Consider a simple case ignoring R a and X l E cos XdId I d E X d cos sin XqIq E E jxdid jxqi q jxdid q jx I d axis q axis I q q d q q I q sin X q d P3 3 I a cos I d I cos abdei cos I sin a q d E Xd Xq P3 3 I cos q Id sin 3 sin 3 sin X X X d d q Approximately, the second item can be ignored: E P3 3 sin X d 17
Power Transformer F=IN=/P Winding resistance Leakage flux Permeability of the core Core losses Ideal Transformer Real Transformer 0 0 0 Infinite (flux is produced even with magnetizing current I P =0) 0 0 (windings do not link the same flux) Finite (magnetizing current I P 0 is needed to produce flux) 0 (including hysteresis losses and eddy current losses due to time varying flux) 18
Real Transformers Ideal Transformer Modeling the current under noload conditions (due to finite core permeability and core losses) When I =I =0 I 1 =I 0 0 I 0 =I m +I c I m is the magnetizing current to set up the core flux In phase with flux and lagging E 1 by 90 o, modeled by E 1 /(jx m1 ) I c supplies the eddycurrent and hysteresis losses in the core A power component, so it is in phase with E 1, modeled by E 1 /(R c1 ) Modeling flux leakages Primary and secondary flux leakage reactances: X 1 and X Modeling winding resistances Primary and secondary winding resistances: R 1 and R E1 I N1 E I N a 19
Exact Equivalent Circuit Ideal Transformer E1 I N1 E I N a I I / a I N N N a 1 N 1 Z 1/a 1/a a a Z Z R jx a Z I a 1/a I N N R j X N 1 1 N Referred to the primary side. 0
Approximate Equivalent Circuits Since 1 E 1, Z 1 can be combined with Z to become an equivalent Z e1 Referred to the primary side: R jx I 1 e1 e1 N R R a R R R 1 e1 1 1 N Referred to the secondary side: N X X a X X X 1 e1 1 1 N R jx I 1 e e R R / a R N R R e 1 1 N1 X X / a X N X X e 1 1 N1 If power transformers are designed with very high permeability core and very small core loss, the shunt branch can be ignored. Then, I 1 =I and I 1 =I. 1
Determination of Equivalent Circuit Parameters Opencircuit (noload) test Neglect (R 1 +jx 1 )I 0 (since R 1 +jx 1 << R c1 //jx m1 ) Measure input voltage 1, current I 0, power P 0 (core/iron loss) P 0 R c1 1 I 1 c P I 0 R m I0 Ic c1 X m1 1 I m
Shortcircuit test Apply a low voltage SC to create rated current I SC Neglect the shunt branch due to the low core flux I SC P SC SC Z e1 P sc sc Re 1 X e1 Ze 1 Re 1 Isc Isc 3
Transformer Performance Efficiency: 95% 99% Given,rated, S =3,rated I,rated (fullload rated A) and PF Actual load is I =n I,rated where n<1 is the fraction of the full load power P out =n S PF P c, rated =3R e I,rated : fullload copper loss (current dependent) P c =3R e I =3R e I,rated n P i, rated : core/iron loss at rated voltage (mainly voltage dependent, almost constant) output power input power n S PF S PF n S PFn P P S PFnP P n c, rated i, rated c, rated i, rated / Maximum efficiency (constant PF) occurs when d di d 0 0 when n dn P P irated, crated, Learn Example 3.4 4
oltage Regulation % change in terminal voltage from noload to full load (rated) nl rated R= 100 Generator rated = nl when I=0 = rated when I=I rated I E rated R= 100 rated E ( R jx ) I a s a Transformer: Utilizing the equivalent circuit referred to one side: Source (generator, transformer, etc.) +, nl, rated R= 100, rated Secondary Primary 1 R= 100 1 R= 100 R jx I 1 e e R jx I 1 e1 e1 5
Three Phase Transformer Connections A bank of three singlephase transformers connected in Y or arrangements Four possible combinations: YY,, Y and Y Yconnection: lower insulation costs, with neutral for grounding, 3 rd harmonics problem (3 rd harmonic voltages/currents are all in phase, i.e. v an3 =v bn3 =v cn3 = m cos3t) connection: more insulation costs, no neutral, providing a path for 3 rd harmonics (all triple harmonics are trapped in the loop), able to operate with only two phases (connection) YY and : H/L ratio is same for line and phase voltages; YY is rarely used due to the 3 rd harmonics problem. Y: commonly used as voltage stepdown transformers Y : commonly used as voltage stepup transformers 6
3 rd harmonics problem with three phase transformers configuration provides a closed path for 3 rd harmonics, or in other words, all triple harmonics are trapped in the loop. The 10 o phase difference between the fundamental harmonic of I 1 an I and I 3 becomes 360 o (in phase) for the 3 rd harmonic Then by KCL, I a (3)=I 1 (3)I 3 (3)=0; also I b (3)=I c (3)=0 Distorted waveform Fundamental 3 rd harmonic, 5, 7, 3, 5, 7 9 3 9, 3, 5, 7 9, 5, 7, 3, 5, 7, 9, 5, 7 7
Y and Y Connections Y and Y connections result in a 30 o phase shift between the primary and secondary linetoline voltages According to the American Standards Association (ASA), the windings are arranged such that the H side line voltage leads the L side line voltage by 30 o E.g. Y (HL) Connection with the ratio of turns a= N H /N X >1 H Side (indicated by H ) in Y connection: = H,L HP, An NH N X, P ab X a = H,P L Side (indicated by X ) in connection: H, L H, P AB 330 An X, L X, P = X,P = X,L H, L X, L AB 3a30 ab (Complex ratio) 8
Per Phase Model for Y or Y Connection Neglect the shunt branch Replace the connection by an equivalent Y connection Work with only one phase (equivalent impedances are linetoneutral values Z ey =Z e /3) All voltages are linetoneutral voltages 1 E.g. for Y Connection, 1 is the phase voltage of the Y side and is the phase (linetoneutral) voltage of the side 9
Autotransformers A conventional twowinding transformer can be changed into an autotransformer by connecting its two coils in series. The connection may use a sliding contact to providing variable output voltage. An autotransformer has ka rating increased but loses insulation between primary and secondary windings (Source: EPRI Power System Dynamic Tutorial) 0MA (115/69k) McGrawEdison Substation AutoTransformer (YY) (Source: http://www.tucsontransformer.com) 30
Autotransformer Model I L 1 I N1 a I N 1 Apparent power: N S I (1 ) I auto 1 1 1 1 N1 1 (1 ) S S S a w w conducted N N (1 a) 1 1 H 1 L L L N N 1 IL I1 I I1 I1 (1 a) IH N N Power rating advantage Transformed power (thru EM induction) Conducted power ( I 1 ) Equivalent Circuit (if the equivalent impedance is referred to the H side) 31
Conventional transformer connected as an autotransformer (Example 11 from ECE35 Wildi s book) 600 H 1 X 1 10 + 600 I I 1 =100A X X 1 H 1 H I =15A + 10 + 480 I 1 =5A H X I =15A S i =600100=60kA S o =48015=60kA I 1 =5A + 600 I +I 1 =150A X 1 X H 1 H + 10 I 1 =5A + 70 + 10 I I 1 =100A S i =600150=90kA S i =10100=1kA S o =7015=90kA S o =4805 =1kA H H 1 X 1 X 600 + I =15A 480 + The maximum apparent power S max =max( E 1, E )( I 1 + I )= ( E 1 + E ) max( I 1, I ) 3