Lecture 06 - Ac Machinery Fundamentals

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Lecture 6: AC machinery fundamentals

Instructor:Dr. Gleb V. TcheslavskiContact:[email protected] Hours:TBD; Room 2030Class web site:http://ee.lamar.edu/gleb/Index.htm

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Preliminary notesAC machines are AC motors and AC generators.There are two types of AC machines:Synchronous machines the magnetic field current is supplied by aseparate DC power source;Induction machines the magnetic field current is supplied by magneticinduction (transformer action) into their field windings.The field circuits of most AC machines are located on their rotors.Every AC (or DC) motor or generator has two parts: rotating part (rotor)and a stationary part (stator).

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The rotating magnetic fieldThe basic idea of an electric motor is to generate two magnetic fields:rotor magnetic field and stator magnetic field and make the stator fieldrotating. In this situation, the rotor will constantly turning to align itsmagnetic field with the stator field.The fundamental principle of AC machine operation is to make a 3-phase set of currents, each of equal magnitude and with a phasedifference of 120o, to flow in a 3-phase winding. In this situation, aconstant magnitude rotating field will be generated.The 3-phase winding consists of 3 separate windings spaced 120oapart around the surface of the machine.

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The rotating magnetic fieldConsider a simple 3-phase stator containingthree coils, each 1200 apart. Such a winding willproduce only one north and one south magneticpole; therefore, this motor would be called a two-pole motor.Assume that the currents in three coils are:

x

x x

. .

.

(6.4.1)

The directions of currents are indicated.Therefore, the current through the coil aa produces the magnetic field intensity

Into the page

(6.4.2)

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The rotating magnetic fieldwhere the magnitude of the magnetic field intensity ischanging over time, while 00 is the spatial angle of themagnetic field intensity vector. The direction of the fieldcan be determined by the right-hand rule.Note, that while the magnitude of the magnetic fieldintensity Haa varies sinusoidally over time, its directionis always constant. Similarly, the magnetic fieldsthrough two other coils are

(6.5.1)

(6.5.2)

The magnetic flux densities resulting from these magnetic field intensities can befound from

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The rotating magnetic field

(6.6.1)

At the time t = 0 (t = 0) :

The total magnetic field from all three coils added together will be

(6.6.2)(6.6.3)(6.6.4)

(6.6.5)

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The rotating magnetic fieldAt the time when t = 900:

The total magnetic field from all three coils added together will be

(6.7.1)(6.7.2)(6.7.3)

(6.7.4)

We note that the magnitude ofthe magnetic field is constantbut its direction changes.Therefore, the constantmagnitude magnetic field isrotating in a counterclockwisedirection.

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The rotating magnetic field: proof

x

x x

. .

.The magnetic flux density in the stator at anyarbitrary moment is given by

Each vector can be represented as a sum of xand y components:

(6.8.1)

(6.8.2)

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The rotating magnetic field: proofWhich can be rewritten in form

(6.9.1)

Finally: (6.9.2)

The net magnetic field has a constant magnitude and rotates counterclockwiseat the angular velocity .

10Relationship between electricalfrequency and speed of field rotation

The stator rotating magnetic field can berepresented as a north pole and a south pole.These magnetic poles complete one mechanicalrotation around the stator surface for eachelectrical cycle of current. Therefore, themechanical speed of rotation of the magnetic fieldequals to the electrical frequency.

(6.10.1)

The magnetic field passes the windings of a two-pole stator in the followingcounterclockwise sequence: a-c-b-a-c-b. What if 3 additional windings will beadded? The new sequence will be: a-c-b-a-c-b-a-c-b-a-c-b and, when 3-phasecurrent is applied to the stator, two north poles and two south poles will beproduced. In this winding, a pole moves only halfway around the stator.

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The relationshipbetween the electricalangle e (currentsphase change) andthe mechanical anglem (at which themagnetic field rotates)in this situation is:

(6.11.1)

Therefore, for a four-pole stator:

(6.11.2)


(6.12.1)

For an AC machine with P poles in its stator:

Relating the electrical frequency to the motors speed in rpm:

(6.12.2)

(6.12.3)

(6.12.4)

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Reversing the direction of field rotationIf the current in any two of the three coils is swapped, the direction ofmagnetic field rotation will be reversed. Therefore, to change the directionof rotation of an AC motor, we need to switch the connections of any twoof the three coils.In this situation, the net magnetic flux density in the stator is

(6.13.1)

(6.13.2)

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Reversing the direction of field rotation

Therefore:

Finally:

The net magnetic field has a constant magnitude and rotates clockwise at theangular velocity . Switching the currents in two stator phases reverses thedirection of rotation in an AC machine.

(6.14.1)

(6.14.2)

(6.14.3)

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15Magnetomotive force and fluxdistribution on an AC machine

In the previous discussion, we assumed that the flux produced by a stator insidean AC machine behaves the same way it does in a vacuum. However, in realmachines, there is a ferromagnetic rotor in the center with a small gap between arotor and a stator.

