Lecture notes on Electrical machines

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Course in Electrical Machines and Systems Year 2 ©2004 J D Edwards 1 INTRODUCTION This course follows on from the Term 3 course Magnetic field plots Electromechanics. Its purpose is to explore in A magnetic field plot is often a useful way of greater depth the AC and DC machines that were picturing the operation of an electromagnetic introduced in the earlier course. The approach is device. Numerous plots have been specially that of the application engineer rather than the prepared for these notes, using the MagNet machine designer, concentrating on the basic electromagnetic simulation software, to develop principles, characteristics, and control. Since the basic concepts with a minimum of induction motors account for more than 90 per cent mathematics. of the motors used in industry, the course gives particular emphasis to these machines. References Course components References to books are listed in section 10, and cited in the text of the notes with the reference The course has three closely linked components: number in square brackets. lectures, problem sheets and laboratories. In addition, there is a design assignment, which introduces some of the basic ideas and problems of Background material design by considering a very simple device: an electromagnet. The course assumes a familiarity with the contents of the Term 3 course Electromechanics, so the Lectures will use video presentation and basic principles covered in that course will not be practical demonstrations. These notes provide repeated. Students are expected to have a copy of support material for the lectures, but they are not a the printed notes for Electromechanics 1, and substitute. Regular lecture attendance is essential. further information will be found in references 2 Problem solving is a vital part of the course. to 4. 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. Introduction 1 Sinusoidal operation 2 TRANSFORMERS If the voltage source is sinusoidal, then the core flux will also be sinusoidal, so we may put: 2.1 Introduction   sint (2-6) m Basic transformer principles were covered in Substituting this expression in equation 2-3 gives: Electromechanics 1, and the main results are given below. Figure 2-1 is a schematic d v  N  N  costV cost (2-7) 1 1 1 m 1m representation of a single-phase transformer with dt two coils on a magnetic core, where the magnetic Thus the maximum primary voltage is: coupling is assumed to be perfect: the same flux  passes through each turn of each coil. V  N   2 fN   2 fN AB (2-8) 1m 1 m 1 m 1 m i i 1 2 where A is the cross-sectional area of the core and B is the maximum flux density in the core. A  m + + + typical value for B is 1.4 T for the silicon steel m + characteristic in figure 2-2. v 1 e e R R Z 1 2 1 2 v 2 2 1.8 1.6 primary secondary 1.4 N turns N turns 1 2 1.2 1 Figure 2-1: Transformer with source and load 0.8 0.6 0.4 Voltage relationships 0.2 0 Kirchhoff’s voltage law applied to the two 0 2 4 6 8 10 12 14 16 18 20 windings gives: Magnetic intensity H, kA/m d v  e  R i  N  R i (2-1) 1 1 1 1 1 1 1 dt Figure 2-2: Silicon transformer steel. d v  e  R i  N  R i (2-2) 2 2 2 2 2 2 2 dt Current relationships If the resistances R and R are negligible, then 1 2 The relationship between the primary and equations 2-1 and 2-2 become: secondary currents can be found by considering the d magnetic circuit of the transformer. From the basic v  N (2-3) 1 1 dt magnetic circuit equation, we have: F  N i  N i  R (2-9) d 1 1 2 2 v  N (2-4) 2 2 dt In a well-designed transformer, the reluctance R is small, so equation 2-9 becomes: Dividing these equations gives the important result: N i  N i  0 (2-10) v N 1 1 2 2 1 1  (2-5) v N 2 2 This gives the counterpart of equation 2-5 for voltage: i N 1 2  (2-11) i N 2 1 2 Electrical Machines and Systems Course Notes Flux density B, T If sinusoidal voltages and currents are represented Unlike the normal transformer with two windings, by phasors, the corresponding forms for the basic known as a double-wound transformer, the auto- voltage and current equations are: wound transformer does not provide electrical isolation between the primary and the secondary, V N 1 1 However, an auto-wound transformer can have a  (2-12) V N much larger apparent power rating than a double- 2 2 wound transformer of the same physical size. I N 1 2 Let N be the number of turns on the upper part 1  (2-13) of the winding in figure 2-3, and N the number of I N 2 2 1 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 Transformer rating two separate windings with a common connection. The maximum voltage at the primary terminals of a Applying Kirchhoff’s law to this circuit gives: transformer is determined by equation 2-8, and is V  V  V (2-14) S 1 2 independent of the current. The maximum primary 2 current is determined by the I R power loss in the I  I  I (2-15) L 1 2 resistance of the transformer windings, which generates heat in the transformer. This power loss As an