer unit (LFT-TET-P1A)

2. Single Phase Transformer unit (LFT-TET-P1B)

3. 1.5 V Battery

4. Push Button Switch.

5. Analog DC Voltmeter

6. Connection Leads

DISCUSSION

The terms primary and secondary refer to source and load sides, respectively (i.e., energy flows from primary to secondary). However, in many applications energy can flow either way, in which case the distinction is meaningless. The terms step up and step down refer to what the transformer does to the voltage from source to load. ANSI standards require that for a two-winding transformer the high-voltage and low-voltage terminals be marked as H1-H2 and X1-X2, respectively, with H1 and X1 markings having the same significance as dots for polarity markings.

Additive and subtractive transformer polarity refer to the physical positioning of high-voltage, low-voltage dotted terminals as shown in Figure 4.1-1. If the dotted terminals are adjacent, then the transformer is said to be subtractive, because if these adjacent terminals (H1-X1) are connected together, the voltage between H2 and X2 is the difference between primary and secondary. Similarly, if adjacent terminals X1 and H2 are connected, the voltage (H1-X2) is the sum of primary and secondary values.

Figure 4.1-1. Transformer polarity (a) Subtractive (b) Additive


LABTECH 55

If the primary voltage is positive at the dotted end of the winding with respect to the undotted end, then the secondary voltage will be positive at the dotted end also. Voltage polarities are the same with respect to the dots on each side of the core.

If the primary current of the transformer flows into the dotted end of the primary winding, the secondary current will flow out of the dotted end of the secondary winding.

PROCEDURES: 1. Prepare all equipment.

2. Build the circuit as shown in Figure 4.1-2a.

240V

220V

0

3A

6A

12A

A

B

C

I

D

E

F

G

H

42V

24V

12V

0

0

0

PB

1.5Vdc

+

+ -DCV

(a)

120V

240V

220V

0

5A

5A

120V

0

0

A

B

C

D

E

F

G

PB

1.5Vdc

+

+ -DCV

(b)

Figure 4.1-2. Polarity Testing

3. Observe Figure 4.1-2a. It shows that terminal A is given (-) polarity and terminal C to (+) polarity. On the secondary winding, terminal E and D are connected to DC voltmeter.

4. Press PB switch momentary while observing the DC voltmeter. What is the response of the voltmeter? Fill the result into table 4.1-1.


LABTECH 56

Table 4.1-1. Polarity Test

No Point Voltmeter deviation (Left/Right) Remark

1 E-D

2 G-F

3 I-H

5. Repeat step 3 and 4. Change the DC voltmeter to G-F and I-H respectively. Write down the result into table 4.1-1.

6. Return the circuit configuration as Figure 4.1-2a.

7. Change the battery polarity to opposite direction. Battery (+) to A and battery (-) to C.

8. Press PB switch momentary while observing the DC voltmeter. What is the response of the voltmeter? Write down the result.

9. Let’s move to Figure 4.1-2b. It shows a single phase transformer as well, but has larger secondary power and voltage ratings. Terminal A is given (-) polarity and terminal C to (+) polarity. On the secondary winding, terminal E and D are connected to DC voltmeter.

10. Press PB switch momentary while observing the DC voltmeter. What is the response of the voltmeter? Fill the result into table 4.1-2.

Table 4.1-2. Polarity Test

No Point Voltmeter deviation (Left/Right) Remark

1 E-D

2 G-F

11. Repeat step 9 and 10. Change the DC voltmeter to G-F. Write down the result into table 4.1-2.

12. Change the battery polarity to opposite direction. Battery (+) to A and battery (-) to C.

13. Press PB switch momentary while observing the DC voltmeter. What is the response of the voltmeter? Write down the result.

14. Draw the transformer block diagram above and mark the polarity based on your experiment result.

15. Remove the jumper connection after completing the experiments. Return the equipment to their respective place.


LABTECH 57

PRIM

ARY

+ 10V/1A

16V/1A

6V/2A

-

+-

PRIM

ARY

PRIM

ARYPower

Source220VAC

PRIM

ARY

+ 10V/1A

4V/1A

6V/2A

-

+-

PRIM

ARY

PRIM

ARYPower

Source220VAC

4.2. SERIES CIRCUIT IN A MULTIPLE WINDING TRANSFORMER.

OBJECTIVE

After completing this experiment you will able to connect a transformer in series winding so that it will either aid or oppose each other.

EQUIPMENT

1. Single Phase- Transformer Experimental Trainer base station. (LFT-TET-B)

2. Single-phase transformer (LFT-TET-P1A)

3. Connection Leads

DISCUSSION

Some transformers are made with more than one primary as well as more than one secondary. Multiple transformer windings can be connected in series or in parallel to change the voltage or current capabilities. Either primary or secondary windings (or both) can be connected in series or in parallel.

For parallel connections, windings should have identical ratings. They must have identical voltage ratings. Before windings are connected in parallel, their phasing must be correct Phasing refers to the instantaneous polarity of the windings. When the secondary winding has identical voltage rating the negative ends of the two secondary windings are connected together, as are the positive ends. With this connections the output voltage is still but the current will double.

Incorrect phasing in parallel winding causes a dead short on the secondaries of the transformer. The voltage of each winding aids the other winding in producing secondary current. Only the resistance of the secondary winding limits the secondary current. Therefore the current becomes very high. If not protected against overload the transformer will soon burn out.

Transformer windings also can be connected in series so that they either aid or oppose each other,

Figure 4.2-1. (a). Series aiding winding connections (b) Series opposing winding connections


LABTECH 58

In Figure 4.2-1 the polarity markings indicate the instantaneous polarities for one half cycle. In the series-aiding configuration (Fig 4.2-1a), the output voltage is the sum of the two secondary voltages. However the current is restricted to the lower rating of the two windings. This because all of the load current flows through both secondaries.

When connected in a series-opposing configuration the two winding produce a voltage equal to the difference between the two voltages. We say the two voltages are bucking one another. Note that the current is limited to the lower current rating of the two windings.

