ELECTRICAL AND ELECTROMECHANICAL CONVERTERS Lecture Notations Authors Ivan Zagradišnik Jožef Ritonja January 2025 Title Electrical and Electromechanical Converters Subtitle Lecture Notations Authors Ivan Zagradišnik (University of Maribor, Faculty of Electrical Engineering and Computer Science) Jožef Ritonja (University of Maribor, Faculty of Electrical Engineering and Computer Science) Review Božo Hribernik (University of Maribor, Faculty of Electrical Engineering and Computer Science) Language editing Shelagh Hedges Technical editors Ivan Zagradišnik (University of Maribor, Faculty of Electrical Engineering and Computer Science) Jan Perša (University of Maribor, University Press) Drawings Denis Dremelj, Viktor Goričan, Miralem Hadžiselimović, Dušan Muršec, Tine Marčič & authost Assistant editors Matej Gajzer, Bojan Slemnik Graphic material Sources are own unless otherwise noted. Zagradišnik, Ritonja (authors), 2025 Cover designer Jan Perša (University of Maribor, University Press) Published by University of Maribor, University Press Slomškov trg 15, 2000 Maribor, Slovenia https://press.um.si, zalozba@um.si Issued by University of Maribor, Faculty of Electrical Engineering and Computer Science Koroška cesta 46, 2000 Maribor, Slovenia https://www.feri.um.si, feri@um.si Edition 1st, translation of the original work from the Slovenian language Zagradišnik, I., Ritonja, J. (2022). Električni in elektromehanski pretvorniki: Zapiski predavanj. Univerza v Mariboru, Univerzitetna založba, doi: 10.18690/um.feri.11.2022 Published at Maribor, Slovenia, January 2025 Publication type E-book Available at https://press.um.si/index.php/ump/catalog/book/937 © University of Maribor, University of Maribor Press / Univerza v Mariboru, Univerzitetna založba Text © Zagradišnik, Ritonja (authors), 2025 This work is licensed under the Creative Commons Attribution 4.0 International License. This license enables reusers to distribute, remix, adapt, and build upon the material in any medium or format for noncommercial purposes only, and only so long as attribution is given to the creator. If you remix, adapt, or build upon the material, you must license the modified material under identical terms. Any third-party material in this book is published under the book’s Creative Commons licence unless indicated otherwise in the credit line to the material. If you would like to reuse any third-party material not covered by the book’s Creative Commons licence, you will need to obtain permission directly from the copyright holder. https://creativecommons.org/licenses/by/4.0/ CIP - Kataložni zapis o publikaciji Univerzitetna knjižnica Maribor 621.313/.314(075.8)(0.034.2) ZAGRADIŠNIK, Ivan Electrical and electromechanical converters [Elektronski vir] : lecture notations / authors Ivan Zagradišnik, Jožef Ritonja ; [language editing Shelagh Hedges]. - 1st ed. - E-knjiga. - Maribor : University of Maribor, University Press, 2025 Način dostopa (URL): https://press.um.si/index.php/ump/catalog/book/937. - Prevod dela: Električni in elektromehanski pretvorniki ISBN 978-961-286-945-8 (PDF) doi: 10.18690/um.feri.1.2025 COBISS.SI-ID 222237955 ISBN 978-961-286-945-8 (pdf) DOI https://doi.org/10.18690/um.feri.1.2025 Price Free copy For publisher Prof. Dr. Zdravko Kačič, Rector of University of Maribor Attribution Zagradišnik, I., Ritonja, J. (2025). Electrical and Electromechanical Converters: Lecture Notations. University of Maribor, University Press. doi: 10.18690/um.feri.1.2025 Page INTRODUCTION 1 Magnetic field 2 Excitation of windings 6 Induced voltage 11 Forces in a magnetic field 12 Energy and power transmission 13 Torque 14 Losses and efficiency 16 Heating and cooling 16 TRANSFORMER 19 Introduction 19 Operation mode 21 Transformer design 21 Transformer tasks 22 Single-phase transformer with iron core 23 Operation of an ideal transformer 24 Real transformer operation 29 Operation of a current-ideal voltage-real transformer 31 Operation of a voltage-ideal current-real transformer 33 Losses in the core 35 No-load current 38 Analytical treatment 41 Transformer tests 44 Transformer operation on a rigid grid 47 Energy balance, losses and transformer efficiency 49 Current transformer 51 Three-phase transformation 54 Asymmetrical load 57 Autotransformer 59 INDUCTION MACHINE 60 Description of construction 60 Induction machine windings 61 Operation mode 67 Analytical treatment 70 Induction machine equivalent circuit 72 Operating an induction machine on a rigid grid 73 Energy balance 76 Torque 77 Starting an induction motor 80 Options for changing the rotation speed 82 Changing the torque 84 Induction generator 89 Single-phase induction motor 89 Description of construction 93 Operation mode 93 Taking on the load 97 Analytical treatment 102 Operating an unsaturated machine on a rigid grid 107 Varying the torque of a synchronous motor 110 Current characteristics for a cylindrical rotor 113 Approximate treatment of a saturated machine 114 Excitation systems for synchronous generators 117 Using permanent magnets for excitation 119 Permanent magnet synchronous motors 121 Permanent magnet synchronous motors with salient poles on the stator 124 COMMUTATOR MACHINE 128 Description of construction 128 Operation mode 128 Commutator machine windings 132 DC machine theory 135 Armature reaction 138 Commutation 143 Commutation assembly 147 Types of DC machines 150 Generators for direct current 152 DC motors 154 Constructing characteristics 157 Starting and changing the speed of the motors 159 Non-stationary operation 161 Special DC machines 161 AC commutator machines 162 Single-phase commutator motor in series connection 165 COMPLEX CALCULATION 168 Mathematical operations of complex quantities 168 Calculation of power 170 The law of induction in complex calculus 173 FOURIER ANALYSIS 175 Examples for excitation curves 175 Example for transformer no-load current 176 LITERATURE 178 LIST OF VARIABLES 179 Greek alphabet 181 Abbreviations 181 INTRODUCTION Electrical and electromechanical converters (EEMCs) are devices that convert energy from one form to another. Types of conversion: electrical to electrical, electrical to mechanical, mechanical to electrical. The conversion is time-limited or unlimited. Electrical machines are also considered as EEMCs. In these, the energy conversion is always mediated by a magnetic field. Examples of energy conversions Electrical → electrical P are losses. l P U , I U , I 1 1 22 TRANSFORMER f f 11 P l P U 1, I1 U2 , I2 f1 f2 = RECTIFIER 0 P l P U 1, I1 U2 , I2 f 1 = INV INVERTER 0 f 2 P l P U , I U , I 1 1 22 (DIRECT ) CON f f CONVERTER 1 2 P l number of branches = number of phases frequency m   m f f 1  2 1 2 Electrical → mechanical P P el m U I , GRID EMC L L is load. m f , MOTOR M M ,  ( ) n L m P l  P − P P el = l = m : 0, 6 . . . 0,98 EMC is an electromechanical converter. P P el el Mechanical → electrical P P m el U I , T EMC GRID T is turbine. M ,  m f , m GENERATOR P l  P el = : 0,95 . . . 0,98 P m Magnetic field The transfer of power or energy in electrical machines is carried out by a magnetic field or magnetic flux. The basic quantities (marked with an asterisk) and derived quantities are used for notation. Basic quantities:  =  B magnetic flux density A  =  B dA magnetic flux  A  = B H magnetic field intensity   = B  permeability H Derived quantities:  =  N = B dA coil flux or magnetic linkage i  i A L  = inductance I F = H dl magnetomotive force (MMF) – excitation m  l K   F = m magnetic conductivity (permeance) or R = resistance (reluctance) m m F  m Examples for the main and leakage mag. fields pole pole column shoe winding  m yoke yoke air gap   tooth σ2  m σ1 yoke winding 1 winding 2  m iron core Basic magnetic field laws 1) Law of current flow (Ampere's law)  H  dl =  J  dA = I N =  ( J is the current density.) l K A discretization H → H ; dl = l l ii  H l = I N i i 2) Flux preservation law Flux tube  = B dA = 0 Ai  Ak A B discretization B k B ; d A → A → i B i i  B A = 0  = B A i i i ii 3) Material law B = f (B ) =  H + B =  H +   H =  H    = =  (1+  ) i 0 i 0 0 0 r 0 B is the magnetic polarization or intrinsic flux density, and  the magnetic susceptibility. i Magnetic curve (B-H curve) The B-H curve is shown in the left Figure for soft iron and a permanent magnet and for the corresponding permeabilities in the right Figure. For a magnetic circuit, the following applies without saturation:  →  ; H → 0 in saturation:  →  Fe Fe Fe0 B  I N I N B ( B = B – saturation) B =  B >B B   Fe sat sat δ 0 δ 0 Fe sat   +lFe B  soft iron soft iron B t sat nt magne permane Bi  0 permanent magnet B B sat H Flux preservation law − application →  =  BFe , HFe , A δ FeFe Law of current flow – application:  = H l = I I N i i   N I  H l i i ˆ = , (if H = B H , , A   H is the peak value). μ 2 N lFe Magnetization characteristic The magnetization characteristic gives the dependence for a magnetic circuit consisting of an air gap and an iron core. The air gap characteristic (AGC) is the tangent to the curve. B  eristic C mag. charact G A Change of measure: B = f( )  B sat Ordinate:  = B A →  = f( ) δ δ δ δ sum   = =   + sum δFe I Abscissa:  = I N → B = f( ) I μ μ      Fe To calculate the magnetic circuit: from E → →  B on part i; from B will be: i δ  B δ =   , = H l ( H ) δ Fe Fe Fe → from B-H curve  Fe 0 I ˆ    2 I N μ sum → = sum = I  = sum = = 2 I N  μ μ μm N 2 N R R m m inductance of the coil   2 N N 1 = = = 2  A L = N  and  = = m m 2 I 2 I R R l μ μ m m Excitation of windings Magnetic field of a concentric (cylindrical) winding a) Example of a transformer The individual windings are spatially close. The excitation is concentrated. The main flux connecting the windings is generated in the iron core. If the Fe core does not have an air gap, it will:  lFe → 0 and  →  R = → 0, L → , L → 0 . Fe m m σ  A Fe Fe L is the magnetizing (main) inductance and L the leakage inductance. m σ b) Example of salient poles for rotating machines In rotating machines, the individual windings are separated by an air gap " "  or equivalent (enlarged) air gap  to take account of the effect of the machine's slotted openings. e  = I N magnetic excitation by direct current of a single pole coil with N turns. c c If the current oscillates, ˆ i = 2 cos( I  t ) = I cos( t) ,  will also oscillate.  p p A x stator IN c − I +I  x a)  e −2I + 2I p rotor B −3I + B 3 I  = IN c −2I +2I b) x −I + I B  e = δe R l x d mxe a) A c) x A B A B A d) B  =  x +3I R l x d mx + 2 I +2I  + I + Imax  ( ) x x x b) x − I −I  −2I −2I Magnetic field of a distributed winding −3I A B A B A   p p a) DC power supply +2I + I +I The left Figure (below) shows the coil excitation (MMF) curve for the developed x rotor perimeter:  I Nc = −2I  A c c) − I −I 2 0  p 2  p and the fundamental harmonic component of the coil excitation  ˆ 4 I N 4 = c  = B A . c1 π 2 π N N 4 I N ˆ are the number of turns of the coil. For p pole pairs are N = and  = . c c c1 p π 2 p   1c12c1 IN c 2 A B    x p x = 0 Fundamental component of sinus excitation The space distribution of the MMF of the coil is: x  = D π  ˆ cos ( / )π , if x c1 (  p ) r π  = = and p 2 p p x = 0 is in the coil symmetry axis. Introducing the angle  = ( / x  )π is  =  ˆ cos . p x c1 For two coils displaced by an angle in adjacent slots of a rotating electromechanical converter, the fundamental harmonic component of the MMF is:  =  +  . res1 1c1 2c1 In general, for a distributed winding, ˆ 4 I N   = f , if f is the 2c1 res1 w w π 2 p  winding factor, i.e. the ratio of the geometric to the arithmetic sum   1c1 res1   of the = = MMF of two or more coils: f . w geom. res1   n  arit. ic1  i 1 = For a uniformly distributed winding of a large number of coils, the MMF equation is a straight line:  x =  for 0   x  x / 2 . (Valid to the right dotted Figure (b) on page 7.) max p  p / 2 b) AC power supply i = 2 cos( I t) MMF ˆ   ( , ) x t  = 0 ( , ) x t =  cos  cos(  t ) → b x t ( , ) (In the air gap, b   .) 1  B ˆ 1  ˆ 1 i −i ˆ 2 x B x 1 b) a) i = 2 cos( I t)  ˆ  ˆ1  ( , ) x t = cos  cos(  t ) = t) cos( t) 1 ( cos(   − +   +) or for b   2 b x t B ˆ 1 ( , ) = (cos(  − t) cos( +  + t )) 2 b x t ( , ) = ( , ) b x t + b x t ( , ) p n direction of motion →  The amplitude of a positive (direct) or negative (inverse) wave travels at the speed obtained from the condition for the value of the argument: x p    t = 0 or π  t = 0 → =  x  t ,  π p v d  2 x = =  p =  p 2 f  =  . p d t π T From the condition v D π f = D n π =  2 f  =  2 f the expression is given: n =  p . 2 p p This is the basic equation for the rotational speed of a rotating magnetic field in an electrical rotating machine. A more practical equation for 60 f 3000 − 1 f = 50 Hz is: n =  =  min (  rot. / min ) . p p The magnetic field of a distributed three-phase winding or of three single-phase windings displaced in space by a mechanical angle  =  / p = (2π / 3) / p  120 / p . fm f developed air gap space phasor diagram time vector diagram  p  p t 1 a)  i i − c b b − c  i − i i c c b a   2 a 2 i a − i i − x i c b a 3   a ia − −  p i a b i  ca ib 2 i i cb  −ia p 3  i a p i a − − i i  b / 2 c a b)   t2 b / 2 x  3 − 2 a c i c ib i i cb −ia The Figure applies to the MMF of a two-pole machine which causes a magnetomotive field in the air gap with phase current phasors according to the time diagrams for  t = 0 and 60 . Using the angle  and the corresponding space and time phase shift, i.e., 120 , we get: phase "a" B ˆ 1 (cos(  − t) cos( +  + t)) = b + b , ap an 2 phase "b" B ˆ 1 cos ( ( ( −120 ) ( − t −120 )) + cos ( ( −120 ) ( + t −120 ))) = b + b , bp bn 2 phase "c" B ˆ 1 cos ( ( (  − 240 ) ( − t − 240 )) + cos ( (  − 240 ) ( + t − 240 ))) = b + b . cp cn 2 The sum of the positive field waves is: b = b + b + b = 3 ˆ cos( B  − t) . The sum of the p ap bp cp 1 2 negative field waves is: b = b + b + b = 0 , because the phasors of the negative wave of the n an bn cn magnetic field of the second and third phases are displaced by 240 and 120 respectively. In general, for a symmetrical m-phase system b = m B   ˆ cos( − p t) , if the field amplitude of each 1 2 phase is: ˆ   ˆ 4 N f m 4 N f 0 1 ˆ ˆw B = = w or MMF  2 I and for m-phases  = 2 . I 1  1 1m π 2 p 2 π 2 p Induced voltage a) Induced voltage of a conductor (bar) due to motion in a magnetic field b e =  ( v B l )  = −   ( B v ) l b e v e = v   B l for B ⊥ v b l b From dA = l x d = −l s d = −l vdt (because dx = −ds ) we prove: vBl t d B A d d e = = − = − . b d t d t d t b) Induced loop voltage due to a time-varying magnetic field Induction law in integral form (Faraday's law):  u e d d =  E  = − d l  B  d A = − . l d t d t e R = 0 K A E is the electromotive force (EMF). l i Example of a transformer: e d d = − N e = − N 1 1 2 2 d t d t e = 1 = 1 K (voltage ratio) U N e N N 1 e N 11 2 N  2 2 The general law of induction, if  = f ( , ) x t : e 2 N2 e d    dx      = − = − + = − + v = +     e e , t r d t   t  x d t   t   x  transformer voltage + rotating or moving voltage. For two windings with mutual inductance L , there will be:  = L i and 12 12 μ e d di  μ dL  = − = − L + i = +  12  e e . 12 μ t r d t  d t d t  B Forces in a magnetic field Force on the current conductor (Lorenz force) i F l B F =   Q ( v B) = q l   ( v B) =   l i ( B) =  ( i B l )  , l F because i = vq will be a force F =   i B l . l i q is the line charge. l Motor: F acts in the direction of v or  . m Generator: F acts against the direction of v or  . m Force on the iron core Energy in the air gap: i and for = W 1 1 2    =   is W =   . m δ δ δ δ δ m δ δ 2 2 F dW d  A 1 1 dl m 2 δ 2 0 δδ = =  = −  δ δ 2 d x 2 d x 2 l d x δ B F A  l  = 0 δ For and dl = −d x is force:  δ δ l δ A B δ   2 A 1 0 δδ F = . 2 2 l δ For   2 2 0 δ 1 B A 1 B B = δ δ δ δ , we get the expression for the force on the iron core: F = = A . δ l 2  2  δ 0 0 Examples of applications: switches, relays, magnets, and step motors. Energy and power transmission Electrical mechanical → P el Stationary operation: P m P − = P P EMC el l m M ,  m EMC – electromechanical converter (motor) P l Electrical electrical → TR Stationary operation: P P 1 2 P − = P P 1 l 2 P l TR – transformer Example of a lossless linear electromechanical converter B Accepted electrical power P = e i = vB l i el P el e F, v Mechanical power output P m P = Fv = iBl v m i Arrows system i P = u i P  0 consumer (motor) u P P  0 producer (generator) Torque There are several ways to calculate the torque. 1) From electrical power P = e i = P = M  ;  = 2π n = v / r (mechanical angular or el m m m circular speed) we get shaft torque: M P m = .  m 2) From induced voltage a) From the induced voltage due to motion, we get: e i = v Bli= viBl = vF = M → M e r v r e r M = iBlr = F r (e – electromagnetic torque) e We need to know the magnetic field distribution (in the air gap). b) The absolute value of the transformer induced voltage gives: e d d = − . e i = i = M  m d t d t  d 1 d = m = (  – mechanical angular velocity) m m d t p d t  = p ( is the mechanical or "space" angel,  – electrical angel) m m M ei d p t d d  d  = = i = pi M = p i N is valid for  = N  e  .   e d t d  d   d  m  We need to know the magnetic field distribution (in the air gap). 3.) From the magnetic field energy  dW field  F =    d x  M = dW dW fieldfield F r = r = e ( for dx = r d ) d x d  The force acts in the direction of increasing the mutual magnetic linkage, or increasing the magnetic conductivity, or decreasing the magnetic resistance. From the equation for energy, for a single excitation winding: M 1 d 1 2 d 1 d 1 dR  m =  = = − 2m or M = −   . e e 2 d  2 d  2 d  2 d  In the equations, we consider    = or   = R . m m For two or more windings, the energy is expressed better in terms of inductances: M = 1 dL 1 dL dL 2 1 2 2 12 i + i + i i . e 1 2 1 2 2 d  2 d  d  L and L are the self-inductances and L the mutual inductance of the two windings. 1 2 12 Losses and efficiency 1) Winding losses: a) joule losses, b) eddy current losses due to skin-effect P = 2 I R or additional losses Cu R = R + R =add 2) Losses in the Fe core: a) hysteresis, b) eddy-current P = x 2 2 = P + P k f B m + k f B m , Fe Feh Fee h Fe e Fe where the exponent for hysteresis losses is x = 1, 6 2,8  . 3) Mechanical losses in rotating machines Efficiency  P P − P P P out = = in l = − l l 1 = − 1 P P P P in in in 1 Heating and cooling Heat transfer 1. The propagation of heat − the heat flux − in a solid body is:   W  =    ( − )  t – specific thermal conductivity t 1 2   m K    thermal conductivity  A  = 2 A – surface ( m ), d – body thickness (m) t d 2. Thermal transmittance by convection   W  =    ( − ) =  A (   − )  t – convection coefficient t 1 2 c c 1 2 c  2   m  K  For air, the approximate equation is:   6,5 0, 05( +  − ) . For natural convection, the c 1 2 convection coefficient for the temperature rise 40 50 K  2 is  = (8, 5  9) W/(m  K) . c 3) Radiative thermal transmittance  =    ( − ) =  A (  − ) t t 1 2 r r 1 2  (radiation coefficient) is a function of temperature, temperature difference and material type. r  = C 1 (  4 −  4 r 1 w a ) Δ   is the absolute wall temperature and  the absolute ambient temperature, w a C =  C is the radiative constant of the surface of a body,  is the absorption ratio 1 (  1 r) , i.e., the ratio of the received (absorbed) to the irradiated radiant energy. C  − W 8 = 5, 67 10 black-body radiation constant r 2 4 m  K Δ    = − =  − the temperature difference between the wall and the ambient 1 2 w a For the temperature rise (40 50) K  , calculate the radiation coefficient   + 5 0, 033(  − ), r 1 2 i.e., 2  = (6, 3  6, 65) W/(m  K) . r EEMC heating Electrical and electromechanical converters are inhomogeneous bodies, but they are treated as homogeneous bodies. In the time differential dt , thermal energy P t d is released, some l of which is stored (temperature rise − first term on the right-hand side of the equation), some of which is dissipated to the surroundings by convection and radiation (second term on the right-hand side). P t d = m c d Δ ( )   + A Δ d t l m c The solution for the time differential is: = t d t d(Δ )  . P l − Δ   t In the equation   = A is the thermal conductivity, the coefficient of conductivity t    = + and the specific heat c W s/(kg K)   . c r ( ) Time = − mc  P  l t ln −   +   C Time constant:   t  t  For t = mc 0 is Δ  = Δ  or 0. T = 0 (a few minutes to a few hours)  A Equation for heating   t −  P  Δ  = l − Δ   1 e − T  +Δ   0   0  t    The equation applies to heating at constant losses and constant cooling conditions. Δ P P l  = = l Δ , if it is Δ  = 0 T , it will be: max   0 Δ Amax t Δ  = Δ 1 e − T  . max   t −     EEMC cooling Δ = Δ , P = 0 0 max l Δ − t = Δ max e T t In general, the time constant T is larger for dimensionally larger EEMC (higher mass m). For forced cooling it will be T smaller because it increases  (the heat transfer coefficient). TRANSFORMER Introduction A transformer is a static power transmission device that converts an AC voltage and current system by electromagnetic induction into another voltage and current system, usually of different magnitudes and the same frequency. Primary Secondary Single-phase system → single-phase system or two-phase system Three-phase system → three-phase system or six-phase system or twelve-phase system In general: m phase system → m phase system (m = m or m  m ) 1 2 1 2 1 2 The most elementary design of a single-phase transformer has two coils (windings): 1) a primary power supply coil, 2) a secondary coil for the power output. The two windings are usually separated galvanically. Example: N = 5 turns, N = 2 turns 1 2 Current i →  = i N →  (6 magnetic lines or density lines) 1 1 1 1 1 i = → 0  = i N = 0 2 2 2 2 i 1 com u e e 1 12 u2 primary secondary N 1 winding winding N 2 coils axis Flux linkage of the primary winding:   6 4  1 = 3  +  2  = 4  (ideal  =  1   5 ) 1 1 1 1  6 6  3 and of the secondary winding   4  1 2 =   = 1 (ideal  =  ). 2   2 1 1 2 1  6  3 Common flux: 4 2   =  =  com 1 1 6 3  2 1  Primary flux:   = +  =  +  1 com σ1 11 2 1  3 3 Leakage flux:  =  =  σ1 1 1 6 3   1 =  = 1  2 com2 1 3    2   1  1 = +  = 5   +  3  = 4  1 com1 σ1   1   11  3   3  3 Only the common flux  is involved in the transformation process. The importance of the com leakage flux, which is not involved in the transformation process, will be explained later. The ideal transformer:  =  = 0 . σ1 σ2 The permeability of the medium (core) around which the winding is wound,  →  , i N → 0, i → 0 . 1μ 1 1μ Operation mode Assume R = R = 0 1 2 Primary winding imposed voltage u . Secondary winding open. 1 → The grid voltage u drives a current through the primary winding: 1 i →  →  →  . 1 1 1 1 The magnetic linkage  must be such that equilibrium is created: 1 u = − = − −  d  d d ( + = = com1 σ1 ) 1 1 e = − 1 1   (e + e . com1 σ1 )  d t  d t d t At the same time, an induced voltage appears in the secondary winding with N turns: 2 − = − (−d sk2 ) d sk2 e = = u . 