example, suppose that N = N . If the 1 2 is independent of the applied voltage. transformer is regarded as ideal (see section 2.3), Consequently, for a given design of transformer, then I = I and V = V . Equations 2-14 and 2-15 1 2 1 2 there is a maximum value for the product V I at 1 1 give: the primary terminals. To a first approximation, V  V  V  2V  2V (2-16) S 1 2 2 L this is also equal to the product V I at the 2 2 secondary terminals. This maximum value does not I  I  I  2I  2I (2-17) L 1 2 1 S depend on the phase angle between the voltage and the current. Transformer ratings therefore specify where V is the voltage across the load and I is the L S the apparent power VI (volt-amperes, VA) rather current supplied by the source. This auto-wound than the real power VI cos  (watts, W). 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. 2.2 Types of power transformer An elegant application of the auto-wound transformer principle is the variable transformer, In addition to the ordinary single-phase power which has a single-layer coil wound on a toroidal transformer, two other types are in common use: core. The output is taken from a carbon brush that auto-wound transformers, and 3-phase makes contact with the surface of the coil; the transformers. brush can be moved smoothly from one end of the coil to the other, thus varying the output voltage. Auto-wound transformers Examples of variable transformers are shown in figure 2-4. 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. I 1 + + V Z 1 L V S + I L I 2 V 2 Figure 2-3: Auto-wound transformer. Transformers 3   cost a m (2-18)   cos(t120) b m   cos(t 240) cos(t120) c m m Figure 2-7 shows flux plots for the transformer at the instants when t = 0, 120º and 240º. (a) Figure 2-4: Variable transformers. (RS Components Ltd) 3-phase transformers In 3-phase systems (see section 4.1), it is common (b) 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. (c) Figure 2-5: 3-phase transformer model. Figure 2-7: 3-phase transformer flux plots: (a) 0º, (b) 120º, (c) 240º. The corresponding fluxes are shown in figure 2-6. There is no requirement for another limb to form a flux return path, because the fluxes  ,  and  a b c sum to zero in a balanced 3-phase system. The proof is as follows. From equation 2-18, the sum is given by:    a b c     a b c m  cost cos(t120) cos(t120) (2-19)  cost 2cost cos120  cost cost 0 Figure 2-6: 3-phase transformer flux. In a balanced system with sinusoidal phase Because the fluxes in the three limbs sum to zero at voltages, the fluxes will be given by: all instants of time, there is no leakage of flux from the core, as the flux plots in figure 2-7 demonstrate. 4 Electrical Machines and Systems Course Notes I I 2.3 Ideal transformer properties 1 2 + If the primary and secondary windings have zero + resistance, and the magnetic core has zero V 1 Z V L 2 reluctance, then the approximate equalities in equations 2-12 and 2-13 become exact equalities. This leads to the concept of an ideal transformer N :N 1 2 element, to accompany the other ideal elements of Figure 2-9: Ideal transformer with a load. circuit theory. Figure 2-8 shows a circuit symbol for the ideal transformer element. The secondary impedance is given by: i i 1 2 V 2 Z  (2-22) L + + I 2 v v 1 2 At the primary terminals, the circuit presents an impedance given by: Figure 2-8: Ideal transformer element N V 1 2 The voltage and current relationships in the time V N 1 2 Z   and frequency domains are given in table 2-1. in N I I 2 2 1 (2-23) Table 2-1: Ideal transformer relationships. N 1 2 2 Time domain Frequency domain  N  V  N  Z 1 2 1 L     Z  L     2 v N V N N I N n 2 2 2 2 2 2 2     (2-20)   n   n v N V N 1 1 1 1 Thus the combination of an ideal transformer of ratio n and an impedance Z can be replaced by an L i N I N 2 1 2 1 2 (2-21) equivalent impedance Z / n .   n   n L i N I N 2 1 2 1 The following properties of the ideal transformer Referred impedances may be deduced from equations 2-16 and 2-17: Figure 2-10(a) shows an ideal transformer with a  The voltage transformation is independent of load impedance Z connected to the secondary. L the current, and vice versa. Another impedance Z is in series with Z . The 2 L  If the secondary is short-circuited, so that input impedance of this circuit is: v = 0, the primary terminals appear to be 2 Z  Z short-circuited since v = 0. 1 2 L Z  (2-24) in 2  If the secondary is open-circuited, so that n i = 0, the primary terminals appear to be open- 2 The input impedance of the circuit in figure 2- circuited since i = 0. 