UEE30811 – Certificate III Electrotechnology Electrician
March 2021
UNIT UEENEEG006A SOLVE PROBLEMS IN SINGLE AND THREE PHASE LOW VOLTAGE MACHINES
Section 1
Please note that within this competency and all others there are required non supervised learning times in this competency there is allocated 42 hours supervised (at TAFE) and 24 in the students time.
Therefore in order to complete all of the activities within this book the student is required to complete those activities directed by their teacher in their own time and seek review at the beginning of the next lesson.
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Gymea College Section 1 – The Ideal Transformer
Lesson sequence
Week 1 Section 1 Transformers
Week 2 Section 2 Transformer Construction
Week 3 Section 3 Transformer operation
Week 4 Section 4 Transformer Impedance and Efficiency
Week 5 Section 5 Transformer Polarity and Paralleling
Week 6 Section 6 Instrument Transformers
Week 7 Test Theory and practical transformers
Week 8 Section 7 & 8 Principles & Construction
Week 9 Section 9 Induction motor load characteristics
Week 10 Section 10 Single phase split phase motors
Week 11 Section 11 Shaded pole and cap start motors
Week 12 Section 12 Series universal motors
Week 13 Section 13 Alternators 1
Week 14 Section 14 Alternators 2
Week 15 Section 15 Synchronous motors V3
Week 16 Section 16 Motor faults protection and testing
Week 17 Test Theory and Practical Motors
Week 18 Resits
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Section 1 – The Ideal Transformer
TRANSFORMERS
PURPOSE: In this section you will learn about mutual induction and the operation of the ideal transformer.
TO ACHIEVE THE PURPOSE OF THIS SECTION:
At the end of this section the student will be able to: • Describe the principle of mutual induction in a transformer. • Describe how an emf is induced in the secondary winding of a transformer. • Explain the meaning of the terms 'step-down' and 'step-up' transformers. • Identify the basic component parts of the transformer. • Calculate the secondary voltage of a single phase transformer given one windings electrical details and turns ratio. • Explain the meanings of the terms 'turns ratio', 'current ratio' and 'voltage ratio'. • Calculate values of voltage, current and turns for a single phase transformer, using transformer ratios.
REFERENCES:
• Hampson, J. & Hanssen, S. Electrical Trade Principles, 2nd Edition, Section 7 • Jenneson, J.R. Electrical Principles for the Electrical Trade, 5th Edition, Pages 335 - 337
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Gymea College Section 1 – The Ideal Transformer
1. INTRODUCTION
Generating stations are the source of electrical energy, but for this energy to be utilised, it must be conveyed from the energy source to the consumer. The best way to provide a reliable and economical supply is via a state-wide transmission line grid fed by large power stations.
The state-wide grid system is dependent upon the operation of the transformer to allow the transmission of electrical power over very long distances. Figure 1 illustrates a simplified layout of the grid. (Teacher to show pictures of the electricity grid)
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Section 1 – The Ideal Transformer
2. THE TRANSFORMER A transformer is a static machine that is it has no moving parts, where electrical energy in one circuit is transferred to electrical energy of the same frequency in another circuit. See figure 2.
Electrical energy
to the load
Electrical energy from the supply
Figure 2
The Australian Standard drawing symbols used to represent transformers are as shown in figure 3.
Complete form Simplified form Figure 3
Transformers are used to - w step down voltage - where the output voltage to the load is lower than the input voltage to the transformer, known as a step down transformer. w step up voltage - where the output voltage to the load is greater than the input voltage to the transformer, known as a step up transformer. w provide electrical isolation - In some applications electrical isolation is required without voltage change, that is the input and output voltage are equal, in this case the transformer is referred to as an isolation transformer.
Figure 4 shows an example of each transformer.
Step down transformer Step up transformer Isolating transformer Figure 4
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Gymea College Section 1 – The Ideal Transformer
3. MUTUAL INDUCTANCE
Consider two coils A and B as shown in figure 5. Coil A is placed within coil B but has no electrical connection with it. A centre zero voltmeter is connected across the terminals of coil B while coil A is connected to a battery through the switch S.
(Teacher to demonstrate this)
0 Voltmete - +
B A A
S
Figure 5
When the switch is closed, the current in coil 0 Voltmeter A builds up a magnetic field, a considerable - + part of which links coil B. The voltmeter, B therefore, gives a momentary deflection which A A indicates that an emf is induced in coil B. See figure 6.
When the switch is closed and the current in coil A begins to build up, the increasing magnetic field produced cuts or links with the S turns of both coil A and coil B, so producing Figure 6 an - ______
When the switch is opened the current in coil A is interrupted, the amount of flux linking 0 Voltmeter - + with coil B is again changed by the collapse of the magnetic field and the voltmeter gives B A A a momentary deflection in the opposite direction to the previous movement. See figure 7.
The collapsing magnetic field cuts or links the turns of both coil A and coil B, again producing an: S Figure 7 • EMF of self induction in coil A and • an EMF of mutual induction in coil B
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Section 1 – The Ideal Transformer
Therefore, irrespective of whether the magnetic field of coil A is increasing or decreasing in value, the induced voltages are always an -
• EMF of self induction in coil A and • an EMF of mutual induction in coil B
The two magnetically linked coils are said to possess the property of mutual inductance.
In the module Applied Electromagnetism it was stated that, in accordance with Lenz's law the direction of the emf of self induction is always such as to oppose the change producing it. The same law applies to the emf of mutual induction, but the mode of action is different.
The emf of self induction in coil A retards a change of flux by either opposing or assisting the applied emf producing the magnetising current, but in coil B the emf of mutual induction establishes a current (if the circuit is closed) which, by an opposing or assisting mmf (magnetomotive force), retards any change in flux from coil A.
If two electric circuits possess the property of mutual inductance they are said to be coupled. The two coils are said to be - • ______or • ______depending on whether all, or only a part, of the magnetic flux links with both coils.
The tightness of the coupling can be varied by - • ______and • ______
In the arrangement shown in figure 7, the induced voltages would only be produced when the current in coil A changes in value, that is at switch on and switch off. If the current in coil A was constant there would be no induced voltages.
0 Voltmete If the battery and switch shown in the figure 7 - + were replaced with an AC supply as shown in B figure 8, you would have a transformer in its A A simplest form. The alternating current in coil A sets up an alternating magnetic flux. This flux links with coil B in which an emf is induced by mutual inductance. This is the principle of operation of the transformer. VAC AC Figure 8
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Gymea College Section 1 – The Ideal Transformer
4. TRANSFORMER OPERATION
In its simplest form the transformer consists of two coils of insulated wire that are insulated from each other and wound on a laminated steel core. The basic parts of a transformer are shown in the elementary diagram of figure 9.
Φ Output Input
the the
Figure 9
The laminated core is made stampings of sheet steel, usually silicon steel, clamped or bolted together. The primary and secondary windings are shown placed on opposite limbs, or legs, of the core. In practice, however, both coils are placed on the same limb to reduce magnetic leakage and thereby increase the tightness of the coupling between the coils. The primary and secondary windings can be identified by their connection - • the primary winding - ______• the secondary winding - ______Thus electrical energy enters the transformer through the primary, and leaves it through the secondary. The arrangement is reversible, that is, either coil may be used as primary or secondary, provided the insulation is adequate to withstand the resulting levels of voltage. Most transformers are designed for widely different primary and secondary voltages,
such as a primary voltage V1 = 240V and a secondary voltage V2 = 12V. The winding with the higher voltage is called the high-tension (H.T.) side and the low voltage winding is called the low-tension (L.T.) side of the transformer. When the high-tension winding is the primary, the transformer is called a ______transformer and when the low-tension winding is the primary, the transformer is called a ______transformer.
The ratio of the primary voltage to the secondary voltage is called the ______V1 ______. V2
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Section 1 – The Ideal Transformer
5. THE IDEAL TRANSFORMER An ideal transformer has no losses, that is - • neither winding has resistance • its core is infinitely permeable • any flux produced by the primary is completely linked by the secondary, and vice versa • all of the input power fed to the primary is transferred via mutual induction to the secondary, for example, if the input power was 1000W the output power would be
Power out
Power in
Figure 11 1000W.
In reality the ideal transformer does not exist. Its operation is considered to allow the understanding of the practical transformer to be achieved a little easier.
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Gymea College Section 1 – The Ideal Transformer
6. IDEAL TRANSFORMER RATIOS - VOLTAGE AND TURNS RATIOS
On no-load and with the effects of winding resistance neglected the ratio of primary to secondary voltage is proportional to the turns ratio, i.e. = V1 N1 V2 N2
V1 N1 N2 V2
Figure 12 The ratio is called the ______. V1 V2 The ratio is called the ______. N1 N2 That is, on no-load the ______.
Example: 1 A single phase transformer is wound with 750 turns on the primary and 100 turns on the secondary. Determine the transformer primary voltage if the secondary voltage is 32V. ______
Example: 2 A 240/20V single phase transformer has 480 turns on the primary. Determine the required number of turns on the transformer secondary ______
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Section 1 – The Ideal Transformer
• CURRENT RATIO
When a load is connected across the terminals of the secondary, as shown in figure 13,
the load current I2 is in the direction shown.
I 1 I 2 V1 N1 N2 V2 Load
Figure 13
The current in each winding can be calculated if the turns ratio is known.
In an ideal transformer the primary ampere-turns (I1N1) will be equal to the secondary
ampere-turns (I2N2), i.e. I1N1 = I2N2
This can also be written in the form of a ratio -
The term is called the transformer ______. I2 Note, the currenI1 t ratio is most accurate when ______.
Example: 3 A single phase transformer has 500 turns on the primary and 250 turns on the secondary. If the current flowing in the primary is 5A, determine the value of the secondary current. ______
By comparing the voltage, turns and current ratios, it can be seen that the current ratio is the inverse of the voltage and turns ratios.
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Gymea College Section 1 – The Ideal Transformer
Example: 4 For the transformer shown in figure 15, determine the - a) secondary current b) transformation ratio c) primary voltage d) primary current
12Ω V1 N1 N 2 V2 = 48V = 400 = 80 I 1 I 2
Figure 15
a) ______
b) ______
c) ______
d) ______
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Tutorial 1 NAME:
THE IDEAL TRANSFORMER SECTION A In the following statements one of the suggested answers is best. Place the identifying letter on your answer sheet. 1. The primary and secondary coils of a transformer are said to be coupled if the magnetic field flux cuts the - a) primary winding only b) secondary winding only c) primary winding on one half cycle and the secondary winding on the other half cycle d) primary and secondary windings simultaneously
2. The secondary voltage of a transformer is produced by - a) electrostatic induction b) current conduction c) mutual induction d) self induction
3. In an isolation transformer the - a) secondary voltage is greater than the primary voltage b) primary voltage is greater than the secondary voltage c) primary is equal to the secondary voltage d) primary and secondary voltages are connected to oppose one another
4. In a step-down transformer the - a) secondary voltage is greater than the primary voltage b) primary voltage is greater than the secondary voltage c) primary is equal to the secondary voltage d) primary and secondary voltages are connected to assist one another
5. In a step-up transformer the - a) secondary current is greater than the primary current b) secondary current is less than the primary current c) primary voltage is greater than the secondary voltage d) primary current is equal to the secondary current
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Tutorial 1 – The Ideal Transformer
SECTION B Blank spaces in the following statements represent omissions. Write the appropriate information on your answer sheet. Two coils having mutual inductance are said to be _____(1)_____. A transformer that has more turns on the secondary winding than the primary winding is called a _____(2)_____ transformer. The voltage induced in the secondary of a transformer is known as the emf of _____(3)_____ induction. If nearly all the flux produced by the primary winding cuts the conductors of the secondary winding, the two windings are said to be _____(4)_____ coupled. In a transformer the primary winding is connected to the _____(5)_____ and the secondary is connected to the _____(6)_____. When the high tension winding is the primary, the transformer is called a _____(7)_____ transformer. When the low tension winding is the primary, the transformer is called a _____(8)_____ transformer. The ratio of primary voltage to secondary voltage is called the ratio of _____(9)_____. The ratio of primary turns to secondary turns is called the _____(10)_____ ratio.
SECTION C 1. The primary winding of a 440/55V transformer has 400 turns. How many turns are there on the secondary winding? (50 turns) 2. A transformer has a turns ratio of 1000:50. Determine the secondary voltage if 240V is applied to the primary. (12V) 3. A 240/32V transformer has a primary current of 0.4A. Calculate the current in the secondary winding. (3A) 4. A single phase transformer steps down from 415V to 32V. Calculate the primary current if the secondary current is 2A. (0.154A)
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Tutorial 1 – The Ideal Transformer SECTION D 1. Redraw the transformer shown in figure 1 then identify the various parts.
Φ Output Input
the the supply
Figure 1
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Section 2
TRANSFORMER CONSTRUCTION
PURPOSE: In this section you will learn about the differences between ideal and practical transformers. In addition the core and winding construction of single and three phase transformers will be covered, along with the no-load operation of the practical transformer.
TO ACHIEVE THE PURPOSE OF THIS SECTION:
At the end of this section the student will be able to: • Describe the differences between ideal and practical transformers. • Describe various types of lamination style and core construction. • Identify different types of winding arrangements. • Identify the voltages and currents associated with the no-load operation of the practical transformer.
REFERENCES:
Hampson, J. & Hanssen, S. Electrical Trade Principles, 2nd Edition, Section 7 Jenneson, J.R. Electrical Principles for the Electrical Trade, 5th Edition, Pages 337, 339-342 and 354
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Section 2 – The Practical Transformer
1. THE IDEAL TRANSFORMER As explained in section 1 the transformer, in its simplest form, consists of two coils of insulated wire that are insulated from each other and wound on a laminated steel core as shown in figure 1.
Core flux Laminated iron core
Φ Output to Input from
the load the supply
Secondary winding Primary winding
Figure 1 In the case of the ideal transformer it is assumed - • the windings have zero resistance • the core is infinitely permeable • all of the flux produced by the primary is mutually coupled with the secondary • all of the input power is transferred by mutual inductance to the output. As a consequence it can be stated - The transformation ratio = turns ratio = current ratio
V N I = = V1 N1 I2
2 2 1
In an ideal transformer primary current (I1) only flows when a secondary current (I2) flows. That is, the ideal transformer has a no-load current equal to ______amperes. This is because the ideal transformer has zero losses - neither winding loss nor core loss.
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Section 2 – The Practical Transformer
2. THE PRACTICAL TRANSFORMER Practical transformers have losses. This can be seen by the fact that when a transformer is operated at no-load it will draw a primary current (I1) from the supply. The losses in a transformer consist of: • winding loss – resistance of the windings • core loss – heating of the iron core and magnetic leakage
3. CORE LOSSES Magnetic leakage Most of the magnetic flux produced by the primary current of a transformer passes through the iron core and links with all the turns in both windings, but a certain amount of flux will take various shorter paths even though of greater reluctance, as shown in figure 2.
P S
Figure 2 Leakage flux serves no useful purpose and has the effect of adding additional reactance to the windings. Magnetic leakage increases as load increases, causing a reduction in the secondary voltage for a given primary voltage, as some voltage is dropped in overcoming the additional reactance, thus changing (decreasing) the ratio of transformation slightly. Except for special purposes, care is taken by designers to keep magnetic leakage to a minimum. In practice both primary and secondary coils are placed on the same limb to reduce magnetic leakage and thereby increase the tightness of the coupling between the coils. See figure 3.
Figure 3
Figure 3
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Section 2 – The Practical Transformer
4. WINDING RESISTANCE
Heat is produced in overcoming the resistance of the windings. This heat is a loss within the transformer.
Losses in a transformer are shown below in figure 4.
Heat (Core loss) Heat Heat (Winding loss) (Winding loss)
Input from Output to the supply Φ the load
Secondary winding Primary winding
Figure 4
5. CORE CONSTRUCTION Transformer cores are usually - • laminated - usually 0.25mm to 0.9mm thick • constructed of a very high grade of silicon steel (approx. 97% iron + 3% silicon) - the silicon content reduces the magnetising losses, particularly hysteresis loss • cold rolled and often specially annealed to orient the grain to provide a very high permeability and low hysteresis in the direction of rolling.
*Teacher to show samples of transformer laminations
Transformer cores are also made from Amorphous steel.
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Section 2 – The Practical Transformer
1. Using amorphous steel in the core of a transformer can: • greatly enhance ______• make the transformers size ______• reduce the transformers ______• lower the transformers ______
To ensure the transformer core works most efficiently- • the number of butt joints are limited • joints are tightly made and laminations interleaved so as to minimize the reluctance of the magnetic circuit • the whole core is built up or stacked to the proper dimension and the laminations are bolted together firmly. As the iron forms a closed circuit, a current will flow. This current is known as an eddy current because it flows or circulates within the iron of the core. The resultant effect of eddy currents is to cause the core to heat. That is energy is taken from the supply that is wasted in heating the core. For this reason eddy currents need to be kept to a minimum. The core is laminated to reduce ______loss. Figure 5 shows the effect on eddy currents of laminating the transformer core.
i i i i i i i Solid core Lightly laminated core Laminating increased Heavily laminated core Eddy currents large Eddy currents reduced Eddy currents small Eddy currents very small Figure 5
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Section 2 – The Practical Transformer
As covered in the module Applied Electromagnetism, a magnetic material such as that used in the core of a transformer, gets warm when it is cyclically magnetised. The energy required to cyclically magnetise a transformer core is called the ______loss, which is proportional to the supply frequency. To minimise hysteresis loss a material with high permeability and low retentivity, such as soft iron, must be used. In practice the material used is ______. Silicon steel laminations are carefully cut, punched where necessary and have any burrs removed by grinding. The surface of each lamination in power transformers are often insulated by coating with an oxide during manufacture. In smaller cheaper transformers each lamination is sometimes treated on one side only with an insulating varnish which resists the action of heat and oil and reduces eddy currents. The cores are built up by interleaving the laminations in order to reduce the core losses to a minimum. Transformers are classified as - • ______type • ______type • ______type, according to the type of core construction used. The laminations for the core-type transformer may be made up of ‘U’ and ‘I’ shaped laminations, as shown in figure 9. Figure 10 shows an example of an assembled single phase, core type transformer.
U and I laminations Figure 9 Figure 10
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Section 2 – The Practical Transformer
The core for the shell-type transformer is normally made up of ‘E’ and ‘I’ shaped laminations, as shown in figure 12. Figure 13 shows an example of an assembled single phase, shell type transformer.
E and I laminations
Figure 12 Figure 13
The toroidal core –
Toroidal transformers *Teacher to show examples of CTs
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Section 2 – The Practical Transformer
6. THREE-PHASE TRANSFORMER Three phase transformers are mostly of the core type, with the concentric primary and secondary windings of each phase on a located on separate legs of the iron core. For very large transformers a five limbed core is used, with the two outer legs providing a parallel magnetic path for the phase fluxes. The construction of the core type of three-phase transformer is shown in figure 3.
A B C
Primary
Secondary
a b c N
Figure 3 The phase of the flux bears the same relationship to the applied voltage in each of the three limbs, consequently the algebraic sum of the fluxes, either positive or negative, is zero at any instant, just as the algebraic sum of the applied voltages is zero at any instant.
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Section 2 – The Practical Transformer
7. TRANSFORMER ON NO-LOAD When a sinusoidal voltage V1 is applied to the primary with the secondary open- circuited, as shown in figure 16, there will be no energy transfer from primary to secondary.
V1 I o N1 N2 V2
Figure 16
The applied voltage V1 causes a small current IO to flow in the primary winding. This no-load current is made up of two components, each having a specific functions - • ______current (Iµ) - magnetises the iron core. - lags the primary voltage V1 by 90º. • ______current (Iη) - provides the energy required to overcome the hysteresis and eddy current losses. These combined losses are known as the core loss.
- is in phase with the primary voltage V1.
The no-load current IO - • is usually a few percent of the rated full-load current of the transformer, approximately 2 to 5%. • the no-load current is equal to the phasor sum of the magnetising current and the iron loss current.
Figure 17 shows the phasor diagram for a transformer operating at no-load. Note, the core flux (Φ) is used as the reference phasor.
Φ
Figure 17 core flux - Φ (the reference) magnetising current - Iµ applied voltage - V1 secondary voltage - V2 iron loss current - Iη Machines Topic 2 Version 2 Page 9 of 16 Owner: Elect, ICT & Design Faculty/Electrical Last Updated 06/01/14 Disclaimer: Printed copies of this document are regarded as uncontrolled.
Section 2 – The Practical Transformer
As the magnetising current is considerably greater than the iron loss current, the no- load current lags behind the applied voltage by a very large angle, usually about 85º. This means that the no-load power factor is very low, approximately 0.087 lag (λ = cosφ).
8. MEASURING NO-LOAD LOSSES The magnetising and iron loss can be determined using a wattmeter as shown below. This is called the open circuit test.
43.2W W 2A 240V 110V
NOTES ******
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NAME: Practical
THE PRACTICAL TRANSFORMER PURPOSE: This practical assignment will be used to examine the no-load operation of a single phase transformer.
TO ACHIEVE THE PURPOSE OF THIS SECTION:
At the end of this practical assignment the student will be able to: • Measure the winding resistances of a transformer. • Measure the primary and secondary voltages of a transformer. • Measure the no-load primary current of a transformer. • Measure the no-load power taken by a single phase transformer. • Calculate the no-load power factor and phase angle of a single phase transformer.
EQUIPMENT: • 1 x variable, single phase AC supply • 1 x 100VA, 42/24V, single phase transformer • 2 x digital multimeters • 2x Prova 23 tong test meters • 4mm connection leads
– REMEMBER – – WORK SAFELY AT ALL TIMES – Observe correct isolation procedures
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Practical 2 The Practical Transformer PROCEDURE
1. WINDING RESISTANCES 1. Using a digital multimeter, measure the winding resistances of the transformer. Record the values in the spaces provided below.
42
Resistance 42V winding = ______
Resistance 24V winding = ______
24
Figure 1
2. NO-LOAD OPERATION 1. Connect the circuit shown in figure 2, using the 42V winding as the primary.
A AC current clamp
± A 1A A W Variable ± V 50V Single Phase V1 V2 AC Supply Digital Multimeter 200V AC N
Figure 2 2. Switch on the ac supply and adjust the variac to give a primary voltage of 42V, as indicated by the digital multimeter V1.
3. Measure V2 (secondary voltage) I1 (primary current) and power taken by the transformer record in table 1. Table 1 Primary Secondary Primary No-Load Transformation Voltage Voltage Current Power Ratio (V1) (V2) (I1) watts V1/V2 volts volts amperes 42V
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The Practical Transformer Practical
TEACHER CHECK POINT - Do not proceed until Progress Table 2 the teacher checks your results. attempt 1 attempt 2 attempt 3 5 2 0
4. Switch off the supply, then disconnect the circuit. 5. Please return all equipment to its proper place, safely and carefully.
3. OBSERVATIONS: 1. Complete table 1 by determining the transformation ratio
Transformation ratio = 𝑉𝑉1 ______𝑉𝑉2 ______2. Determine the no-load power factor of the transformer tested. = =
𝑃𝑃 𝑃𝑃 ______𝜆𝜆 𝑆𝑆 𝑉𝑉 𝑥𝑥 𝐼𝐼 ______
3. Determine the no-load phase angle of the transformer tested. φ = cos-1λ ______
4. Was the transformer used in this practical an ideal or practical transformer? Explain your answer. ______
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Tutorial 2 NAME:
THE PRACTICAL TRANSFORMER SECTION A Place the identifying letter of the best suggested answer on your answer sheet.
1. Practical transformers are different to ideal transformers because practical transformers have - a) a perfect transformation ratio b) losses c) zero hysteresis loss d) a better iron loss
2. The core of a transformer is laminated and insulated from each other to - a) reduce hysteresis loss b) reduce eddy current loss c) enhance the coupling between windings d) make core construction simpler
3. In a transformer, sections of both primary and secondary windings are usually wound on each limb to reduce - a) magnetic leakage b) iron losses c) hysteresis losses d) the amount of wire
4. Silicon steel is used for transformer cores because it - a) reduces hysteresis loss b) keeps the iron loss to a minimum c) is cheaper than ordinary steel d) has low resistance
5. Transformer laminations are prevented from shorting together by -
a) a varnish or oxide layer on each lamination b) a layer of insulation between each lamination c) filling the transformer with insulating oil d) using spacing blocks to provide air gaps
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Tutorial 2 – The Practical Transformer
SECTION B Blank spaces in the following statements represent omissions. Write the appropriate information on your answer sheet.
1. The power factor of a transformer on no-load is very ____(1) ____ . 2. Eddy currents in a transformer core produce _____(2)_____ in the iron core. 3. Joints in the laminations of a transformer core are staggered so as to minimise _____(3)_____. 4. If a transformer was constructed using a solid iron core the _____(4)_____ loss would be very large and the transformer would run very hot. 5. Flux established by the primary that does not cut the conductors of the secondary is known as _____(5)_____ flux. 6. If a transformer was constructed using a mild steel core the _____(6)_____ loss would be very large and the transformer would run very hot.
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Section 3
TRANSFORMER OPERATION
PURPOSE:
In this section you will learn about how transformers are rated and the operation of the practical transformer when delivering load current.
TO ACHIEVE THE PURPOSE OF THIS SECTION:
At the end of this section the student will be able to:
• State the method used to rate transformers.
• Calculate the maximum current that may be delivered by a transformer.
• Describe the relationship between transformer rating and cooling.
• List methods used for natural and forced cooling of transformers.
• Interpret the phasor diagram of a loaded transformer, including voltage and current components.
• Explain the reasons for variation in transformer secondary voltage.
• Calculate the voltage drop of a transformer
REFERENCES:
• Hampson, J. & Hanssen, S. Electrical Trade Principles, 2nd Edition, Section 9
• Jenneson, J.R. Electrical Principles for the Electrical Trade, 5th Edition, Pages 335 - 337
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Section 3 – Transformer Operation
1. RATING OF TRANSFORMERS
Read the section headed ‘Rating of transformers’ on page 337 of your textbook and answer the following questions:
Power Transformers are generally rated in ______. This rating informs us of the maximum ______the transformer can deliver to a load without overheating. Other important rating information given is the Secondary terminal______.
For example, 100VA at 24V, 50Hz or 500kVA at 415V, 50Hz.
This means that transformers are designed to carry a certain load safely, with temperature rise being the limiting factor. The heat developed in the windings of a transformer depends upon the load current that has to be supplied, regardless of the power factor; hence the rating is usually expressed in volt-amperes (VA), that is -
Single phase transformer Three phase transformer
where: S = apparent power in VA S = apparent power in VA
V = secondary terminal voltage in volts V = secondary line voltage in volts
I = load current in amperes I = secondary line current in amperes
Example: 1
A single phase 1000/250V, transformer is rated at 10kVA. Determine the full-load secondary current.
______
What would happen if the transformer in example 1 was called on to deliver a secondary current of 50A over an extended period of time.
The winding designed to deliver a maximum current of 40A, would ______.
Consequently, transformers are rated in terms of VA for current limiting reasons, since heating effect is proportional to current squared (I2) and is independent of power factor.
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Section 3 – Transformer Operation
Example: 2
A three phase, 11000/415V transformer is rated at 500kVA. Determine the full-load secondary current.
______
Example: 3
The 42/24V single phase transformers used in the practical assignments of this module are rated at 100VA. Determine the full load secondary current, given the transformers are designed to deliver 24V at 50Hz.
______
Example: 4
Determine the full load secondary current of the 33/11kV, 1000kVA 3 phase transformer shown. ______
2. TRANSFORMER COOLING
A transformer delivering load current generates heat in the core and the windings. The extent to which the temperature of a transformer increases with load is an important operating factor. The temperature of a transformer should not rise more than 50ºC above the surrounding air temperature.
Transformer rating can be increased if the windings are cooled by some external means. There are two commonly used cooling mediums are -
• ______and/or ______.
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Section 3 – Transformer Operation
Water cooling is used but to a lesser extent.
In practice transformers are allocated symbols which indicate the type of cooling used.
Cooling Medium Type of Circulation Air A Natural N
Oil O Forced F
Water W Forced directed FD
The order in which cooling symbols are arranged is as shown below.
1st Letter 2nd Letter 3rd Letter 4th Letter The cooling medium The cooling medium in contact
in contact with the windings with the external cooling system Kind of medium Circulation type Kind of medium Circulation type
Figures 2, 3, and 4 show examples of various air and oil transformer cooling methods.
Air Natural Air forced Oil natural, air natural Type AN Type AF Type ONAN
Figure 2
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Section 3 – Transformer Operation
Oil forced air forced Oil natural air forced Type OFAF Type ONAF
Figure 3
Oil forced water forced Type OFWF Figure 4
Read the first 2 paragraphs of the section titled ‘Mineral oil” on page 350 in your textbook and answer the following question: 1. List 4 functions of transformer mineral oil in a transformer. ______Transformer oil is tested periodically to ensure it is serviceable. Oil testing is also used to give an indication of the condition of the transformer winding insulation in power transformers. See the section titled ‘Oil testing’ in your textbook.
