Resistor. Inductor. i = C dv dt I = jwc V. Capacitor. v = L di dt V = jwl I. Passive Elements and Phasor Diagrams V = R I.

Transcription

1 assve Elements and hasor Dagrams Resstor R v v R R v nductor v v v d dt jw Capactor C v C dv dt jwc v Threephase systems 1

2 deal Transformer 1 v1 N1:N2 2 v2 Transformer feedng load: a a v1 2 N1 v2 1 N2 1 2 N1 2 1 N Z 2 1/a 2 2/Z 1 2/a 2 1 Assumng a R load connected to secondary and deal source to prmary 1 2 Threephase systems 2

3 Two Wndng Transformer Model The lnear equvalent model of a real transformer conssts of an deal transformer and some passve elements 1 v1 N1:N2 2 v2 Threephase systems 3

4 AC Generators and Motors AC synchronous generator nglephase equvalent AC synchronous motor nglephase equvalent AC nducton motor (rarely used as generator) Threephase systems 4

5 teadystate oluton n snusodal steadystate a crcut may be solved usng phasors R vs jw θ 0 π 2π R jw (R jw) (R jx) Z Z m ax 0 Z θ m ax θ x cosθ θ y snθ Rectangular form olar form θ x j y max θ F ro m re c ta n g u la r fo rm to p o la r fo rm : 2 2 x y M a g n tu d e θ ta n 1 y x A n g le o r p h a s e F ro m p o la r fo rm to re c ta n g u la r fo rm : x c o sθ R e a l p a rt y sn θ R e a c tv e o r m a g n a ry p a rt Threephase systems 5

6 nglephase ower Defntons (t) m sn (wtθ) amps v(t) m sn(wtθv) volts w2πf oad: any R,,C combnaton w: angular frequency n rad/sec f: frequency n Hz nstantaneous power [ m θv ][ m θ ] p( t ) v( t ) ( t ) sn wt sn( wt ) 1 p( t ) 2 m m cos( v ) cos( wt v ) { θ θ 2 θ θ } Average ower (or REA OWER) T 1 1 p( t ) dt 2 m m cos rms rms cos T θ θ 0 Apparent ower rms rms ower Factor REA OWER pf AARENT OWER For ths crcut, the power factor s pf cos rms rms θ rms rms cosθ Threephase systems 6

7 ower Trangle θ cosθ sn θq Real ower cos θ cos θ watts Reactve ower Q sn θ sn θ vars C om plex o w er θ j Q cos θ j sn θ f θ θ θ and assum ng a reference θ 0 then θ θ therefore [ θ θ ] [ θ θ ] cos( ) jsn( ) cos( ) jsn( ) * v v T he m agntude s called A pparent ow er: volt am peres ( A ) Threephase systems 7

8 ower Consumpton by assve Elements mpedance: Z R jx Z θ Ω Z R R 0 cos0 R / R watts Q sn 0 0 vars o o 2 2 o Resstve oad A resstor absorbs Z jw jx X 90 o urely nductve oad cos(90 ) 0 watts o 2 Q sn(90 ) X / X var s An nductor absorbs Q o 2 Z 1 jwc urely Capactve oad jx X 90 cos(90 ) 0 watts C o C o 2 Q sn(90 ) X / X var o 2 s A capactor absorbs negatve Q. t supples Q. Threephase systems 8

9 Advantages of Threephase ystems Creaton of the threephase nducton motor Threephase nducton motor nglephase nducton motor tartng torque yes no needs auxlary startng crcutry teady state torque Constant Oscllatng causng vbraton Effcent transmsson of electrc power 3 tmes the power than a snglephase crcut by addng an extra cable v nglephase oad va vb vc a b c Threephase oad p v p va a vb b vc c avngs n magnetc core when constructng Transformers Generators Threephase systems 9

10 Threephase oltages va vb vc va(t) m sn wt vb(t) m sn (wt 2π/3) m sn (wt 120 ) vc(t) m sn (wt 4π/3) m sn (wt 240 ) or vb(t) m sn (wt 2π/3) m sn (wt 120 ) volts volts volts volts w 2 π f w: angular frequency n rad/sec f : frequency n Hertz c a Threephase systems 10 b

11 tar Connecton (Y) Yconnected oltage ource a c cn n an bn b ne to neutral voltages an, bn, cn. (phase voltages for Y connecton) same magntude: an bn cn ne to lne voltages ab, bc, ca same magntude: ab an bn 3 Threephase systems 11

12 Delta Connecton ( ) connected oltage ource c ca a ab bc b ne to lne voltages ab, bc, ca. (phase voltages for connecton) same magntude: hase currents ab, bc, ca. same magntude: ne currents a, b, c. same magntude: 3 Threephase systems 12

13 Yconnected oad a a a c cn n an c a bn b Zc Za n' b a Zb Balanced case: Za Zb Zc Z a b c 0 b a 120 c a 240 Threephase systems 13

14 connected oad a a a c cn n an c a bn b Zca Zbc b a Zab Threephase systems 14

15 Y Equvalence Za Zc n' Zb Zca Zbc Zab Balanced case: Za Zb Zc Zy Z 3Zy Zab Zbc Zca Z 3Zy Threephase systems 15

16 ower n Threephase Crcuts Threephase voltages and currents: ( θ ) sn( θ ) ( θ 120 ) sn( θ 120 ) ( θ 240 ) sn( θ 240 ) v sn wt wt v sn wt wt v sn wt wt The threephase nstantaneous power s: p( t) p v v v p a m v a m b m v b m v p c m v c m 3φ 3φ m m a a b b c c ( wt θv ) ( wt θ ) ( wt θv 120 ) ( wt θv 120 ) sn( wt θ 240 ) sn( wt θ 240 ) sn sn sn sn v Ths expresson can easly be reduced to: 3φ ( θ θ ) 3 cos 2 m m v nce the nstantaneous power does not change wth the tme, ts average value equals ts ntantaneous value: p 3φ 3φ 3φ 3 cosθ where: m m v 2 2 θ θ θ Threephase systems 16