2 2 d t d t The primary leakage flux  , or magnetic linkage  , creates an inductive voltage drop σ1 σ1 across the primary winding: − d σ1 e = and do not participate in the transformation process. σ1 d t Transformer design A two-coil transformer without an iron core (air transformer) has a large leakage flux  and σ1  a small common flux  = −   . σ2 com 1 σ1 Therefore, a core of ferromagnetic material, i.e., an oriented transformer sheet, is used to guide the flux (   4000 40000,  maximum). rFe Design of single-phase transformer cores There is a core (column) type and a shell type. In the core type, the winding is mounted on one or two columns connected by a yoke (Figures a and b). The same applies to the cutting ribbon core (Figure c). In the shell type (Figure d), the flux in the yoke is half that in the column (half the cross section of the yoke → the lower the height of the transformer). 1 2 2 2 1 2 1 1 1 1 a) b) c) d) Transformer tasks Energy transmission and U U voltage adjustment 1 2 I I 1 2 transmission → U (high) 2 U distribution → U (low) 2 U 1 U 2 U phase rotation 1 2 m = 6 2 Industry – phase multiplier m = 1 3 m 1 m2 U 1 U 2 U U 1 2 Electrical circuits impedance adjustment Z 2 = K Z 1 U 2 Z Z 1 2 K – voltage ratio K U U Measurement technique U 1 I K I 1 current voltage converter K converter U I 2 I 1 U 1 I = U 2 = U 2 K 2 I K U K − current ratio I Single-phase transformer with iron core Direction arrows They are adapted as follows: Both coils (windings) are wound clockwise. The direction of the voltage drops across winding 1 is positive and so is the current. Since the magnetization of both windings are in the same direction, this determines the direction of the current in winding 2, and also the voltage drop across winding 2.  i 1 u i 11 e P u N 1 1 1 1 i2 i u 2 P 22 N 2 u 2 e 2  com The positive direction also applies to the magnetic coupling through the coils ( )  , or the magnetic flux in the core (  ) . The positive current in the coils generates positive ampere-com turns for the integration path in the direction of the arrow through the coils. Operation of an ideal transformer The properties of the materials used are: 1) permeability of the magnetic circle  =  , Fe 2) the electrical conductivity of the magnetic circuit  = 0 , Fe 3) the permeability of air  = 0 , air 4) electrical conductivity of the conductors  =  , con 5) the magnetic circuit has no air gaps  = 0 . gap For a loop (winding) with ohmic resistance R and voltage drop Ri , the voltage equation is: u d = R i e − = R i + . dt Since R = 0 , it applies to windings 1 and 2: u d 1 = − = 1 2 e , 1 1 d t u = − = d 2 u 1 u e . N  2 1 N 2 2 2 d t Magnetic linkage of the two windings:  = N  1 1  = N  2 2 This is true for an ideal transformer, because all the flux in the positive direction of the arrows passes through windings 1 and 2. It is therefore   = =  =  . 1 2 com side 1 side 2 In the absence of eddy currents, ampere-turns i operate along the integration path 1 u i N 11 u 12 i N and i N . 1 1 2 2 i N i 2 2 2  =  → iN = =  0 Fe  integration path i N + i N = 0 1 1 2 2 For the sinusoidal form of the primary applied voltage U (rigid grid) and the magnetic 1 linkage ˆ jt (   = e ) the complex notation of the ideal transformer equations is introduced: U   1 = − E = j  = j  N , 1 1 1 2 2 U = −   =  2 E j = j  N , 2 2 2 2 2 I N + I N = 0 . 1 1 2 2 The first two equations give the voltage ratio: U U N U 1 11 j (u1−u2 ) N = → e =1 . N U N 2 2 2 2 K U N 11 = = for  =  U u1u2 U N 2 2 The voltage U or U dictates the magnitude (amplitude) and phase position of the flux. 1 2  ˆ 2 U 2 U 2 U 2 U ˆ = 1 = 2  = 1 =2  N  N  N  N 1 2 12  π π π π  2π π / 2  =  − =  − =  + =  + 4, 44 = = 4 =  4 1,11 Φ u1 u2 e1 e2   2 2 2 2  2 2  E  ˆ ˆ =  N = 4, 44 f N  is the RMS value of the induced voltage in the primary and 1 1 1 2 E  ˆ ˆ =  N = 4, 44 f N  in the secondary winding. 2 2 2 2 The phasor diagram shows the position U , E and .  The third equation gives the current ratio 1 1 K : I I I N I N 1 = − 2 → 12 K = = for  =  − π . I i2 i1 N I N 2 1 2 1 The current phasor diagram is shown in the middle Figure for an example N  N . 1 2 The right Figure shows the ampere-turns for which: I N = −I N . 1 1 2 2 U Re Re 1 I N 11  I1 i1   i2 I N 22 E I 1 2 No-load of an ideal transformer U = U U (grid voltage) 1 1 grid U 2 I = 0 2 U N N  2 2 = − E = U = − E = j  N 2 2 1 1 2 N N  2 1 1 After I = 0 follows from I N + I N = 0 → I = 0 . 2 1 1 2 2 1 A phasor diagram is shown for the example N  E 2 1 N . 2 E1 Load of an ideal transformer 1 U = U 2 L 2  I L L U I 1 U I Z 1 I 2 L L 2  U U I 2 L = U Z = L 2 L I L E 2 = − I I L 2 Two current phasors appear on the secondary side of the diagram, whose phase position is shown in the example Figure: Z = R + jX . L L L According to the power flow theorem: Re( * * U I )  0 and Re( U I )  0 . L L 2 2 The magnitude and phase position of the voltage and flux are unaffected by the loading of the ideal transformer. Valid: N N  2 2 U = − E = U = − E = j  N . 2 2 1 1 2 N N 1 1 2 From the equilibrium condition  N I N + I N = =2 0 follow: I = − I . 1 1 2 2 1 2 2 N 1 At  =  or R = 0 any deviation from the value  = 0 would cause the flux and induced Fe m voltage to increase beyond all limits (  =  / R ) . m m The following Figure shows a complete phasor diagram with the imposed voltage on the primary side U and the load Z = R + j X on the secondary side. 1 L L L Energy balance: U U1 I L  * N N  = 1 2 * = U P Re( U I ) = − Re U I = 2 L 1 1 1  2 2  I 1  N N 2 1   = − * * Re( U I 2 2 ) = − = P Re( U I ) = P 2 L L L E In an ideal transformer, power is transmitted without losses, 2 only the voltage level changes according to the turn ratio: I E1 2 N U 21 U = U = . 2 1 N K 1 U Transformed (reduced) quantities It is not transparent for N N 11  1 or  1 the phasor diagram. N N 2 2 Therefore, transformed quantities are introduced for voltage and current. For voltage it is: U N1  = U = K U . For the current, it is from the equal excitation condition: 2 2 U 2 N 2  ˆ N 1 ˆ  = 2 I N  = 2 I N =  → 2 I  = I = K I = I . 2 2 1 2 2 2 2 2 I 2 2 N K 1 U The power must also remain the same and valid: U I   = N N 12 U I = U I → apparent power S = S and therefore it is P = P . 2 2 2 2 2 2 2 2 2 2 N N 2 1 For an ideal transformer, the following equations now apply: U   U = U U 1 2 = 1 = j  = j  L N , 1 1 2 2 I 1 = I  L  N N   1 2 1 U  = j  = j  N = j  N . 2 2 1 N 2 2 N 2 2 2 The difference of the two equations gives: U −U  = 0 . 1 2  For magnetic flux linkage:  N I  2 1  =  = K  →   −  = 0 . 2 2 U 2 1 2 N 2 From the equation for the ampere-turns I N + I N = 0 , the sum of I + I  = 0 . 1 1 2 2 1 2 Under load, it is possible to write: U  = 1  2 L 1 2 U and I = I → Z  = = =   Z K Z . L L L L L L U L N N N U 2   N  2 N N 1 I  2  L The above Figure shows the corresponding phasor diagram of an ideal transformer with transformed quantities of the secondary side to the primary side. Real transformer operation Because: 1.    → H  dl = 2 I N + 2 I N  0 , Fe  Fe 1 1 2 2 K 2. hysteresis loop → hysteresis losses in the iron P , Feh 3. B = f (H ) (non-linear dependence between B and H ) → higher harmonic components, 4.   0 → eddy current → eddy current losses in the iron P , Fe Fee 5.  =   0 → leakage magnetic linkage (   −  =  0) , air 0 1 2 σ 6.    → ohmic resistance of the conductors → voltage drops ( R I and R I ) → c 1 1 2 2 joule losses in the windings = 2 2 P R I + R I , Cu 1 1 2 2 7.  → eddy current in the massive conductors → σ additional eddy current losses P , Cue 8. ohmic and leakage reactance voltage drops  → U − U = 0 . 1 2 General equations The crosses and dots indicate the direction of the eddy current I . e The following equations apply: 1 2 1 d  U = + 1 R I , 1 I 1 1 2 dt I N e  U 1 1 U 1 d 1 2 I N 2 2 2 U = R I + 2 , 2 2 2 dt I N  + I N + I = . 1 1 2 2 e 2 In order to facilitate the treatment of a real transformer and given the specific operating condition, the transformer is treated as: a) a current-ideal voltage-real transformer, b) a voltage-ideal current-real transformer. Operation of a current-ideal voltage-real transformer A real transformer is current-ideal for:  =  and  = 0, Fe Fe  =   0 and   . air 0 c No excitation is required to magnetize iron, and therefore I N + I N = → 0 I + I  = 0 . 1 1 2 2 1 2 As a result of the leakage, the  U − U  0 . 1 2 General equations describe the state of a current-ideal real transformer with transformed quantities (if the   = j t ˆ e ): U  1 = R I + j  , 1 1 1 2 U     =  2 R I + j , 2 2 2 2 I + I  = 0 . 1 2 The equal loss condition gives: 2 2 I R   = I R → 2 2 2 2 →  2 2 R =   2 I  2 R = K R  = −  , because the third equation implies I I .   2 U 2 2 1 I 2  The voltage difference from the first two equations is: U − U  = (R + R I  ) + 1 j (    −  ) / 2 = U +U . 1 2 1 2 1 2 r σ The U = ( R + R I  ) = RI part (ohmic voltage drop) is in phase with the current. r 1 2 1 1 The U = j (     − ) / 2 part (inductive voltage drop) overtakes the leakage magnetic σ 1 2 linkage by 90 . Since  I = − I , the ampere-turns in the core column are opposed, and the difference in 1 2 magnetic linkage −    is present only in the space between the windings. In the Figure, 1 2 each winding is shown with only one thin turn wound around the core in the same direction (usually the direction of the right-hand screw). column axis  B B     I 2 is Ix 1 an  m I I 21 lu co N N 21 instantaneous direction of leakage flux The winding always has a finite thickness. Therefore, the leakage in the cross-section has the shape of a trapezoid. The leakage linkage is written with the total inductance:    − = L I 2 and, hence, voltage drops U 1 2 − U  = U + U = (R + jX )I . σ 1 1 2 r σ σ 1 This equation corresponds to the following phasor diagram, where U is the imposed voltage 1 and the load Z = R + jX . L L L U −U  is the hypotenuse of a rectangular − Kappa triangle. 1 2 Energy balance U 1 jX I = j (   −  ) / 2  1 1 2 If there is I = −I  and U  = U  , it will be: 1 2 2 L P = Re * U I = Re 1 U  I * + (R + R I I  * ) 1 1 L L 1 2 1 1 ( ( ) ( )U −U  RI )1 = 1 2 U  2 = 2 2 P + R I + R I , L 1 1 2 2 I1 P = P + P . 1 L Cu (    ) 1 − 2 The received power is used for the power at the consumer and the losses in the transformer winding.  I 2 In a short-circuit current-ideal and voltage-real transformer at the imposed grid voltage U1 U = jX I   1 U = U and U = 0 will be: 1 grid 2 U = U + U = (R + jX )I . 1 r σ σ 1 I 1 Short-circuit current: U 1 I = − I  = , Z = + R jX , U = RI 1 2 sc σ r I  1 Z 2 sc (R = R = R + R ) , ( X = X = X + X  ) . sc 1 2 σ sc σ1 σ2 Operation of a voltage-ideal current-real transformer A real transformer is voltage-ideal for: U − U  = 0 on condition 1 2  = 0,  =  (no voltage drops or losses P ) air c Cu   ,   0 (needs magnetizing current and has losses P ) Fe Fe Fe Because  = 0 will be:  = N  and  = N  . air 1 1 2 2 Voltage equations: d u = N 1 1 d t u d  N = 2 2 d t For a voltage of cosine (sine) form u = U ˆ cos( t +  ) to be applied, the flux must also 1 1 u1 be of sine (cosine) form:  U  ˆ cos( t +  − π/2) = 1 u1 =  ˆ sin( t +  ) .  u1 N 1 Magnetization phenomena in operation on a rigid grid The magnetic curve of the core material B = f (H ) with the dimensions of the core and the definitions of  = B  dA and  = H dl gives us the magnetization characteristic l    A K = f ( )  of the magnetic circuit. B  = H dl l  K H  = B dA  A The sinusoidal voltage determines the sinusoidal flux with phase shift 90 . Taking into account the non-linear magnetization characteristic (hysteresis loop)  = f ( )  results in excitation ampere-turns that are not sinusoidal in shape.     1     t  h According to Fourier, a function  = f ( ) t has odd-order components ( = 3, 5, 7, .....) in addition to the fundamental harmonic component ( = 1) .  overtakes the flux  by an 1 angle of ˆ  . Between ˆ  and  a non-linear dependence applies. h 1  ˆ  1 h   ˆ 1 Losses in the core They are divided into hysteresis losses (due to the alternating magnetization) and eddy current losses (due to the eddy current in the core lamellae). Hysteresis losses P 1 ˆ =  t T + t T +  Feh   d = f   d = f ( , f 2 ) . Approximately valid P  f  . Feh T t t Eddy current losses In a short-circuit loop (core lamella) the equation applies: 0  = R I − E . Because it is E = − j  , it will be: e e e e 2 I E   e = = − j e R R 2 e e The eddy current contributes to the ampere-turns on the integration path through the magnetic core ( N = 1): e  = (I N + I N + I ) 2 . 1 1 1 2 2 e  1fic Combining − 2 I e  and I gives 1 e  1 the total (fictitious) ampere-turns:  I  e h  E e =  − 2 I = 2 I N + 1fic 2 I N . 1 e 1 1 2 2 Eddy current losses in the core  2 ˆ     P = 2 2 2 ˆ  R = = f ( , f  Fee ) e I R  = 2 ˆ e e    R 2 2R e e massive core laminated core e a ll e n insulation " " lam   / , n nR , E / n , R , E Fe e Fe e The eddy current losses in the massive core are: P = E I , if the current is Fee e e I d 1 d(  / ) n 1 1 d 1 = − and I = − = − . e en 2 d t R d t nR n d t R Fe Fe Fe For an n-times laminated core thickness d , there are eddy current losses: P = E I P 1 e e Fee   2 n = → P = f = f ( d ) . Fe(en) 2 2 Fe(en)  2  n n n  n  Total magnetization losses in the core: P ˆ = P + P = f ( , f  ) . Fe Feh Fee The losses in the iron core are given as specific losses per kilogram of mass for a given thickness of steel: p = x 2 2 k f B + k f B (W/kg) . Fe h e The exponent for hysteresis losses x  1, 6 2,8  and depends on the value of B . Electrical steel manufacturers give the losses as curves p = f ( ) B for f = 50 Hz , or at Fe magnetic flux densities of 1 T and 1,5 T. The specific losses for any one B are calculated using the equation: x p  x p . Fex  2 B  Fe    B  Example of a curve B = f (H ) with losses data The curve for the magnetic flux density is given as a function of the magnetic field intensity in two different scales, marked I and II. Steel sheet thickness d = 0,35 mm . The value for H can be given in RMS values and measured, e.g., at 50 Hz, (AC curve), or in peak values when magnetizing with a direct current (DC curve). B T ( ) II 1,8 I 1, 6 1, 4 1, 2 1, 0 p1 = 1, 3 W kg 0,8 p = 3,1 W kg 1,5 d = 0, 35 mm 0, 6 0, 4 I 500 1000 1500 2000 H A m () II 5000 10000 15000 20000 No-load current The no-load current i is calculated from , which contains higher harmonic components.  10 i  i 1e i i = − (fundamental harmonic of 10 10,1 N N 1 1 the no-load current) i 10,1 i = ( i i 10 1 v 3 5  ) (  − +  +  + + ...  / N1 ) 10,3 i = i + i + i + ... + i 10 10,1 10,3 10,5 10,  i 10,5 t Fourier analysis gives the amplitude of the currents ˆ I ˆ 10 I and, hence, I = . 10  10  2 The common no-load current will be: I 2 2 2 2 = I + I + I + + ... I . 10 10,1 10,3 10,5 10, If the higher harmonic components are neglected, only the fundamental frequency magnitudes remain. Therefore, a complex notation of the equations is used. U = −   1 E = j  = j  N 1 1 1 2 2 U   2 = − E = = 2 2 j  j  N 2 2 2 Using transformed values, it is valid: U    2  = − E  = j  = j  N 2 2 1 2 2 and follows U − U  = → 0    − = 0 (no leakage magnetic fields). 1 2 1 2 The equation of equilibrium of ampere-turns: I N +  1 I N = − I 1 1 2 2 e 2 and with transformed values I  1     1 1fic + I = − I =  I 1 2  e  (the higher harmonic components are neglected). 10 N 1  2  2 N 1 The deviation from ideal conditions is represented by the ampere-turns  required for 1 excitation and reverse action I in the core. e No-load voltage-ideal current-real transformer I = 0 → the voltage on the primary side is equal to the grid voltage. 2 U = 2U 1 U = U  U →  = 1 2 1 grid j  N 1 The energy balance: U  1fic  =  U , I I =  2 1 1 101fic 2 N 1 I  10 10 I 10w I 10   = P P = P = P + P 10 mag Fe Feh Fee P = Re * U I 10 1 10 ( ) P E = E  = U I cos  1 2 10 1 10 10 Load of a voltage-ideal current-real transformer U , Z = load impedance 1 L I = − U U N U L = − = − 2 2 1 I = − 2 L Z Z N Z L L 1 L Even under load, the imposed (applied) voltage U dictates the magnitude and phase of the 1 flux  = 2U / ( j N ). So the  stays the same as in no-load. 1 1 1fic Load with  1fic I → I N → I → I N + I N = 2 2 2 1 1 1 2 2 2  →  → E → U + E = 0 1fic 1 1 1 By introducing transformed secondary values on the primary side, it is valid: I  I +  = 1fic → I + I  = I . 1 2 1 2 10 2 N 1 Every change I causes a change I , so that I = const. 2 1 10 U 1 2 I N 11 U = U  = U  1 2 L I 1 U I 21 2 I N 22  I  2 1fic I  10  I I  2 2 Energy balance P = * * * ( ) ( Re U I = Re U I + U I = P + P = P + P 1 1 1 L L 1 10 ) L mag L Fe The accepted power P covers the power of the consumer and the losses in the iron, but not 1 the losses in the transformer windings. Analytical treatment General equations for stationary operation The derivation will be carried out for linear transformer theory, where:  = const., → no hysteresis loop and  = → 0 hence P = 0 Fe Fe Fe  =  = const. and  = const. air 0 c The linear theory implies linear relationships B  between the flux linkage and the currents.  = L 2 I + L 2 I 1 1 1 12 2  = L 2 I + L 2 I 2 H  21 1 2 2 It is also true that L = L . Now write the two voltage equations: 12 21 U =  1 R I + j  = R I + jX I + jX I , 1 1 1 1 1 1 1 12 2 2 U  2 = R I + j  = R I + jX I + jX I . 2 2 2 2 2 21 1 2 2 2 By introducing transformed quantities for winding 2 with already known relationships U  = K U , I = K I ,   = K  , continue to 2 U 2 2 I 2 2 U 2 X  = 2 2 X  = K X , X  = K X and R  = K R 12 21 U 12 2 U 2 2 U 2 it is:  1 U = R I + j  = R I + jX I + jX I   , 1 1 1 1 1 1 1 12 2 2 U   2 R I   =  + j  = R I   + jX I  + jX I   . 2 2 2 2 2 21 1 2 2 2 The difference of the two voltage equations gives the deviation from the ideal situation. U −U  = R I − R I   j (  +    − ) / 2 1 2 1 1 2 2 1 2 Transformer equivalent circuit A transformer with two electrically isolated windings, i.e., isolated circuits, is converted into a circuit with electrically coupled circuits. This is achieved by adding the first voltage equation  jX I  and the second equation  jX I   , and combining the individual terms in 12 1 21 2 a meaningful way. U = R I + j( X − X  )I + jX  (I + I  ) = R I + j( X − X  )I − E 1 1 1 1 12 1 12 1 2 1 1 1 12 11 U  = R I   + j( X  − X  )I  + jX  (I + I  ) = R I   + j( X  − X  )I  − E 2 2 2 2 21 2 21 1 2 2 2 2 21 22 The following circuit corresponds to these two equations. U − U  1 2 R j(X − X  ) j(X  − X  ) 1 12 2 21 R 1 2 I I  I  1 2 L U E jX  12 1 I + I Z  1 1 2 U  L 2 Deviation from voltage-ideal transformer → voltage drop U − U  across longitudinal 1 2 links, deviation from current-ideal transformer → magnetizing current I = + I I  μ 1 2 flowing across the transverse link. Further, ( X − X  ) is the total leakage reactance between the two windings (no-load 1 12 winding of side 2) for windings 1 and 2 on the same column, and, analogously, ( X  − X  ) 2 21 is the total leakage reactance (no-load winding of side 1). If the two windings are separated, then: X − X  = X and X  − X  = X  . Taking into account the losses in the iron 1 12 σ1 2 21 σ2 P = P (I = + I I  ) , the equivalent resistance for the losses in the iron R is added in 10 Fe 0 1 2 Fe the equivalent circuit in parallel to the magnetizing (main) reactance X  = X . 12 m I + I  U 1 1 = I jX I 2 0 1 1  I I R I w μ 1 1 E1 RFe jX  = jX 12 m jX  2 I  2 − E = − E  1 2   R I 2 2 U  2 Now it is possible to draw a complete I 1 phasor diagram of a loaded transformer  L I 0 −I  = I  2 L with impedance Z = R + jX . L L L I w I  Approximate situation If the transformer is excluded from the observation during no-load and at low loads, it will be I + I  = → 0 I  = − I and the difference of the voltage equations: 1 2 2 1 U −  U = (R + R I  ) + j (X X ) ( ) 1 2 1 2 1 ( −  + X  − X  1 12 2 21) I , 1 U  − U = (R + jX )I . 1 2 σ 1 The equation U − U is illustrated by a 1  R jX  2 simplified equivalent circuit. U I1 U  1 2 Transformer tests They are carried out at no-load and in a permanent short-circuit. They are used to determine the actual operating conditions of the transformer. No-load test For an I = 0 → I = I  I → transformer is the voltage-ideal. The measurements 2 1 10 1N will give: 1) nominal voltage ratio U E N 1 11 K =  = and U U E N 2 2 2 2) the no-load characteristic. The sinusoidal voltage U is said to dictate the sinusoidal flux, because U + E = 0 , 1 1 1 → ˆ 2U ˆ  = 1 generated by  = 2I N . 1fic  10 1 N 1 Therefore, due to ˆ ˆ ˆ ˆ E  U = f ( )  ,  = f( )  and  = f (I ) , the no-load characteristic 1 1 1fic 10 U = f (I ) is similar to the magnetizing characteristic or the magnetic curve, if the air gap 1 10 can be neglected. A 1 2 U W1 U1N U V V 1 I10N I10 3) Iron losses Since I  I , the losses in the primary winding are negligible ( I  0, 01  0 02 ,I ). 10 N 10 N Measurement: P  P (at U = U ) gives the magnetization losses. From the losses it is w Fe0 N possible to calculate the equivalent resistance R , I , from I still I and X  = X . Fe 0w 0 0μ 12 m Short circuit test The switching scheme measures the current I and the power P at the variable voltage 1sc 1sc at the transformer terminals. 1 2 A W U V short-circuit 1 The measurements give: 1) the short-circuit voltage at which the rated current flows through the transformer windings. Relative value for the short-circuit voltage: u = U U scsc * or u = 100 . sc sc U U N N In the short circuit, the hypotenuse of the Kappa triangle U = U and the scN 1 U = U cos  and U = U sin  . U r scN sc σ scN sc  U = U 1 scN Permanent short-circuit current I U I = I 1sc1N 2 2 = N , if Z = R + X .  sc sc σ sc Z U r sc Relative value of the permanent short-circuit current i * I U U 1 1 sc = = N N = = → I = I sc sc N I Z I U * * u u N sc N sc sc sc Relative value for the ohmic and inductive voltage drop: u R I R U N * = = = N r , where = Z − is the rated impedance. r N U Z I N N N Also applies * N N CuN * u r = = = R I 2 R I P = = p r (relative losses in the windings), and CuN U U I S N N NN u * σ N σ N σN * X I 2 X I Q = x = = = = q σ (the relative reactive power of the leakage field). σ σN U U I S N N NN 2) Losses in the windings From the short-circuit equivalent circuit it can be seen that the induced voltage is approximately half of the applied voltage: E  U  1 N →   → I  I . 1 scN 10sc 10N U E , 2 2  1 1 U  U U 11N = 1 E 10 2 2 R1 j(X − X  ) j(X  − X  ) R 1 12 2 21 2 E  1 1 U1 2 U E1 jX  1 12 U  = 0 2 I I 10sc10N I10 In a short circuit, when the voltage U =  (4 16 %)U is reduced by I = I and sc N 1sc 1N I = I for:    → B  B → P  0 ; P = (P )  P = 2sc 2N sc N sc N Fesc w 1sc I = 1 I N CuN = 2 I (R + R ) (rated losses in the windings). 1N 1 2 A wattmeter shall be used to measure the total short-circuit losses in the windings, i.e., joule losses and eddy current losses in the case of massive conductors. Since   N  → I  0 → X    a simple short-circuit equivalent circuit is valid, sc μ 12 2 drawn for approximate conditions (p. 44). The short-circuit impedance can also be obtained from U and I : sc sc Z U sc = R + j X = . (R = R + R and X  X ) sc sc sc sc 1 2 sc σ I sc Transformer operation on a rigid grid For a rigid grid: U = const. 