electrical machines and systems course notes by j d edwards_very good figures

Course in Electrical Machines and Systems

Year 2

2004 J D Edwards

ELECTRICAL MACHINES AND SYSTEMS COURSE NOTES

These notes were prepared for a Year 2 course module in the Department of Engineering and Design at the University of Sussex. The module runs for ten weeks, and absorbs 25% of student time.

This introductory course in electrical machines follows on from the Year 1 course Electromechanics, with a similar philosophy. It avoids the traditional mathematical derivation of the theory of AC and DC machines, and makes extensive use of field plots generated with MagNet to explain the principles. With this physical background, the theory is developed in terms of circuit models and phasor diagrams.

Field plots are particularly useful for explaining armature reaction and compensating windings in DC machines, and for demonstrating the action of variable-reluctance and hybrid stepper motors. They also show the essential unity of the conventional rotating machines; there are similar plots for the stator and rotor field components and the resultant field in DC, synchronous and induction machines.

Electrical Machines and Systems Course Notes Copyright 2004 J D Edwards

CONTENTS

1 INTRODUCTION 1

2 TRANSFORMERS 2 2.1 Introduction 2

2.2 Types of power transformer 3

2.3 Ideal transformer properties 4

2.4 Circuit model of a transformer 6 2.5 Parameter determination 7

2.6 Transformer performance 9

2.7 Current transformers 11 2.8 Transformer design 12

3 DC MACHINES 15 3.1 Introduction 15

3.2 DC machines in practice 16 3.3 Characteristics and control 18 3.4 Series motors 21

4 INTRODUCTION TO AC MACHINES 24 4.1 Review of 3-phase systems 24

4.2 Rotating magnetic field 25

4.3 Multi-pole fields 28

5 SYNCHRONOUS MACHINES 30 5.1 Introduction 30

5.2 Characteristics 31 5.3 Salient-pole machines 35 5.4 Linear synchronous motors 37

6 INDUCTION MACHINES 39 6.1 Introduction 39

6.2 Characteristics 40 6.3 Losses and efficiency 46

6.4 Parameter determination 48

6.5 Single-phase induction motors 50 6.6 Dynamic conditions 52

6.7 Linear induction motors 54

7 STEPPER MOTORS 56 7.1 Introduction 56

7.2 Variable-reluctance principle 56

7.3 Variable-reluctance stepper motors 58

7.4 Hybrid stepper motors 59

7.5 Stepper motor characteristics 61 7.6 Stepper motor control 64

8 POWER ELECTRONIC CONTROL 66 8.1 AC/DC Converters 66 8.2 DC motor control 68 8.3 DC/AC Inverters 68 8.4 AC motor control 70

9 REFERENCES 76

10 APPENDICES 77 10.1 Induction motor 2-axis equations 77

10.2 List of formulae 80

Introduction 1

1 INTRODUCTION This course follows on from the Term 3 course Electromechanics. Its purpose is to explore in greater depth the AC and DC machines that were introduced in the earlier course. The approach is that of the application engineer rather than the machine designer, concentrating on the basic principles, characteristics, and control. Since induction motors account for more than 90 per cent of the motors used in industry, the course gives particular emphasis to these machines.

Course components

The course has three closely linked components: lectures, problem sheets and laboratories. In addition, there is a design assignment, which introduces some of the basic ideas and problems of design by considering a very simple device: an electromagnet. Lectures will use video presentation and practical demonstrations. These notes provide support material for the lectures, but they are not a substitute. Regular lecture attendance is essential. Problem solving is a vital part of the course. Problem sheets will be issued at the first lecture each week, and methods of solving the problems will be discussed in each lecture. The laboratory runs from week 4 to week 9, with three 3-hour experiments: EMS1: Speed control of induction motors. EMS2: Characteristics of a power transformer. EMS3: Control of a stepper motor. Experiment EMS1 is a sequel to the simple DC motor-control experiment in Electromechanics. EMS2 introduces some important electrical measurement techniques as well as exploring the properties of a transformer. EMS3 explores a stepper motor and controller of the kind widely used in industry.

Magnetic field plots

A magnetic field plot is often a useful way of picturing the operation of an electromagnetic device. Numerous plots have been specially prepared for these notes, using the MagNet electromagnetic simulation software, to develop the basic concepts with a minimum of mathematics.

References

References to books are listed in section 10, and cited in the text of the notes with the reference number in square brackets.

Background material

The course assumes a familiarity with the contents of the Term 3 course Electromechanics, so the basic principles covered in that course will not be repeated. Students are expected to have a copy of the printed notes for Electromechanics [1], and further information will be found in references [2] to [4].

2 Electrical Machines and Systems Course Notes

2 TRANSFORMERS

2.1 Introduction Basic transformer principles were covered in Electromechanics [1], and the main results are given below. Figure 2-1 is a schematic representation of a single-phase transformer with two coils on a magnetic core, where the magnetic coupling is assumed to be perfect: the same flux passes through each turn of each coil.

~ Zv1+ + +

v2e1 e2

i1 i2

R1 R2

primaryN1 turns

secondaryN2 turns

+

Figure 2-1: Transformer with source and load

Voltage relationships

Kirchhoffs voltage law applied to the two windings gives:

1111111 iRdtdNiRev (2-1)

2222222 iRdtdNiRev (2-2)

If the resistances R1 and R2 are negligible, then equations 2-1 and 2-2 become:

dtdNv 11 (2-3)

dtdNv 22 (2-4)

Dividing these equations gives the important result:

2

1

2

1NN

v

v (2-5)

Sinusoidal operation

If the voltage source is sinusoidal, then the core flux will also be sinusoidal, so we may put:

tm sin (2-6) Substituting this expression in equation 2-3 gives:

tVtNdtdNv mm

coscos 1111 (2-7)

Thus the maximum primary voltage is:

mmmm ABfNfNNV 1111 22 (2-8) where A is the cross-sectional area of the core and Bm is the maximum flux density in the core. A typical value for Bm is 1.4 T for the silicon steel characteristic in figure 2-2.

00.20.40.60.8

11.21.41.61.8

2

0 2 4 6 8 10 12 14 16 18 20Magnetic intensity H, kA/m

Flux

densi

ty B,

T

Figure 2-2: Silicon transformer steel.

Current relationships

The relationship between the primary and secondary currents can be found by considering the magnetic circuit of the transformer. From the basic magnetic circuit equation, we have:

RF 2211 iNiN (2-9) In a well-designed transformer, the reluctance R is small, so equation 2-9 becomes:

02211 iNiN (2-10) This gives the counterpart of equation 2-5 for voltage:

1

2

2

1NN

ii (2-11)

Transformers 3

If sinusoidal voltages and currents are represented by phasors, the corresponding forms for the basic voltage and current equations are:

2

1

2

1

NN

VV

(2-12)

1

2

2

1

NN

II

(2-13)

Transformer rating

The maximum voltage at the primary terminals of a transformer is determined by equation 2-8, and is independent of the current. The maximum primary current is determined by the I2R power loss in the resistance of the transformer windings, which generates heat in the transformer. This power loss is independent of the applied voltage. Consequently, for a given design of transformer, there is a maximum value for the product V1I1 at the primary terminals. To a first approximation, this is also equal to the product V2I2 at the secondary terminals. This maximum value does not depend on the phase angle between the voltage and the current. Transformer ratings therefore specify the apparent power VI (volt-amperes, VA) rather than the real power VI cos (watts, W).

2.2 Types of power transformer In addition to the ordinary single-phase power transformer, two other types are in common use: auto-wound transformers, and 3-phase transformers.

Auto-wound transformers

A transformer can have a single coil with an output taken from a portion of the coil, as shown in figure 2-3. This is known as an auto-wound transformer or auto-transformer.

VS V2

I1

IL

ZL +

+ I2

V1 +

Figure 2-3: Auto-wound transformer.

Unlike the normal transformer with two windings, known as a double-wound transformer, the auto-wound transformer does not provide electrical isolation between the primary and the secondary, However, an auto-wound transformer can have a much larger apparent power rating than a double-wound transformer of the same physical size. Let N1 be the number of turns on the upper part of the winding in figure 2-3, and N2 the number of turns on the lower part. The conventional transformer equations 2-12 and 2-13 apply to these parts of the winding, since they are equivalent to two separate windings with a common connection. Applying Kirchhoffs law to this circuit gives:

1 2S V V V (2-14)

1 2L I I I (2-15) As an example, suppose that N1 = N2. If the transformer is regarded as ideal (see section 2.3), then I1 = I2 and V1 = V2. Equations 2-14 and 2-15 give:

1 2 22 2S L V V V V V (2-16)

1 2 12 2L S I I I I I (2-17) where VL is the voltage across the load and IS is the current supplied by the source. This auto-wound transformer behaves as a step-down transformer with a ratio of 2:1, and the current in each winding is equal to half of the load current. An elegant application of the auto-wound transformer principle is the variable transformer, which has a single-layer coil wound on a toroidal core. The output is taken from a carbon brush that makes contact with the surface of the coil; the brush can be moved smoothly from one end of the coil to the other, thus varying the output voltage. Examples of variable transformers are shown in figure 2-4.