In summary, it is possible to obtain four voltages from two secondary windings:

- The voltage of secondary 1

- The voltage of secondary 2

- The sum of the voltages of secondaries 1 and 2

- The difference between the voltages of secondaries 1 and 2

PROCEDURES:

1. Prepare all equipment.

2. Ensure the power is OFF. Make circuit connections as shown in Figure 4.2-2.

3. Ask your instructor to recheck your wiring.

240V

220V

0

3A

6A

12A

A

B

C

I

D

E

F

G

H

42V

24V

12V

0

0

0

RCCB

MCB

N

L

GND

GND

V1

V2

CAUTION: HIGH VOLTAGE!!

Figure 4.2-2. Series aiding winding connections

4. Turn on the RCCB and MCB.

5. Measure the voltage at the primary and secondary points. Write down the result into table 4.2-1.

6. Turn off the MCB and RCCB.

7. Remove the secondary connection leads.


LABTECH 59

8. Connect point G-H.

9. Turn on the power.

10. Measure the voltage at point F-I. Write down the result into table 4.2-1. Table 4.2-1. Series aiding winding connections

Step no V1 (Vac) V2 (Vac)

5

10

11. Turn off the power. Remove the secondary connection leads.

12. Build the circuit as shown in Figure 4.2-3.

13. Ask your instructor to recheck your circuit.

240V

220V

0

3A

6A

12A

A

B

C

I

D

E

F

G

H

42V

24V

12V

0

0

0

RCCB

MCB

N

L

GND

GND

V1

V2


Figure 4.2-3. Series opposing winding connections


15. Measure the voltage (V1) and (V2). Fill the result into table 4.2-2.

16. Turn off the power.

17. Change the secondary connections. Connect terminal G to terminal I, and voltmeter (V2) connected to terminal H-F.

18. Turn on the power. Measure the voltage (V1) and (V2). Fill the result into table 4.2-2.




LABTECH 60

Table 4.2-2. Series opposing winding connections

Step no V1 (Vac) V2 (Vac)

15

18


LABTECH 61

4.3. OPEN CIRCUIT TESTING

OBJECTIVE

After completing this exercise you will be able to measure and determine the excitation impedance of a transformer.

EQUIPMENT

1. Single Phase Transformer Experimental Trainer base station (LFT-TET-B)

2. Single phase transformer (LFT-TET-P1A)

3. Single phase transformer (LFT-TET-P1B)

4. Connection Leads

BASIC INFORMATION

It is possible to experimentally determine the values of the inductances and resistances in the transformer model. An adequate approximation of these values can be obtained with only two tests, the open-circuit test and the short-circuit test.

In the open-circuit test, a transformer secondary winding is open-circuited, and its primary winding is connected to a full-rated line voltage. Look at the equivalent circuit in Figure 4.3-1. Under the conditions described, all the input current must be flowing through the excitation branch of the transformer. The series elements RP and XP are too small in comparison to RC and XM to cause a significant voltage drop, so essentially all the input voltage is dropped across the excitation branch.

Figure 4.3-1 a) The transformer model referred to its primary voltage level.

b) The transformer model referred to its secondary voltage level.


LABTECH 62

The open-circuit test connections are shown in Figure 4.3-2. Full line voltage is applied to the primary of the transformer, and the input voltage, input current, and input power to the transformer are measured. From this information, it is possible to determine the power factor of the input current and therefore both the magnitude and the angle of the excitation impedance.

The easiest way to calculate the values of RC and XM is to look first at the admittance of the excitation branch. The conductance of the core-loss resistor is given by

Gc = CR

1

and the susceptance of the magnetizing inductor is given by

BM = MX

1

Since these two elements are in parallel, their admittances add, and the total excitation admittance is

YE = Gc - jBM

= CR

1 – MX

j 1

Figure 4.3-2. Connection for transformer open-circuit test.

The magnitude of the excitation admittance (referred to the primary circuit) can be found from the open-circuit test voltage and current:

EY = OC

OC

VI

The angle of the admittance can be found from a knowledge of the circuit power factor. The open-circuit power factor (PF) is given by

PF = cos θ


LABTECH 63

=OCOC

OC

IVP

and the power-factor angle θ is given by

θ = cos –1

OC

OC

VP

The power factor is always lagging for a real transformer, so the angle of the current always lags the angle of the voltage by θ degrees. Therefore, the admittance YE is

YE = OC

OC

VI ∠ -θ

= OC

OC

VI ∠ cos –1 PF

PROCEDURES 1. Prepare all equipment required.

2. Ensure the power is OFF. Build circuit as shown in Figure 4.3-3.

240V

220V

0

3A

6A

12A

A

B

C

I

D

E

F

G

H

42V

24V

12V

0

0

0

RCCB

MCB

N

L

GND

GND

V1 V

WATTMETER

AA1


Figure 4.3-3. Open circuit testing


4. Turn on the RCCB and MCB.

5. Observe the voltage, current and power

Primary Voltage (Voc) = ……….V

Primary Current (Ioc) = ……….A

Power (Poc) = ……….W


LABTECH 64

6. Calculate the power factor of the transformer

PF = cos θ

=OCOC

OC

IVP

7. Calculate the admittance of the transformer

Y = OC

OC

VI ∠ cos –1 PF

8. Turn off the power and remove the connection leads.

9. Build a circuit as shown in Figure 4.3-4. It uses the single phase transformer (LFT-TET-P1B).



12. Measure the voltage, current and power.

Primary Voltage (Voc) = ……….V

Primary Current (Ioc) = ……….A

Power (Poc) = ……….W

RCCB

MCB

N

L

GND

V1 V

WATTMETER

AA1

GND

120V

240V

220V

0

5A

5A

120V

0

0

A

B

C

D

E

F

G

CAUTION : HIGH VOLTAGE!!

