AC Electric Machines - University of...

AC Electric Machines

Objectives1. To understand what the meant by the term “ac circuit.”

2. To understand how to analyze ac circuits.

3. To understand the basic construction and operation of an ac machine.

4. To understand how to analyze an ac machine.

Introduction• We have studied the dc machine. One problem with using dc voltages and currents is that electricity is normally distributed using time-variant sinusoidal values. This means that when you plug an electrical device into a wall socket you are connecting your device to a sinusoidal source.

• As a result we need to consider the ac (alternating current) machine. AC machines, also known as

ECE 3010 ELEMENTS OF ELEC. MACHINES AND DIGITAL SYSTEMS CHAPTER 5

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induction machines, are the most widely used type of electric machine. The basic operation of the ac machine is similar in concept to the dc machine where the interaction of magnetic fields will be tied to the rotation of the machines rotor.

• To study the ac machine we will need to extend our basic understanding to include the behaviour of circuits under sinusoidal conditions rather than constant conditions.

A basic ac electric machine

• The basic ac machine involves multiple magnetic fields interacting with one another. The magnetic fields are created using electromagnets connected to sinusoidal voltages and currents. Like the dc machine, it is the interaction of these magnetic fields which form a connection between the electrical and mechanical portions of these machines. However, the use of



sinusoidal voltages and currents will affect how we model and analyze these machines.

AC Electric Circuits Review• Up to this point, the circuit analysis techniques presented have been applied to circuits which only contained time invariant sources. The addition of time varying sources does not change any of the fundamentals which we use. However, the equivalent circuits for our machines contain reactive component, capacitors and or inductors, and since any equations describing voltages or currents which use derivatives will not simplify to 0, we therefore cannot treat them as simple open or short circuits.

• If we consider that the time varying quantities we are interested in are limited to sinusoidal values, we can introduce techniques which allow us to remove the need to perform any calculus.



• A sinusoidal voltage or current is governed by a function of the form Asin(ωt + ø). Where: A is the amplitude of the waveform; ω is the radian frequency of the voltage or current in units of radians/sec, and ø is the phase angle by which the waveform may be shifted in time.

A

2/

Acos(t+ø°)

t

ø

If we consider a simple circuit consisting of a sinusoidal current source and an inductor we can see the impact of dealing with sinusoidal quantities

is(t)=cos(2t)A L

iL(t)+

vL(t)

–



• Examining the expression for voltage across the inductor, and the inductor current, the only differences between the parameters are that the magnitudes, and phase angles are different. In effect, when determining voltage or current we do not need to keep track of the function or frequency, we only need to find the new magnitude and phase angle. We can keep track of the magnitude and phase angle information by using a two dimensional vector.

• Rather than graphically using vectors, we can map the magnitude and phase on to a complex number which can be expressed as



Cˆ

!C ˆ

|C | ˆ

ˆ C = ˆ C ∠θC

whereˆ C = magnitude of vector ˆ C

∠θC = angle of vector ˆ C

• We can use this representation to simplify dealing with the complex values, however, we need to think about how to represent a circuit such that we can take advantage of the complex number/vector math.

• If we consider the previous circuit with the current source and the inductor, and represent the source via its magnitude and phase representation, and think of the inductor as if it were a complex valued resistor we have the following:

is(t)=cos(2t)A L

iL(t)+

vL(t)

–

I=1!0°A

Rcomplex

ˆ

+

VL=2L!90° V

–

ˆ

IL=1!0°Aˆ



Consider what Rcomplex needs to be such that if we were to apply Ohm’s law we get the correct value for the voltage, it turns out that Rcomplex = jωL.

ˆ V = ˆ I Rcomplex ⇒ Rcomplex =ˆ V ˆ I

= 2L∠90°1∠0°

= 2L∠90° = j2L = jωL Ω

• This means that if we replace inductors with complex resistors of value jωL we can apply dc circuit analysis techniques to determine the magnitude and phase value of any sinusoidal current or voltage for an inductor.

• We can extend this transformational analysis which we have performed on an inductor to resistors and capacitors, and formalize this technique into what is known as the phasor method of circuit analysis.

The phasor method• The phasor method of circuit analysis allows us to analyze the response of a circuit to sinusoidal sources by applying a transform to the circuit. This removes the need to perform any calculus when dealing with reactive



components, and allows for the use of simpler dc circuit analysis techniques.

