# ELECTRIC MACHINERY FUNDAMENTALS CHAPMAN 4TH EDITION PDF

IN Libmry of Co n ~ress Gltalo~in~-in-l'ublic:ltion Data Chapman. Stephen J. Electric machinery fundamentals / Stephen Chapman. - 4th ed. p. em. Includes. Stephen J. Electric machinery fundamentals / Stephen Chapman. - 4th ed. p. em Stephen J. Chapman received a B.S. in Electrical Engineering from Louisiana. instructor's manual to accompany chapman electric machinery fundamentals fourth edition stephen chapman bae systems australia instructor's manual to.

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Instructor's Manual to accompany. Chapman. Electric Machinery Fundamentals. Fourth Edition. Stephen J. Chapman. BAE SYSTEMS Australia. Stephen J. Electric machinery fundamentals / Stephen Chapman. - 4th ed. Electric Machinery Fundamentals, 4th Edition (McGraw-Hill Series in Electrical and Computer pellwillfigalus.gq Oh Crap! Potty Training. Electric Machinery Fundamentals 4th ED (Chapman) - Ebook download as PDF File .pdf) or view presentation slides online. Electric Machinery Fundamentals.

Segment cd The velocity of the wire is tangential to the path of rotation, while B points to the right. The quantity v x B points into the page, which is the same direction as segment cd. Segment da same as segment bc, v x B is perpendicular to l. To determine the magnitude and direction of the torque, examine the phasors below: The force on each segment of the loop is given by: Segment ab The direction of the current is into the page, while the magnetic field B points to the right.

Segment bc The direction of the current is in the plane of the page, while the magnetic field B points to the right. Segment cd The direction of the current is out of the page, while the magnetic field B points to the right. Segment da The direction of the current is in the plane of the page, while the magnetic field B points to the right. An alternative way to express the torque equation can be done which clearly relates the behaviour of the single loop to the behaviour of larger ac machines.

Examine the phasors below: If the current in the loop is as shown, that current will generate a magnetic flux density Bloop with the direction shown. The magnitude of Bloop is: The area of the loop A is 2rl and substituting these two equations into the torque equation earlier yields: The Rotating Magnetic Field Before we have looked at how if two magnetic fields are present in a machine, then a torque will be created which will tend to line up the two magnetic fields.

If one magnetic field is produced by the stator of an ac machine and the other by the rotor, then a torque will be induced in the rotor which will cause the rotor to turn and align itself with the stator magnetic field.

How do we make the stator magnetic field to rotate? It is a 2-pole winding one north and one south. Assume currents in the 3 coils are: It produces the magnetic field intensity: Refer again to the stator in Figure 4. Assume that we represent the direction of the magnetic field densities in the form of: We know that: The Relationship between Electrical Frequency and the Speed of Magnetic Field Rotation The figure above shows that the rotating magnetic field in this stator can be represented as a north pole the flux leaves the stator and a south pole flux enters the stator.

These magnetic poles complete one mechanical rotation around the stator surface for each electrical cycle of the applied current. The mechanical speed of rotation of the magnetic field in revolutions per second is equal to electric frequency in hertz: In this winding, a pole moves only halfway around the stator surface in one electrical cycle.

To prove this, phases B and C are switched: However, the flux in a real machine does not follow these assumptions, since there is a ferromagnetic rotor in the centre of the machine with a small air gap between the rotor and the stator. The rotor can be cylindrical a nonsalient-pole , or it can have pole faces projecting out from its surface b salient pole. The reluctance of the air gap in this machine is much higher than the reluctances of either the rotor or the stator, so the flux density vector B takes the shortest possible path across the air gap and jumps perpendicularly between the rotor and the stator.

To produce a sinusoidal voltage in a machine like this, the magnitude of the flux density vector B must vary in a sinusoidal manner along the surface of the air gap.

The flux density will vary sinusoidally only if the magnetizing intensity H and mmf varies in a sinusoidal manner along the surface of the air gap. A cylindrical rotor with sinusoidally varying air-gap flux density The mmf or H or B as a function of angle in the air gap To achieve a sinusoidal variation of mmf along the surface of the air gap is to distribute the turns of the winding that produces the mmf in closely spaced slots around the surface of the machine and to vary the number of conductors in each slots in a sinusoidal manner.

The number of conductors in each slot is indicated in the diagram. The mmf distribution resulting from the winding, compared to an ideal transformer. The distribution of conductors produces a close approximation to a sinusoidal distribution of mmf. The more slots there are and the more closely spaced the slots are, the better this approximation becomes. In practice, it is not possible to distribute windings exactly as in the nC equation above, since there are only a finite number of slots in a real machine and since only integral numbers of conductors can be included in each slot.

Induced Voltage in AC Machines The induced voltage in a Coil on a Two-Pole Stator Previously, discussions were made related to induced 3 phase currents producing a rotating magnetic field. Now, lets look into the fact that a rotating magnetic field may produce voltages in the stator. The Figures below show a rotating rotor with a sinusoidally distributed magnetic field in the centre of a stationary coil. Assume that the magnetic of the flux density vector B in the air gap between the rotor and the stator varies sinusoidally with mechanical angle, while the direction of B is always radially outward.

The magnitude of the flux density vector B at a point around the rotor is given by: In this case, the wire is stationary and the magnetic field is moving, so the equation for induced voltage does not directly apply. The total voltage induced in the coil will be the sum of the voltages induced in each of its four sides.