Figure 2-4: Variable transformers. (RS Components Ltd)

3-phase transformers

In 3-phase systems (see section 4.1), it is common practice to use sets of three single-phase transformers. It is also possible, however, to make 3-phase transformers with three sets of windings on three limbs of a core, as shown in figure 2-5.

Figure 2-5: 3-phase transformer model.

The corresponding fluxes are shown in figure 2-6.

a b c

Figure 2-6: 3-phase transformer flux. In a balanced system with sinusoidal phase voltages, the fluxes will be given by:

)120cos()240cos()120cos(

cos

tt

t

t

mmc

mb

ma

(2-18)

Figure 2-7 shows flux plots for the transformer at the instants when t = 0, 120 and 240.

Figure 2-7: 3-phase transformer flux plots: (a) 0, (b) 120, (c) 240.

There is no requirement for another limb to form a flux return path, because the fluxes a, b and c sum to zero in a balanced 3-phase system. The proof is as follows. From equation 2-18, the sum is given by:

0coscos120coscos2cos

)120cos()120cos(cos

tt

tt

tttm

cba

(2-19)

Because the fluxes in the three limbs sum to zero at all instants of time, there is no leakage of flux from the core, as the flux plots in figure 2-7 demonstrate.

(a)

(b)

(c)

Transformers 5

2.3 Ideal transformer properties If the primary and secondary windings have zero resistance, and the magnetic core has zero reluctance, then the approximate equalities in equations 2-12 and 2-13 become exact equalities. This leads to the concept of an ideal transformer element, to accompany the other ideal elements of circuit theory. Figure 2-8 shows a circuit symbol for the ideal transformer element.

+ +v1 v2

i1 i2

Figure 2-8: Ideal transformer element

The voltage and current relationships in the time and frequency domains are given in table 2-1.

Table 2-1: Ideal transformer relationships.

Time domain Frequency domain

nNN

v

v

1

2

1

2 n

NN

1

2

1

2

VV

(2-20)

nNN

ii

1

2

2

1 n

NN

1

2

2

1

II

(2-21)

The following properties of the ideal transformer may be deduced from equations 2-16 and 2-17: The voltage transformation is independent of

the current, and vice versa. If the secondary is short-circuited, so that

v2 = 0, the primary terminals appear to be short-circuited since v1 = 0.

If the secondary is open-circuited, so that i2 = 0, the primary terminals appear to be open-circuited since i1 = 0.

The output power is equal to the input power, so there is no power loss in the element.

Impedance transformation

The ideal transformer has the important property of transforming impedance values in a circuit. Consider an ideal transformer with an impedance ZL connected to its secondary terminals, as shown in figure 2-9.

V1 V2

I1 I2

ZL

N1:N2

+ +

Figure 2-9: Ideal transformer with a load.

The secondary impedance is given by:

2

2IVZ L (2-22)

At the primary terminals, the circuit presents an impedance given by:

2

2

2

1

2

22

2

1

1

22

2

21

1

1

nNN

NN

NNN

N

LL

in

ZZIV

I

V

IVZ

(2-23)

Thus the combination of an ideal transformer of ratio n and an impedance ZL can be replaced by an equivalent impedance ZL / n2.

Referred impedances

Figure 2-10(a) shows an ideal transformer with a load impedance ZL connected to the secondary. Another impedance Z2 is in series with ZL. The input impedance of this circuit is:

22

n

Lin

ZZZ (2-24)

The input impedance of the circuit in figure 2-10(b) is:

22 n

Lin

ZZZ (2-25)

The two expressions for Zin will be identical if:

22

2n

ZZ (2-26)


ZL

N1:N2

Z2

Zin

(a)

ZL

N1:N2

Z'2

Zin

(b) Figure 2-10: Referred impedance 1.

The impedance Z'2 is termed the secondary impedance Z2 referred to the primary. In a similar way, a primary impedance Z1 can be referred to the secondary, as shown in figure 2-11. In this case, the referred impedance is given by:

12

1 ZZ n (2-27)

ZL

N1:N2

Z1

Zin

(a)

ZL

N1:N2

Zin

Z''1

(b) Figure 2-11: Referred impedance 2.

The concept of referred impedance is often a useful device for simplifying circuits containing transformers, as will be shown in the next section. It is conventional to use a single prime (') to denote quantities referred to the primary side, and a double prime (") to denote quantities referred to the secondary side.

2.4 Circuit model of a transformer In a practical transformer, the winding resistances and the core reluctance are not zero. In addition, there will be some power loss in the core because of eddy currents and hysteresis in the magnetic material. All of these effects can be represented by the equivalent circuit [3, 4] shown in figure 2-12.

N1:N2

R1 R2jx1 jx2

jXm RcV1 V2

+ +

I1 I2I2 = nI2I0

I0m I0c

Figure 2-12: Transformer equivalent circuit.

This circuit is based on the ideal transformer element, with additional circuit elements to represent the imperfections. The resistances R1 and R2 represent the physical resistances of the windings, and Rc represents the power lost in the core. The reactance Xm, known as the magnetising reactance, allows for the current required to magnetise the core when the reluctance is not zero. Reactances x1 and x2, known as leakage reactances, represent the leakage flux that exists when the magnetic coupling between the primary and the secondary is not perfect. Figure 2-13 shows the leakage flux when the core has an artificially low relative permeability of 10, and one winding at a time is energised. In practice, the leakage is much less than this, so the leakage reactances are normally very much smaller than the magnetising reactance Xm.

(a)

Transformers 7

(b) Figure 2-13: Transformer leakage flux:

(a) left coil energised, (b) right coil energised. In the circuit of figure 2-12, the current I2 is the effective value of the secondary current as seen from the primary side of the transformer. It is also known as the secondary current referred to the primary. The current I0 is the no-load current, which is the current taken by the primary when there is no load connected to the secondary. It has a component I0m, known as the magnetising current, which represents the current required to set up the magnetic flux in the core. The current I0c is the core loss component of the no-load current.

Approximate equivalent circuit

For power transformers with ratings above 100 VA, the values of the series elements R1 and x1 are generally much smaller than the shunt elements Rc and Xm. Under normal working conditions, the voltage drop in R1 + jx1 is much smaller than the applied voltage V1. Similarly, the no-load current I0 is much smaller than the load current I1. It follows that the shunt elements can be moved to the input terminals, as shown in figure 2-14(a), with very little loss of accuracy. The secondary elements R2 and jx2 can be replaced by equivalent elements R2 = R2 / n2 and x2 = x2 / n2 on the primary side (see section 2.3), giving the circuit shown in figure 2-14(b). Finally, the series elements can be combined to give an effective resistance Re and leakage reactance xe, as shown in figure 2-14(c), where the values are:

22

1n

RRRe , 22

1n

xxxe (2-28)

N1:N2

R1 R2jx1 jx2

jXm Rc

(a)

N1:N2

R1 R2jx1 jx2

jXm Rc

(b)

N1:N2

Re jxe

jXm Rc

(c) Figure 2-14: Approximate equivalent circuit.

2.5 Parameter determination The parameters of the approximate equivalent circuit (figure 2-14) can be determined experimentally from two tests: An open-circuit test, where the secondary is

left unconnected and the normal rated voltage is applied to the primary.

A short-circuit test, where the secondary terminals are short-circuited and a low voltage is applied to the primary, sufficient to circulate the normal full-load current.


Open-circuit test

With the secondary unconnected, I2 = 0, so the equivalent circuit reduces to the form shown in figure 2-15.

jXm Rc

I1oc

V1oc

+

Figure 2-15: Open-circuit test.

To a very close approximation, the current I1oc supplied to the primary is equal to the no-load current I0 in figure 2-12. The values of the elements Rc and Xm can be determined from measurements of the input voltage V1oc, current I1oc and power P1oc as follows. The input power is entirely dissipated in the resistance Rc, giving:

oc

occ P

VR1

21 (2-29)

The input impedance of the circuit is given by:

mc

oc

ococ

jXR11

11

11

IVZ (2-30)

In terms of magnitudes, equation 2-30 becomes:

221

11 11

1

mc

oc

ococ

XR

IV

Z

(2-31)

Re-arranging equation 2-31 gives the value of Xm:

2

2

1

1 1

1

coc

oc

m

RVI

X

(2-32)

From figure 2-14(c), it follows that the turns ratio is given by:

oc

oc

VV

NN

n1

2

1

2 (2-33)

Short-circuit test

If the secondary terminals are short-circuited, the ideal transformer in figure 2-14 can be replaced by a short circuit, so the equivalent circuit takes the form shown in figure 2-16(a). In a typical power transformer, the shunt elements Rc and Xm are at least 100 times larger than the series elements Re and xe. Consequently, the shunt elements can be neglected, and the circuit reduces to the form shown in figure 2-16(b).

jXm Rc

jxeRe

(a)

jxeReI1sc

V1sc

+

(b)

Figure 2-16: Short-circuit test.