Figure 4.3-4. Open circuit testing

13. Calculate the power factor of the transformer

PF = cos θ

=OCOC

OC

IVP

14. Calculate the admittance of the transformer


LABTECH 65

Y = OC

OC

VI ∠ cos –1 PF




LABTECH 66

4.4. SHORT CIRCUIT TESTING

OBJECTIVE

After completing this experiment you will able determine the current ratio and impedance of a transformer

EQUIPMENT

1. Single Phase Transformer Experimental Trainer base station (LFT-TET-B)


3. Single phase transformer (LFT-TET-P1B)

4. Variable Power Supply

5. Connection Leads

DISCUSSION In the short-circuit test, the secondary terminals of the transformer are short-circuited, and the primary terminals are connected to a fairly low-voltage source, as shown in Figure 3.15. The input voltage is adjusted until the current in the short-circuited windings is equal to its rated value. (Be sure to keep the primary voltage at a safe level. It would not be a good idea to burn out the transformer’s windings while trying to test it.) The input voltage, current, and power are again measured.

IIIIIII

Figure 4.4-1. Short circuit testing

Since the input voltage is so low during the short-circuit test, negligible current flows through the excitation branch. If the excitation current is ignored, then all the voltage drop in the transformer can be attributed to the series elements in the circuit. The magnitude of the series impedances referred to the primary side of the transformer is

SEZ = SC

SC

IV

The power factor of the current is given by

PF = cos θ


LABTECH 67

=SCSC

SC

IVP

and is lagging. The current angle is thus negative, and the overall impedance angle θ is positive:

θ = cos-1 SCSC

SC

IVP

PROCEDURE

1. Prepare all equipment required.

2. Ensure the power and meter switch are OFF. Build circuit as shown in Figure 4.4-2.

240V

220V

0

3A

6A

12A

A

B

C

I

D

E

F

G

H

42V

24V

12V

0

0

0

RCCB

MCB

N

L

GND

GND

V1 V

WATTMETER

AA1

AUTOTRANSFORMER

A2


Figure 4.4-2. Short circuit testing.

3. Set the autotransformer to minimum position.

4. Turn on the RCCB and MCB. Wait a moment and then turn on the meter switches.

5. Adjust the autotransformer gently until the ammeter (A2) shows 6 Amperes.

CAUTION: DO NOT EXCEED THE MAXIMUM CURRENT RATING!!

6. Read the meter measurement.

V1 (Vsc) = ……….. Vac

A1 (Isc) = ……….. Vac

W (Psc) = ……….. W

A2 = 6 A

7. Set the autotransformer to minimum position and turn off the power and meter switches. Remove the cable jumper connections.

8. Prepare the single phase transformer unit LFT-TET-P1B.


LABTECH 68

9. Build the circuit as shown in Figure 4.4-3. RCCB

MCB

N

L

GND

V1 V

WATTMETER

AA1

GND

120V

240V

220V

0

5A

5A

120V

0

0

A

B

C

D

E

F

G

A2


Figure 4.4-3. Short circuit testing.


11. Adjust the autotransformer gently until the ammeter (A2) show 5 Ampere. Read the meter measurement.

V1 (Vsc) = ……….. Vac

A1 (Isc) = ……….. Vac

W (Psc) = ……….. W

A2 = 5 A

12. Set the autotransformer to minimum position and turn off the power and meter switches. Remove the cable jumper connections.


LABTECH 69

4.5. TURNS RATIO OF PRIMARY AND SECONDARY WINDING OF A TRANSFORMER.

OBJECTIVES After completing this experiment you will able to determine the voltage values according to the turns ratio and calculate the turn ratio by determining the voltage values.

EQUIPMENT 1. Single Phase- Transformer experimental Trainer base station. (LFT-TET-B)


3. Connection Leads

4. Variable Transformer

DISCUSSION

A transformer can either step up or step down a voltage. If the primary voltage is greater than the secondary voltage, the transformer is stepping the voltage down. If the voltage in the secondary exceeds the voltage in the primary the transformer is a step-up transformer. Some transformers with multiple secondaries have one or more step-up or step-down secondaries.

Whether a secondary is a step-up or step down winding is determined by the primary to secondary turns ratio. When the primary turns exceed the secondary turns and the coupling is 100 percent, the transformer steps down the voltage. In fact, with 100 percent coupling, the turns ratio and the voltage ratio are equal. Mathematically we can write:

Vpri Npri _______ = ________ Vsec Nsec

In this formula, N is the abbreviation for the number of turns. This formula can be rearranged to show that:

Npri Nsec _______ = ________ Vpri Vsec

In this new arrangement, the relationship between voltage and turns is very informative. It shows that the turns-per-volt ratio is the same for both the primary and the secondary. Furthermore, the turns-per-volt ratios of all secondary windings in a multiple-secondary transformer are equal. Thus, once you determine the turns-per-volt ratio of any winding, you know the ratios of all other windings.

In this experiment we will use a single phase step-down transformer that consist of a primary winding and three secondary windings with the following specifications:


LABTECH 70

Primary: Input voltage : 240 Volt

Turns : 377

Φ : 1 mm

Connection Terminals : A - C

Secondary: Output voltage : 42 Volt Turns : 67 Max. current : 3 Amp. Φ : 1.25 mm Connection Terminals : D - E

Output voltage : 24 Volt Turns : 36 Max. current : 6 Amp. Φ : 1.8 mm Connection Terminals : F - G

Output voltage : 12 Volt

Turns : 18 Max. current : 6 Amp. Φ : 2.5 mm Connection Terminals : H - I

PROCEDURES:

1. Prepare all equipment.

2. Ensure the RCCB, MCB and meter switches are OFF. Build a circuit as shown in Figure 4.5-1.


4. Adjust autotransformer to minimum.

5. Turn on the RCCB and MCB. Wait a moment then turn on the Voltmeter switch (V1) and Ammeter switch (A1).

6. Observe the Voltmeter and Ammeter reading and record the result in Table 4.5-1.

7. By using Voltmeter (V2) measure the voltage on terminals D - E. Observe and record the result in Table 4.5-1. Also measure the voltage on terminal F – G and H – I.