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Section 3 – Transformer Operation
4. TRANSFORMER AUXILIARY EQUIPMENT
CONSERVATOR TANKS
Some coolant filled transformers have a conservator tank or expansion tank. Coolant flows into and out of the tank as the main tank expands and contracts due to temperature changes.
HV Bushing
Conservators have various uses which include acting as a sump to collect impurities in the oil and to remove oxygen and moisture from the oil.
BUCHHOLZ RELAY
Mounted near the conservator you will generally find a “Buchholz Relay”. These relays are designed to detect air or gas in the oil and will trigger an alarm which will shut down the
transformer before major damage can occur.
*Teacher to show sample Buchholz relay
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Section 3 – Transformer Operation
BREATHER – see Figure 7.72 on page 359 in your text book.
Also mounted near the conservator is a “Beaded silica gel breather”. This breather is designed to remove moisture from the transformer oil before it causes damage. When no moisture is present the beads appear as a blue colour and they change to pink or orange as they draw the moisture from the oil and incoming air. The beads can be removed and dried with heat and reused once they are blue again.
SURGE PROTECTION A power surge or an increase in voltage supplying a transformer can have detrimental effect on its operation and can even destroy a transformer.
Transformers need to be protected from power surges and this is done by the use of valve type surge diverters which consist of two series connected elements. The first element is a series of ‘spark gaps’ followed by a valve element consisting of voltage dependant resistors or VDR. When a high voltage occurs it flashes over the spark gaps and discharges to ground through the VDR.
Figure 15 shows an example of a valve type diverter
Figure 15
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Section 3 – Transformer Operation
5. TRANSFORMER ON LOAD
When a load is applied to the secondary of a transformer a current I2 flows in the secondary windings, as shown in figure 5. The value of this current, and its phase position, depends upon the nature of the load.
V1 I 1 N1 N2 V2 I 2 Load
Figure 5
Under load the transformer secondary delivers energy to the load, therefore additional energy must be taken by the transformer primary. As the primary applied voltage is constant, the primary current must increase to supply the additional energy required.
Shown below is an equivalent circuit for a transformer which shows the losses that occur when load is applied.
X1 R1 Ideal transformer X2 R2
V1 V2 Load
The total primary current taken by the transformer under load conditions is made up of two components, the -
• no-load current (IO) - magnetises the core and overcomes core loss (hysteresis and eddy current loss)
• load component of primary - supplies the energy required by the load.
current (I1')
The load component of primary current is determined using transformer ratios, as follows –
= V x I = V x I ∴ V1 I2 ′ V2 I1 ∴ 1 1 2 2
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Section 3 – Transformer Operation
= N x I = N x I ∴ N1 I2 ′ N2 I1 ∴ 1 1 2 2
The total primary current taken by a transformer supplying a load is equal to the phasor sum of the no-load current and the load component of primary current –
Figure 7 below shows the phasor diagram for a transformer with a voltage ratio of 1:1 operating on inductive load and illustrates the phase relationship between the –
* core flux - Φ, which is taken as the reference * secondary current – I2
* applied voltage – V1 * no-load current - IO
* secondary voltage – V2 * primary current – I1
* load component of primary current – I1’
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Section 3 – Transformer Operation
I 1 V1 I 1 '
Io Φ
I 2 V2
Figure 7
Figure 8 shows the phasor addition of the no-load current and the load component of primary current to determine the transformer primary current when loaded. Figure 8
I1 V1 I1 '
Io Φ
I2 V2
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Section 3 – Transformer Operation
6. TRANSFORMER VOLTAGE DROP ON LOAD.
When transformer load is applied or changed, the secondary terminal voltage (V2) changes. See figure 1.
V1 I 1 N1 N2 V2 I 2 Load
Figure 1 The change in voltage is caused by the resistance and reactance of the two windings; that is the impedance. Figure 2 shows the transformer equivalent circuit.
Z 1 Z 2 X1 R1 Ideal transformer X2 R2
V1 V2 Load
Figure 2
Assuming the applied primary voltage (V1) is constant and load is increased -
• secondary current (I2) - ______
• load component of primary (I1') - ______
• primary current (I1) - ______
• voltage drop across primary impedance (ZP) - ______• voltage applied to primary of ideal transformer - ______• voltage induced in secondary of ideal transformer - ______
• voltage drop across secondary impedance (ZS) - ______
• secondary terminal voltage (V2) - ______. Therefore, in this case, as load is increased secondary terminal voltage decreases.
The voltage regulation of a transformer is the change in the secondary terminal voltage when a load of stated current and power factor is applied or removed. This is usually expressed as a percentage of the open circuit secondary voltage.
At no-load the voltage ratio of a transformer is equal to the ratio of primary turns to secondary turns, but as load is applied to the secondary, the terminal voltage decreases because some voltage is dropped across the impedance of the transformer windings.
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Section 3 – Transformer Operation
The change in secondary voltage is almost directly proportional to the change in secondary load current. V- drop under load = No load Secondary voltage – Secondary voltage on Load
Where: VNL = secondary no-load voltage VL = secondary load voltage
Example: 1 The no-load secondary voltage of a transformer is 300V and at full-load the secondary voltage drops to 293V. Determine the transformers voltage drop under load. ______Example: 2 The open circuit secondary voltage of a transformer is 250V and at full-load the voltage drops to 244.5V. Determine the transformers voltage drop under load. ______Example: 3 A 6600/415V step down transformer has a voltage drop of 13.1V at full-load. What is the secondary terminal voltage of the transformer at this load? ______
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Section 3 – Transformer Operation
7. TAP CHANGERS Most distribution and large power transformers are fitted with tap changing devices which keep the secondary terminal voltage practically constant. The Tap changer does this by: • Adding turns to the secondary winding or • cutting out turns from the primary winding as the load current increases.
That is, a tap changer adjusts the turns ratio , which in turn adjusts the 𝑁𝑁1 transformation ratio - �𝑁𝑁2� 𝑉𝑉1 • for an increase in� 𝑉𝑉load2� current -
Tap changer operates
I 2 V2 V2
• for a decrease in load – Tap changer operates I 2 V2 V2 There are two broad categories of tap changer - • ______tap changer • ______tap changer.
Figure 3 below shows an off load tap changer that would be fitted to a three phase, 11kV/415V distribution transformer. The tap changer would be connected to the primary winding.
Figure 3
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Section 3 – Transformer Operation
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NAME: Practical TRANSFORMER OPERATION
PURPOSE:
This practical assignment will be used to examine the no-load and full load operation of a single phase transformer.
TO ACHIEVE THE PURPOSE OF THIS SECTION:
At the end of this practical assignment the student will be able to:
• Measure voltages, currents and powers associated with a transformer.
• Using test results, determine the load component of primary current.
• Verify the primary current of a transformer is equal to the phasor sum of the no- load current and the load component of primary current.
• Using test results draw the phasor diagram for a transformer identifying the associated voltages and currents.
EQUIPMENT:
• 1 x variable, single phase, 50Hz AC supply
• 1 x 100VA, 42/24V, single phase transformer
• 1 x 100VA transformer load panel
• 1 x wattmeter (1A, 50V)
• 2 x digital multimeters
• 1 x AC Prova 23 tong test meter
• 4mm connection leads
– REMEMBER – – WORK SAFELY AT ALL TIMES –
Observe correct isolation procedures
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Practical 3 Transformer Operation PROCEDURE
1. NO-LOAD OPERATION
1. Connect the circuit shown in figure 1, using the 42V winding as the primary.
A AC current clamp
± A 1A A W Variable ± V 50V Single Phase V1 V2
AC Supply Digital Multimeter 200V AC N
Figure 1
2. Switch on the ac supply and adjust the variac to give a primary voltage of 42V, as indicated by the digital multimeter V1.
3. Measure the secondary voltage (V2), primary current (I1) and power taken by the transformer. Record your readings in table 1.
Primary Secondary Primary No-Load
Voltage (V1) Voltage (V2) Current (I1) Power volts volts amperes watts 42V
4. Do not proceed until the teacher checks your results and completes the progress table.
5. Switch off the supply. Progress Table 1
attempt 1 attempt 2 attempt 3 5 2 0
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Transformer Operation Practical 2. FULL LOAD OPERATION
± A 5A A W Variable ± V 25V Single Phase V1 V2 100VA Tx AC Supply Digital Load panel Multimeter 200V AC N Figure 2
1. Connect the transformer as shown in figure 2.
2. Switch on the ac supply and adjust the variac to give a primary voltage of 42V, as indicated by the digital multimeter V1.
3. Adjust the load panel for a full load current of 4.17A.
*Note, be sure to maintain the primary voltage of 42V.
4. Measure the full load primary current (I1) and record in table 2.
Table 2 Primary Voltage Primary Current
(V1) volts (I1) amperes 42V
5. Measure the secondary voltage (V2) and current (I2) and power supplied to the load Record your readings in table 3 below.
Table 3 Secondary Secondary Load Current Voltage Power amperes volts watts 4.17A
Do not proceed until the teacher checks your results.
6. Switch off the supply then disconnect the circuit.
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Practical 3 Transformer Operation OBSERVATIONS:
1. NO LOAD
a) Using the results recorded in table 1, calculate the no-load primary power factor and record in table 1.
λ = =
P P ______S V x I ______
b) Determine the primary no-load phase angle of the transformer. Record in table 1.
Φ = cos λ −1 ______
2. FULL LOAD
a) Using the results recorded in table 3, calculate the transformer full load secondary power factor and record in table 3.
λ = =
P P ______S V x I ______
b) Determine the full load, secondary phase angle, then record in table 3.
Φ = cos λ −1 ______
c) From your observations, what happened to the power factor of the transformer when the load was applied? ______
d) Power factor of a transformer at no load is poor / good e) A fully loaded transformer supplying a resistive load the power factor is poor / good
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Tutorial 3 NAME:
TRANSFORMER OPERATION
SECTION A - Place the identifying letter of your answer on your answer sheet.
1. The no-load power factor of a transformer is approximately -
a) 0.1- 0.2 b) 0.9 - 1.0 c) 0.8 - 0.9 d) 0.707
2. Transformers are rated in terms of the -
a) true power (kW) out b) true power (kW) in c) apparent power (kVA) out d) apparent power (kVA) in
3. When constructing an oil filled transformer tank to obtain maximum cooling effect the transformer tank should be -
a) plain b) fitted with radiator fins c) fitted with tubes to allow the oil to circulate d) painted in a light colour
4. An oil filled transformer which is cooled by means of a fan blowing across the radiators on the tank is termed -
a) ONAN b) OFAF c) OFAN d) ONAF
5. Transformer oil is used to -
a) aid cooling of the transformer b) enhance the level of insulation c) keep the windings free of moisture d) aid cooling and enhance the level of insulation Machines Topic 3 Version 2 Page 19 of 21 Owner: Elect, ICT & Design Faculty/Electrical Last Updated 31/03/2021 Disclaimer: Printed copies of this document are regarded as uncontrolled.
Tutorial 3 – Transformer Operation
SECTION B - Write the missing word answer on your answer sheet.
The cooling mediums used in transformers are designated with the letters A, O and W. The letter A stands for _____(1)_____, the letter O _____(2)_____ and the letter W _____(3)_____.
Cooling methods used in conjunction with transformers are designated by the letters N and F. The letter N stands for _____(4)_____ and the letter F _____(5)_____.
The rating of a transformer is determined by its ability to dissipate ____(6)______under load conditions.
A transformer that is cooled using only the surrounding air and oil in and around the windings, would be classified as _____(7)_____.
If a 100kVA transformer is required to deliver 110kVA for an extended period, it is most likely that the transformer windings would _____(8)_____.
The purpose of the conservator on an oil filled power transformer is to allow for the oil to ____(9)______and _____(10)______as the oil heats and cools.
SECTION C
1. A 33/11kV, 20MVA transformer is used in a zone substation. Neglecting any losses determine the -
a) full load secondary line current (1049.7A) b) full load primary line current (349.9A)
2. A single phase transformer supplies a load of 10A at voltage of 230V. A wattmeter connected to the load measures 1610W. Calculate the power factor of the load.
3. Explain the purpose of a Buchholz relay.
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SECTION D
1. What methods of cooling are used on the transformer shown in figure 1?
Figure 1
2. Is the transformer shown below, a single or three phase transformer?
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Section 4
TRANSFORMER IMPEDANCE AND EFFICIENCY
PURPOSE: In this section you will learn about how the internal impedance of a transformer affects its operation
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this section the student will be able to: • Explain the meaning of the terms impedance and percentage impedance, as applied to a transformer. • Calculate the percentage impedance of a transformer. • Describe the tests used to determine impedance of a transformer. • Describe the power losses that occur in a transformer. • Describe the tests used to determine the losses in a transformer. • State typical values of transformer efficiency. • Calculate the efficiency of single and three phase transformers. • Define the all day efficiency of a transformer.
REFERENCES:
• Electrical Principles for the Electrical Trades. 5th Edition. Jenneson J.R. Pages 339, 347-348.
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Section 4 – Transformer Imp. Volt drop. & Eff. 1. TRANSFORMER IMPEDANCE In determining the performance of a transformer, the impedance of the transformer caused by its core and windings, must be considered. The impedance of a transformer causes: • A drop in the secondary voltage of the transformer under load conditions. • Limit the short circuit current in the event of a fault. Transformers can be designed to: • have as little impedance as possible to limit the voltage drop on load or • have a higher impedance to limit short circuit current in the event of a fault.
Transformer impedance Voltage drop under load Fault current Low Low High High High Low The nameplate on a transformer gives the impedance, expressed as a percentage of its rated voltage. Transformer impedance can be determined using the ______test. If, as shown in figure 7, a transformer is operated with its secondary short circuited and the primary voltage is slowly increased until rated secondary current flows. Figure 7
A A
Variable Short I N N I AC supply 1 1 2 2 circuit
Using the results of the short circuit test, transformer percentage impedance can be calculated by:
where: %Z = percentage impedance of the transformer
VSC = short circuit voltage applied to the transformer V = rated voltage of transformer winding to which the test voltage is applied
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Section 4 – Transformer Imp. Volt drop. & Eff.
The test for determination of the percentage impedance may be carried out on either winding, that is - • voltage applied to the primary, with the secondary shorted • voltage applied to the secondary, with the primary shorted
Student exercise: Calculate the rated full load secondary current of the 42/24V 100VA transformer used in class. ______
*Teacher to demonstrate a short circuit test using a 100VA transformer.
A A
Variable Short I N N I AC supply 1 1 2 2 circuit
*Student to repeat the short circuit test and record values below
100 VA transformer short circuit test
V VSC I1 I2 (Rated voltage of (Test Voltage applied to (Primary (Secondary winding test supply is cause full load current to current) current) connected to) flow)
Example: 1 Calculate the percentage impedance of the transformer used ______
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Section 4 – Transformer Imp. Volt drop. & Eff. Example: 2 A 3 phase Delta/Star 5MVA, 33kV/11kV transformer is tested with the secondary short circuited. A voltage of 1.65kV is required to cause full load current to flow. Determine the transformer - a) rated secondary current b) percentage impedance ______
Example: 3 A single phase 5kVA, 240V/415V step up transformer required 10V applied to the primary to cause rated current to flow in the short circuited secondary. Determine the - a) rated primary current b) percentage impedance. ______
2. PROSPECTIVE SHORT CIRCUIT CURRENT If during the short circuit test, the voltage was raised to the rated voltage (100%) the transformers short circuit current would flow. So we can use the short circuit test and the percentage impedance to calculate the prospective short circuit current of a transformer. ie: If during a short circuit test, 5% of rated voltage causes rated current to flow, 100% will cause short circuit current to flow. I = x I 100 SC Rated Z%
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Section 4 – Transformer Imp. Volt drop. & Eff.
Example: 4 Using the results from the short circuit test on the 42/24V 100VA transformer calculate the: a) rated full load secondary current b) the prospective short circuit current of the transformer. ______
Example: 5 A 3 phase 11kV/430V delta/star transformer is rated at 450kVA and has a Z% of 5%. a) Find the rated secondary line current. ______b) Find the prospective short circuit current of the transformer. ______
3. SHORT CIRCUIT TEST FOR COPPER LOSSES The copper losses of a transformer can be obtained from what is called the short circuit test. For this test the secondary winding is short circuited through an ammeter. A wattmeter, an ammeter and a voltmeter are connected in the primary circuit as shown below.
M L A 1 W V1 V2 Variable low voltage V1 P S A 2 AC supply
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Section 4 – Transformer Imp. Volt drop. & Eff. The short circuit test is carried out in accordance with the following - • a low voltage at the correct frequency is applied to the primary • the primary voltage is slowly increased until the ammeters show full-load values • the wattmeter then indicates the full-load copper losses for the transformer This method gives the full load copper loss of the transformer.
*Using the 100VA transformer carry out a short circuit test for copper losses.
Procedure.
1. Adjust the variac for minimum output. 2. Connect the transformer as shown in figure 2, using the 42V winding as the primary and with a 4mm lead shorting the secondary.
AC current clamp ± A 5A A W ± V 25V Variable Short Single Phase V1 V2 AC Supply Circuit Digital Multimeter 200V AC N Figure 2
3. Switch on the AC supply and slowly adjust the variac until full-load current (4 .2 A) flows in the transformer secondary. 4. Measure the power taken by the transformer and record. The power taken equals the copper loss of the transformer.
V1 Power I2 (Supply voltage) (Copper loss) (Secondary current)
Transformer copper loss = ______Watts
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Section 4 – Transformer Imp. Volt drop. & Eff.
4. OPEN CIRCUIT TEST FOR IRON LOSSES If the secondary of a transformer is open circuit and the primary supplied with rated voltage and frequency, a wattmeter connected in the primary as shown in figure 10 will indicate the iron losses.
M L A 1 W V1 V2 Rated Secondary primary V P S 1 open circuit voltage
Figure 10 When the secondary of a transformer is open circuit, a small current flows in the primary winding. This current is called the no-load current, and is the current required to: • magnetise the core to the desired flux density and • provide the energy to overcome the losses in the iron core. Because of the small value of this no-load current, the copper loss in the open circuit test is negligible. Therefore the whole of the power input, as indicated by the wattmeter, is taken to represent the iron losses. The iron losses of a transformer remain practically constant at all loads if the - • ______and • ______remain constant.
If the applied voltage to the transformer is constant, the transformer voltages will be constant, hence the core flux and the iron losses are very nearly constant at all loads.
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Section 4 – Transformer Imp. Volt drop. & Eff. *Using the 100VA transformer carry out an open circuit test for iron losses. Procedure. 1. Connect the circuit shown in figure 1, using the 42V winding as the primary.
AC current clamp ± A 1A A W ± V 50V Variable Single Phase V1 V2 AC Supply Digital Multimeter 200V AC N Figure 1 2. Switch on the AC supply and adjust the variac to give a primary voltage of 42V, as indicated by the digital multimeter V1. 3. Measure the no-load input power and record. The power taken equals the iron loss of the transformer. 4. Measure the no-load primary current and record 5. Switch off the supply.
V1 Power I1
(Supply voltage) (Iron loss) (Primary current)
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Section 4 – Transformer Imp. Volt drop. & Eff.
5. TRANSFORMER EFFICIENCY Figure 8 illustrates the flow of power through a transformer.
Figure 8 The output of a transformer is always less than the input because of the losses in the iron core and the resistance of the windings. From the arrangement shown in figure 8 it can be seen -
Input power, PIN = Output power, PO =
All losses cause the temperature of a transformer to rise. Transformer losses are due to- • ______- Known as the iron losses. Considered to be constant, irrespective • ______- of load • ______- varies with load. To determine the input and output powers of single and three phase transformers - Single phase
Input power, PIN = Output power, PO =
Three Phase
Input power, PIN = Output power, PO =
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Section 4 – Transformer Imp. Volt drop. & Eff. The efficiency of a transformer is given by the equation -
where: η% = transformer percentage efficiency
PO = output power in watts
PIN = input power in watts Example: 8 A three phase distribution transformer supplies 675kW at full load. If the input power to the transformer at this load is 695kW, determine the efficiency of the transformer. ______
The usual method of determining the efficiency of a transformer is to obtain the losses by testing, using the - • open circuit test - to determine the ______
• short circuit test - to determine the ______.
Typically, the efficiency of transformers is very high. In the case of distribution and power transformers, in the range of 95% to 98%.
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Section 4 – Transformer Imp. Volt drop. & Eff.
6. MAXIMUM EFFICIENCY OF A TRANSFORMER The efficiency of a transformer is high, typically in the range 95 to 98% for high output transformers. This implies that the transformer losses are as low as 2 to 5% of the input power. The core loss in a transformer is a constant loss for all load conditions. The copper loss varies proportionally to the square of the load current. It is therefore a simple matter to plot the efficiency versus load current curve for transformer, see figure 11.
Percentage Efficiency
Load
Figure 11
Because of the relationship between the iron and copper losses, it can be shown that the efficiency of a transformer is at a maximum when: - the fixed losses are equal to the variable losses.
In other words, the maximum efficiency of a transformer occurs at the load that makes the – • ______equal to the ______
7. ALL DAY EFFICIENCY The all day efficiency of a transformer averages out the efficiency over a 24 hour period, by comparing the total energy output to the total energy input. A transformer running on light load for extended periods reduces the all day efficiency and so costs money. It is important that transformers are not allowed to operate for extended periods of time on either no-load or light load. If this can be avoided, the all day efficiency will be improved.
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Section 4 – Transformer Imp. Volt drop. & Eff. The best all day efficiencies are achieved when transformers are operated at or near full-load for the entire day.
8. HIGH EFFICIENCY TRANSFORMERS
Power transformers must comply with MEPS (Minimum Energy Performance Standards) which specify minimum levels of efficiency allowable. Amorphous steel is a more efficient core material than silicon steel and is sometimes used in high efficiency transformers. This is because Amorphous steel cored transformers use less energy at light loads and uses less energy to magnetise the core.
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NAME: Practical TRANSFORMER IMPEDANCE, VOLTAGE DROP AND EFFICIENCY PURPOSE: This practical assignment will be used to determine the iron loss, copper loss and the efficiency of a single phase transformer
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this practical assignment the student will be able to: • Carry out the transformer open circuit test to determine iron loss. • Carry out the transformer short circuit test to determine copper loss. • Carry out the transformer load test. • Determine the voltage drop of a transformer on load. • Determine the percentage impedance of a transformer. • Determine the ohmic impedance of a transformer. • Determine the prospective short circuit current of a transformer. • Determine the efficiency of a transformer.
EQUIPMENT: • 1 x variable, single phase, 50Hz AC supply • 1 x 100VA, 42/24V, single phase transformer • 1 x 100VA transformer load panel • 2 x digital multimeters • 2 x AC Prova 23 tong test meter • 4mm connection leads
– REMEMBER – – WORK SAFELY AT ALL TIMES – Observe correct isolation procedures
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Practical 4 Transformer Imp, Reg & Eff
PROCEDURE
1. LOAD TEST 1. Connect the circuit as shown in figure 3, using the 42V winding as the primary.
AC current clamp ± A ± A A 5A 5A W W ± V 50V ± V 25V Variable Single Phase V1 V2 100VA Tx AC Supply Digital Load panel Digital Multimeter Multimeter 200V AC 200V AC N Figure 3 2. Make the following adjustments on the load panel - coarse control to position 1 - fine control fully anticlockwise. These settings will give no-load operation. 3. Switch on the AC supply and adjust the variac to give a primary voltage of 42V, as indicated by the digital multimeter V1. 4. Measure the primary input power and record in table 1. 5. Adjust the load panel for a secondary current of 1A. *Be sure to maintain the primary voltage at 42V. Measure the primary input and secondary output powers and the secondary output voltages, record in table 1.
6. Repeat the procedure for each value of load current listed in table 1.
Table 1 Load Current Input Power Output Power Output Volts amperes watts watts 0
1
2
3
4.2 (Full load)
4.5
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Transformer Imp, volt drop & Eff Practical 4
7. Do not proceed until the teacher checks your results and completes the progress table.
8. Switch off the supply and disconnect the Progress Table 3 circuit. attempt 1 attempt 2 attempt 3 5 2 0 9. Please return all equipment to its proper place, safely and carefully.
2. OBSERVATIONS: 10. Using the results from the Load Test, determine the losses of the transformer at full load. P = P P
Loss______In − Out ______
11. Using your results from the load test, determine the transformer efficiency at each value of load current. = x 100 PO η%______PIN ______
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Practical 4 Transformer Imp, Reg & Eff
12. On the axis shown in figure 4, plot the efficiency curve for the transformer tested.
100
90
80
70 %
- 60
50
40
Percentage Efficiency 30
20
10
0 0 1.0 2.0 3.0 4.0 Load Current - amperes Figure 4 13. From your efficiency curve, determine the load current at which maximum efficiency occurred? ______14. If the load on the transformer was increased beyond full-load, what would happen to the transformers efficiency? Explain why this would be the case. ______15. What is there about the operation of the transformer that identifies it has losses? ______16. Using your results from the load test, procedure 3, determine the secondary voltage drop of the transformer at full load. ______
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Transformer Imp, volt drop & Eff Practical 4
17. What causes the voltage output of a transformer to fall as load current increases? ______
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Tutorial 4 NAME: TRANSFORMER IMPEDANCE, VOLTAGE DROP AND EFFICIENCY SECTION A In the following statements one of the suggested answers is best. Place the identifying letter on your answer sheet. 1. The efficiency of a transformer - a) is constant over the load range b) has a maximum of 90% c) varies with load d) varies with the iron losses
2. The impedance of a transformer - a) causes secondary voltage to drop when load current increases b) limits fault current when secondary is short circuited c) is determined by the core material and winding resistance d) all of the above
3. The iron loss of a transformer at rated voltage and frequency is - a) proportional to the load current b) proportional to the square of the load current c) practically constant at all times d) dependent on the power factor of the load
4. If the load on a transformer is doubled the iron losses are - a) doubled b) halved c) constant d) decrease slightly
5. The short circuit test on a transformer is used to determine - a) ohmic impedance b) percentage impedance c) copper losses d) all of the above
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Tutorial 4 Transformer Imp, Volt drop & Eff
SECTION B Blank spaces in the following statements represent omissions. Write the appropriate information on your answer sheet. Maximum efficiency of a transformer occurs at the load which makes the _____(1)_____ equal to the _____(2)_____. The short circuit test on a transformer is used to determine the _____(3)_____ losses. Copper losses in a transformer vary as the _____(4)_____ of the current. The _____(5)_____ losses in a transformer are constant for all loads at rated voltage and frequency. When measuring the iron losses of a transformer, the secondary winding must be _____(6_____. To achieve the best all day efficiency, a transformer should be operated at or near _____(7)_____ load for the entire day. Transformer output voltage drop between no load and full load is caused by the _____(8)_____ of the core and the windings of the transformer.
SECTION C 1. A transformer when tested using the open circuit test had an iron loss of 3700W and when tested using the short circuit test had a copper loss of 2100W. Determine the total transformer loss. (5800W)
2. Determine the full load efficiency of a transformer supplying full load output of 15kW at unity power factor, if the transformer has iron losses of 400W and copper losses of 800W when tested at full load. (92.6%)
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Transformer Imp, volt drop & Eff Tutorial 4 SECTION D 1. Re-draw the components in figure 1 to show the connections required to carry out a short circuit test on the transformer.
M L W V1 V2 Variable low voltage V1 A 1 P S A 2 AC supply
Figure 1
2. Re-draw the components in figure 2 to show the connections required to carry out an open circuit test on the transformer.
M L W V1 V2 Rated primary V1 A 1 P S voltage
Figure 2
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TRANSFORMER POLARITY & PARALLELING
PURPOSE: In this section you will learn about the common winding connections used in three phase transformers. In addition you will examine the need for parallel operation of single and three phase transformers.
TO ACHIEVE THE PURPOSE OF THIS SECTION:
At the end of this section the student will be able to: • State the method used to label transformer windings. • List the advantages of using a single three phase transformer over three single phase transformers. • State the meaning of the term polarity as applied to a transformer. • Add polarity marks to transformer windings. • Explain the need for parallel operation of transformers. • List the conditions required for transformers to be paralleled. • State the significance of transformer vector groups. • Calculate the load sharing of transformers. • State the significance of transformer vector groups in relation to the parallel operation of three phase transformers. • State the phase shift of a transformer given the transformer's vector group. • Describe the effects of harmonics on transformers and the methods used to reduce harmonics. • Interpret transformer nameplate information. • Describe the safety precautions to be observed when working with high voltages associated with transformers.