The values of the elements Re and xe can be determined from measurements of the input voltage V1sc, current I1sc and power P1sc as follows. The input power is entirely dissipated in the resistance Re, giving:

21

1

sc

sce I

PR (2-34)

The input impedance of the circuit is given by:

ee

sc

scsc jxR

1

11 I

VZ (2-35)

In terms of magnitudes, equation 2-35 becomes:

22

1

11 ee

sc

scsc xRI

VZ (2-36)

Transformers 9

Re-arranging equation 2-36 gives the value of xe:

22

1

1e

sc

sce RI

VX

(2-37)

In practice, the open-circuit test is usually made on the low-voltage side of the transformer to minimise the value of Voc, and the short-circuit test is made on the high-voltage side to minimise the value of Isc. The resulting parameter values are then referred to the primary side of the transformer.

2.6 Transformer performance Consider a transformer with an impedance ZL connected to the secondary. With the approximate equivalent circuit, this may be represented by the circuit diagram of figure 2-17.

N1:N2

Re jxe

jXm Rc ZL =RL + jXL

Figure 2-17: Transformer with a load.

The load impedance can be referred to the primary side of the ideal transformer element, giving the circuit shown in figure 2-18.

Re jxe

jXm RcZL =ZL / n2 V2 = V2 / n

+V1

+

I1 I2 = nI2

I0

Figure 2-18: Circuit with referred impedance.

This circuit is easily solved for the currents I0 and I2. The referred secondary voltage is given by V2 = ZLI2, and the secondary terminal quantities are given by V2 = nV2, I2 = I2 / n.

Voltage regulation and efficiency

When a load is connected to the secondary of a transformer, there will be a voltage drop in the series elements Re and xe, so the secondary terminal voltage will change. The voltage regulation is defined as:

nl

flnlV

VV

2

22 (2-38)

where V2nl is the magnitude of the no-load secondary terminal voltage, and V2fl is the corresponding full-load voltage. Power is lost as heat in the windings and core of the transformer, represented by the resistance elements Re and Rc in the equivalent circuit. The efficiency is defined in the usual way as:

in

out

PP

(2-39)

where Pin is the power input to the primary and Pout is the power output from the secondary. The power lost as heat in the transformer is:

outinloss PPP (2-40) so we have the following alternative forms of equation 2-39:

lossout

loss

lossout

out

in

loss

in

lossin

in

out

PPP

PPP

PP

PPP

PP

1

1 (2-41)

Large transformers are very efficient. Even a 2 kVA transformer can have an efficiency of about 95%. Above 25 kVA, the efficiency usually exceeds 99%. It is very difficult to make an accurate measurement of efficiency by direct measurement of Pout and Pin, since this would require an accuracy of measurement of the order of 0.01%. Instead, the normal practice is to determine the losses from measurements, and calculate the efficiency from one of the alternative expressions in equation 2-41. The losses can be calculated with high accuracy from the equivalent-circuit parameters determined from tests on the transformer.

Maximum efficiency

The power loss in a transformer has two components: the core loss, given by V12 / Rc, and


the I2R loss, given by I'2Re. The core loss will be constant if the primary voltage V1 is constant, but the I2R loss will vary with the secondary current. When the current is low, the output power will be low, but the core loss remains at the normal value. Consequently, the efficiency of the transformer will be low under these conditions. It may be shown that the efficiency is a maximum when the secondary current is such that the variable I2R loss is equal to the fixed core loss. This result also applies to other devices where the losses have fixed and variable components. Transformers are usually designed to have maximum efficiency at the normal operating value of secondary current, which may be less than the maximum rated current.

Power relationships

When calculating the transformer performance, the following power relationships can be useful. The complex power S is given by [2]: S = P + jQ = VI* (2-42) where V is the voltage phasor, I* is the complex conjugate of the current, P is the real power, and Q is the reactive power. If is the phase angle, then:

*)Re(coscos VI VISP (2-43)

*)Im(sinsin VI VISQ (2-44)

22 QPVIS (2-45) If the voltage phasor V is chosen as the reference quantity, and defined to be purely real (V = V + j0), then the power relationships take a simple form:

)Re(cos IVVIP (2-46)

)Im(sin IVVIQ (2-47)

Worked example 2-1

A 2 kVA, 50 Hz, power transformer has the following equivalent-circuit parameter values referred to the primary: Re = 0.682 , xe = 0.173, Rc = 1080, Xm = 657, N2 / N1 = 0.472. If the primary is connected to 230 V 50 Hz supply, and a load impedance (6.0 + j2.5) is connected to the secondary, determine: (a) the secondary current magnitude,

(b) the secondary terminal voltage magnitude, (c) the primary current magnitude, (d) the voltage regulation, (e) the efficiency of the transformer. (a) Secondary current The load impedance referred to the primary is:

2.119.26)472.0(5.20.6

22 jj

n

LL

ZZ

The secondary current referred to the primary is:

A 70.794.212.7

A 94.212.74.116.27

0230)2.119.26()173.0682.0(

0230

2

12

jI

jjj

jjj

Le ZZVI

The secondary current magnitude is:

A 3.16472.070.72

2

n

II

(b) Secondary voltage The secondary voltage magnitude is:

V 0.1065.63.165.20.63.1622

jZIV L

(c) Primary current The no-load current is:

A 350.0213.0657

02301080

0230110

jj

jjjXR mc

VVI

The primary current is:

A 03.829.333.7A 29.333.7

)94.212.7()350.0213.0(

1

201

jIj

jjIII

(d) Voltage regulation On no load, the secondary voltage is:

V 6.108230472.012 nVV

Transformers 11

The voltage regulation is therefore:

%33.26.108

0.1066.1082

22

nl

flnlV

VV

(e) Efficiency The output power is:

W 15970.6)31.16( 222 Lout RIP The power loss is:

W 4.89

682.0)699.7(1080

)0.230( 22

22

21

ec

loss RIRVP

The input power is therefore:

W 16864.891597 lossoutin PPP

Alternatively, from equation 2-46 the input power is:

W 1686330.70.230)Re( 11 IVPin The efficiency is therefore:

%70.9416861597

in

out

PP

2.7 Current transformers The use of transformers for measuring current has been introduced in Electromechanics [1], where the danger of open-circuiting the secondary has been explained. This section introduces the important topic of measurement errors.

Current transformer errors

In a well-designed current transformer, the core flux density is low and the core is made from a high-quality magnetic material. Under normal operating conditions, the core loss will be negligibly small, so the core loss resistance Rc can be omitted from the equivalent circuit. A circuit model for a current transformer connected to a load therefore takes the form shown in figure 2-19.

N1:N2

R1 R2jx1 jx2I1

I2

I2 = nI2

I0

jXm ZL

Figure 2-19: Current transformer circuit model.

Only the relationship between currents is of interest, so the primary impedance (R1 + jx1) can be disregarded. It is now convenient to refer the primary quantities to the secondary side, giving the circuit model shown in figure 2-20.

R2jx2I1I2I0 = I0 / n

jXm = jn2Xm ZL

Figure 2-20: Modified circuit model.

The circuit acts as a current divider, where the current in the secondary branch is given by:

mL

m

XjXj

ZZ

II2

12 (2-48)

where I1 is the primary current referred to the secondary, and Z2 = R2 + jx2. In a well-designed transformer, the secondary impedance Z2 is very small in comparison with the referred magnetising reactance Xm, so this term introduces very little error. Equation 2-48 shows that it is desirable to keep the load impedance ZL as small as possible if the error is to be minimised. In practice, current transformers are designed for a specified maximum secondary voltage at the rated secondary current. This defines a maximum apparent power for the secondary load, or burden. Typically, a small current transformer will have a rated secondary burden of 5 VA. With the usual secondary current rating of 5 A, this implies that the maximum secondary voltage is 1 V, and the maximum impedance magnitude is 0.2 .


Worked example 2-2

A current transformer has 10 turns on the primary and 100 turns on the secondary. It has a rated secondary current of 5 A, the magnetising reactance referred to the secondary is 10 , and the maximum burden is 5 VA. If the primary current is 50 A, determine the secondary current, and hence the percentage error in the current measurement, if the secondary load is (a) purely resistive, (b) purely inductive. The transformer secondary impedance may be neglected. (a) Resistive load Since the burden is 5 VA and the secondary current is 5 A, the secondary voltage is 1 V, and the resistance is 1 / 5 = 0.2 . The turns ratio is n = 100 / 10 = 10, so the primary current referred to the secondary is 50 / 10 = 5.0 A. The secondary load current is:

0.102.00.50.101

2 jj

XjRXj

mL

m

II

The magnitude is given by:

A 999.4002.10

0.50.100.102.0

0.50.102

jI

The percentage error is thus:

%02.00.5

999.40.5

e

(b) Inductive load Since the impedance magnitude is the same as before, the load reactance is 0.2 . The secondary current is now:

0.102.00.50.101

2 jjj

XjjXXj

mL

m

II

The magnitude is given by:

A 902.40.102.00.50.10

2

I

The percentage error is thus:

%96.10.5

902.40.5

e

2.8 Transformer design The majority of single-phase transformers use the shell type of construction shown in figure 2-21.

Figure 2-21: Shell-type transformers. (RS Components Ltd)

Normally the core laminations are made in two parts, termed E and I laminations, as shown in figure 2-22.