LABTECH 71

240V

220V

0

3A

6A

12A

A

B

C

I

D

E

F

G

H

42V

24V

12V

0

0

0

RCCB

MCB

N

L

GND

GND

V1

A1

AUTOTRANSFORMER

V2


Figure 4.5-1. Turn ratio experiment

8. Increase primary input voltage by adjusting the variable Transformer output until voltmeter V1 shows 50Vac, and then measure the voltage on terminals D – E; F – G; and H – I. Nsec Vsec = _______ x Vpri

Npri

Note:

Vsec = Secondary voltage on points; D – E; F – G; and H – I.

Vpri = Primary voltage on point A-C

Nsec = Secondary winding

N pri = Primary winding

9. Repeat step 8 above, for primary input voltage: 100V; 150V; 200V and 220V. Record the result in Table 4.5-1

10. Set the autotransformer to minimum position, turn off the power and meter switches. Remove the cable jumper connections.


LABTECH 72

Table 4.5-1. Turn ratio measurement

Measurement (Volt) Primary Secondary

Calculation (Volt) Variable Transformer

Output A - C D - E F - G H - I D - E F - G H - I

0V

50V

100V

150V

200V

220V


LABTECH 73

4.6. IMPEDANCE TRANSFORMATION

OBJECTIVES

In this exercise you will practice how a transformer can change the value of a resistance

EQUIPMENT

1. Single Phase- Transformer experimental Trainer base station (LFT-TET-P1B)


3. Connection Leads

4. Variable load unit (LFT-TET-01)

DISCUSSION

For ideal transformer, there is no exciting current, and the resistance of its windings should be zero. Also it will no leakage flux, and leakage reactance. The voltage on the secondary terminals E2 is given by the voltage ratio equation:

Es = (Ns / Np) Ep

Since it is no exciting current, the impedance between terminals A – B is infinite. This corresponds to the infinite impedance between secondary terminals C – D.

If a resistor R is connected across the secondary terminals C and D, the impedance between the primary terminals A and B is no longer infinite. It is equal to the value R multiplied by the square of the turns ratio.

Zp = (Np / Ns)2 R

This is an example of impedance transformation. When the resistance R is connected to the secondary terminals, the resulting secondary current Is is:

Is = (Es / R)

The secondary current causes a current flow in the primary given by the current ratio equation:

Ip = (Ns / Np) Is

The impedance ZAB seen between terminals A and B is:

ZAB = (Ep / Ip)

We see that the impedance between the primary terminals is equal to the resistance across the secondary terminals multiplied by the square of the turns ratio. Thus the transformer has the remarkable property of being able to transform the value of a resistance.


LABTECH 74

PROCEDURES 1. Prepare all equipment required in this experiment.

2. Ensure the RCCB and MCB and meter switches are OFF.

3. Build circuit as shown in Figure 4.6-1.

240V

220V

0

3A

6A

12A

A

B

C

I

D

E

F

G

H

42V

24V

12V

0

0

0

RCCB

MCB

N

L

GND

GND

V1

V2

A1

A2


Figure 4.6-1. Impedance measurement.

4. Calculate the theoretical value of the impedance that should appear between terminals A and B when the load resistor = 15 Ω :

Zp = (Np / Ns)2 R = ………… Ω

5. Set the variable load unit (Rheostat) to 15 Ω.

Note: You are not recommended to turn on the power while the rheostat is set to minimum position (below 1Ω). It will burn the rheostat unit.

6. Turn on the RCCB and MCB. Wait a moment then turn on all of the meter switches.

7. With load = 15 Ω measure primary current on ammeter A1. Calculate the value of the impedance Zp that appears across terminal A and B:

Zp = Vp/Ip = …….. Ω

8. With the same way as you have done on above steps, change the load as shown in Table 4.6-1 and write your measurement results.

9. After completing this experiment, turn off the MCB and RCCB. Remove cable connections and return the equipment to their place.


LABTECH 75

Table 4.6-1

Primary Secondary

Ep Ip Zp Zp Es Is Zs Load (H-I)

(V1) (A1) (Ep/Ip) Calculation (V2) (I2) (Es/Is) Resistor 15 Ohm Resistor 20 Ohm Resistor 30 Ohm Resistor 50 Ohm Resistor 100 Ohm


LABTECH 76

4.7. VOLTAGE REGULATION

OBJECTIVES

After completing this experiment, you will be able to measure the voltage regulation of a transformer

EQUIPMENT

1. Single Phase- Transformer experimental Trainer base station (LFT-TET-B)


3. Single-phase transformer. (LFT-TET-P1B)

4. Connection Leads

5. Variable load unit (LFT-TET-01)

DISCUSSION

The measure of how well a power transformer maintains constant secondary voltage over a range of load currents is called the transformer's voltage regulation. It can be calculated from the following formula:

“Full-load” means the point at which the transformer is operating at maximum permissible secondary current. This operating point will be determined primarily by the winding wire size and the method of transformer cooling.

Incidentally, this would be considered rather poor (or “loose”) regulation for a power transformer if less than 3%. Inductive loads tend to create a condition of worse voltage regulation, so this analysis with purely resistive loads was a “best-case” condition.

PROCEDURE

1. Prepare all equipment required in this experiment.




LABTECH 77

240V

220V

0

3A

6A

12A

A

B

C

I

D

E

F

G

H

42V

24V

12V

0

0

0

RCCB

MCB

N

L

GND

GND

V1

V2

A1

A2


Figure 4.7-1. Voltage regulation measurement

4. Set the variable load unit (Rheostat) knob to middle position.

Note: You are not recommended to turn on the power while the rheostat is set to minimum position (below 1Ω). It will burn the rheostat unit.