## Electric Machinery Fundamentals 4th Edition (Stephen J Chapman) Part 2.pdf

These are determined as follows: Segment bc The voltage is zero, since the vector quantity v x B is perpendicular to l. Segment da The voltage is zero, since the vector quantity v x B is perpendicular to l. Therefore total induced voltage: This derivation goes through the induced voltage in the stator when there is a rotating magnetic field produced by the rotor. The Induced Voltage in a 3-Phase Set of Coils If the stator now has 3 sets of different windings as such that the stator voltage induced due to the rotating magnetic field produced by the rotor will have a phase difference of o, the induced voltages at each phase will be as follows: Induced Torque in an AC Machines In ac machines under normal operating conditions, there are 2 magnetic fields present - a magnetic field from the rotor circuit and another magnetic field from the stator circuit.

The interaction of these two magnetic fields produces the torque in the machine, just as 2 permanent magnets near each other will experience a torque, which causes them to line up.

A simplified ac machine with a sinusoidal stator flux distribution and a single coil of wire mounted in the rotor.

Its resultant direction may be found in the diagram above. How much torque is produced in the rotor of this simplified ac machine? This is done by analyzing the force and torque on each of the two conductors separately: Since the total magnetic field density will be the summation of the BS and BR, hence: Synchronous Generator Construction 2. The Speed of Rotation of a Synchronous Generator 3. The Equivalent Circuit of a Synchronous Generator 5.

The Phasor Diagram of a Synchronous Generator 6. Power and Torque in Synchronous Generator 7. Measuring Synchronous Generator Model Parameters 8. Parallel operation of AC Generators - The conditions required for paralleling - The general procedure for paralleling generators - Frequency-power and Voltage-Reactive Power characteristics of a synchronous generator. Synchronous Generator Construction A DC current is applied to the rotor winding, which then produces a rotor magnetic field.

The rotor is then turned by a prime mover eg. Steam, water etc. This rotating magnetic field induces a 3-phase set of voltages within the stator windings of the generator. Generally a synchronous generator must have at least 2 components: Salient Pole b. Non Salient Pole b Stator Windings or Armature Windings The rotor of a synchronous generator is a large electromagnet and the magnetic poles on the rotor can either be salient or non salient construction.

Non-salient pole rotors are normally used for rotors with 2 or 4 poles rotor, while salient pole rotors are used for 4 or more poles rotor. Non-salient rotor for a synchronous machine Salient rotor A dc current must be supplied to the field circuit on the rotor. Since the rotor is rotating, a special arrangement is required to get the dc power to its field windings. The common ways are: One end of the dc rotor winding is tied to each of the 2 slip rings on the shaft of the synchronous machine, and a stationary brush rides on each slip ring.

If the positive end of a dc voltage source is connected to one brush and the negative end is connected to the other, then the same dc voltage will be applied to the field winding at all times regardless of the angular position or speed of the rotor. Some problems with slip rings and brushes: Small synchronous machines — use slip rings and brushes. Larger machines — brushless exciters are used to supply the dc field current. A brushless exciter is a small ac generator with its field circuit mounted on the stator and its armature circuit mounted on the rotor shaft.

The 3-phase output of the exciter generator is rectified to direct current by a 3-phase rectifier circuit also mounted on the shaft of the generator, and is then fed to the main dc field circuit. By controlling the small dc field current of the exciter generator located on the stator , we can adjust the field current on the main machine without slip rings and brushes.

Since no mechanical contacts occur between the rotor and stator, a brushless exciter requires less maintenance. A brushless exciter circuit: A small 3-phase current is rectified and used to supply the field circuit of the exciter, which is located on the stator. The output of the armature circuit of the exciter on the rotor is then rectified and used to supply the field current of the main machine.

To make the excitation of a generator completely independent of any external power sources, a small pilot exciter can be used. A pilot exciter is a small ac generator with permanent magnets mounted on the rotor shaft and a 3-phase winding on the stator.

It produces the power for the field circuit of the exciter, which in turn controls the field circuit of the main machine. If a pilot exciter is included on the generator shaft, then no external electric power is required. The permanent magnets of the pilot exciter produce the field current of the exciter, which in turn produces the field current of the main machine.

Even though machines with brushless exciters do not need slip rings and brushes, they still include the slip rings and brushes so that an auxiliary source of dc field current is available in emergencies. The Speed of Rotation of a Synchronous Generator Synchronous generators are by definition synchronous, meaning that the electrical frequency produced is locked in or synchronized with the mechanical rate of rotation of the generator.

Hence, the rate of rotation of the magnetic field in the machine is related to the stator electrical frequency by: The Internal Generated Voltage of a Synchronous Generator Voltage induced is dependent upon flux and speed of rotation, hence from what we have learnt so far, the induced voltage can be found as follows: The Equivalent Circuit of a Synchronous Generator The voltage EA is the internal generated voltage produced in one phase of a synchronous generator.

If the machine is not connected to a load no armature current flowing , the terminal voltage will be equivalent to the voltage induced at the stator coils. This is due to the fact that there are no current flow in the stator coils hence no losses. These differences are due to: We will explore factors a, b, and c and derive a machine model from them.

The effect of salient pole rotor shape will be ignored, and all machines in this chapter are assumed to have nonsalient or cylindrical rotors. Armature Reaction When the rotor is spun, a voltage EA is induced in the stator windings.

If a load is attached to the terminals of the generator, a current flows. But a 3-phase stator current flow will produce a magnetic field of its own.