Figure 2-22: E and I laminations.

The centre limb is twice the width of the outer limbs because it carries twice the flux, as shown by the flux plot in figure 2-23.

Figure 2-23: Flux plot: shell-type transformer.

Transformers 13

The coils are wound on a bobbin that fits the centre limb of the core, and the core is assembled by inserting E laminations alternately from each side and adding matching I laminations. Dimensions are chosen so that two E and two I laminations can be punched from a rectangular steel sheet without any waste, as shown in figure 2-24.

Figure 2-24: Punching E and I shapes.

Current flowing in the resistance of the transformer windings will produce heat, which must escape through the surface of the windings. In addition, there will be power loss in the core, which also appears as heat. The power output from a given size of transformer is governed by the rate at which heat can be removed. Large transformers are usually cooled by circulating oil, but small transformers rely on natural convection cooling in air. A simple design approach for small transformers is given below.

Thermal model

The rate of cooling depends on the exposed surface area of the transformer and the temperature rise above ambient. An exact calculation is complex, since it needs to take account of temperature gradients within the transformer as well as the cooling conditions on different surfaces. A simple thermal model ignores temperature gradients, and the power loss in the core. It just considers the I2R loss in the windings, and assumes that this heat escapes through the exposed surfaces of the windings. It is assumed that the temperature rise is proportional to the power loss per unit area:

s

lcAPkT (2-49)

where T is the temperature rise above ambient, Pl is the total power loss in the windings, As is the exposed surface area, and kc is a cooling coefficient with a typical value of 0.04 Km2/W. Figure 2-25 shows the side view and top view of a shell-type transformer.

2a

b

mean turn

aa 2a a a

6a

3a 5a

a

Figure 2-25: Shell-type transformer dimensions It is assumed that the core is made from laminations punched as shown in figure 2-24. If a is the width of each outer limb of the core, then the width of the centre limb is 2a, and the other dimensions are as shown in figure 2-25.

Winding resistance

The total cross-sectional area of the two windings is the window area of height 3a and width a. Each winding occupies half of this area, so the conductor cross-sectional area for each winding is:

25.1 akA sc (2-50) where ks is the conductor space factor, which allows for insulation and space between the turns. For simplicity, it will be assumed that the primary and secondary windings are placed side-by-side on the core, and that they have the same number of turns N. From figure 2-25, the mean turn length of each winding is:

babaalm 2)4(24 (2-51)


If the winding has N turns, then the total length of wire is Nlm, and the cross-sectional area of the wire is Ac / N. The winding resistance is therefore:

2

2

5.1}2)4{(

/ akbaN

NANl

AlR

sc

m

(2-52)

Temperature rise

If the RMS current in one winding is I, the power loss is I2R. The cooling surface area of the winding, from figure 2-25, is:

)410(4)24(5.1

2

22

a

aaaaaAs (2-53)

Substituting in equation 2-49 gives:

)25(3}2)4{(

4

222

akbaINk

ARIk

APkT

s

c

s

c

s

lc (2-54)

Thus, the current is given by:

}2)4{()25(3

2

42

baNkTak

Ic

s

(2-55)

From equation 2-10, the number of turns is:

mm

m

fabBV

fABVN

42

2 (2-56)

Worked example 2-3

A transformer has a primary wound for 230 V, and the core measures 60 50 30 mm. The maximum winding temperature is 90C, the ambient temperature is 30C, the winding space factor is 0.4, the cooling coefficient is 0.04 Km2/W, and the resistivity of copper at 90C is 21.9 nm. If the maximum flux density in the core is 1.4 T and the frequency is 50 Hz, determine (a) the number of turns on the primary, (b) the maximum current in the primary.

Solution From figure 2-26, a = 10 mm and b = 30 mm. From equation 2-56, we have:

12304.110301010504

2302

42

33

mfabBVN

Equation 2-55 gives:

}2)4{()25(3

2

42

baNkTakI

c

s

A 216.0}103021010)4{(

)1230(109.2104.0)3090)(25()1010(4.03

}2)4{()25(3

33

29

43

2

4

baNkTakI

c

s

Thus, I = 216 mA, so the transformer rating is 230 0.216 VA = 49.7 VA.

Rating and size

A relationship between the apparent power rating of the transformer and the dimensions can be obtained by substituting for N from equation 2-56 in equation 2-55:

}2)4{()25(32 3

bakTkbaB

VIc

sm

(2-57)

If it is assumed that the core depth b is proportional to the dimension a, and other quantities remain constant, then equation 2-57 gives the following relationship:

5.3aVI (2-58) From equation 2-58, if the dimensions of the transformer are doubled, the rating will increase by a factor of 11.3. A similar result is obtained for the increase in the power output of a DC machine when the dimensions are doubled see section 3.2.

DC Machines 15

3 DC MACHINES

3.1 Introduction Basic DC machine principles were covered in Electromechanics [1], and the main results are given below. Brushless DC machines, which were mentioned briefly in [1], are beyond the scope of this course but are covered in the Year 3 course Electrical Machine Drives.

Basic equations

The generated voltage and the developed torque are given by:

[V] rfa Ke (3-1)

[Nm] afd iKT (3-2)

where K is the armature constant, f is the field flux, and ia is the armature current. Note that the rotor angular velocity r must be in radians per second (rad/s) and not in rev/min. If the rotational speed is nr rev/s or Nr rev/min, then:

[rad/s] 60

22 rrrN

n (3-3)

In a permanent-magnet machine, the field flux f is fixed, but in a wound-field machine, it is a function of the field current if.

Magnetisation characteristic

From equation 3-1, if the speed r is held constant, the flux is proportional to the armature generated voltage ea. A graph of ea against if is known as the magnetisation characteristic of the machine, and a typical curve is shown in figure 3-1 for a 3 kW motor.

050

100150200250300350

0 0.2 0.4 0.6 0.8 1 1.2Field current (A)

Arm

atu

re vo

ltage

(V

)

Figure 3-1: Magnetisation characteristic.

In this example, the relationship is almost linear up to the rated field current of 0.5 A, but there is significant non-linearity above this value, when parts of the magnetic circuit saturate. There is also a small residual flux when the field current is zero, giving a corresponding generated voltage. For the initial part of the magnetisation characteristic, it is approximately true that f if. Equations 3-1 and 3-2 then become:

rfa iKe (3-4)

afd iiKT (3-5)

Armature equation

Figure 3-2 shows a symbolic representation of a DC machine, with the armature connected to a voltage source va.

if

iarva

+ea

+

Figure 3-2: DC machine with voltage source.

If the armature has a resistance Ra, then Kirchhoffs voltage law gives the armature voltage equation:

rfaaaaaa KiReiRv (3-6)


3.2 DC machines in practice

Slotted armature

The elementary theory of DC machines assumes that conductors are on the surface of the armature, so that the simple expressions e = Blu and f = Bli are applicable. In practice, the armature conductors are placed in slots, as shown in figure 3-3.

Figure 3-3: DC motor armature.

Figure 3-4 shows a simple model of a machine with a slotted armature, and figure 3-5 shows the corresponding flux plot when there is no armature current.

Figure 3-4: DC machine model.

Figure 3-5: DC machine field flux.

This is similar to the flux plot when conductors are on the surface, but with one important difference: most of the flux lines pass between the conductors, indicating that the flux density in the slots is very low. Consequently the force on the conductor, given by f = Bli, is very small. Most of the force is exerted on the armature iron and not on the conductors. The basic equations 3-1 and 3-2 are not affected by the location of the conductors. In particular, the generated voltage is not affected, although a direct application of e = Blu appears to contradict this. See reference [1] for a discussion of induced voltage in such situations.

Armature reaction

Current flowing in the armature conductors will also create a magnetic field in the machine, known as the armature reaction field. Figure 3-6 shows a flux plot of this field when the main field flux is absent.

Figure 3-6: Armature reaction flux plot.

DC Machines 17

Note that the axis of the armature reaction field is at right angles to the axis of the main field. When currents flow in the field and armature conductors, the two component magnetic fields combine to give a resultant field of the form shown in figure 3-6. The bending of the field lines indicates that the stator exerts a counter-clockwise torque on the rotor. Another way of picturing the electromagnetic action is to consider the magnetic poles representing the field and armature flux components, as shown in figure 3-7.

Figure 3-7: Total flux plot.

An important effect of armature reaction, which is evident in figure 3-7, is to increase the flux density at one side of a field pole and decrease it at the other side. Figure 3-8 shows a shaded plot of the flux density magnitude, where the colour range from blue to red represents the flux density range from minimum to maximum.

Figure 3-8: Flux density magnitude.

High values of flux density may result in local saturation of the steel, increasing the reluctance of the magnetic circuit, and reducing the value of the

field flux. This has important consequences for the motor characteristics see section 3.3.