• The phasor method replaces functions of time with functions of frequency. Replacing the sinusoidal sources with complex valued sources, and passive devices with complex valued representations, results in currents and voltages in the form of complex numbers. These complex values need to be transformed back into the time domain once the analysis is complete.

• Ohm’s law still applies if we have a complex voltage and current passing through a device. In order to make it clear that we are dealing with complex valued components, rather than refer to the ratio of the voltage/current as resistance we will refer to it as an impedance denoted by the letter ‘Z.’

• The term impedance still indicates the resistance of a device to current flow for a specific voltage, however, it



can also be thought of as a “complex resistance” or “frequency-dependent resistance.”

• Impedance has two parts. The real portion of impedance is resistance ‘R’, and the imaginary portion is the reactance ‘X.’

Z = R + jX Ω

• We can determine the impedance of the resistor and the capacitor.

is(t)=cos(2t)A R

iR(t)+

vR(t)

–

I=1!0°Aˆ

+

VR

–

ˆ

IR=1!0°Aˆ

ZR

iR (t) = is (t) = cos(2t) = 1cos(2t + 0°)AvR(t) = Ri(t) = Rcos(2t + 0°)V

ˆ V = ˆ I ZR ⇒ ZR =ˆ V ˆ I

= R∠0°1∠0°

= R∠0° = R Ω



is(t)=cos(2t)A C

iC(t)+

vC(t)

–

I=1!0°Aˆ

+

VC

–

ˆ

IC=1!0°Aˆ

ZC

iC (t) = is (t) = cos(2t) = 1cos(2t + 0°)A

vC (t) = 1C

i(t)∫ dt = 1C

cos(2t + 0°)∫ dt = 12C

sin(2t + 0) = 12C

cos(2t − 90°)V

ˆ V = ˆ I ZC ⇒ ZC =ˆ V ˆ I

=1

2C∠− 90°1∠0°

= 12C

∠− 90° = − j 12C

= − j 1ωC

Ω

• Notice that the impedance of both the inductor and capacitor are purely imaginary.

• The phasor method can be summarized by describing the required transformation as follows:

v(t)⇔VM∠θV

i(t)⇔ IM∠θ I

R ⇔ RL ⇔ jωL Ω

C ⇔− j 1ωC

Ω

Note that the impedance of a resistor is a real number without an imaginary component. The lack of an


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imaginary component means there is only a difference in magnitude between the voltage and current. In addition, since the impedance of the inductor and capacitor are purely imaginary values, we can see that not only will the magnitudes be different, the voltage and current associated with reactive components will be out of phase with one another by ±90°.

Kirchhoff’s laws using phasors• The simplest way to analyze an ac circuit is to perform the phasor transformation and to treat the phasor circuit in exactly the same way as one would a resistive circuit.

• It should be noted that the voltages and currents within a phasor circuit will match the type of the source which drives the circuit. To properly apply the phasor method all the sinusoids need to be converted into the same function, either sine or cosine.


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• Both KVL and KCL remain valid when used with phasor voltages and currents.

Example - Using the phasor method

• Given the following circuit solve for all the voltages and currents

C=1/8F R=3!

2sin4tV

L=3H

+

–

Instantaneous power and average power

• When dealing with sinusoids, the power associated with any device can be calculated by taking the product of current and voltage, and will therefore be a function of time. We can calculate the flow of power at any instant in time using


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p(t) = v(t)i(t)p(t) = V cos(ωt +θV )× I cos(ωt +θ I )

=V I2cos(θV −θ I )+

V I2cos(2ωt +θV +θ I )Watts

Note that the expression for instantaneous power consists of two parts, a time independent constant and a sinusoid with twice the frequency of the original waveforms. If we consider the average value the instantaneous power over time, the sinusoidal term will be zero leaving only the constant term. As a result, it is more useful to consider only the first term which represents the average power.

• Recall that power represents the rate of absorption of energy. If we apply the concept of average power to the basic passive components we get the following:

Resistor: θV = θI Pave ≠ 0

Inductor: θV = θI + 90° Pave = 0

Capacitor: θI = θV + 90° Pave = 0


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• This means that the only passive component which can absorb energy is the resistor. The fact that the inductor and capacitor do not on average absorb energy is because for half the period of the sinusoid they will store energy and during the other half they release the previously stored energy. In other words the periodic component of the instantaneous power represents the power which is entering and leaving the reactive components but never being absorbed.