17 Threephase ower n a Yconnecton 3 3φ 3 cosθ 3 cosθ 3 cosθ 3 n a connecton 3 3φ 3 cosθ 3 cosθ 3 cosθ 3 Regardless of the connecton (for balanced systems), the average power (real power) s : 3 cosθ watts 3φ mlarly, reactve power and apparent power expressons are: Q 3φ 3 snθ vars 3φ 3 A Threephase systems 17

18 er unt modellng ower lnes operate at klovolts (K) and klowatts (KW) or megawatts (MW) To represent a voltage as a percent of a reference value, we frst defne ths BAE AUE. Example: Base voltage: 120 K Crcut voltage ercent of value er unt value 108 K 90% K 100% K 105% K 50% 0.5 actual quantty per unt quantty quantty oltage_ p. u. ** The percent value and the per unt value help the analyzer vsualze how close the operatng condtons are to ther nomnal values. Threephase systems 18

19 Defnng s 4 quanttes are needed to model a network n per unt system: : voltage BAE : current BAE : power BAE Z: mpedance ZBAE pu pu actual pu actual Z pu actual Z Z actual Gven two s, the other two quanttes are easly determned. f voltage and pow er are know n: 100 K, 100 M A then, current and m pedance are: Z 100, 000, A. 100, 000 A nother w ay to express m pedance s: Z Z ( ) Real pow er and reactve pow er are: 100 M W Q 100 M A R 100, Ω Threephase systems 19

20 Three phase s n threephase systems t s common to have data for the threephase power and the lnetolne voltage. 3Φ 3 1Φ N 3 The current and mpedance for the three phase case are: 3Φ Φ Z 3 3Φ 3 ( ) 3Φ 2 n per unt, lne to neutral voltage lne to lne voltage N(pu) (pu) why? Wth Wth p.u. p.u. calculatons, threephase values values of of voltage, voltage, current current and and power power can can be be used used wthout wthout undue undue anxety anxety about about the the result result beng beng a factor factor of of 3 3 ncorrect!!!!!! Threephase systems 20

21 Example The followng data apply to a threephase case: 300 MA 100 K (threephase power) (lnetolne voltage) a b c Threephase load 270 MW 100 K pf0.8 Normally, we'd say: 3 cos θ 3 pf 3 pf 6 270x A. 3 (100x10 ) ( 0. 8) Usng the per unt method: p.u p.u. pf then pf p. u. ( 10. )( 08. ) nglephase equvalent: p.u. 1 p.u. Ths current s 12.5% hgher than ts value! To check: 1.125x (1.125) 300, x A Threephase systems 21

22 Transformers n per unt calculatons Wth an deal transformer j 2.5 ohms Hgh oltage Bases 1 5 KA / A Z / Ω 2400:120 5 KA ow oltage Bases 2 5KA / A Z 2 120/ Ω From the crcut: /a1/ n per unt: 11.0 p.u p.u The load n per unt s: Z(5 30 )/Z p.u. The current n the crcut s: (1.0 0 )/ ( ) p.u. The current n amperes s: rmary: x A. econdary: x 2 24 A. Threephase systems 22

23 One lne dagrams A one lne dagram s a smplfed representaton of a multphasephase crcut. TRANFORMER Transmsson lne TRANFORMER GENERATOR GENERATOR Transmsson lne Transmsson lne OAD Threephase systems 23

24 Nodal Analyss uppose the followng dagram represents the snglephase equvalent of a threephase system z13j2 p.u. z1j1 p.u. z3j2 p.u. z12j0.5 p.u. z23j0.5 p.u. 1 1 p.u. z210 p.u. 3 j1 p.u. Fndng Norton equvalents and representng mpedances as admttances: y13j0.5 p.u. 1 y12j2 p.u. 2 y23j2 p.u. 3 y1j1 p.u. y20.1 p.u. y3j0.5 p.u. 1 j1 p.u p.u. 1y1 1 y12(12) y13(13) 0 y12 (21) y2 2 y23(23) 3y13(31) y23(32) y3 3 n matrx form: y 1 y 12 y13 y12 y13 y12 y 12 y 2 y23 y23 y y y y y j35. j2 j j2 01. j4 j2 2 j0. 5 j2 j3 3 j solvng p. u. Threephase systems 24

25 General form of the nodal analyss The system of equatons s repeated here to fnd a general soluton technque: y 1 y 12 y13 y12 y13 y12 y 12 y 2 y23 y23 y y y y y or Y Y Y Y Y Y Y Y Y J1 2 J2 3 J n general: Y N j1 Y y j j y j 12,... N 12,... N; j 12,... N; j J (from current sources flowng nto the node) 12,... N Once the voltages are found, currents and powers are easly evaluated from the crcut. We have solved one of the phases of the threephase system (e.g. phase a ). Quanttes for the other two phases are shfted 120 and 240 degrees under balanced condtons. Actual quanttes can be found by multplyng the per unt values by ther correspondng s. Threephase systems 25