cos sin XqIq cos sin V X Consider a simple case ignoring R a and X l d axis q axis V q I q V d I d Approximately, the second item can be ignored:

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1 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

2 Power Transformer F=IN=/P Winding resistance Leakage flux Permeability of the core Core losses Ideal Transformer Real Transformer 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

3 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

4 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

5 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

6 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

7 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

8 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

9 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= R= 100 R jx I 1 e e R jx I 1 e1 e1 5

10 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

11 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, , 3, 5, 7 9, 5, 7, 3, 5, 7, 9, 5, 7 7

12 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

13 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

14 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: 30

15 Autotransformer Model I L 1 I N1 a I N 1 Apparent power: N S I (1 ) I auto 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

16 Conventional transformer connected as an autotransformer (Example 11 from ECE35 Wildi s book) 600 H 1 X I I 1 =100A X X 1 H 1 H I =15A I 1 =5A H X I =15A S i =600100=60kA S o =48015=60kA I 1 =5A I +I 1 =150A X 1 X H 1 H + 10 I 1 =5A 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 I =15A The maximum apparent power S max =max( E 1, E )( I 1 + I )= ( E 1 + E ) max( I 1, I ) 3