Efficiency

The efficiency of a motor is defined in the usual way:

in

out

PP

(3-7)

where Pin is total electrical power input to the motor terminals, and Pout is the mechanical power output from the motor shaft. The power lost as heat in the motor is:

outinloss PPP (3-8) so we have the following alternative forms of equation 3-7:

lossout

loss

lossout

out

in

loss

in

lossin

in

out

PPP

PPP

PP

PPP

PP

1

1 (3-9)

As with transformers (see section 2-5), the efficiency of a large motor is not usually determined from equation 3-7 by direct measurement of the input and output power. Instead, losses are determined from several tests, and the efficiency calculated from equation 3-9. The output power is given by

)( ldrrout TTTP (3-10) where Tl is the rotational loss torque. It follows that the rotational power loss is rTl. The rotational loss has two components: mechanical loss, which is also known as the windage and friction loss, and core loss in the armature and the field poles resulting from the rotation of the armature. The total power loss includes in addition the I2R loss in the field and armature windings, and the brush contact loss, which results from the voltage drop between the brushes and the commutator segments. For further information, see references [3, 4].

S S

N N


Power output and size

The developed torque may be calculated from the equation f = Bli, even though the armature conductors are in slots. For this purpose, the machine will be represented by a simple model with the conductors on the surface, as shown in the flux plot of figure 3-9.

Figure 3-9: Model for torque calculation.

It is useful to define a current loading A as the current in amperes per metre length of circumference on the surface of the armature. The current in an element of length ds is then di = A ds. The force on this element is:

dsBlAdiBldf (3-11)

where l is the axial length of the armature. The corresponding contribution to the torque is:

dsrBlAdfrdTd (3-12) where r is the radius of the armature. If Bav is the average value of the flux density at the armature surface, then the total torque is just:

AVBAlrBrlArBT avavavd 222.2 (3-13)

where V is the volume of the armature. The maximum value of Bav is limited by the saturation of the magnetic material of the armature, and the value of the current loading A is limited by the I2R heating of the armature conductors. If it is assumed that the maximum value of A is independent of the machine size, equation 3-13 shows that the developed torque is proportional to the rotor volume. The treatment of the effects of scale in reference [1] also gives this result. The gross power output is given by:

AVBTP avrdr 2 (3-14) Thus, the maximum power output of a DC motor is roughly proportional to the product of the armature volume and the armature speed. In practice, the current loading A also increases with size, so the power output of large machines is further increased. To quantify this, assume that the power dissipation per unit surface area is constant. Let d be the radial depth of a conducting layer representing the armature conductors. The resistance of an element ds of this layer is:

dsdldR (3-15)

The cooling surface area of the element is dS = l ds, so the power dissipation per unit area is:

dA

dsldsdldsA

dsldRdi

dSdP 2

22 )()(

(3-16)

If dP / dS is constant, then A d, and if d r then A r. For geometrically similar machines, the axial length l is proportional to the radius r. If the rotational speed r is constant, it follows from equation 3-13 that the power output is proportional to r3.5. Thus, if the dimensions are doubled, the power output will increase by a factor of 11.3. A similar result was obtained for transformers (section 2.8), and it holds for other types of machine.

3.3 Characteristics and control

Speed control

In a large motor, the total power loss is small. Since the armature power loss Raia2 is only a part of the total loss, this term must be small in comparison with the armature input power vaia. It follows that:

aaa viR (3-17) So Raia may be neglected in equation 3-6, giving:

rfa Kv (3-18)

r ds

DC Machines 19

The rotational speed is therefore:

fa

r Kv

(3-19)

Equation 3-19 shows that the speed is independent of the torque, provided inequality 3-17 holds. Equation 3-19 is the basis of speed control. If the field flux f is constant, the rotational speed is proportional to the applied voltage. The normal values of va and f define the base speed r0, and the speed can be reduced to zero by varying va. Electronic controllers for DC motors deliver an adjustable voltage to the motor armature by phase-controlled rectification of the AC mains supply, using thyristors: see section 8. This is the basis of DC variable-speed drive systems, which are widely used in industry. In a wound-field motor, it is possible to increase the speed above the base speed by reducing f known as field weakening. Only a limited speed increase is possible, for the following reason. The torque is related to the armature current through equation 3-2:

afd iKT [3-2]

If f is reduced, there will be a compensating increase in ia to maintain the torque, and there is a risk of exceeding the current rating of the machine.

Small motors

In small motors, with power ratings below 1 kW, inequality 3-17 does not hold, so it is not permissible to neglect the Raia term. The rotational speed then depends on the developed torque, as may be seen by substituting for ia in terms of Td in equation 3-6:

rffda

rfaaa

KK

TR

KiRv

(3-20)

Thus, the speed is given by:

2)( fda

fa

r KTR

Kv

(3-21)

A graph of speed against torque is a straight line, as shown in figure 3-10. The no-load speed, which is the speed when the torque is zero, is given by:

fa

r Kv

0 (3-22)

r

Td

slope Ra

r

Figure 3-10: Speed-torque characteristic.

Effect of armature reaction

In section 3.2, it was noted that the armature reaction field could cause local saturation of the field poles, thereby reducing the value of the field flux f. This effect increases with the armature current ia, and therefore with the developed torque Td. From equation 3-19, a decrease in f will cause the speed r to rise. Armature reaction therefore has the opposite effect to armature resistance, which causes r to fall with increasing torque load. There is an important difference, however. The effect of resistance is linear, as illustrated in figure 3-8, but the effect of armature reaction is non-linear. At low values of armature current, the uneven distribution of flux density in the field pole is insufficient to cause saturation, so there is hardly any reduction in the field flux. At high values of current, on the other hand, there may be a significant reduction. It is possible, therefore, for the speed of a motor to fall with increasing load when the armature current is low, but to increase with load when the current is high. This increase in speed can be very undesirable, leading to instability with some kinds of load. A large DC motor generally includes some form of compensation for armature reaction. A simple method of compensation is to provide a second winding on the field poles, connected in series with the armature (figure 3-11), to increase the field MMF when the armature current increases. The resulting motor is known as a compound motor [3, 4]. However, this cannot


compensate for the non-linear nature of the armature reaction effect.

Figure 3-11: DC compound motor.

A better method, which is frequently used in high-power DC drives, is to use a compensating winding [4]. This takes the form of conductors embedded in slots in the field pole faces, connected in series with the armature. These conductors carry current in the opposite direction to the armature conductors, thereby cancelling the armature reaction flux.

Compensating winding

Figure 3-12 shows a model of a DC machine with a compensating winding, figure 3-13 shows the resulting flux plot, and figure 3-14 shows a shaded plot of the flux density magnitude. These plots should be compared with figures 3-7 and 3-8 for the machine without a compensating winding.

Figure 3-12: Compensating winding model.

Figure 3-13: Total flux plot.

Figure 3-14: Flux density magnitude.

With a compensating winding, the flux distribution in each field pole is symmetrical and uniform, so the effect of the armature reaction has been cancelled in this part of the machine. The torque exerted on the armature is not affected, however, because this depends on the interaction of the armature currents and the field flux.

Starting of DC motors

When the speed is zero, the armature generated voltage is also zero. From equation 3-6, the corresponding armature current is given by:

a

aa R

vi 0 (3-23)

S S

N N

DC Machines 21

This is the stalled armature current, which is much larger than the normal running current. In a small motor, it is permissible to connect the armature directly to a constant-voltage supply. The armature can withstand the stalled current for a short time, and it will accelerate rapidly. As it does so, the generated voltage will rise and the current will fall to its normal value. However, a large motor must not be started in this way because the armature resistance Ra is very low, and the stalled current would be large enough to cause serious damage. An electronic controller for speed control will limit the starting current to a safe value. If this is not available, a variable resistance must be connected in series with the armature and the value progressively reduced to zero as the armature accelerates.

Transient conditions

When the voltage applied to a DC motor is changed, the speed will not change instantly because of the inertia of the rotating system. If J is the moment of inertia and TL is the mechanical load torque, Newtons second law gives:

Ldr TT

dtdJ (3-24)

Mechanical loss torque is assumed included with TL. Substituting for Td in terms of ia from equation 3-2, and ia from equation 3-6, gives:

La

rfafr TR

KvKdt

dJ

)(

(3-25)

The solution of equation 3-25 gives the speed as a function of time after a change in the applied voltage. It should be noted that the derivation of equation 3-25 has neglected the inductance of the armature circuit. See reference [10] for the difference this makes to the transient behaviour. As an example, suppose that TL = 0 and the motor starts from rest with a suddenly applied voltage va = V. Equation 3-25 becomes:

fr

r

fa

KV

dtd

KJR

2)( (3-26)

The solution of equation 3-26 is:

/0/ 11 tr

t

fr eeK

V (3-27)

where r0 is the no-load speed and the time constant is given by

2)( fa

KJR

(3-28)

The armature current can be determined by substituting for r in equation 3-6:

)1( /

taa

rfaa

eViR

KiRV

(3-29)

/0

/ ta

t

aa eieR

Vi (3-30)

Figure 3-15 shows graphs of the normalised current ia / ia0 and the normalised speed r / r0 against the normalised time t / .

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5Time

Curr

ent,

Speed

Figure 3-15: Starting performance.