1 U or U  is a function of the size and type (character) of the load Z , as shown in the 2 2 L Figures below. RI 1 jX I RI jX I  1 1  1 U RI jX I 1 1 U 2 U  1  U 2 2 U 1 U 1 I  = I L1  L I 1 I  1L R L C a) b) c) Kappa diagram For the Kappa diagram it is not derived from U = const. , but I = const. , U  = f ( ) , and, 1 1 2 L here, the position of the Kappa triangle is unchanged. From the equation U − U  = (R + jX )I express U  : 1 2 σ 1 2 U  = − (R + jX )I + U . 2 σ 1 1 ohmic − inductive ohmic − capacitive load U load In the complex plane, draw:  = 0 U L 1  U  2 1 I I 1) − ( R + j X ) I , σ 1 1 = 1 0 =1 I I I 1 1N 1N  2) circles of radius U for different  1  =  = − L  L L − j X I 2  I , e.g., I = I and I = 0 on the 1N 2 1 1 1N 1 0 − RI 1N starting line.  0  2 starti I  2 ng For any phase angle  is line L obtained ΔU respectively U  . 2 External characteristic U = f ( ) I 2 2 U2 ohmic - capacitive U 2N load ohmic ohmic - inductive 0 1 / 4 2 / 4 3 / 4 4 / 4 I N The Kappa diagram gives for arbitrary values of current I , e.g. I / I = 1/ 4 , 1/ 2 … 2 2 2N and  = const., external (load) characteristics U = f ( ) I at U = const. L 2 2 1 Energy balance, losses and transformer efficiency Energy balance - power flow diagram input power P 1 P magnetizing losses mag − P conductors winding losses Cu − (joule losses) P additonal losses add − P 2 output power Losses Total losses are: P = P + P + P . l Fe Cu add The additional losses are divided into: those in the iron − voltage-dependent − which are captured in the no-load losses P , and those in the conductors − load-current-dependent − 0 (eddy-current losses in massive conductors in higher-power transformers), and are captured in the short-circuit losses P . sc For any load, i.e., y = I I / , the losses are: N  2 I  P = 2 I S P + P =  P + y P for y = =  l , 0N scN 0N scN  IN  I S NN where S is the apparent transformer power. Introduce relative losses: p* l 0N N scN N 2 0N * P * P S P S p = = + y = + l p y . scN S S S S S y N N At a given load (a given current  I  I ), for pi N y * = y the total losses ( p) are the lowest opt l ( * * 0,15 p / y = p y ) and the efficiency is the 0N scN  p l highest.  0,10 p y scN 0, 05 p  0N y Efficiency P P − 0 1/4 2/4 3/4 4/4 y It is defined as:  = 2 P  1 l y or = .opt P P 1 1 For 2 P = y S cos  and P = P + = P yS cos  + P + y P it will be: 2 N 1 2 l N 0N scN  P y S cos  = 2 = N → =  f ( , cos ) y  . 2 P y S cos  + P + y P 1 N 0N scN Example: S = 20 kVA, y = 1, cos = 1, 0 → =  0,97 N cos = 0,8 → =  0,963 Current transformer It operates at an imposed current I which is supposed to be constant. Of course, it is not 1 possible to assume that the current in a circuit will remain unchanged if a current transformer is connected to it in series. It is subject to the same laws as a voltage transformer, i.e., the same equivalent circuit and the same phasor diagram. I R j(X − X  ) j(X  − X  ) R  1 1 12 2 12 2 I  2 1 Z j X12  Z  1 L Ideal current transformer Valid: N I 1 1 I N + I N = 0 → I = − I = − or I + I  = 0 → I  = −I 1 1 2 2 2 1 1 2 21 N K 2 I The load current I = −I causes a voltage drop across the complex resistance Z of the L 2 L two-terminal U = Z I = −Z I = U . L L L L 2 2 The voltage drops across a current transformer feed back to the circuit. As it is considered to be the voltage of an ideal transformer 2 U = K U and Z  = K Z 1 , U 2 L U L it will be: = 2 U = K U K ( − Z I ) = K Z K I = K Z I = Z  I . 1 U 2 U L 2 U L U 1 U L 1 L 1 U is the voltage on the primary side, and, at the same time, the voltage drops across the 1 transformer. Therefore, the impedance Z must be small relative to the impedance of circuit L 1, which determines the current I . 1 Phasor diagram of a loaded ideal current transformer U I 21 I = I  1 L I L U = U  U  1 2 = L U 1   I 2 I − I  = 0 1 L E I  2 2 Analytical treatment of a real transformer From the voltage drop across the secondary winding (p. 42) U = R I + jX I + jX I 2 2 2 21 1 2 2 for U = −Z I = −(R + jX )I is the current: 2 L 2 L L 2 I jX 21 = − I ; I  I and are no longer displaced by 180 . 2 1 2 1 R + j X + Z 2 2 L Voltage drops across the primary winding: U  2 X  12 = R I + j X I + j X I = + + =  R j X  I Z I . 1 1 1 1 1 12 2 1 1 + + 1 1 1  R j X Z 2 2 L  The complex resistance Z is, according to the alternative circuit in the Figure, the resistance 1 between the input terminals on the primary side of the current transformer. Since U = f ( ) I will be I = 0, U = 0 and U = 0. Therefore, the current transformer is not 1 1 1 1 2 used as a voltage source, but for measurements. Measuring current transformer For a real transformer, it is: I + I  = I  0 or I − I  = I . 1 2 0 1 L 0 Deviations from the ideal conditions result in ratio error and angle error. E , Z =  20 L I 1  B, ,  E I I  = −I  2 L I Z , E L2 0 I I = I 0N 01 I 0 Ratio error: e = ΔI CT 100 (%) I 1 Angle error:  → in degrees or minutes To minimize these two errors, the current transformer operates in the linear part of the magnetization characteristic (B = 0,08 1 T).  Accuracy class (0,1 – 0,2 – 0,5 – 1 – 3 – 5) defines both errors, e.g., the class 1 → e = 1 % and  = 1 =  60 . CT Since the ratio error changes as the load increases above the nominal value, the overcurrent number is also a known quantity. The overcurrent number is that multiple of the rated current at which the ratio error reaches the value −10 %. Danger of open secondary terminals It's normal: I + I  = I . 1 2 0N For Z =  → I = 0, I  I →  = x → B = xB and L 2 1 0 0 0 P ( )2 2 = k B / B = x P → core heating. Due to  = x → E = xE (increase of Fe 0 FeN 0 20 2 voltage at the terminals). Therefore, after the load is removed, the secondary terminals must be short circuited. Three-phase transformation The three-phase system is the most widespread in the world. One reason for this is that the power in a symmetrical three-phase system does not have an AC share, as is the case in a single-phase system for example. The three-phase system also makes it possible to take advantage of the phenomenon of rotating magnetic fields in rotating machines. Three-phase voltage transformers connect individual three-phase systems of different voltage levels. Options: a) three single-phase transformers b) three-phase transformer Basic winding connections There are three typical connections: a) star Y b) delta D c) zigzag Z For star and zigzag connection  = I 0 , if the winding has no zero conductor. In a delta connection, the phase Z and line currents can be asymmetrical. A B C D Y Phase shift This occurs between the primary and secondary actual or imagined value of a phase voltage. The phase shift is: n  30 electric ( n =  0 12) . Typical shifts are: n = 0, 5, 6, 11. It is given as the phase lag of the lower voltage phasor against the higher voltage phasor. The phase shift is identical to the displacement of the hour hands by whole hours. Vector group 12 Example of marking: Dy5 n 0 = 5 11 1 Dy5 A Capital letter: – delta connection of the primary side; 10 2 c lowercase letter: – star connection on the secondary side; 9 3 b Number: a C B – phase shift ( 5 30 8 4  = 150 ). 7 5 The direction of the secondary phase voltage (U ) 6 a is equal to the direction of the primary line voltage (U ) between phase A and phase B. AB The Dy5 vector group can be changed to Dy11, and vice versa, by cyclic permutation of the two terminals on the primary side and the next two terminals on the secondary side. The Yy0 vector group can be changed to Yy6, and vice versa. This is done by swapping the starts and ends of the windings that are connected to the neutral terminal of the star on the primary or secondary side. Example of use: generator transformer Dy5 distribution transformer Yy0, Yd5 local grid Dy5, Yz5 Types of cores Three-phase shell transformer UA  A A It is obtained by adding three 2 single-phase transformers.  UB C  B B  2  C  B 2 A 1 2 2 UC    C A ( C  +  ) B 2 C B 2 2  1 ( +  ) B A B 2 The yoke of a shell transformer has half the cross-section of the column if the winding on the middle column is wound in the opposite direction to the outermost two, or the winding connection sequence on the middle column (phase) is reversed. Valid  =  / 2 +  / 2 or  / 2 + / 2 . y A B C B Three-phase core transformer It is made up of three single-phase transformers and is technically usable as an asymmetrical design (Figure c). symmetrical B 3-level core B A  med C C A A B C a) l = l = l b) FeA FeBFeC c) It has two magnetic nodes in which the condition  + + = 0. A B C The Figures show the flux waveform for  =  ˆ and  = 0. A A B t t   A A   C B  C  B A B C Asymmetrical load I I I A B C a) There are no problems in the YNyn connection with zero conductors on the primary and secondary sides. b) In the Yyn connection with a zero conductor only I a on the secondary side, asymmetry occurs. Worst case: pure single-phase load. 0 a b c Z L Because it is  + +  = 0 and  =  =  , shall be  =  =  . A B C A B C A B C With asymmetrical loading, the balance of ampere-turns on the columns will be: 0 I N + I N = I N a 2 A 1 x1 1  → I = I  B a + I N = I N 3  2 B 1 x 1 I = −  ( I + I ) = − I A B Ca 1 3 0 + I N = I N → I = I   C 1 x 1 C a 3  I N + + + = =  UAph A ( I I I ) N 3 I N 3 / 2 , because it is a 2 A B C 1 x 1 air in the star I + I + I = 0 . In-phase excitation A B C  1 1 N 1 2 = 2 I N = 2 I N → I = I = I UCA UAB air x 1 a 2 x a a 3 3 N 3 I 1 a  UCph UBph →  → E → U (additional) → air zero shift air air add Uadd  C B U A +  B +  C =  BC air Yzn connection A B C I A I B  I a air c) In the Yzn connection with no zero conductor on the primary I a side there is no problem (only two phases are loaded). 0 a b c d) There is also no problem in the Dyn connection, because Z L the load current on the primary side flows only in one phase. Autotransformer If the two windings of a normal transformer are combined, part of the winding will be common to both voltage sides. Common – parallel winding → difference (differential) currents Series winding →load current of primary or secondary Use → to save material (iron and conductors). This saving is maximum for the case of K  1 . In the three-phase version, the Y connection is used. U a) p − parallel b) (common) winding I N s Ns s − series winding s s I = I I = I s 2 U U 1 s = I + I I = I I 1 s p 2 s + p p p =+U U=+ U 2 1 N I 1 = U I N U = U p p p p p 2p U U For an ideal single-phase autotransformer (connection a): I N − I N = 0 → s s p p → N p U U −U N s 2 1s I = I and = = . s p N U U N s p 1 p The transformer's inherent or typical power, i.e., the power for which the transformer is built, is: S = U I = U I = (U −U I ) or for connection b) S = (U − U I ) . t p p s s 2 1 2 t 1 2 1 Passing power: S = U I = U I pas 1 1 2 2 The equation for the typical power is: =  U   U  1 − = − 1 S U I 1 S 1  t 2 2   pas   . ( S S ) t pas  U 2   U 2  The reverse would be true for the connection b): =  U   U  2 2 S U I 1 − = S 1 − . t 1 1   pas    U   U 1 1  INDUCTION MACHINE An induction machine (IM) has some similarities with a transformer. V1 stator Primary winding - on the stator  Secondary winding - on the rotor rotor cage bar There is an air gap  between them. cage ring shaft Number of phases on the stator m s U1 Number of phases on the rotor m W1r (m  m ) r s The stator currents induce rotating ampere-turns, which excite a rotating flux that induces a voltage in both windings. The frequency of the induced voltage in the stator f = f . s Frequency of induced voltage in the rotor f  f or f  f . The rotor at standstill operation r s r s is equivalent to a short-circuit of the transformer; the rotor winding is short-circuited. The resultant rotating magnetic field and currents in the short-circuited rotor winding produce a force that rotates the rotor. The rotating magnetic field rotates with synchronous speed n , the rotor with speed n n s − hence asynchronously. s  Description of construction Stator – laminated sheet, with slots in which the m phase winding is located. s Rotor – laminated sheet with slots. We distinguish between: a) a wound rotor I r b) a rotor with a short-circuit squirrel cage I rb c) massive iron rotor The stator and rotor can also be cylindrical or disc-shaped. Induction machine windings Three-phase belt windings: are single, double, and combined layer windings. Single-layer winding: II I perimeter 1 diameter winding (  =  18 ) c p distribution 17 2 16 3 15 4 III 14 5 III 13 6 12 7 11 8  10 9 =  c p I II 3 Each phase has a belt:  360 360 = = = b 60 and there are q slots in it. 2m 6 For Q slots on the perimeter of the machine: Q = Q / (2 ) p the number of slots per pole, and p 2 p = 2 the number of slots per pole and phase. The pole-arc is Q = 18 slots, or the width of the coil expressed by the number of slots. The voltages of the individual bars (or conductors) of the slots (or their phasors) are offset from each other by an electrical angle  360 = p  = Q p and form a slot star. Q Three-phase single-layer winding: calculation of the data for 2 p = 2 and Q = 18 slots. The number of slots per pole is Q = Q 18 = = p 9 , the number of slots per pole and phase 2 p 2 q = Qp 9 360 360 = = 3 , the electrical angle between the slots  = p =  1 = 20 . m 3 Q 18 E E Eb1 Eb2 b18 E potential b3 b17 circle  r belt Ebelt  E b q  / 2  / 2 1) Distribution factor f geom.        E b 2 sin r  q  sin  q   2   2  = = = d  E       b q r 2 sin   q sin   arit.   2   2 For our example, the distribution factor: sin    q   sin 3 10  2 ( )   0, 5 f = = = = 0, 96 . d    3sin 10 q ( ) 3 0,17365  sin     2 Diameter winding: the width of the Y coils expressed in the number of slots Y = Q = 9 . Q Q p Next, determine the winding pitch. The latter is defined as the expression: pitch = − + 1 ( 1 Y ) , i.e., for the slots from the initial (first) slot onwards. The pitch is: Q 1 (1 − +Y ) = − + = 1 (1 9) = − 1 10 , i.e., from the first to the tenth slot. Q  =   Y = Q c p Qp 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 W2 U1 U2 V1 V2 W1 The start of the first phase is marked U1 and the end U2. In the Figure we see that the start of the second phase is in slot 7, which is shifted by 6 20  = 120 (electrical) with respect to slot 1, and analogously, the start of the third phase is in slot 13 by 12 20  = 240 . The phase shifts correspond to the condition for the generation of a circular rotating field. The direction of the current in the slots of the first phase changes twice. This corresponds to two poles. The same argument also applies to the three phases taken together. The directions of the currents in the Figure correspond to the position of the timeline  t = 60 , when the currents of the first and second phases are positive and of half amplitude, and the current of the third phase is negative and maximum. In the case of twice as many poles (2 p = 4) , the number of slots is also twice as large, i.e., Q = 36 . The number of slots / pole is Q = 36/4 = 9 , the diameter width of the coils Y = Q = 9 , the p Q p number of slot / pole / phase q = 3 , the electrical angle between the slot  360 360 = p =  2 = 20 . Q 36 Thus, practically all the data are the same as for the two poles on 18 slots. The winding pitch (of the coils) will also be the same 1 (1 − + Y ) 1 (1 9) 1 10 = − + = − . According to the Figure Q below (drawn for the first phase only), it can be seen that the group of the first three coils and the other three coils are connected in series, i.e., the end in the 12th slot with the beginning of the second group of coils in the 19th slot. The 19th slot is shifted with respect to the 1st slot by 360 (electrical). 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 U1 U2 2) Pitch factor In practice, it proves useful to shorten the winding, i.e., to shorten the pitch (fractional pitch winding), thereby shortening the end winding connecting the individual slots. This also saves on material (copper). In the end winding, which are outside the slots of the stator pack (in the air), no voltage is induced due to the weak field. Only a two-layer winding can be shortened by x slots (x  → 1    or Y  Q ) so that the c p Q p width of the coils is Y = Q − x . A two-layer winding has two coils in the slots with half the Q p number of turns. Therefore, there are twice as many coils as in a single-layer winding. Example for a two-layer winding: Q = 12, 2 p = 2, m = 3 , q = 2,  = 30 , Q = 12 / 2 = 6 , p Y = Q = 6 and the winding pitch 1 (1 − + Y ) = − + 1 (1 6) = − 1 7 . Q p Q a) Diameter pitch 1 7 − 1 2 4 5 3 E 6 7 8 9 10 11 12b1 E b2 −E b7 E t  c =  p E b1 Turn (loop) voltage E b7 E = E − E . t b1 b7 a) b) Shorted pitch 1−6 (Y = Q − = − = 1 6 1 5) Q p 1 −E 2 E b1 3 4 5 6 7 8 9 10 11 12 b6 E t  c   p     c c  p E p E b6  b1 b) Turn voltage E = E − E . t b1 b6 pitch factor: = geom.  E   1  c 2 E sin π t b     2     p  π   Y  = c Q π f = sin   = sin p        E 2 E  2 Q 2 t b  p   p  arit. winding factor: f = f f ( f  1) w d p w Example: Q = 24, 2 p = 4, m = 3, q = 2,  = 30 (electric angle), Y = Q = 24 / 4 = 6 Q p 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 1U1 1U2 2U1 2U2 a) Single-layer winding: pitch 1 7 − , drawn for the first phase 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 W2 U1 V1 W1 U2 V2 b) Single-layer winding: pitch 1 7 − (The directions of the currents are drawn for t = 60 .) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 U1 U2 c) Two-layer winding with shortened pitch: pitch 1−6, drawn for the first phase. Operation mode a) Rotor winding − open (transformer no-load) Amplitude of the stator MMF caused by the magnetization current: stator I m I m m 4 N f ˆ = s sws  2 I , s m 2 π s 2 p generates a rotating (main) flux rotor B A  ˆ 2 2 ˆ ˆ = = B  l . m 1 δ 1 p r π π 2 / π is the factor of mean value for a sinusoidal magnetic field. U s The rotating flux induces voltages: I m m  m ˆ E = 4, 44 f N f  , r r wr m  p =  B ˆ ˆ E = 4, 44 f N f  , s s ws m 1 frequencies f f = f . When the rotor is at s r E a standstill, E = E is the induced voltage of r ro r E the open terminals of the (wound) rotor. s The reduction factor or voltage ratio is: E N f s sws K = = . U E N f r r wr b) Rotor winding − short-circuit The amplitude of the rotor's rotational ˆ m 4 N f r rwr MMF is:  = 2 I . r r 2 π 2 p Due to the equilibrium condition, U = −E , to remain the same  as at no-load, the s s m current in the stator winding must increase from I to I . m s   + =  − r s r m U  s s I + I  = I s r m Due to the compensation of the ampere-turns: −I  r I s  r Current ratio  ˆ m 4 N f  = s sws ˆ 2 I  =  . r r r 2 π 2 p I m  m  m  I r K r m N f m 1 I r r wr r = I  E r I = = r I m N f m K r s s ws sU  Es r Rotating machine operation The notion of a slip  ( → v ) s s 1) s = n stator n − n f rel = s → n = n (1 − = s ) (1 − s ) − at rest s n n s s p rotor − rotates n =  0 n induction motor ( s =  1 0) , s ( → v )  ( → v ) r rel 2) n =  +  n induction generator (s =  − 0) , s 3) n =  − 0 brake, the rotor rotates against the rotating magnetic field (s =  +  1) . Slip is the difference between the synchronous speed of the MMF n and the rotor speed n . s Frequencies of the induced voltages: synchronous speed of the rotating magnetic field v = D n π s s rotor speed v = D n π relative speed v = − = v v Dπ (n − n) = D n π = D n s π = v s rel s s rel s s e = Blv (induced stator voltage − in the bar or conductor) sb s e = Blv = Bl s v = e s (induced rotor voltage − in the bar) rb rel s sb f = f = pn (frequency of the induced stator voltage) s s f = pn = pn s = s f = s f (frequency of the induced rotor voltage) r rel s s X = 2π f L = 2π s f L = s X (change of the mutual reactance) sr r sr s sr sr 0 Force and torque of a short-circuit rotor winding The occurrence of a force on a short-circuited rotor winding (squirrel cage): rotating magnetic field of the air gap B → E = E → I . 