1 10(b) is:  The output power is equal to the input power, Z so there is no power loss in the element. L  Z  Z  (2-25) in 2 2 n Impedance transformation The two expressions for Z will be identical if: in Z The ideal transformer has the important property of 2  Z  (2-26) 2 2 transforming impedance values in a circuit. n Consider an ideal transformer with an impedance Z connected to its secondary terminals, as shown L in figure 2-9. Transformers 5 Z 2.4 Circuit model of a transformer 2 In a practical transformer, the winding resistances and the core reluctance are not zero. In addition, Z in Z L 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 N :N 1 2 (a) the equivalent circuit 3, 4 shown in figure 2-12. ' jx jx I R I 1 I = nI 2 1 2 2 2 2 R Z 1 2 I 0 + + I I 0m 0c Z in Z L V jX V R 1 m 2 c N :N 1 2 (b) N :N 1 2 Figure 2-10: Referred impedance – 1. The impedance Z' is termed the secondary 2 Figure 2-12: Transformer equivalent circuit. impedance Z referred to the primary. 2 This circuit is based on the ideal transformer In a similar way, a primary impedance Z can 1 element, with additional circuit elements to be referred to the secondary, as shown in figure represent the imperfections. The resistances R and 1 2-11. In this case, the referred impedance is given R represent the physical resistances of the 2 by: windings, and R represents the power lost in the c 2 core. The reactance X , known as the magnetising m Z n Z (2-27) 1 1 reactance, allows for the current required to magnetise the core when the reluctance is not zero. Z 1 Reactances x and x , known as leakage 1 2 reactances, represent the leakage flux that exists Z in when the magnetic coupling between the primary Z L 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 N :N 1 2 (a) a time is energised. In practice, the leakage is much less than this, so the leakage reactances are '' Z 1 normally very much smaller than the magnetising reactance X . m Z in Z L N :N 1 2 (b) Figure 2-11: Referred impedance – 2. The concept of referred impedance is often a useful (a) 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. 6 Electrical Machines and Systems Course Notes jx 1 jx R 2 R 1 2 jX R m c (b) N :N 1 2 Figure 2-13: Transformer leakage flux: (a) left coil energised, (b) right coil energised. (a) In the circuit of figure 2-12, the current I is 2 the effective value of the secondary current as seen jx jx 1 R R 2 1 2 from the primary side of the transformer. It is also known as the secondary current referred to the primary. The current I is the no-load current, which is 0 jX m R c the current taken by the primary when there is no load connected to the secondary. It has a component I , known as the magnetising current, 0m which represents the current required to set up the N :N 1 2 magnetic flux in the core. The current I is the 0c core loss component of the no-load current. (b) jx e R e Approximate equivalent circuit For power transformers with ratings above 100 VA, the values of the series elements R and x 1 1 jX R m c are generally much smaller than the shunt elements R and X . Under normal working conditions, the c m voltage drop in R + jx is much smaller than the 1 1 applied voltage V . Similarly, the no-load current I 1 0 N :N 1 2 is much smaller than the load current I . It follows 1 (c) that the shunt elements can be moved to the input Figure 2-14: Approximate equivalent circuit. terminals, as shown in figure 2-14(a), with very little loss of accuracy. The secondary elements R and jx can be 2 2 2 replaced by equivalent elements R = R / n and 2 2 2.5 Parameter determination 2 x = x / n on the primary side (see section 2.3), 2 2 The parameters of the approximate equivalent giving the circuit shown in figure 2-14(b). Finally, circuit (figure 2-14) can be determined the series elements can be combined to give an experimentally from two tests: effective resistance R and leakage reactance x , as e e  An open-circuit test, where the secondary is shown in figure 2-14(c), where the values are: left unconnected and the normal rated voltage R x 2 2 is applied to the primary. R  R  , x  x  (2-28) e 1 e 1 2 2 n n  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. Transformers 7 Open-circuit test Short-circuit test With the secondary unconnected, I = 0, so the If the secondary terminals are short-circuited, the 2 equivalent circuit reduces to the form shown in ideal transformer in figure 2-14 can be replaced by figure 2-15. a short circuit, so the equivalent circuit takes the form shown in figure 2-16(a). In a typical power I 1oc transformer, the shunt elements R and X are at c m least 100 times larger than the series elements R e + and x . Consequently, the shunt elements can be e neglected, and the circuit reduces to the form V 1oc jX R m c shown in figure 2-16(b). jx e R e Figure 2-15: Open-circuit test. jX m R c To a very close approximation, the current I 1oc supplied to the primary is equal to the no-load current I in figure 2-12. The values of the 0 (a) elements R and X can be determined from c m measurements of the input voltage V , current I 1oc 1oc jx e and power P as follows. The input power is I 1 1oc sc R e entirely dissipated in the resistance R , giving: c + 2 V 1oc R  (2-29) c P V 1sc 1oc The input impedance of the circuit is given by: (b) V 1 1oc Z   (2-30) 1oc 1 1 I 1oc  Figure 2-16: Short-circuit test. R jX c m The values of the elements R and x can be e e In terms of magnitudes, equation 2-30 becomes: determined from measurements of the input voltage V , current I and power P