Speed r / r0

Current ia / ia0

t /


3.4 Shunt and series motors So far, it has been assumed that the field flux f is independent of the conditions in the armature. This is the case in a permanent-magnet motor, and in a wound-field motor where the field winding is supplied from a separate voltage source. The latter is sometimes termed a separately excited motor. There are two ways of introducing constraints between the armature and field of a wound-field motor. In a shunt motor, the field winding is connected in parallel with the armature. In a series motor, the field winding is designed to carry the full armature current, and it is connected in series with the armature.

Shunt motor

If the field winding is connected to the same supply voltage as the armature, the field current will be proportional to the armature voltage. Usually, a variable resistance is connected in series with the field winding to provide some measure of control for the field current [3, 4, 13]. The characteristics of a shunt motor may be deduced from equation 3-19:

fa

r Kv

[3-19]

Provided there is a linear relationship between the field flux and the field current, we have:

f af f f

f

k vk i

R (3-31)

where Rf is the total resistance of the field circuit and kf is a constant. Substituting in equation 3-19 gives:

far

f f

RvK Kk

(3-32)

Thus, the speed of a shunt motor is, to a first approximation, independent of the applied voltage.

Series motor

Figure 3-16 shows the connection of a series motor to a voltage source. The characteristics are readily deduced if the following assumptions are made: The field flux is proportional to the current. The resistance of the windings is negligible.

v

iea

+

r+

Figure 3-16: Series motor.

Equations 3-4 and 3-5 become:

2iKTd (3-33)

ra iKev (3-34) Eliminating i gives the relationship between speed and torque:

ddr TK

v

KT

K

v

iKv

(3-35)

A graph plotted from equation 3-33 for a small motor is shown in figure 3-17.

0

50

100

150

200

0 5 10 15 20 25Developed torque, Nm

Speed,

ra

d/s

Figure 3-17: Speed/torque characteristic.

DC Machines 23

Two features of the graph should be noted: The motor can develop a very large torque at

low speeds. When the developed torque is low, the speed is

very high. The high torque at low speeds is a useful feature, but the high speed associated with low torque can be dangerous. Consider what happens if the external load torque is removed. The developed torque then just supplies the mechanical losses in the motor. In a small motor, the mechanical losses are usually high enough to limit the no-load speed to a safe value. However, in motors with power ratings above about 1 kW, the losses are proportionately smaller, and the no-load speed would be dangerously high. Such motors must never be run without a load.

Effect of saturation

From equations 3-2 and 3-5, the torque developed by a series motor is given by:

2iKiKT fd (3-36) This equation is valid only when the current is low enough to avoid saturation of the magnetic circuit, so that the flux is proportional to current. When the current is very high, the field flux will tend towards a constant value fm, and the torque is then given by

iKT fmd (3-37) Under these conditions, the torque is proportional to i instead of i2. The performance of motors such as car starter motors must therefore be measured under actual operating conditions, where equation 3-36 would give very inaccurate results.

Applications

A correctly designed series motor will work on AC as well as DC [1, 3, 4]. Such motors are termed universal. They are widely used in products such as portable power drills, and in domestic appliances such as food processors and vacuum cleaners. In applications such as power tools, the speed/torque characteristic of the series motor is advantageous. An important reason for their use is the high speeds that are possible with series motors: speeds of up to 10 000 rev/min (1050 rad/s) are common in small motors. It was shown in section 3.2 that the maximum power output of a motor is roughly proportional to the product of the rotational speed and the armature (or rotor) volume. Thus, a small motor can deliver a large amount of power if the speed is high. Induction motors (see section 6) are limited to speeds below 3000 rev/min (314 rad/s) when operated from the 50 Hz mains supply. The power per unit volume for a series motor is therefore about three times the value for an induction motor. Disadvantages include the higher manufacturing cost, and the need for maintenance of the commutator and brushes.


4 INTRODUCTION TO AC MACHINES

4.1 Review of 3-phase systems Industrial AC motors use 3-phase alternating current to generate a rotating magnetic field from stationary windings. The 3-phase supply may be taken from the AC mains, or it may be generated electronically with an inverter. In either case, the requirement is a symmetrical set of sinusoidal currents with relative phase displacements of 120. Formally, 3-phase sets of currents and voltages may be defined as follows in the time domain:

)120cos()240cos()120cos(

)cos(

tItIitIitIi

m

mc

mb

ma

(4-1)

)120cos()240cos()120cos(

)cos(

tVtVvtVvtVv

m

mc

mb

ma

(4-2)

The quantities in equations 4-1 and 4-2 correspond to a positive phase sequence, in which the quantities reach their maximum values in the sequence a b c. If the connections to any two phases are interchanged, the effect is to reverse the phase sequence. For example, if b and c are interchanged, and primes denote the new phase labels, we have:

)120cos()120cos(

)cos(

tVvvtVvvtVvv

mbc

mcb

maa

(4-3)

These quantities reach their maximum values in the sequence c b a, corresponding to a negative phase sequence. Figure 4-1 shows the star connection of a 3-phase AC source to three lines, where the time-varying quantities of equation 4-1 are represented by complex (phasor) quantities in the usual way.

VaVb

Vc

1

23

+

++

V12+

Figure 4-1: 3-phase star connection.

The corresponding voltage phasor diagram is shown in figure 4-2.

VaVb

Vc

V12

V23

V31

Figure 4-2: Star connection: voltage phasors.

Figure 4-3 shows the delta connection of a 3-phase AC source to three lines, and figure 4-4 shows the corresponding current phasor diagram.

1

23

Ic

Ia

Ib

Figure 4-3: 3-phase delta connection.

IaIb

Ic

I1

I2

I3

Figure 4-4: Delta connection: current phasors.

Introduction to AC machines 25

From these diagrams, we have the relationships between the magnitudes of line and phase quantities given in table 4-1:

Table 4-1

Star Delta

phaseline VV 3 phaseline VV

phaseline II phaseline II 3

4.2 Rotating magnetic field In AC motors, the stationary part is termed the stator, and the rotating part is the rotor. Figure 4-5 shows the stator of a small AC motor with a 3-phase winding.

Figure 4-5: 3-phase AC motor winding.

Coils are arranged in slots in a laminated steel core, rather like the armature coils in a DC motor. The function of the winding is to route currents to slots from the three phases, as shown in the simplified diagram of figure 4-6. This diagram also shows a steel cylinder in the centre, representing the rotor of the motor.

Figure 4-6: 3-phase conductor groups.

The groups labelled a, b, c carry currents in the positive direction, into the plane of the diagram, and the groups labelled a', b', c' carry currents in the negative direction. All coils have the same shape, with one coil side at the bottom of a slot and the other side at the top of a slot, as shown by the pair of circles in figure 4-6. If phase a is energised on its own with positive current, the resulting magnetic field pattern is shown in figure 4-7(a). The corresponding patterns for phases b and c are shown in figures 4-7(b) and 4-7(c).

(a)

(b)

(c)

Figure 4-7: AC machine flux plots: (a) phase a, (b) phase b, (c) phase c.

a

a'

c

c'

b

b'


When all three phases are energised, the individual phase fields combine to give a resultant field. Consider the instant when t = 0. From equation 4-1, the currents in the three phases are:

mcmbma IiIiIi 2121 ,,

Since the currents in phases b and c are negative, the component fields are in the directions shown in figure 4-8. They combine to give a resultant in the same direction as for phase a, so a flux plot of the resultant field is the same as for phase a alone. As time advances, the currents in the phases change, and the resultant pattern changes. Figure 4-9 shows the patterns when t = 30, 60 and 90.

Figure 4-8: Component fields when t = 0.

(a)

(b)

(c)

Figure 4-9: Flux plots for values of t: (a) 30, (b) 60, (c) 90.

Notice that the magnetic field pattern has rotated by the same angle as the time phase of the currents. We have a rotating magnetic field, which makes one revolution for each cycle of the alternating current. This field is capable of exerting a torque on a suitably designed rotor. Two kinds of AC motor exploit the rotating field effect. In synchronous motors, the rotor has magnetised poles, which lock in with the rotating field, so that the rotor moves in synchronism with the field. With induction motors, the rotating field induces currents in conductors on the rotor, so the rotor must move more slowly than the rotating field.

a

c

b


Sinusoidal fields

The rotating magnetic field will induce voltages in the phase windings. These voltages should be as nearly sinusoidal as possible, which implies that the flux density should vary sinusoidally with angular position. This in turn requires that the currents producing the field should be distributed sinusoidally. A sinusoidal distribution is also required if the resultant field is to rotate at a uniform speed. The conductor distribution for each phase shown in figure 4-6 approximates to a sinusoidal distribution. It is a better winding than it appears, because harmonics that are multiples of 3 are missing from the Fourier series for the waveform. This is one of the many advantages of using three phases. Practical windings generally use more slots than this simple model, and give a better approach to the ideal sinusoidal distribution. Assume that the three phases produce component fields in the airgap with magnitudes Ba, Bb and Bc as follows:

)120cos()120cos(

cos

cc

bb

aa

kiBkiBkiB

(4-4)

where k is a constant and is the angle from a reference axis, as shown in figure 4-10 for the component field Ba.

Figure 4-10: Angular position definition.