RMS Values

• The average power delivered to a load by an ac source is a constant value, we can view this as the power which would be delivered if a constant dc source were used. The value for this dc source is known as its effective or RMS (Root-Mean-Square) value. For a sinusoid:


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x(t) = X sin(ωt +φX )

Xrms = 1T

X sin(ωt +φX )[ ]2dt0

t

∫

= X2

For sinusoidal voltages and currents, we can calculate the average power delivered to an element as:

Pave =VrmsIrms cos(θV −θ I )

=V2

I2cos(θV −θ I )

=V I2cos(θV −θ I )Watts

Which is exactly the same equation we derived earlier.

Since we are only able to transfer power to a resistor. To determine the power transferred to an electrical load by measuring the voltage/current with a multimeter:

θV = θ I ⇒ Pave =V I2cos 0( ) = V I

2=V2

I2=VrmsIrms

This is why handheld meters provide rms values when reading ac values.


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Complex power

• It is useful to define average power in terms of phasors. We can define complex power as

ˆ S = ˆ V rmsˆ I rms

* =ˆ V I *

2 (when dealing with sinusoids)

= ˆ V rms ∠θV( ) ˆ I rms ∠−θI( )= ˆ V rms

ˆ I rms ∠(θV −θI )

= ˆ V rmsˆ I rms cos(θV −θI ) + j ˆ V rms

ˆ I rms sin(θV −θI )ˆ S = Pave + jQ VA

The magnitude of the of the complex power is referred to as apparent power and has the units of Volt-Amps (VA). The real portion of the complex power is the average power in Watts (W). The average power is real power which will be dissipated as heat (thermal) or converted to useful work (kinetic). The imaginary portion is referred to as the reactive power which has the units of Volt Amps Reactive (VAR). The VAR represents energy stored in the reactive elements.


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• As is the case when dealing with power in the time domain, conservation of energy also applies when dealing with complex power.

ˆ S ii=1

n

∑ = ˆ V rmsˆ I rms

* =i=1

n

∑ Pave + j Q =i=1

n

∑i=1

n

∑ 0 ⇒ Pave = 0, Q =i=1

n

∑i=1

n

∑ 0

Mutual inductance / the ideal transformer

• The flow of current produces a magnetic field which circulates around the path of the current. The magnetic field is external to a coil, and can affect a second coil via electromagnetic induction. Induction machines take advantage of this phenomena to energize the rotor without requiring an electrical connection.

• We can examine the simplest device which takes advantage of electromagnetic induction, the transformer. The transformer can be considered a static (non-rotating) electric machine.


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• If we have a loop of wire in which a varying current is creating a varying magnetic field, and we introduce a second loop into the vicinity of the first loop, the lines of flux from the first loop will also pass through the second loop and induce a voltage on the ‘non-powered’ loop. The coupling of the two coils is via a mechanism referred to as mutual inductance.

• The circuit symbol for two magnetically coupled coils which form an ideal transformer is

+

v1(t)

–

+

v2(t)

–

i1(t) i2(t)

• The polarity of the induced voltage is denoted by a ‘dot convention’ where if the current in one inductor enters the ‘dotted’ terminal the resulting induced voltage on the second inductor is positive at its ‘dotted’ terminal. The use of this convention is necessary since the direction of the windings will affect the resulting voltage.


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• Three characteristics of an ideal transformer are:

1. All the magnetic flux produced by one coil will pass through the other coil. The magnetic flux is contained and guided by the core on which the coils are wound.

2. It is a lossless device. p1(t) = v1(t)i1(t) = -p2(t) = v2(t)i2(t), i.e. p1(t)+ p2(t) = 0. All the power put into one side comes out the other side.

3. The self-inductances are considered to be infinite but their ratio is finite. Therefore we can relate the input and output voltages and currents as

v1(t)v2 (t)

= N1N2, and − i2 (t)

i1(t)= N1

N2

• The negative sign in the current expression is required so that the power entering must equal the power exiting the transformer.


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• The expressions relating the voltages and the currents to the ratio of turns in each coil allows the transformer to act as a machine which can be designed to increase or decrease output values simply by altering the number of turns for each coil.