5. Turn on the RCCB and MCB. Wait a moment then turn on the meter switches.

6. Set the rheostat until the ammeter (A2) show 3A. Read the voltage at voltmeter (V2).

V2 : ……………Vac (Full-load voltage)

7. Set the rheostat to maximum (500 Ω). This condition can be said no load condition, because the current will be very small. Read the voltage at voltmeter (V2)

V2 : ……………Vac (No-load voltage)

8. Calculate the regulation percentage based on your experiment data.

Vfull-load – Vno-load Regulation percentage = x 100 % Vno-load

9. After completing this experiment, turn off the MCB and RCCB. Remove cable connections and return the equipment to their place.


LABTECH 78

4.8. FULL WAVE RECTIFIER

OBJECTIVE

In this experiment you will practice how measure voltage characteristic of a fullwave rectifier with and without load.

EQUIPMENT

1. Single Phase- Transformer experimental Trainer base station (LFT-TET-B)


3. Single-phase transformer (LFT-TET-P1B)

4. Connection Leads

5. DC Voltmeter

6. Oscilloscope

DISCUSSION Figure 4.8-1 is the circuit diagram of a transformer-fed bridge rectifier. The high-voltage secondary winding of transformer T supplies four silicon rectifiers, D1 to D4. Operation of the circuit is as follows: Assume that during the positive alternation (alternation 1) of the input sine wave, point C is positive with respect to D (the voltages at the opposite ends of a transformer winding are 180° out of phase). This makes the anode of D1 positive with respect to its cathode and D1is therefore forward-biased. Similarly the cathode of D3, connected to point D, is negative relative to its anode. Hence, D3 is forward-biased. It is evident also that D2 and D4 are reverse-biased during alternation 1. Thus, in a circuit Dl and D3 will conduct during alternation 1 while D2 and D4 will be cut off.

Figure 4.8-1. Full wave rectifier

Figure 4.8-2a shows that during the positive alternation there is a complete path for current for rectifiers D1 and D3, which are connected in series with the load resistor RL. Current flows through RL, through D1; through winding CD, and through D3, with the polarity shown.

Figure 4.8-2b shows the positive-voltage waveform developed during alternation 1 across RL. During the negative alternation (alternation 2), D1 and D3 are reverse-biased and are cut off. If D2 and D4 were not in the circuit, D1 and D3 would act as a half-wave rectifier.


LABTECH 79

Figure 4.8-2c shows that during the negative alternation (alternation 2), that is, when point C is negative relative to point D, the anode of D2 is positive with respect to its cathode, and the cathode of D4 is negative with respect to its anode. Hence, rectifiers D2 and D4 are forward-biased, while D1 and D3 are reverse-biased.

Figure 4.8-2. Action of bridge rectifier (a, b) on positive alternation;(c,d) on negative alternation

Now D2 and D4 conduct, permitting current through RL. The polarity across RL is the same as in Figure 4.8-2d.

Thus D1 in series with D3 rectifies during the positive alternation of the input, while D2 in series with D4 rectifies during the negative alternation. A bridge rectifier is therefore a full-wave rectifier. The center tap (CT) of the secondary is not connected in the bridge rectifier. In a conventional circuit rectifier, the CT acts as the common return, and the voltage across each diode is one-half the voltage across the transformer. Hence, if the same transformer is used, the output voltage of a conventional full-wave rectifier is only one-half that of a bridge circuit.

The same type of filter arrangement can be used with a bridge rectifier as with any other rectifier circuit. For the bridge rectifier the voltage rating of the filter capacitors must be at least twice for the full-wave rectifier using the same transformer.

PROCEDURE 1. Prepare all equipment required in this experiment.



4. Set the rheostat to 20 Ω.



LABTECH 80

6. Observe the waveform at the input and output of the bridge rectifier. Compare the waveform. Draw the waveform into table 4.8-1.

7. Read the voltage and current measurement at V2, A2, and DC Voltmeter (Vdc). Fill the result into table 4.8-1.

8. Repeat step 6-7 for other load values as listed on table 4.8-1.

240V

220V

0

3A

6A

12A

A

B

C

I

D

E

F

G

H

42V

24V

12V

0

0

0

RCCB

MCB

N

L

GND

GND

V1

V2

A1

A2

VDC

OSCILLOSCOPE

Figure 4.8-3. Fullwave rectifier

Table 4.8-1

LOAD A2 (A) V2 (Vac) VDC Waveform

20 Ω

25 Ω

50 Ω

100 Ω

9. Turn off the RCCB, MCB and meter switches.

10. Assemble the circuit as shown on Figure 4.8-4.


12. Read the voltage and current measurement at V2, A2, and DC Voltmeter (VDC). Fill the result into table 4.8-1.

13. Repeat step 12 for other load values as listed on table 4.8-2.


LABTECH 81

V2

A2

VDC

RCCB

MCB

N

L

GND

V1

A1

GND

120V

240V

220V

0

5A

5A

120V

0

0

A

B

C

D

E

F

G

Figure 4.13. Rectifier with load

Table 4.8-2

LOAD A2 (A) V2 (Vac) VDC

120 Ω

150 Ω

200 Ω

500 Ω


LABTECH 82

THREE PHASE TRANSFORMER

4.9. Y-Δ CONNECTION

OBJECTIVE

After completing this experiment you will be able to construct three phase transformer and able to verify the voltage and current relationships in a wye-delta transformer.

EQUIPMENT 1. Three Phase Transformer Experimental Trainer base station (LFT-TET-B)

2. Three Phase Transformer Unit (LFT-TET-P3)

3. Three Phase Load Unit/Rheostat (LFT-TET-06)

4. Connection Leads

DISCUSSION

Three phase voltages can be transformed either by a single three phase transformer or by three single phase transformer. The end is the same, all three of the phase voltages are changed.

The structure of a three-phase transformer is illustrated in Figure 4.9-1.