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This stator magnetic field will distorts the original rotor magnetic field, changing the resulting phase voltage. This effect is called armature reaction because the armature stator current affects the magnetic field, which produced it in the first place. Refer to the diagrams below, showing a two-pole rotor spinning inside a 3-phase stator. This stator magnetic field BS and its direction are given by the right-hand rule.

The stator field produces a voltage of its own called Estat. If X is a constant of proportionality, then the armature reaction voltage can be expressed as: In series with RF is an adjustable resistor Radj which controls the flow of the field current. The rest of the equivalent circuit consists of the models for each phase. If the 3 phases are connected in Y or I, the terminal voltage may be found as follows: If it is not balanced, a more in-depth technique is required.

The per-phase equivalent circuit: The phasor diagrams are as follows: Unity power factor Lagging power factor Leading power factor For a given phase voltage and armature current, a larger internal voltage EA is needed for lagging loads than for leading loads.

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Alternatively, for a given field current and magnitude of load current, the terminal voltage is lower for lagging loads and higher for leading loads. Power and Torque in Synchronous Generators A generator converts mechanical energy into electrical energy, hence the input power will be a mechanical prime mover, e. Regardless of the type of prime mover, the rotor velocity must remain constant to maintain a stable system frequency.

The power-flow diagram for a synchronous generator is shown: Stray losses, friction and windage losses, core loss Converted power: Copper losses Output: This gives a phasor diagram as shown: Based upon the simplified phasor diagram: Measuring Synchronous Generator Model Parameters There are basically 3 types of relationship which needs to be found for a synchronous generator: This plot is called open-circuit characteristic OCC of a generator.

With this characteristic, it is possible to find the internal generated voltage of the generator for any given field current. Open-circuit characteristic OCC of a generator At first the curve is almost perfectly linear, until some saturation is observed at high field currents. The unsaturated iron in the frame of the synchronous machine has a reluctance several thousand times lower than the air-gap reluctance, so at first almost all the mmf is across the air-gap, and the resulting flux increase is linear.

When the iron finally saturates, the reluctance of the iron increases dramatically, and the flux increases much more slowly with an increase in mmf.

The linear portion of an OCC is called the air-gap line of the characteristic. During the short circuit analysis, the net magnetic field is very small, hence the core is not saturated, hence the reason why the relationship is linear.

SCC is essentially a straight line. To understand why this characteristic is a straight line, look at the equivalent circuit below when the terminals are short circuited. When the terminals are short circuited, the armature current IA is: Since the net magnetic field is small, the machine is unsaturated and the SCC is linear. Therefore, an approximate method for determining the synchronous reactance XS at a given field current is: Find XS by applying the equation above.

Problem with this method: EA is taken from the OCC whereby the core would be partially saturated for large field currents while IA is taken from the SCC where the core is not saturated at all field currents. Hence the value of XS is only an approximate. Hence to gain better accuracy, the test should be done at low field currents which looks at the linear region of the OCC test. To find out on the resistive element of the machine, it can simply be found by applying a DC voltage to the machine terminals with the rotor stationary.

Value obtained in this test RA may increase the XS accuracy. Short Circuit Ratio Definition: Ratio of the field current required for the rated voltage at open circuit to the field current required for rated armature current at short circuit. Example A kVA, V, 50Hz, Y-connected synchronous generator with a rated field current of 5A was tested, and the following data were taken: When a dc voltage of 10V was applied to two of the terminals, a current of 25A was measured.

Find the values of the armature resistance and the approximate synchronous reactance in ohms that would be used in the generator model at the rated conditions. The synchronous generator operating alone The behaviour of a synchronous generator under load varies greatly depending on the power factor of the load and on whether the generator is operating alone or in parallel with other synchronous generator.

The next discussion, we shall disregard RA and rotor flux is assumed to be constant unless it is stated that the field current is changed. Also, the speed of the generator will be assumed constant, and all terminal characteristics are drawn assuming constant speed.

Load increase: An increase of load is an increase in real and reactive power drawn from the generator. Such a load increase increases the load current drawn from the generator. If EA is constant, what actually varies with a changing load?? Initially lagging load: This change may be seen in the phasor diagram. The effect of an increase in generator loads at constant power factor upon its terminal voltage — lagging power factor.

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The effect of an increase in generator loads at constant power factor upon its terminal voltage — unity power factor. Initially leading load: The effect of an increase in generator loads at constant power factor upon its terminal voltage — leading power factor.

However, in practical it is best to keep the output voltage of a generator to be constant, hence EA has to be controlled which can be done by controlling the field current IF. This generator has a synchronous reactance of 0. At full load, the machine supplies A at 0. Under full-load conditions, the friction and windage losses are 40kW, and the core losses aree 30kW.

Ignore any field circuit losses. How much power is supplied to the generator by the prime mover? How much field current would be required to keep VT at V? Its full-load armature current is 60A at 0. This generator has friction and windage losses of 1. Since the armature resistance is being ignored, assume that the I2R losses are negligible. The field current has been adjusted so that the terminal voltage is V at no load. It is loaded with the rated current at 0. It is loaded with the rated current at 1.

How large is the induced countertorque? Conditions required for Paralleling The figure below shows a synchronous generator G1 supplying power to a load, with another generator G2 about to be paralleled with G1 by closing switch S1.

## Chapman Electric Machinery Fundamentals Fourth Edition Solutions Manual

What conditions must be met before the switch can be closed and the 2 generators connected? If the switch is closed arbitrarily at some moment, the generators are liable to be severely damaged, and the load may lose power.