1 rb r rb A force is acting on a single bar (cage) in a rotating magnetic field on a current-carrying conductor according to Lorenz's law: F B ˆ 1 = I l , b rb 2 where I is the current in each rotor bar, and l is the length of the conductor. rb In the equation for force, we have to account for the RMS value of the fundamental harmonic of the flux density. Considering the equation for the frequency in the rotor, the equation for the induced voltage in each rotor bar is z = 1 (half turn N = z / 2 1/ 2 = ): r r r E =  ˆ ˆ s r  B m = 1 E = sE s 2π f N f = l . r rb sb r wr 2 p 2 The winding factor of each rotor bar is taken into account ˆ ˆ f = 1 ,  = (2 / π) B  l and pole wr m 1 p pitch  = D π / (2 ) = π / p r p . Express B = f (E ) and the equation for the force takes a new p 1 r form: F p E I E I E I r rb r rb r rb = = = , b s  r s  r s v ms s where is the electrical angular frequency,   the angular frequency of the stator MMF s ( = ) and  =  / p = 2π f / p is the mechanical synchronous angular velocity. s ms s s The force is applied to each bar of the rotor with radius r = D / 2 . For the Q bars in the r rotor, we obtain the final expression for the torque from the internal power ( P ): int M = Q E I m E I m E I P r r r r r rb s r r int Q F r = = = = , r b s  s  s   ms ms ms ms if = m Q rr I I = I . (Transforming multiphase bar current to a three-phase rotor current.) r rb rb m m s s In the case of a squirrel cage, the number of phases m is equal to the number of slots in the r rotor Q , or equal to the number of bars (Q = m ) . The equation for the torque fails when r r r s = 0 . (For s = → 0 E = 0, I = 0 and get 0 / 0 .) r r Analytical treatment Voltage equations: U  m = R I + j X I + j (  N f ) = R I + jX I − E s s s σs s s ws s s σs ss 2 U  m = R I + j s X I + j s  ( N f ) = R I + js X I − sE r r r σr r r wr r r σr rr 2  j0 =  ˆ e generate the resulting ampere-turns:  =   + . m m res s r The magnetic flux density is given by: stator b ˆ    ( , ) x t = B cos ( x / )π t res res (  −   − s p 0 ) . eres rotor For ˆ ˆ  →  0 , an equivalent (smooth) air gap is considered: B =   . Fe res  res e For x = 0 , i.e., in the winding symmetry of phase "a", the resulting MMF is given by: s  ˆ =  cos   t + res s (is ) ˆ +  cos   r ( t + . ir )  2 2  3 4 1 0 =  l B =  l (N f )I + (N f )I m p res ( s r ) 2 p  s ws r wr π π 2 π 2 p e Now insert the equation for the main flux on the left-hand side of the two equations for the stator and rotor voltages on the previous page, and they take the following form: U 2  3 4  N f N f  0 = R I + j X I + j  ( N f )  s ws r wr l I + I , s s s σs s s ws p  r  π  s 2 π 2  p 2 p e  U = 2  3 4  N f N f  0  )  s ws r wr R I + j sX I + j s N f l +  I I  . r r r σr r ( r wr p s r π  2 π 2 p 2 p e   By introducing magnetic or air gap reactance  2 2  3 4 ( N f ) X   0 s ws = L = l are the rearranged voltage equations: m m p π  2 π 2 p e N f  2 N f  U r wr r wr = R I + j s X I + j s X I + j s X  r I . r  r σr r m s m r N f N f s   ws s ws The stator's own reactance is introduced: X = X + X . s σs m In the voltage equations, the term N f X r wr m X = = X = X is equal to the mutual or m sr rs N f K s ws U magnetizing reactance between the stator and the rotor: X = X K = X  m sr U sr Introduce the rotor's own reactance: = + +  m sr X X X = X = X + r σr m  . σr 2 σr   2 N f  r X X wr N f K K s ws  UU The result is two basic voltage equations: U = R I + jX I + jX I s s s s s sr r  U = R I + +  r jsX I jsX I r r rs s r r Induction machine equivalent circuit 1) By introducing reduced values and for the same number of phases, m = m is: s r U I r 2 2  = U K , I  = , R  = R K , X  = X K and X  = X K = X . r r U r r r U σr σr U sr sr U m K U 2) Add the first equation for voltage  jX I  and the second equation  j   X I . sr s rs r 3) The equation for the rotor voltage is divided by the slip to give: U = (R + jX )I + jX (I + I  ) , s s σs s m s r U  / s = (R / s + jX  )I  + jX (I + I  ) , r r σr r m s r where jX (I + I  ) = E = E  . These equations are matched by an equivalent circuit. m s r s r R s jX jX  R  / s  s  r r U I I  s I + I  r s r j X s m U  / s r For a squirrel cage rotor winding U = 0 , respectively U  = 0 , the equivalent circuit applies: r r R  s j X   s j X R / s  r r U I I  s I + r I  s s r j X m Approximate situation a) No-load run The assumption (R + jX ) I = 0 gives an expression for U : s σs m s U = U s + I  = j X ( I ) j X I → I = . s m s r m m m U j X s m I s b) Load The voltage drop due to magnetizing current is neglected, I = −I  s r s  and U / s = I 0 are taken. r r The voltage equations take the form: I m U = −(R + jX )I  + jX (I + I  ) , s s σs r m s r I j(X + X  ) R + R / s s  s  r s r 0 = (R  / s + jX  )I  + jX (I + I  ). r σr r m s r Subtract the two equations to get: U I  r I j X s m m U = − ( s ( R + R / ) s + j( X + X  ) s r σs σr) I  . r This equation corresponds to the simplified equivalent circuit above. For the slip s = 1 , the simplification applies: R = R + R and X = X = X + X  . sc s r sc σi σs σr Operating an induction machine on a rigid grid The rotor winding is short-circuited (U = 0) . r The second basic voltage equation gives: I R ( + js X ) = − js X I , divided by slip " " s r r r sr s I = − jX sr I and inserted into the first voltage equation r s R / s + j X r r U  2  X sr = U =  R + j X +  I = Z I . s s  s s  + s s s R / s j X  r r  Calculate the current in the stator winding: U U ss I = = . s 2 Z X s R + sr j X + s s R / s + j X r r The no-load current for = U s s = 0 : I . s0 R + j X s s In an ideal short circuit, the current for U U ss s =  : I = = . ki  2 R + j  X X  s s sr R + j X 1 − s s    X X s r  Stator current curve of IM For an approximate situation: U s I = − I  = I − I  . s r m r j X m I  = − U s , where X is the leakage reactance ( X  X + X  ) r σi σi σs σr R + R s  / + j X s r σi for ideal short-circuit i.e., common leakage of the stator and rotor. The stator current is now: U U ss I = + . s j X R + R  / s + j X m s r σi We look at the characteristic values of the current for s = 0, 1 and  . For R + R / s → 0 , the current is purely inductive. s r The second term in the equation for stator current represents a circle of diameter: Us I = . Φ j X σi The circle is a graphical process, and is obtained by inversion of the complex quantities. Us s = r 0 R + + jX s circle R i Us P  1 s R r r P + j X 1 s + U 2s  i X s  P R  line s = 1 R R 1 −  1 R s =   P   i −  s  r s = 1 U  s 2 X R P s s =  i  + 0 j − j X I U s =  X  i i P P = R R s r ctg  I = U and P P   = ctg  I − P P  = U , if it is Φ   Φ s  Φ  2 1 1 Φ s 2 X X σi σi ctg = R R + R s s r  and ctg  = .  1 X X σi σi Now move the circle from the coordinate origin for the magnitude of the magnetizing current I = − jU / X to the right to obtain the current curve I = f ( ) s . m s m s U s P , 1s = 1 U s P 1 P 0  I s ne P, s r li I we P s =  s P po  I  r e line  s  I torqu r s P 0 I 0 P  I M I P , s = w 0 m 0 M P  − j 0 I m − j Taking into account the losses in the iron, the friction and windage losses, the no-load current has a watt component I . w Energy balance Power flow diagram P 1 P P = P + P + P (P = P ) 1 Cus Fe δ Fe Fes stator P + P Cus Fe P  (air gap power) δ P = P + P + P (P = 0, f → 0) P P + P δ Cur fw Cur fw Fer r P n rotor = P the power delivered or mechanical m shaft power at the motor shaft. Input stator power is: P = P + P + P + P + P . stator 1 Cus Fe Cur fw Power calculation for the m-phase stator (m = 3) s Input electrical power delivered to the stator: P = m U I cos  1 s s ss Joule losses in the stator winding: 2 P = m I R Cus s s s Power delivered to the rotor: = 2 R P r Cur P m I = δ s r s s Joule losses in the rotor winding: 2 2 P = m I R   = m I R Cur s r r r r r Output mechanical power (P = P) at the shaft: P = P − P = m I R   2  1  − = 1 2 δ Cur s r r    s  = 2 1 − s   m I R   = P (1 − s) s r r   δ  s  Motor torque At " " n rotor rotations  = 2π n = 2π (1 n− = s )  (1− s) and P =  M m s ms δ ms δ M = P m I R   2 P = = s r r 1− s P = δ = M δ   (1 − s )  (1 − s ) s  m ms ms ms M = M is valid only if the friction and windage losses are P = 0 and M = 0 , respectively. δ fw fw Operating areas on the circle diagram Depending on the points on the circle, the following areas of operation are obtained for different slip values: a) rotor lagging behind the rotating field s = 0 to 1, " motor area " ( n =  n 0 ); s b) the rotor rotates against the rotating field s = 1 to +  , " braking area " ( n =  − 0 ); c) the rotor rotates faster than the rotating field s = 0 to − , " generator area " ( n =  +  n ). s Torque The torque M = f ( ) s is obtained graphically (from a circle diagram) or analytically. a) Torque waveform M = f ( ) s for the slip area s =  0 2 M M = f ( ) s M b M s M ib s = 0 s s = 1 s = 2 s b The labels in Figure M = f ( ) s have the following meanings: M is the breakdown, b M is the starting (pull-up) torque, s M is the initial braking torque. ib b) Analytical derivation from power P δ M = P   2 m I R δ s r r U =s for I  = (p. 74) is obtained:   r s 2 2 ms ms ( R + R  / ) s + X s r σi m U r m U 1 s 2 2 R  / s M s s s = = .  2 2 2 2 ( R + R  / ) s + X  ( R + X ) / s R + R / s + 2R ms s r σi ms s σi r r s The maximum (breakdown) torque is obtained from the condition:   R R M = r r 0 , which gives a value for s =    if ( R X ), and  b s σi s 2 2 X R + X σi s σi = m U 2 2 1 3 U   M s s s M . b b 2 2   ms R 2 + R + 2 X X s ms σi s σi Effect of changing R and X on the torque curve r σi a) Increase R ( R ) r r R  M → R increase r R  r(b) R r(a)r (R ) (R ) r r(a) (Rr(b) ) As R increases, the point s rb moves along the circle towards M b higher values of the slip. It is M s(b) even possible that s  1. M b s(a) M s s = 0 sb sb(a) s s=1 s b(b) b) Change X = X + X  σi σs σr X ib X  X  X M  ia  iib X i (X )  ib (X )  i X ia ( X )  ia s = 0 s = 1 s The Kloss equation M It represents the ratio , which is obtained from the equation for M and M respectively: b M b M M 2(1 + s R / R  ) b = s r . s s / + s / s + 2 s R / R  b b b b s r If R  R  and s  1, it is obtained for R → 0 . s r b s The Kloss ratio: = M 2 M s s / + , b s / s b b where the breakdown slip is R  2 r 3 U s  s and M  . b b X σi 2  X ms σi For s → M 2s 0 , applicable s s /  s / and s = = b k s (equation of a line). b 1 M s bb For M 2s 1 b s → 1 , applicable s s /  s / and s = = k (hyperbola). b b 2 M s s b Starting an induction motor Induction motor starting is the process that takes a certain amount of time for the rotor to reach speed from n = 0 to n . Ideally: ratio M / M is as large as possible and I / I as small as possible. s N s N a) Starting the motor with a starter During start-up, apply additional stator ohmic resistors via the slip rings (applies only to the wound rotor). R rotor + R  R + R  a r b r  R + R  R c rr starter M I M N d 3 c b 2 a 1 M M smin smax I I Ismax N smin M M N N s = 0 s = 1 s s = 0 s = 1 s n = ns n = 0 Maximum permissible starting current I  I smaxs Minimum permissible starting current I  I sminN b) Starting the motor with the switch Y / Δ In a star connection, the voltage across the winding is: U = Y U / 3 = U . s The current in the terminals for a star U U YL I connection compared to a delta  connection is: I Y L  U I 3 I I I I I Δ LΔ LΔ = I = = = . LY Y 3 3  3 3 N − turns Display current (and torque) curves N − turns by switching from Y in Δ I The winding voltage is U / 3 → I N M M 5 I L  B  = Δ →Δ B = . Y Y switching Y in  4 3 3 N 1, 5 2, 5 Starting torque 3 M  2, 0 B I Δ M = k B I =Δ 2 I LY k Y Y Y 3 3 1, 0 1 0, 5 M Y M I N 1 LN M = M Y Δ 3 n = n 0 n n N s c) Start-up transformer (autotransformer) Ratio = U I 1 1N  L K 1 and K = =  1 , stator phase current of the motor I = K I , U I sx I s U I K 1x s U grid (line) current = 2 I K I = 2 K I and M = K M . Lx I s x I s x I Special versions of squirrel cages sc a) Double cage Above is the starting cage (index "sc"). The lower one is the operating cage (index "oc"). oc    → X  X ; X = s X (ss − standstill) σsc σoc σoc σsc σr σr(ss) h b) Deep slots We take advantage of the impact of the ejection current. shaft The height or depth of the slot h is increased by the inclination of the slot. Shapes of torque curves of different rotor M winding designs: a) double cage, a b) deep slots, b c) normal cage, c d) wound rotor. d Options for changing the rotation speed s = 1 s = 0 n = 0 n = ns n = f n (1 − = s) (1 − s s) p 1) by changing the frequency, 2) by changing the number of pole pairs, 3) by changing the slip. 1) Frequency-changing From U U x U = k f  it applies: = = k = const. (Applies up to the value U  U .) x N f f x 2) Pole pairs (poles) changing a) With multiple windings for different numbers of poles b) Dahlander connection ratio by 2:1. This uses a Δ / YY or Y/YY circuit combination. The first connection is used mostly, the second only to drive the fans.  p4 p2 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1U 2V 1V 1U 2V 1V Directions of currents for 2 p = 4. Directions of currents for 2 p = 2. Circuit illustration for connection T 1W 1W T R 2W 2U 2W 2U 1U 1V 1U 1V 2V 2V R S S 3) Slip-changing (for a motor with slip rings) Since P δ M = , is M = const. for P = const. and with the resistance R in the rotor is: δ  add ms P = 2 r = 2 r add  R R + R R + R r m I m I  = add s s . δ s rN s rN x N s s R  N x r With higher slip, the losses P = P s in the rotor circuit increase. This is not economical. Cur δ x It is more economical to use a cascade, i.e., a rectifier − inverter ( U , , f U are variable) r r = or converter cascades (U , f are variable, U = const.) in the rotor circuit. r r = U r , fr Ur , fr U U == L Changing the torque The torque is proportional to the power of the air gap or the power at the resistor according to the equivalent circuit R / s : r M = 3  2  R r 3 = 2 Rr I I ,  r r s  s ms ms where the rotor current I r I  is reduced to the stator I  = I K = 2 and R  = K R r r r I r Ur K U (for m = m ) . s r Introduce a voltage drop across the rotor at the replacement resistor R / s (E = I R / ) s . The r Rr r r rotor current I or I  is opposite to the stator current, and therefore the voltage drop r r E  = I R  / s is of opposite magnitude to the induced voltage due to the rotation magnetic Rr r r field E = E = E  . m s r M = 3 p 3 p E I   =  Rr E I r Rr r  In an induction motor, torque variation (control) is achieved by varying the voltage E , i.e., Rr by varying the current I . (The phase angle between E and I is zero.) For this purpose, r Rr r we modify the usual equivalent circuit. The stator and rotor leakage reactance are expressed as the difference of the own and mutual reactance between the stator and rotor. The transformation constant (in the circuit) can be chosen to be any value except zero. For the  case  = K , i.e., equal to the ratio of the effective stator and rotor turns, we obtain again U the usual equivalent circuit. R s j( X −  2 X ) j(  X − X ) s sr r sr U I I r s K U  I 2 Rr m  s   s j X sr From the condition  2 X −  X = 0 we obtain:  = X / X , i.e., the transformation r sr sr r constant at which the leakage reactance in the rotor circuit of the equivalent circuit vanishes. The new rotor current is now 1 / -times the actual (three-phase) rotor current ( ) I , and the r new rotor voltage is -times the original rotor voltage (voltage drop E ) across the Rr  resistance. This new voltage (voltage drop * E ) is applied to the terminals of the new Rr magnetic (main) reactance and is, thus, related directly to the flux of the rotor, which causes the voltage * E ( E =  E ) . Rr Rr Rr By taking the product of 2  X = X / X sr , we obtain a new equivalent circuit. sr r R j  2  X I  sr X s −  r    X s r   U I I s s 2 X sr 2 Rr *  s j E =  Rr E Rr X s r This equivalent circuit is reformulated by introducing a transient machine parameter. This is defined as the stator leakage reactance of the whole machine, and given by equation: X = − = sr u m X X 1 − = X 1 − =     X σs s s  s s . 2  X sr X X K  2 2 2 2    X X X K X X  r  s r u   s r  The asterisk index does not mean relative, but means the changed stator reactance. Similarly, an asterisk indicates the changed magnetizing reactance: = X 2 2 2 K X = X X =  sr U m  m . sr X 2 K X  r U r X K X R 2 2  X R Considering that: sr r U m 2 r m  = =  K and =  , we obtain a finitely X 2 U 2 K X  s X  r s U r r transformed equivalent circuit in which all quantities are transformed to the stator. R * j X s  s U I I I s = − r s  I sM 2   X R m r * j s X *  m E Rr  2 X s r The stator current is thus divided into two (mutually orthogonal) current components, namely the reactive component of the current I = I sin  (I = (K / ) I , i.e., the sΨ s sΨ U m excitation current) and the working component of the current I = I cos  ; this represents sM s the torque. The working component may also be marked I . sT Varying the torque with the current I and I sΨ sM The induced voltage, or the stator-reduced voltage drop * R I E s s Rr * jX 2 I across the rotor substitution resistance   s s R / s , is equal to the r time variation of the rotor magnetic linkage: E * − * * E = j X I Rr m s = −  * * j / 2 = − j X I . Rr r m sΨ U s Reduced values can also be omitted from the equation, and it is: E = − j / 2 . Rr r I sM I s According to a redesigned circuit, the reactive component of  s the excitation current I is given by equation: sΨ  = − E *  E E = − = − I Rr Rr Rr s . * I s jX j X j  L  *  m sr sr r From the last equation, the induced voltage is E =  L I . Rr sr sΨ I = −  I r sM Product L I in the induced voltage equation E sr sΨRr represents a rotor magnetic linkage  = L I (RMS value because the current is not r sr sΨ multiplied by * * 2 ), and the reduced value of this is:  =   =  L I = L I r . r sr sΨ m sΨ The torque component of the current I is obtained directly from the circuit as the expression: sM I I r = I = → I =  I . sM sM  r sM The final equation for the torque of the machine, if the expression for the voltage E and Rr the current I (without the negative sign) is substituted in, reads: r M 3 p * = (  L I )(  I ) = 3 p L I I  = 3 p L I I . sr sΨ sM sr sΨ sM m sΨ sM  The equation represents an expression that is a function of the two components of the current. The basic equation for the torque takes the form (with * *  = L I and I = I cos  r ): m sΨ sM s M = * * 3 p  I cos  = 3 p I , r s r sM i.e. the product of the rotor magnetic linkage and the working component of the stator current. Given an equivalent circuit, we obtain the working component of the stator current I : sM I  E 1 sE X sE L sE = − Rr = − Rr = − r Rr = − r Rr . sM  2 R r / s  R r X sr R r L sr R r Combining the last equation with the previous equation for E gives the relationship Rr between the two stator current components I and I : sM sΨ I j s I  = L r sM . sΨ R r This connection is due to the fact that the voltages on the new magnetizing (mutual) reactance and on the replacement (fictitious) rotor resistance are the same. The slip angular frequency is also determined by the two current components. Rearranging the last equation gives the expression for the slip angular frequency: s  R I r R I 1 I sM sMsM = = r = .  L I L I  I r sΨ r sΨ r sΨ  = L / R is the electrical time constant of the change in all rotor quantities. When both r r r components of the current are selected (in stationary machine operation − motor), only one value of the slip in this equation gives us the corresponding rotor magnetic linkage  and r torque M . Induction generator Operating on a rigid grid U Us  P 0 I sr − j induction driving I I s sw generator machine P n The grid provides the reactive power (Q = 3U I sin  = 3U I ) and the driving machine Gs s s s sr provides the working power. An IM works as a generator if it is a slip s = (0  −1) . Then the load drives the rotor faster than the speed of the rotating magnetic field and the generator emits electrical power. For the slip s  − 1, the machine works in the generator braking area and does not emit any power. Single-phase induction motor main winding (MW) They are building up to 2, 2 kW . It has a main winding (usually occupying 2 / 3 of the stator slots) and an auxiliary winding, the axis MW axis of which is shifted by 90 (electrical) degrees auxiliary with respect to the axis of the main winding winding (AW) (usually occupying the remaining 1/ 3 of the axis AW stator slots). a) Operating mode if only the main winding is connected. The sinusoidal voltage U dictates the sinusoidal flux  . The fundamental harmonic s m coordinates, or the circumferential angle =  s ( component of the magnetic flux density is a function of the local distribution in stator x /  π in the air gap, and time: s p ) b ˆ MW = f ( , ) x t = B cos  cos(  t ) = 1 s 1 s I r = B ˆ I 1 ( cos (   − t + cos   + t s ) ( s ) )s . U 2s In the rotor coordinate system, the fundamental harmonic component of the field for   f = + p t =  + p (2π ) n t , taking into account n = n (1− = s s) (1 s) and r m r − s p   = +2π pnt =  + 2π pn (1− s t ) =   + (1− s t ) , is: s r r s r b B ˆ ˆ B 1 = f ( , ) x t = cos (  −  ) + 1 s t cos (2 s) t 1 r r (  + −  r ) . 2 2 In the rotor, the induced voltage of frequency s f and (2 − s f ) , i.e., the induced voltage is obtained of the positive and negative components of the field. At s = 1 the two rotating magnetic fields are equal and develop the same torque. Such a motor therefore does not start by itself. M M p s s = 1 s = 2 s = 0 M M n b) Start An auxiliary phase − winding (AW) is required to start the motor. The current in the auxiliary phase has to be shifted in time and an asymmetric (elliptical rotating magnetic field) is obtained. The capacitor gives the largest displacement. U Torque ratio: 1) Z = R M / M =  1 1,3 s N Sw 2) Z = jX M / M  0,3 axis MW L s N I I s3 s1 e 3a) Z = jX M / M = 1, 6 2,1  C s N Z 3b) Z = jX M / M  0,5 C s N s L R C I r 3a) starting capacitor, axis AW b) operating capacitor, e s c) combination 3a) + 3b). Switch "Sw" switches off the start capacitor by n  0, 75n . s Phasor diagram drawn for example 3b) In this case, we have two-phase asymmetrical  U = U (MW) s1 1 excitations    . s1 s3 U C U  U Cs I cos   1 s1 I s3 1 The capacitor motor rotates from the auxiliary to the   1 main phase. To reverse the direction of rotation, the U (AW) s3 terminals, i.e., the start (s) and end (e) of the auxiliary or main phase, must be reversed. Torque curves for starting with auxiliary winding for cases 3a), 3b) and 3c) M M M M s M b 0 n 0 n 0 n n s n s ns s = 1 s = 1 s = 1 a) b) c) Shaded poles motor These types of motors are built for low power (a few tens of watts). A short-circuit turn (ring) is mounted on part of the pole shoe. The main flux  is divided into a part m  m going directly into the rotor  and a m1 shaded part  passing through the short- m3 circuit ring. The latter is time delayed   towards  and smaller. An elliptical m3 m1  m1 rotating magnetic field is created. Due to the m1 delay (  relative to  ) the rotor will m3 m1 rotor with always rotate from the unshaded to the  short-circuit m3 squirrel cage shaded part. To reverse the direction of rotation, the rotor must be reversed  m single-phase (reassembled). concentric shaded pole turn winding SYNCHRONOUS MACHINE Description of construction There are two designs, a machine with salient poles on the rotor, and a machine with non-salient poles or a machine with a cylindrical rotor. For both designs, the primary winding is on the rotor and the secondary winding is on the stator. The stator is identical to the IM stator. For the salient poles, the air gap is  = f ( ) x , non-salient pole  = konst. stator V1 V1 stator (armature) salient cylindrical (armature) poles rotor   F2 I f n ns transverse s − transverse − F1 q-axis q-axis I I U1 W1 U1 W1 longitudinal longitudinal − − d-axis d-axis For the salient poles, the air gap " "  in "d" and "q" axis is different (magnetic conductivity    ). The excitation − field winding is wound concentric on the salient pole body, but d q belt (usually spread over 2/3 of the circumference) at a cylindrical rotor. Operation mode The rotor is supplied with a DC excitation current I or I . To generate the excitation rotating f fd ampere-turns (magnetomotive force − MMF) of the  , rotate it by n = f / p rotation. For f s a cylindrical rotor it is:  ˆ 4 N f = fwf I . f f π 2 p The salient pole winding factor ˆ 4 f = 1 , turns/pole N = N / (2 ) p and  = N I . wf f p f f fp f π No-load  ˆ   ˆ ˆ 0f0 ˆ → B = = B magnetic flux density of the fundamental harmonic component or f0 δ 1  e the mean value of the main air gap flux ˆ 2 2 ˆ ˆ  = B A = B  l m 1 δ 1 p π π ( is the pole pitch, l the length of the stack,  the equivalent air gap of the machine). p e At no-load, the current in the stator winding or armature winding is equal to I = I = = I 0 . s a The main flux induces a no-load voltage in the stator winding which, when excited by I = I , is equal to the grid voltage: f f 0 U ˆ = 4, 44 f N f  . s ws m  =  =  is the flux of the salient pole due to the field winding concentrically wound m m0 p0 around the salient pole body, or arranged on a cylindrical rotor. Load When loaded, the stator current I generates stator ampere-turns  in the three-phase a winding (reaction or armature ampere-turns):  ˆ 3 4 N f = sws 2 I . a 2 π 2 p Together with the field ampere-turns  (of the cylindrical rotor), the resultant ampere-turns are: f  =  + . res f a The ampere-turns of the reaction are converted to the excitation side of the machine from the condition of equality of the reduced and original ampere-turns with the current ratio K . I  ˆ 4 N f 3 N f = f wf ˆ I =   I = sws 2 I = K 2 I and applies to fa fa a fa I π 2 p 2 N f f wf resultant excitation: I = I + I . fres f fa The effect of the armature reaction is captured in the magnetizing reactance of the machine X 1/  (p. 104), because the armature reaction acts through the air gap on the excitation m e side of the synchronous machine. For  =  = const. (U = d-axis const.) m m0 we need to change the excitation that will ˆ  =  or    B . fdmax B res f 0 f f0fd n - pole field, s f 0  x e   p f  res  n s  lin. non-lin. a ns armature reaction field,  d-axis  a At no-load, the amplitude of the main flux ˆ  lies in the d-axis of the pole, while, under m0 load, the flux is displaced with respect to the d-axis of the pole. Its position then depends on the size and character of the load, i.e., the size and direction of the armature reaction. The magnetizing characteristic is a non-linear curve in saturation and only in the initial (unsaturated) part is it linear and the fluxes can be summed geometrically. For a linear theory (  = const.) , it is: Fe  p stator   ˆ ji ˆ → =  e 2 I →  a a aa rotor  ˆ   j  0 ˆ  → =  e I →  f  a f p p f p m result.  