as follows. The 1sc 1sc 1sc V 1 1oc Z   (2-31) input power is entirely dissipated in the resistance 1oc I 1 1 1oc R , giving:  e 2 2 R X c m P 1sc R  (2-34) e 2 Re-arranging equation 2-31 gives the value of X : m I 1sc 1 The input impedance of the circuit is given by: X  (2-32) m 2   I 1 1oc V    1sc   Z   R  jx (2-35) 2 1sc e e V R  1oc  c I 1sc From figure 2-14(c), it follows that the turns ratio In terms of magnitudes, equation 2-35 becomes: is given by: V 2 2 1sc Z   R  x (2-36) N V 1sc e e 2 2oc n  (2-33) I 1sc N V 1 1oc 8 Electrical Machines and Systems Course Notes Voltage regulation and efficiency Re-arranging equation 2-36 gives the value of x : e When a load is connected to the secondary of a 2   V 1sc 2 transformer, there will be a voltage drop in the   X   R (2-37) e e   I series elements R and x , so the secondary terminal e e  1sc  voltage will change. The voltage regulation is In practice, the open-circuit test is usually defined as: made on the low-voltage side of the transformer to V V minimise the value of V , and the short-circuit test 2nl 2 fl oc   (2-38) is made on the high-voltage side to minimise the V 2nl value of I . The resulting parameter values are then sc where V is the magnitude of the no-load referred to the primary side of the transformer. 2nl secondary terminal voltage, and V is the 2fl corresponding full-load voltage. 2.6 Transformer performance Power is lost as heat in the windings and core of the transformer, represented by the resistance Consider a transformer with an impedance Z L elements R and R in the equivalent circuit. The e c connected to the secondary. With the approximate efficiency is defined in the usual way as: equivalent circuit, this may be represented by the circuit diagram of figure 2-17. P out   (2-39) P jx in e R e where P is the power input to the primary and P in out is the power output from the secondary. The power Z = lost as heat in the transformer is: L jX m R c R + jX L L P  P  P (2-40) loss in out so we have the following alternative forms of equation 2-39: N :N 1 2 Figure 2-17: Transformer with a load. P P  P P out in loss loss     1 P P P in in in The load impedance can be referred to the primary (2-41) side of the ideal transformer element, giving the P P out loss   1 circuit shown in figure 2-18. P  P P  P out loss out loss jx e I R e I = nI 1 2 2 Large transformers are very efficient. Even a I 2 kVA transformer can have an efficiency of about 0 95%. Above 25 kVA, the efficiency usually + + exceeds 99%. It is very difficult to make an Z = L V 1 jX 2 m R c Z / n accurate measurement of efficiency by direct L V = V / n 2 2 measurement of P and P , since this would out in 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 Figure 2-18: Circuit with referred impedance. efficiency from one of the alternative expressions in equation 2-41. The losses can be calculated with This circuit is easily solved for the currents I and 0 high accuracy from the equivalent-circuit I . The referred secondary voltage is given by 2 parameters determined from tests on the V = Z I , and the secondary terminal quantities 2 L 2 transformer. are given by V = nV , I = I / n. 2 2 2 2 Maximum efficiency The power loss in a transformer has two 2 components: the core loss, given by V / R , and 1 c Transformers 9 2 2 the I R loss, given by I' R . The core loss will be (b) the secondary terminal voltage magnitude, e constant if the primary voltage V is constant, but 1 (c) the primary current magnitude, 2 the I R loss will vary with the secondary current. (d) the voltage regulation, When the current is low, the output power will be (e) the efficiency of the transformer. low, but the core loss remains at the normal value. (a) Secondary current Consequently, the efficiency of the transformer will be low under these conditions. It may be The load impedance referred to the primary is: shown that the efficiency is a maximum when the 2 Z 6.0 j2.5 secondary current is such that the variable I R loss L  Z    26.9 j11.2 L 2 2 is equal to the fixed core loss. This result also n (0.472) applies to other devices where the losses have The secondary current referred to the primary is: fixed and variable components. Transformers are usually designed to have maximum efficiency at V 1 the normal operating value of secondary current,  I  2 Z  Z which may be less than the maximum rated current. e L 230 j0  (0.682 j0.173) (26.9 j11.2) Power relationships 230 j0 When calculating the transformer performance, the   7.12 j2.94 A 27.6 j11.4 following power relationships can be useful. The complex power S is given by 2: I  7.12 j2.94  7.70 A 2 S = P + jQ = VI (2-42) The secondary current magnitude is: where V is the voltage phasor, I is the complex  I 7.70 2 conjugate of the current, P is the real power, and Q I    16.3 A 2 n 0.472 is the reactive power. If  is the phase angle, then: P S cos  VI cos  Re(VI) (2-43) (b) Secondary voltage The secondary voltage magnitude is: Q  S sin  VI sin  Im(VI) (2-44) V  I Z  16.3 6.0 j2.5 2 2 L 2 2 S  VI  P  Q (2-45)  16.3 6.5 