AC Machines• AC machines, also known as induction machines, use ac (alternating current i.e. sinusoidal) voltages and currents to establish the required magnetic fields, and utilize ac values at its terminals.

• Induction machines are the most widely used type of electric machine, and can range in size from small sub one-horsepower machines to large many thousands of horsepower machines. The larger ac machines use what is known as three phase power and will be the type of machine we will focus on.


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• Three phase induction machines consists of two magnets and their associated magnetic fields. The induction machines consist of a stator, which is constructed with a three phase winding which generates a rotating magnetic field, and the rotor which also consists of a three phase winding. The term “three phase” refers to having three sets of coils each with its own set of voltages and currents, each of which has a phase component 120° apart from the others.

Stator construction

• The key feature to a three phase induction machine is that there are three sets of windings. However, first consider a single winding with a single sinusoidal source connected to it

Stator

Air Gap

Rotor

N(magnetic field)

Stator winding

Stator winding


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A varying magnetic field is produced as the current in the single stator winding follows a sinusoid:

t

i(t)

• Adding the other two phase windings to our stator

Winding a

Winding a'

Winding bWinding c

Winding c'Winding b'

Field a

Field b

Field c

t

i(t)

phasea

phaseb

phasec


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The magnetic field which will interact with the rotor is the cumulative effect of the three individual magnetic fields

• The resulting magnetic field, which interacts with the rotor, varies in direction. The key result is that we have a rotating magnetic field which is generated without the use of any mechanical moving parts.

• By using a system with three-phase windings we have created a rotating two pole magnetic field, and the resulting machine would be referred to as a 2-pole machine.

• If we were to place a permanent magnet into the core of the stator to act as the rotor, it would be dragged around by the rotating magnetic field. The rate of rotation for the rotor would be tied to the speed of the rotating magnetic field, which we will refer to as ωsync. For a complete revolution of the magnetic field requires the electrical


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sources to complete one cycle. If we isolate phase a, the rotor will follow the field for phase a as it goes positive, then negative, then back to positive. In other words the rotor will go from N, to S, and back to N. This progression is a function of the number of magnetic poles on the stator, which in turn is a function of how the windings are formed for the stator coils. The following figure shows one phase of a 4-pole machine

S N

N S

• Notice that when compared to the two pole machine the single phase winding is distributed over two separate coils where each coil encompasses only a quarter of the stator rather than completely encircling the rotor. As a result we end up with two coils being connected to one


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phase. In a three phase machine the other two phases would be configured in a similar manner.

• The net result of adding the additional two poles is that as the sinusoidal source completes one 360° cycle the rotor will only move 180° (from N, to S, to the next N), and the mechanical speed ωmech will be slower than the electrical speed/frequncy ωe.

• As the movement of the rotor is due to the rotating magnetic field we can relate the speed of rotation for magnetic field around the stator (the synchronous speed) to the electrical speed ωe (the frequency of the sinusoid) via

ωsync = 2no. of poles

ωe = 2Pωe rad / s ⇒ nsync = 2

Pωe ×

602π

rpm

From this expression we can see that for a 4-pole motor the magnetic field is rotating at half the frequency of the supplies to the motor.


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• Since the frequency of the sinusoids supplying energy to the stator is fixed (60 Hz in North America), as a result it is easier to select a machine with more than 2 poles to slow the rotational speed of a machine.

Rotor construction

• The rotor consists of a coil of wire which will move through the magnetic field provided by the field winding on the stator. However, since we are dealing with ac values, the stator coils will induce a voltage and current onto the rotor via the same means as observed with the transformer.

• If we consider a single phase on the stator and the rotor and hold the rotor stationary we will have

EstatorErotor

= NstatorNrotor

• Recall that energizing the non-powered coil in a transformer relies on the phenomena of mutual induction which requires a time varying magnetic field


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passing through the non-powered coil. As a result, once the rotor begins to rotate, the relative speed between the rotor windings and the rotating magnetic field will be reduced, which will reduce the rate of change in the magnetic flux linking the stator and the rotor. This will reduce the voltage induced on the rotor coil, and therefore reduce the resulting magnetic effects. The result of this dynamic interaction is that the rotor will never reach the same speed as the rotating magnetic field on the stator. If the speed of rotation of the rotor matches the rotating magnetic field, the induced voltage and current will drop to zero, which will eliminate the magnetic field of the rotor, and result in any torque acting on the rotor to be eliminated.