ˆ j eres ˆ  →  = I →  res m m fresm  res  =  ˆ  jres ˆ =  = j0  ˆ +   j e  e + ei m m p a p a   a is the phase position of the pole flux for  t = 0,  0 the phase i  a position of the current and  the phase position of the main flux. res Machine operation on a rigid grid (U = const., f = const.) Three-phase machine (m = 3) a) Induced voltages I U  E = − j (  N f m ) 0 s ws 2 N with angle  =  − π / 2 0 res , R , X s s  s and with flux components  and  p a I N E = E + E . , R f f 0 p a f For R = 0 and X = 0 it is valid −E = U , or by the components of the pole wheel voltage s σs 0 s U and the armature reaction voltage U . Ep a U = −E and U = −E Ep p aa U = U +U , U  I / 2 and U  I s Ep a Ep f a Sw b) Synchronization to the grid An excited synchronous machine (generator) with phase "a" U Us n revolutions is connected to the grid via the s switch (Sw) if zero conductor U = − U U = 0 . Therefore, Sw s I = 0 , i.e., at U = U and U = U . s Ep c b An excited synchronous machine is therefore synchronized to the grid in an unloaded state. The two conditions for synchronization are: U = U = U s Ep 1) U U U s = U = U = U and  =  =  Ep , s Ep us up u 2) f  m =  f p s  =  (grid), n = n = = . s s p p = E E 0 p Taking on the load 1) Reactive load acceptance at R = 0 and X = 0 s σs 2) Working load acceptance at R = 0 and X = 0 s σs It starts from synchronism I = 0. Under load, if the voltage drop is neglected U = U , the s flux will be  = const. =  + . m p a 1) Reactive load acceptance To change the reactive power input to a machine (generator or motor), change I with f respect to the no-load value ( I ) (assume here that P = 0 , because is R = 0 and R = 0 ). f 0 l s f The size of the excitation can be I  I or I  I . f f0 f f0 a) Synchronized b) Over-excitation c) Sub-excitation I = 0, I I  0, I  I I  0, I  I f 0 f f0 ff0 U U U U U U U Ep Ep U Ep d-axis I  I   p a p  = m  p0    m a m  p   p0  p   p0 E E E 0 00 a) b) c) b) I is capacitive for the grid, inductive for the machine. c) I is inductive for the grid, capacitive for the machine. 2) Taking on the workload This is based on the assumption that ˆ ˆ ˆ ˆ I = I ,  =  and U = U ; force the f f 0 res p0 Ep Ep0 generator to take up the power by connecting a mechanical load (the motor), or by connecting Z to the generator terminals. Let's leave it that I = const. L f a) Synchronous motor Mechanical load, M = M  0 , causing the rotor to lag behind the rotating magnetic field L d  . Therefore, the flux  lags behind the original position  =  and is − E  U . m p p0 m 0 The voltage difference U −U drives a current I and this causes a reaction  or flux Ep a  . The condition for the resultant magnetic field applies: a  2 U s +  =  = . a p m j (  N f ) s ws  ib Space Figure and phasor a  a diagram of motor MMF  ( 2 p = 2 ) res  B ia    res n If  f  f Phasor diagrams i c 1) Synchronized machine 2) Loaded machine I = 0, n = n I  0 , load angle   0 s − U U U U Ep  U Ep I d-axis   m =  p0  m   a E 0  p d-axis When a load is applied, a load angle  occurs. The load angle  is the angle between the rotor d-axis (flux  ) and the resultant rotating field (flux  ). The synchronous motor draws p m operating power from the grid (I  0,   0) and develops a torque +M: M m  =s U I cos  . For number of phases m 3, ,  s =  = =  (grid) will be: ms s p ms M 3 p  3 p  ˆ ˆ = N f  I cos  = c  I cos  c = N f , (  = − ) . s ws m M m  M s ws  u i 2  2  3) Stability limit  = − b π P b 2 U I In stationary operation: M + M = 0. (M = M − load) L L The M of the motor will be maximum at  π = − b . π  U Ep b  = − 2 At  2     , the machine falls out of synchronism. m b b b) Synchronous generator  a  p If the driving machine M  0 accelerates the rotor (pole d wheel),  P b overtakes with respect to position  p . p0 d-axis This makes a difference U −U , Ep   f f which caused the current I →  →  . a a ib n The following condition applies:  2 If    res U i +  =  = s a .  B  a p m  res j ( N f ) s ws The current increases by enough to keep it in stationary operation:  a i c  a − + M M = 0 . d The current has a negative watt component. −M = c  I cos M m Supplying mechanical power to the shaft delivered electrical power to the grid (load angle →   0 ). Phasor diagrams 1) Synchronized machine 2) Loaded machine U U Ep +  U U U Ep xis d-axis  p d-a   m =   p0a   m E 0 I 3) Stability limit  π = + bd 2 d-axis Before the load angle corresponding to a given U load (angle  ) is established, the pole wheel,  p  π  = and  oscillate around position  − mechanical p 0 2  a oscillation. That is why it needs to be throttled. This  is achieved by a short-circuit cage at the rotor poles. b U n  n induction motor  Ep s m n  n induction generator s The cage is also used to start the motor. I  P p b  ( ) +  ( ) −  p0  p0  p Analytical treatment For stationary operation, it is −1 n = n = s f / (s ) p . The derivation is performed for non- salient poles. For the salient poles, we need to take into account    . The magnetizing d q characteristic of a synchronous machine is linear (  = const.) . Fe Stator voltage equation: =  + m U U = R I j  L I + j (  N f ) s s σs s ws 2 a)  complex value of rotational (main) flux m Resulting ampere turns ˆ ˆ ˆ (  =  +  ) in the air gap  → B →  . res f a res res m In stator coordinates, the main flux: =   ˆ cos    − t m m ( − . In symmetry of the un-s res ) shortened coil x = 0 or  = j ( ( x /  )π = 0 the main flux is equal  =  ˆ eres s s s p ) . m m The main flux consists of the flux of the pole wheel and the flux of the armature reaction. b)  pole wheel flux p MMF ˆ  c =    ( ) =  p cos  causes a magnetic flux density in f r f ( ) r the air gap ˆ b (  ) = B cos , written in rotor coordinates f r f r ( = (x / )π r r p) xs 0  e 0 or pole wheel flux  =  j x r  ˆ e 0 p p and this voltage U  +π/2) x = j( Δ U e 0 . Ep Ep  Δx 0 = π is the rotor displacement of t = 0 with respect to the armature winding 0  p symmetry (phase "a"). In stator coordinates, the excitation is =   ˆ cos    t f f ( − − + π / 2 . s up ) The change in coordinates of the rotor rotating with the n revolutions is obtained from the s following operations: x = − x Δx = − − x v t Δx . r s s 0 Let's insert an expression for v =  = =  = r D n π 2 p  and derive: ms p p f p  π x −  p π  π π  = x  t − Δ x  ,  =  − =  + −  r s 0  up π  0 u  p  2 2      π  π  + = = − t − + .   =  +  r s up  0 up u  2  2  We usually take the voltage position  π    = 0 + =  j U = U e . u  0  , and it will be: Ep Ep  2  The induced voltage is calculated using the well-known equation: ˆ U = 4, 44 f N f  . Ep s ws p c)  armature reaction flux a This flux is caused by load currents i i , and . i In phase ''a'', the moment value of the a b c current is i i = 2 I cos(   t + ) or complex I = = I I e a . a j  i a In the stator coordinate system:  ˆ ( ) x =  cos (    − t − ) , which causes a magnetic flux density in the air gap a s a s i b = B ˆ cos(   − t − ) . a a s i In a rotary coordinate system, it will be:  ˆ ( ) x =  cos  + (  + − π / 2) a r a ( r) , if it is   = −  . u i The position of the ampere-turns of the armature reaction  is f ( , )   , i.e., a function of a the phase angle and the load angle of the machine. Machine with cylindrical rotor The excitation and reaction of the armature create a resultant excitation in the machine's air gap (  =  + ) . For x = 0 , i.e., in the coil symmetry of phase "a", there is a resultant res f a s excitation:  =  ˆ cos(  t + ) . res res res The resultant excitation creates a resultant magnetic flux density in the air gap: b x ( ) = b x ( ) + b x ( ) . s f s a s The amplitude of the magnetic flux density of the fundamental harmonic component in the air gap ˆ  0 ( B = B = ˆ B ) is calculated from equation B =  and the peak-mean value 1 δ res 1  res e of the pole flux ˆ 2  2 ˆ  = B   =  l0 l . The complex flux value is:  . m 1 p m p  res π π e The stator voltage equation will now be given by the equations for excitation amplitude ˆ f and armature reaction ˆ  : a U 2  4  N f I j( −π/2) 3 N f  0 f wf f up s ws = R I + j X I + j (  N f )  l e +  I  . s s σs s ws p π  π 2 e  p 2 2 2 p  If the magnetizing (main) reactance is = 2 3 4 ( )2  N f    0 s ws X L = l (i.e., the m m p π  2 π 2 p e same as for the induction machine p. 71), will be: U = 2 N f I j( −π/2) f wf v up R I + j X I + j X e + jX I . s s σs m m 3 N f 2 s ws Expression 2 N f X f wf m X = = X is the mutual reactance between the stator and the rotor. m avd 3 N f K s ws I The stator voltage equation takes the following form when the order of the last two terms is reversed: U = I j f +up R I j X I + j X I + X e . s s σs m afd 2 Introduce synchronous reactance X = X + X ( X = K X ) and the voltage of the pole d m σs m I afd wheel U = X I / 2 , to give the final form of the stator voltage equation: Ep afd f U = U = R I + jX I + U . s s d Ep To this equation, we fit the equivalent circuit of a synchronous machine with a cylindrical rotor. Pole wheel voltage ˆ ˆ U = − E  I and its position  =  +  . Ep p f up u As it is R  X , take R = 0 and get a phasor diagram of a generator or motor with a s d s cylindrical rotor. Synchronous reactance is usually given in the relative terms: X U d Nf x = and Z = . d N Z I N Nf x = 0,8 2,5  d jX I U d Ep jX I d R s jX d I U U U Ep U E p ~  I    I Machine with salient poles The longitudinal or d-axis and transverse or q-axis are distinguished. X  X and X  X . d q d q X = X + X and X =  X d σs ad ad adm X = X + X and X =  X q σs aq aq aqm  and  are the pole shape factors in the d and q-axes. ad aq U Ep "q" axis U "d" axis  xis d-a  I d I q I q-axis The voltage equation applies: U = R I + jX I + jX I +U . s d d q q Ep Machine phasor diagram with salient poles jX I d d Valid for R = 0 . U s Ep jX I q q To draw a phasor diagram, the following must be given: U U U , and I or Ep U I , , X , X and load angle   d . q  I d I q Permanent short circuit (U = 0) I For R = 0 will: s 0 = jX I + jX I +U . d d q q Ep The two components of the short-circuit current are: jX I = → 0 I = 0 and q q q jX I + U U Ep Ep U = → 0 I = I = − = d j . d Ep k dk jX X d d The effective value of the permanent short-circuit current is: I = U Ep k . X d The value X is determined from the measurements: d a) of the no-load U = U no load voltage, Ep b) of the short-circuit I short circuit current. k X U Ep U  no-load voltage  = = at the same excitation. d =   I I short-circuit current k k   Operating an unsaturated machine on a rigid grid Derive the equations for a machine with salient poles. I = + j = −up jup I I , I j e I and I = I e d q d d q q  =  +  for R = 0 and  = 0 will be  =  and obtained from the previous up u s u up voltage equation: U = j j j − j  X I e + jX I e + d d U e . Multiply the equation by e . q q Ep The real part of the equation: U U U cos  Ep cos  = X I + U → I = − d d Ep d X X d d The imaginary part of the equation: − U sin  U sin  = X I → I = − q q q X q This is put in the initial current equation I = I + I , and the solution for the armature current d q is: I U Ep U j U  j = j − j e cos  − e sin  . X X X d dq Taking into account 1  1 j − j  j − j cos  = (e + e ) and j sin  = (e − e ) calculate the current: 2 2 I U     Ep U 1 1  U 1 1 j 2 = j + j  −  e − j  +  .     X 2 X X 2 X X d  q d   d q  For a cylindrical rotor U −U U −U Ep Δ X = EpU X : I = j = = . d q X j X j X d d d Torque: M P  ( ) = → * P = Re U I and for m = 3 ,  = we get:  s ms p ms M = = Re U I = U I for U = U . w  p P 3 p * 3 p ( )   Using the equation for the current, we derive: M 3  2 UU    p Ep U 1 1 = −  sin  +  −  sin 2   .      X 2 X X  d  q d   synchronous reactance (reluctance) torque torque For a cylindrical (turbo) rotor X = X and only synchronous torque is obtained: d q M = − 3 p U U Ep sin , M = f ( , ). I   f X d The maximum (breakdown) torque in both motor and generator operation is for a machine with salient poles at angle  − =   π / 2 and for a cylindrical rotor  − =  = π / 2 . b b+ b b+ Torque Figures: salient poles  π π  cylindrical rotor  = b b 2 2 − 3 motor M M ( )  M M motor b p UU Ep sin  M ( )   X 2 d   3 p U 1 1 −  −  sin(2 )     2 X X  q d  M N  b+  b+   b −  b −  generator generator I f If0 motor M 1,5 1,0 For a given torque, the smaller the excitation −M L current, the greater the load angle 0,5   . If the  = b 2 excitation current is too low (dashed curve),    = − b  1,5 the motor falls out of synchronism or out of 2  1 step. We are talking about static stability. generator Static stability The machine is able to take the load if it is loaded slowly. The limit of stability is determined by the breakdown torque. For a machine with a cylindrical rotor, the following applies: M and for M = = 3 U U p NEpN 3 p U I cos  b N N N  X  d gives the ratio for the relative 2,5 breakdown torque M b cos 2,0 = 0,8 M M M N cos 1,5 = 1 U b = EpN . 1, 0 X I cos  N d N 0,5 U = f (X , cos )  , i.e., the voltage 0,8 1, 2 1, 6 2,0 2, 4 xd Ep d drops across X ( X I ) and cos  . d d It is for various cos M b  : = f (X ) . d M N For stable operation, e.g., for a motor loaded with M , the following applies: L The increased load will be d M = −M  0 . L  dM  Δ M =  Δ   d  M / d    0 .  0 For small changes in angle  , a linear M ( )  Δ M Δ M M L relationship holds at the point of operation M = π f ( )  . 2 π Δ  −π π  −  d M  0 2 M =  Δ   ,  d   δ 0 if the operating angle  is before the load change. 0 Varying the torque of a synchronous motor A synchronous motor's torque is varied (controlled) by the armature voltage, or by excitation. In frequency inverter operation, the motor is varied simultaneously in voltage and frequency and thus in speed (U =  0U , f =  0 x f , x is a multiple of f ). If the torque is varied N N N by excitation, only the load angle is varied. With permanent magnet excitation, the excitation cannot be corrected directly, but indirectly by varying the angle of the stator current. Referring to the phasor diagram in Figure a) for a motor with a cylindrical rotor, we can introduce the internal angle  , i.e., the angle between the pole wheel voltage U and the Ep armature current I :     = + = − . up i The equation for the synchronous torque of the machine can be rearranged: M 3 UU p Ep 3 p ˆ = − sin  = U I cos  = c  I cos ,   Ep M m X d where −U sin  = I X sin(90 −  ) = I X cos  and U =  N f  ˆ / 2 . d d Ep s ws m The torque is at its maximum when  = 0 , i.e., in the case of  = − . Then the armature current is in phase with the pole wheel voltage. jX I d U jX I d U U Ep  − 90  I U Ep I       a)  d b)  d In the case of larger synchronous motors this is not very favorable, because the current lags behind the grid voltage (Figure a). In the case of larger motors, we usually want the current to overtake the grid voltage (reactive power generation). In this case, the angle is  = 40  60 and the current overtakes the voltage (Figure b). Stationary operation of a cylindrical rotor motor in a d-q model The armature current can be decomposed into two components (d, q): I = I sin  and I = I cos  . d q The angle  is the space angle between the q-axis, where the voltage U is located, and Ep the stator current. In conventional theory,  is the (time) phase angle. The Figure shows a phasor diagram of a synchronous machine with a cylindrical rotor in d, q components. The voltage of a pole wheel can also be expressed as a "voltage drop": U =  d j X I = j  L I = j  . Ep afd f afd f 2 X = X / K is the mutual reactance between the armature winding with index "a" and afd m I the excitation winding in the d-axis, i.e., the magnetizing (main) reactance of the motor (stator) divided by the current ratio (p. 104); and ˆ  = L I the value of the magnetic d afd f leakage in the d-axis of the rotor. Pole wheel voltage U = f ( , ) f I . Ep f The equation for the synchronous torque can be transformed into the form: M 3 p 3 p  L I afd f = U I cos  = I = 3 p I .  Ep  q d q 2 It can be seen that only the "q" component of the stator current, which is perpendicular to the excitation axis and in phase with the voltage U , contributes to the torque. In permanent Ep magnet excitation (  = const. ), the torque varies only with the "q" component of the stator d current, the "d" component of the stator current (in the axis of jX I d s q axi excitation) affects only the excitation (armature reaction), but q- also, indirectly, the magnitude of the torque. Therefore, if we jX j U d X I I d d want to vary the torque, we need to control the stator current in amplitude and phase. The stator current is increased by the U = jX I Ep afd f voltage across the armature terminals, but this is only possible up to the nominal value. If the frequency f increases beyond the nominal value f , the input reactance of the machine also N increases, and, therefore, at constant voltage, as the frequency increases, the armature current I q I I ( ) decreases, and so does the q  I torque. The situation is similar to the magnetic field weakening in an induction motor. d-axis I d I f Current characteristics for a cylindrical rotor I f For the armature current of a cylindrical rotor, I f0 the equation (p. 108) applies: + U 1,5 I U U e Ep 1,0 = − j motor  j + j . X X dd 0,5 − U The stator current characteristics j I as a function X d of angle  (circles in the complex plane) are +j  obtained for U = I const. (I = Ep const.) and load U fEp j j angle  e  const. X The domain in which the motor d generator or generator operates is the operating diagram. It covers only a part of the full-field domain. stability limits over-excitation sub-excitation Approximate treatment of a saturated machine The superposition of magnetic field components in the air gap is no longer valid: B  B + B . Synchronous reactance is important for stationary operation ( X = X + X ) , res f a d σs m the main proportion of which is determined by the equivalent air gap  ( X = f (1/  )) . e m e The size of  is affected by saturation. In saturation, there is no longer any proportionality e between the current and the magnetic field of the air gap. We will consider saturation for a machine operating at no-load and short-circuit roughly. No-load and short-circuit No-load: I = I and I = 0 or I = I and I = 0 for the air gap characteristic (AGC). f f0 f fδ U =  ˆ m =  I U NLC E ( N f ) k , 0 s ws AGC 2 U N U = E = f ( I ) 0 f No-load characteristic (NLC): U = E = f ( ) I 0 f measured at n = const. C SC = I Short circuit: E  I E k N I = k f (If ) 0k 0  I k1   mk due to armature reaction I mkδ Short circuit characteristic (SCC): I = I f ( ) Ifδ If0 Ifk If k f The excitation current I is important. It is that excitation current at which, in a three-phase fk permanent short circuit, the armature current will be equal to the rated current (I = I ) . k N The value of the unsaturated synchronous reactance is considered by IEC Standard 60034-4 to be the ratio of the no-load voltage U = U to the short-circuit current I : Ep kδ X U N = with the same excitation (I = I ) . d f fδ I kδ The relative value of the synchronous reactance is: X X I U I I I d d N N N Nfk x = = = = = . d Z U I U I I N N kδ N kδ fδ As a measure of armature reversibility, the IEC gives the ratio of the excitation current I f 0 to I f0 I , defined as the short-circuit ratio of the saturated machine K = . fk c I fk A large ratio means that a larger short-circuit current I is required to compensate for the k1 field of the excitation current I (i.e., a small influence of the armature reaction). f 0 a) Method for determining the excitation current of a saturated machine For a given load at terminal voltage U and current I , I and the load angle  must be f determined. Mathematically, it is: j j  res(up −π/2) j I e = I e + I ei . ( I = K 2 I , p. 95) fres f fa fa I For R = 0 , we obtain the voltage equation: s U  ˆ   m = j  j L I + j (  m N f ) and with E = − j (  N f ) erez . σs s ws 0 s ws 2 2 Valid: U = jX I − E (bottom Figure) or: U − jX I = −E (Figure p. 116). σs 0 σs 0 Graphical method for determining the excitation current for a cylindrical machine at a given load I and voltage U (dire − jXσs I E E 0 NLC E0 cti U on U −/2) Ep up j( ) e I f j I ei fa I j res e fres I I I fresf From the known no-load characteristic (NLC), we determine I for E . We also need to fres 0 know I to obtain I and the load angle  or direction U . fa f Ep b) Internal characteristics U = f ( ) for II = const., cos  = 0 ( = −π / 2) (overexcited machine) f For j −jπ/2 res   = = − π / 2 and  =  + =  0 valid I e = ( I − I )e i up u fres f fa E U E ,0 = f ( If ) 0 − − jX I X I E σs σs 0 X I p U I fa U = f (I ) f  = − π / 2 I j E I = konst. = X I 0k σs  e f f I j res j I fres e I ei I I I fa fa fresf If I cos  = 0 fk A triangle with sides X I and I travels along the NLC for I = const. and we get σs fa U = f( ) I . In a short circuit, U = 0 and E = X I . As the excitation increases from I to f 0k σs f0 I , the leakage between the poles in the rotor increases and the voltage across the terminals fN will be lower. We are talking about Potier reactance X  X , so the actual curve U I ( ) p σs f will be lower − the dashed line in the Figure. c) American diagram (ASA) The ASA diagram is used to determine I at U , I and at any cos  . For the design we fN N N need the no-load characteristic (NLC) and the short-circuit characteristic (SCC), I at U , fN N I , cos  = 0 (ind.) and X (Potier reactance). N p Determination of the Potier reactance: From SCC (I = f ( )) I → I for I = I . The distance 0B corresponds to I , cos  = 0 (ind.). f fk N fN Subtract from point A I  AF . At point F, draw a parallel to the tangent to the NLC, and fk obtain the relative Potier reactance x = HG . NLC and SCC are drawn in relative units. p ASA diagram Apply perpendicularly to I from the end of U the value of the voltage drop ΔU = 3I X N N p Np (for the Y connection) and obtain the Potier voltage U . The projection of the voltage U onto p p the ordinate gives us, between the AGC and NLC characteristics, the value of the increase in excitation due to saturation ΔI . The ASA diagram for the total excitation current is: f I = (I + I ) Δ +I . fN fδ fk f U U I C NLC NLC SC E U I N N AGC 3 I X  H´ H N p I I = v f ( I ) f A G 1,0 U p F I U N fk If I N I fk 0 D  BN  N I I I fk fNv I cos  0 = 0 fa I I f fp IfN If Excitation systems for synchronous generators We use DC self-exciting generators to excite small synchronous generators. For larger synchronous machines, we use three-phase synchronous exciters or thyristor rectifiers (static converters). The excitation power is about 1 % of S . N 3~ 1 3~ 1 G G 3 2 3~ 2 G G regulator regulator 1 three-phase synchronous generator 2 three-phase synchronous exciter 2 DC excitation generator 3 fully controlled thyristor rectifier 4 2 2 4 3 3 6 1 3~ 5 G 3~ G 1 5 6 7 regulator regulator 2 transformer block 5 half-controlled thyristor rectifier 3 own-use transformer 6 auxiliary transformer 4 foreign grid 7 auxiliary rectifier 5 excitation transformer 6 fully controlled thyristor rectifier Using permanent magnets for excitation It is possible to choose between ceramic permanent magnets and permanent magnets made of metal alloys. Ceramic permanent magnets are, e.g., barium or strontium ferrite. Among the metal alloys, the most known are AlNiCo magnets and rare earth-based alloys, e.g., samarium with cobalt, or, more recently, neodymium-iron-boron. Magnetic hysteresis In immaterial space, the magnetic flux density will be B = −7  H . =  4π 10 V s / (A m)  () 0 0   B varies in the ferromagnetic material. B =  H + B , 0 i where B is the magnetic polarization (unit T or mT). The magnetic polarization is: i B =   H .  