106.0 V If the voltage phasor V is chosen as the reference (c) Primary current quantity, and defined to be purely real (V = V + j0), The no-load current is: then the power relationships take a simple form: V V 230 j0 230 j0 1 1 P VI cos  V Re(I) (2-46) I     0 R jX 1080 j657 c m Q  VI sin V Im(I) (2-47)  0.213 j0.350 A The primary current is: Worked example 2-1  I  I  I 1 0 2 A 2 kVA, 50 Hz, power transformer has the  (0.213 j0.350) (7.12 j2.94) following equivalent-circuit parameter values  7.33 j3.29 A referred to the primary: I  7.33 j3.29  8.03 A R = 0.682 , x = 0.173, R = 1080, 1 e e c X = 657, N / N = 0.472. m 2 1 (d) Voltage regulation If the primary is connected to 230 V 50 Hz supply, and a load impedance (6.0 + j2.5)  is connected On no load, the secondary voltage is: to the secondary, determine: V  nV  0.472 230 108.6 V 2 1 (a) the secondary current magnitude, 10 Electrical Machines and Systems Course Notes The voltage regulation is therefore: jx jx 1 I I = nI 2 R 1 2 2 R 2 1 V V 108.6 106.0 2nl 2 fl I     2.33% I 2 0 V 108.6 2nl jX m Z L (e) Efficiency The output power is: 2 2 N :N P  I R  (16.31)  6.0 1597 W 1 2 out 2 L The power loss is: Figure 2-19: Current transformer circuit model. 2 V Only the relationship between currents is of 1 2  P   I R loss 2 e interest, so the primary impedance (R + jx ) can be R 1 1 c disregarded. It is now convenient to refer the 2 (230.0) 2 primary quantities to the secondary side, giving the   (7.699)  0.682 1080 circuit model shown in figure 2-20.  89.4 W jx 2 R I 2 1 The input power is therefore: I I = I / n 2 0 0 P  P  P  1597 89.4 1686 W in out loss 2 Alternatively, from equation 2-46 the input power jX = jn X Z L m m is: P  V Re(I ) 230.0 7.330 1686 W in 1 1 The efficiency is therefore: Figure 2-20: Modified circuit model. P 1597 out     94.70% The circuit acts as a current divider, where the P 1686 in current in the secondary branch is given by: jX I m 1 I  (2-48) 2 Z  Z  jX 2 L m 2.7 Current transformers where I is the primary current referred to the 1 The use of transformers for measuring current has secondary, and Z = R + jx . In a well-designed 2 2 2 been introduced in Electromechanics 1, where transformer, the secondary impedance Z is very 2 the danger of open-circuiting the secondary has small in comparison with the referred magnetising been explained. This section introduces the reactance X , so this term introduces very little m important topic of measurement errors. error. Equation 2-48 shows that it is desirable to keep the load impedance Z as small as possible if L Current transformer errors the error is to be minimised. In practice, current transformers are designed In a well-designed current transformer, the core for a specified maximum secondary voltage at the flux density is low and the core is made from a rated secondary current. This defines a maximum high-quality magnetic material. Under normal apparent power for the secondary load, or burden. operating conditions, the core loss will be Typically, a small current transformer will have a negligibly small, so the core loss resistance R can c rated secondary burden of 5 VA. With the usual be omitted from the equivalent circuit. A circuit secondary current rating of 5 A, this implies that model for a current transformer connected to a load the maximum secondary voltage is 1 V, and the therefore takes the form shown in figure 2-19. maximum impedance magnitude is 0.2 . Transformers 11 Worked example 2-2 2.8 Transformer design A current transformer has 10 turns on the primary The majority of single-phase transformers use the and 100 turns on the secondary. It has a rated shell type of construction shown in figure 2-21. 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: jX I j10.0 5.0 m 1 I   Figure 2-21: Shell-type transformers. 2  R  jX 0.2 j10.0 (RS Components Ltd) L m The magnitude is given by: Normally the core laminations are made in two parts, termed E and I laminations, as shown in 10.0 5.0 10.0 5.0 figure 2-22. I    4.999 A 2 0.2 j10.0 10.002 The percentage error is thus: 5.0 4.999 e  0.02% 5.0 (b) Inductive load Since the impedance magnitude is the same as before, the load reactance is 0.2 . The secondary current is now: Figure 2-22: E and I laminations.   jX I j10.0 5.0 m 1 I   The centre limb is twice the width of the outer 2  jX  jX j0.2 j10.0 L m limbs because it carries twice the flux, as shown by the flux plot in figure 2-23. The magnitude is given by: 10.0 5.0 I   4.902 A 2 0.2 10.0 The percentage error is thus: 5.0 4.902 e  1.96% 5.0 Figure 2-23: Flux plot: shell-type transformer. 12 Electrical Machines and Systems Course Notes The coils are wound on a bobbin that fits the centre k P c l T  (2-49) limb of the core, and the core is assembled by A s inserting E laminations alternately from each side and adding matching I laminations. Dimensions are where T is the temperature rise above ambient, P l chosen so that two E and two I laminations can be is the total power loss in the windings, A is the s punched from a rectangular steel sheet without any exposed surface area, and k is a cooling coefficient c 2 waste, as shown in figure 2-24. with a typical value of 0.04 Km /W. Figure 2-25 shows the side view and top view of a shell-type transformer. b 2a mean turn 6a a Figure 2-24: Punching E and I shapes. 