Primaryphase 1

Primaryphase 2

Primaryphase 3

Secondaryphase 1

Secondaryphase 2

Secondaryphase 3

AB

G

H

I

J

K

L

CD

EF

Figure 4.9-1. Illustration of a 3 phase transformer construction

The flux in the phase 1 lag is equal to the phase 2 flux plus phase 3. Phase 2 flux equals phase 3 flux plus phase 1 flux etc. This is caused by the flux, like the current, of each phase is displaced by 120°.

In Figure 4.9-2 the flux of each phase is assumed to be in step with the current in the phase. (This assumption implies that the core has no hysteresis loss). At instant A, the phase 1


LABTECH 83

Primary Secondary

Primary

Secondary

current is zero. Therefore, the phase 1 flux is also zero. At instant B, current in phase 1 and phase 3 are both positive. At this same time phase 2 current is negative and of twice the value of either phase 1 current or phase 3 current. Thus the phase 2 lag of the core has twice as much flux as either phase 1 or phase 3.

Phase 1

A B C D

Time

Curre

nt

Phase 2

Phase 3

Figure 4.9-2. Core flux in a three phase transformer.

The primary and secondary windings of a three phase transformer may be either wye connected or delta connected. The secondary does not have to have the same configuration (wye or delta) as the primary. Figure 4.9-3 shows four possibility ways to connect a three phase transformer. The dots on one end of each winding indicate the beginning of each winding. Notice that all windings are wound in the same direction (counterclockwise) when you start at the dotted end of the winding. Identifying the start of the windings is necessary before they can be properly phased.

a) Delta-delta (Δ −Δ)

b) Delta – wye (Δ − Y)

Secondary

Primary

c) Wye – wye (Y − Y)


LABTECH 84

Primary

Secondary

d) Wye - Delta (Y − Δ)

Figure 4.9-3. Three phase transformer connections

The diagram in Figure 4.9-3 shows four ways of connecting the windings to obtain correct phasing. With a wye connection, correct phasing can also be obtained by connecting all the dotted ends to the star point. In the delta connection all three windings can be reversed; just be sure that two dotted ends are not connected together. On a transformer, the dotted (start) of a winding is identified by the manufacturer. The identification may be made in several ways. It may be a colored strip wrapped around the insulation of the start lead or may be a number on a diagram mounted on the transformer.

Incorrect phasing of the primary, in either the wye or the delta configuration, causes excessively high primary current. If not protected against overload, the incorrectly phased primary can be destroyed by the excess current. Improper phasing of a delta connected secondary also causes excessive, destructive current.

The Y-Δ connection of three-phase transformers is shown in Figure 4.9-4. In this connection, the primary line voltage is related to the primary phase voltage by VLP = √3VθP, while the secondary line voltage is equal to the secondary phase voltage VLS = VθS. The voltage ratio of each phase is

S

P

VV

θ

θ = a

so the overall relationship between the line voltage on the primary side of the bank and the line voltage on the secondary side of the bank is

LS

LP

VV = 3

S

P

VV

θ

θ

S

P

VV

θ

θ = 3 a Y - Δ

Primary Secondarya c’

b b’

c a’

VLP

VLS

V Pθ

V Sθ

NP2NS2

NP3NS1

NS3

NP1


LABTECH 85

a c’

b b’

c

n

a’

V LP V Pθ V Sθ

NP2 NS2

NP3

NS1

NS3

NP1

PRIMARY SECONDARY

Figure 4.9-4.

The Y-Δ connection has no problem with third-harmonic components in its voltages, since they are consumed in a circulating current on the Δ side. This connection is also more stable with respect to unbalanced loads, since the Δ partially redistributes any imbalance that occurs.

PROCEDURES 1. Prepare all equipment required in this experiment.


3. Make connections between base station and transformer module as shown in Figure 4.9-5.



6. Measure the voltage and current as list on Table 4.9-1.


LABTECH 86

A

D

G

B

E

H

C

F

I

380V

380V

380V

415V

415V

415V

0

0

0

GND

GND

L1

L2

L3

N

A1

V1

V1

V1

V2

V2

V2

A1

A1

J

K

L

M

N

O

P

Q

R415V

415V

415V

41.5V

41.5V

41.5V

0

0

0

ILP1

ILP2

ILP3


Figure 4.9-5. Wye-Delta Configuration

Note: The ammeter (A1), voltmeter (V1, V2) are used interchangeable. Turn off the MCB when you change the ammeter and voltmeter position.

Table 4.9-1. Wye-Delta measurement result

Description Measurement Result

VLine pri

VF-I V

VC-F V

VC-I V

Vphase pri

VC-G V

VF-G V

VI-G V

ILine pri


LABTECH 87

ILP1 A

ILP2 A

ILP3 A

VLine sec

VJ-L V

VM-O V

VP-R V

7. Calculate the line-to line voltage at the primary winding.

VLLP = (VC-F + VF-I + VC-I) / 3 = ………….. V 8. Calculate the ratio of VLine and Vphase on the primary side.

VC-F / VC-G : ………… VF-I / VF-G : ………………

VC-I / VI-G : ………………

Does the comparison close to √3?

9. Measure the power which is drawn by the transformer without load. The wattmeter has been wired internally.

P = ……… W Note: The wattmeter has been defaulted to measure the power on the primary winding