If the voltages are not exactly the same in each conductor being tied together, there will be a very large current flow when the switch is closed. To avoid this problem, each of the three phases must have exactly the same voltage magnitude and phase angle as the conductor to which it is connected.

Thus, paralleling 2 or more generators must be done carefully as to avoid generator or other system component damage. Conditions are as follows: If the generators were connected in this manner, there would be no problem with phase a, but huge currents would flow in phases b and c, damaging both machines.

This is done so that the phase angles of the incoming machine will change slowly with respect to the phase angles of the running system. General Procedure for Paralleling Generators Suppose that generator G2 is to be connected to the running system as shown below: Using Voltmeters, the field current of the oncoming generator should be adjusted until its terminal voltage is equal to the line voltage of the running system.

Check and verify phase sequence to be identical to the system phase sequence. There are 2 methods to do this: Alternately connect a small induction motor to the terminals of each of the 2 generators. If the motor rotates in the same direction each time, then the phase sequence is the same for both generators.

If the motor rotates in opposite directions, then the phase sequences differ, and 2 of the conductors on the incoming generator must be reversed. Another way is using the 3 light bulb method, where the bulbs are stretched across the open terminals of the switch connecting the generator to the system as shown in the figure above. As the phase changes between the 2 systems, the light bulbs first get bright large phase difference and then get dim small phase difference.

If all 3 bulbs get bright and dark together, then the systems have the same phase sequence. If the bulbs brighten in succession, then the systems have the opposite phase sequence, and one of the sequences must be reversed. Using a Synchroscope — a meter that measures the difference in phase angles it does not check phase sequences only phase angles. Check and verify generator frequency to be slightly higher than the system frequency.

This is done by watching a frequency meter until the frequencies are close and then by observing changes in phase between the systems. Once the frequencies are nearly equal, the voltages in the 2 systems will change phase with respect to each other very slowly. The phase changes are observed, and when the phase angles are equal, the switch connecting the 2 systems is shut.

All prime movers tend to behave in a similar fashion — as the power drawn from them increases, the speed at which they turn decreases.

The decrease in speed is in general non linear, but some form of governor mechanism is usually included to make the decrease in speed linear with an increase in power demand. Whatever governor mechanism is present on a prime mover, it will always be adjusted to provide a slight drooping characteristic with increasing load.

The speed droop SD of a prime mover is defined as: Most governors have some type of set point adjustment to allow the no-load speed of the turbine to be varied. A typical speed vs.

Since mechanical speed is related to the electrical frequency and electrical frequency is related with the output power, hence we will obtain the following equation: Example Figure below shows a generator supplying a load. A second load is to be connected in parallel with the first one.

The generator has a no-load frequency of Load 1 consumes a real power of kW at 0. Operation of Generators in Parallel with Large Power Systems Changes in one generator in large power systems may not have any effect on the system. A large power system may be represented as an infinite bus system.

An infinite bus is a power system so large that its voltage and frequency do not vary regardless of how much real and reactive power is drawn from or supplied to it. The power-frequency characteristic and the reactive power-voltage characteristic are shown below: We shall consider the action or changes done to the generator and its effect to the system. When a generator is connected in parallel with another generator or a large system, the frequency and terminal voltage of all the machines must be the same, since their output conductors are tied together.

Thus, their real power-frequency and reactive power-voltage characteristics can be plotted back to back, with a common vertical axis.

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Such a sketch is called a house diagram, as shown below: A synchronous generator operating in parallel with an infinite bus The frequency-power diagram house diagram for a synchronous generator in parallel with an infinite bus.

This is shown here: Suppose the generator had been paralleled to the line but instead of being at a slightly higher frequency than the running system, it was at a slightly lower frequency. In this case, when paralleling is completed, the resulting situation is as shown here: At this frequency, the power supplied by the generator is actually negative. Assume that the generator is already connected, what effects of governor control and field current control has to the generator?

Governor Control Effects: In theory, if the governor set points is increased, the no load frequency will also increase the droop graph will shift up.

Since in an infinite bus system frequency does not change, the overall effect is to increase the generator output power another way to explain that it would look as if the generator is loaded up further.

Hence the output current will increase. As the governor set points are further increased the no-load frequency increases and the power supplied by the generator increases.

If the governor is set as such that it exceeds the load requirement, the excess power will flow back to the infinite bus system. The infinite bus, by definition, can supply or consume any amount of power without a change in frequency, so the extra power is consumed. Field Current Control Effects: Increasing the governor set point will increase power but will cause the generator to absorb some reactive power. The question is now, how do we supply reactive power Q into the system instead of absorbing it?

This can be done by adjusting the field current of the generator. Power into the generator must remain constant when IF is changed so that power out of the generator must also remain constant. Now, the prime mover of a synchronous generator has a fixed-torque speed characteristic for any given governor setting.

This curve changes only when the governor set points are changed. Since the generator is tied to an infinite bus, its speed cannot change. In other words, increasing the field current in a synchronous generator operating in parallel with an infinite bus increases the reactive power output of the generator. Hence, for a generator operating in parallel with an infinite bus: Note that these effects are only applicable for generators in a large system only. Operation of Generators in Parallel with Other Generators of the Same Size When a single generator operated alone, the real and reactive powers supplied by the generators are fixed, constrained to be equal to the power demanded by the load, and the frequency and terminal voltage were varied by the governor set points and the field current.