REFERENCES:
• Hampson, J. & Hanssen, S. Electrical Trade Principles, 2nd Edition, Section 7 • Jenneson, J.R. Electrical Principles for the Electrical Trade, 5th Edition, Pages 335 - 337
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Section 5 – Transformer Polarity & Paralleling
1. TRANSFORMER TERMINAL MARKING Australian Standard AS2374.1 - 1997, Power Transformers, sets out the standards for the marking of terminals on transformers. The method described uses a combination of letters and numbers to identify the starts and finishes of the windings. For single phase transformers - • the high tension winding is identified with the letter ‘A’ and the starts and finishes with the subscript numbers 1 and 2
For example, A1 and A2 • the low tension winding is identified with the letter ‘a’ and the starts and finishes with the subscript numbers 1 and 2.
For example, a1 and a2
HT
LT
Figure 1 In the case of three phase transformers - • the high tension windings are identified with the letters A, B, C and the starts and finishes with the subscript numbers 1 and 2
For example, A1, B1, C1 • the low tension windings are identified with the letters a, b, c and the starts and finishes with the subscript numbers 1 and 2.
For example, a1, b1, c1
HT HT HT
LT LT LT
Figure 2
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Section 5 – Transformer Polarity & Paralleling
2. THREE-PHASE TRANSFORMER CONNECTIONS The polarity markings and connections of a core type three-phase transformer is shown in figure 3 below.
A B C
Primary
Secondary
a b c N
Figure 3 Three-phase transformers windings can be connected in star, delta
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Example: 1 A three phase transformer has a transformation ratio of 6.6kV/240V. Draw the connections required to supply a 415V delta connected load.
6.6kV
Figure 5
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Section 5 – Transformer Polarity & Paralleling
3. THREE PHASE V THREE SINGLE PHASE TRANSFORMERS A three phase transformer - • costs less • occupies less floor space • is lighter • is more efficient than three separate single-phase transformers of the same total output. These advantages are offset by the fact that a complete three-phase transformer is often necessary as a spare in case of a breakdown, while with three single-phase transformers a single unit (of one-third total output) is sufficient as a spare. The bank of three transformers, if delta-delta connected, has an advantage in case of a breakdown. Part of the load can be carried temporarily by using two of the transformers in V-V or open delta.
4. PARALLEL OPERATION OF TRANSFORMERS Transformers are connected in parallel to provide additional ______, that is, to have the capacity to supply a greater current to a single load or to supply a greater number of individual loads. That is, transformers are paralleled when the load becomes too high for one transformer to supply. In such cases transformers are connected in parallel and the parallel group shares the load. Load sharing by transformers of equal rating is dependent upon the respective impedances of the transformers and may be determined in the following manner -
where: Load TxA = load taken by transformer A Load TxB = load taken by transformer B LoadT = overall load to be supplied %ZA = impedance of transformer A %ZB = impedance of transformer B
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Example: 2 Two transformers A and B of equal rating share a total load current of 300A. If transformer A has an impedance of 5.1% and transformer B an impedance of 4.3%, determine the load taken by each transformer. ______
Example: 3 Two delta/star transformers are required to supply a load of 2000kVA. If transformer A has an impedance of 5.4% and transformer B an impedance of 6.7%, determine the load taken by each transformer. ______
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Section 5 – Transformer Polarity & Paralleling
5. CONDITIONS FOR PARALLEL TRANSFORMER OPERATION For transformers to operate satisfactorily in parallel the following conditions must be fulfilled.
Single-phase transformers: • Same secondary voltage • same polarities • Same percentage regulation.
Three-phase transformers • Same secondary voltage • Same polarities / vector grouping must be identical • Same percentage impedances • phase sequences must be identical • same voltage change per tap on tap changers
6. POLARITY OF TRANSFORMERS The term polarity when used with reference to the parallel operation of electrical machinery is understood to refer to a certain relationship existing between the direction of the voltages in two or more units. In the case of the primary and secondary windings of a transformer, two terminals are said to have the same polarity if at the same instant in time the two terminals have the same voltage polarity, that is, both positive or both negative. There are two methods used to mark the polarity of the terminals of a transformer - • ______• ______Figure 6 shows a transformer with the polarity of its terminals marked using the lettering system and figure 7 shows a transformer with the polarity of its terminals marked using the dot method.
P S P S
Figure 6 Figure 7
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When the primary and secondary induced terminal voltages are in the same direction at every instant the transformer is said to have - ______polarity. Alternately, if the induced terminal voltages are in opposite directions, the transformer is said to have - ______polarity. Polarity marks indicate the direction of current into the primary and out of the secondary at the same instant.
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Section 5 – Transformer Polarity & Paralleling
7. POLARITY TESTS FOR TRANSFORMERS Two single phase transformers may be paralleled by joining together similarly marked terminals, as shown in figure 8.
High tension
A1 A2 A1 A2 Primaries
Secondaries
a1 a2 a1 a2 Low tension
Figure 8 If there is any doubt in regard to the internal connections of the transformers they must be tested for polarity. The following is a simple method of testing. One pair of adjacent primary and secondary terminals are joined, as in figure 4, and an alternating voltage is applied to one winding while a voltmeter is connected across the other pair of terminals.
VAC
A1 A P V a1 S a2
Figure 9 As shown in figure 9 the voltmeter reading will depend on the respective polarities of the two terminals A2 and a2. There are two possibilities – • if the voltage indicated by the voltmeter is less than the supply voltage the transformer is said to have - ______- which means that the primary and
secondary induced voltages are in the same direction and the two terminals A2 and a2 have the same polarity. • if the voltage indicated by the voltmeter is greater than the supply voltage the transformer is said to have - ______- which means that the primary and secondary induced voltages are in opposite directions and the two terminals A2 and a2 have opposite polarity.
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Example: 4 For the transformers shown in figures 10 and 11, determine whether they have additive or subtractive polarity and correctly label the secondary terminals.
VAC = 100V VAC = 100V
A1 A2 A1 A2 P P V 60V V 140V S S
V = 40V 2 V2 = 40V
Figure 10 Figure 11 ______*Conduct winding resistance and polarity test on an unknown transformer. As a final check, to ensure the transformers are being correctly paralleled the following test should be carried out. Connect the two transformers as shown in figure 12, with a voltmeter connected between the two open ends of the secondaries.
High tension
P P
S S
Low tension
Figure 12 If these terminals have similar polarity the voltmeter will indicate ______. If this is the case, the voltmeter is disconnected and the transformer secondaries can be connected as shown in figure 13.
High tension
P P
S S
Low tension
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Section 5 – Transformer Polarity & Paralleling
If the terminals are of opposite polarity the voltmeter will indicate______. If this is the case, connect the transformers as shown in figure 14.
High tension
P P
S S
Low tension Figure 14
8. POLARITY TEST - THREE PHASE TRANSFORMER Before paralleling three phase transformers the polarities of the windings should be tested if there is any doubt about the internal connections of the transformers. This is done by connecting the primaries to the supply and joining one pair of secondary terminals (usually the neutral). The other secondary terminals are tested with voltmeters as shown in below in figure 15. Figure 15 A B C Tx 1 A B C Tx 2 A B C 1 1 1 1 1 1
A2 B2 C2 A2 B2 C2
a 2 c 2 a2 c2 b2 b 2
a 1 b1 c 1 a1 b 1 c 1 V V a V n a b c
If the terminals are of - • the correct polarity, ______voltage will be indicated. • reversed polarity, ______the phase voltage will be shown.
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9. NAMEPLATE INFORMATION
The following information is given on a power transformer nameplate: • kind of transformer (transformer, auto-transformer, booster transformer) • the Standard to which the transformer complies • manufacturer's name • manufacturer's serial number • year of manufacture • number of phases • rated power in kVA or MVA • rated frequency • rated voltages (in V or kV) and tapping range • rated current (in A or kA) • connection symbol (diagram showing winding connections and phase displacement) • percentage impedance • type of cooling (for example, ONAN) • total mass and mass of insulating oil.
Example of a transformer nameplate.
Figure 19
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Section 5 – Transformer Polarity & Paralleling
10. HIGH VOLTAGE SAFETY PROCEDURES AS/NZS 3000 defines High voltage as any voltage above 1000 V ac or 1500 V dc AS/NZS 3000 devotes Section 7.6 to High Voltage Installations.. High voltage may be encountered in • power stations • transmission systems • substations • high rise office blocks or large shopping complexes (on site sub station) • heavy industrial sites or large factory complexes (on site sub station) • transformers to supply the above.
The regulation that covers safe working practices on high voltage installations is the Electricity (Workers' Safety) Regulation 1992 under the Electricity Act 1945.
The minimum safe working distances for trained and authorised persons.
Voltage level Minimum safe working distance
up to 650V 500mm 650V to 11kV 700mm
11 kV to 66kV 1000mm
66kV to l32kV 1500mm l32kV to 220kV 2500mm
220kV to 330kV 3000mm
above 330kV 4000mm
*Greater distances apply for untrained persons
The safety procedure for working on or near high voltage exposed conductors requires that an employee must be authorised to do so by an ‘Access Authority’. For an access authority to be issued, the following procedure must be followed. • conductors and/or equipment isolated • isolation must be locked out and danger tagged • conductors and/or equipment proved to be de-energised by an approved method • conductors shorted together and earthed • identify safe work area by taping (tape barrier) • issue the access
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Tutorial 5 NAME:
TRANSFORMER POLARITY & PARALLELING SECTION A In the following statements one of the suggested answers is best. Place the identifying letter on your answer sheet. 1. Transformers are paralleled when - a) a greater output voltage is required b) the load becomes too large for one transformer c) transformer regulation must be minimised d) the effect of the vector groups must be reduced
2. Transformer polarity marks indicate the direction of current - a) into the primary and out of the secondary b) out of the primary and into the secondary c) into the primary and into the secondary d) out of the primary and out of the secondary at the same instant
3. When the primary and secondary terminal voltages are in the same direction at every instant, the transformer is said to have -
a) additive polarity b) negative polarity c) subtractive polarity d) positive polarity
4. One advantage of three phase transformers over single phase transformers is: a) They are smaller and more efficient than 3 single phase units of the same size b) They operate at a higher power factor on full load c) They are easier to maintain d) They make less noise when running
5. When paralleling two power transformers to supply a common load the transformer with the highest impedance will:
a) take a higher share of the load b) take a smaller share of the load c) share the load evenly with the other transformer d) take no load at all
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Tutorial 5 Transformer Polarity & Paralleling
SECTION B Blank spaces in the following statements represent omissions. Write the appropriate information on your answer sheet. When two transformers of equal rating are paralleled, the transformer with the _____(1)_____ impedance will take the greater share of load. If two transformers of equal rating and impedance are paralleled, the transformers will share the load _____(2)_____. Identical single phase transformers that are to be paralleled must have the same _____(3)_____, _____(4)_____ and _____(5)_____. When the primary and secondary terminal voltages of a transformer are in opposite directions at every instant, the transformer is said to have _____(6)_____ polarity. Three phase transformers that are to be paralleled must have identical _____(7)_____, _____(8)_____, _____(9)_____, _____(10)_____ and _____(11)_____.
SECTION C 1. Three single phase 5:1 transformers have their primaries connected in star to a 415V supply. The delta connected secondaries supply three 6Ω star connected elements. Determine - a) primary phase voltage (240V) b) secondary phase and line voltages (48V, 48V) c) load phase current (4.62A) d) secondary line current (4.62A) e) secondary phase current (2.67A) f) primary phase and line current. (0.53A, 0.53A) 2. Two transformers of equal rating are to be paralleled to supply a load of 3MVA. If transformer A has an impedance of 5% and transformer B an impedance of 4.5%, determine how the transformers will share the load. (Tx A = 1.42MVA and Tx B = 1.58MVA)
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Transformer Polarity & Paralleling Tutorial 5
SECTION D 1. What would be the expected voltmeter readings in the circuit of figure 1, if the transformers were - a) correctly paralleled b) incorrectly paralleled.
High tension 415V
P P
S S V
Low tension 240V
Figure 1 2. What would be the expected voltmeter readings in the circuit of figure 2, if the transformers were - a) correctly paralleled b) incorrectly paralleled.
Figure 2
A B C Tx 1 Tx 2
A1 B 1 C1 A1 B 1 C 1
A2 B2 C2 A2 B2 C2 a2 c2 b2 a2 c2 b2
b1 c 1 a 1 b1 c 1 a1 V V V n a bb c
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INSTRUMENT & AUTO-TRANSFORMERS
PURPOSE: In this section you will learn about the construction, operation and application of instrument and auto-transformers.
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this section the student will be able to: • Identify potential, current and auto-transformers from their winding diagrams. • List the advantages and disadvantages of the auto-transformer. • Describe the construction of potential, current and auto-transformers. • Calculate voltages and currents associated with auto-transformers. • State the ratings of potential and current transformers. • List the precautions to be taken with the connection and disconnection of instrument transformers. • State applications for potential, current and auto-transformers. • Test a transformer to determine its suitability for connection to the supply.
REFERENCES:
• Hampson, J. & Hanssen, S. Electrical Trade Principles, 2nd Edition, Section 7 • Jenneson, J.R. Electrical Principles for the Electrical Trade, 5th Edition, Pages 349 - 352
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Section 6 – Instrument & Auto Transformers
1. AUTO-TRANSFORMERS An auto-transformer is one which has a part of the winding common to both primary and secondary circuits. In general, the two coils of a transformer are connected together only by the magnetic field, but in an auto-transformer, as well as the magnetic coupling, there is an electrical connection. The winding of an auto-transformer is provided with one or more taps, from which the secondary voltage is obtained, in the case of a step down auto-transformer, or across which the primary voltage is applied if the auto-transformer is of the step up type. Figure 1 shows the connections for a step down auto-transformer and figure 2 those of a step up auto-transformer.
Load Load
Step down auto-transformer Step up auto-transformer
Figure 1 Figure 2
As is the case with the double wound transformer, if losses are neglected the transformation ratio of the auto-transformer is equal to the turns ratio, and also equal to the current ratio -
Since the direction of current in the secondary of a transformer is opposite to that of the primary current, it follows that in an auto-transformer the part of the winding common to the primary and secondary carries the difference of the two currents.
This condition permits the common part of the winding to be of much smaller cross section than the other part, which saves copper and reduces size and cost.
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Section 6 – Instrument & Auto Transformers Example: 1 For the step down auto-transformer shown in figure 3 determine the -
a) secondary voltage I 1 b) secondary current I c) primary current 2 600 1200V d) current in the common section of turns 550 55Ω the winding turns
______Example: 2 Determine the current in the common section of the winding of a 240/220V auto- transformer supplying a 1kW load at unity power factor. ______Figure 4 shows an example of one type of auto-transformer.
Figure 4 There are three main applications for auto-transformers - • ______on long transmission lines and • to reduce voltage for ______• variable ac supplies - ______.
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2. ADVANTAGES AND DISADVANTAGES OF AUTO-TRANSFORMERS By combining the primary and secondary of a transformer into one winding, a saving of material and a reduction in losses can be effected. As the primary and secondary currents are 180º out of phase, that part of the winding which carries both currents will actually carry their difference. The copper losses will be lower and a smaller iron core can be used for a given output, thus reducing the iron loss. Therefore, the advantages are - • ______Especially where the ratio • ______of transformation is low • ______Auto-transformers are restricted in their application, since the primary and secondary circuits are electrically connected. *Read clause 4.14.4 of AS3000, for details regarding the installation of auto- transformers and answer the following question: Equipment connected to auto-transformers must be rated at or above ______
3. INSTRUMENT TRANSFORMERS Instrument transformers are generally used with AC instruments when high voltages or high currents have to be measured. Instrument transformers are of two types - • ______• ______.
4. POTENTIAL TRANSFORMERS Voltage or potential transformers (PT's) are used in conjunction with - • voltmeters • protection relays • the potential coils of wattmeters and kilowatt-hour meters, and do not differ materially from the constant voltage power transformers already discussed, except that their power rating is low. Since only instruments or relays are connected to their secondaries, such transformers have ratings of 5VA to 200VA. Potential transformers are very carefully designed to maintain an extremely accurate transformation ratio, which changes very little with load.
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Section 6 – Instrument & Auto Transformers The standard secondary voltage of a potential transformer is ______. This arrangement enables high voltage to be kept away from instruments, thus eliminating the danger of electric shock or a breakdown of insulation in meter windings. Figure 5 shows examples of potential transformers used on single and three phase systems.
Three phase capacitive voltage transformer
Single phase voltage transformer Figure 5
Figure 6 shows the Australian Standard drawing symbols for the potential transformer.
Complete form Simplified form
Figure 6 Figure 7 shows the connection of a potential transformer to allow the measurement of the voltage on an 11kV supply line. The primary is connected across the 11kV line, with a 110V voltmeter connected across the transformer secondary. The voltmeter would be scaled to read 0-11kV.
11kV supply
110V V
Figure 7
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5. CURRENT TRANSFORMERS Current transformers (CT's) are used - • when the current to be measured is greater than can be carried safely by a meter connected directly into the circuit • where an instrument is to be used at some distance from the load circuit • where it is necessary to isolate the windings of instruments or relays from high voltage circuits. The current transformer has a primary, usually of a few turns only, wound on an iron core and connected in series with the line. When the primary has a high current rating it may consist of a straight conductor passing through the centre of a ring shaped laminated core. The secondary consists of the required number of turns wound on the core. Figure 8 below shows how a current transformer works.
Figure 8
Examples of current transformers.
Figure 9 shows the Australian Standard drawing symbols for the current transformer.
Complete form Simplified form Figure 9
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Section 6 – Instrument & Auto Transformers The current transformer is designed to work with a very low flux density in the core, and the secondary has a low impedance. This arrangement makes it possible to obtain a current in the CT load which is practically proportional to the line current. The standard values of rated primary currents are - 10, 12.5, 15, 20, 25, 30, 40, 50, 60 and 75A and their decimal multiples or fractions. The preferred values are those underlined. The standard values of rated secondary currents are 1, 2, 5, but the preferred value is ______. Figure 11 shows the connection of a current transformer and an ammeter, which are used to measure the line current taken by a three phase motor. The current transformer is used to step the current down to a value which can be safely measured using an ammeter. In this case the ammeter used would be a 5A ammeter, with the scale calibrated to read 0 - 200A.
M 3φ 200:5 CT
shorting switch or link 5A ammeter A scaled 0 - 200A
Figure 11 The secondary circuit of a potential transformer may be opened safely and the load removed, but in the case of a current transformer, the secondary circuit must be short circuited before the load is removed. If there were no secondary current to produce a demagnetising effect, the flux would rise above normal value, inducing a voltage in the secondary which may damage the insulation or create a shock hazard. The increased iron losses may also cause such a rise in temperature that the quality of the iron may be altered permanently and the accuracy of the transformation ratio affected.
6. PROTECTION OF INSTRUMENT TRANSFORMERS Instrument transformers are always given a VA load rating and these ratings should not be exceeded if accuracy is desired. An excessive current in the secondary circuit of a potential transformer produces an excessive current in the primary circuit. Consequently, a potential transformer can and should be protected by fuses in either the primary or secondary circuit. The currents in the windings of a current transformer are determined only by the line current and therefore cannot become excessive when a fault occurs in the transformer or ammeter. Consequently, fuses are not required and must not be used to protect current transformers.
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Example: 3 Connect the components shown in figure 12 to allow the load voltage, current and power to be measured. (current and voltage circuits must be kept separate)
CT
M L 11kV P V W A supply S Load V1 V2 PT
Figure 12 Example: 4 A CT with a ratio of 200/5A is used to measure the current taken by an induction motor. If the motor current is 175A determine the current flowing in the CT secondary. ______Example: 5 A PT with a ratio of 11kV/110V is used to supply the potential coil of a maximum demand indicator. If the line voltage drops to 10450V, determine the voltage supplied to the maximum demand indicator. ______
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Section 6 – Instrument & Auto Transformers 7. TRANSFORMER TESTING Transformers must be tested according to Australian standards to determine whether they are in safe working order. Tests which electricians need to do before commissioning or as part of maintenance are listed below.
WINDING RESISTANCE TEST Winding resistance for small transformers can be checked using an ohm-meter. This test proves each winding is continuous and can also be used to identify HV & LV windings on an unmarked transformer. INSULATION RESISTANCE TEST One of the tests carried out on a transformer to determine its suitably for connection to the supply is the measurement of the transformers - • ______. Insulation resistance is measured between - • ______and • ______. Theoretically the insulation resistance should be ______.
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The connections required to measure the insulation resistance between windings are as shown in figure 13.
P S
M Ω
Figure 13 The connections required to measure the insulation resistance between the windings and earth are as shown in figure 14.
P S
M Ω
Figure 14
With double wound transformers it is particularly important that no breakdown occurs between primary and secondary windings, as this could lead to a dangerous situation and damaged equipment. To provide additional protection, some transformers are fitted with an earth shield between the primary and secondary windings.
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Tutorial 6 NAME: INSTRUMENT & AUTO-TRANSFORMERS SECTION A In the following statements one of the suggested answers is best. Place the identifying letter on your answer sheet. 1. An auto transformer has a primary current of 10 amperes and a secondary current of 20 amperes. The current in the common part of the winding would be — a) 30 amperes. b) 20 amperes. c) 15 amperes. d) 10 amperes.
2. The current in the common section of the winding of an auto transformer, when on load is equal to— a) the phasor sum of the primary and secondary currents b) the phasor difference of the primary and secondary currents c) the sum of the primary and secondary currents d) the difference of the primary and secondary currents
3. An auto transformer compared to a double wound transformer of the same rating - a) has a larger physical size b) requires less material to manufacture c) has a lower efficiency d) has higher losses
4. The rated secondary current of a standard current transformer is - a) 0.6A b) 5A c) 10A d) 25A
5. The rated standard secondary voltage of a potential transformer is - a) 415V b) 240V c) 120V d) 110V
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Tutorial 6 Instrument & Auto-Transformers
SECTION B Blank spaces in the following statements represent omissions. Write the appropriate information on your answer sheet. Auto transformers are transformers with a _____(1)_____ winding. A short circuit between the turns in an auto transformers is dangerous because high ______(2)______can be transferred to the load connected to the transformer. Two types of instrument transformers are _____(3)_____ transformers and _____(4)_____ transformers. Instrument transformers are usually rated in terms of their _____(5)_____ output. The secondary of a _____(6)_____ transformer must always be short circuited before disconnecting its associated meter. The windings of _____(7)_____ instrument transformers should always be protected by fuses. When the primary of a current transformer is energised and the secondary is open circuited, a high _____(8)_____ is produced at the secondary terminals. The minimum acceptable insulation resistance for a transformer between windings is _____ (9)_____ and between windings and earth _____(10)_____.
SECTION C 1. An auto transformer is used to step down from 300 volts to 200 volts. The complete winding consists of 600 turns and the secondary current is 30 amperes. Determine:
a) secondary turns (400 turns) I 1 b) primary current (20A) 30A c) current in common portion of 600 300V winding, neglect all losses turns 200V (10A).
2. An auto transformer is used to step up from 200 volts to 250 volts. The primary winding consists of 400 turns and the secondary current is 20 amperes. Determine: a) secondary turns (500 turns) b) primary current (25A) c) current in common portion of winding, neglecting all losses (5A). 3. A 500/5A CT has a primary current of 450A, what is the secondary current? (4.5A) 4. A 415/110V potential transformer has a primary applied voltage of 425V, what is the secondary voltage? (112.65V)
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Instrument & Auto-Transformers Tutorial 6 SECTION D 1. Re-draw the symbols shown in figure 1 and connect the ammeter, voltmeter and wattmeter to measure the line current and line voltage of the three phase motor. The instruments are to be connected via instrument transformers. Note, a bar type CT is used.
11kV Supply
11kV three phase 50A 11kV motor
5A 110V
A V
Figure 1
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Topic 1 OPERATING PRINCIPLES OF THREE PHASE INDUCTION MOTORS
PURPOSE: This section introduces the purpose, construction and operation of the squirrel cage three phase induction motor.
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this section the student will be able to: • Determine circuit operation characteristics by using the right hand (grip) rule for conductors and solenoids and Fleming’s left and right hand rules • State the characteristics of the magnetic field produced by a single, two and three phase winding. • Calculate the speed of rotation of the rotating magnetic field. • Describe the relationship between the rotor speed, slip and rotor frequency • Describe the basic principle of operation of an induction motor. • Reverse the direction of rotation of a three phase induction motor.
REFERENCES:
• Hampson, J. & Hanssen, S. Electrical Trade Principles, 2nd Edition, Section 6 • Jenneson, J.R. Electrical Principles for the Electrical Trade, 3rd Edition, Pages 223 - 228
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Topic 1 – Operating Principles
1. INTRODUCTION A typical squirrel cage, three phase induction motor is shown in figure 1. The basic function of a motor of this type is to carry out energy conversion, that is - the conversion of ______energy to ______energy.
Figure 1 Figure 1
Student exercise *Read the first paragraph of the section titled ‘Three phase motors’ on page 248 in your textbook and answer the following question: Three phase motors are the most ______used in industry. They are ______, frequency dependant constant speed machines, ______in construction and easy to ______.
The squirrel cage induction motor consists of two main parts -
• A stationary part called the ______
Figure 2
Figure 2 • A revolving part called the ______
Figure 3
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Topic 1 – Operating Principles
Stator construction The stator consists of three main parts, as shown in figure 4, -
Windings
Frame
Figure 4 Core
Figure 4
Frame - is of cast or fabricated steel. - are manufactured in various forms.
The type of frame used is governed by the environmental conditions the motor is to operate under, for example - • Open type - which allows free ventilation to take place. • Drip proof - which has a closed upper half, while allowing ventilation through the lower half. • Totally enclosed - which prevents the exchange of air between the inside and outside of the frame.
Stator core - slotted on the inside circumference to carry the stator winding. - made of low loss silicon steel to reduce hysteresis loss. - laminated to reduce eddy current loss, see figure 5 - laminations are insulated from one another to further reduce eddy current loss - is pressed directly into the frame.
Figure 5
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Topic 1 – Operating Principles
Windings - consists of three identical phase windings which are displaced by 120 electrical degrees around the stator. - each phase winding consists of a number of series connected coils which produce the required number of stator poles. - the windings are generally connected in delta, but may be connected in star. In either case the phase conductors must be of sufficient csa area to carry the line current when star connected and of the line current 1 when delta connected. √3
Figure 6
2. ROTOR CONSTRUCTION The squirrel cage motor uses a rotor known as the squirrel cage rotor, or cage motor. The name is derived from the arrangement of the rotor conductors. See figure 7. The squirrel cage rotor consists of - • a shaft Figure 7 • a laminated silicon steel core having partly closed slots,. • a squirrel cage winding - which consists of solid copper alloy or aluminium bars embedded in the rotor slots. Each bar is short circuited by end rings. See below:
Figure 9
Figure 8
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Topic 1 – Operating Principles
In smaller motors the complete rotor winding, including bars, end rings and fan, is a one piece casting. In larger motors the conductors are formed bars, generally copper, inserted in slots in the rotor laminations, which are then welded to the short circuiting end rings.
Rotor slots are generally "______" to reduce any magnetic locking between the stator and the rotor and to reduce magnetic noise when the motor is running. The rotor is supported on bearings so that it can rotate freely. The air gap clearance between the rotor and the stator is typically about "0.5mm", but varies depending upon the rating of the motor.
3. OVERALL CONSTRUCTION Figure 10 shows and exploded view of a general purpose squirrel cage induction motor.
Figure 10 The circuit diagram for the squirrel cage motor would W W U2 be as shown in figure 11. 1
Figure 11 W2 U1 V V1 V2 U
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Topic 1 – Operating Principles
4. ROTATING MAGNETIC FIELD PRODUCED BY THREE PHASE CURRENTS *Teacher to show the rotating magnetic field using coke can in 3 phase ELV stator. The rotating magnetic field of a three phase induction motor is produced by supplying current to three single phase windings on the stator. The three windings must be: • 120 electrical degrees apart, irrespective of the number of poles Figure 12 shows a simplified diagram of the winding arrangements for a three phase induction motor.
Figure 12 Figure 13 illustrates the phase relationship between the currents in a three phase system, in which the current in B phase lags the current in A phase by an angle of 120° and the current in C phase lags that in B phase by 120°.
+I IA IB I C 0.866 0.5
0 time
-0.5 -0.866 -I 0º 60º 120º 180º 30º 90º 150º
Figure 13
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Topic 1 – Operating Principles
5. SPEED OF ROTATING MAGNETIC FIELD The speed of rotation of the rotating magnetic field is given a specific name, that is, the • ______. The speed of rotation of the rotating magnetic field depends on the - • ______and the • ______.
The speed of rotation of the rotating magnetic field is known as the synchronous speed and is given by -
where: ns = synchronous speed in rpm f = supply frequency in hertz P = number of stator poles Example: 1 Determine the synchronous speed of a 2 pole stator which is supplied from a 50Hz supply. ______
Example: 2 A 4 pole induction motor is supplied from a 50Hz supply, determine the synchronous speed. ______
Example: 3 Determine the number of stator poles required for a three phase induction motor to operate with a synchronous speed of 1000 rpm when supplied from a 50Hz supply. ______
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Topic 1 – Operating Principles
6. INDUCTION MOTOR ACTION The effort required to rotate a shaft about its centre is called "______" or "______".
In an induction motor, torque is developed by the interaction of the - • rotating magnetic field established by the stator windings and the • magnetic field set up by the rotor currents.