3a 5a Current flowing in the resistance of the transformer windings will produce heat, which must escape through the surface of the windings. In a a a 2a a addition, there will be power loss in the core, which also appears as heat. Figure 2-25: Shell-type transformer dimensions The power output from a given size of transformer is governed by the rate at which heat It is assumed that the core is made from laminations punched as shown in figure 2-24. If a can be removed. Large transformers are usually is the width of each outer limb of the core, then the cooled by circulating oil, but small transformers width of the centre limb is 2a, and the other rely on natural convection cooling in air. A simple dimensions are as shown in figure 2-25. design approach for small transformers is given below. Winding resistance Thermal model The total cross-sectional area of the two windings The rate of cooling depends on the exposed surface is the window area of height 3a and width a. Each area of the transformer and the temperature rise winding occupies half of this area, so the conductor above ambient. An exact calculation is complex, cross-sectional area for each winding is: since it needs to take account of temperature 2 A  1.5k a (2-50) c s gradients within the transformer as well as the cooling conditions on different surfaces. where k is the conductor space factor, which s A simple thermal model ignores temperature allows for insulation and space between the turns. gradients, and the power loss in the core. It just For simplicity, it will be assumed that the 2 considers the I R loss in the windings, and assumes primary and secondary windings are placed side- that this heat escapes through the exposed surfaces by-side on the core, and that they have the same of the windings. It is assumed that the temperature number of turns N. From figure 2-25, the mean turn rise is proportional to the power loss per unit area: length of each winding is: l a 4a 2b (  4)a 2b (2-51) m Transformers 13 If the winding has N turns, then the total length of Equation 2-55 gives: wire is Nl , and the cross-sectional area of the wire m 4 3k a (5 2 )T 2 s is A / N. The winding resistance is therefore: c I  2 k  N (  4)a 2b c 2 Nl l N (  4)a 2b m R   (2-52) 2 A A / N 1.5k a c 4 s 3k a (5 2 )T s I  2 k  N (  4)a 2b c 3 4 Temperature rise 3 0.4 (1010 ) (5 2 )(90 30)  9 2 If the RMS current in one winding is I, the power 0.04 21.910  (1230)  2 loss is I R. The cooling surface area of the 3 3 (  4)1010  2 3010 winding, from figure 2-25, is:  0.216 A 2 2 A  1.5a(4a 2 a) 4a  a s Thus, I = 216 mA, so the transformer rating is (2-53) 2  a (10 4 ) 230  0.216 VA = 49.7 VA. Substituting in equation 2-49 gives: Rating and size 2 2 2 k P k I R k  N I (  4)a 2b c l c c T    (2-54) 4 A relationship between the apparent power rating A A 3k a (5 2 ) s s s of the transformer and the dimensions can be Thus, the current is given by: obtained by substituting for N from equation 2-56 in equation 2-55: 4 3k a (5 2 )T 2 s I  (2-55) 3 2 2 B ba 3k (5 2 )T k  N (  4)a 2b m s c VI  (2-57) k (  4)a 2b c From equation 2-10, the number of turns is: If it is assumed that the core depth b is proportional V 2V m N   (2-56) to the dimension a, and other quantities remain 2fAB 4fabB m m constant, then equation 2-57 gives the following relationship: 3.5 VI  a (2-58) Worked example 2-3 From equation 2-58, if the dimensions of the A transformer has a primary wound for 230 V, and transformer are doubled, the rating will increase by the core measures 60  50  30 mm. The maximum a factor of 11.3. A similar result is obtained for the winding temperature is 90ºC, the ambient increase in the power output of a DC machine temperature is 30ºC, the winding space factor is 2 when the dimensions are doubled – see section 3.2. 0.4, the cooling coefficient is 0.04 Km /W, and the resistivity of copper at 90ºC is 21.9 nm. 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: 2V N  4fabB m 2 230   1230 3 3 4501010  3010 1.4 14 Electrical Machines and Systems Course Notes 3 DC MACHINES 350 300 3.1 Introduction 250 200 Basic DC machine principles were covered in Electromechanics 1, and the main results are 150 given below. Brushless DC machines, which were 100 mentioned briefly in 1, are beyond the scope of 50 this course but are covered in the Year 3 course Electrical Machine Drives. 