10. Turn off the power and meter switches. Make circuit as shown in Figure 4.9-6.

A

D

G

B

E

H

C

F

I

380V

380V

380V

415V

415V

415V

0

0

0

GND

GND

L1

L2

L3

N

A1

V1

V1

V1

V2

A2

A2

A2

V2

V2

A1

A1

J

K

L

M

N

O

P

Q

R415V

415V

415V

41.5V

41.5V

41.5V

0

0

0

ILP1

VLP1

VLP2

VLP3

ILP2

ILP3

ILS1

VLS1

VLS2

VLS3

ILS2

ILS3

Figure 4.9-6. Wye-delta connections with load


LABTECH 88

11. Set the Rheostat to 400 Ω. 12. Turn on the power. Wait a moment and then turn on the meter switches.

13. Measure the voltage and current as shown in Table 4.9-2. Table 4.9-2. Wye-Delta with load measurement result


VLS1 V

VLS2 V

VLS3 V

ILS1 A

ILS2 A

ILS3 A

IPS1 A

IPS2 A

IPS3 A

14. Turn off the RCCB, MCB and meter switches.

15. Make circuit as shown in Figure 4.9-7.

16. Turn on the power. Wait a moment and then turn on the meter switches.

17. Measure the current as shown in Table 4.9-2.

A

D

G

B

E

H

C

F

I

380V

380V

380V

415V

415V

415V

0

0

0

GND

GND

L1

L2

L3

N

J

K

L

M

N

O

P

Q

R415V

415V

415V

41.5V

41.5V

41.5V

0

0

0

IPS1

IPS2

IPS3

A2

A2

A2

Figure 4.9-7. Current measurement


LABTECH 89

18. Calculate ratio of the phase current and line current of the secondary winding.

ILS1 / IPS1 : ………

ILS2 / IPS2 : ………

ILS2 / IPS2 : ………

Does the ratio close to √3 ?

Note that in the delta connections, Vline will be same with the Vphase.

19. Turn off the MCB and RCCB.

20. Remove all cable connections and return the equipment to their respective place.


LABTECH 90

4.10. Y-Y CONNECTION OBJECTIVE After completing this experiment you will be familiar with three phase transformer and able to verify the voltage and current relationships in a wye-wye transformer.

EQUIPMENT 1. Three Phase Transformer experimental Trainer base station (LFT-TET-B)

2. Three Phase Transformer (LFT-TET-P3) 3. Three Phase Load Unit (LFT-TET-06)

4. Connection Leads

5. Digital Multimeter

DISCUSSION

The Y-Y connection of three-phase transformers is shown in Figure 4.10-1. In a Y-Y connection, the primary voltage on each phase of the transformer is given by VøP = VLP / √3. The primary-phase voltage is related to the secondary-phase voltage by the turns ratio of the transformer. The phase voltage on the secondary is then related to the line voltage on the secondary by VLS = √3VØs. Therefore, overall the voltage ratio on the transformer is

SLS

LP

VPV

VV

θ

θ

33

= = a Y - Y

The Y-Y connection has two very serious problems: 1. If loads on the transformer circuit are unbalanced, then the voltages on the phases of the

transformer can become severely unbalanced. 2. Third-harmonic voltages can be large. If a three-phase set of voltages is applied to a Y—Y transformer, the voltages in any phase will be 120º apart from the voltages in any other phase. However, the third-harmonic components of each of the three phases will be in phase with each other, since there are three cycles in the third harmonic for each cycle of the fundamental frequency. There are always some third-harmonic components in a transformer because of the nonlinearity of the core, and these components add up. The result is a very large third-harmonic component of voltage on top of the 50- or 60-Hz fundamental voltage. This third-harmonic voltage can be larger than the fundamental voltage itself.


LABTECH 91

Figure 4.10-1. Three-phase transformer connections and wiring diagrams Y-Y Both the unbalance problem and the third-harmonic problem can be solved using one of two techniques: 1. Solidly ground the neutrals of the transformers, especially the primary winding’s neutral.

This connection permits the additive third-harmonic components to cause a current flow in the neutral instead of building up large voltages. The neutral also provides a return path for any current imbalances in the load.

2. Add a third (tertiary) winding connected in Δ to the transformer bank. If a third Δ -connected winding is added to the transformer, then the third-harmonic components of voltage in the A will add up, causing a circulating current flow within the winding. This suppresses the third-harmonic components of voltage in the same manner as grounding the transformer neutrals.

The Δ -connected tertiary windings need not even be brought out of the transformer case, but they often are used to supply lights and auxiliary power within the substation where it is located. The tertiary windings must be large enough to handle the circulating currents, so they are usually made about one-third the power rating of the two main windings. One or the other of these correction techniques must be used any time a Y—Y transformer is installed. In practice, very few Y—Y transformers are used, since the same jobs can be done by one of the other types of three-phase transformers.


LABTECH 92

PROCEDURES

1. Prepare all equipment required in this experiment.





A

D

G

B

E

H

C

F

I

380V

380V

380V

415V

415V

415V

0

0

0

GND

GND

L1

L2

L3

N

A1

V1

V1

V1

V1

V1

V1

V2 V2

V2 V2V2

V2

A1

A1

J

K

L

M

N

O

P

Q

R415V

415V

415V

41.5V

41.5V

41.5V

0

0

0

ILP1

VLP1

VLP2

VLP3

VPP1

VPP2

VPP3

ILP2

ILP3

VPS1 VLS1

VLS2 VLS3VPS2

VPS3

Figure 4.10-2. Y-Y Connections

6. Measure the voltage, current and power according to Table 4.10-1. Table 4.10-1. Y-Y Connections Measurement


VLP1 V

VLP2 V

VLP3 V

VPP1 V

VPP2 V

VPP3 V


LABTECH 93

VLS1 V

VLS2 V

VLS3 V

VPS1 V

VPS2 V

VPS3 V

ILP1 A

ILP2 A

ILP3 A

7. According to your result in step 5 and 6, calculate the ratio of:

VLP1 / VPP1 = ………….

VLP1 / VPP2 = ………….

VLP1 / VPP3 = ………….

VLS1 / VPS1 = ………….

VLS2 / VPS1 = ………….

VLS3 / VPS1 = ………….

Is the ratio close to √3 ?

8. Measure the power which is drawn by the transformer without load.

P = ………….. Watt Note: The wattmeter has been defaulted to measure the power on the primary winding

9. Calculate the average value of line to line voltage on the primary and secondary side

VLLP = (VLP1 + VLP1 + VLP1 ) / 3 = ………….. V

VLLS = (VLS1 + VLS2 + VLS3) / 3 = ………….. V

10. Turn off the power and meter switches. Make circuit connections as shown in Figure 4.10-3.

11. Set the rheostat to 500 Ω.

12. Ask your instructor to recheck your wiring. 13. Turn on the RCCB and MCB. Wait a moment and then turn on the meter switches.