The development of torque in an induction motor can be explained by considering the interaction of the fluxes set up by the stator and rotor windings, as illustrated in figure 14. This diagram shows a section of the stator, air gap and rotor of a squirrel cage induction motor.
S Rotation of rotating magnetic N field
S
Figure 14
Due to the interaction of the stator and rotor magnetic fields, forces are exerted on the rotor conductors. As the rotor is mounted on bearings and free to rotate, a torque or turning effort is developed which causes the rotor to spin - "______" direction as the rotating magnetic field.
To maintain an induced rotor field, the rotor must spin at a speed that is "______" than that of the rotating magnetic field.
This difference in speed is known as the "______" of the motor
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Topic 1 – Operating Principles
The previous operation may be summarised as follows. When the stator winding of a three phase motor is energised the following events occur - • the stator sets up a rotating magnetic field that rotates at ______speed • the rotating stator flux crosses the air gap and ______an emf in the rotor conductors • as a result of the induced rotor emf, rotor ______flows • these high rotor currents establish a rotor ______• rotor flux interacts with the stator flux which causes the rotor to ______• the rotating force exerted on the rotor is known as the ______of the motor • the rotor will accelerate until it approaches ______speed. • acceleration of the rotor ______when the induced rotor current is of sufficient value to develop the necessary load torque.
If the load torque increases, the rotor has to slow down to induce a greater rotor emf and hence a greater rotor current to accommodate the extra load. The torque developed by the induction motor is proportional to the product of the - • air gap flux and • component of rotor current in phase with it.
7. LOAD VARIATION If the torque required by the load increases - • the rotor speed ______• a ______emf is induced in the rotor • the rotor currents ______• the forces exerted on the rotor conductors ______, that is, additional torque to accommodate the extra load • the current taken by the stator ______to provide the additional energy required to drive the additional load.
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Topic 1 – Operating Principles
8. SLIP To produce torque, there must be a rotor flux caused by current flowing in the rotor conductors. If the rotor could run at synchronous speed, there would be - • no relative motion between the ______and ______. Consequently, there would be no - • ______• ______• ______• ______
and so the rotor would slow down. Therefore, an induction motor cannot run at synchronous speed. With the rotor running just below synchronous speed, sufficient torque is developed to keep the rotor turning. The difference between the synchronous speed of the rotating field and the actual speed of the rotor is called the - "______". Slip speed can be determined from the expression -
where: ns = synchronous speed in rpm n = rotor speed in rpm nslip = slip speed in rpm Slip is generally expressed as a percentage of the synchronous speed and is known as the - "______".
Percentage slip can be calculated using the equation -
where: s% = percentage slip ns = synchronous speed in rpm n = rotor speed in rpm
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Topic 1 – Operating Principles
Practical values of slip are - • large motors, rated at or greater than 1,000kW, rarely exceeds 0.5% • small motors, rated at 10kW or less, seldom exceeds 3%. Example: 4 A 4 pole 50Hz induction motor operates with a rotor speed of 1440 rpm. Determine the - a) synchronous speed b) slip speed c) percentage slip
______
Example: 5 An induction motor operates with a 2% slip. If the synchronous speed of the motor is 1500 rpm, determine the actual rotor speed. ______
9. TERMINAL BLOCK MARKINGS The overall winding of the basic three phase induction motor consists of three individual windings, one per phase. These windings - • are identical in every respect they - - are wound using the same size and type of U1 U2 winding wire
- have the same number of turns. V1 V2
• should be thought of as three individual, highly W1 W2 inductive, iron cored coils. See figure 15.
Figure 15
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Topic 1 – Operating Principles
The international standard for labelling each phase winding is U, V and W - • with the starts of the windings labelled U1, V1 and W1 • and the finishes U2, V2 and W2. The purpose of marking the windings in this fashion is to assist in their correct connection. The windings may be connected internally in either star or delta. Motors that are internally connected in star or delta are fitted with a terminal block having three connection studs.
U1 U2
V1 V2
W1 W2
Figure 16 Many three phase motors the windings are not internally connected in either star or delta. Instead, a connection is taken from each end of the phase windings to an individual connection stud on the terminal block. As a result, the terminal block on a motor using this arrangement would be as shown in figure 17. The connection studs in the terminal box would be labelled as shown in figure 18.
U1 U2 U1 V1 W1
V1 V2 V2 W2 U2 W1 W2
Figure 17 Figure 18 The provision of a terminal block fitted with six connection studs allows the motor to be - • ______connected; or • ______connected; or • started via a ______starter.
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Topic 1 – Operating Principles
Within the six stud terminal block, the terminals are arranged for easy connection in either star or delta by the connection of terminal block links. The type of connection, that is, star or delta can be identified by the placement of these links. Figure 19 shows the placement of links for a star connected motor and figure 20 the placement of links for a delta connected motor.
L 1 L 2 L 3 L 1 L 2 L 3
U1 V1 W1 U1 V1 W1
V2 W2 U2 V2 W2 U2
Figure 19 Figure 20
When a motor is started via a star-delta starter there will no links used in the terminal block. The reason being that six leads have to be connected from the motor to the starter to achieve the required operation. In this case the connections at the terminal block would be as shown in figure 21.
L 3 W1 U2 L 6
to to L V1 W2 L starter 2 5 starter
L 1 U1 V2 L 4
Figure 21
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Topic 1 – Operating Principles
10. REVERSAL OF ROTATION The direction of rotation of the rotating magnetic field depends upon the order in which the successive phase currents reach their maximum positive values, that is, the- ______. Since the rotor follows the rotating magnetic field set by the stator, the direction of rotation of the motor may be reversed by reversing the phase sequence of the supply to the motor. This is achieved by reversing any two incoming supply leads. Figure 22 shows the connections to the terminal block of a squirrel cage motor. Figure 23 shows the connections required to reverse the direction of rotation of the motor.
L 1
L 2
L 3
Figure 22
L 1
L 2
L 3
Figure 23
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Topic 1 – Operating Principles
11. NAMEPLATE DETAILS Motor nameplates provide information from which you can - • determine wiring requirements • obtain performance data • obtain the information required to achieve appropriate motor replacement. The type of information found on motor nameplates may include –
manufacturer frequency rating
type of motor number of phases serial and model number full load speed frame size winding connection method output rating duty full load current rating ambient temperature voltage rating
Figure 24 shows the layout of a typical induction motor nameplate.
Motor 3~ 50Hz IEC 34-1 No. 15kW 1440 rpm Cl. F cos φ = 0.85 415V 29A ∆ Cat. No. 0410E IP55 kg
Figure 24
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Topic 1 – Operating Principles
NOTES *******
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SQUIRREL CAGE MOTOR CHARCTERISTICS PURPOSE: To investigate the operating characteristics of the squirrel cage induction motor.
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this practical assignment the student will be able to determine by test and measurement the effect of increasing the load on an induction motor on: • line current. • rotor speed. • motor slip. • motor output power.
EQUIPMENT:
• 1 x 41.5/24V three-phase, 50Hz, AC supply • 1 x 41.5V three-phase induction motor + double machine bed • 1 x eddy current load • 1 x variable DC power supply • 1 x digital multimeter • 1 x AC current clamp • 1 x optical tachometer • 4mm connection leads
– REMEMBER – – WORK SAFELY AT ALL TIMES – Observe correct isolation procedures
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Practical 1 – Operating Principles
PROCEDURE
1. MOTOR OPERATION 1. Connect the motor in delta as shown in figure 1 with the dynamometer attached.
R L 1 W L 2 BL 3 Dynamometer R 41.5V 3φ 3φ W U1 V1 W1 M U1 V1 W1 Supply B AC current clamp
U2 V2 V2 W2 U2W 2 A Variable DC Supply Figure 1 2. Do not proceed until the teacher checks your circuit and completes the progress table. Progress Table 1 attempt 1 attempt 2 attempt 3 5 2 0 3. Switch on the supply and adjust the DC power supply to the dynamometer for a motor torque of 0.1Nm. 4. Using the optical tachometer measure the rotor speed. Also measure the line current. Record both values in table 1.
Table 1 Torque Rotor Speed Line Current Slip Power Output
Nm rpm amperes % watts 0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
5. Increase to DC supply to the dynamometer for a motor torque of 0.15Nm.
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Practical 1 – Operating Principles
6. Measure motor speed and line current. Record in table 1. 7. Repeat the procedure for each of the torque values shown in table 1. 8. Do not proceed until the teacher checks your results and completes the progress table. Progress Table 2 attempt 1 attempt 2 attempt 3 5 2 0 9. Switch off the supply, then disconnect the circuit. 10. Please return all equipment to its proper place, safely and carefully. 11. The nameplate details for the motor tested are shown in figure 2.
Motor 3~ 50Hz IEC 34-1 Poles 4 No. FF42 50W 1420 rpm Cl. F cos φ = 0.7 41.5V 2A ∆ Cat. No. 0410E IP55 kg
Figure 2
2. OBSERVATIONS: 1. What is the effect of increased load on the line current taken by an induction motor? ______
2. If the load on an induction motor is increased, is the motor required to deliver more or less torque? ______
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Practical 1 – Operating Principles
3. On the axes shown in figure 3 draw the speed-torque curve for the motor tested.
0.45 0.4
0.35 Nm 0.3 - 0.25 0.2 Torque Torque 0.15 0.1 0.05
0
0 900 700 800 1500 1300 1200 1400 1000 1100 Rotor Speed - rpm Figure 3
4. Based on your observations and the graph in figure 3, what effect does increased load have on the speed of an induction motor? ______
5. For each value of torque determine the percentage slip of the motor and record the values in the appropriate column of table 1. Show the complete calculation for the first value in the space below.
Use the equation S% = x 100 𝜂𝜂𝑆𝑆 − 𝜂𝜂 ______𝜂𝜂𝑆𝑆 ______
6. Based on your observations, what effect does increased load have on the slip of an induction motor? ______
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Practical 1 – Operating Principles
7. For each value of torque determine the power output of the motor and record the values in the appropriate column of table 1. Show the complete calculation for the first value in the space below.
Use the equation P = 2πηT ______OUT 60 ______
8. Based on your results, what effect does increased load have on the output power of an induction motor? ______
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Practical 1 – Operating Principles
NOTES *******
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Tutorial 1 – Operating Principles
OPERATING PRINCIPLES SECTION A In the following statements one of the suggested answers is best. Place the identifying letter on your answer sheet. 1. The rotor current in a three phase induction motor is:- a) zero, since no supply is connected to the rotor circuit; b) supplied by the d.c. connected to the rotor terminals; c) supplied by the a.c. connected to the rotor terminals; d) induced by the stator field cutting the rotor conductors.
2. A three phase winding will produce an electromagnetic field which:- a) rotates at a constant speed; b) reverses direction each cycle; c) reverses direction each half cycle; d) is stationary and constant in strength.
3. Increasing the frequency of supply to a three phase stator winding will:- a) cause the magnetic field to rotate faster; b) cause the magnetic field to rotate slower; c) increase the strength of the magnetic field; d) increase the number of poles in the stator winding.
4. To reverse the direction of rotation of a rotating magnetic field you must:- a) reverse the connections to alternate pole windings; b) reverse the phase sequence of the supply; c) reverse the connections to the rotor winding; d) reverse the connections to all pole windings.
5. The rotor speed of an induction motor is:- a) always slightly higher than the speed of the rotating magnetic field; b) always slightly lower than the speed of the rotating magnetic field; c) always the same as the speed of the rotating magnetic field; d) dependant only on the size of the load the motor is driving.
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Tutorial 1 – Operating Principles
SECTION B Blank spaces in the following statements represent omissions. Write the appropriate information on your answer sheet. As load on an induction motor increases the speed of the motor will _____(1)_____ and the slip speed of the motor will _____(2)_____. The speed of the rotating magnetic field in a three phase induction motor depends on the number of poles in the winding and the _____(3)_____ of the supply . When an induction motor is driving a load the speed of the motor cannot reach_____(4)_____ speed . The stator core of a three phase induction motor is laminated to reduce _____(5)_____ loss and the laminations are made from silicon steel to reduce _____(6)_____ loss. The stator winding of a three phase induction motor consists of _____(7)_____ identical windings placed at_____(8)_____ electrical degrees from each other. To change the direction of a three phase induction motor the connections to any _____(9)______of the supply conductors must be changed. Either the rotor slots or the stator slots are _____(10)_____ to reduce the noise from a three phase induction motor.
SECTION C 1. A six pole three phase induction motor is connected to a 60Hz supply and runs at full load at 1050r.p.m. Determine:- a) the synchronous speed of the motor; (1 200 r.p.m.) b) the slip speed of the motor. (150 r.p.m.)
2. Answer review questions 1-4 on page 257 of your textbook.
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Tutorial 1 – Operating Principles
SECTION D 1. Figure 1 represents the stator windings of a three phase induction motor and it's terminal block:- a) Connect the windings to the terminal block using the international standard; b) Connect the terminals of the terminal block so that the motor windings are connected in delta; c) Connect the terminal block to the supply terminals. d) Connect the terminal block of Figure 2 to the supply terminals so that the direction of rotation of the motor in Figure 1 is reversed.
L1 W1 U2 . . L2 V1 W2 . . L3 U1 V2 . .
Supply Terminal Block Windings
Figure 1.
L1 W1 U2
L2 V1 W2
L3 U1 V2
Supply Terminal Block Figure 2.
Tutorial answer sheet Name: ______
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Tutorial 1 – Operating Principles
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COMPLEX CAGE AND SLIP RING INDUCTION MOTORS
PURPOSE: This section introduces the behaviour and characteristics of three phase slip-ring induction motors. In addition, the torque-speed characteristics of the squirrel cage motor are considered.
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this section the student will be able to: • Identify the components parts of a slip-ring induction motor. • Interpret information from the torque-slip curves of the slip-ring induction motor. • Reverse the direction of rotation of a slip-ring induction motor. • List the operating characteristics of slip-ring induction motors. • List the advantages and disadvantages of both squirrel cage and slip-ring induction motors. • Calculate the output power of induction motors given the shaft speed and torque.
REFERENCES:
• Hampson, J. & Hanssen, S. Electrical Trade Principles, 2nd Edition, Section 6 • Jenneson, J.R. Electrical Principles for the Electrical Trade, 5th Edition, Pages 223 - 232
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Section 8 – Slip Ring Ind. Motors
1. SQUIRREL CAGE MOTOR CHARACTERISTICS
Rotor Impedance and Torque *Read the first paragraph of the section titled rotor impedance and Torque on page 253 of your textbook and answer the following question: Student exercise Maximum torque of an induction motor occurs when the rotor ______is equal to the rotor ______.
In a squirrel cage motor the resistance and inductance of the rotor is set when the motor is designed and can’t be altered.
The performance characteristics of the squirrel cage motor are governed by - • ______and • ______. Squirrel cage induction motors are manufactured in three general types shown below. Table 1 Motor type Rotor construction General purpose motor - low resistance - normal starting torque - low inductive reactance - low starting current - normal slip.
Moderate starting torque - low resistance -higher starting torque - high inductive reactance -lower starting current -higher slip
High starting torque - high resistance - highest starting torque - low inductive reactance - lowest starting current - highest slip.
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Section 8 – Slip Ring Ind. Motors
Typical torque-speed performance curves for the three types of motor are shown in figure 1. 30 0 3 25 Motor type Curve 0 2 20 General purpose 1 0 1 15 Moderate starting torque 2 0
10 High starting torque 3 0 que 5 Percentage Full LoadPercentage Full Tor 0 0 0 1 2 3 4 5 6 7 8 9 10 0 0 0 0 0 0 0 0 0 0 Percentage Synchronous
Figure 1
Table 2 details the characteristics of the three motors whose torque-speed curves are shown in figure 1. Table 2 Curve Starting Torque Starting Percentage Slip Applications Current General purpose motor
where load power Approx. Normal - less Normal - approx. requirements are not 1 6 x IFL that 5% 1.5 x TFL severe. For example, fans, blowers and machine tools.
Used where the motor
is required to start at High - approx. Approx. Increased - full load. Examples 2 to 2.5 x TFL 6 x IFL around 5% 2 include, driving conveyors, reciprocating pumps and compressors.
High - approx. Approx. High - Flywheel mounted 3 3 x TFL 5 x IFL up to 10% machinery such as presses and punches.
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Section 8 – Slip Ring Ind. Motors
2. DUAL CAGE ROTORS With the object of achieving the features of - • high efficiency • small speed variation • high starting torque and low starting current.
The dual cage rotor is fitted with two squirrel-cages in specially shaped slots as shown in figure 2.
Higher resistance outer cage bars
Low resistance High inductive reactance inner cage conductors
Figure 2 At the moment of starting:
• The rotor frequency is high (eg. 50Hz) causing rotor XL to be high (XL = 2πfL)
• The higher resistance outer cage combines with the high reactance (XLR) of the inner cage (rotor XL = rotor R) • A high starting torque is produced with a low starting current.
At normal operating speed - • The rotor frequency is so low that the reactances of both cages are negligible. • Rotor current flows in the low resistance inner cage and the outer cage, giving the motor the running characteristics of a low-resistance squirrel-cage motor. The torque-speed characteristic for an induction motor fitted with a double cage rotor is shown in figure 3. 300
250
200
150
100 50 Percentage Full LoadPercentage Full Torque 0 Figure 3 0 10 20 30 40 50 60 70 80 90 100 Percentage Synchronous Speed Page 4 of 22 Version 2 Machines Section 8 Last Updated 31/03/2021 Owner: Elect, ICT & Design Faculty/Electrical Disclaimer: Printed copies of this document are regarded as uncontrolled.
Section 8 – Slip Ring Ind. Motors
3. MOTOR CURRENT
Starting Current
When power is first applied to a stationary motor, the stator windings act as the primary of a transformer and the rotor behaves as a shorted secondary winding. As a result of this action: • a ______circulating current flows in the rotor and • a ______starting current is taken by the stator windings.
Running current As the rotor accelerates in the direction of the rotating magnetic field - • the difference between rotor speed and the rotating magnetic field ______• the generated rotor voltage ______• causing the rotor circulating currents to ______• which in turn ______stator current. The typical relationship between the stator current and the rotor speed is shown in figure 4.
% Current 100 0 0 Speed
Figure 4 The initial circulating current in the rotor is affected by the - • ______• ______• ______That is, the rotor current is limited by the impedance of the rotor circuit.
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Section 8 – Slip Ring Ind. Motors
4. TORQUE - SPEED CURVES As discussed, the torque a motor develops depends upon its speed, but the relationship between the two cannot be expressed in the form of an equation. Consequently, the relationship is shown in the form of a curve, known as the "torque-speed curve".
Torqu e
0 0 Speed 100%
Figure 5
Student exercise *See Figure 6.17 on page 253 of your textbook.
Starting torque is typically = ______full load torque
Breakdown torque is typically = ______full load torque
3. Starting torque is sometimes referred to as ______torque.
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Section 8 – Slip Ring Ind. Motors
5. SLIP-RING MOTOR CONSTRUCTION Many industrial applications require three phase motors with high starting and acceleration torque. The slip-ring induction motor, also called the wound rotor induction motor, meets these needs.
Figure 6 – exploded view of slip ring motor
• Stator - the same stator construction and winding arrangement as the squirrel- cage induction motor. Star connected coils • Wound Rotor 120 electrical degrees apart Carbon brushes ride on these Laminated steel core slip rings
Slip Rings Figure 7
*See Figure 6.12 in your textbook
6. TERMINAL MARKINGS The terminal markings used in conjunction with slip-ring motors are as follows - • the stator leads are labelled U, V and W. This is the same marking system used with squirrel-cage induction motors. • the rotor leads are marked K , L and M.
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Section 8 – Slip Ring Ind. Motors
The circuit diagram and standard terminal marking of the slip-ring motor are shown below in figure 8.
Stator Rotor
Figure 8
The Australian Standard drawing symbol for the slip-ring motor is shown in figure 9.
U K
V L Figure 9 W M
7. OPERATION Three different operating conditions for the slip ring motor will be considered- • slip-rings open circuit • slip-rings short circuited • resistance added to the rotor circuit via slip-rings and brushes.
Slip-rings Open Circuit: The motor circuit is arranged as shown in figure 10.
U
K L M V
W Stator Rotor
Figure 10
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Section 8 – Slip Ring Ind. Motors
*Student exercise
1. Draw the delta Connections on the motor terminal block below.
L 1 L 2 L 3
U1 V1 W1
V2 W2 U2
Figure 11
2. Connect the motor in delta as shown in figure 12, with the slip-rings open circuit. R K Slip rings 41.5V 3φ 3φ W open M L circuit Supply B AC current clamp M
A
Figure 12 3. Switch on the supply and note any effects associated with the motor (Does the motor run?) ______
4. Measure the stator line current, stator line voltage and rotor line voltage. Record your results in table 3.
Table 3 Stator Line Stator Line Rotor Line Current Voltage Voltage amperes volts volts
5. Why is rotor voltage present in the slip-ring motor when the slip-rings are open circuit? ______
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Section 8 – Slip Ring Ind. Motors
Conclusions As the rotating field travels at the synchronous speed, it cuts the wound-rotor windings and - • ______voltages in the rotor windings • as the rotor winding is open circuit ______rotor currents flow • with no rotor current ______torque is developed. Therefore, with the slip-rings open circuit, the stator draws current from the supply but the motor does not ______.
Slip-Rings Short Circuited: The circuit diagram for the motor with the slip-rings short circuited is as shown in figure 13.
U
K L M V
W Stator Rotor
Figure 13
*Student exercise 1. Connect the motor as shown in figure 14. 2. Connect the stator again in delta with the slip-rings short circuited.
R 41.5V K 3φ 3φ W M L Supply B AC current clamp M
Slip rings A short circuited Figure 14 3. Switch on the supply and run the motor on no load.
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Section 8 – Slip Ring Ind. Motors
Measure the line current taken by the motor and the speed of rotation. Record your results in table 4. Table 4 Line Current Speed amperes rpm
4. How does the operation of the slip-ring motor, with slip-rings short circuited, compare to that of the squirrel cage motor? ______Conclusions - Under these operating conditions - • the stator establishes a ______magnetic field • the rotating field ______voltages into the rotor windings • induced rotor voltages cause rotor ______to flow • the interaction between the rotating field and the rotor field results in the development of ______• the rotor ______and accelerates up to final operating speed. With the slip-rings short circuited the motor operates exactly the same as the squirrel cage motor.
Added Rotor Resistance: The operating characteristics of the motor may be changed by the insertion of resistance into each phase of the rotor circuit. Resistance may be added to the rotor circuit via the slip-rings and brushes. It is normal practice to use three resistors, one per rotor phase. The three resistors are star connected and connected to the brushes as shown in figure 15.
U R R K K L M R V
M L
W Stato Roto
Figure 15
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Section 8 – Slip Ring Ind. Motors
Student exercise – Connect the circuit as shown in figure 16, with a 1Ω, 10W resistor added to each phase of the rotor. Note: Connect the motor to the inertia wheel.
Rotor resistance R K 1Ω 41.5V 3φ 3φ W M L 1Ω Supply B AC current clamp M 1Ω
A Figure 16
1. Switch on the supply, then measure the stator current and the rotor speed. Record the values in table 5.
Table 5 Added Rotor Stator Line Rotor Resistance Current Speed ohms amperes rpm 1Ω
2.2Ω
4.7Ω
6.8Ω
2. Replace the three 1Ω rotor resistors with three 2.2Ω, 4.7Ω and then 6.8Ω.resistors and repeat the procedure. 3. What effect does added rotor resistance have on the operating speed of the slip- ring induction motor? ______
Conclusions To examine the effect of the additional resistance, consider the motor is running at a particular speed and the resistance is then added - • added resistance causes rotor current to ______• for an instant decreased current causes a ______in torque • reduced torque causes the rotor speed to ______• decreased rotor speed causes ______rotor emf • increased rotor emf causes ______rotor current • motor torque returns to the value required to drive the load but at ______
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Section 8 – Slip Ring Ind. Motors
Slip-ring motors have the advantage that maximum torque can be made to occur at starting and at any speed by adding resistance to the rotor circuit to match the rotor reactance. ( max torque when Rotor XL = Rotor R)
The circuit used to achieve this operation is shown below in figure 17.
U K
V L
W M
8. TORQUE PERFORMANCE In the slip-ring motor rotor circuit resistance is varied via the rotor rheostat to achieve a torque-speed characteristic which suits the requirements and nature of the load. The torque-speed curves shown in figure 18 illustrate the variations that can be achieved with the slip-ring motor by variation of the rotor resistance.
Curve 1 high resistance added
2 Curve 2 lower resistance added
Curve 3 resistance removed
1 - 3
Torque Torque Nm
0 Figure 18 0 Speed - rpm
Curve 1 - rotor resistance equals rotor reactance at standstill, develops ______starting torque. Curve 2 - rotor resistance is lower than standstill reactance (rotor resistance equals rotor reactance during acceleration) Curve 3 - no added rotor resistance, operation is the same as the squirrel cage motor Curve 4 - the starting characteristic that is achieved by varying the external resistance as the motor accelerates, maximum torque is developed throughout acceleration.
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Section 8 – Slip Ring Ind. Motors
Secondary liquid resistance starter varies rotor external resistance as motor accelerates.
Figure 19
9. TORQUE AND OUTPUT POWER If a force (F) be applied to the shaft of a machine at some distance (r) from the centre as shown in figure 20, the tendency to turn the shaft is given by the product of the force (F) and the radius (r). ______
Torque is the effort required to rotate a shaft, flywheel, armature or any other device about its centre.
Figure 20
In the case of a motor the torque developed within it causes the motor shaft to rotate.
In an induction motor this torque is developed by the reaction of the rotor currents with the rotating magnetic field of the stator.
The output power delivered by an induction motor to a load depends on the - • ______• ______.
Power is ‘the rate at which work is done’.
In a motor, the work done per minute is given by -
Work done per revolution per minute = 2 x π x n x T
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Section 8 – Slip Ring Ind. Motors
Since the mechanical unit of power, the watt, is the rate of doing work, the power developed by the motor is -
Where: POUT = motor output power in watts n = motor speed in rpm T = torque developed by motor in Newton metres
Example: 1 A three phase induction motor operates with a rotor speed of 1440 rpm and develops a torque of 66.32Nm at full load. Determine the full load power output. ______
Example: 2 A 3.75kW induction motor operates with a full load speed of 1465 rpm. Determine the torque developed by the motor. ______
Since maximum torque can be maintained throughout the acceleration, the slip-ring motor is used where high starting torque and running torque is required.
Typical applications for this motor include - • cranes • large compressors • elevators • guillotines • rock crushers • pumps and presses. This type of motor is also used in applications requiring adjustable speed.
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Section 8 – Slip Ring Ind. Motors
10. REVERSAL OF ROTATION The direction of rotation of a wound-rotor induction motor can be reversed. To do this, interchange the connections of any two of the three line leads feeding the stator windings. Figure 21 shows the connection changes that are required to reverse the direction of rotation. Note, there is no reversal in the direction of rotation of the motor when any of the leads feeding from the slip-rings to the speed controller are interchanged.
U K U K
V L V L
W M W M
11. COMPARISON OF SQUIRREL CAGE AND SLIP RING MOTORS Table 6 highlights the advantages and disadvantages of the squirrel cage and slip ring induction motors.
Table 6 Motor Advantages Disadvantages • Simplicity and • Relatively poor ruggedness of construction. starting torque.
• • Squirrel cage No sliding electrical Large starting motor contacts, hence no sparking and current. possible use in explosive • Fixed atmospheres. characteristics. • Wide range of speed control when used with electronic variable speed drives.
• High Starting • High production cost compared torque. to the squirrel cage motor.
• • Slip-ring Relatively low Brushes and brush gear must be motor starting current. used causing the possibility of • Smooth speed sparking and increased control over a wide range. maintenance. • Inefficient speed control.
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NAME: Practical 8 *If required use practical 10 ‘induction motor starting’ from previous work book.
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Tutorial 8 NAME:
SLIP-RING INDUCTION MOTORS SECTION A In the following statements one of the suggested answers is best. Place the identifying letter on your answer sheet. 1. An advantage of wound rotor induction motors is:- a) high starting current and torque; b) low starting torque with low current; c) low starting current and high starting torque; d) high starting current and low starting torque.
2. The rotor windings of a slip ring induction motor are connected to an external:- a) source of a.c. supply; b) source of d.c. supply; c) variable resistance; d) star delta starter.
3. The rotor and stator windings of a slip ring induction motor are normally connected:- a) rotor in star and stator in delta; b) rotor in delta and stator in delta; c) rotor in star and stator in star; d) rotor in delta and stator in star.
4. Resistance is added to the rotor circuit of a slip induction motor to:- a) increase torque at starting b) increase torque during acceleration c) reduce current during starting; d) all of the above.