When a generator is operating in parallel with an infinite bus, the frequency and terminal voltage were constrained to be constant by the infinite bys, and the real and reactive powers were varied by the governor set points and the field current. What happens when a synchronous generator is connected in parallel not with an infinite bus, but rather with another generator of the same size?

What will be the effect of changing governor set points and field currents? The system is as shown here: In this system, the basic constraint is that the sum of the real and reactive powers supplied by the two generators must equal the P and Q demanded by the load.

The system frequency is not constrained to be constant, and neither is the power of a given generator constrained to be constant. The power-frequency diagram for such a system immediately after G2 has been paralleled to the line is shown below: As a result, the power-frequency curve of G2 shifts upward as shown here: The total power supplied to the load must not change. At the original frequency f1, the power supplied by G1 and G2 will now be larger than the load demand, so the system cannot continue to operate at the same frequency as before.

In fact, there is only one frequency at which the sum of the powers out of the two generators is equal to Pload. That frequency f2 is higher than the original system operating frequency. At that frequency, G2 supplies more power than before, and G1 supplies less power than before.

Thus, when 2 generators are operating together, an increase in governor set points on one of them 1. What happens if the field current of G2 is increased? The resulting behaviour is analogous to the real-power situation as shown below: When 2 generators are operating together and the field current of G2 is increased, 1.

The system terminal voltage is increased. The reactive power Q supplied by that generator is increased, while the reactive power supplied by the other generator is decreased.

Example shows how this can be done. Example G1 has a no-load frequency of The 2 generators are supplying a real load totalling 2. The resulting system power-frequency or house diagram is shown below. What would the new system frequency be, and how much power would G1 and G2 supply now? When 2 generators of similar size are operating in parallel, a change in the governor set points of one of them changes both the system freq and the power sharing between them.

How can the power sharing of the power system be adjusted independently of the system frequency, and vice versa? Therefore, to adjust power sharing without changing the system frequency, increase the governor set points of one generator and simultaneously decrease the governor set points of the other generator. Shifting power sharing without affecting Shifting system frequency without affecting system frequency power sharing Reactive power and terminal voltage adjustment work in an analogous fashion.

To shift the reactive power sharing without changing VT, simultaneously increase the field current on one generator and decrease the field current on the other. Shifting reactive power sharing without Shifting terminal voltage without affecting affecting terminal voltage reactive power sharing It is very important that any synchronous generator intended to operate in parallel with other machines have a drooping frequency-power characteristic.

If two generators have flat or nearly flat characteristics, then the power sharing between them can vary widely with only the tiniest changes in no-load speed. This problem is illustrated below: Rated frequency will depend upon the system at which the generator is connected. Voltage Ratings: Generated voltage is dependent upon flux, speed of rotation and mechanical constants.

However, there is a ceiling limit of flux level since it is dependent upon the generator material. Hence voltage ratings may give a rough idea on its maximum flux level possible and also maximum voltage to before the winding insulation breaks down. Apparent Power and Power Factor Ratings Constraints for electrical machines generally dependent upon mechanical strength mechanical torque on the shaft of the machine and also its winding insulation limits heating of its windings.

For a generator, there are 2 different windings that has to be protected which are: The heating effect of the stator copper losses is given by: And since we can find the maximum field current and the maximum EA possible, we may be able to determine the lowest PF changes possible for the generator to operate at rated apparent power.

Figure below shows the phasor diagram of a synchronous generator with the rated voltage and armature current. The current can assume many different angles as shown. Notice that for some possible current angles the required EA exceeds EA,max. If the generator were operated at the rated armature current and these power factors, the field winding would burn up. It is possible to operate the generator at a lower more lagging power factor than the rated value, but only by cutting back on the kVA supplied by the generator.

Synchronous Generator Capability Curves. Based upon these limits, there is a need to plot the capability of the synchronous generator. This is so that it can be shown graphically the limits of the generator. The capability curve can be derived back from the voltage phasor of the synchronous generator.

Assume that a voltage phasor as shown, operating at lagging power factor and its rated value: Note that the capability curve of the must represent power limits of the generator, hence there is a need to convert the voltage phasor into power phasor.

The powers are given by: The length corresponding to EA on the power diagram is: The final capability curve is shown below: Any point that lies within both circles is a safe operating point for the generator. It has a synchronous reactance of 1 ohm per phase. Assume that this generator is connected to a steam turbine capable of supplying up to 45kW.

The friction and windage losses are 1. Why or why not? Synchronous Motor In general, a synchronous motor is very similar to a synchronous generator with a difference of function only. Steady State Operations A synchronous motor are usually applied to instances where the load would require a constant speed. Hence for a synchronous motor, its torque speed characteristic is constant speed as the induced torque increases. Since, v ind? If load exceeds the pullout torque, the rotor will slow down.

Due to the interaction between the stator and rotor magnetic field, there would be a torque surge produced as such there would be a loss of synchronism which is known as slipping poles.

Also based upon the above equation, maximum induced torque can be achieved by increasing Ea hence increasing the field current. Effect of load changes Assumption: A synchronous generator operating with a load connected to it.

The field current setting are unchanged.

Varying load would in fact slow the machine down a bit hence increasing the torque angle. Due to an increase to the torque angle, more torque is induced hence spinning the synchronous machine to synchronous speed again. The overall effect is that the synchronous motor phasor diagram would have a bigger torque angle f. In terms of the term Ea, since If is set not to change, hence the magnitude of Ea should not change as shown in the phasor diagram fig.