14. Measure the voltage, current and power according to Table 4.10-2.


LABTECH 94

A

D

G

B

E

H

C

F

I

380V

380V

380V

415V

415V

415V

0

0

0

GND

GND

L1

L2

L3

N

A1

V1

V1

V1

V1

V1

V1

V2 V2

V2 V2

A2

A2

A2

V2

V2

A1

A1

J

K

L

M

N

O

P

Q

R415V

415V

415V

41.5V

41.5V

41.5V

0

0

0

ILP1

VLP1

VLP2

VLP3

VPP1

VPP2

VPP3

ILP2

ILP3

ILS1

VPS1 VLS1

VLS2 VLS3VPS2

VPS3

ILS2

ILS3

Figure 4.10-3. Y-Y Connections with Load

Table 4.10-2. Y-Y Connections with Load Measurement Result


VLP1 V

VLP2 V

VLP3 V

VPP1 V

VPP2 V

VPP3 V

VLS1 V

VLS2 V

VLS3 V

VPS1 V

VPS2 V

VPS3 V

ILP1 A

ILP2 A

ILP3 A

ILS1 A


LABTECH 95

ILS2 A

ILS3 A

15. Measure the power which is drawn by the transformer with load.

P = ……….Watt Note: The wattmeter has been defaulted to measure the power on the primary winding

16. Compare with your calculation.

P = √3 VLLP x ILLP

17. Turn off the RCCB and MCB. Remove the jumper cable connections.


LABTECH 96

4.11. THREE PHASE RECTIFIER

OBJECTIVE After completing this experiment you will be able to measure the characteristic of a three phase rectifier

EQUIPMENT 1. Three Phase Transformer experimental Trainer base station (LFT-TET-B)

2. Three Phase transformer (LFT-TET-P3)

3. Single Phase Load Unit (LFT-TET-01)

4. Connection Leads

5. Digital Multimeter

DISCUSSION Parallel connection via interphase transformers permits the implementation of rectifiers for high current applications. Series connection for high voltage is also possible, as shown in the full-wave rectifier of figure 4.14. With this arrangement, it can be seen that the three common cathode valves generate a positive voltage respect to the neutral, and the three common anode valves produce a negative voltage. The result is a dc voltage twice the value of the half wave rectifier. Each half of the bridge is a three-pulse converter group.

This bridge connection is a two-way connection, and alternating currents flow in the valve-side transformer windings during both half periods, avoiding dc components into the windings, and saturation in the transformer magnetic core. These characteristics made the also called Graetz Bridge the most widely used line commutated thyristor rectifier. The configuration does not need any special transformer, and works as a six-pulse rectifier.

Figure 4.11-1


LABTECH 97

PROCEDURE 1. Prepare all equipment required in this experiment.





A

D

G

B

E

H

C

F

I

380V

380V

380V

415V

415V

415V

0

0

0

GND

GND

L1

L2

L3

N

A1

V1

V1

V1

V2

A2

A2

A2

V2

V2

A1

A1

J

K

L

M

N

O

P

Q

R415V

415V

415V

41.5V

41.5V

41.5V

0

0

0

ILP1

VLP1

VLP2

VLP3

ILP2

ILP3

ILS1

VLS1

VLS2

VLS3

ILS2

ILS3

V1

V1

V1

VPP2

VPP1

VPP3

Vo

Figure 4.11-2.

6. Measure the voltage and current according to Table 4.11-1.

Table 4.11-1


VLP1 V

VLP2 V

VLP3 V

VPP1 V

VPP2 V

VPP3 V

VLS1 V

VLS2 V


LABTECH 98

VLS3 V

ILS1 A

ILS2 A

ILS3 A

*Note that VLS will be same with VPS

7. Measure the output of the bridge diode using dc voltmeter.

Vo = ………….. Vdc

8. Turn off the power and meter switches. 9. Connect output circuit in Figure 4.11-2 with single phase load unit as shown in Figure

4.11-3.

10. Set the load to 500 Ω.


A

D

G

B

E

H

C

F

I

380V

380V

380V

415V

415V

415V

0

0

0

GND

GND

L1

L2

L3

N

A1

V1

V1

V1

V2

A2

A2

A2

V2

V2

A1

A1

J

K

L

M

N

O

P

Q

R415V

415V

415V

41.5V

41.5V

41.5V

0

0

0

ILP1

VLP1

VLP2

VLP3

ILP2

ILP3

ILS1

VLS1

VLS2

VLS3

ILS2ILS1

ILS2

ILS2

ILS3

V1

V1

V1

VPP2

VPP1

VPP3

Vo

Rheo

stat

A2

A2

A2

Figure 4.11-3

12. Measure the voltage, current, and power. Write down into Table 4.11-2.

13. Change the load as list on Table 4.11-2. Repeat step 12.


LABTECH 99

Table 4.11-2.

Load Description

500Ω 400Ω 300Ω 250Ω

VLP1 V V V V

VLP2 V V V V

VLP3 V V V V

VPP1 V V V V

VPP2 V V V V

VPP3 V V V V

VLS1 V V V V

VLS2 V V V V

VLS3 V V V V

ILS1 A A A A

ILS2 A A A A

ILS3 A A A A

ILP1 A A A A

ILP2 A A A A

ILP3 A A A A

Vo Vdc Vdc Vdc Vdc

P W W W WNote: The wattmeter has been wired to measure the power on the primary winding

VLS will be same with VPS

14. Calculate the secondary current ratio

ILP1 / ILS1 : ………

ILP2 / ILS2 : ……… ILP3 / ILS3 : ………

Does the ratio close to √3?

15. Calculate the power and compare with measured power!

P = √3 x VLP x ILP

16. Turn off the RCCB and MCB. Remove the jumper cable connections.

3. Lft-tet, Transformer Trainer Exp Manual, 9082b, Imtac

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