is magnetic susceptibility. i 0 It follows: B =  (1+  )H =   H , where the relative permeability is  = + 1  . 0 0 r r In saturation,  = 1 will be and the course B H ( ) will be a straight line. The relative r permeability is given for a transformer or dynamo sheet. For permanent magnets the recoil permeability is given    1 1,1. (Permeabilis is Latin for permeable.) p The magnetic polarization B can also be expressed as a function of the magnetic field intensity H : i B = − B  H . i 0 This is the intrinsic magnetic flux density, shown in the right Figure below. For good quality permanent magnets, the point of remanent magnetic flux density is the same in both Figures (B ) . (Remanere is Latin for remnant.) r 2. quadrant 1. quadrant B i B B Br r H B =  H 0 Hc c H H H ci H ci B =  H 0 3. quadrant 4. quadrant The characteristic points for both curves are: H = → 0 B (remnant magnetic flux density) r B = → 0 H (coercive magnetic field intensity) (Coercere is Latin for restraint.) c B = → 0 H (coercive magnetic field intensity polarization) i ci For excitation with permanent magnets, we use quadrant II (IV). This part of the hysteresis curve is called the demagnetization curve. For excitation with permanent magnets, it is essential that B and H are as large as possible. What matters is the stored magnetic r ci energy, i.e., the product of BH . ( ) (BH ) = max. the value at the knee of the demagnetization curve. Working line The magnet is embedded in a magnetic circuit (soft iron), which usually has an air gap. Ignoring leakage, it is:  = B A = B A . m m δ δ Ignoring the MMF drop in iron, the following is true: soft magnetic Fe  = −H l = H l m m δ δ air and follows l gap m magnet B = lm  H = −  lδ H . δ 0 δ 0 m l δ The magnet's operating point will now be: b a mm b aδ A A l δ B = δ B = −  δ m H m δ 0 m A A l m m δ and angle of the working line B rve on cu B izati r  − B   A l  gnet  m δ m dema = arctg = arctg      . 0  H  A l m  m δ  P B m P  line from point P to point P g li . This must not be ne The reaction of the armature moves the working workin below the knee of the curve, otherwise the magnet  is demagnetized (weakened) irreversibly. −H H H 0 c m Permanent magnet synchronous motors They are characterized by their high economic importance, due to their higher efficiency than induction motors. Synchronous motors of special designs are used in household and technical applications. From an engineering point of view, the synchronous torque is of particular importance. Stators are built differently: • an annular coil with claw-shaped poles, • side coils with clawed or toothed poles, • individual coil poles, • a normal three-phase stator − identical to the stator of an induction motor (larger units). Permanent magnet motors are considered to have high synchronous torque, but they do not start on their own. Synchronous and reluctance torque We will first establish the relationship between the synchronous torque and the magnet data, assuming that the stator (2) in Figure a) generates excitation (  with a sinusoidally e distributed magnetic field). I  m breakdown  M M S b 1 N 2 S N  90 180 a) b) When moved by a (mechanical) angle  from the longitudinal direction, the electrical m excitation  (sinusoidal in shape) will cause a change in the MMF in the permanent magnet el on the rotor (1) in Figure a). This results in a change in energy ΔW : Δ 1 W =   cos( p ) δ . el m 2 In the equation there is:  = B A ,  = I N and p is the number of pole pairs. δ δ δ el From what is written follows: M = dW 1 = p   sin  d δ el 2 and the maximum, i.e., the breakdown torque at an angle of  = p  = 90 (Figure b) m M = 1 p   b . δ el 2 For the stator, we use ordinary laminated soft iron. The magnet in the iron poles builds up an opposing magnetic field and so the reluctance (adhesive) torque opposes the rotor twist. The Figure on the right marks the reluctance torque M 3 M with 1, the synchronous (electrical) torque M rel M bel 2 resulting from the electrical supply to the stator with 2 and the resultant torque with 3. M br 1 For the motor to start, it must be M  M . It is el r  generally accepted that M  3M . 90 180 el r The equation for the breakdown torque can also be written in another way: M = 1 1  el   =   p p . b δ el δ p 2 2  p In the equation, the flux of the air gap  is equal to the flux of the permanent magnet δ (    ) , if we ignore the dissipation in the rotor. The excitation of the permanent magnet δ p is  . For  = B A ,  = H l , the specific energy w = B H / 2 and the magnet p p m m p m m p m m volume V = A l , the breakdown torque will be given by the energy: m m m M = p w V   / b p m ( elp ) proportional to the magnet volume and dependent on the magnet quality, i.e., the stored energy of the magnet. For this reason, synchronous servomotors today use mainly rare earth magnets, e.g., neodymium-iron-boron (B =  1 1, 21 T and H = 690 920 kA/m)  . r ci Three-phase synchronous motors Given a three-phase winding on the stator, the magnet rotates at a mechanical (synchronous) angular velocity  =  / p and induces a pole wheel voltage ms E =  N f  ˆ / 2 . p s ws δ For small motors, the ohmic resistance of the stator winding must not be neglected, which is shown in the phasor diagram (Figure a), i.e., in the Kappa triangle of voltage drops and in the equivalent circuit (Figure b). U jX I k  I Z I R I s jX R ks U Ep E p    U tan  = R X s k a)  d-axis b) p For the synchronous (short-circuit) reactance in Figure b), an approximate equation applies: X  E / I . k p k Permanent magnet synchronous motors with salient poles on the stator The permanent magnet synchronous motors are abbreviated in the English literature as PMSMs (permanent magnet synchronous motors). The peculiarity of this type of motors is that the armature winding is not distributed in the stator slots (Figure a) for a 4-pole PMSM with distributed winding for the Q = 12 , (q = 1) , but concentrically mounted on the teeth s s of the stator or, according to German literature, wound on the salient poles of the stator. It is considered that the adhesive torque can be reduced at zero flux if such motors have a number of slots on the stator Q  2 p . The number of slots on the stator is: Q = 2 p + 2k and the s s factor k = 0,5, 1,   2... For all motors with stator pole windings, the number of slots per pole and phase shall be q  0,5 . For example, in Figures b) and c): Q = 6 and 2 p = 4 s s (k = 1) . We have one coil per phase for a single-layer winding, and two coils per phase for a two-layer (two-phase) winding. In general, there are many combinations for Q and 2 p : s 3/2, 3/4, 6/4, 6/8, 9/8, 9/10, 9/12, 12/10, 12/14, 24/16, 24/22, ... 36/42, etc. The number of periods of adhesive torque at zero flux is equal to the product between Q / p and 2 p . s overlapping end winding end winding end winding A A C' B' C' C B C' A' A' A B A' C A A' B C C B B' B' B' C' a) b) c) PMSMs motors have 2 p poles. Thus, the stator winding for Q  2 p has a relatively large s winding factor for the 2 p period. In the case of Q = 6 and 2 p = 4 , this factor for a con-s centrated winding with respect to the width of the coil is equal to sin(120 / 2) = 3 / 2 = 0,866 . With such a winding, a virtually sinusoidal induced voltage can be achieved in the stator phases. These motors do not have a cage in the rotor, and cannot be connected directly to the AC grid. They depend on a frequency-variable power supply with power electronics. Even if they have a cage, stator excitation has very strong harmonic components due to Q  2 p . s The field shape for the example of a motor with permanent magnets on the rotor surface (Figure a) below with the "d" and "q" axes marked) and windings on the salient stator poles for q = 0,5 is shown in Figure b). s The arc of permanent magnets is usually equal to the arc of the stator slot width  = =  D π / Q to reduce the adhesive torque. m s s d-os surface permanent magnets q-os N b non-magnetic spacer S massive yoke S x N thin magnetically and N S shaft  p 2 p ele trically c non-conducting coat S ( b) for high speed ) N a) 2 p = 4 Calculating torque Assume that the magnetic coupling of the permanent magnets on the stator, with two coils per phase, varies sinusoidally. The maximum magnetic linkage for two coils with N turns will t be as follows:  = ˆ 2N  B A , p t δe m where B is the magnetic flux density in the equivalent air gap (  ) and the area A of the δe e m magnet in the direction of the air gap. The maximum magnetic linkage will occur in the d-axis, i.e., the symmetry of the permanent magnets, while the q-axis is the symmetry between the magnets (Figure a). For a sinusoidal arrangement of magnetic linkage: ˆ   ( ) =  sin  . p r p r  is the electrical angle and is p-times the mechanical angle ( = p ). r r rm B is given by the equation: δe B B l rm  . δe 1 + k l +  r m e The factor k = 0,1 0, 2  and takes into account the edge flux. l is the thickness of the r m magnet. The torque in the case where the armature current is in phase with the induced pole wheel voltage U =  ˆ / 2 , i.e., in the case for I = 0 , where we have only the "q" component Ep p d of the current I = I , is calculated by the well-known equation: q M  ˆ p = 3 p I . 2 In the case where the current is in phase with the induced voltage ( I = 0 ), the phasor d diagram in Figure a) is valid, in which we consider the synchronous inductance L and the s stator resistance R . In Figure b) I  0 , the and reaction reduces the excitation. s d R I j L I max s d s d j j L I  L I max sq s q U U U Ep R I s q R I s U Ep I I I q = I q I d a)  p b)  p The synchronous inductance in this case is not only equal to the sum of the magnetizing L m and leakage inductances L (i.e., L for classical machines), but also to the mutual inductance σs d between the adjacent phases L  L / 3 . In the case where the component of the current I 12 md is in the opposite direction to the permanent magnet excitation (for control at higher rotor speeds    ), the phasor diagram in Figure b) applies. max N COMMUTATOR MACHINE Description of construction A commutating machine is an electrical machine with a commutator in the secondary. The commutator assembly (collector − brushes) can be replaced by electronics. A DC or AC machine can be distinguished depending on the voltage applied. The main components are: a stator and rotor with a commutator. The stator consists of: Y Y – a massive stator yoke, ps – a pole with a pole shoe, 1 ps C ew – an excitation winding. NZ − v + NZ The rotor consists of: A B ps A – armature electrical steel 2 ew (dynamo sheet metal) N (1, 2 – armature winding), C – commutator (collector), Sketch of a two-pole DC machine B – stationary brushes. Operation mode a) Generator The rotor coil is rotated in a DC magnetic field. A voltage is induced in each side of the coil − the conductor (bar) according to equation: e = vB l . b This is identical in shape to the magnetic field. N N N B Z Z n  N Z 0 90 180 270 360 S 360 0 e A B Coil connected to two slip rings Magnetic field distribution of the salient poles e B B e v v t  e B Induced conductor voltage Induced coil voltage The induced voltages in the two sides of the coil are oriented oppositely. The geometric sum is twice. The voltage is alternating in time. Such a voltage is also obtained on the slip rings. Instead of two slip rings, we take just one slip ring, which we cut, i.e., two lamellas. The beginning of the coil is connected to one lamella and the end to the other. N e B n  t a) e S t b) e = A( ) B(+) a) Space and temporal Figure of an induced coil voltage b) Voltage on the brushes A collector (commutator) is a mechanical rectifier. The induced voltage will always be in the same direction (−) at terminal A, always (+) at terminal B. A single coil has a large voltage ripple. Usually we have at least two (Figure) or more coils. e e the two coil voltages summed e the voltage of the second coil e the voltage of the first coil emax e min t b) Motor B Y S F  t 1 ps − v M C a) + D NZ NZ B ps 2 ew F t  b) N a) Magnetic flux density b) Torque The rotor is connected to a voltage and a current I flow in it. A force is acting on a conductor in a magnetic field of magnetic flux density B: F = IB l and a pair of forces on both sides of the coil, creating a torque: M = D D F + F = F D . 2 2 Each motor can act as a generator, and vice versa. In a generator, there is also a force opposing the force (torque) of the driving machine. Voltage magnitudes: generator U  E motor E  U Commutator machine windings These are usually two-layer windings. The number of coils is equal to the number of slots. Coil design Arrangement of the coil in the slots The start and end are connected to adjacent lamellas. Therefore, the number of lamellas K is equal to the number of coils or the number of slots Q: K = Q . Vector star and induced voltage polygon The induced voltage of the individual slots is represented by a vector (phasor). All the conductors in a slot have an induced voltage of the same direction and magnitude. The electrical angle in degrees between the slots will be:  360 = p = p  Q , Q where the mechanical angle between the slots is  360 = Q . Q Example: Q = 3 8 , 2 p = 2 , =  45 2 1 4 2 8 1 45 3 7 5 6 4 8 5 6 7 If we connect the end of the first coil with the beginning of the second, etc., we get a polygon. For Q =  , this is a circle. If we place the brushes at two diametral points of the circle (commutator), the winding breaks into two parallel branches. The voltage across the brushes will be equal to the diameter of the circle. Winding versions Depending on the connection of the individual coils, we distinguish between lap and wave windings (or hairpin winding). Y 1 Y 1 Y2 Y Y 2 Y Lap winding Wave winding In a lap winding, the end of the previous coil is connected to the beginning of the next coil. In a wave winding, we skip some adjacent coils. A winding, lap or wave, is always self-connected or self-linkage. For a lap winding, the width of the scroll (winding) is: Y = − Y Y . 1 2 For a wave winding, the width of the scroll (winding) is: Q Q  1 → Y = + Y Y  2 Q = Y = . 1 2 p p p Y is the width of the coils Y  Q Q = 1 . ( Q is the number of slots per pole.) 1 p p 2 p Y is the connection width. 2 Example of a lap winding Example of a wave winding Q Q + 1 = 6, 2 p = 2, Y = Q = 3, Y = 1 Q = 8, 2 p = 2, Y = Q = 4, Y = = 9 1 p 1 p p The pitch is 1 (1 − + Q ) = − 1 4 , Y = 2 . The pitch is 1 (1 − + Q ) = − 1 5 , Y = − = Y Y 5. p 2 p 2 1 1 2 3 4 5 6 1 2 3 4 5 6 7 8 6 1 2 3 4 5 3 4 5 6 7 8 1 2 3 In the pictures we can see that the commutator winding is closed in on itself. The thickly drawn coils are short-circuited via the brushes. DC machine theory Induced voltage Current in the excitation winding: I →  = I N → B →  f f f f δm neutral zone B x B ( ) S  p No-load magnetic field image for No-load magnetic field distribution Figure pole pitch for pole pitch  p  One pole flux:  = l B x ( ) dx m  m 0  = b l B =  l B m pe δ p B B  b is the equivalent peripheral pole width. pe b pe  Flux is represented as the volume of a geo- p metric body (quad) with sides for the mean zone zone value B , dimensions l and  neutral neutral p . The mean value of the induced voltage in a conductor (bar): E = 2 pD π v B l = 2 pn  , if it is v = D n π = = 2 p b n . m p 2 p A winding with " z " conductors (z = 2 ) N connected in series and with " 2a " parallel branches, where " " a is the number of parallel branches of half the armature (rotor), will have an induced voltage: E = z z N E = 2 p n = 4 pn  = 4 pn N  b , m m a m 2a 2a 2a where N = N / (2 ) a is the effective number of turns, i.e., the number of turns of the parallel a branch, and the product " pn" is the frequency of the induced voltage in the armature (rotor). With the introduction of the design voltage constant c = p z e : a E = c n = C  , e m E m and  c  e = C is the magnetic flux coefficient for calculating the induced voltage C  . If E  E m   2π   = const., C is also constant. The voltage at the generator terminals is: m E U = − E I R − ΔU . a a c The voltage is smaller for the voltage drop across the armature resistance I R and for the a a voltage drop across the commutator brushes ΔU . The reverse is true for the motor: c U = + E I R + ΔU . a a c Torque It is calculated from the mechanical power: P = M  = E I = P , which is equal to the m m a i internal power of the machine (if friction and windage losses can be neglected). This statement is proved by the following derivation: P I p z a = M  = F D 2π n = N Bl  4 p n = I n = I c n  = E I , m m p a m a e m a 2 a a M ce =  I = c  I = C I m a M m a M a 2π where c = c / (2π) is the design torque constant. The coefficient C (or C ) is the M e M T magnetic flux coefficient for calculating the torque. Thus, C is the same in value as C ME for the induced voltage if the torque of the rotor friction and ventilation losses can be neglected. The unit of the coefficient is (N m / A)  for torque and (V s / rad.)  for induced voltage. For servomotors is C usually given at 1000 rpm. The coefficients for flux and E torque are also denoted by the letter K. DC machines a) generator b) motor U  fm U  fm E I E I U  E + − U  E + − RL U = − E IR E = − U IR = c n a a em U U − IR a = c n  − IR n = e m a c  e m P 2 = U I = E I − I R M = c  I (Index "i" is intrinsic. ) el a i M m P = E I = = P P + P P = M  = E I = P m i el Cu m i m i P = − P P P = P − P el i Cu m el Cu Armature reaction In a loaded machine, a load current I flows in the armature. I →  → B (the magnetic a a a a field of the armature reaction). Together with the magnetic field of excitation, we obtain the resulting magnetic field. The physical picture of the action changes. The whole phenomenon and its consequences are called the armature reaction. S S  N N a) b) c) The magnetic field of a DC machine in and near an air gap: a) no-load − excites the excitation winding, b) unexcited machine − excites only the armature winding (armature reaction), c) sum of a) and b). The difference between a) and c) is twofold: 1. the magnetic field in the air gap is not distributed homogeneously, 2. the neutral zone is displaced by angle  from the symmetry between the poles. For generator operation, the spatial distribution of the no-load magnetic field is denoted by B , 0 the angular reaction magnetic field by B and the resultant magnetic field by B. Due to a saturation, the increase in the MMF is not equal to the decrease and  will be smaller. m b b pp NZ N S NZ NZ  p B0 a B B a x B   The consequences of the armature reaction are as follows: 1. reduction of induced voltage E , 2. an increase in iron losses, 3. a shift of the neutral zone, 4. an increase in the inter-lamellar voltage. 1. Reduction of induced voltage due to armature reaction a) Generator E   . The main flux  falls due to the reaction of the armature, and, m m hence, E . b) Motor U = const. → E = const. (neglecting voltage drop).  drops due to the armature m reaction and the revolution increase, so it remains E  const. Graphical method for determining the effect of armature reaction We start from the known no-load characteristic (NLC) E = f ( ) I . f  Distance KD = E at excitation  0 m S d  = N I E 2 S 1 e I → b K c 1 a a a a h Resulting excitation: g e S  C B 2 2 S   =    / 2 1 res 0 a a KB = E E = aH = − E e 2 2  E = dG = + E e 1 1  a a 2 2 e = ab and e = cd 2 1 H I F D Gf According to Simpson:  V  0 sat E − + e 4E + + E e e − e 2 Δ = − 1 21 E E = . 6 6 Graphical procedure to reduce the induced voltage: The triangle with nodes a, b, K has surface S , 1 and the triangle with nodes K, c, d has surface S . 2 We travel along the NLC from a point K to a point C, which is determined by the fact that the area of the triangles with vertices a, h, C and C, g, d is (S  = S  ) . CF is the reduced 1 2 induced voltage by a distance KB due to the armature reaction. This voltage reduction is due to a drop of the excitation V = FD (sat − saturation). Therefore, we need to increase sat the excitation due to the armature reaction when the load is applied by  = V . sat sat When the load changes:  2 I  V a  V   sat satN .  IaN  2. Increase in iron losses due to the armature reaction Due to the armature reaction (B  const.) , the losses in the iron, i.e., mainly in the rotor δ teeth, increase by approximately 2 ( B / B ) . tmax t0 3. Neutral zone displacement due to the armature reaction NZ NZ The voltage on the brushes is reduced due to the B displacement of the NZ. The commutating coil + comes under the influence of the main magnetic field. A voltage is induced in it, which deteriorates − the commutation. x 4. Increasing the voltage between the lamellas Mean voltage between the lamellas of a commutator with " K " lamellas: EK = 2 pU . K E K = − 16 20 V is permissible. Due to the deformation of the magnetic field this increases E and the sparking too. K Reducing the effects of an armature reaction can be achieved by: a) moving the brushes, b) a compensating winding. Depending on the movement of the neutral zone, move the brushes at: 1. the motor in the opposite direction of rotation, 2. the generator in the direction of rotation. N 0 compensating winding Gen c Mot   a NZ NZ S The armature reaction is removed by a compensating winding through which the load current flows. The direction of magnetization shall be opposite to the direction of the armature reaction. Voltage at the machine terminals E No-load: E = E A 0 U E0 B E  Distance A represents the reduction C E in voltage due to the movement of U the brushes out of the geometric neutral zone. A = Distance B illustrates the effect of influence  c B saturation. = influence  sat C = I R a + Uc Distance C illustrates the effect of the I voltage drops. N I Commutation I I caca When a coil passes from one pole region through the t = 0 a) NZ to the other pole region, the direction of the 1 induced voltage changes, and so does the direction 2 v c of the current. This is commutation. I At the time of the current change, the commutating I coil is short-circuited across the adjacent lamellas ca i I ca t = t and the brush (Figure b). b) 1 2 c v If the current does not drop to zero at the moment the number 2 lamella leaves the brush, a spark I (electric arc) occurs. This maintains the short circuit I until the change in current is complete. I ca ca t = T c c) 1 2 c v I bt Linear commutation The time T waveform of the current during commutation is unknown (dashed line). c The branch current is considered to be I = I / (2 ) a , T – the duration of the commutation; ca c and T − time of constant current. i The time course of the current is given by Kirchhoff's laws. R c Rc R c R c + I i ca i Ica Ica i R R i 1 con con2 − 1 S S 3 I 1 2 2 v t ca c T bt Rbc T T c i c t T − t c T c Depending on the labels, the equation for the commutation time is: T = b b D t = t a , c v v D c a c where D is the rotor diameter, D the commutator diameter, b the tangential width of the a c t brush, and "v " the corresponding speeds. c According to Kirchhoff's first law: i = I − i , 1 ca i = I + i , 2 ca where " " i is the instantaneous value of the current, " " i and " " i are the input currents at 1 2 lamellas 1 and 2. In the loop, we neglect the resistance of the coil which commutates R and c the resistance of the connection conductors R . con The brush-to-commutator contact resistance R is taken into account. bc According to Kirchhoff's second law: i R − i R = 0 . 