Since the angle of d changes, the armature current magnitude and angle would also change to compensate to the increase of power as shown in the phasor diagram fig.

Effect of field current changes on a synchronous motor Assumption: The synchronous generator is rotating at synchronous speed with a load connected to it. The load remains unchanged. As the field current is increased, Ea should increase. Unfortunately, there are constraints set to the machine as such that the power requirement is unchanged.

Therefore since P is has to remain constant, it imposes a limit at which Ia and jXsIa as such that Ea tends to slide across a horizontal limit as shown in figure Ia will react to the changes in Ea as such that its angle changes from a leading power factor to a lagging power factor or vice versa. This gives a possibility to utilise the synchronous motor as a power factor correction tool since varying magnetic field would change the motor from leading to lagging or vice versa.

This characteristic can also be represented in the V curves as shown in figure Synchronous motor as a power factor correction Varying the field current would change to amount of reactive power injected or absorbed by the motor. Hence if a synchronous motor is incorporated nearby a load which require reactive power, the synchronous motor may be operated to inject reactive power hence maintaining stability and lowering high current flow in the transmission line.

Starting Synchronous Motors Problem with starting a synchronous motor is the initial production of torque which would vary as the stator magnetic field sweeps the rotor. As a result, the motor will vibrate and could overheat refer to figure for diagram explanations. There are 3 different starting methods available: Stator magnetic field speed reduction The idea is to let the stator magnetic field to rotate slow enough as such that the rotor has time to lock on to the stator magnetic field.

This method used to be impractical due to problems in reducing stator magnetic field. Now, due to power electronics technology, frequency reduction is possible hence makes it a more viable solution. Using a prime mover This is a very straightforward method. Motor Starting using Amortisseur windings This is the most popular way to start an induction motor. Amortisseur windings are a special kind of windings which is shorted at each ends. Its concept is near similar to an induction motor hence in depth explanation can be obtained in the text book page The final effect of this starting method is that the rotor will spin at near synchronous speed.

Note that the rotor will never reach synchronous speed unless during that time, the field windings are switched on hence will enable the rotor to lock on to the stator magnetic field. Induction Motor Construction 2. The Equivalent Circuit of an Induction Motor. Powers and Torque in Induction Motor. Starting Induction Motors 8. No dc field current is required to run the machine. Induction Motor Construction There are basically 2 types of rotor construction: Wound rotor are known to be more expensive due to its maintenance cost to upkeep the slip rings, carbon brushes and also rotor windings.

Cutaway diagram of a wound rotor induction motor. Basic Induction Motor Concepts The Development of Induced Torque in an Induction Motor When current flows in the stator, it will produce a magnetic field in stator as such that Bs stator magnetic field will rotate at a speed: This rotating magnetic field Bs passes over the rotor bars and induces a voltage in them.

The voltage induced in the rotor is given by: And this rotor current will produce a magnetic field at the rotor, Br. Hence the interaction between both magnetic field would give torque: An induction motor can thus speed up to near synchronous speed but it can never reach synchronous speed.

This can be easily termed as slip speed: Apart from that we can describe this relative motion by using the concept of slip: But unlike a transformer, the secondary frequency may not be the same as in the primary. If the rotor is locked cannot move , the rotor would have the same frequency as the stator refer to transformer concept. Another way to look at it is to see that when the rotor is locked, rotor speed drops to zero, hence by default, slip is 1. But as the rotor starts to rotate, the rotor frequency would reduce, and when the rotor turns at synchronous speed, the frequency on the rotor will be zero.

Example 7. The Equivalent Circuit of an Induction Motor An induction motor relies for its operation on the induction of voltages and currents in its rotor circuit from the stator circuit transformer action. This induction is essentially a transformer operation, hence the equivalent circuit of an induction motor is similar to the equivalent circuit of a transformer.

The Transformer Model of an Induction Motor A transformer per-phase equivalent circuit, representing the operation of an induction motor is shown below: The transformer model or an induction motor, with rotor and stator connected by an ideal transformer of turns ratio aeff. As in any transformer, there is certain resistance and self-inductance in the primary stator windings, which must be represented in the equivalent circuit of the machine.

They are - R1 - stator resistance and X1 — stator leakage reactance Also, like any transformer with an iron core, the flux in the machine is related to the integral of the applied voltage E1. The curve of mmf vs flux magnetization curve for this machine is compared to a similar curve for a transformer, as shown below: This is because there must be an air gap in an induction motor, which greatly increases the reluctance of the flux path and thus reduces the coupling between primary and secondary windings.

The higher reluctance caused by the air gap means that a higher magnetizing current is required to obtain a given flux level. Therefore, the magnetizing reactance Xm in the equivalent circuit will have a much smaller value than it would in a transformer. The primary internal stator voltage is E1 is coupled to the secondary ER by an ideal transformer with an effective turns ratio aeff. The turns ratio for a wound rotor is basically the ratio of the conductors per phase on the stator to the conductors per phase on the rotor.

It is rather difficult to see aeff clearly in the cage rotor because there are no distinct windings on the cage rotor. ER in the rotor produces current flow in the shorted rotor or secondary circuit of the machine. The primary impedances and the magnetization current of the induction motor are very similar to the corresponding components in a transformer equivalent circuit.

The Rotor Circuit Model When the voltage is applied to the stator windings, a voltage is induced in the rotor windings. In general, the greater the relative motion between the rotor and the stator magnetic fields, the greater the resulting rotor voltage and rotor frequency. The largest relative motion occurs when the rotor is stationary, called the locked-rotor or blocked-rotor condition, so the largest voltage and rotor frequency are induced in the rotor at that condition.