A rotor can be cylindrical(such machines are said tohave non-salient poles), orit may have pole facesprojecting out from it(salient poles). We willrestrict our discussion tonon-salient pole machinesonly (cylindrical rotors).


The reluctance of the air gap is much higher thanthe reluctance of either the rotor or the stator;therefore, the flux density vector B takes theshortest path across the air gap: it will beperpendicular to both surfaces of rotor and stator.To produce a sinusoidal voltage in this machine,the magnitude of the flux density vector B mustvary sinusoidally along the surface of the air gap.Therefore, the magnetic field intensity (and themmf) will vary sinusoidally along the air gapsurface.

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One obvious way to achieve a sinusoidal variationof mmf along the air gap surface would be todistribute the turns of the winding that producesthe mmf in closely spaced slots along the air gapsurface and vary the number of conductors ineach slot sinusoidally, according to:

(6.17.1)where Nc is the number of conductors at the angleof 00 and is the angle along the surface.

Ideal mmf

mmf resultingfrom the winding

n0

However, in practice, only a finite number of slotsand integer numbers of conductors are possible.As a result, real mmf will approximate the idealmmf if this approach is taken.

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Induced voltage in AC machines

Just as a 3-phase set of currents in a stator can produce arotating magnetic field, a rotating magnetic field canproduce a 3-phase set of voltages in the coils of a stator.

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19The induced voltage in a single coilon a two-pole stator

Assume that a rotor with a sinusoidally distributedmagnetic field rotates in the center of a stationary coil.We further assume that the magnitude of the flux density Bin the air gap between the rotor and the stator variessinusoidally with mechanical angle, while its direction isalways radially outward.Note, that this is an ideal fluxdistribution.

Flux density in a gap

The magnitude of the fluxdensity vector at a pointaround the rotor is

(6.19.1)

Where is the angle from thedirection of peak flux intensity.

statorcoil


Since the rotor is rotating within the stator at an angular velocity m, themagnitude of the flux density vector at any angle around the stator is(6.20.1)

The voltage induced in a wire is(6.20.2)

Here v is the velocity of the wire relative to the magnetic fieldB is the magnetic flux density vectorl is the length of conductor in the magnetic field

However, this equation was derived for a moving wire in a stationary magneticfield. In our situation, the wire is stationary and the magnetic field rotates.Therefore, the equation needs to be modified: we need to change reference suchway that the field appears as stationary.

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The total voltage induced in the coil is a sum of the voltages induced in each of itsfour sides. These voltages are:1. Segment ab: = 1800; assuming that B is radially outward from the rotor, theangle between v and B is 900, so

2. Segment bc: the voltage will be zero since the vectors (v x B) and l areperpendicular.

3. Segment cd: = 00; assuming that B is radially outward from the rotor, theangle between v and B is 900, so

4. Segment da: the voltage will be zero since the vectors (v x B) and l areperpendicular.

(6.21.1)

(6.21.2)

(6.21.3)

(6.21.4)


Therefore, the total voltage on the coil is:

Since the velocity of the end conductor isThen:The flux passing through a coil is

Therefore:Finally, if the stator coil has NC turns of wire, the total induced voltage in the coil:

(6.22.1)

(6.22.2)

(6.22.3)

(6.22.4)

(6.22.5)

(6.22.6)

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23The induced voltage in a 3-phase setof coils

In three coils, each of NC turns, placed around the rotor magnetic field, theinduced in each coil will have the same magnitude and phases differing by 1200:

(6.23.1)

A 3-phase set of currents can generatea uniform rotating magnetic field in amachine stator, and a uniform rotatingmagnetic field can generate a 3-phaseset of voltages in such stator.

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The rms voltage in a 3-phase statorThe peak voltage in any phase of a 3-phase stator is:

For a 2-pole stator:

Thus:The rms voltage in any phase of a 2-pole 3-phase stator is:

The rms voltage at the terminals will depend on the type of stator connection: ifthe stator is Y-connected, the terminal voltage will be . For the deltaconnection, it will be just EA.

(6.24.1)

(6.24.2)

(6.24.3)

(6.24.4)

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Induced voltage: ExampleExample 6.1: The peak flux density of the rotor magnetic field in a simple 2-pole3-phase generator is 0.2 T; the mechanical speed of rotation is 3600 rpm; thestator diameter is 0.5 m; the length of its coil is 0.3 m and each coil consists of 15turns of wire. The machine is Y-connected.a) What are the 3-phase voltages of the generator as a function of time?b) What is the rms phase voltage of the generator?c) What is the rms terminal voltage of the generator?The flux in this machine is given by

The rotor speed is

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Induced voltage: Examplea) The magnitude of the peak phase voltage is

and the three phase voltages are:

b) The rms voltage of the generator is

c) For a Y-connected generator, its terminal voltage is

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Induced torque in an AC machineIn an AC machine under normal operating conditions two magnetic fields arepresent: a field from the rotor and a field from the stator circuits. The interaction ofthese magnetic fields produces the torque in the machine.Assuming a sinusoidal stator flux distributionpeaking in the upward direction