The total field is the sum of the component fields. If the currents are given by equation 4-1, then substituting in equation 4-4 gives the result:

)cos()cos(23

tB

tkIBBBB

m

mcba

(4-5)

Equation 4-5 represents a sinusoidal field with a constant maximum value, rotating in a counter-clockwise direction with a constant angular velocity r, illustrated in figure 4-10. The axis of the field makes an angle t with the horizontal axis. At an angle , the displacement from the field axis is t, so the magnitude of the flux density is Bm cos( t).

Figure 4-11: Rotating magnetic field.

The rotating field represented by equation 4-5 will induce a sinusoidal voltage in any coil on the stator. Consider a single-turn coil on the stator, with one side at = 0 and the other side at = 180, as shown in figure 4-12.

Figure 4-12: Single-turn coil.

If the direction of the flux density in the airgap is assumed radial, it may be shown that the magnetic flux through this coil is:

tlrBm sin2 (4-6) where r is the radius and l is the axial length. The induced voltage in the coil is thus:

tlrBdtd

e m

cos2 (4-7)

See references [3, 4] for further information about sinusoidal fields.

t


Speed of the rotating field

At any instant, there are two effective magnetic poles for the field shown in figure 4-8, so this is termed a 2-pole field. It makes one revolution in one AC cycle, giving the relationships between the field and the current shown in table 4-2.

Table 4-2: 2-pole field properties.

Current waveform 2-pole field Frequency f [Hz] Rotational speed ns = f [rev/s] Rotational speed Ns = 60f [rev/min] Angular frequency Angular velocity s = [rad/s] As an example, if f = 50 Hz, ns = 50 rev/s, Ns = 3000 rev/min and s = 2f 314 rad/s

4.3 Multi-pole fields Coils in an AC machine winding can be arranged to produce fields with more than two poles. For example, figure 4-13 shows the simple winding of figure 4-6 re-arranged to give a 4-pole field, with the resulting field pattern shown in figure 4-14 at the instant when t = 0.

Figure 4-13: 4-pole winding.

Figure 4-14: 4-pole flux plot.

For every 30 advance in the time phase of the currents, the field pattern rotates 15. In one cycle of the supply, the field pattern moves through two pole pitches an angle corresponding to two poles. With any number of poles, the field moves through two pole pitches in one cycle of the supply. If there are p pairs of poles, this corresponds to a fraction 1 / p of a revolution. Thus, the general expressions for the speed of the rotating field are:

[rad/s] 2pf

ps (4-8)

[rev/s] pf

ns (4-9)

[rev/min] 60p

fN s (4-10)

Table 4-3 gives the field speeds at 50 Hz for different numbers of poles.

Table 4-3: Field speeds at 50 Hz.

Poles p s ns Ns rad/s rev/s rev/min 2 1 314 50 3000 4 2 157 25 1500 6 3 105 16.7 1000 8 4 78.5 12.5 750 10 5 62.8 10 600

a

a'

c

c'

b

b'

b'

b

c

a

c'

a'


Reversing the direction of rotation

To reverse the direction of rotation, all that is required is to interchange the connections to any two of the three phases. For example, suppose that the connections to phases b and c are interchanged. Figure 4-15(a) shows the original current pattern and figure 4-15(b) shows the new pattern.

(a)

(b)

Figure 4-15: Interchange of connections: (a) original, (b) b and c interchanged.

The new pattern is a mirror image of the old, so the resultant magnetic field will progress in the opposite direction as time advances. It follows that the direction of rotation depends on the phase sequence of the 3-phase supply, since interchanging any pair of phases has the effect of reversing the phase sequence (see section 4-1). Therefore, it is important to know the phase sequence of the supply before connecting the motor.

a

a'

c

c'

b

b'

b'

b

c

a

c'

a'

a

a'

b

b'

c

c'

c'

c

b

a

b'

a'


5 SYNCHRONOUS MACHINES

5.1 Introduction In a synchronous machine, the rotor is magnetised and it runs at the same speed as the rotating magnetic field. Permanent-magnet rotors are common in small machines, so the machine structure is similar to that of the brushless DC motor shown in figure 3-4. Figure 5-1 is a simplified model of the structure of this type of machine, where the rotor has surface-mounted segments of permanent-magnet material. Other forms of rotor with embedded magnets are also possible.

Figure 5-1: 2-pole PM synchronous machine.

In larger sizes, a synchronous machine has a field winding on the rotor instead of permanent magnets. Direct current for the rotor excitation can be supplied through sliprings and brushes, but large machines normally have a brushless excitation system [3, 4]. The AC stator of a synchronous machine is termed the armature. It handles the main electrical power, so it has a similar function to the rotating armature of a DC machine.

Motors and generators

As with DC machines, there is no fundamental difference between a synchronous motor and a synchronous generator. In a motor, the magnetic axis of the rotating magnetic field is ahead of the magnetic axis of the rotor, resulting in a positive torque that depends on the displacement between

the two axes. The flux plots in figure 5-2 show (a) the field produced by stator currents at a particular instant of time, (b) the field produced by the rotor magnets, (c) the resultant field when both sources are active.

(a)

(b)

(c)

Figure 5-2: Synchronous motor flux plots: (a) stator field, (b) rotor field, (c) resultant field.

Synchronous Machines 31

In this example, the displacement angle is 90, which gives the maximum torque for a given current. The current has been chosen so that the axis of the resultant field is at approximately 45 to the magnetic axis of the rotor. In a synchronous generator, the displacement is reversed: the magnetic axis of the rotor is ahead of the magnetic axis of the rotating field, so the torque is negative. This is illustrated in figure 5-3.

(a)

(b)

(c)

Figure 5-3: Synchronous generator flux plots: (a) stator field, (b) rotor field, (c) resultant field.

Most of the AC generators in electric power systems are synchronous machines. High-speed turbine generators normally have two poles. The rotor is made from a cylindrical steel forging, with the field winding embedded in slots machined in the steel. Apart from the slots for conductors, the active surfaces of the stator and rotor are cylindrical, so these are uniform-airgap or non-salient machines. Low-speed hydro generators have many poles. These are salient-pole machines, where the poles radiate like spokes from a central hub. Synchronous generators will not be considered in detail in this course, but will be studied in the context of power systems in year 3. Large synchronous motors are widely used as high-efficiency constant-speed industrial drives, where they can also be used for plant power factor correction: see section 5.2. These are normally salient-pole machines, and they generally have four or more poles. The main features of synchronous machine operation can be deduced from the simple theory of non-salient machines. An introduction to salient-pole machines is given in section 5.3.

5.2 Characteristics

Circuit model

A non-salient synchronous machine can be represented by a simple equivalent circuit [3, 4], shown in figure 5-4.

jXsRaI

V

+

E+

V

+

Figure 5-4: Equivalent circuit. This circuit represents one phase of a 3-phase machine. The voltage V is the phase voltage at the machine terminals, and the current I is the corresponding phase current. Other elements in the circuit have the following significance.


The voltage E is termed the excitation voltage. It represents the voltage induced in one phase by the rotation of the magnetised rotor, so it corresponds to the magnetic field of the rotor shown in figure 5-2(b) or 5-3(b).

The reactance Xs is termed the synchronous reactance. It represents the magnetic field of the stator current in the following way: the voltage jXsI is the voltage induced in one phase by the stator current. This voltage corresponds to the magnetic field of the stator shown in figure 5-2(a) or 5-3(a).

The voltage V = E + jXsI represents the voltage induced in one phase by the total magnetic field shown in figure 5-2(c) or 5-3(c).

The resistance Ra is the resistance of one phase of the stator, or armature, winding.

The resistance Ra is usually small in comparison with the reactance Xs, so it may be neglected in most calculations from the equivalent circuit. The resulting approximate equivalent circuit is shown in figure 5-5, described by the equation:

EIV sjX (5-1)

jXsI

V

+

E+

Figure 5-5: Approximate equivalent circuit.

Phasor diagram

Figure 5-6 shows a phasor diagram corresponding to the equivalent circuit of figure 5-5, for the machine operating as a generator. The terminal voltage V lags the excitation voltage E by an angle . This is known as the load angle, because it varies with the torque load. In terms of the magnetic field, represents the angle between the axis of the rotor field in figure 5-3(b) and the axis of the resultant field in figure 5-3(c). The phase angle is greater than 90, so the power given by VI cos is negative, indicating a flow of electrical power out of the machine.

I

E

jXsI

V

I

Figure 5-6: Phasor diagram: generator.

In power system studies, it is customary to reverse the reference direction for the current, so the current is then represented by the phasor I in figure 5-6. For operation as a motor, V leads E by an angle , as shown in figure 5-7. The phase angle is now less than 90, indicating a flow of electrical power into the machine.

I

EjXsI

V

M

N

Figure 5-7: Phasor diagram: motor.