The smallest voltage and frequency occur when the rotor moves at the same speed as the stator magnetic field, resulting in no relative motion. The magnitude and frequency of the voltage induced in the rotor at any speed between these extremes is directly proportional to the slip of the rotor.

Therefore, if the magnitude of the induced rotor voltage at locked-rotor conditions is called ER0, the magnitude of the induced voltage at any slip will be given by: The rotor resistance RR is a constant, independent of slip, while the rotor reactance is affected in a more complicated way by slip.

The reactance of an induction motor rotor depends on the inductance of the rotor and the frequency of the voltage and current in the rotor. With a rotor inductance of LR, the rotor reactance is: The resulting rotor equivalent circuit is as shown: The rotor circuit model of an induction motor. The rotor current flow is: The rotor circuit model with all the frequency slip effects concentrated in resistor RR.

Based upon the equation above, at low slips, it can be seen that the rotor resistance is much much bigger in magnitude as compared to XR0. At high slips, XR0 will be larger as compared to the rotor resistance. The Final Equivalent Circuit To produce the final per-phase equivalent circuit for an induction motor, it is necessary to refer the rotor part of the model over to the stator side.

In an ordinary transformer, the voltages, currents and impedances on the secondary side can be referred to the primary by means of the turns ratio of the transformer. Power and Torque in Induction Motor Losses and Power-Flow diagram An induction motor can be basically described as a rotating transformer. Its input is a 3 phase system of voltages and currents.

The secondary windings in an induction motor the rotor are shorted out, so no electrical output exists from normal induction motors.

Instead, the output is mechanical. The relationship between the input electric power and the output mechanical power of this motor is shown below: The input power to an induction motor Pin is in the form of 3-phase electric voltages and currents. Then, some amount of power is lost as hysteresis and eddy currents in the stator Pcore. The power remaining at this point is transferred to the rotor of the machine across the air gap between the stator and rotor.

This power is called the air gap power PAG of the machine. After the power is transferred to the rotor, some of it is lost as I2R losses the rotor copper loss PRCL , and the rest is converted from electrical to mechanical form Pconv. The remaining power is the output of the motor Pout. The core losses do not always appear in the power-flow diagram at the point shown in the figure above.

Because of the nature of the core losses, where they are accounted for in the machine is somewhat arbitrary. The core losses of an induction motor come partially from the stator circuit and partially from the rotor circuit.

Since an induction motor normally operates at a speed near synchronous speed, the relative motion of the magnetic fields over the rotor surface is quite slow, and the rotor core losses are very tiny compared to the stator core losses. Since the largest fraction of the core losses comes from the stator circuit, all the core losses are lumped together at that point on the diagram. These losses are represented in the induction motor equivalent circuit by the resistor RC or the conductance GC.

If core losses are just given by a number X watts instead of as a circuit element, they are often lumped together with the mechanical losses and subtracted at the point on the diagram where the mechanical losses are located. The higher the speed of an induction motor, the higher the friction, windage, and stray losses. On the other hand, the higher the speed of the motor up to nsync , the lower its core losses.

Therefore, these three categories of losses are sometimes lumped together and called rotational losses. The total rotational losses of a motor are often considered to be constant with changing speed, since the component losses change in opposite directions with a change in speed. The stator copper losses are 2kW, and the rotor copper losses are W.

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The friction and windage losses are W, the core losses are W, and the stray losses are negligible. The input current to a phase of the motor is: The stator copper losses in the 3 phases are: Thus, the air-gap power: The power converted, which is called developed mechanical power is given as: This torque differs from the torque actually available at the terminals of the motor by an amount equal to the friction and windage torques in the machine.

Hence, the developed torque is: Hence it may be useful to separate the rotor copper loss element since rotor resistance are both used for calculating rotor copper loss and also the output power.

The core loss is lumped in with the rotational losses. For a rotor slip of 2. Induced Torque from a Physical Standpoint The magnetic fields in an induction The magnetic fields in an induction motor under light loads motor under heavy loads No-load Condition Assume that the induction rotor is already rotating at no load conditions, hence its rotating speed is near to synchronous speed. The net magnetic field Bnet is produced by the magnetization current IM.

The magnitude of IM and Bnet is directly proportional to voltage E1. If E1 is constant, then Bnet is constant. In an actual machine, E1 varies as the load changes due to the stator impedances R1 and X1 which cause varying volt drops with varying loads.

However, the volt drop at R1 and X1 is so small, that E1 is assumed to remain constant throughout. At no-load, the rotor slip is very small, and so the relative motion between rotor and magnetic field is very small, and the rotor frequency is also very small.

Since the relative motion is small, the voltage ER induced in the bars of the rotor is very small, and the resulting current flow IR is also very small. Since the rotor frequency is small, the reactance of the rotor is nearly zero, and the max rotor current IR is almost in phase with the rotor voltage ER.

The rotor current produces a small magnetic field BR at an angle slightly greater than 90 degrees behind Bnet. The stator current must be quite large even at no-load since it must supply most of Bnet. The induced torque which is keeping the rotor running, is given by: Since the rotor speed is slower, there is now more relative motion between rotor and stator magnetic fields.

Greater relative motion means a stronger rotor voltage ER which in turn produces a larger rotor current IR. With large rotor current, the rotor magnetic field BR also increases. However, the angle between rotor current and BR changes as well. Therefore, the rotor current now lags further behind the rotor voltage, and the rotor magnetic field shifts with the current.