• Since the mechanical speed of the rotor will be less than the speed of the rotating magnetic field we can define the quantity slip which relates the mechanical and magnetic field speeds as being

s =ωsync −ωmechanical

ωsync% =

nsync − nmechanicalnsync

%


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Since the rotor is moving relative the to the frequency of the rotating magnetic field, the frequency of the induced voltage and current on the rotor (ωrotor) will be a result of the relative speed (nsync-nmech) between the two. Therefore

s =ωsync −ωmechanical

ωsync⇒ sωsync =ωsync −ωmechanical =ωrotor rad / s

• If we are calculating the voltage output of an induction machine being used as a generator, the output voltage needs to be scaled by the slip, Erotor@slip=s = sErotor@ω=0.

• The difference in frequency between the rotor circuit and the stator circuit will become important when we develop the circuit model for the induction machine. Since we will need to consider the various coils used in the construction of the machine, the frequencies involved will affect how we will calculate the impedances for those components.

Example - Basic ac machine calculations


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• Given a 3-phase induction machine, with two poles, connected to a 60 Hz source, with a slip of 3% when operating under its rated load, determine: a) The rotational speed of the rotating magnetic field; b) The frequency of the rotor voltage and current; c) The rotational speed of the rotor.

AC machine equivalent circuit model

• To be able to analyze the ac machine we need an equivalent circuit for the machine.

• To form a model of the machine we need to consider: that wire used in the coils will have a nonzero resistance; the coils are essentially inductors; because of the non-constant voltages and currents, the stator and rotor coils cannot be replaced with short circuits, and will interact with each other in a manner similar to a transformer; and finally the frequencies on the rotor side of the circuit are not the same as the frequencies on the stator side of the circuit which will affect the impedance values which are used.


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• We can consider a single phase of the three phase machine.

+

Vphase

–

Stator Rotor

R1 jX1 jsX2 R2

RC jXM

+

E1

–

+

sE2

–

I1 I1-IM I2

IM

The R1 and jX1 terms represent the resistance and inductance of the stator coil, the RC and jXM terms represent the transformer core losses, i.e. the losses due to the fact we are not using an ideal transformer, and the R2 and jsX2 terms represent the resistance and inductance of the rotor coil. Note that the frequency dependent parameters on the rotor side of the transformer are scaled by the slip since the slip will affect the frequency of the waveforms on the rotor.

• If we manipulate the quantities on rotor side we can separate the power associated with the resistive losses and the mechanical power


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I2 = sE2R2 + jsX2

= E2R2

s + jX2= E2R2 + R2 (1−s)

s( ) + jX2

With this equation we have separated the resistive losses, Presistive, and the mechanical power used to drive the rotor, Pmechanical.

Protor(total ) = (I2rms )2 R2

s = I22

2R2s = Presistive + Pmechanical

Presistive = (I2rms )2R2 = I2

2

2 R2Watts

Pmechanical = (I2rms )2 R2 (1−s )

s( ) = I22

2R2 (1−s )

s( ) Watts

• We can also represent the various power losses in the machine with the following power flow diagram

Pin

electricalinput

Statorresistivelosses (R1)

Statorcorelosses (RC)

Powertransferredacross airgap

Rotorresistivelosses (R2)

Mechanicalfrictionlosses

Pout

rotationaloutput


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• If we are given the values for the various components we can apply Thevenin’s theorem generate the Thevenin equivalent

+

Vphase

–

Stator Rotor

R1 jX1 jX2 R2/s

RC jXM

+

E1

–

+

E2

–

I1 I1-IM I2

IM

jX2 R2/s

+

VThev

–

RThev jXThev

P(air gap)

I2

Where the Thevenin parameters can be calculated via

VThev =VphaseXM

R12 + X1 + XM( )2

≈ VphaseXM

X1 + XM=VphaseKThev

ZThev = RThev + jXThev ≈ KThev2 R1 + jX1

• Using this model we can calculate the power across the air gap as


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Pairgap = (I2rms )2 R2

s = I22

2R2s Watts

and applying the same reasoning used in the original circuit model we can split R2/s into the electrical and mechanical portions

jX2 R2(1-s)/s

+

VThev

–

RThev jXThev

P(air gap)

I2 R2

Pmechanical = (I2rms )2 R2 (1−s )

s( ) = (1-s)Pairgap Watts

Note that this is the combined power for the mechanical rotation and to overcome any mechanical friction losses.