1 1 2 2 According to the Figure, the contact resistance of the individual parts of the brush is proportional to the areas between the brush and the lamella: R = S T S T c c cc R = R , R = R = R , 1 bc bc 2 bc bc S t S T − t 1 2 c where the surface of the brush S = S + S or c 1 2 S = c t S , S = S . i 1 c 2 c T t − T I ca i 1 c T c t i 2 The shape of the current in the commutating coil is: i =  t  − T I  1 2  . c ca  T c  The equation represents a straight line, and we call this type of commutation linear. Influence of self-induced voltage i (by the influence of its According to classical theory, the commutation self-induced voltage) process is the switching on and off of an +I ca inductive circuit, i.e., a commutating coil. According to Faraday's law: −I ca commutation) i (linear t e d di σc = − N = − T L , c c σcc d t d t where  is the leakage flux of the commutating coil (with the N turns) and L is its σc c σc leakage inductance. Due to the self-induced voltage, the current lags behind the linear commutation. This degrades the commutation (increasing arcing on the brushes, and thus radio interference in the surroundings). Ways to reduce the impact of self-induced voltage: 1. a reduction in d / d it , i.e., a reduction in current and rotational speed, 2. reducing inductance L by reducing N , σc c 3. commutating auxiliary poles that induce a voltage in the opposite direction (e  e ) . cp c These are used on larger machines. Commutation auxiliary poles S S N N S main 1 field 2 + I ca 3 0 armature 4 field − 5 I ca commutation 6 pole field 7 0 resulting field Place them in the neutral zone. The poles shall be narrow and shall correspond to the width of the brush. The Figure is drawn for a generator. For the motor, the opposite direction of rotation or the opposite arrangement of the commutation poles applies. The load current flows through the windings of the commutator poles and compensates for the effect of the armature reaction. Depending on the dimensioning of these poles, different current curves are obtained: 1 and 2 sub-commutation, 4 and 5 optimal state and 6 and 7 over-commutation. Commutation assembly Classic Commutation Assembly spring This assembly includes the following parts: commutator, brushes, and brush holders. The holder picture shows a simple commutation assem- brush bly used in small machines. The brushes and commutator brush holders are mounted in holders on the lamella stator and are stationary. The commutator is mounted on the rotor shaft and rotates with it. insulation There are two different designs of commutators: 1. for small machines where the lamellas are bonded with plastic, 2. for larger machines, we know the swallowtail version. insulation (plastic) connector commutator lamella insulation airiness F c F F F 1 In addition to these designs, there is a disc design for robotic motors, where the lamellas are mounted radially and the brushes in the direction of the machine shaft, and a turbo commutator for high-speed machines. Depending on the type of material and the manufacturing process, the brushes are divided into: carbon, carbon-graphite, graphite, electro graphite, metal-graphite and resin-bonded graphite. They must have certain properties, e.g., for carbon graphite: specific resistance 30 800  (μΩ m)  , current density 5 7  2 (A/cm ) , peripheral speed up to 20 (m/s) , voltage drop across the brush pair  2,8 V , pressure 21 (kPa) . Applications: small DC and universal motors t t The dimensions of the brushes are tagged a a according to the IEC recommendations: r r t a r   , where it is t – tangential, a – longitudinal (axial), and r – radial dimension. The connecting conductor to the brush holder is made of (fine) copper braid. The brush holders are of different designs. They are usually radial, but skewed (reaction) brush holders are also possible. The picture shows the holders for universal motors. Electronic commutation assembly The commutator-brush assembly replaces a stationary switch. The excitation is on the rotor (permanent magnet) and the armature winding is on the stator. The English abbreviation for these converters is a BLDC motor and stands for brushless (BL) DC motor. The sensors report the position of the rotor (magnet poles) and provide a signal to control the electronic switches. The types of sensors are: Hall elements, photo diodes, or photo transistors and inductive encoders. The armature winding is: a single, two, three and four phase system. Single and two phase → high torque pulsation system The Figure shows a scheme of a three-phase system with constant polarity (unipolar). The torque is f ( )  . Hall element magnet I a) operating diagram 1 H1 N3 b) motor cross-section H3 N1 N c) torque P1 P3 S P2 H2 I IB 3 N2 EB I 2 Tr1 Tr2 Tr3 a) phase I II III M  120 120 c) b) Types of DC machines Types of excitations Classical electromagnet excitation, i.e., one or more excitation windings on the pole → shoes. Another option permanent magnets. → Different machine properties are obtained, depending on the dependence of the excitation on the physical quantities (I, U). Depending on the connection, we distinguish: 1. foreign excitation, 2. parallel excitation, 3. serial excitation, 4. compound excitation. Excitation by electromagnets is the first circuit. The second circuit is the winding of the armature and any auxiliary poles and the compensating winding. Different types of machines are distinguished according to the connection of the two circuits. No-load characteristic (NLC)  m E0 NLC This is a basic characteristic. E0 E = f (  ) and  =  / R → E = f ( )  or m m m   = f ( )  m m The magnetic resistance R depends on the saturation. m (n0 = const.) For NLC we can change the measure for  = f( )  , m because the excitation is I f I N  = →  = f ( ) I . f f m f  Excitation can be a function of voltage, armature current, or a combination of the two. The characteristics of the machines depend on the type of excitation and the connection of the circuits. External characteristic by the "I" line method Generator U = → f ( ) I n = const. Motor → n = f (M ) U = const. It is a geometric representation of physical quantities at different drive states. The process is not analytical, such as the circuit diagram of an induction machine or the current characteristics of a synchronous machine, but graphical. Example for a foreign excitation generator: the starting point is the no-load point P (distance 0 P A = E ) of the NLC, i.e., E = f ( )  . Distance: 0A =  = I N . Then draw: 0 0 0 f 0 f E NLC AB = ΔU = IR and BC =  = V = I N . a sat sat fa f U P 0 ( I = K I and K is the current ratio of the fa P I I machine.)  =   − = 0A − BC or res 0 sat U I ( ) I = I − I . res f0 fa As the load current I changes, ΔU and  change sat Ifa and the extreme point C travels in the direction of the B C IR I af abscissas and describes the "I" line. The distance 0 A  PC = U I ( ) is the voltage of the generator terminals. The "I" line thus represents a combination of the geometrical locations of the points of the resultant ampere-turns, and the ohmic voltage drop of a DC machine as a function of the change in load current; hence the name "I" line. Due to the non-linear influence of the armature reaction (saturation), the "I" line is not really a straight line, but a curve. Generators for direct current Generator with foreign excitation We need two separate voltage sources. P N E N P a U E 0 E b c U A F1 F2 B (F1) I = f const. n = const. I I f (F2) I R U a + c d (A1) (A2) 0 I N I (B1) (B2) When loaded with I is E  E . E = f( ) I is an internal characteristic. The distance ab is N 0 due to the reduction in induced voltage due to the armature reaction. Taking into account ΔU = bc = I R + ΔU we get U = f( ) I . The voltage drops and at I , there is U = 0 , E = ΔU a c k . I f For U = const. we obtain a control I = f( ) I f curve I = f ( ) I . This takes into I f f0 account the influence of the armature reaction and voltage drops. I Parallel excitation generator The excitation voltage is equal to the rotor voltage (U = U ) and I = + I I . It is also called f a f self-exciting if there is remanent magnetism. E → I = E / R , E is increased rem f rem f gradually up to P0. N P E P R0 fcr NLC E i I E AE BE (E1) Rf If I a (E2)  (A1) E (A2)rem (B1) (B2) i fi I I f0f R defines a line at an angle , tan  E  = = f R . f If The process is possible if R  R (critical). f fcr An instantaneous excitation current E raises the voltage difference. U E Δ difi E = E − i R = L i fi f f d t U The external characteristic is softer than for a foreign-excited machine. U I  I I a =  U const. k N k I f Generator with series excitation It is not used as a voltage source. U is highly variable with the current I. It is usually a series winding combined with a parallel winding in compound generators, where it compensates partly for the effect of the armature reaction. DC motors Each generator can work like a motor. Only the direction of current at the rotor terminals is changed for the same direction of rotation. Motor with foreign excitation We are interested in the external characteristic n = f (M ) and the current characteristic I = f (M ) . n  load characteristics f( = I = f(M ) → M = f( ) I  M ) → = n f ( ) I  P n N P  I N n0 n M ( ) A F1 F2 B (F1) I f I I M ( ) (F2) (A1) (A2) (B1) (B2) M In stationary operation n = 1 E c  . e m For I = const. , E will be such that E = − U IR − ΔU holds, and will be: a c n = 1 U − I R − ΔU ac . c  e m For  = const. is M = c  I and M  I . m M m Due to the armature reaction  decreases and the speed increases from a certain load, i.e., m an unstable operating range. The unstable operating region occurs when the influence of the voltage drops becomes less than the influence of the armature reaction (which varies with the square of the armature current at higher load). Parallel excitation motor If the grid is sufficiently rigid, a parallel excitation motor has the same characteristics as a foreign excitation motor. P n N I n0 n M ( ) I AE BE (E1) I M ( ) If I (E2) a I = I f0 (A1) (A2) (B1) M (B2) Difference: grid current I = I + I  I = − I I a f af The difference can be seen in the current characteristic I = f (M ) , because the current has an initial value I = I , whereas in a foreign excitation motor it is zero. f 0 Series excitation motor The series motor is considered to be   I , and thus: m M = 2 1 c  I = c I → = I M , M m 1 c 1 n = 1 E E = c . 3 c I 2 M P N n I n M ( ) A D (D1) Ik I Ia (D2) I I M ( ) (A1) M (A2) M k (B1) (B2) In start-up (short-circuit) I = = I I , M = M and n = 0 . s k k For low loads (no-load): M → 0, n →  Compound excitation motor In most cases the series winding supports a parallel winding. The winding that has the greater influence determines the shape of the characteristic. The motor has a harder characteristic (curve a) if the parallel winding dominates, and a softer characteristic if the series winding dominates (curve c). P n N I AE DE n0 (E1) (D1) If I a I a (D2) a I (E2) b c (A1) (A2) (B1) (B2) M Constructing characteristics Constructing the external characteristics of generators U(I) Assume: the "I" line is a straight line, choose, e.g., I = (1/4  5/4) I . N a) Generator with foreign excitation E NLC U P 0 C U I ( ) U N "I" line B  (4/4)I N 0 2 4 I I f0 I 0 f I N I N 4 4 Point B on the 'I' line is defined for the rated current, i.e., the armature current I = (4/4)I . N The magnitude of the voltage for a given load, e.g., I = I , is obtained from the relation: N U = E  BC / P I . 0 0 f0 b) Parallel generator E P U 0 C U I ( ) NLC ne U N B li U N ng iti xc A lf-e se "I" line  (4/4)I N 0 I f0 If 2 4 I 0 I N I N 4 4 Direction of self-excitation (angle) :  sin  = P I / P 0 = E / P 0 and U = E  AC / P 0 . 0 f0 0 0 0 0 0 Constructing motor rotational speed characteristics n(I) NLC n E P 0 2 4 I n N0 D "I" line C E 4 4 I R aN a nN B I faN 6 4 I N D 0 I A I I f0 f 0 (2 / 4) I (4/4) I (6 / 4) I N N N a) Foreign-excited and parallel motor The characteristic designs ( ) n I for the foreign-excited and parallel motors are identical. The revolutions for a given load, e.g., I = I , are given by: n = n AB / AC . From point E on aN 0 the curve ( ) n I is the unstable operating range. b) Serial motor E NLC n P 0 3 3 I N "I" line P C B nN nN 0 I 1 2 3 4 I fres A I f 0 I I I I N N N N 3 3 3 3 The rotations for an arbitrary load are obtained from the relation: n = AB n , because NLC N AC is valid for n . N Starting and changing the speed of the motors Starting: n = U 0 → E = 0 and I = , M = k C I . k M k R a The current is reduced by adding an additional (starting) resistor, or by reducing U. Changing the rotational speed of foreign-excited motors:  U − I R U R M R M U  M  a = a a a 2π n = = − =  − =  1 − m m0 m0   . C C C C C C U E E E M E M  M k  Options for varying the speed of foreign-excited engines: 1. armature voltage control U , 2. magnetic field control  = f (U ) , m f 3. with additional resistance R . add M I M ( ) m U  U U N = UN I = I I  I f fN ffN U Ia (  m ) If I a  m  m0  m C controlling field E armature voltage weakening From 0 to  , vary U ( I = const. ,  = const. , M = const.). We weaken the magnetic m0 field from  (U = U ) onwards. For   0 (M  M ) , we switch to generator m0 N m L braking. The speed control of the series engine is a particular problem. A change in U causes a change in I . Therefore, we vary the speed by applying a shunt to the excitation winding. f Non-stationary operation Changes are possible after a transient phenomenon determined by the excitation time constant T = L / R and armature T = L / R T   f f f a a a ( (5 20)T . This is electrical inertia. f a ) Mechanical inertia is determined by the masses of the rotor and the load, or the total moment of inertia J . Basic equations for non-stationary operation (drive) u di f = R i + L f f f f d t u = dia R i + + e L (induced voltage e = C  ) a a a E m d t J d m = + m m (torque m = C i ) For the motor m  0 . L M a L d t The familiar equations in stationary operation apply for i = I and i = I is d/dt = 0 . f f a a The following block scheme shows the physical picture of the effects of quantities in a foreign excited machine. U  dif L − E f = 0  + compare − R I Scheme is valid for E  d t a a E m d  C =  ia L R d  a = 0 t a I a CM U I J d M = C I  M a m + f d d t − J f 1 / R C f E  m d t + M L Special DC machines Most of them have only historical significance. One of the remaining is amplidyne, which served as an exciter in synchronous generators. Amplidyne It's a power amplifier. It has two pairs of brushes. The first − normal brushes (q-q) are short circuited, and a rotor excitation greater than the fundamental excitation is obtained. The consumer is connected to the second pair of brushes (d-d). U1 q I 1  c U d d  I d , , I   q 1 q I RL U , I , q q2 Amplification is: U I  100U I  10000U I . d q q 1 1 Because of the rotor (armature) reaction, it has a compensating winding (  compensates c the flux of the armature reaction   ). AC commutator machines Implementation → motors only Only a single-phase commutator motor of small power remains. Use → for household appliances and power hand tools. Also known as a universal motor. Induced voltages of a single-phase commutator motor U (alternating) →  (alternating) → E (alternating) f m r E is in phase with  . r m  m  m e r n t Er ( ) − ( ) + If N = N / (2 ) a = z / (4 ) a is the number of effective turns of the armature, then the rotational a or motion voltage will be: E ˆ = 2 2 p nN  . r a m The alternating magnetic field also results in a transformer voltage: E 2π ˆ = f N f  . t a w m 2 The winding factor is given by f = 2r 2 = w (where r is the radius of the potential circle) πr π and therefore the transformer voltage will also be ˆ E = 2 2 f N  . The space position of t a m the two can be seen in the Figures below. For pn = f , will E = E . r t E The voltage across the brushes in the t neutral zone is equal to E = 0 . t E r The corresponding transformer voltage is also induced in the excitation winding. Calculation of torque For alternating ( ) i t and  ( ) t valid: M t ( ) = c  ( ) ( ) t i t . M m If it is ( ) ˆ i t = I 2 sin(  t ) , it will be  ( ) t =  sin( t − ) , m m where  is the phase shift between the current and magnetic flux. Taking the trigonometric transformation into account, we obtain: M t = cM ˆ ( )  I  − cos(2 t − ) m ( cos ) . 2 The first component is the mean torque: M c M ˆ ~ =  I cos , m 2 around which the torque oscillates at double the frequency, the mean value of which is zero over one period.  ˆ at AC supplied voltage shall m ~ M t be equal to the flux at the DC ( ) I t ( ) supplied voltage  m ( ) t M ( ˆ B = B ) . m ~ m= t That's why: M cos  = M . ~ = 2 From the equation for rotational voltage E , we express  , and note that r m N = N / (2 ) a = z / (4 ) a and the constant c = pz / (2π ) a . The torque value is then: a M M = Er I cos and P = M = E I cos . m  r m Commutation of a single-phase commutator machine Physically, the events during commutation are the same as in a DC machine. For the commutation time: T  = T 1/ f . c The current remains alternating after commutation. The problem is the transformation voltage E . This is greatest in the coil that commutates. It therefore worsens the commutation t process. i t T T T c cc Single-phase commutator motor in series connection The torque is maximum for cos  = 1 ( = 0) , if I and  are in phase, i.e., in series m connection. Application example: small universal motors in household appliances or hand tools with power 5 1000 W  and up to 20000 rpm. The advantage is the high starting torque. It has no compensating winding and no commutating poles. The name universal is used because it can be connected to DC or AC voltages of the same peak value. Current I = I = I →  a ff The excitation winding with N turns excite the main flux: f  ˆ  ˆ L I 2  L I 2 X I 2 = f = f = f ˆ = f ,  → E and E . m m r t N N  N  N f f f f Equivalent scheme U ˆ ~ Taking the equation for  , it will be: m N E = 2 pn N n a X I = c I , r f π f f N n f s where c is the constant of the unsaturated machine  = c 2 Na X . m f π N f E is in phase with I, and is, therefore, E = R I  . R = cn n / is the equivalent (fictitious) r r s resistance. The result is a similar replacement circuit as for the rotor of an induction machine. The equivalent resistance R also I R = R + R jX = j(X + X ) f a f a represents the internal mechanical power: P 2 = I E = I R . m r U Er The current is through an equivalent R  = I circuit: I = U . R c + n + jX n s For n = 0 , i.e., short-circuited, the following applies: I = U k . R 2 2 + X Such a machine, like an induction machine, has a maximum current value at n = 0 , i.e., at a standstill. Phasor diagram We draw it for two examples: a) n = 0 (short-circuited motor), b) n  0 (any number of revolutions). I R U U j I X E = n IR  = r I c ns  I R I  I = I k k j I X   m mk a) b) For n = U 0 it is also E = 0 , and so I = I = r . k Z k For n  0 , E will be in phase with I (  ) for I  I and cos  cos . The power factor r m k k cos   0,95 . N 168 COMPLEX CALCULATION The time-sine quantity of magnitude " v ", whose waveform is shown in the Figure, can be expressed as a function of time t or as a function of the argument  t as follows: v v = ˆ cos(  t + ) . v The amplitude ˆ v represents the maximum value of a v sinusoidal quantity. The proportionality factor before v ˆ time in the argument of the cosine function is the  v t  electric angular frequency  . During the period T , the T argument increases by the value of the angle 2  . From v the condition t =  2 we get: v ˆ 2  v t  = =  2 f . T 2 Frequency is defined in the equation as: f = 1 . T The phase angle  gives the negative displacement of the maximum of the cosine quantity v from the origin of the time coordinate. The cosine function is taken to use complex calculus. Using the Euler notation jx e = cos x + jsin x , the complex equation can be expressed as: v = Re ˆ j(   t + ev) = ˆ j ev j v t Re v e . ( ) ( ) There are now three equal quantities in the equation: the amplitude ˆ j  v v , the phase factor e and the frequency factor e j t . Of interest are the amplitude and the phase position, which, together, represent a complex quantity: v v ˆ e j =v . The corresponding instantaneous value of the complex quantity is given by the following basic equation using the previous equation: v Re j = v t e + j . v ( ) ˆ v The Figure on the right shows a complex  quantity v v as a phasor in the complex plane. + Mathematical operations of complex quantities Multiplying a sine quantity by a constant For multiplication of a complex quantity, the equation: v = v j t jt Re e = a v = Re a v e . 1 ( 1 ) ( ) 169 From it we get v = a v , respectively 1 v ˆ j  ev1 ˆ j = a v ev . 1 We can see that multiplying a complex quantity by a + j constant, changes its amplitude and preserves its phase v = 1 av position. Multiplication by a constant is illustrated in v the Figure on the right. + Adding two sinusoidal quantities This is an example of the law of nodes or the law of the loop. Using the basic equation, we derive v j =t = + jt jt  t Re v e v v = Re v e + v e Re ( )e 1 ( ) ( 1 ) = ( v v + j 2 2 1 2 ) , it is: v v v = + . 1 2 With the introduction of real and imaginary + j v v = 1 + v2 parts, it applies: Re( ) v v Im(v ) 2 v + jIm( v = Re( ) Re( v + v ) + 2 2 1 2 Im( v ) 2 ( + j Im( ) Im( v + v ) . 