Torque characteristic

Consider the line MN in figure 5-7, which is perpendicular to V. The length of MN can be expressed in terms of E and from the left-hand triangle, and in terms of XsI and from the right-hand triangle. We have:

cossin IXEMN s (5-2) Multiplying both sides of equation 5-2 by the voltage magnitude V and dividing by Xs gives:

cossin VIX

VEs

(5-3)


The right-hand side of equation 5-3 is the electrical power input to one phase of the motor. Since energy is conserved, and there are no electrical losses the resistance has been neglected the total electrical power input must equal the work done by the rotating magnetic field. If s is the synchronous speed in rad/s and Td is the developed torque, then:

sds

TVIX

VE cos3sin3 (5-4)

The developed torque is thus given by:

[Nm] sin3ss

d XVET

(5-5)

This has a maximum value when = 90, given by:

ss

d XVET

3

max (5-6)

Equation 5-5 may therefore be written in normalised form:

sinmax

d

d

TT

(5-7)

A graph of this normalised torque is shown in figure 5-8. Positive values of represent motor operation; negative values represent generator operation.

-1

-0.5

0

0.5

1

-180 -90 0 90 180Load angle, degrees

Norm

alis

ed to

rque

motor

generator

Figure 5-8: Synchronous machine torque.

If the mechanical load on a synchronous motor exceeds Tdmax, the rotor will be pulled out of synchronism with the rotating field, and the motor will stall. Tdmax is therefore known as the pullout torque. The torque characteristic shown in figure 5-8 has an important practical consequence. If the rotor

loses synchronism with the rotating field, the load angle will change continuously. The torque will be alternately positive and negative, with a mean value of zero. Synchronous motors are therefore not inherently self-starting. Induction machines, considered in section 6, do not have this limitation, and the induction principle is generally used for starting synchronous motors.

Excitation control

In a wound-field synchronous machine, the excitation voltage E can be controlled by varying the excitation current in the rotor field winding. This has consequences for the machine characteristics, as will now be shown for the case of a synchronous motor. If the torque load on a synchronous motor is held constant, the power output will be constant, and if losses are neglected, there will be a constant electrical power input per phase given by:

cosVIP (5-8) In normal operation the terminal voltage V is constant, so the quantity I cos must be constant. In figure 5-9, the line AB is drawn parallel to the imaginary axis, and its distance from the axis is I cos. This line is the locus of the current phasor I. The length of the line MN is XsI cos, which is also constant, so the line CD, parallel to the real axis, is the locus of the excitation voltage phasor E.

I

EjXsI

V

M

N

A

B

C D

Figure 5-9: Locus diagram for constant load.


If the value of E is progressively increased, by increasing the current in the rotor field winding, the point M will move towards D, and the angle will decrease. Figure 5-10 shows the condition where = 0. This is the unity power factor condition, which gives the minimum value for I. Let E0 be the value of E that gives this condition, and 0 the corresponding value of .

I

E0jXsI

V

M

N

A

B

C D

Figure 5-10: Unity power factor condition.

When E < E0, is negative, and > 0. The machine takes a lagging current, as shown in figure 5-9. In this condition, the machine is said to be under-excited. As the value of E is progressively reduced, the values of I and increase, until the limit of stability is reached when = 90. This condition is shown in figure 5-11.

I

E jXsI

V

A

B

C D

Figure 5-11: Stability limit.

When E > E0, is positive, and < 0. The machine now takes a leading current, as shown in figure 5-12. In this condition, the machine is over-excited. If continuous operation at a leading power factor is required, as in power factor correction (discussed below), then the machine must be

specially designed for the higher values of rotor and stator current.

I

E

jXsIV

M

N

A

B

C D

Figure 5-12: Leading power factor condition.

If the mechanical load is removed from an over-excited synchronous motor, then = 0, and the phasor diagram takes the form shown in figure 5-13. The phase angle is now 90, so the machine behaves as a 3-phase capacitor. In this condition, it is known as a synchronous compensator. The magnitude of the current, and hence the effective value of the capacitance, depends on the difference between E and V.

I

EjXsIV

Figure 5-13: Synchronous compensator.

Power factor correction

The ability of a synchronous motor to operate at a leading power factor is extremely useful. It will be shown in section 7 that induction motors always operate at a lagging power factor. Many industrial processes use large numbers of induction motors, with the result that the total load current is lagging. It is possible to compensate for this by installing an over-excited synchronous motor. This may be used to drive a large load such as an air compressor, or it may be used without a load as a synchronous compensator purely for power factor correction.


When a synchronous motor is used in this way, the rotor excitation is normally controlled automatically to keep the total current nearly in phase with the voltage, so that the plant as a whole operates at a power factor close to unity. It may not be economic to give full correction, if this would require a very large synchronous machine. Figure 5-14 shows a phasor diagram for the general case of power factor correction. The synchronous machine current Is combines with the load current Il from the rest of the plant to give a total current It which is more nearly in phase with the voltage V.

Is

V

Il

It

Figure 5-14: Power factor correction.

The relationships between the currents can be expressed in terms of power, as follows. We have:

slt III (5-9) The complex power is given by:

slslsltt SSVIVIIIVVIS **** )( (5-10) Equating the real and imaginary parts of equation 5-10 gives:

slt PPP (5-11)

slt QQQ (5-12) For an over-excited synchronous machine, Qs will be negative, so equation 5-12 shows that the overall reactive power Qt will be less than the reactive power Ql of the original plant. For complete power factor correction, we require Qt = 0, and therefore Qs = Ql. If the synchronous machine is run as a synchronous compensator, without a mechanical load, then Ps 0, and the real power demand of the plant is unchanged.

5.3 Salient-pole machines The theory given in section 5.2 is only valid for non-salient machines. This section gives a brief introduction to salient-pole machines; for further information, see references [3, 4]. When the rotor has salient poles, the airgap is not uniform, and there is a preferred direction for magnetic flux in the machine. Figure 5-15 shows a simple model of a salient-pole machine, where part of the rotor steel has been removed to accommodate the field winding.

Figure 5-15: Salient-pole machine model.

The rotor has a path of easy magnetisation, known as the direct-axis or d-axis, where the reluctance is low. This is also the axis of the magnetic flux produced by the rotor field current. An axis at right angles, known as the quadrature-axis or q-axis, has a higher reluctance because of the larger airgap. Figure 5-16 shows the armature flux in a non-salient machine produced by a particular set of currents in the armature (stator). The flux axis is at 45 to the rotor d-axis.

Figure 5-16: Non-salient armature flux.

d-axis

q-axis


Figure 5-17 is a flux plot for same current pattern in the salient-pole model. Saliency has the effect of moving the axis of the armature flux towards the d-axis.

Figure 5-17: Salient-pole armature flux.

Voltage harmonics

In a non-salient machine, the flux distribution shown in figure 5-16 is approximately sinusoidal (see section 4.2). In a salient-pole machine, however, figure 5-17 shows that the flux density distribution is non-sinusoidal. It follows that the voltages induced in the armature phases will also be non-sinusoidal. The voltages may be represented by Fourier series:

...3cos)120cos(...)120(3cos)120cos(

...3cos)120cos(...)120(3cos)120cos(

...5cos3coscos

31

31

31

31

531

tVtVtVtVv

tVtVtVtVv

tVtVtVv

mm

mmc

mm

mmb

mmma

(5-13) If the windings are connected in star (figure 4-1), the third harmonics will cancel. For example, the voltage between lines 1 and 2 is:

...)120cos(cos...3cos)120cos(

...3coscos

11

31

31

12

tVtVtVtV

tVtVvvv

mm

mm

mm

ba

(5-14)

However, if the windings are connected in delta (figure 4-2), the third harmonic voltages will accumulate round the delta:

...3cos3...3cos)120cos(...3cos)120cos(

...3coscos

3

31

31

31

tVtVtVtVtV

tVtVvvvv

m

mm

mm

mm

cba

(5-15)

This can result in a large third-harmonic circulating current. Similar considerations apply to higher harmonics where the harmonic number is a multiple of 3; these are known as triplen harmonics. Therefore, it is normal practice to connect the windings of a synchronous machine in star. It is possible to eliminate any harmonic from the voltage waveform by using short-pitched coils in the stator winding. Short-pitched coils span less than a pole pitch; the coils shown in figure 4-6 are of this form. It may be shown that, if the coil is short-pitched by a fraction 1/n of the pole pitch, then the nth harmonic will be eliminated from the voltage waveform. For example, in a 2-pole machine, if the coils span 120 instead of 180, so that they are short-pitched by 1/3, the third harmonic will be eliminated. In this particular case, it is permissible to connect the windings in delta. It is more usual, however, to short-pitch by 1/6, as shown in figure 4-6, since this reduces the fifth and seventh harmonics to low levels. If the windings are star connected, all the triplen harmonics are eliminated, so the resulting voltage waveform is a good approximation to a sine wave.

Salient-pole machine theory

It is possible to analyse the salient-pole machine by resolving the armature flux into two components: a component acting along the rotor d-axis, and a component acting along the q-axis [3, 4]. This is represented in the phasor diagram by resolving the current I into components Id and Iq. The component Id acts on the d-axis, where the low reluctance gives a correspondingly high reactance Xd. The component Iq acts on the q-axis, where the


high reluctance gives a low reactance Xq. The resulting phasor diagram is shown in figure 5-18, and the phasor equation is:

qqdda jXjXR IIIEV (5-16)

RaI

E

V

Iq

jXqIq

IId jXdId

Figure 5-18: Salient-pole ph

electrical machines and systems course notes by j d edwards_very good figures

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