• Electrically, a three phase machine can have the individual phases connected in one of two configurations, a Delta (Δ) configuration, and a Wye (Y) configuration. In the Δ configuration we have the following


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+Vc

–

R1

jX1

jsX2

R2

RC

jXM

+

E1–

+sE2

–

I1

I1-I

MI2

IM

+

Vb

–

R1

jX1

jsX2

R2

RC

jXM

+

E1

–

+

sE2

–

I1

I1-IM

I2

IM

+

Va

–

R1

jX1

jsX2

R2

RC

jXM

+

E1

–

+

sE2

–

I1

I1-IM

I2

IM

we need to realize that we are applying each input voltage to an identical circuit, and can analyze one phase of the circuit and apply the result to the other two branches.

• The other possible configuration is the Y configuration where we connect the three phases as


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Stato

r

Roto

r

R1

jX1

jsX2

R2

RC

jXM

+E1

–

+

sE2

–

I1

I1-IM

I2

IM

– Va +

+

V

b

–+ Vc –

• The key difference between the Y and the Δ configuration is that the input voltage in the Y configuration is not applied across the inputs to a single phase as it was in the Δ configuration, it is being applied


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across two separate equivalent circuits. The result of this is that we can still perform calculations on a single equivalent circuit, however, we need to modify the input voltage we are using for our analysis with the circuit model. It turns out that

VΔ = VY3

which means that if we have a Y connected machine, we can use the input voltage to the machine divided by √3 and apply the result to the input of our equivalent circuit to perform any calculations.

Expressions for the developed torque

• From a mechanical perspective torque is defined as

• Applying this expression to the Thevenin equivalent circuit we have


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Tind = 3×I2rms2 R2

s

ω sync

= 3×

VThevrmsRThev+

R2s( )+ j XThev+X2( )

⎡⎣⎢

⎤⎦⎥

2R2s

2Pω e

Nm

Note that ωe represents the frequency of the voltage applied to circuit, and the factor 3 is a result of the fact the power calculation represents the power for only one of the three phases.

Example – Power Calculations

• Given a 3-phase, Δ configured, 4-pole, 60 Hz induction machine which is operated at 1600 rpm, with an input power of 15 kW and is drawing 22 Arms/ϕ. Determine the mechanical output power if the stator winding resistance is 0.2Ω and we ignore the core losses and any mechanical friction losses.

+Vc–

R1

jX1

jsX2

R2

RC

jXM

+E1–

+sE2

–

I1I1-IM

I2

IM+

Vb

–

R1

jX1

jsX2

R2

RC

jXM

+

E1

–

+

sE2

–I1

I1-IM

I2

IM

+Va

–

R1

jX1

jsX2

R2

RC

jXM

+E1

–

+

sE2

–

I1

I1-IM

I2

IM

Stator Rotor

0.2Ω jX1 jX2 R2/s

RC jXM

+

E1

–

+

E2

–

I1=22A

I1-IM I2

IM=0Pin=5kW

j0.209Ω 0.144Ω/0.02

+

122.38V

–

0.273Ω j0.503Ω

P(air gap)

I2


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Pairgap = 15000 − 3 2222 × 0.2⎡⎣ ⎤⎦ = 14854.8W

This Pairgap is the total power absorbed by the rotor. It includes the electrical power dissipated as heat in the resistive part of the rotor as well as mechanical power (power to overcome mechanical losses and available to drive a mechanical load on the rotor)

Pmechanical = 1− s( )Pairgap

s =ω sync −ωmech

ω sync

% =nsync − nmech

nsync%

ω sync = 2Pω e = 2

4 × 60 × 2π = 60πrad / sωmech = 1600

60 × 2π = 53.33πrad / s

∴s = 60π − 53.33π60π

= 11.11%

Pmech = 1− 0.11( )×14854.8 = 13204.27WThis Pmechanical is power left after accounting for the electrical losses in the rotor. As a result we could calculate the rotor parameters if given some additional information.