5. In a squirrel cage induction motor with dual cage rotor:- a) the inner cage has the higher resistance and carries the greater current at starting; b) the outer cage has the higher resistance and carries the greater current at starting; c) the inner cage has the higher resistance and carries the least current at starting; d) the outer cage has the higher resistance and carries the least current at starting.
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Tutorial 8 Slip-Ring Induction Motors SECTION B Blank spaces in the following statements represent omissions. Write the appropriate information on your answer sheet. Adding resistance to the rotor circuit of the slip ring induction motor at starting _____(1)_____ the starting current taken by the motor and _____(2)_____ the available starting torque. If two of the three leads connecting the slip rings of the slip ring induction motor to it's starting resistors are reversed the direction of rotation of the motor will be _____(3)_____. The unit of torque is the _____(4)_____. The direction that a three phase induction motor rotates depends on the _____(5)_____ of the supply currents. The _____(6)______induction motor has a wound rotor winding which is always _____(7)_____ connected to allow the connection of _____(8)_____ during starting. Dual cage rotor motors are designed to give ______(9)______starting toque combined with _____(10)_____ slip.
SECTION C Explain the difference between a general purpose and a moderate starting torque squirrel cage rotor. (You may use a drawing if you like)
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Slip-Ring Induction Motors Tutorial 8 SECTION D 1. Figure 1 represents the incomplete circuit diagram of a three phase slip ring induction motor and it's supply:- a) Connect the stator windings in delta; b) Connect the rotor windings in star; c) Complete the required connections, including the supply and starting resistors, for the motor to start and run.
L L L 1 2 3 L1 . . U1 . . U2 L2 V1 V2 . . .U1 V1 W1. L3 . . W1. . W2 V2 W2 U2 Stator Windings Supply Isolating Switch
K1 K2 . .
L1 L2 . .
M1 M2 . .
Slip Rings Rotor Windings
.
.
. Starting Resistors Figure 1.
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Tutorial 8 Slip-Ring Induction Motors Tutorial answer sheet Name: ______
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Tutorial 9 NAME: INDUCTION MOTOR LOAD CHARACTERISTICS
PURPOSE: To examine the load characteristics of the three phase induction motor and the effect of supply voltage variation on torque developed.
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this section the student will be able to: • Interpret information from the load characteristic curves of an induction motor. • Describe the flow of power through the induction motor. • Calculate the percentage efficiency of an induction motor. • State the losses associated with an induction motor. • State the components of induction motor stator current. • State the effects that a change of load has on the following quantities associated with an induction motor -
o speed o rotor and stator current o power input and output o power factor o efficiency. • Calculate the torque of an induction motor given the motor operating voltage, rated voltage and rated torque.
REFERENCES: Electrical Trade Principles 2nd Edition J Hampson & S Hanssen. Section 6 Electrical Principles for the Electrical Trades. 5th Edition. Jenneson J.R. Pages 229-231.
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Tutorial 9 Induction Motor Load Characteristics
1. POWER FLOW & LOSSES Figure 1 below shows the power flow and losses in an induction motor
Figure 1
2. MOTOR RATING AND EFFICIENCY Motors convert electrical energy to mechanical energy, as a result motors are always rated in terms of the ______power. The flow of power through a three phase induction motor is shown in figure 2.
Figure 2 Losses
Electrical power input Mechanical power output
The power input to the three phase motor is determined using -
where: VL = motor line voltage
IL = motor line current λ = motor power factor
Induction Motor Load Characteristics Tutorial 9
The output power from the motor is determined using -
where: n = rotor speed in rpm T = motor torque in Nm
Motor efficiency varies with changes in load conditions and can be determined by taking the ratio of the output power to the input power -
where: POUT = motor output power in watts
PIN = motor input power in watts
Note: output power is always less than the input power due to the losses that occur in the energy conversion process.
Tutorial 9 Induction Motor Load Characteristics
Example: 1 Calculate the full load efficiency of a three phase induction motor given - rating = 110kW voltage = 400V
λ = 0.86 lag IFL = 194A ______
Example: 2 Calculate the power output and full load line current of a three phase induction motor, given the following - Efficiency = 89.3% voltage = 400V torque = 49.3Nm Speed = 1450rpm λ = 0.81 lag ______
Induction Motor Load Characteristics Tutorial 9
3. EXAMPLE: 3 A 3 phase 4-pole, 11kV, 50Hz induction motor operates with a rotor speed of 1485 rpm and takes a line current of 44.6A at a power factor of 0.93 lag when delivering its full load output of 750kW. Determine the - a) power input b) efficiency c) total losses d) torque developed. ______
Review points • speed of a motor ______fairly uniformly as the power output increases. • rotor current ______as the load increases. • stator current depends upon the rotor current, therefore it ______as the load increases • power-factor ______as the load increases • power-input ______as the load increases. • efficiency ______as the load is increased, reaching a ______at approximately full-load, but ______when the motor is overloaded.
Tutorial 9 Induction Motor Load Characteristics
INDUCTION MOTOR LOAD CHARACTERISTICS SECTION A In the following statements one of the suggested answers is best. Place the identifying letter on your answer sheet.
1. Copper loss in an induction motor is due to:- a) hysteresis in the stator and rotor cores; b) eddy currents in the stator and rotor core; c) resistance of the stator and rotor windings; d) friction and windage loss in the motor.
2. The mechanical losses on no load in an induction motor include:- a) hysteresis in the stator and rotor cores; b) eddy currents in the stator and rotor core; c) resistance of the stator and rotor windings; d) friction and windage loss in the motor.
3. The difference between the input power to a motor and the output power from a motor is the:- a) total loss given off as heat; b) electrical loss given off as resistance; c) mechanical loss given off as friction; d) magnetic loss given off as inductance.
4. An increase in rotor current in an induction motor:- a) reduces stator current to keep stator flux constant; b) reduces power factor due to rotor inductance; c) reduces motor efficiency due to increased losses; d) causes stator current to increase to maintain stator flux.
5. As the load on an induction motor increases:- a) both the efficiency and the power factor improve; b) efficiency increases but power factor decreases; c) efficiency decreases and power factor decreases; d) efficiency decreases but power factor increases.
Induction Motor Load Characteristics Tutorial 9
SECTION B Blank spaces in the following statements represent omissions. Write the appropriate information on your answer sheet. Copper losses in a squirrel cage induction motor _____ (1) _____ with an increase in load. Iron losses are made up of ______(2)______and ______(3) ______losses and remain relatively _____ (4) _____. Mechanical losses in a motor occur due to _____ (5) _____ and _____ (6) _____.
Maximum efficiency occurs when the _____(7)_____ losses equal the _____(8)_____ losses while maximum torque occurs when the _____(9)_____ equals the _____(10)_____ .
SECTION C
1. A 4 pole three phase 415 volt 50Hz squirrel cage induction motor runs at 1460 r.p.m. while delivering 9 kilowatts to its load at a maximum efficiency of 88 percent. If the power factor of the motor at this load is 0.86 lag determine:- a) the input power to the motor; (10 227W.) b) the line current taken by the motor; (16.55A.) c) the torque delivered to the load; (58.9Nm.) d) the losses in the motor; (1 228W.) e) the copper losses in the motor. (614W.)
Tutorial 9 Induction Motor Load Characteristics
SECTION D
1. Figure 1 is a set of performance curves for a 5kW, 6 pole 415 volt squirrel cage induction motor. From the curves determine:- a) Line current, speed and efficiency at rated load; b) Input power, line current and power factor at no load; c) Speed and efficiency when stator current is 8 amperes.
1000 60 1.0 15 100 Speed
Power Factor
Efficiency %
Power Input
- Rotor Current
-
Speed rpm Power Factor Stator Current
Input Power Power Input kW
-
0 Percentage 0 Efficiency 0 0 0 1 2 3 4 5 6 7 8 Power Output - kW
Stator & Rotor Current Stator Rotor Current & Amperes Figure 1
The Single Phase Split Phase Motor Section 10
SINGLE PHASE SPLIT PHASE MOTOR
PURPOSE: This section introduces the single phase split phase induction motor.
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this section the student will be able to: • Describe the principle of operation of the single phase split phase motor. • Outline the construction and basic characteristics of the single phase split phase induction motor. • List at least three applications of the split phase induction motor. • Connect, run and reverse a split phase induction motor. • List typical faults associated with split phase induction motors. • Name the tests and describe the testing procedures for the identification of faults associated with split phase induction motors.
REFERENCES:
Electrical Trade Principles 2nd Edition J. Hampson & S. Hanssen. Section 6 pages 247 -323
Electrical Principles for the Electrical Trades. 5th Edition. Jenneson J.R. Pages 232-235.
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Section 10 The Single Phase Split Phase Motor
1. INTRODUCTION Single phase motors may be divided into two general classes, and in turn, these classes can be broken down into the individual types of motor. Figure 1 illustrates this type of motor classification. Split phase motors Those with squirrel cage rotors Capacitor Motors Shaded Pole Motors
Single Phase Motors
Repulsion motors Those with wound rotors Repulsion start, induction run motors Universal or series motors
Figure 1
In a single phase motor a normal single phase winding on the stator - • does not produce any ______torque, • because it cannot develop a ______magnetic field.
The single phase magnetic field simply: • ______or alternates in direction with the changes in the current, • And will not ______the rotor
Single phase motors constructed on the rotating magnetic field principle must have some auxiliary help to enable a rotating magnetic field to be produced. Such auxiliary help may be in the form of phase ______, or using a ______coil.
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The Single Phase Split Phase Motor Section 10
2. SINGLE PHASE INDUCTION MOTOR A single phase winding supplied from an alternating current source produces a field that is alternating in magnitude and direction, but with a constant axis. This pulsating field will not produce a torque.
Positive half cycle Negative half cycle Figure 2 The reason for this can be seen in figure 3. If a rotor is placed in the magnetic field produced by a single phase field system, the squirrel cage winding acts as the secondary of a transformer and emfs are induced in the rotor conductors by transformer action. These induced emfs cause rotor currents to flow which are at all times in such a direction as to oppose any change in the magnetic field of the stator (Lenz's Law). For the arrangement shown in figure 3, the rotor currents form two bands in opposite directions, one to the left and one to the right of vertical, which establish a rotor field along the axis of the stator field.
Positive half cycle Negative half cycle Figure 3 The net effect of this action is that the rotor is locked in position by the magnetic attraction of the primary and secondary fields. Thus not only is there no starting torque developed by the motor, but also a torque must be applied to cause the rotor to rotate.
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Section 10 The Single Phase Split Phase Motor
*Teacher demonstration: Using the 24V single phase split phase motor: 1. Connect the main winding only and turn the supply on. Does the motor run? 2. Spin the motor by hand Does the motor continue to run?
When the rotor is started mechanically by spinning the shaft, the rotor conductors cut the stator magnetic field and cause emfs to be induced. The currents circulating in the rotor bars as a result of these induced emfs produce a field with an axis at right angles to the stator field. This is shown in figure 4.
Rotation Rotation
Figure 4
Positive half cycle Negative half cycle
The combination of the two magnetic fields at right angles results in a rotating flux which serves the same purpose as that produced by the stator windings of a three phase induction motor. The result of these two fields is a rotating magnetic field which, when applied to the rotor, maintains the rotor in motion. The speed at which this field rotates is known as the "synchronous speed".
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The Single Phase Split Phase Motor Section 10
TORQUE SPEED CURVES Figure 5 below shows the typical torque-speed curve for a single phase stator winding. Although the starting torque is zero, the motor develops a reasonable value of torque as it approaches synchronous speed. As a result, once a single phase motor has started to rotate, the motion is maintained and the rotor develops a torque that can drive a load.
25
10
Torque % Torque
0 0 Speed 100% Figure 5 The various modifications of single phase motors are designed essentially to create the initial torque that can start the rotor spinning.
3. SPLIT PHASE MOTOR To produce a starting torque in a single phase motor, a rotating field must be created. This is done by fitting two windings to the stator, as shown in figure 6. These windings are known as the - • ______
• ______
Figure 7
Figure 6
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Section 10 The Single Phase Split Phase Motor
The start winding is placed near the tops of the slots and is displaced from the main or run winding by 90 electrical degrees. See figure 7. To make the two windings electrically dissimilar they are wound with different gauge wires, with a different number of turns and are physically positioned at different depths in the stator slots. The windings are named by the function that they serve with respect to the operation of the motor. It is important to note that the two windings are always wound to produce the same number of poles. That is, if the main winding is wound with four poles the auxiliary winding will also be wound with four poles.
4. WINDING CHARACTERISTICS
Table 1 highlights the characteristics of the main and auxiliary windings of the split phase motor.
Table 1 Main Winding Auxiliary Winding
(run or working winding) (start winding) heavy gauge wire fine gauge wire Physical small number of turns Characteristics large number of turns wound deep in stator slots wound in the top of stator slots
Electrical Characteristics
The phase angle between the start and run windings is achieved by each winding having a different electrical characteristic. I.E: more resistive / less resistive, more inductive / less inductive. Often this results in the start winding having a higher resistance than the run winding, but this is not always the case.
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The Single Phase Split Phase Motor Section 10
V
Since the winding currents are out of phase (split) by approximately 30 electrical degrees, their resultant magnetic fields will also be out of phase by the same amount. The phasor addition of the two fields will produce a non-uniform, rotating magnetic field during the starting period. This non-uniform field acting on the rotor produces sufficient starting torque for the rotor to commence rotation and accelerate up to speed. After the motor has run up to speed, the auxiliary winding is open circuited. The vibration and humming noise associated with split phase motors when they are started is due to the non-uniformity of the rotating magnetic field.
Because of the characteristics of the auxiliary winding, high resistance wire and low inductance, the winding cannot be left connected to the supply for an extended period. If it was, the winding would overheat and burn out.
To eliminate this problem, the auxiliary winding is connected to the supply only during the starting period and once the rotor has reached: • approximately ______the winding is disconnected • via a ______. The circuit diagram for the split phase motor is shown in figure 8.
U1 U 2
Z 1 Z 2 n
AC Supply
Figure 8
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Section 10 The Single Phase Split Phase Motor
5. CONSTRUCTION The constructional breakdown of the split phase motor is shown in figure 10.
Figure 10
6. STARTING There are two methods that are commonly used to disconnect the auxiliary winding from the supply once the motor has started - • ______• ______
CENTRIFUGAL SWITCH As shown in figure 11, the centrifugal switch is made up of basically two parts - • the switch contacts - which are usually mounted on the inside of the end shield • the centrifugal device - which is mounted on the shaft of the rotor.
Figure 11
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The Single Phase Split Phase Motor Section 10
When the rotor is at standstill the force exerted by the springs on the centrifugal device hold a sliding collar back away from the rotor and the contacts are closed. See figure 12.
Figure 12 Figure 13 When the motor is started, the rotor accelerates towards synchronous speed. At approximately 75% of rated speed the forces exerted on the weights of the centrifugal device overcome the spring tension, causing the weights to fly out from the shaft and the collar to slide along the shaft toward the rotor and open the contacts. See figure 13. When the motor is turned off, the switch closes at approximately 25% of rated speed.
CURRENT RELAYS The alternative to the centrifugal switch is the current relay, which consists of a few turns of heavy gauge winding wire as shown in figures 14 and 15.
Figure 14 Figure 15 The coil of the current relay is connected in series with the main winding and the contacts of the current relay are connected in series with the auxiliary winding.
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Section 10 The Single Phase Split Phase Motor
Figure 16 below shows the connections for a current relay.
U1 U2
Z 1 Z 2
AC Supply Figure 16
The large starting current of the main winding causes the relay to close its contacts, connecting the auxiliary winding to the supply and allowing the motor to start. With the motor started the main winding current decreases and the relay contacts open, disconnecting the auxiliary winding.
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The Single Phase Split Phase Motor Section 10
7. SPLIT PHASE MOTOR CHARACTERISTICS The characteristics of the split phase motor are as detailed below - • typically the starting current is ______the rated current. • starting current is ______and ______as the motor accelerates up to speed. • typically the starting torque is ______the rated full load torque. • typical output power ratings are ______. • the number of starts/unit time is limited due to heating of the ______winding. A generalised torque-speed characteristic for the split phase motor is shown in figure 17.
250 Torque Torque %
100
0 0 100% Speed Figure 17 The main advantage of this type of motor is that it is operated from a single phase supply, whereas the main disadvantages are - • ______and • ______.
Practical applications for single phase split phase motors include - • washing machines • blowers • buffing and polishing machines • grinders • machine tools.
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Section 10 The Single Phase Split Phase Motor
8. SPLIT PHASE MOTOR REVERSAL The split phase motor will rotate in a direction governed by the instantaneous direction of current flow in the main and auxiliary windings. To reverse a split phase motor it is necessary to reverse the connections to the - • ______or the • ______, but not both. Figure 18 shows the connections for a split phase motor to operate in the forward direction. Main U1 U2
Auxiliary CS Z 1 Z 2 n
AC Supply Figure 18
Example: 1 Draw the necessary connections on figures 19 and 20 to reverse the direction of rotation of the motor, firstly by reversing the auxiliary winding and secondly the main winding.
Main Main U1 U 2 U1 U2
Auxiliary Auxiliary CS CS Z 1 Z 2 Z 1 Z 2 n n
AC Supply AC Supply Figure 19 Figure 20
Note - because of the low starting torque produced, the split phase motor does not lend itself to reversal from a running condition.
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The Single Phase Split Phase Motor Section 10
9. OVERTEMPERATURE PROTECTION Small single phase motors are often too small to use separate 3 terminal thermal overload protection devices. Protection is often accomplished by the use of thermistors or 'Klixons'.
THERMISTORS Thermistors are made from semiconducting metal oxides and are embedded into the stator windings of a motor to detect temperature change. The thermistors have a positive temperature co-efficient (PTC), that is, as their temperature goes up, so does their resistance. These devices are connected to external control equipment and are designed to stop the motor should the operating temperature exceed 150% of normal. Thermistors are ideally suited to motors that have poor ventilation.
KLIXONS Klixons are placed on the end shield of a motor and detect the heat being radiated out from the windings. These devices are wired in series with the main winding and disconnect supply should the motor get too hot. Klixons can be of the manual or automatic reset type. Klixons are usually used for motors rated 550W or less.
10. TESTING AND FAULT FINDING The insulation resistance between windings and between windings and earth must be at least equal to or greater than 1MΩ. The motor frame and associated exposed metal parts must be suitably earthed. Given that the auxiliary winding is designed for intermittent operation only, it is the winding which is generally affected by fault conditions. This comes about primarily because of the fine gauge wire with which it is wound, and the associated faults are usually caused by overheating. Table 2 lists some of the common faults associated with split phase motors.
Table 2 Damage to Auxiliary Winding Possible Causes may be due to for Not Starting contacts of the centrifugal switch fused loss of supply current relay contacts fused thermal cut-out needs resetting too frequent start/stop actions oxidation of centrifugal switch contacts overload - causing prolonged acceleration open circuit in either winding incorrectly connected windings locked rotor low supply voltage
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Practical 10 NAME:
SINGLE PHASE SPLIT PHASE MOTOR PURPOSE: To develop the students ability to carry out the testing of a single phase motor to correctly connect its windings, and determine whether the motor can be safely connected to the supply.
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this practical assignment the student will be able to: • measure the resistance and inductance of the start and run windings of a single phase split field motor. • correctly connect the motor to run in a specified direction. • carry out tests to determine whether the motor is safe to connect to the supply.
EQUIPMENT:
• 1 x 24 volt split phase induction motor + single bed • 1 x insulation resistance tester • 2 x digital multimeters • 1 x Prova 23 tong test meter • 1 x 240V single phase split phase display motor (cutaway mounted on plywood) • 4mm connection leads
NOTE:
This practical segment is to be completed by students on an individual basis.
The time given per student is to be no longer than 40 minutes at the bench.
– REMEMBER – – WORK SAFELY AT ALL TIMES – Observe correct isolation procedures
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Practical 10 The Single Phase Split Phase Motor
PROCEDURE:
1. MOTOR TESTING 1. Connect the windings of the motor as shown in figure 2. Do not connect the motor to the supply.
A U1 n
Main Z 1 Auxiliary
U2 Z 2 N Figure 2 2. Use appropriate instruments to carry out insulation resistance and continuity tests on the motor and supply leads. Record all results.
Insulation resistance: active and neutral (joined) to earth ______ohms.
Winding continuity: active to neutral resistance ______ohms.
3. Check that the shaft rotates freely by hand. YES / NO (tick one)
Yes No
4. Do the results of procedures 2 and 3 show that the motor is in a suitable condition to be connected to the supply? YES / NO (tick one) Yes No
5. If the answer to 3 is YES, proceed to Motor Operation, if the answer is NO, check with your teacher.
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The Single Phase Split Phase Motor Practical 10
2. MOTOR OPERATION 1. Connect the motor to the single phase 24V supply, switch on and record the motor direction of rotation looking at the drive end.
Drive end direction of rotation = ______2. Switch off the supply and disconnect the motor from the supply. 3. To reverse the motor direction of rotation, reconnect as shown in figure 3.
A U1 n
Main Z 1 Auxiliary U2 Z 2 N Figure 3 4. Reconnect the motor to the single phase 24V supply, switch on and record the direction of rotation.
Drive end direction of rotation = ______5. Do not proceed until the teacher checks your results and completes the progress table. Progress Table 2
attempt 1 attempt 2 attempt 3 5 2 0
6. Switch off the supply and then disconnect the circuit. 7. Please return all equipment to its proper place, safely and carefully.
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Practical 10 The Single Phase Split Phase Motor
Observations: 1. Based on the values determined during this lesson, compare the electrical characteristics of the two windings of the split phase motor tested? That is, compare the resistance and inductance or phase angle of the two windings. ______
2. Which of the two windings of a split phase motor has the lower power factor? ______
3. What determines the direction of rotation of the split phase induction motor? ______
4. Given the split phase motor tested was a 4 pole machine, determine the motors synchronous speed, when supplied from a single-phase 50Hz, AC supply. ______
5. What tests need to be carried out to ensure a split phase motor is safe to connect to the supply? ______
6. Explain how to reverse the direction of rotation of a split phase motor. ______
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Tutorial 10 NAME:
SINGLE PHASE SPLIT PHASE MOTOR SECTION A In the following statements one of the suggested answers is best. Place the identifying letter on your answer sheet. 1. A single phase winding produces:- a) a stationary magnetic field; b) a rotating magnetic field; c) a steady magnetic field; d) an alternating magnetic field.
2. To develop a rotating magnetic field a split phase induction motor simulates a:- a) two phase motor; b) three phase motor; c) series universal motor; d) shaded pole motor.
3. The angle of phase displacement between the start and run winding currents of a split phase induction motor is approximately:- a) 10 degrees; b) 30 degrees; c) 90 degrees; d) 120 degrees.
4. If the centrifugal switch on a split phase motor goes permanently open circuit:- a) the motor will not start; b) the start winding will burn out; c) the start capacitor will burn out; d) starting torque will drop to about half of normal value.
5. The auxiliary winding switch should open when:- a) rotor speed is about 25 percent of rated speed; b) rated speed is about 25 percent of synchronous speed; c) rotor speed is about 75 percent of synchronous speed; d) slip speed is about 75 percent of synchronous speed.
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Tutorial 10 The Single Phase Split Phase Motor
SECTION B Answer review questions 1-7 on page 281 in your textbook.
SECTION C 1. A single phase 240 volt 50 hertz 4 pole split phase motor runs at rated speed of 1425 r.p.m. For full load determine:- a) the synchronous speed of the motor; (1 500 r.p.m.) b) the slip speed; (75 r.p.m..) c) the slip percent; (5%.) d) the rotor frequency. (2.5Hz)
SECTION D. 1. Figure 1 represents the torque speed curves for a single phase split phase induction motor. a) Which curve represents the torque speed characteristic for the main winding only?
______b) If the centrifugal switch opens when the slip is 25 percent trace on the curves the cpmplete torque speed characteristic for the motor showing the transition from start to run condition.
Curve B 250 Torque Torque %
Curve A 100
0 0 100% Speed %
Figure 1
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CAPACITOR & SHADED POLE MOTORS
PURPOSE: This section introduces the single phase capacitor motors and the shaded pole motor.
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this section the student will be able to: • Identify the four single phase induction motors listed - . capacitor start . capacitor start - capacitor run . permanently split capacitor . shaded pole • Explain the principle of operation for each type of motor. • List the operating characteristics and typical applications for each type of motor. • Connect, run and reverse each of the motors listed. • List common motor faults, and demonstrate testing procedures to identify these faults for each type of motor.
REFERENCES:
Electrical Trade Principles 2nd Edition J. Hampson & S. Hanssen. Section 6 pages 277 - 286.
Electrical Principles for the Electrical Trades. 5th Edition. Jenneson J.R. Pages 235-238.
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Section 11 Capacitor & Shaded Pole Motors
1. THE CAPACITOR START MOTOR Design limitations restrict the split phase motor to a maximum phase angle between the main and auxiliary winding currents, at start, to approximately ______. To increase this angle and produce improved starting characteristics a capacitor is connected in series with the auxiliary winding, as shown in figure 1.
Main
Auxiliary CS
n
AC Supply
Figure 1 As shown in figure 2, if the correct size capacitor is selected then the two currents will be - • approximately ______and • improved ______is obtained.
V
Figure 2
The 90º angle of phase displacement between IM and IA provides a more uniform strength of stator flux during starting. Due to this more even field strength, the starting torque - • is ______than the equivalent sized split phase motor • can be up to ______the rated full load torque of the motor.
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Capacitor & Shaded Pole Motors Section 11
• Figure 3 shows the torque speed characteristic for a capacitor start motor.
400
300
200
Torque Torque % 100
0 0 100% Speed
Figure 3 It can be seen that there is a large increase in starting torque due to the addition of the capacitor, while the torque produced is the same as the split phase motor after the switch has operated. In a similar fashion to the split phase motor the centrifugal switch operates at approximately - • ______.
As with any induction motor, the initial starting current of the capacitor start motor is high and decreases as the motor accelerates to its operating speed. The addition of the capacitor in series with the auxiliary winding requires the centrifugal switch to break a large current at the instant of opening. The breaking of this large starting current can be detrimental to the contacts of the switch. To overcome this problem current relays are often used. Current relays are designed to switch large values of current and are generally more reliable.
Figure 4 shows the connection of a current relay in a capacitor start motor.
AC Supply
Figure 4 *See Figure: 6.59 on page 278 of your textbook for the connections of a capacitor start motor.
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Section 11 Capacitor & Shaded Pole Motors
To obtain optimum starting torque the capacitance of the starting capacitor must be - • matched to the electrical characteristics of the auxiliary winding • selected to prevent series resonance. Motor manufacturers specify the capacitance required for a particular motor rating. Typical values include - • 180W motor - ______• 250W motor - ______. Motor starting capacitors are usually AC electrolytic capacitors which are rated for a short duration connection time, typically - • ______. Since the starting torque of the capacitor start motor is much higher than that of the basic split phase motor, it is suitable for uses requiring - • ______and/or • ______. For example - • washing machines, dishwashers • fans and blowers • pumps and compressors. The capacitor start motor will rotate in a direction governed by the instantaneous direction of current flow in the main and auxiliary windings. To reverse a capacitor start motor it is necessary to reverse the connections to the - • ______or the • ______, but not both. Figure 5 shows the connections for a capacitor start motor to operate in the forward direction. Main
Auxiliary CS
n
AC Supply
Figure 5
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Capacitor & Shaded Pole Motors Section 11
Draw the necessary connections on figures 6 and 7 to reverse the direction of rotation of the motor, firstly by reversing the auxiliary winding and secondly the main winding.
Main Main
Auxiliary Auxiliary CS CS
n n
AC Supply AC Supply
Figure 6 Figure 7
2. CAPACITOR START - CAPACITOR RUN MOTORS The capacitor start motor just discussed still has a relatively low starting torque, although as we have seen, it is considerably better than the split phase motor. For many applications this does not present a serious limitation. In cases where high starting torques are required, best results will be obtained if a large value of capacitance is used at starting which is then decreased as the speed increases. In practice, two capacitors are used for starting and one is switched out of circuit by a centrifugal switch once a certain speed is reached, usually at about 75% of rated speed. As shown in figure 8, the capacitor start - capacitor run motor has two windings Main permanently connected across the supply. These windings are known as the - • main winding Auxiliary • auxiliary winding. The starting capacitor is of a fairly high capacity, usually of the order of 10 to 15 times n the value of the run capacitor. AC Supply
Figure 8
*See Figure: 6.67 on page 280 of your textbook for the connections of a capacitor start capacitor run motor.
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Section 11 Capacitor & Shaded Pole Motors
The combined effects of the two windings permanently connected to the supply and the varying values of capacitance result in a rotating magnetic field which tends to be more constant in strength, so producing - • ______• ______• ______. Consequently, a motor of this type is used to drive heavy duty loads which require substantial starting torque and quietness in running, for example - • wall mounted air conditioning units • compressors • industrial floor polishers. The important features of the start and run capacitors are listed in table 1.