(where BS is the magnitude of the peak fluxdensity) and a single coil of wire mounted onthe rotor, the induced force on the firstconductor (on the right) is

The torque on this conductor is (counter-clockwise)

(6.27.1)

(6.27.2)

(6.27.3)

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Induced torque in an AC machine

The torque on this conductor is (counter-clockwise)(6.28.1)

(6.28.2)

The induced force on the second conductor (on the left) is

Therefore, the torque on the rotor loop is(6.28.3)

We may notice the following:1. The current i flowing in the rotor coil produces its own magnetic field HR, whosemagnitude is proportional to the current and direction can be found via the RHR.2. The angle between the peak of the stator flux density BS and the peak of themagnetic field intensity HR is .

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Induced torque in an AC machineFurthermore,

(6.29.1)(6.29.2)

Therefore, the torque on the loop is(6.29.3)

Here K is a constant dependent on themachine design. Therefore:

(6.29.4)

Since (6.29.5)

(6.29.6)

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Induced torque in an AC machineAs before, in (6.29.5) k = K/ is a constant dependent on the machine design.The equation (6.29.5) can be applied to any AC machine, not just to simple one-loop rotors. Since this equation is used for qualitative studies of torque, theconstant k is not important.Assuming no saturation, the net magnetic field is a vector sum of rotor and statorfields: (6.30.1)Combining the last equation with (6.29.5), we arrive at

Since the cross-product of any vector with itself is zero:(6.30.2)

(6.30.3)

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Induced torque in an AC machine

(6.31.1)

Assuming that the angle between the rotor BR and stator BS magnetic fields is :

Assume that the rotor of the AC machine isrotating counter-clockwise and theconfiguration of magnetic fields is shown.The combination of (6.30.3) and the RHRshows that the torque will be clockwise, i.e.opposite to the direction of rotation of therotor. Therefore, this machine must be actingas a generator.

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Winding insulation in AC machinesWinding insulation is of critical importance. If insulation of amotor or generator breaks down, the machine shorts out andthe repair is expensive and sometimes even impossible.Most insulation failures are due to overheating.To limit windings temperature, the maximum power that canbe supplied by the machine must be limited in addition to theproper ventilation.ROT: the life expectancy of a motor with a given type ofinsulation is halved for each 100C rise above the ratedwinding temperature.

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33AC machine power flows andlosses

The efficiency of an AC machine is defined as

Since the difference between the input and output powers of a machine is dueto the losses occurring inside it, the efficiency is

(6.33.1)

(6.33.2)

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AC machine power lossesLosses occurring in an AC machine can be divided into four categories:

1. Electrical or Copper lossesThese losses are resistive heating losses that occur in the stator (armature)winding and in the rotor (field) winding of the machine. For a 3-phase machine, thestator copper losses and synchronous rotor copper losses are:

(6.34.1)

(6.34.2)

Where IA and IF are currents flowing in each armature phase and in the fieldwinding respectively. RA and RF are resistances of each armature phase and ofthe field winding respectively. These resistances are usually measured at normaloperating temperature.

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AC machine power losses2. Core losses

These losses are the hysteresis losses and eddy current losses. They vary as B2(flux density) and as n1.5 (speed of rotation of the magnetic field).

3. Mechanical lossesThere are two types of mechanical losses: friction (friction of the bearings) andwindage (friction between the moving parts of the machine and the air inside thecasing). These losses are often lumped together and called the no-load rotationalloss of the machine. They vary as the cube of rotation speed n3.

4. Stray (miscellaneous) lossesThese are the losses that cannot be classified in any of the previous categories.They are usually due to inaccuracies in modeling. For many machines, straylosses are assumed as 1% of full load.

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The power-flow diagramOn of the most convenient technique to account for powerlosses in a machine is the power-flow diagram.AC generator:

AC motor:

The mechanical power is input,and then all losses but cupperare subtracted. The remainingpower Pconv is ideally convertedto electricity:

(6.36.1)

Power-flow diagram is simplyreversed.

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Voltage regulationVoltage regulation (VR) is a commonly used figure of merit for generators:

(6.37.1)

Here Vnl and Vfl are the no-load full-load terminal voltages of thegenerator. VR is a rough measure of the generators voltage-currentcharacteristic. A small VR (desirable) implies that the generators outputvoltage is more constant for various loads.

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Speed regulationSpeed regulation (SR) is a commonly used figure of merit for motors:

Here nnl and nfl are the no-load full-load speeds of the motor. SR is arough measure of the motors torque-speed characteristic. A positive SRimplies that a motors speed drops with increasing load. The magnitude ofSR reflects a steepness of the motors speed-torque curve.

(6.38.1)

(6.38.2)

Lecture 06 - Ac Machinery Fundamentals

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