0 0 0.2 0.4 0.6 0.8 1 1.2 Field current (A) Basic equations The generated voltage and the developed torque Figure 3-1: Magnetisation characteristic. are given by: In this example, the relationship is almost e  K ù V (3-1) a f r linear up to the rated field current of 0.5 A, but there is significant non-linearity above this value, T  K i Nm (3-2) when parts of the magnetic circuit saturate. There d f a is also a small residual flux when the field current where K is the armature constant,  is the field f is zero, giving a corresponding generated voltage. flux, and i is the armature current. Note that the a For the initial part of the magnetisation rotor angular velocity  must be in radians per r characteristic, it is approximately true that   i . f f second (rad/s) and not in rev/min. If the rotational Equations 3-1 and 3-2 then become: speed is n rev/s or N rev/min, then: r r e  Ki ù (3-4) a f r 2ð N r ù  2ðn  rad/s (3-3) r r 60 T  Ki i (3-5) d f a In a permanent-magnet machine, the field flux  is fixed, but in a wound-field machine, it is a f function of the field current i . f Armature equation Figure 3-2 shows a symbolic representation of Magnetisation characteristic a DC machine, with the armature connected to a voltage source v . a From equation 3-1, if the speed  is held constant, r the flux is proportional to the armature generated i a  voltage e . A graph of e against i is known as the r a a f + + magnetisation characteristic of the machine, and a e v a a typical curve is shown in figure 3-1 for a 3 kW i f motor. Figure 3-2: DC machine with voltage source. If the armature has a resistance R , then a Kirchhoff’s voltage law gives the armature voltage equation: v  R i  e  R i  K ù (3-6) a a a a a a f r DC Machines 15 Armature voltage (V) 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 3-5: 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 Figure 3-3: DC motor armature. exerted on the armature iron and not on the Figure 3-4 shows a simple model of a machine conductors. The basic equations 3-1 and 3-2 are with a slotted armature, and figure 3-5 shows the not affected by the location of the conductors. In corresponding flux plot when there is no armature particular, the generated voltage is not affected, current. 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 3-4: DC machine model. Figure 3-6: Armature reaction flux plot. 16 Electrical Machines and Systems Course Notes Note that the axis of the armature reaction field is field flux. This has important consequences for the at right angles to the axis of the main field. When motor characteristics – see section 3.3. currents flow in the field and armature conductors, the two component magnetic fields combine to Efficiency give a resultant field of the form shown in figure 3- 6. The bending of the field lines indicates that the The efficiency of a motor is defined in the usual stator exerts a counter-clockwise torque on the way: rotor. Another way of picturing the P electromagnetic action is to consider the magnetic out   (3-7) poles representing the field and armature flux P in components, as shown in figure 3-7. where P is total electrical power input to the in motor terminals, and P is the mechanical power out output from the motor shaft. The power lost as heat in the motor is: P  P  P (3-8) loss in out N so we have the following alternative forms of S N equation 3-7: S P P  P P out in loss loss     1 P P P in in in (3-9) P P out loss   1 P  P P  P out loss out loss Figure 3-7: Total flux plot. As with transformers (see section 2-5), the efficiency of a large motor is not usually An important effect of armature reaction, which is determined from equation 3-7 by direct evident in figure 3-7, is to increase the flux density measurement of the input and output power. at one side of a field pole and decrease it at the Instead, losses are determined from several tests, other side. Figure 3-8 shows a shaded plot of the and the efficiency calculated from equation 3-9. flux density magnitude, where the colour range The output power is given by from blue to red represents the flux density range from minimum to maximum. P  T  (T  T ) (3-10) out r r d l where T is the rotational loss torque. It follows that l the rotational power loss is  T . The rotational loss r l 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 2 I R 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. 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 DC Machines 17 P T  2 B AV (3-14) r d r av Thus, the maximum power output of a DC motor is Power output and size roughly proportional to the product of the armature The developed torque may be calculated from the volume and the armature speed. equation f = Bli, even though the armature In practice, the current loading A also increases conductors are in slots. For this purpose, the with size, so the power output of large machines is machine will be represented by a simple model further increased. To quantify this, assume that the with the conductors on the surface, as shown in the power dissipation per unit surface area is constant. flux plot of figure 3-9. Let d be the radial depth of a conducting layer representing the armature conductors. The resistance of an element ds of this layer is: l dR (3-15) d ds ds r The cooling surface area of the element is dS = l ds, so the power dissipation per unit area is: l 2 ( Ads) 2 2 dP (di) dR d ds  A    (3-16) dS l ds l ds d If dP / dS is constant, then A  d, and if d  