Table 1 Run Capacitor Start Capacitor
180W motor ≈ 135µF Capacitance 10 - 20µF 370W motor ≈ 150-200µF 550W motor ≈ 180-240µF Oil bath Type AC electrolytic Metallised polypropylene film Soggy foil (high voltage) Only during starting Connection Permanent Limited connection time ≈ 3 seconds Limited number of starts ≈ 20/hour To reverse a capacitor start - capacitor run motor it is necessary to reverse the connections to the - • ______or the • ______, but not both.
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Capacitor & Shaded Pole Motors Section 11
3. PERMANENTLY SPLIT CAPACITOR MOTOR (PSC) The permanently split capacitor motor also has both windings permanently connected across the supply, with a capacitor in series with one of the windings as shown in figure 9. Winding 1 A
Winding 2 B
AC Supply
Figure 9 Both windings are identical and because the AC electrolytic capacitor is connected in series with one winding, the current in that winding leads the current in the other, providing the necessary phase displacement to produce a stator rotating magnetic field. However, the phase displacement between the two currents and hence the two fluxes is small, and so the starting torque is low. By interchanging the supply connection from terminal A to terminal B in figure 9, the capacitor is then in series with winding 1 instead of winding 2. The current in winding 1 leads that in winding 2 and the rotor runs in the reverse direction. The permanently split capacitor motor has the following advantages - • ______• ______• ______• ______. Motors of this type are used to drive low torque loads which may require frequent reversal or speed control, for example - • ceiling fans • dampers for regulating air flow in air conditioning systems.
*See a torque speed curve for a permanent split capacitor motor in Figure: 6.64 on page 279 of your textbook.
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Section 11 Capacitor & Shaded Pole Motors
4. SHADED POLE INDUCTION MOTORS In the shaded pole motor, a short circuited turn of copper, called a "shading coil", is fitted around one tip of each pole of the motor as shown in figure 10. A squirrel cage rotor is used. The alternating magnetic flux in the poles, produced by the alternating current in the stator windings, induces emfs in the shading coils. Because of the low resistance path of the shading coil, the currents due to these induced emfs are relatively high and according to AC Supply Lenz's Law, in such a direction as to oppose the flux Figure 10 change causing them. Consequently the fluxes in the shaded portions of the poles lag behind the fluxes in +I the unshaded parts and an imperfect rotating magnetic B C field is produced. Consider the operation during the positive half cycle 0 of the supply as shown in figure 11. A D Figure 11
When the supply current rises from A to B, an induced voltage is established in the shading coil. This current produces a flux which opposes the build up of the main flux. As a result the main flux is concentrated in the unshaded section of the pole, as in figure 12.
Figure 12 Between B and C there is little current change and very little induced voltage in the shading coil. Consequently, practically no current nor flux is produced in the shading coil. The main flux is at this time nearly always at a maximum value, and is uniformly distributed over the whole pole face, as shown in figure 13. Figure 13
When the current decreases from C to D, an induced voltage is established in the shading coil. The current in the shading coil produces a flux which opposes the collapse of the main flux. The concentration of flux therefore occurs in the shaded section of the pole, as shown in figure 14.
Figure 14
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Capacitor & Shaded Pole Motors Section 11
The magnetic field shifts across the pole face, from the - • ______section to the ______section of the pole. This shifting flux is similar to a rotating magnetic field, and produces a small torque, causing the rotor to rotate in the direction of the flux. The advantages and disadvantages of the shaded pole motor are listed in table 2.
Table 2 Advantages Disadvantages simple construction very low starting torque low cost low efficiency (5 - 35%) high reliability not easily reversed low maintenance limited speed control simple speed control
* REVERSING THE ROTATION OF A SHADED POLE MOTOR. *Read the 2 paragraphs below figure 6.54 on page 277 of your textbook and answer the following questions. Generally shaded pole motors are not used in applications which require the motor to be reversed. However, if reversal is required it is possible to have ______sets of shading coils or: • ______• ______
*Note: It is not usually practical to dismantle and / or turn the motor end for end.
*See a torque speed curve for a shaded pole motor in Figure: 6.56 on page 277 of your textbook and answer the following questions. Shaded pole motors develop ______starting torque and have______efficiency. They are used for low power applications where starting torque and motor efficiency is not a concern.
Applications of the shaded pole motor include - • fans • small pumps
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Section 11 Capacitor & Shaded Pole Motors
5. CAPACITOR MOTOR FAULTS The faults that occur in split phase motors can also occur in capacitor motors. However, the capacitor motor has an additional component which can fail, that is, the capacitor. Capacitor failure will dramatically alter the performance of all capacitor motors. The capacitor may fail due to being - • ______• ______• ______.
6. SHADED POLE MOTOR FAULTS. Due to their simple construction there is not a great deal that can go wrong with shaded pole motors. Even if the motor stalls it is not uncommon to find that the stator winding is unaffected. The most common fault is dry or sticky bearings which stall the rotor, since the developed torque is very low.
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NAME: Practical
CAPACITOR START MOTOR PURPOSE: To compare the operating characteristics of the split phase and capacitor start motors.
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this practical assignment the student will be able to: • use appropriate test instruments to test capacitors and determine their serviceability. • connect a capacitor in series with the start winding of a split phase motor to produce a capacitor start motor. • measure the starting torque, starting current and full load speed of a split phase motor and a capacitor start motor. • describe the effect of increasing the value of capacitance connected in series with the start winding of a capacitor start motor.
EQUIPMENT:
• 1 x 24 volt split phase induction motor + double machine bed • 1 x eddy current load + 1 x inertia wheel • 1 x insulation resistance tester • 2 x digital multimeters + 1 x AC current clamp • 1 x optical tachometer + 1 x stopwatch • 4mm connection leads
NOTE:
This practical segment is to be completed by students on an individual basis. The time given per student is to be no longer than 40 minutes at the bench.
– REMEMBER –
– WORK SAFELY AT ALL TIMES – Observe correct isolation procedures
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Practical 11 Capacitor & Shaded Pole Motors
PROCEDURE:
1. SPLIT PHASE MOTOR CHARACTERISTICS 1. Connect the 24V split phase motor as shown in figure 1.
A Dynamometer U1 A n 1φ Z 1 M Main N Auxiliary AC current clamp
U2 Z N 2 A
Figure 1 2. Couple the dynamometer to the motor shaft and lock the dynamometer shaft.
NOTE: The following step must be carried out very quickly. The motor has a locked rotor and may be damaged due to excessive current.
3. Switch on the supply and quickly measure and record the locked rotor torque and starting current. Record in table 2.
Table 2 Starting Torque Starting Current Nm amperes
4. Switch off the supply. 5. Replace the dynamometer with the inertia wheel. 6. Switch on the supply and allow the motor time to accelerate to its final operating speed. Measure and record the speed and current. Record in table 3.
Table 3 Full-load Speed Full-load Current rpm amperes
7. Switch off the supply. 8. Measure and record the time taken for the motor to reach full load speed and current. Accelerating time = ______
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Capacitor & Shaded Pole Motors - Practical
9. Do not proceed until the teacher checks your results and completes the progress table. Progress Table 2 attempt 1 attempt 2 attempt 3 5 2 0
2. CAPACITOR START MOTOR CHARACTERISTICS 1. Connect the 24V capacitor start motor as shown in figure 2. Note: The capacitor in series with the auxiliary winding has a value of 560µF.
A Dynamometer U1 560µF A n 1φ Z 1 M Main N Auxiliary AC current clamp
U2 Z N 2 A
Figure 2
NOTE: The following step must be carried out very quickly. The motor has a locked rotor and may be damaged due to excessive current.
2. Switch on the supply and quickly measure and record the locked rotor torque and starting current. Record in table 4.
Table 4 Starting Capacitor Starting Torque Starting Current µF Nm amperes 190-240µF + 240- 320µF = 560µF 560µF + 320-400µF = 960µF 2500µF
3. Switch off the supply and discharge the capacitor. 4. Repeat steps 1-3 using capacitance values of 960µF and 2500µF. 5. Switch off the supply.
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Practical 11 Capacitor & Shaded Pole Motors
6. Replace the dynamometer with the inertia wheel. 7. Using a starting capacitance of 960µF, switch on the supply and allow the motor time to accelerate to its final operating speed. Measure and record the speed and current. Record in table 5.
Table 5 Full-load Speed Full-load Current rpm amperes
8. Switch off the supply. 9. Measure and record the time taken for the motor to reach full load speed and current. Accelerating time = ______10. Note the direction of rotation of the motor looking at the drive end. Drive end direction of rotation = ______11. To reverse the motor direction of rotation, reconnect as shown in figure 3.
A 960µF U1 n
Z Main 1 Auxiliary U2 Z 2 N
Figure 3 12. Reconnect the motor to the single phase 24V supply, switch on and record the direction of rotation. Drive end direction of rotation = ______
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Capacitor & Shaded Pole Motors - Practical
13. Do not proceed until the teacher checks your results and completes the progress table.
Progress Table 3 attempt 1 attempt 2 attempt 3
5 2 0 14. Switch off the supply, then disconnect the circuit. 15. Please return all equipment to its proper place, safely and carefully.
3. OBSERVATIONS: 1. Compare the starting torques of the split phase and capacitor start motors. ______
2. Compare the starting currents of the split phase and capacitor start motors. ______
3. Compare the accelerating times of the split phase and capacitor start motors. ______
4. Compare the full-load speeds and currents of the split phase and capacitor start motors. ______
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Practical 11 Capacitor & Shaded Pole Motors
5. What is the effect of increasing the value of starting capacitance on the starting torque developed by a capacitor start motor? ______
6. Explain how to reverse the direction of rotation of a capacitor start motor. ______
7. Based on the results of this practical assignment, which motor, split phase or capacitor start, would be best to drive a load requiring a high starting torque? ______
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Tutorial 11 NAME:
CAPACITOR & SHADED POLE MOTORS SECTION A In the following statements one of the suggested answers is best. Place the identifying letter on your answer sheet. 1. The capacitor start single phase induction motor has a capacitor connected:- a) in series with the auxiliary winding during starting; b) in series with the running winding during starting; c) in parallel with the main winding during starting; d) in parallel with the start winding during starting.
2. The single phase induction motor that is commonly used to drive cooling fans in small appliances is the:- a) permanently split capacitor motor; b) shaded pole motor; c) series universal motor; d) split phase induction motor.
3. A capacitor start induction motor has an open circuited capacitor. The motor will:- a) start with reduced torque; b) fail to start; c) start normally but stop when the centrifugal switch operates; d) start in the reverse direction.
4. If the centrifugal switch on a split phase motor goes permanently open circuit:- a) the motor will not start; b) the start winding will burn out; c) the start capacitor will burn out; d) starting torque will drop to about half of normal value.
5. The shading coils on a shaded pole motor are used to:- a) cause the flux to move across the pole face; b) reduce the noise of the motor; c) prevent the rotor "poling" or "cogging" with the stator; d) allow the motor to be reversed easily.
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Tutorial 11 Capacitor & Shaded Pole Motors
6. A starting switch is not required in:- a) a capacitor start, capacitor run motor; b) a split phase motor; c) a capacitor start motor; d) a shaded pole motor.
7. The single phase motor which would produce the highest starting torque when compared to other motors of a similar rating is the:- a) split phase capacitor start; b) shaded pole; c) split phase; d) universal.
8. On a capacitor start, capacitor run induction motor the start capacitor may be identified as having:- a) the lower capacitance and a continuous rating; b) the higher capacitance and a continuous rating; c) the lower capacitance and a short term rating; d) the higher capacitance and a short term rating.
SECTION B Answer review questions 8-11, 14 & 16 on page 282 of your textbook.
SECTION C 1. A single phase 240 volt 50 hertz capacitor start induction motor has a run winding which takes 4 amperes at 0.6 lag power factor at start while the start winding/capacitor takes 3 amperes at 0.8 lead power factor. Determine the phase angle of each current and the angle between them; (53.1Olag, 36.9O lead, 90O)
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Capacitor & Shaded Pole Motors Tutorial 11
SECTION D. 1. This question relates to the motor illustrated in Figure 1 below. a) Identify the type of motor illustrated in Figure 1. b) Identify the parts of the motor labelled A, B, C and D. c) Is the direction of rotation of the motor clockwise or anti-clockwise?
C
A D
B
Figure 1 2. Figures 2(a) and (b) show the windings and capacitor for a permanent split capacitor motor. Complete each circuit for different directions of rotation.
Winding 1 Winding 1
Winding 2 Winding 2
AC Supply AC Supply
Figure 2(a) Figure 2(b)
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NAME: Section 12
SERIES UNIVERSAL MOTORS
PURPOSE: This section introduces the series universal motor.
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this section the student will be able to: • Explain the principle of operation of the series universal motor. • Identify by name and state the function of each of the basic parts of a series universal motor. • List the operating characteristics and typical applications for the series universal motor. • Connect, run and reverse the series universal motor. • List common universal motor faults and their possible causes.
REFERENCES: • Electrical Trade Principles 2nd Edition J Hampson & S Hanssen. Section 6 Pages 280 & 281. • Electrical Principles for the Electrical Trades. 4th Edition. Jenneson J.R. Pages 241-242.
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1. INTRODUCTION Universal motors are small, series wound motors which can be operated from either a DC or a single phase AC supply at approximately the same speed. The motor performance is the same in either application, hence the term "universal". Figure 1 shows a small universal motor. Figure 1 Since universal motors are series wound, their performance characteristics are very similar to those of the series wound DC motor. Universal motors are usually small motors but can be up to 2kW. The very low fractional horsepower universal motors, with ratings up to 5W are used in such equipment as - • sewing machines • fans • electric shavers • hair dryers. The higher ratings, 5W to 2kW, are found in - • drills • sanders • circular saws • vacuum cleaners • food processors.
There are several types of universal motor in use today. The most popular type is a small, two pole series motor fitted with concentrated field poles, as shown in figure 2. Another type of universal motor has a field winding distributed in slots, much the same as the split phase motor. Note, the treatment of the universal motor in this module will concentrate on the type of motor fitted with concentrated field poles. The same operation is equally applicable to the motor fitted with a distributed field winding. Figure 2
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Series Universal MotorsCollege Section 12
2. CONSTRUCTION The main parts of the concentrated field universal motor are the - . Frame - is of rolled steel, aluminium or cast iron and large enough to hold the field core laminations snugly. Very often the frame is constructed to form an integral part of the machine it supports. . Field core - is constructed of laminations that are tightly pressed together and held by rivets or bolts. As shown in figure 3, the laminations are designed to contain both field poles of a two pole motor.
Figure 3 . Field coils - nearly all universal motors are two pole machines and therefore have two field coils. See figure 4. Consist of relatively few turns of wire, generally several hundred turns, and are fitted into the field core as shown in figure 5. The field coils when fitted to the field core form field poles which are known as "salient poles". The field coils are connected in series with the armature winding.
Figure 4 Figure 5
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. Armature - is similar to that of a small DC motor. It consists essentially of a laminated core Winding having either straight or skewed slots and a Slots commutator to which leads of the armature Laminated winding are connected. core
Both the core and commutator are pressed on the shaft. See figure 6. Figure 6
. End plates - as in other motors, the end plates are located on the ends of the frame and held in place by screws or bolts. The end plates house the bearings, usually of the ball or sleeve type, in which the armature shaft revolves. Many universal motors contain an end plate that is cast as part of the frame. Only one end plate can be removed from this type of motor. The other end plate has brush holders bolted to it and can be removed, see figures 7 and 8. Brush holders and brushes provide the sliding electrical contact between the field poles and the armature.
Figure 7 Figure 8
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Series Universal MotorsCollege Section 12
*Student exercise See figure 6.70 and 6.71 on page 280 in your textbook for more details of series universal motor construction and answer the following questions. 1. The field coils in a series universal motor are connected in ______with the armature. 2. If the brushes, armature or field coils became open circuited the motor would ______.
The series universal motor derives it name from the fact that the field windings are connected in _ with the armature winding. The connection between the rotating armature winding and the stationary field windings is achieved via the commutator and brush gear. Figure 10 shows the circuit diagram for the universal motor.
AC Supply
Figure 10
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3. OPERATION The universal motor is so constructed that when the armature and field coils are connected in series and an appropriate supply voltage applied, the magnetic lines of force created by the fields will react with the lines of force created by the armature and cause rotation. This is true regardless of whether the current is alternating or direct. Assume that during the positive half cycle of the supply current flows through the motor as shown in figure 11.
I AC Supply Figure 11 The current flowing through the two field poles will establish two magnetic poles, one north the other south. The same current flows through the armature winding, therefore the armature also establishes a magnetic field. The two magnetic fields interact causing a torque to be developed. As the armature is mounted on bearings, the developed torque will make the armature rotate. The direction of rotation of the armature can be determined by the application of Fleming's left hand rule. When the supply current reverses, that is during the negative half cycle, the polarities of the field poles are reversed as shown in figure 12. The current through the armature also reverses, so reversing the armature flux. Again, by application of Fleming's left hand rule, it will be found that the torque developed will cause the armature to rotate in the same direction.
I AC Supply Figure 12
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Series Universal MotorsCollege Section 12
Therefore, in spite of the polarity reversal of motor current the torque developed is unidirectional, showing that a series machine will run as a motor when fed from an AC supply. To keep the armature continuously turning in the one direction, a commutator is used. The purpose of the commutator is to - • ______of the current flowing in the armature coils as the coils move from the influence of one field pole to the other. As soon as the armature begins to rotate, the armature conductors cut the magnetic lines of force established by the stator field poles. This causes a voltage to be induced in the armature conductors which opposes the supply voltage. This induced armature voltage is known as the motor - • ______.
*Teacher demonstration Set up a series universal motor coupled to the dynamometer and measure the voltage across the armature as the motor accelerates. The voltage measured across the armature______and is called ______. Figure 13 shows the equivalent circuit for the universal motor, in which -. V = supply voltage I = motor current Z = motor impedance Eg = armature back emf I
V
By applying the laws of series circuits it is found that the supply voltage is given by the equation -
From the voltage equation the current taken by the motor can be determined -
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The equation for the determination of motor current illustrates one of the important factors in the operation of the universal motor. That is:
The generated back emf developed in the armature is the main current limiting factor as the motor impedance is a low value.
4. UNIVERSAL MOTOR CHARACTERISTICS At the instant of starting the back emf developed by the motor is equal to ______. Under these conditions the current taken by the motor is given by-
Therefore the starting current is very ______. As a result of the very high starting current, the magnetic fields developed by the stator and the armature will be strong, resulting in a starting torque which is - • ______. As the motor accelerates the - • back emf ______• motor current ______• stator and armature magnetic fields ______• torque developed ______.
The motor will continue to accelerate until it reaches a speed at which the developed torque equals the torque required by the load. The motor will then run at a constant speed unless there is a change in load.
If the load on the motor is increased the - • motor speed ______• armature back emf ______• motor current ______• stator and armature magnetic fields ______• torque developed ______to drive the additional load. If the load on the motor is decreased the - • motor speed ______
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Series Universal MotorsCollege Section 12
• armature back emf ______• motor current ______• stator and armature magnetic fields ______• torque developed ______to meet the requirements of the load.
From the previous examination of motor operation it should be seen that the universal motor has the torque-speed characteristics typical of series motors. That is - • a very high speed on light load, • rapidly falling speed when load is applied, • reaching a very low value, or even zero, on overload.
Overall, the speed regulation of the universal motor is poor, that is, there are large variations in speed with changes in load. This is illustrated in figure 14 which shows the typical torque-speed characteristics for the universal motor.
Torque Torque % 100
0 0 Speed Figure 14 From the characteristic shown in figure 14 it can be seen that under no load conditions the armature can operate at extremely high speeds with the possibility of runaway. This is true for universal motors of higher power ratings and it is normal practice to couple these motors to the load via a built-in gear train so that the motor is always effectively loaded. In the case of motors with lower power ratings the problem does not occur, because the loading applied due to the fan, bearing and windage friction is sufficient to load the motor.
Overall, the universal motor is considered to be a motor which - • operates at relatively high speed (up to 15000rpm in some appliances)
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• has good starting torque characteristics • has good running torque characteristics considering its small size These characteristics make the series universal motor especially suitable for portable hand tools and appliances.
5. UNIVERSAL MOTOR REVERSAL The direction of rotation is changed by reversing the flow of current through either the • ______or the • ______. The usual method is to interchange the leads on the brush holders. See figures 15 and 16.
Forward rotation
AC
Figure 15
Reverse rotation AC Figure 16
Some portable power tools are fitted with double pole reversing switches. Figure 17 illustrates the connection and operation of these switches.
F R A Supp
Figure 17
6. PERMANENT MAGNET DC MOTOR
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Series Universal MotorsCollege Section 12
The permanent magnet motor is a DC-only commutated motor that uses permanent magnets to provide the stationary field poles. Power is applied only to the DC rotor (armature) through a set of carbon brushes.
7. NEW TECHNOLOGY
Brushless DC and 3 phase inverter drive motors are also being used in industry for their reliability and speed control characteristics.
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8. MOTOR FAULTS The universal motor is not as robust or as simple in construction as the induction motor. Table 1 lists common motor faults and the symptoms they produce
Table 1 Symptom Possible Cause shorted field poles
open circuit armature coils short circuited armature coils high mica between commutator segments worn bearings motor not designed for reversal
worn bearings or excessive load
dry bearings short circuited armature coils short circuited field coils
short circuited armature
short circuited field overload incorrect voltage
short circuited armature coils
short circuited field coils worn bearings
worn brushes
open circuit field coil open circuit armature coils faulty switch No supply
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NAME: Practical
SERIES & SHADED POLE MOTORS PURPOSE: To observe the operation of single phase series and shaded pole motors.
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this practical assignment the student will be able to: • connect the single phase series motor to both AC and DC supplies for both forward and reverse directions of rotation. • describe the effect of variation of load on the series motor. • connect a the shaded pole motor. • describe the effect of variation of load on the shaded pole motor. • draw graphs of speed and current versus torque for both motors.
EQUIPMENT:
• 1 x 24 volt series motor + double machine bed • 1 x 24 volt shaded pole motor • 1 x dynamometer • 1 x 24 volt single phase supply + 1 x variable DC power supply • 1 x AC/DC current clamp • 1 x rectifier panel • 1 x optical tachometer • 4mm connection leads
NOTE: This practical segment is to be completed by students on an individual basis.
The time given per student is to be no longer than 40 minutes at the bench.
– REMEMBER – – WORK SAFELY AT ALL TIMES – Observe correct isolation procedures
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Procedure:
1. SERIES MOTOR - DC OPERATION 1. Connect the series motor as shown in figure 1 and couple the dynamometer and motor shafts.
Dynamometer A ~ Rectifier 1φ A ~ M 24V AC Supply N ~ DC current clamp N ~ A
DC current clamp
A
Figure 1 2. Switch on the AC supply and record the direction of rotation of the motor, looking at the drive end. Drive end direction of rotation - ______3. Switch off the supply. 4. Reverse the DC supply to the motor circuit terminals as in Figure 2.
Rectifier A ~ 24V AC Supply N ~
DC current clamp
A
Figure 2 5. Switch on the supply and record the direction of rotation for this connection. Again, looking at the drive end. Drive end direction of rotation - ______
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Series Universal MotorsCollege Section 12
6. Do not proceed until the teacher checks your results and completes the progress table.
Progress Table 1 attempt 1 attempt 2 attempt 3 5 2 0 7. Connect the dynamometer terminals to a DC power supply making sure that the power supply is off and its voltage control is at zero. 8. Switch on the AC supply but not the DC power supply - this will give the no load condition. Record speed and motor current in table 1 for this load.
Table 1 Motor Torque Motor Speed Motor Current Nm rpm A No load
0.1
0.2
0.3
zero Breakdown torque = ______Nm 9. Adjust the DC power supply voltage to produce motor torque values as shown in table 1. For each of these values record the current and speed of the motor. 10. Breakdown torque will occur when the motor stalls.
Read this value quickly and switch of the supply so the motor will not overheat
11. Switch off the AC supply, reduce the DC power supply voltage to zero and switch it off. 12. Do not proceed until the teacher checks your results and completes the progress table.
Progress Table 2 attempt 1 attempt 2 attempt 3
5 2 0
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2. SERIES MOTOR - AC OPERATION 1. Reconnect the motor circuit directly to 24 volts AC supply as in figure 3.
Dynamometer A 1φ A M 24V AC Supply N N AC current clamp A
AC current clamp
A
Figure 3 2. Switch on the AC supply and note the direction of rotation. Drive end direction of rotation - ______3. Step 3 is the no load condition, record current and speed for this load in table 2.
Table 2 Motor Torque Motor Speed Motor Current Nm rpm A No load
0.1
0.2
0.3
zero Breakdown torque = ______Nm
4. Adjust the DC power supply voltage to the dynamometer to produce the motor torque values shown in table 2. For each of these values record the current and speed of the motor. 5. Breakdown torque will occur when the motor stalls.
Read this value quickly and switch of the supply so the motor will not overheat
6. Switch off both the AC and DC power supplies.
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Series Universal MotorsCollege Section 12
7. Do not proceed until the teacher checks your results and completes the progress table.
Progress Table 2 attempt 1 attempt 2 attempt 3
5 2 0
3. SHADED POLE MOTOR 1. Connect the shaded pole motor directly to 24 volts AC supply as in figure 4.
Dynamometer A 24V 1φ AC Supply M N AC current clamp
A
Figure 4 2. Switch on the AC supply and note the direction of rotation. Drive end direction of rotation - ______3. Step 3 is the no load condition, record current and speed for this load in table 3.
Table 3 Motor Torque Motor Speed Motor Current Nm rpm A No load
0.1
0.2
0.3
zero Breakdown torque = ______Nm 4. Adjust the DC power supply voltage to the dynamometer to produce the motor torque values shown in table 2. For each of these values record the current and speed of the motor. 5. Breakdown torque will occur when the motor stalls.
Read this value quickly and switch of the supply so the motor will not overheat
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6. Do not proceed until the teacher checks your results and completes the progress table.
Progress Table 3 attempt 1 attempt 2 attempt 3
5 2 0
7. Switch off the AC supply, reduce the DC power supply voltage to zero and switch it off 8. Disconnect the circuit. 9. Please return all equipment to its proper place, safely and carefully.
4. OBSERVATIONS: 1. Use the information recorded in tables 1 and 2 to complete the graph of figure 5.
10
9
8
7
6
amperes 5
- rpm x 1000 rpm
-
4 Current Current Speed
3
2
1
0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Torque - Nm Figure 5
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Series Universal MotorsCollege Section 12
2. When operated as an AC motor, what effect does increasing the load have on:- • motor speed: ______
• motor current: ______
3. How do the characteristics listed above compare with the operation of the squirrel cage motor? ______
4. From the curves drawn in figure 5, compare the AC and DC operation over the tested load range by commenting on:- • motor speed: ______
• motor current: ______
• breakdown torque: ______
5. What effect did changing the polarity of the DC supply have on the operation of the series motor? ______
6. What was the effect of changing the relative direction of currents through the field and armature of the series motor when operating on the AC supply? ______
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7. In your opinion do the results show that the tested motor can be considered to be a universal motor? (that is, able to be operated on either AC or DC supply). ______
8. If the answer to question 7 is no, explain or give reasons for this. If the answer to 7 is yes, which supply will give the best operating characteristics, AC or DC. Use results to justify, this answer. ______
9. Use the information recorded in table 3 to complete the graph of figure 6 for the shaded pole motor.
1500
1350
1200
1050
900
750
-
600 Speed rpm 450
300
150
0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Torque - Nm Figure 6
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Series Universal MotorsCollege Section 12
10. From the results obtained, comment on the advantages and disadvantages of the shaded pole motor. ______
11. What is the effect of the shading coil when used in starting induction motors? ______
12. Explain how the shaded-pole motor is reversed. ______
13. For what applications is the shaded-pole motor most suitable? ______
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Tutorial 12 NAME:
SERIES UNIVERSAL MOTORS SECTION A In the following statements one of the suggested answers is best. Place the identifying letter on your answer sheet. 1. A single phase motor rated at 240 volt, 500 watt, 6000 r.p.m., 3 amperes, 50 Hz would be:- a) a split phase motor; b) a shaded pole motor; c) a series universal motor; d) a permanent capacitor motor. 2. The series universal motor is reversed by:- a) reversing the supply connections; b) reversing the armature and field connections; c) physically reversing the rotor in the field; d) reversing the armature connections. 3. The motor used in most mains powered portable hand tools is the:- a) shaded pole motor; b) split phase motor; c) capacitor start motor; d) series universal motor. 4. A series universal motor is easily identified by its:- a) Universal rating b) Commutator c) series winding d) nameplate
5. The most commonly used motor for a 240 volt single-phase vacuum cleaner is:- a) split phase type; b) universal type; c) capacitor start type; d) shaded pole type.