1 2 ) Im( ) v 1 v Im( ) 1 v 1 + Re( v ) 2 The Figure shows the phasors in the complex Re( ) v Re(v ) 1 2 plane, which are summed geometrically (vectorially). Differentiating a sinusoidal quantity by time This is the case for the law of induction. Using the basic equation, we derive: v Re ( jt e ) dv  d  t    = v  = = Re ( v j e ) = Re j (  v jt   e ) d t  d t  and it is v  = j v . Differentiating by time in the domain of instantaneous values means, in the complex domain, multiplying by j = j /   . Taking into account j e and v j v e =  ˆ j( ev+ )  v / we get the ratio between the amplitudes  ˆ v  = v ˆ and phase angles   =  +   / . v v The differentiated quantity overtakes the original quantity by 90 . 170 + j Corresponding to the derivative equation, draw a phasor diagram of the derivative v = j v v and the v original quantity v . + Time integration of a sine quantity Such a use case occurs in a voltage − current coupling on a capacitor such as: u = (1 / C) i dt .  The basic equation gives: v = t 1 ( j     t Re e ) = d = Re ( j t e d ) 1 v  v  v = j t t Re v e 1   and it is j    v 1 = v 1 . j Integrating over time in the domain of instantaneous values means, in a complex domain, dividing by j −   j / . For 1 / j = e , the equation can also be represented as: v ˆ j 1  v1 = ˆ j( ev− ) / e v . 1  The following is the relationship between the amplitudes  + j 1 v ˆ = v ˆ and phase angles   =  −   / . 1  v1 v v The integrated quantity lags the original quantity by 90 . Plot the + v 1 = v corresponding equations on the phasor diagram of the integrated 1 j quantity v and the original quantity v . 1 The feasibility of a computational operation in the complex domain fails in the case of multiplication of two sinusoidal quantities corresponding to the expression, as required in the case of determining the instantaneous power value. The cause of the failure is conditional on the term: Re ( j t ) ( j v t ) ( jt j t e Re v e  Re v e v e . 1 2 1 2 ) To determine the instantaneous value, we need to consider the individual instantaneous values. Calculation of power Calculating single-phase power in complex calculus The power flowing into the terminals of the converter at the applied voltage u = 2U sin(  t + ) and current i = 2 sin( I  t + ) is calculated using the equation: u i p = u i = U I cos(   − ) −U I cos(2  t + +  ) u i u i 171 The power oscillates at twice the frequency of the voltage or current around the mean value P = U I cos  and is known as the working (watt) power. The phase angle between voltage and current is   = −  . u i u i p The time waveform of power when the p u S current and voltage are sinusoidal quantities of the same frequency is i P shown in the Figure for the case where t the current lags voltage. Without taking into account the phase shift of the current with respect to the voltage, we obtain an expression for the total or apparent power: S = U I . In addition to the total or apparent power, there is, quite formally, the reactive power, which is calculated as: Q = U I sin  . It is not possible to determine the instantaneous power value from the complex voltage and current, but the power components can be determined. Complex power S is introduced by multiplying the complex voltage by the conjugate complex current. This is the case: S = * j(   t +u ) −j(  t + ei ) j(  − = ) U I = U I e U I e u i = U I cos + jU I sin  = + P jQ . It is necessary to emphasize that this complex power is a complex quantity of a different kind, and behind it is not subject to the basic equation. Calculation of three-phase working power for different winding connections The three-phase (symmetrical) winding is connected in a delta or star connection. A zigzag or broken star is also used in transformers. In all three cases, the equation for the working power applies: P = 3U I cos  , ph ph where U and I are the phase quantities, i.e., the RMS values of the voltages and currents in the ph ph winding. According to the Figure for a three-phase winding connected in a delta circuit, the currents in the terminals (lines) are considered to be the difference of two adjacent phase currents, depending o n the sequence of the phases. Therefore, the equations for the currents in the supply lines apply: I = I − I = 2I cos 30 = I 3 , I = I − I = I 3 and I = I − I = I 3 . A   C   B    C  C   C 172 A B C −I ΔC I A U I Δ I = I C 30 I C I B I = I ph C ΔC I = I = I / 3 I ph  C  The equations apply to a symmetrical system where the currents in the individual phases are equal to each other: I = I = I = I respectively I = I / 3 . ph  A   C ph In a delta connection, U = U gives the equation for the working power in a three-phase ph system, expressed in terms of the effective magnitudes at the terminals: P = 3 U I cos  U B = A UAB I ph = I Y U a U U U = U U ph = CAAB 3 U c 30 A U U ph =b C C B U BC In a symmetrical star or zigzag connection, the relationship for the voltages is: U = U = U = U = 2U cos 30 = U 3 or U = U / 3 . AB BC CA ph ph ph The currents in a star or zigzag connection are the same in the windings as in the supply (I = I ) . ph Taking the values for the phase quantities gives the equation for the working power: P = 3U I cos  . We find that the working power in a three-phase system can also be calculated from the input (line) quantities using the same equation, i.e., independently of the type of winding connection. Calculation of losses for different winding connections Assume a symmetrical three-phase winding, connected in a delta, star or zigzag connection. The resistances are given between the supply terminals R = R = R = R and the current t AB BC CA in the supply terminals I = I = I = I . The winding losses are calculated using the equation A B C P = 2 3 I R , i.e., from the phase quantities. For a delta circuit, the phase currents are Cu ph ph assumed to be given by the preceding derivations: I = I / 3 . ph The phase resistance is obtained from the resistance between the terminals with respect to the parallel connection of one phase and the two series connected resistances of the adjacent phases according to the equation: 173 1 1 1 3 3 = + = → R = ph R and the losses for a delta connection are AB R R 2R 2R 2 AB ph ph ph P = 2 2 3 I R = 3 R = =  2 I 3   2 Cu  1, 5 I R 1, 5 I R . ph ph AB AB t 3   2 For a star or zigzag, I = I and R = R / 2 = R / 2 apply, and we calculate the losses using ph ph AB t equation: P = 2 R 2t 2 3 I R = 3 I = 1,5 I R . Cu ph ph t 2 The same equation (independent of the connection) is used to calculate the winding losses as a function of the input quantities. The power losses in a three-phase winding in the complex region are calculated by multiplying the complex current (in the terminals) by the conjugate complex current value. That is: P ( ) * = 1, 5 Re I I R . Cu t The law of induction in complex calculus The induced voltage in the turn is governed by Faraday's law of electromagnetic induction:  d e = − = −  . dt As a rule of thumb, the magnetic field (or flux) is usually given as the "maximum", i.e., the peak value of the time sine function:   = ˆ e jt . Considering the differentiation − derivative equation, we derive: e = − j  . Flux multiplied by j − , means that the induced voltage lags the flux by an angle of 90 . The effective value of the induced voltage of one turn (the turn voltage) will thus be: E E ˆ ˆ ˆ  2    / 2    ˆ ˆ ˆ ˆ = =  = f  = 4 f  = 4, 44 f  = 1,11 = 1,11 . T 2 2 2 2 / 2  t The induced voltage is equal to twice the change in flux over the half-period ( T / 2 ). In the equation 1,11 =    ( / ) 2 =  ( )  ( / / ) (1 / 2) is the form factor for alternating quantities of the sinusoidal shape. The ratio of the mean value to the peak value is /   , 2 the ratio of the peak value to the RMS value. (Note: ˆ  is the peak mean flux.) 174 Calculating the mean and RMS value of an alternating quantity To calculate the mean value of an alternating quantity of sinusoidal shape (amplitude) for half the period (Figure), the equation is used: Y = 1 1  2 Y sin d x x = Y ( cos ) − x =   0 Y .   0  Y ˆ   x = t 2 ˆ Y = Y  sinus quantity y = Y ˆ sin x To calculate the RMS value of a sinusoidal quantity from the peak value, use the following derivation: 1 1 1 2    ˆ   Y ˆ = ˆ 2  1  Y ˆ ( ) ( sin ) d Y xx = Y 1 cos(2 ) d − xx = Y x −  sin(2 ) x =   . 2   2   0 0 0 To add the RMS values of the sine magnitudes of the fundamental and higher order frequencies, we follow the rule: Y = 2 + 2 2 2 Y Y + Y + ......+ Y . 1 3 5  In electromechanical converters, the order  (usually) contains only odd (unpaired) higher harmonic components (  = 1, 3, 5, ....∞), because the quantities are (usually) symmetrical with respect to the y-axis . Higher harmonic components occur, e.g., in the magnetizing current of an electromechanical converter as a result of saturation. Note: To calculate the RMS value of the induced voltage, you would have to write the mean value of the flux (which is in fact the peak mean value) in the equation for the induced voltage in the form ˆ ˆ  = (2 / )  . We do not write it this way, but omit the sign for the mean, and write only the sign for the amplitude or peak value of the flux amplitude. It is essential to emphasize the peak flux value, which is important for calculating the correct value of the magnetic flux density, and, hence, the excitation required. It is also a rule that alternating electrical quantities (voltage, current, etc.) are given as RMS values, and magnetic quantities (flux, magnetic flux density, etc.) are given as peak or peak- mean values. 175 FOURIER ANALYSIS Examples for excitation curves In rotating electromechanical converters, a given periodic excitation curve  = f ( ) x must be replaced by a period 2    2 of exactly or approximately the trigonometric sum: p s x ( ) = 1 a + a cos x + a cos(2 ) x + + a cos(   x) 0 1 2  2 + b sin x b + sin(2 ) x + + b sin(  x) . 1 2  The best approximation of s x ( ) to the curve f ( ) x is to choose for the coefficients a and b   (  = 0, 1, 2, ) the Fourier coefficients of the given function: 2   a = 1 0 f ( ) d xx , 2 0 2   a = 1 f ( ) cos( x   x) dx ,  0 2   b = 1 f ( ) sin( x   x) dx .  0 In the case of rotating electromechanical converters, the function f ( ) x is odd, i.e. f (− = − x ) f ( ) x (type II symmetry), and, in addition, symmetrical with respect to the x-axis f (x +  = − ) f ( ) x (type III symmetry), so this is a type IV symmetry. In this case, a = b = 0 and k 2k /2  b = 4 f ( ) sin( x   x) dx .  0 ( =  2k +1 ) and it is k = 0, 1, 2, . We see that only sinusoidal unpaired higher harmonic components are obtained. In the following examples we will learn about some of the most typical excitation curve functions and their mathematical solutions. Figure a) shows the most typical excitation curve of a single coil with excitation amplitude y = A for 0    x . y y A A 0  2 x 0  2 x a) b) The evolution of the function into a trigonometric series for the first seven higher harmonic components is as follows: 176 y 4  1 1 1  = A sin x +  sin(3 ) x + sin(5 ) x + sin(7 ) x +  .   3 5 7  For an infinite number of coils, uniformly distributed around the circumference of the machine, with the excitation amplitude of y = A for − / 2    x / 2 in Figure b), we obtain the solution: y 8  1 1 1  = A − + −  .  sin x sin(3 ) x sin(5 ) x sin(7 ) x   2  2 2 2 3 5 7  In the case that a winding with an infinite number of coils is distributed over only part of the circumference of the converter and y = A for     − x  (Figure c), the solution is: y 4 A  1 1  = sin sin  x + sin(3 ) sin(3 )  x + sin(5 ) sin(5 )  x + .    2 2   3 5  y y A A 0  2  x 0  2 x   c) d) Normally, the winding is only distributed over 2 / 3 of the circumference of the machine ( =  / 3) , giving the equation the form: y 6 3  1 1  = A 2  sin x − sin(5 ) x + sin(7 ) x .    2 2 5 7  All unpaired harmonic components divisible by three are dropped. The last example in Figure d) shows the excitation of a coil for salient poles, where the pole width is narrower than the pole pitch and y = A is by     − x  . The solution is given by the equation: y 4  1 1  = A cos sin  x + cos(3 ) sin(3 )  x + cos(5 ) sin(5 )  x + .     3 5  If the pole width is 2 / 3  and  =  / 6 (usually the pole width is slightly larger), the equation takes the form: y 2 3  1 1  = A  sin x − sin(5 ) x − sin(7 ) x .    5 7  Example for transformer no-load current The shape of the no-load current of a real transformer is shown in Figure a) (next page). Since the current is a periodic function of time with period T , the value on the abscissa axis is x = t . The magnetizing current is symmetrical in shape with respect to the x-axis and asymmetrical 177 with respect to the y-axis. The function is odd because it is ( f− = − x ) f x ( ) , i.e. symmetry of the second kind and at the same time it is ( f x T + / 2) = − f x ( ) , i.e., symmetry of the third kind. For this form of current, only unpaired higher harmonic components (  = 2k +1 ) are obtained for k = 0, 1, 2, . For respectively k = 0 or  = 1 , i.e. for the harmonic component of the fundamental frequency, the transformer has the sine and cosine components of the current. One of the two components, the larger one, is the magnetizing current (the reactive component) and the other, the smaller one, is the losses in the iron (the working component). For all other higher harmonic components, according to the Figure, only the sinusoidal harmonic components are obtained, i.e. the magnetizing current of the higher harmonic components. For the case of a shift of the x coordinate system, only the cosine higher harmonic components are obtained, while the fundamental harmonic component of the current always has a sine and a cosine term. I I    x T T x   4 2 a) b) If the effect of hysteresis loop losses and eddy current in the iron core can be neglected, the no-load current picture of a transformer normally operating in saturation, i.e. at the knee of the magnetic curve, is perfectly symmetrical. Such a simplified form of the current is shown in Figure b). The current is symmetrical with respect to the x and y axes, i.e. a type IV symmetry. For the magnetizing current in Figure b), we obtain only the cosine or only the sine of the fundamental component and the unpaired higher harmonic components of the transformer magnetizing current. 178 LITERATURE 1 K. Reichert, Elektromagnetische Energiewandler, ETH Zürich, Institut für Elektrische Maschinen, lecture extracts, Zürich 1983/85 2 P. Jereb, Fundamentals of electrical machines, University of Ljubljana, Faculty of Electrical Engineering, University Printing House Ljubljana, (Slovenian), Ljubljana 1975. 3 G. Müller, Elektrische Maschinen, VEB Verlag Technik, Berlin 1970 4 G. Gotter, Erwärmung und Kühlung elektrischer Maschinen, Springer-Verlag Berlin/Göttingen/Heidelberg, 1954 5 A. Dolenc, Transformers, University of Zagreb, (Croatian), Zagreb 1961 6 J. Ritonja, I. Zagradišnik, Transformers, University of Maribor, press.um.si, textbook, (Slovenian), Maribor 2020 7 A. Dolenc, Asynchronous machine, Faculty of Electrical Engineering, University of Ljubljana, (Slovenian), Ljubljana 1969 8 I. Zagradišnik, B. Slemnik, Electric rotating machines, Printing House of Technical Faculties Maribor, textbook, 6th edition, (Slovenian), Maribor 2012 9 B. Jurković, Z. Smolčić, Collector machines, University of Zagreb, School Book – Zagreb, (Croatian), Zagreb 1986 10 T. Kenyo, S. Nagamori, Permanent-Magnet and Brushless DC Motors, Clarendom Press, Oxford 1985 11 IEC Standard, Letter symbols to be used in electrical technology, Part 4: Symbols to be used for rotating electrical machines, Publication 27-4, © IEC 1985, Genève, Suisse 12 IEC Recommendation, Recommendation for rotating electrical machinery, Part 4: Methods for determining synchronous machine quantities from tests, Publication 60034-4, © IEC 1985, Genève, Suisse 13 IEC Recommendation, Rotating electrical machines, Part 8, Terminal markings and direction of rotation of rotating machines, Publication 60034-8, Genève, 1972, Suisse 14 IEC Handbook, Letter symbols, © IEC 1983, Genève, Suisse 15 IEC 60050 (421), International Dictionary of Electrical Engineering, Chapter 421, Power transformers and chokes, 1990-10 16 IEC 60050 (411), International Dictionary of Electrical Engineering, Chapter 411, Rotating machines, 1996-06 179 LIST OF VARIABLES a number of parallel branches p. 136; (m) a longitudinal dimension of brushes p. 148 A 2 (m ) area or cross-section p. 2 b T 2 ( =  V s / m = N / (A m) ) instantaneous value of the magnetic flux density p. 8 b (m) [11] width code p. 7 or pole arc (width) p. 135 B T =  2 V s / m = N / (A m)  magnetic flux density p. 3 ( ) c J / (kg K  ( ) specific heat p. 18; c design constant p. 99; c ( )  motor constant p. 166 C ( 2 4 W / (m  K )) radiant constant p. 17; C N m/A or V s/rad (  ) coefficient p. 136; C F =  ( A s / V) capacitance p. 170 d mathematical operator of the derivative of a function (differential) p. 3; (m) d thickness p. 16 D (m) diameter (stator bores) p. 8; delta winding connection p. 54 e (V) induced voltage p. 11; e  2.71828 number of p. 18; (%) e ratio error p. 53 E (V / m) electric field intensity p. 11; (V) E induced voltage p. 15 f −1 (Hz = s ) frequency p. 1; f ratio − winding factor p. 8 and p. 65 F (A) magnetomotive force p. 3; F (N) force (in magnetic field) p. 12 H (A / m) magnetic field intensity p. 3 i (A) instantaneous value (of current) p. 6 I (A) RMS current p. 1; DC value p. 93 j j /2  = e phase shift in the complex plane p. 25 J 2 2 A / m 2 or (A / mm ) current density p. 4; J kg m  Moment of inertia p. 161 ( ) ( ) k factor p. 127; constant p. 136 K ratio (transformation ratio) p. 12; constant p. 99; K number of commutator lamellas p. 132; K ( N m / A  =  V s) ) coefficient p. 137 l (m) length p. 3, longitudinal (axial) length (machine package) p. 67 L (H =  Ω s) inductance p. 3 m number of phases of AC machine winding p. 1; m (kg) mass p. 16 M (N m)  torque (force torque) p. 2 n −1 −1 (s ) or (min ) speed p. 2; N number of turns p. 4 p number of pole pairs p. 8 180 P (W) working or loss power p. 1 q (A s / m)  line charge p. 12; q relative reactive power p 46; q number of slots per pole and phase p. 62 Q (A s)  charge p. 12; Q number of slots p. 62; (V A) Q  reactive power p. 46 r relative resistance p. 46; (m) r radius p. 14 or radial dimension of brushes p. 148 R H ( −1 −1 =  ( s) ) magnetic resistance p. 3; ( ) R  ohmic resistance p. 16 s slip (induction machine) p. 68 S 2 (V A)  apparent power p. 50; S (m ) area or cross-section p. 145 t (s) time p. 8; (m) t tangential dimension of brushes p. 148 T (s) repeat period p. 9, time constant p. 18 u (V) voltage p. 14; u relative voltage p. 46 U (V) RMS voltage value p. 1 v (m / s) speed p. 9 V 3 (m ) volume p. 127; (A) V drop of excitation p. 140 w 3 (J/m ) specific magnetic energy p. 127 W (J) energy p. 13 x (m) coordinate (abscissa) p. 7; x relative value of reactance p. 46 X ( )  inductive resistance − reactance p. 32 y actual to nominal value ratio factor p. 50 Y three-phase star winding connection p.54; Y width of winding coils p. 63 z number of conductors (coils) p. 136 Z ( )  impedance (complex resistance) p. 40; Z three-phase zigzag winding connection p.54;  (rad.) electrical or mechanical angle p. 8  2 W / (m K) coefficient (heat) p. 16,  transformation constant p. 85 ( )  pole form factor p. 105,  (rad) electrical or mechanical angle p. 138  2 (S m / mm )  specific electrical conductivity p 30; (rad.)  phase angle p. 87  (rad.) wheel or pole angle p. 99  (m) air gap p. 6;  (m) equivalent air gap p. 7 e  mathematical sign of the difference p. 17; delta winding connection p. 81  absorption ratio (heat) p. 17  efficiency p. 2  (rad.) electric space angle p. 8;  ( C) temperature p. 16 181  (A) instantaneous value of the magnetic voltage (excitation) p.8  (A) magnetic voltage (excitation) p. 4;  (K) absolute temperature p. 17  magnetic susceptibility p. 4  W / (m K) ( ) specific thermal conductivity p. 16  H ( =  Ω s) magnetic conductivity p. 3;  (W / K) thermal conductivity p. 16  −7 relative permeability p. 4;  =  4 10 V s/(A m) vacuum permeability 0 (   )  spatial order of higher harmonic components p. 38  (degrees or minutes) angle error p. 53  =  Ludolf's number p. 7  field leakage factor p. 86  mathematical sum operator p. 4  (m) arc on the perimeter of the machine p. 7; (s)  time constant p. 88  (rad.) phase angle or angle of displacement p. 25  (Wb =  V s) magnetic flux (instantaneous value) − flux p. 11  (Wb =  V s) magnetic flux − flux p. 2;  (W) heat flow p. 16 t  (V s)  magnetic linkage (instantaneous value) p. 12  (V s)  magnetic linkage (RMS value) p. 3  −1 (rad. / s) or (s ) electrical angular frequency (circular speed) p. 8  −1 (rad. / s) ali (s ) mechanical angular velocity (circular velocity) p. 2 Greek alphabet  ,  alpha  ,  iota  ,  rho  ,  beta  ,  kappa  ,  ,  sigma  ,  gamma  ,  lambda  ,  tau  ,  delta  ,  mu  ,  upsilon  ,  epsilon  ,  nu  ,  ,  phi  ,  zeta  ,  xi  ,  chi  ,  eta  ,  omicron  ,  psi  ,  ,  theta  ,  pi  ,  omega Abbreviations AC alternating current, DC direct current, IM induction machine, MMF magnetomotive force, MMF magnetomotive field, EMF electromotive force, NZ neutral zone, RMS root mean square E DOI LECTRICAL AND https://doi.org/ 10.18690/um.feri.1.2025 ELECTROMECHANICAL ISBN 978-961-286-945-8 CONVERTERS: LECTURE NOTATIONS IVAN ZAGRADIŠNIK, JOŽEF RITONJA University of Maribor, Faculty of Electrical Engineering and Computer Science, Maribor, Slovenia ivan.zagradisnik@guest.um.si, jozef.ritonja@um.si This publication presents the basics of electromechanical conversion and Keywords: fundamentals of the four basic electrical machines: the transformer, the induction machine, electromagnetics, transformer, the synchronous machine and the commutator machine. The publication induction machine, is divided into five chapters. The content of each chapter is presented synchronous machine, commutator machine below. Introduction: magnetic field, excitation of windings, induction of voltage, forces and torque, conversion of electrical power into electrical or mechanical power, losses, efficiency, heating and cooling. Transformer: construction elements, ideal and real single-phase transformer, three-phase transformer, special transformer designs. Induction machine: description of construction with windings and mode of operation, starting motors and varying speed and torque, induction generator, single-phase induction motors. Synchronous machine: description of construction and operation, operation on a rigid grid, approximate treatment of a saturated machine, excitation systems and the use of permanent magnets for excitation, and permanent magnet synchronous motors. Commutator machine: description of the construction with windings and mode of operation, armature reaction, commutation problems, characteristics of different stator-rotor winding connections, variation of rotational speed, AC (universal) commutator machines.