PR2 =14709.6 −13075.2

3= 544.8Watts Phase =

I22R22


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Example – operating point calculations

• Given a 3-phase, Y-connected, 220 V, 7.5 kW, 60 Hz, 6-pole machine with the following equivalent circuit parameters: R1=0.294Ω; R2=0.144Ω; X1=0.503Ω; X2=0.209Ω; XM=13.25Ω, and a mechanical friction loss of 400 W (independent of load). For a slip of 2%, determine the speed, the output torque, and the output power of the machine when it is operated at rated voltage and frequency.

VThev ≈ VphaseXM

X1 + XM= 220

313.25

0.503+13.25= 220

3×0.963 =122.37V

ZThev = KThev2R1 + jX1 = (0.963)2 ×0.294 + j0.503 = 0.273+ j0.503Ω

j0.209! 0.144!/0.02

+

122.37V

–

0.273! j0.503!

P(air gap)

I2


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ωmech = ωsync − sω sync = 26×60 ×2π − 0.02 × 2

6×60 ×2π = 39.2π rad / s

⇒ nmech = ωmech2π

×60 =1176rpm

j0.209! R2'= 0.144!

+

122.37V

–

0.273! j0.503!

P(air gap)

I2

R2'= 0.144!(1-s)/s

+

Vx

–

Example – vehicle drive motor

• Given a 3-phase, ∆-connected, 12 V, 7.5 kW, 60Hz, 6-pole machine with the following equivalent circuit parameters: R1=0.294Ω; R2=0.144Ω; X1=0.503Ω;


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X2=0.209Ω; XM=13.25Ω, and mechanical loss are negligible. If the motor is being used to accelerate a vehicle determine the slip and output torque at rotor speeds of 1rpm; 10rpm; 100rpm; 1000rpm; 1200rpm; 2000rpm.

For 1rpm:


ω sync

=2Pω e −

1rpm60 2π

2Pω e

=26 60 × 2π − 1rpm

60 2π26 60 × 2π

= 0.9992

jX2 R2(1-s)/s

+

VThev

–

RThev jXThev

P(air gap)

I2 R2

VThev ≈ VphaseXM

X1 + XM

= VphaseKThev = 1213.25

0.503+13.25= 11.56V

ZThev = KThev

2 R1 + jX1 = 0.273+ j0.503Ω

R21− ss

= 0.1441− 0.99920.9992

= 120 ×10−6Ω

I2 =VThevZTotal

= 11.56∠0°0.417 + j0.712

= 11.56∠0°0.825∠59.64°

= 14.011∠− 59.64°A


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SR2 1−ss = VI*

2=14.011∠− 59.64°( )× 120 ×10−6( )× 14.011∠59.64°( )

2= 0.012∠0°VA

T = Pmechωmech

= 3× 0.012160 2π

= 0.338Nm

For 10rpm:


ω sync

=26 60 × 2π − 10rpm

60 2π26 60 × 2π

= 0.992

jX2 R2(1-s)/s

+

VThev

–

RThev jXThev

P(air gap)

I2 R2

R21− ss

= 0.1441− 0.9920.992

= 1.21×10−3Ω

I2 =VThevZTotal

= 11.56∠0°0.418 + j0.712

= 11.56∠0°0.826∠59.64°

= 14.003∠− 59.58°A

Smech = SR21−ss

= VI*

2= 0.119∠0°VA = 0.119W + j0VAR

T = Pmechωmech

= 3× 0.1191060 2π

= 0.340Nm


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For 100rpm:


ω sync

=26 60 × 2π − 100rpm

60 2π26 60 × 2π

= 0.917

R21− ss

= 0.1441− 0.9170.917

= 0.013Ω

I2 =VThevZTotal

= 11.56∠0°0.430 + j0.712

= 11.56∠0°0.832∠58.87°

= 13.90∠− 58.87°A

Smech = SR21−ss

= VI*

2= 1.265∠0°VA = 1.265W + j0VAR

T = Pmechωmech

= 3×1.265100060 2π

= 0.362Nm

For 1000rpm:


ω sync

=26 60 × 2π − 1000rpm

60 2π26 60 × 2π

= 0.167

R21− ss

= 0.1441− 0.1670.167

= 0.72Ω

Smech = SR21−ss

= VI*

2= 26.740∠0°VA = 26.740W + j0VAR


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T = Pmechωmech

= 3× 26.73100060 2π

= 0.766Nm

For 1200rpm:


ω sync

=26 60 × 2π − 1200rpm

60 2π26 60 × 2π

= 0

R21− ss

= 0.1441− 00

= ∞Ω ⇒ I2 = 0

Smech = SR21−ss

= VI*

2= 0∠0°VA = 0W + j0VAR

T = Pmechωmech

= 3× 0120060 2π

= 0Nm

For 2000rpm:


ω sync

=26 60 × 2π − 2000rpm

60 2π26 60 × 2π

= −0.666

R21− ss

= 0.1441+ 0.666−0.666

= −0.36Ω ⇒ P = I 2R < 0 ⇒ Generator

I2 =VThevZTotal

= 11.56∠0°0.057 + j0.712

= 11.56∠0°0.71∠85.43°

= 16.19∠− 85.43°A


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VR2 1−ss = I2 R2 1−ss( ) = 16.19∠− 85.43° −0.36( ) = −5.83∠− 85.43°V

VR2s= I2

R2s( ) = 16.19∠− 85.43° −0.216( ) = −3.496∠− 85.43°V

Srotor =VI2

*

=−3.496∠− 85.43°( ) 16.19∠85.43°( )

2= −28.29∠0°VA

Smech = SR21−ss

= VI*

2= −47.16∠0°VA = −47.16W + j0VAR

Note that |Srotor|<|Smech|, this makes sense since Srotor represents the power that crosses the air gap. Some of the mechanical energy that is applied will be lost to the resistive rotor losses and will therefore not be available to cross the air gap to deliver energy to whatever is connected to the stator.

T = Pmechωmech

= 3× −47.16200060 2π

= −0.675Nm

Note that the torque has a sign opposite to the motor cases, which makes sense as in the motor cases the mechanical load is pushing back against the motor and in the generator case it is the machine pushing back.


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Example

• Given a 3-phase, ∆-connected, 12V, 7.5 kW, 60Hz, 6-pole machine with the following equivalent circuit parameters: R1=0.294Ω; R2=0.144Ω; X1=0.503Ω; X2=0.209Ω; XM=13.25Ω, and mechanical loss are negligible. If the rotor is to turn at 1000rpm, what is the output power and torque if the machine is replaced with a 4-pole or a 2-pole machine with identical resistance and inductance parameters?

For the 4-pole machine:


ω sync

=24 60 × 2π − 1000rpm

60 2π24 60 × 2π

= 0.444

jX2 R2(1-s)/s

+

VThev

–

RThev jXThev

P(air gap)

I2 R2

R21− ss

= 0.1441− 0.4440.444

= 0.18Ω

Smech = SR21−ss

= VI*

2= 13.94∠0°VA = 13.94W + j0VAR ⇒ Pave4Pole < Pave6 pole


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T6 pole =Pmechωmech

= 3× 26.73100060 2π

= 0.766Nm

T = Pmechωmech

= 3×13.94100060 2π

= 0.40Nm ⇒ T4 pole < T6 pole

For the 2-pole machine:


ω sync

=22 60 × 2π − 1000rpm

60 2π22 60 × 2π

= 0.722

R21− ss

= 0.1441− 0.7220.722

= 0.055Ω

Smech = SR21−ss

= VI*

2= 5.07∠0°VA = 5.07W + j0VAR ⇒ Pave2Pole < Pave6 pole

T = Pmechωmech

= 3× 5.07100060 2π

= 0.145Nm ⇒ T2 pole < T4 pole < T6 pole

Poles s Power Torque

6 16.7% 26.7W 0.766Nm

4 44.4% 13.94W 0.4Nm

2 72.2% 5.07W 0.145Nm

Therefore power is best transferred when the desired speed of rotation is close to the synchronous speed.


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Conclusion• This chapter provided a review of basic ac circuits and introduced the principles behind the operation of ac machines. In addition it provided a review of the phasor method of analysis; and analysis methods which can be applied to the study of ac machines. Upon completing this chapter you should have mastered the following learning objectives:

1. Be able to apply KVL, KCL, and Ohm’s Law to solve for any voltage, and current parameters within an ac circuit.

2. Be able to determine the power associated with any circuit element.

3. Be able to use the phasor technique to analyze circuits subject to sinusoidal conditions.

4. Understand the basic operation and configuration of an ac machine.

5. Be able to analyze a given ac electric machine and determine its external parameters both electrical and mechanical.


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AC Electric Machines - University of...

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