6. A starting switch is not required in:- a) a capacitor start, capacitor run motor; b) a split phase motor; c) a capacitor start motor; d) a series universal motor.
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Tutorial 12 Series Universal Motors
SECTION B Blank spaces in the following statements represent omissions. Write the appropriate information on your answer sheet. As the universal motor accelerates the back emf _____(1)_____ causing the motor current to _____(2)_____ . The motor will continue to accelerate until it reaches a speed at which the developed _____(3)_____ equals that required by the load. If the load on the motor increases the motor speed _____(3)_____ , back emf _____(4)_____ , motor current _____(5)_____ , producing _____(6)_____ flux and _____(7)_____ torque to meet the increase in load. The speed of the series universal motor is _____(8)_____ on no load and very _____(9_____ on heavy loads. In the universal series motor both the armature and field are laminated to reduce _____(10)_____ loss and made from silicon steel to reduce _____(11)_____ loss. The field windings and armature in a universal motor are connected in _____(12)_____ .
SECTION C 1. A 240 volt series universal motor has a total winding/armature impedance of 60 ohms. Determine:- a) the current taken by the motor at starting; (4A) b) the current taken by the motor when armature back emf is 120 volts. (2A.) c) the armature back emf when the motor takes rated current of 1.5 amperes; (150V)
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Series Universal Motors Tutorial 12
SECTION D 1. The performance curves for a 240V, 50Hz, 45 watt single phase motor are shown below in Figure 1. From these curves estimate the following:- a) starting torque; 5 1 10 b) full load torque; 90 c) full load current; 4 1. 80
d) full load speed; 70 Spe e) breakdown 60 3 0 Curr SP EE D R.P torque; 50 CU RR EN T A MP OU TP UT PO WE R W Po f) number of poles; 2 0 40
g) slip speed at full 30
load. 1 0 20
10
0 0 0. 0. 0. 1. TORQUE NEWTON Figur
2. For the current directions given in the series universal motor of Figure 2 determine and show:- a) the polarity of the field poles; b) the direction of the armature flux and pole location; c) the direction of rotation of the motor.
I AC Supply Figure 2.
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ALTERNATORS – PART 1
PURPOSE: In this section you will learn the construction operation and rating of single and three phase alternators.
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this section the student will be able to: • State the basic function of an alternator. • Identify the main parts of an alternator. • Describe the principles of operation of an alternator. • List the methods used to provide dc excitation to alternators. • Calculate the synchronous speed of an alternator given the number of poles and the required frequency. • Draw the no-load (open circuit) characteristics of an alternator. • State the factors to be considered when rating an alternator and calculate the apparent power output of an alternator.
REFERENCES: Electrical Trade Principles, 2nd Edition J Hampson & S Hanssen. Section 6 Pages 313 - 323 Electrical Principles for the Electrical Trades, 5th Edition. Jenneson J.R. Pages 250-253.
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Section 13 Alternators Part 1
1. INTRODUCTION The primary function of an alternator is to convert ______energy to ______energy, see figure 1. When carrying out this energy conversion process the alternator generates an alternating voltage at a specific frequency, for example 240V at 50Hz.
INPUT
Three Phase OUTPUT Supply
Figure 1 Commercial alternators are available in a wide range of output voltages and power ratings. In addition, there are many styles and there exists an enormous variation in physical size. An example of a physically small alternator is that found in a motor car. An alternator of this type generates 14V and has a maximum rated output of approximately 500VA. Single phase portable generating sets as used on building sites are designed to generate 240V, and are available in a range of power ratings. Figure 2 shows one type of portable generating set, in this case, the unit can deliver 3kVA at 240V and is driven by a small diesel engine. Figure 2 Figure 3 shows a large scale generating set as used in a hydro- electric power station. In this case, the alternator generates a three phase line voltage of 11kV and can deliver 95MVA.
In thermal power stations, alternators having exceptionally high output power ratings can be found. For example, a generated line voltage of 17kV and a maximum power rating of 640MVA. An alternator of this rating can deliver a maximum current of 21,736A. Irrespective of the differences, all alternators use the same principle of operation - ______.
Figure 3
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Alternators Part 1 Section 13
2. GENERATOR ACTION The process of generation relies on the fact that if there is ______between a conductor and a ______an emf will be induced in the conductor. The direction in which the induced emf acts can be determined by the application of Fleming's right hand rule. See figure 4. The magnitude of the induced emf is be governed by the - • ______• ______• ______• ______. Figure 4
3. AC VOLTAGE GENERATION Figure 5 shows a basic alternator. In this arrangement we have a simple loop of wire rotating in the magnetic field between two poles, one north the other south. The ends of the loop terminate in two metal slip-rings which are insulated from the driving shaft (not shown), and the conductors make rubbing contact with these rings through carbon brushes. Thus whatever the position of the loop, the conductor ab is always connected to the upper conductor of the external circuit, and similarly, the lower conductor of the circuit is, in effect, a continuation of the side cd of the loop.
b d N N 90º 90º a c
0º d 0º b 360º 360º
c a 270º 270º S S
conductor ab at an angle of 90º to the magnetic field conductor ab at an angle of 270º to the magnetic field
Figure 5
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Section 13 Alternators Part 1
If the emf induced in the loop is plotted for one complete revolution, the waveform shown in figure 6 would result.
As shown, the emf within the loop has EM F 270º 360º 0 undergone a complete series of changes 0º 90º 180º during the revolution. Such a waveform is termed an alternating emf. If the circuit were closed, the resulting current would follow the variations of Figure 6 emf and an alternating current would flow in the loop and the external circuit. By the use of suitable construction methods, practical alternators can be manufactured to produce an output voltage that takes the form of a sinusoidal waveform. Some of the constructional methods used to obtain a sinusoidal waveform relate to the arrangement of the windings within the alternator and are beyond the scope of this module. From the action previously described it should be seen that a single phase alternator consists essentially of two major parts - • ______- normally carries the alternating current winding (called the armature) • ______- carries the exciting current and produces the necessary magnetic field. Figure 7 shows a simplified view of the construction of a single phase alternator.
A S N N
Figure 7 The output waveform of a single phase alternator would be a sinusoidal voltage as shown in figure 8. When a load is applied to the alternator output a sinusoidal current flows, the phase angle of this current being determined by the type of load, that is resistive, inductive or capacitive.
+V
180º 270º 360º 0 time 0º 90º
-V
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Alternators Part 1 Section 13
Consider three coils, or phase windings, spaced 120º apart as in the simple alternator shown in figure 9.
A S N B C N
Figure 9 It should be seen from figure 9 that whatever occurs at a certain instant in phase A will occur 120º later in phase B and 240º later in phase C. Thus each of the three emfs produced in this alternator, has a phase difference of 120º between it and the other two, and the combination produces a "three phase system". The emfs produced by the elementary three phase alternator of figure 9 are represented by the sinusoidal voltage waves shown in figure 10.
+V
0 time
-V 120º 240 360º 0º º
Figure 10
The same voltage waves may be represented by the phasor diagram shown in figure 11.
Figure 11
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Section 13 Alternators Part 1
4. FREQUENCY AND NUMBER OF POLES Alternators are designed to generate a specified voltage at a definite frequency. A definite relationship exists between the speed (n), the number of poles, and the frequency (f).
If an alternator has P poles on the field system then the emf goes through cycles per P revolution or cycles per second. 2 n x P 2 x 60
Thus, the frequency of the generated emf can be determined using the equation - where: f = frequency in hertz (Hz) P = number of field poles n = speed of rotation in revolutions per minute (rpm) Alternately, if the desired generating frequency is known, the required speed of rotation can be determined -
Example: 1 An alternator used in a hydro-electric scheme has 48 field poles and rotates at a speed of 125rpm. Determine the frequency of the generated emf. ______
Example: 2 Determine the required speed of rotation for an alternator which has 2 field poles and generates at a frequency of 50Hz. ______
Therefore, in order for a given alternator to generate at a fixed frequency (eg. 50Hz), it must be driven at a ______speed, known as the ______speed.
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Alternators Part 1 Section 13
5. ALTERNATOR CONSTRUCTION The construction of all alternators, whether single or three phase, can be broken into two main parts, the - • AC winding in which the induced emf will be generated, known as the ______• DC winding which establishes the magnetic field, known as the ______. From a theoretical point of view it does not matter whether the armature rotates between the field poles or whether the poles rotate inside a stationary armature. Consequently, there are two distinct types of alternator construction - • ______type - the armature rotates between the field poles • ______type - the poles rotate inside a stationary armature.
6. ROTATING ARMATURE TYPE OF ALTERNATOR Some small alternators have the same type of construction as DC generators. The field system is fixed and the armature rotates. However, instead of the armature winding being connected to a commutator, it is connected to slip-rings. The construction of a small alternator armature is shown in figure 12.
A
B
C
N
Figure 12 Figure 13
The armature coils are laid in slots in a laminated core. The slip-rings, usually made of bronze, are mounted on the shaft from which they are insulated. Carbon brushes make contact with the slip-rings and when a load is connected conduct current from them to the load. The circuit diagram for the rotor shown in figure 12, would be as shown in figure 13.
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Section 13 Alternators Part 1
The field system of the rotating armature type of alternator is shown in figure 14. The field poles are excited by means of current Field supplied to the field coils form a source of Supply direct current. Note, in one complete revolution of the field system, successive field poles are connected to produce poles of opposite magnetic polarity. The DC supply used to establish the magnetic fields of the field system is Figure 14 generally known as the - ______.
7. ROTATING FIELD TYPE OF ALTERNATOR The rotating field type of alternator is now standard practice, except for some very small machines. An alternator of this type is arranged so that the field poles rotate inside a stationary armature. The armature and rotating field poles are commonly referred to by alternative names - • the armature is known as the ______• the rotating field poles are known as the ______.
The principle of construction of the rotating field type of alternator is illustrated in figure 15. In the diagram shown in figure 15, the field winding terminals A and B are connected to two slip-rings. Since the field poles are magnetised by direct current they are not usually laminated but are made of solid steel. The stator conductors have alternating emfs induced in them, consequently the stator core is made up of silicon steel laminations. The stator winding is carried in slots in the stator core. Figure 15
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Alternators Part 1 Section 13
Rotor construction. Rotor construction depends largely upon how the machine is to be ______and the ______at which it must run.
*Student exercise: Read the first paragraph below figure 6.134 on page 313 of your textbook and answer the following question. 1. There are two types of rotor constructions. ______rotors are used where alternators driven by high speed prime movers such as ______. 2. ______rotors are used where alternators are driven by low speed prime movers such as ______.
Salient pole rotors
o Have projecting poles, with cores bolted to a heavy cast iron or steel ring of good magnetic quality. See figure 16.
o used in alternators that operate at low and moderate speeds (up to 1500rpm)
o used in small alternators and large low speed engine or water turbine driven alternators. Figure 16
o the pole faces of salient pole rotors are shaped so that the air gap is shorter at the pole centres than near the edges, as shown in figure 15. This helps to produce a sinusoidal waveform of induced emf.
Cylindrical rotors.
o used in large, high speed (3000rpm), steam turbine driven alternators. See figure 17.
o the cylindrical type rotor has slots machined in parts of the circumference.
o The field coils are wound in the slots and the main part of the flux passes through the unslotted portion. Rotors are designed to produce a sinusoidal waveform of induced emf.
Figure 17
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Section 13 Alternators Part 1
o the cylindrical type rotor has several advantages over the salient pole type - . it can withstand the ______set up by high speeds of rotation
. mechanical ______of the rotor is not as difficult
. the smoother surface reduces running ______. *See also Figure 6.135 on page 313 of your textbook.
Stator construction. The stator of an the rotating field type of alternator, as shown in figure 18, consists of - • a laminated silicon steel core formed with slots • an armature winding which is laid in the stator slots • phase windings made up of a number of series connected coils forming the same number of poles as the rotor • In the case of three phase alternators the armature is always ______connected.
Figure 18
8. EXCITATION Alternators have to be excited with direct current to provide the rotor field. The excitation current is often supplied by a DC generator called an ______. The exciter can be mounted on the end of the alternator shaft, see figure 19, or can be driven by a separate motor. The excitation current is fed to the rotor field winding via brushes and slip-rings.
Figure 19
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Alternators Part 1 Section 13
The circuit diagram of figure 20 shows the connections between the exciter and alternator. The exciter output is fed into the field windings of the alternator. By adjusting the rheostat in the exciter field circuit, the strength of the magnetic field in the rotor can be varied. Hence, the level of output voltage from the alternator may be controlled. A Field Rheostat
Brush Gear Stator Windings G and Slip-rings B
Rotor Field N Exciter Field C
Figure 20 Another arrangement is the ______exciter, where the excitation current is obtained from a separate AC winding placed on a separate rotor but connected to the main rotor. This system is shown in figure 21. A Rheostat Three Phase Exciter Windings Stator Windings
B
Rotor Field N Exciter Field C
External DC Supply Figure 21
The AC voltage developed by the three phase exciter is fed through a three phase, full wave bridge rectifier mounted on the end of the exciter rotor, and converted to DC. The resulting DC voltage is in turn fed into the rotor windings of the alternator. By varying the current through the exciter field, the rotor field is varied and so controls the Figure 22 value of generated voltage produced by the alternator Figure 22 shows a salient pole rotor which incorporates brushless excitation. *See also Figure 6.145 on page 317 of your textbook
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Section 13 Alternators Part 1
The value of generated voltage developed by an alternator depends on the - • strength of the ______• ______at which the field flux cuts the armature conductors. Because the speed must be constant to maintain a fixed generating frequency, the level of generated emf is determined by the strength of the field flux. The phase voltage developed by an alternator may be determined using the following equation -
where: V = phase voltage in volts φ = magnetic flux per pole in webers f = frequency of generated voltage in hertz N = number of armature turns per phase k = a constant related to the arrangement of the stator windings
Example: 3 A three phase star connected, 4 pole alternator is driven at 1500rpm. The alternator has 224 turns per phase and operates with a flux of 76.6mWb per pole. If the alternator has a winding constant of 1, determine the - a) frequency of the generated voltage b) generated phase voltage c) alternator line voltage ______
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Alternators Part 1 Section 13
Example: 4 If the field flux in the alternator of example 3 was reduced to 70mWb per pole, determine the - a) frequency of the generated voltage b) generated phase voltage c) alternator line voltage ______
9. ALTERNATOR EQUIVALENT CIRCUIT An alternator can be considered to consist of - • an AC generating source • a resistor - representing winding resistance, iron and copper losses • an inductor - representing the inductance of the windings and magnetic leakage. Any load placed on the alternator must be assumed to be in series with these components. Figure 23 shows the equivalent circuit for a single phase alternator or one phase of a three phase alternator.
L
R
G ~
Figure 23
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Section 13 Alternators Part 1
10. ALTERNATOR OPEN-CIRCUIT CHARACTERISTIC CURVE The open-circuit characteristic curve of an alternator shows the relationship between the - • ______current and • ______, when operated on open circuit, that is, no- load. To obtain this curve the alternator is driven at - • its rated speed • while the field current is increased in steps from zero to its maximum value. At each value of field current, the corresponding value of open-circuit terminal emf is measured, this being plotted against field current in the form of a curve as shown in figure 24. Under these conditions, the terminal voltage and the generated emf are the same because no voltage drop is incurred in the armature circuit. The curve of figure 24 is called the "open-circuit characteristic" of the alternator. It is also called the magnetisation curve because it indicates the degree of magnetisation of the pole cores of the field system.
- 0 0 Field Current - amperes Generated emf volts Figure 24 The straight (linear) part of the curve shows that over this range the generated emf is directly proportional to the field current. Beyond this the curve flattens out, showing that the field poles are becoming saturated. It should be noted from figure 24 that when the field current is zero, the generated emf is not zero, but is a small percentage of the normal voltage; this is due to the residual magnetism remaining in the magnetic circuit after the pole cores have once been magnetised. The normal operating voltage of the machine is just below the knee of the curve.
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Alternators Part 1 Section 13
11. RATING OF ALTERNATORS The main factors to be specified in the rating of an alternator are - • terminal voltage at full load • VA output at full load (that is, the output in terms of apparent power) • speed • frequency • the number of phases. To determine the VA output for an alternator - Single Phase Three Phase
where: S = apparent power in volt-amps V = terminal voltage in volts (line voltage for a three phase machine) I = load current in amperes
Example: 5 At full-load a single phase alternator delivers 25A at 240V. What is the VA rating of the alternator? ______
Example: 6 At full-load, a three phase 11,000V alternator supplies a load current of 2,624.32A. Determine the rated output of the alternator. ______
Example: 7 What is the full-load current of a three phase 100kVA, 415V alternator? ______
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Section 13 Alternators Part 1
Example: 8 A 240V single phase alternator is rated at 800VA. What is the full-load current? ______
Example: 9 A three phase 630MVA, 17kV, 50Hz alternator has two poles. Determine the alternators - a) synchronous speed b) rated current. ______
12. INDUCTION GENERATORS
*Student exercise: Read the 2 paragraphs below figure 6.142 on page 315 of your textbook and answer the following questions.
1. How can an induction motor be used as an alternator? ______
2. What differences are there in the design of a purpose built alternator and an induction motor used as an alternator? ______
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Tutorial 13 NAME:
ALTERNATORS – PART 1 SECTION A In the following statements one of the suggested answers is best. Place the identifying letter on your answer sheet. 1. Alternators are generally run at a constant speed to maintain:- a) a constant output voltage; b) a constant load current; c) maximum efficiency; d) a constant output frequency. 2. Low speed alternators use:- a) salient pole rotors with many poles; b) cylindrical rotors with many poles c) salient poles rotors with two poles d) cylindrical rotors with two poles 3. High speed alternators use:- a) salient pole rotors with many poles; b) cylindrical rotors with many poles c) salient poles rotors with two poles d) cylindrical rotors with two poles 4. In commercial alternators, the output voltage is adjusted by:- a) adjusting the field current supplied to the alternator. b) adjusting the speed of the alternator; c) adjusting the three phase alternator supply voltage; d) adjusting the number of turns in the alternator field. 5. Most three phase alternators have their AC stator or armature windings:- a) connected to a d.c. supply for excitation; b) connected to an a.c. supply for excitation. c) connected in star to allow earthing of the star point; d) connected in delta to increase the generated output voltage. 6. Large alternators have the high voltage AC armature winding on the stator. This is because - a) this winding is larger than the field winding and the stator has more room; b) high voltage windings are easier to insulate if they are not rotating/vibrating; c) this allows the use of two low current slip rings rather than four high current ones; d) all of the above.
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Tutorial 13 Alternators Part 1
SECTION B Blank spaces in the following statements represent omissions. Write the appropriate information on your answer sheet. The AC armature windings of three phase alternators are spaced _____ (1)_____ electrical degrees apart. The field windings of an alternator are supplied with ____ (2) _____ current. The open circuit characteristic of an alternator shows the change in _____(3)_____ voltage when the _____(4)_____ current is changed. Small alternators may be of the rotating _____ (5)_____ type while large alternators are normally of the rotating _____(6)_____ type. If a star connected alternator in a power station is generating 15kV in each winding the output line voltage will be _____(7)_____ kV. The winding in the alternator that generates the output voltage is termed the _____(8)_____ winding, regardless of whether it is on the rotating or stationary part of the machine.
SECTION C 1. At what speed must a 24 pole 50Hz alternator in a Hydro-electric power station be driven? (250 r.p.m.) 2. How many poles would be required on a 25Hz alternator running at 375 r.p.m.? (8 poles) 3. Is it possible to design a 50Hz alternator that runs at 1200 r.p.m.? Explain your answer. 4. A three phase star connected 50Hz alternator is to be used as an emergency supply with an output line voltage of 11kV. What voltage must be generated in each phase winding? (6.35kV).
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Section 14
ALTERNATORS – PART 2
PURPOSE: In this section you will learn about the alternator load characteristics, voltage regulation, efficiency and synchronising.
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this section the student will be able to: • Identify alternator load characteristics for resistive, inductive and capacitive loads. • Describe why the terminal voltage of an alternator varies with load. • Define alternator voltage regulation. • Calculate the percentage voltage regulation of an alternator. • Describe the losses that occur within and alternator and calculate alternator percentage efficiency. • State the conditions necessary to allow satisfactory paralleling of alternators. • Describe the process of synchronising alternators using either lamps or a synchroscope. • Describe the effect of a change in excitation, when alternators are paralleled. • State the purpose of amortisseur windings. • State the advantages gained by the parallel operation of alternators
REFERENCES: Electrical Trade Principles 2nd Edition J Hampson & S Hanssen. Section 6 Pages 313 - 323 Electrical Principles for the Electrical Trades. 5th Edition. Jenneson J.R. Pages 253-258.
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Section 14 Alternators Part 2
1. ALTERNATOR LOAD CHARACTERISTIC CURVE The load characteristic of an alternator is obtained by determining the relationship between the - • ______and the field current and speed being kept constant • ______. The load increased in steps up to 125 percent of full load (i.e., 25 percent overload). At each step, the terminal voltage is measured and plotted against the armature or stator current as one of the points on a curve. Figure 1 shows typical load characteristic curves for an alternator supplying the following loads:- • unity power factor. • power factor 0.8, lagging current. • power factor 0.8, leading current.
0 0 Stator Current - amperes
- Figure 1 Terminal Voltage Voltage Terminal volts
For a load of unity power factor the load characteristics is a slightly drooping curve. For a load of lagging current it droops considerably, whereas for a load of leading current the curve may rise, showing that the terminal voltage increases with in increase of load. The change of terminal voltage with an increase in load is due to a number of causes, the chief of which are:- • the impedance of the armature or stator windings and • Armature reaction.
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Alternators Part 2 Section 14
2. WINDING IMPEDANCE The stator winding of an alternator has resistance and reactance, and the joint effect of these two quantities produces the impedance. This causes a voltage drop equal to the product of the stator current and its impedance. See figure 2.
L
R Load
G ~
Figure 2
3. ARMATURE REACTION Armature reaction is the effect of the magnetic field produced by the armature current on the main field and varies depending upon the type of load Armature reaction either weakens or strengthens the main field depending on the power factor of the load. See table below
Armature reaction Load P.F Effect on main field Effect on generated voltage Lagging P.F load Weakens main field Fall in generated voltage Leading P.F load Strengthens main field Rise in generated voltage Unity P.F load No effect No effect
Note: Armature reaction is a voltage change in addition to voltage drop caused by alternator winding impedance.
VT = Eg – Vdrop +/- Armature reaction Voltage change
VT = Alternator terminal voltage
Eg = Alternator generated voltage
Vdrop = Voltage drop across impedance of alternator windings
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Section 14 Alternators Part 2
4. VOLTAGE REGULATION When an alternator is required to supply a varying load, the terminal voltage varies with the load and the amount of this change determines the regulation of the alternator. The regulation of an alternator is defined as -
"The percentage rise in voltage when full load is switched off, the excitation current being adjusted initially to give normal voltage".
The percentage regulation depends upon the - • design of the machine • the power factor of the load. Practical values are of the order of 8 to 10% at unity power factor and at a power factor of 0.8 lag, 25 to 35% is quite normal. It is usual to design an alternator with a considerable amount of internal reactance, because this limits the current if a short-circuit develops in the alternator or the load circuit. The high internal reactance causes a high voltage drop inside the machine or, in other words, a high value of regulation. For all normal loads, the terminal voltage of large power units can be maintained approximately constant by employing automatic voltage regulators.
Note: For standalone alternators output voltage is kept constant by varying DC excitation current. For grid connected alternators the output voltage is locked to the grid voltage. Varying excitation simply varies the P.F at which the alternator delivers Current
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Alternators Part 2 Section 14
5. ALTERNATORS IN PARALLEL Practically all modern power plants are designed for operation at constant voltage and usually employ several alternators which are operated singly or in several parallel combinations to supply a common load. Reasons for paralleling alternators are: • Ability to supply peak electricity supply network loads. These loads vary widely throughout each twenty-four hour period. A number of machines are connected in parallel when peak loads occur. • Continuity of service. In the operation of a power distribution system alternators are connected in parallel so that if one alternator needs to be taken out of service no interruption of supply occurs.
When alternators need to be paralleled there are four main factors to consider - • The method of connecting an alternator to bus-bars that are already alive; the process being called ______. • The division of load between the alternators, that is, the sharing of load. • The effect produced when the excitation of any given alternator is changed. • The method of disconnecting an alternator from the line. These factors will now be discussed in the order listed above.
6. SYNCHRONISING Before a three-phase alternator is connected in parallel with another alternator, or with live bus-bars, the following conditions must be fulfilled - • The two r.m.s values of the ______are approximately equal at the terminals being connected. • The two ______are the same. • The two voltages are directly opposite in phase. • The ______of the two voltages are the same. • The two voltages have the same ______.
Alternator characteristic to be adjusted Method of control Voltage output Adjust field excitation Output frequency Adjust speed of prime mover Phase sequence Check and change connections Wave form Design of machine
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Section 14 Alternators Part 2
Once the conditions of paralleling are met, the alternator to be paralleled can be synchronised with the power station bus-bars and / or other alternators.
7. SYNCHRONISING WITH LAMPS The connections generally used for synchronising low voltage alternators by means of lamps are shown in figures 6 and 7. The phase sequence is first checked by connecting a lamp across each phase of the paralleling switch as shown in figure 6.
Paralleling switch
Incoming alternator
Bus-bars L
L
L
Figure 6
If the alternator is connected correctly, then the three lamps should all glow and darken simultaneously. If they glow and darken in sequence, it means that the phase sequence of the incoming alternator is opposite that of the bus-bars, consequently the connections to two of the phases of the alternator must be interchanged.
Paralleling switch
Incoming alternator
A B C L Bus-bars 3 L2
L1
The lamps are then connected as shown in figure 7.
Figure 7
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Alternators Part 2 Section 14
It can be seen from figure 7 that - • lamp L1 is connected across C phase of the paralleling switch • lamps L2 and L3 are cross-connected, L2 being connected between the switch terminals A and B, and L3 between B and A. The three lamps provide the following indications - • when the alternator is in synchronism with the bus-bars - o lamp L1 is dark o L2 and L3 glow with equal brightness. • if the speed of the incoming alternator is too slow - the lamps glow in the order L3-L2-L1 • if the speed of the incoming alternator is too fast - the lamps glow in the order L1-L2-L3. This means, that if the three lamps are placed in the form of a triangle, the light will travel around it in a direction depending upon whether the incoming alternator is fast or slow. See figure 8.
L1 L1
L3 L2 L3 L2
Figure 8 The paralleling switch is closed when the changes in light are very slow, and when the lamp L1 is dark, indicating that the incoming alternator is running at nearly synchronous speed. The voltage across lamp L1 is zero, and it can be proved that the voltage across the lamps L2 and L3 is 3 times the phase voltage. The maximum voltage occurring across the lamps during synchronising is 3 times the phase voltage, √ therefore the lamps should be capable of withstanding this voltage. √
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Section 14 Alternators Part 2
8. THE ROTARY SYNCHROSCOPE Synchronising by means of lamps is not a very accurate method, as a considerable amount of experience and judgment are called for on the part of the operator, and in large machines even a small angle of phase displacement causes a certain amount of shock to the machines as well as causing line voltage disturbances. For this reason an instrument known as a "rotary synchroscope" is used.
Figure 9 One type of rotary synchroscope consists of a small motor which rotates when the alternator and bus-bar frequencies are different. The field winding of the motor is excited by a single-phase supply from the main bus-bars. The armature takes a single- phase supply from the incoming machine (usually through synchronising bus-bars) and this is converted into two-phase currents in the armature by a phase-splitting circuit. The speed of rotation depends on the difference in frequency between the incoming alternator and main bus-bar voltages, whereas the direction of rotation depends on whether the incoming alternator is running fast or slow. The synchroscope is usually connected between two of the main bus-bars and two synchronising bus-bars, through potential transformers, as shown in figure 10.
Figure 10
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Alternators Part 2 Section 14
The incoming alternator is synchronised by closing the synchronising switch Sw., this connects the synchroscope between two terminals of the alternator and the main bus-bars. If the frequency of the alternator voltage is not the same as that of the bus-bars, the pointer of the synchroscope rotates. The speed of the incoming alternator is varied until the synchroscope pointer remains stationary in a vertically upward position. This indicates that the incoming alternator is in synchronism with the bus-bars; the main switch is now closed and the synchronising switch opened. By the use of synchronising bus-bars, any one of a group of alternators may be synchronised by means of one synchroscope. Before synchronising newly installed machines, however, the phase sequence must first be checked by either lamps or a phase sequence indicator. The synchroscope does not check the phase sequence since it operates only on single-phase supply.
9. DIVISION OF LOAD BETWEEN TWO OR MORE ALTERNATORS The prime movers of alternators are usually fitted with governors that can be adjusted while the machines are in normal operation. • If an alternator is required to carry more load, the governor is adjusted slightly causing it to admit more input power. This will tend to cause the machine to increase in speed, and it is found that only a very slight increase in speed is required to cause one alternator to take a greater share of the load when operating in parallel with others. • If it is desired to reduce the load on a given machine, the governor is adjusted to admit less input power to the prime mover.