This effect is known as pullout torque. Modelling the torque-speed characteristics of an induction motor Looking at the induction motor characteristics, a summary on the behaviour of torque: Current flow will increase as slip increase reduction in velocity b The net magnetic field density will remain constant since it is proportional to E1 refer to equivalent induction motor equivalent circuit.

Since E1 is assumed to be constant, hence Bnet will assume to be constant. Adding the characteristics of all there elements would give the torque speed characteristics of an induction motor. Graphical development of an induction motor torque-speed characteristics The Derivation of the Induction Motor Induced-Torque Equation Previously we looked into the creation of the induced torque graph, now we would like to derive the Torque speed equation based upon the power flow diagram of an induction motor.

Hence there is a need to derive PAG. By definition, air gap power is the power transferred from the stator to the rotor via the air gap in the induction machine. Based upon the induction motor equivalent circuit, the air gap power may be defined as: The easiest way is via the construction of the Thevenin equivalent circuit. Since Pconv may be derived as follows: Based upon the maximum power transfer theorem, maximum power transfer will be achieved when the magnitude of source impedance matches the load impedance.

Since the source impedance is as follows: By adding more resistance to the machine impedances, we can vary: At what speed and slip does it occur? What is the new starting torque?

S decrease, PAG increase, and efficiency increase. Use a wound rotor induction motor and insert extra resistance into the rotor during starting, and then removed for better efficiency during normal operations. But, wound rotor induction motors are more expensive, need more maintanence etc. Solution - utilising leakage reactance — to obtain the desired curve as shown below A torque-speed characteristic curve combining high- resistance effects at low speeds high slip with low resistance effects at high speed low slip.

Thus, if the bars of a cage rotor are placed near the surface of the rotor, they will have small leakage flux and X2 will be small. Deep-Bar and Double-Cage rotor design How can a variable rotor resistance be produced to combine the high starting torque and low starting current of Class D, with the low normal operating slip and high efficiency of class A?? Deeper in the bar, the leakage inductance is higher.

The impedances of all parts of the bar are approx equal, so current flows through all the parts of the bar equally. The resulting large cross sectional area makes the rotor resistance quite small, resulting in good efficiency at low slips. Since the effective cross section is lower, the rotor resistance is higher.

Thus, the starting torque is relatively higher and the starting current is relatively lower than in a class A design. It is similar to the deep-bar rotor, except that the difference between low-slip and high-slip operation is even more exaggerated. Hence, high starting torque. However, at normal operating speeds, both bars are effective, and the resistance is almost as low as in a deep-bar rotor.

They are low starting torque machines. Starting Induction Motors An induction motor has the ability to start directly, however direct starting of an induction motor is not advised due to high starting currents, which will be explained later.

In order to know the starting current, we should be able to calculate the starting power required by the induction motor. The Code Letter designated to each induction motor, which can be seen in figure , may represent this. The starting code may be obtained from the motor nameplate. For a wound rotor type induction motor, this problem may be solved by incorporating resistor banks at the rotor terminal during starting to reduce current flow and as the rotor picks up speed, the resistor banks are taken out.

Reducing the starting terminal voltage will also reduce the rated starting power hence reducing starting current. One way to achieve this is by using a step down transformer during the starting sequence and stepping up the transformer ratio as the machine spins faster refer figure below.

A magnetic motor starter of this sort has several built in protective features: There is a finite delay before the 1TD contacts close. During that time, the motor speeds up, and the starting current drops. And finally 3TD contacts close, and the entire starting resistor is out of the circuit. Speed Control of Induction Motor Induction motors are not good machines for applications requiring considerable speed control.

Varying slip may be achieved by varying rotor resistance or varying the terminal voltage. Consider one phase winding in a stator. By changing the current flow in one portion of the stator windings as such that it is similar to the current flow in the opposite portion of the stator will automatically generate an extra pair of poles. In terms of torque, the maximum torque magnitude would generally be maintained. This method will enable speed changes in terms of 2: Multiple stator windings have extra sets of windings that may be switched in or out to obtain the required number of poles.

Unfortunately this would an expensive alternative. Read More Posted on March 8, Read More Posted on May 16, Workshop Instructors: Would you like to know more about how to use the formatting features in Microsoft Word? Research Commons staff will help you with your questions about the nuts and bolts of formatting: This includes a special section about copyright and your thesis. Graduate students at any stage of the writing process are welcome, but some prior knowledge of Microsoft Word is recommended.

To keep up-to-date with all of the workshops, consults, and events subscribe to the UBC Library Research Commons monthly newsletter. Workshop Facilitators: Amir and David Workshop Description: This workshop provides hands-on experience with NVivo for literature reviews, focusing on coding, advanced queries, and visualization methods.

It is recommended to take this workshop after completing NVivo Parts 1 and 2. If possible, bring articles that you plan to use for your own research.

This workshop is offered as part of our Literature Reviews workshop series. Searching and Keeping Track -Literature Reviews: Analyzing with NVivo -Literature Reviews:Now, connect a load to the terminals of the machine, and a current will flow in its armature windings. Research Commons staff will help you with your questions about the nuts and bolts of formatting: You've reached the end of this preview.

In other words, the concept of magnetic permeability corresponds to the ability of the material to permit the flow of magnetic flux through it. These are determined as follows: The transformer has Np turns of wire on its primary side and Ns turns of wire on its secondary sides.

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