r then A  r. For geometrically similar machines, Figure 3-9: Model for torque calculation. the axial length l is proportional to the radius r. If the rotational speed  is constant, it follows from r It is useful to define a current loading A as the equation 3-13 that the power output is proportional current in amperes per metre length of 3.5 to r . Thus, if the dimensions are doubled, the circumference on the surface of the armature. The power output will increase by a factor of 11.3. A current in an element of length ds is then di = A ds. similar result was obtained for transformers The force on this element is: (section 2.8), and it holds for other types of df  Bl di BlAds (3-11) machine. where l is the axial length of the armature. The corresponding contribution to the torque is: 3.3 Characteristics and control dT  r df  rBlA ds (3-12) d Speed control where r is the radius of the armature. If B is the av In a large motor, the total power loss is small. average value of the flux density at the armature 2 Since the armature power loss R i is only a part of a a surface, then the total torque is just: the total loss, this term must be small in 2 T  rB lA.2 r 2 B Alr  2B AV (3-13) comparison with the armature input power v i . It d av av av a a follows that: where V is the volume of the armature. R i  v (3-17) a a a The maximum value of B is limited by the av saturation of the magnetic material of the armature, So R i may be neglected in equation 3-6, giving: a a and the value of the current loading A is limited by 2 v  K ù (3-18) the I R heating of the armature conductors. If it is a f r 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: 18 Electrical Machines and Systems Course Notes v The rotational speed is therefore: a ù  (3-22) r0 K f v a ù  (3-19) r K f  r Equation 3-19 shows that the speed is independent of the torque, provided inequality 3-17 holds.  r Equation 3-19 is the basis of speed control. If the field flux  is constant, the rotational speed is f  slope R a proportional to the applied voltage. The normal values of v and  define the base speed  , and a f r0 the speed can be reduced to zero by varying v . a Electronic controllers for DC motors deliver an adjustable voltage to the motor armature by phase- controlled rectification of the AC mains supply, T d using thyristors: see section 8. This is the basis of DC variable-speed drive systems, which are widely Figure 3-10: Speed-torque characteristic. used in industry. In a wound-field motor, it is possible to increase the speed above the base speed by Effect of armature reaction reducing  – known as field weakening. Only a f In section 3.2, it was noted that the armature limited speed increase is possible, for the following reaction field could cause local saturation of the reason. The torque is related to the armature field poles, thereby reducing the value of the field current through equation 3-2: flux  . This effect increases with the armature f T  K i 3-2 d f a current i , and therefore with the developed torque a T . From equation 3-19, a decrease in  will cause d f If  is reduced, there will be a compensating f the speed  to rise. Armature reaction therefore r increase in i to maintain the torque, and there is a a has the opposite effect to armature resistance, risk of exceeding the current rating of the machine. which causes  to fall with increasing torque load. r There is an important difference, however. The effect of resistance is linear, as illustrated in figure Small motors 3-8, but the effect of armature reaction is non- In small motors, with power ratings below 1 kW, linear. At low values of armature current, the inequality 3-17 does not hold, so it is not uneven distribution of flux density in the field pole permissible to neglect the R i term. The rotational a a is insufficient to cause saturation, so there is hardly speed then depends on the developed torque, as any reduction in the field flux. At high values of may be seen by substituting for i in terms of T in a d current, on the other hand, there may be a equation 3-6: significant reduction. v  R i  K ù It is possible, therefore, for the speed of a a a a f r motor to fall with increasing load when the (3-20) R T a d   K ù armature current is low, but to increase with load f r K f when the current is high. This increase in speed can be very undesirable, leading to instability with Thus, the speed is given by: some kinds of load. A large DC motor generally v R T includes some form of compensation for armature a a d ù   (3-21) r 2 reaction. K (K ) f f A simple method of compensation is to provide A graph of speed against torque is a straight line, a second winding on the field poles, connected in as shown in figure 3-10. The no-load speed, which series with the armature (figure 3-11), to increase is the speed when the torque is zero, is given by: the field MMF when the armature current increases. The resulting motor is known as a compound motor 3, 4. However, this cannot DC Machines 19

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