10. EFFECT OF CHANGE OF EXCITATION • If the bus-bar voltage is low, the field currents of ALL the alternators must be increased. Increasing the excitation of one machine only causes it to acquire reactive load. • If the frequency of the system is low, the power input to all the prime movers must be increased.
11. DISCONNECTING AN ALTERNATOR FROM THE LINE When it is desired to remove an alternator from the line: • The load should first of all be removed gradually from the machine by reducing the power input to the prime mover. • The circulating current can be removed by reducing the field excitation current until the alternator is merely floating on the bus-bars. The main switch controlling the alternator may now be opened, thus disconnecting the machine from the line, after which the prime mover may be shut down.
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Section 14 Alternators Part 2
12. HUNTING It has already been shown that when two alternators are operating in parallel, their rotors maintain the same relative position, and the voltages of the two machines are in direct opposition as far as their local circuits are concerned. Any variation of the driving torque exerted by the prime mover of either machine tends to disturb this condition of stability and establishes a synchronising current tending to pull the two alternators back to the correct phase position.
In reciprocating engines, the turning torque varies during each stroke, and this variation is only partly corrected by the use of fly-wheels. The varying torque gives rise to a regular motion of the rotor of the alternator on either side of the normal position, thus producing a synchronising current. This phenomenon is referred to as "hunting."
The oscillations produced by hunting, particularly if reciprocating engines are used as prime movers, may become large enough to produce excessive synchronising current and even to cause the machines to fall out of step.
To increase the damping effect of eddy currents, special windings are sometimes placed in slots in the pole shoes. These windings are known as damping grids or "amortisseur windings", and consist of a number of heavy copper bars or rods, one in each slot, securely joined at the ends to a common connecting bar to form a short- circuited grid arrangement, as shown in figure11.
The rotor pole faces can therefore be considered to be situated in a magnetic field which oscillates back and forth and consequently have emfs induced in them. Now
Figure 11
these emfs will produce currents in any closed circuit in which they are able to act and thus tend to damp out the motion which produces them (Lenz's law) and so the pole face eddy currents result in smoother running.
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Tutorial 14 NAME:
ALTERNATORS – PART 2 SECTION A In the following statements one of the suggested answers is best. Place the identifying letter on your answer sheet. 1. The efficiency of an alternator is the ratio of:- a) kVA output to kVA input; b) kW output to kW input; c) kVA output to kW input; d) kW output to kVA input;
2. The voltage output of a three phase 50 hertz alternator is adjusted to the required value by means of:- a) altering the field excitation; b) changing the speed c) using a tapped winding; d) adjusting the number of poles.
3. Alternators are connected in parallel to:- a) increase the output voltage supplied to the load; b) increase the output current supplied to the load; c) allow two alternators to be driven by one prime mover; d) because two small alternators are more efficient than one large one.
4. Alternators are rated in terms of:- a) speed and output voltage; b) current and output voltage; c) Output voltage and kVA; d) Output voltage and kW.
5. An alternator designed with a high winding impedance better:- a) controls output voltage; b) limits short circuit current; c) reduces losses in windings; d) limits open circuit speed.
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Tutorial 14 Alternators Part 2
6. If the terminal voltage on an alternator increases as load increases the type of load connected to the alternator is a:- a) modern highly efficient load; b) lagging power factor load; c) unity power factor load; d) leading power factor load.
7. The load characteristic curve of an alternator shows the how the:- a) excitation varies with load; b) speed varies with excitation; c) t.p.d. varies with load; d) t.p.d. varies with excitation.
8. Armature reaction in an alternator causes:- a) a reduction in torque due to load current; b) a change in field flux for leading and lagging power loads; c) an increase in armature speed as load increases; d) a decrease in armature speed as load increases.
SECTION B Blank spaces in the following statements represent omissions. Write the appropriate information on your answer sheet. The percentage rise in terminal voltage of an alternator when full load is removed is called it's _____(1)_____. The impedance of the windings of an alternator causes the t.p.d. to _____(2)_____ when a load is connected to the alternator. When operating alternators in parallel the load share of one alternator may be increased by increasing the power input to the alternators_____ (3) _____ . Alternators are rated in kVA rather than kW because the voltage and kVA rating can be used to determine the maximum _____(4)_____ that the alternator can deliver. Automatic voltage regulation equipment is used to keep the _____(5)_____ of an alternator constant as load current changes.
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Alternators Part 2 Tutorial 14
SECTION C 1. A three phase 6 600 volt alternator supplies a current of 2 200 amperes at full load. Determine:- a) the rated kVA output; (25.15MVA or 25 150kVA) b) the output power at 0.8 lagging power factor. (20.1MW or 20 100kW)
SECTION D 1. For the alternator load characteristic curves shown in Figure 1:- a) indicate whether curve A, B and C are lagging, leading or unity power factor; A:- ______power factor; B:- ______power factor; C:- ______power factor; b) for the rated current shown, determine the open circuit and full load voltage of the alternator for the leading, lagging and unity power factors shown.
Full Load Current 300 C
TPD B
200 A
V
100
0 Load Current - Amperes 0
Figure 1
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SYNCHRONOUS MOTORS
PURPOSE: In this section you will learn about the construction and operation of three phase synchronous motors.
TO ACHIEVE THE PURPOSE OF THIS SECTION: At the end of this section the student will be able to: • Describe the construction of three phase synchronous motors. • Describe the operation of the three phase synchronous motor. • List methods used to start three phase synchronous motors. • Calculate the synchronous speed of a three phase synchronous motor, given the supply frequency and number of poles. • Describe the effect of varying the dc excitation on the operation of a three phase synchronous motor. • Identify and interpret the V-curves of a three phase synchronous motor. • List practical applications of three phase synchronous motors.
REFERENCES: • Electrical Trade Principles 2nd Edition J Hampson & S Hanssen. Section 6 • Electrical Principles for the Electrical Trades. 5th Edition. Jenneson J.R. Pages 259-262.
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Section 15 Synchronous Motors
1. INTRODUCTION An alternator, when supplied with electrical energy at rated voltage and frequency, will operate satisfactorily as a motor. If a synchronous machine operates as a motor, it is called a "synchronous motor". As the name implies, synchronous motors run in synchronism with a rotating magnetic field. The speed of rotation is therefore tied to the frequency of the supply. Because the frequency is fixed, the motor speed • ______, irrespective of the load. However, synchronous motors are used not so much because they run at constant speed, but because they possess other unique electrical properties. These properties will be dealt with in this section. Most synchronous motors are rated between - • ______Consequently these machines are and operate at speeds ranging from mainly used in heavy industry • ______. At the other end of the power spectrum, you will find tiny single phase synchronous motors used in control devices and electric clocks. The operation of these smaller synchronous motors will be treated at the end of this module.
2. TYPES OF SYNCHRONOUS MOTORS There are three types of synchronous motors in common use. All three motors have the same type of construction for their stators and they differ in their rotor construction. Reluctance Synchronous Motor: The rotor of this type of motor contains low reluctance material which allows the concentration of magnetic flux so that the rotor can lock into the rotating magnetic field produced in the Stator windings. Hysteresis Synchronous Motor: The rotor of this motor is constructed of a very hard, highly permeable permanent magnet alloy material supported on a non-magnetic shaft. This rotor is smooth without windings or slots and this design causes the motor to develop a constant torque through every revolution of the rotor. These motors are generally rated at less than 1kW. Permanent Magnet Synchronous Motor: The rotor of this motor is a permanent magnet without any external excitation required. The number of poles on the rotor must equal the number of poles on the stator. The amount of torque developed by these motors varies according to the machines construction parameters.
* See images of these motors on pages 304 -308 of your textbook
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Synchronous Motors Section 15
3. CONSTRUCTION As with alternators and induction motors, synchronous motors may be single, two or three phase. The constructional details of a three phase synchronous motor are essentially the same as that of a three phase alternator, as discussed in section 1. Therefore, depending upon the required application the synchronous machine maybe used to convert - • mechanical energy to electrical energy - ______operation • electrical energy to mechanical energy - ______operation. Synchronous Synchronous machine machine Exciter Exciter
Mechanical Mechanical input power output power
Electrical Electrical output power input power
Figure 1
In construction, the synchronous motor is almost identical with the alternator. However, better synchronous motor characteristics can be obtained if slight modifications are made, for example, in length of air-gap and number of field ampere-turns. In general, synchronous motors are of the salient pole type, since this type gives greater stability of operation. As shown in figure 2, the "stator" is composed of a slotted magnetic core which carries a three phase winding. The winding is identical to that of a three phase induction motor. The stator is supplied with electrical energy from a three phase source. The combination of three phase windings displaced by 1200E and supplied with three phase currents which are also displaced by 1200E results in the production of a - • ______Figure 2 Figure 2
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Section 15 Synchronous Motors
The "rotor" has a set of salient poles that are excited by DC current, see figure 3. The exciting coils are connected in series to two slip rings, and the DC current is fed into the winding from an external source, usually an exciter.
Figure 3 Slots are punched out along the circumference of the salient poles. They carry a squirrel cage winding similar to that in a three phase induction motor, see figure 4. This so called damper winding serves to start the motor and improve operating stability.
Figure 4
The general arrangement of the three phase synchronous motor is shown in figure 5. The stator winding is supplied from a three phase source. DC Supply The rotor winding is supplied with direct current from an external source, usually Three Phase Supply an exciter. Figure 5
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Synchronous Motors Section 15
4. PRINCIPLE OF OPERATION The stator of the synchronous motor is similar to the stator of the alternator and the induction motor. Consequently, if polyphase currents are supplied to the stator, a rotating magnetic field is produced. This field travels at synchronous speed. The speed is determined by the same factors that govern the synchronous speed of induction motors. The synchronous speed is found using the equation -
where: ns = synchronous speed in rpm f = supply frequency in hertz P = number of poles N Figure 6 illustrates such a rotating magnetic field and the four pole field produced is assumed to be rotating in a clockwise S S direction. If a stationary salient pole rotor having four poles is placed in this field, figure 7, it will be found that the rotor N • ______. Figure 6
N However, if the rotor is run up to nearly synchronous speed, the south poles of the rotor will be attracted to the north poles of the S stator and the north poles of the rotor attracted to the south poles S N N S of the stator. The rotating field of the stator, thus being locked S with the rotor poles, will drag the rotor round at N • ______. Figure 7
5. METHODS OF FIELD EXCITATION The rotor field windings are supplied with low voltage DC through slip rings, in the following ways - • from a small DC generator, called an exciter, direct coupled to the shaft of the synchronous motor. • from an independent DC supply.
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Section 15 Synchronous Motors
6. LOAD ON A SYNCHRONOUS MOTOR When load is applied to the shaft of a synchronous motor, its average speed cannot decrease because the motor must operate at constant average speed, that is, synchronous speed. The rotor will still continue to run steadily at synchronous speed, but the position of the rotor lags continuously behind the position which it would occupy on no load. This means there is an • ______between the rotor and the stator rotating magnetic field.
C D C N N
S S
O O Figure 8 Figure 9
7. EFFECT OF VARYING FIELD EXITATION The most suitable excitation is that which requires the stator current to be a minimum for a particular load. Any increase or decrease in the exciting current from this normal value will result in an increase in stator current and a change in the power factor of the motor stator current.
*Teacher demonstration:
Using the setup shown in the practical, demonstrate the effect of changes to field excitation: • on stator current • on Motor power factor Record what you observe in table 1 below
Field excitation Stator current Power factor Normal Below normal Above normal
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Synchronous Motors Section 15
• V-CURVES V-curves are curves showing the relationship between the stator current, field excitation current and power factor of a synchronous motor, for various loads. Full load 1.0 Stator C A current 0.8
No load Full load 0.6 Stator Power current D factor
0.4
0.2 No load Power Factor Power Stator Stator Current B Lagging Leading factor 0 0 0 Under Over Field Excitation Current Figure 15 If the power input to a synchronous motor is held constant and the excitation current varied, it is found that - • for small values of excitation current the stator takes a large lagging stator current and the power factor is low. This is shown by curves A and B in figure 15, which represents stator current and power factor, respectively, for a synchronous motor on no load. • as the excitation current is increased, the stator current falls until a minimum value is reached with a power factor of unity. • further increases in excitation current cause the stator current to increase and become leading while the power factor falls.
The excitation current that gives the minimum stator current and unity power factor is termed - ______for the particular load considered. If power factor improvement is not required, then normal excitation is the best excitation to use. If the excitation is less than normal, the motor is said to be ______. If the excitation is greater than normal, the motor is said to ______. If the power output is increased to its full load value and kept constant, the stator current (curve C) and the power factor (curve D) are obtained. Notice that the normal excitation is increased by an increase in load.
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Section 15 Synchronous Motors
It should be kept in mind that when under excited the motor acts as an inductance, in that it takes a lagging current. With over excitation it takes a leading current and behaves as a capacitor. Over excitation is desirable for the following reasons - • it provides a stronger working field and so increases the pull out torque. • it reduces the liability to fall out of step through hunting. • the motor works with a leading stator current and so compensates for lagging currents in other parts of the system.
8. HUNTING Hunting, or phase swinging, is an effect produced in alternators and synchronous motors which interferes with their smooth working. It is the oscillation of the rotor above and below normal synchronous speed, that is, a swinging movement on each side of the true synchronous position. The hunting effect can be detected by ear, in some cases, because of the pulsating sound set up by the pendulum motion. In all cases, the hunting effect is shown on instruments connected in the stator circuit, unless the meters are very highly damped. The pointers of analogue instruments sway to and fro in time with the oscillations of the machine. Hunting in a synchronous motor is caused in the following ways - • sudden change of load on the motor. • sudden change of supply frequency. • periodic pulsation of the load on the motor, usually when reciprocating machinery is being driven. • sudden changes in the excitation current caused by loose connections or chattering brushes. It is possible for hunting to cause such an amount of oscillation that the magnetic coupling stretches beyond the limit of its elasticity and the motor stops. Hunting is prevented or reduced in the following ways - • by running the motor with strong excitation. • by using "amortisseur windings" or "damping windings" on the pole shoes. The swaying to and fro of the flux across the pole face cuts the conductors of the amortisseur winding, inducing emf's in them and as the circuit is closed a current is produced, which by Lenz's Law, will tend to damp out the motion which causes it and thus produce smoother running. If the rotation of the rotor is perfectly uniform there is no relative movement between the flux and amortisseur windings and so no currents are set up in them.
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Synchronous Motors Section 15
9. STARTING SYNCHRONOUS MOTORS Since the ordinary synchronous motor is not self starting, it is necessary to start it by auxiliary means and bring it up to about synchronous speed. The stator and the field poles are then energised and the motor falls into step and continues to rotate at synchronous speed. The following methods are commonly employed to start synchronous motors -
• using the exciter as a motor - The exciter is sometimes used as a motor to bring the synchronous motor up to synchronous speed, if a suitable DC supply is available. The motor is then synchronised by connecting its stator winding to the AC supply and its field to the DC supply. After synchronisation, the motor has its field supplied from the exciter. • using a pony motor - An alternative method of starting a synchronous motor is to use a small auxiliary direct coupled induction motor called a pony motor. If a pony motor is connected in parallel with the main motor it is designed with two poles less to enable it to reach the synchronous speed of the synchronous motor. • using the damping winding - Three phase synchronous motors can be started by the addition of a special damping winding on the field system. The stator currents set up a rotating magnetic field which cuts the damping winding and establishes currents in the short circuited bars or grids. A torque is developed and the machine runs up as an induction motor to a speed slightly less than synchronous speed. A gradually increasing direct current is then supplied to the field winding until the two sets of poles lock together with a magnetic coupling between them. Because of the large starting current required by this method it is usual to start on a reduced voltage, as with the squirrel cage induction motor. This reduced voltage may be obtained from tappings on an auto transformer. The DC field circuit should be disconnected in several places during the earlier part of the starting period, because of the high voltages induced in it by transformer action when the speed is low. For the same reason also, it is usual to insulate the field windings for a voltage much higher than the excitation voltage, when this method of starting is used. • using a variable frequency drive - A variable frequency supply may be used to reduce the supply frequency to allow the rotor field to lock onto the stator rotating magnetic field. Once started the supply frequency is increased to accelerate the motor to rated speed. The supply voltage is reduced when frequency is reduced to compensate for the reduction in stator winding inductive reactance at the lower frequency.
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Section 15 Synchronous Motors
10. APPLICATIONS OF SYNCHRONOUS MOTORS Synchronous motors have three important characteristics which dictate their application - • ______for all loads (zero speed regulations) • ______power factor • ______operating efficiency (85% to 96%). Applications for synchronous motors include - • power factor correction • transmission line voltage control • constant speed drives, such as fans, air compressors, pumps, rock crushers and mill rolls • variable speed drives, such as extruder and mill drives, the advantage being higher efficiency at slower speeds than an induction motor.
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NAME: Practical 15
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Tutorial 15 NAME:
SYNCHRONOUS MOTORS SECTION A In the following statements one of the suggested answers is best. Place the identifying letter on your answer sheet. 1. The rotor of a synchronous motor spins:- a) At Synchronous speed; b) Slightly slower than Synchronous speed; c) Slightly faster than Synchronous speed; d) at constant speed regardless of frequency and load; 2. The operating power factor of a synchronous motor:- a) is affected by phase sequence of the supply: b) is affected by field excitation; c) is constant regardless of any changes; d) improves as load increases. 3. "Normal excitation" of a synchronous motor at full load:- a) is the rated field current on the nameplate; b) gives minimum power factor and maximum current; c) gives minimum power factor and minimum current; d) gives unity power factor and minimum stator current. 4. Synchronous motors develop a torque because of:- a) electromagnetic induction between stator and rotor; b) mutual induction between stator and rotor; c) attraction between stator and rotor fields; d) stator field hunting the rotor field. 5. The advantages of operating a synchronous motor with "over excitation" are:- a) increased pull out torque and increased power factor; b) decreased line current and a leading motor power factor; c) decreased line current and a leading motor power factor; d) increased pull out torque and a leading motor power factor.
6. Damper windings or amortisseur windings are used in synchronous motors to:- a) start the motor and reduce hunting on reciprocating loads; b) reduce the amount of direct current required in the field windings; c) reduce the amount of current taken from the supply during starting; d) bring the motor to a stop quickly after being turned off.
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Tutorial 15 Synchronous Motors
SECTION B Blank spaces in the following statements represent omissions. Write the appropriate information on your answer sheet. The term "synchronous capacitor" is used to describe a synchronous motor which has been _____(1)_____ excited. If the exciter is used to bring the synchronous motor up to speed it is necessary to have a separate _____(2)_____ supply. A three phase synchronous motor may be started as an induction motor if the motor is fitted with _____(3)_____ windings. If a pony motor is used to bring a synchronous motor up to speed it is necessary to start the motor on _____(4)_____ load or have a clutch or eddy current drive between the motor and ____(5)_____. A three phase synchronous motor would have _____(6)_____ slip rings.
SECTION C 1. A four pole synchronous motor is connected to a 60Hz supply. Determine the speed of the motor. (1 800 r.p.m.) 2. A three phase four pole 415 volt synchronous motor is connected to 50Hz supply. The motor takes a current of 75 amperes at full load with normal excitation while driving a 50kW load. Determine:- a) the input power to the motor; (53.91kW) b) the efficiency of the motor under these conditions; ( 92.75%) c) the speed of the motor; (1 500r.p.m.) d) the torque delivered to the load at normal excitation; (318.3Nm)
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Synchronous Motors Tutorial 15
SECTION D The questions below relate to the V curves for a three phase synchronous motor shown in Figure 1:- 1. Use the labels on the curves (A, B, C, and D) to indicate which curve is:- a) the curve of power factor vs field current for no load; b) the curve of power factor vs field current for full load; c) the curve of stator current vs field current for no load; d) the curve of stator current vs field current for full load.
2. What is the value of field current for the motor at normal excitation on full load? ______
3. What is the value of stator current taken by the motor at normal excitation on no load? ______
4. If the field current is 4 amperes while the motor is driving full load:- a) What is the stator current taken? b) What is the power factor? c) Is the power factor leading, lagging or unity? d) Is the motor under, over or normally excited?
1.0 50 A C 0.8 40
0.6 30 D
0.4 20
10 Power Factor 0.2 B 0 0 0 1 2 3 4 5 6 Field Excitation Current Amperes Stator Current Stator Current Amperes Figure 1
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Section 16
Motor Protection, Faults & Maintenance
PURPOSE:
This section introduces the direct on line motor starter and the AS/NZS 3000 requirements for starting three phase induction motors.
TO ACHIEVE THE PURPOSE OF THIS SECTION:
At the end of this section the student will be able to: • Understand the reasons why motor protection is required. • List the requirements of the AS/NZS3000 Wiring rules with regards to motor protection. • Understand the operating principles of microtherm devices, thermal and magnetic motor protection devices. • Understand the electrical features of motor protection HRC fuses. • Understand the effects of under voltage and over voltage on motors and motor circuits. • Understand the effects of repetitive starting and/or reversing on motors. • Understand the special requirements for motor protection, in high humidity or moist environments, • Understand the requirements for high temperature areas and corrosive atmospheres. • Understand the operating principles of phase failure protection. • select suitable protective devices for a given motor and starter combination • List the steps involved in dismantling and re-assembling a motor. • Test insulation resistance of a three-phase induction motor prior to connection to the supply • Test winding resistance (ohmic value and continuity) of a three-phase induction motor prior to connection to the supply
REFERENCES:
• Electrical Trade Principles 2nd Edition J Hampson & S Hanssen. Section 6
• Electrical Principles for the Electrical Trades. 5th Edition. Jenneson J.R. Pages 297-298, 304-305, 313-318.
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Section 16 Motor protection, faults and testing
1. MOTOR PROTECTION – See ‘Three phase motor protection’ on page 257 of your text book and answer the following questions: Motor protection is required to safeguard the motors ______equipment and the ______conductors against excessive ______. Motor protection also safeguards the motor windings against overheating.
2. AS/NZS 3000 REQUIREMENTS
The requirements laid down in AS/NZS 3000 relating to motors include -
• isolation and switching
• overload protection
• over temperature protection
• restarting or reversal.
These are found in Clause 4.3 of AS/NZS 3000/ 2007
3. MOTOR PROTECTIVE DEVICES
Motor protective devices include:
• Motor start HRC fuses
• Motor type thermal magnetic circuit breakers
• Magnetic overload relays
• Under and overvoltage relays
• Thermal overload relays
• Over-temperature protection
Motors also need protection against moisture.
*Teacher to show and discuss motor protective devices.
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Motor faults, protection and testing Section 16
4. HRC FUSES How are motor start HRC fuses different from standard HRC fuses? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………….
5. THERMAL MAGNETIC CIRCUIT BREAKERS What is the difference between a type C and a Type D thermal magnetic circuit breaker? …………………………………………………………………………………………… …………………………………………………………………………………………… ……………………………………………………………………………
6. MAGNETIC OVERLOAD RELAYS *See Section titled magnetic overload relays on page 258 of your textbook and answer the following question. Electromechanical magnetic overload relays are also known as ______or ______relays.
These relays provide ______and ______protection and are fitted with pistons and dashpots to prevent ______tripping. *See figure 6.26 on page 258 of your textbook.
7. UNDER AND OVERVOLTAGE PROTECTION *See the table below for the effects of over and under voltage.
Speed Current Effect on motor without protection Over increases Increases damage to insulation / burn out voltage Under decreases Increases Windings overheat / burn out voltage to drive the load *See under voltage relay shown in figure 6.27 on page 259 of your textbook.
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Section 16 Motor protection, faults and testing
8. THERMAL OVERLOAD RELAYS *See Section titled thermal overload relays on page 259of your textbook and answer the following question. Thermal overload relays simulate the ______effect in the motor windings by sending the motor ______through ______elements in the relay. In your own words explain how a thermal overload relay trips. …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………… Most thermal overloads are fitted with an ______current setting, for example 4A to 8A, or 45A to 56A. It is the responsibility of the installing ______to correctly set the overload current setting to the appropriate value. This value is usually the full load rated current of the motor.
*See a motor control circuit shown in figure 6.33 on page 265 in your text book. Discuss with your teacher the wiring of a thermal overload relay.
9. OVER TEMPERATURE PROTECTION PTC Thermistors Thermistors are designed to sense and protect motors from damage caused by excessive temperature. The table below shows the maximum operating temperature of motor winding insulation.
Source : http://www.cooperindustries.com/content/dam/public/bussmann/Electrical/Resou rces/solution- center/technical_library/BUS_Ele_Tech_Lib_Motor_Insulation_Class_Life.pdf
*See PTC thermistors in figure 6.29 on page 260 of your text book.
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Motor faults, protection and testing Section 16
The table below shows the use of PTC thermistors embedded in Stator windings for protection against over temperature in ABB AC motors.
Source: HTTP://WWW05.ABB.COM/GLOBAL/SCOT/SCOT234.NSF/VERITYDISPLAY/11C6E0AF95BF9782C1257C0100 36A39A/$FILE/GENERAL%20PERFORMANCE%20IE2%20MOTORS%20EU%20MEPS%209AKK105789%20 EN%2001_2013%20REV%20A.PDF
Klixons
Klixons are devices containing a bi-metal strip which open circuits when excessive temperature of the motor housing is reached. They can be manually or automatic reset type.
10. EFFECTS OF REPETITIVE STARTING OR REVERSAL Motor current during starting or reversal is much higher than running current. This increased current generates excessive heat and can damage motor insulation. Motors therefore are limited to a number of starts per hour. *Read the last paragraph in the section titled starting and reversing on page 261 -262 and fill in the missing words below.
Standard 3 phase motors Number of starts per hour Below 1kW Above 30kW
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Section 16 Motor protection, faults and testing
11. MOTORS OPERATING IN HUMID OR MOIST ENVIRONMENTS. Moisture can cause corrosion and a breakdown of insulation in electric motors. Motors are constructed with different types of enclosures to suit the environment they will be working in. Read the section titled ‘Enclosures’ on page 261 of your textbook and answer the following questions. ODP stands for ______. These motors are designed to allow water dripping on the motor to run off and are cooled by a fan on the non- drive end of the motor shaft. TEFC stands for ______. These motors are completely sealed and are cooled by a fan mounted on the non-drive end of the shaft outside of the sealed enclosure. TENV stands for ______. These motors are completely sealed and rely on large cooling fins to radiate heat away from the motor. Motors suitable for moist environments are IP rated. *See Ingress protection on page 261 of your textbook
XPRF stands for ______.
12. DISMANTLING AND RE-ASSEMBLING A MOTOR
When it is necessary to dismantle a machine, it is important to follow a set procedure. Wherever possible use the manufacturer’s data as a guide. Failure to do so could cause damage to machine components and unnecessary expense. Inspection of the machine before dismantling helps determine which parts are satisfactory, which parts need repair or replacement. Inspection should be done during dismantling as well.
*See pages 293 & 297 in your text book.
Reassembly usually follows the reverse procedure to disassembly. Care needs to be taken when assembling bearings, gaskets, seals etc. to avoid damage. Of course cleanliness of the work area is most important to avoid contamination of the assembly.
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MOTOR TESTING
PURPOSE:
To test a squirrel cage induction motor.
TO ACHIEVE THE PURPOSE OF THIS SECTION:
At the end of this practical assignment the student will be able to understand and perform: • Stator Winding resistance tests • Earth resistance tests • Insulation resistance tests • Measure motor line current when running.
EQUIPMENT:
• 1 x 41.5/24V three-phase, 50Hz, AC supply
• 1 x 41.5V three-phase squirrel cage induction motor + single machine bed
• Motor fault panels
• 1 x digital multimeter
• 1 x insulation tester
• Prova 23 tong test meter
• 4mm connection leads
– REMEMBER – – WORK SAFELY AT ALL TIMES – Observe correct isolation procedures
Machines Section 16 Version 2 Page 1 of 9 Owner: Elect, ICT & Design Faculty/Electrical Last Updated 31/03/2021 Disclaimer: Printed copies of this document are regarded as uncontrolled Practical 16 Motor protection, faults and testing PROCEDURE - 3 PHASE INDUCTION MOTOR FAULT PANEL EXERCISE
1. Attach the Motor fault panel to the front of the 3 phase induction motor 2. Determine the correct numbers for each of the 3 phase windings using an ohmmeter. Record your results in the table below. 3. Connect the wires you have identified as the phase windings to the terminals U1,U2, V1, V2, W1, W2 on the motor terminal block. 4. Measure the resistance of each winding and record the readings in the table below 5. Switch on fault 1 and Test the motor for winding continuity, insulation resistance and earth continuity. 6. Repeat tests for each fault switch 2,3,4 & 5. 7. Record any faults found in the table below. Panel no: Numbers R Numbers R Numbers R (Ω) Numbers R (Ω) (Ω) (Ω) Winding numbers pairs
Fault 1
Fault 2
Fault 3
Fault 4
Fault 5
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