ECE 320 Energy Conversion and Power Electronics Dr. Tim Hogan. Chapter 1: Introduction and Three Phase Power

Transcription

1 ECE 3 Energy Conerson and Power Electroncs Dr. Tm Hogan Chapter : ntroducton and Three Phase Power. eew of Basc Crcut Analyss Defntons: Node - Electrcal juncton between two or more deces. Loop - Closed path formed by tracng through an ordered sequence of nodes wthout passng through any node more than once. Element Constrants: Ohm s Law Capactor Equaton nductor Equaton d C dt d L dt Connecton Constrants: Krchhoff s Current Law - The algebrac sum of currents enterng a node s zero at eery nstant n tme. k Node (. Krchhoff s oltage Law - The algebrac Sum of all oltages around a loop s zero at eery nstant n tme. (. Loop k Passe Sgn Conenton: Wheneer the reference drecton of current nto a two termnal dece s n the drecton of the reference oltage drop across the dece, then the power absorbed (or dsspated s poste. _ Fgure. Crcut element and passe sgn conenton. p ( t ( t ( t (.3 -

2 When the aboe conenton s used, p(t > for absorbed power, and p(t < for delered power. Tme aryng Sgnals Although a number of exceptons can be found throughout the world, the predomnance of electrc power follows 6Hz or Hz frequences. North Amerca, part of Japan, and shps at sea use 6Hz whle most of the rest of the world uses Hz. The hstorcal reasons for these two frequences stem from the dfferences n lghtng (flaments n acuum or flaments n a gas atmosphere. The lower frequences caused an annoyng flcker for lghts hang flaments n a gas atmosphere, and thus a hgher frequency was adopted n part of the world that ntally used such lghtng. Whle not strctly adhered to wthn your textbook and these notes, an attempt to use the followng conentons has been made. Scalar tme aryng sgnals (examples: sn ωt ( max ( ( t cos( ωt max ( 8 sn( 377 t (k t cos π t ( ( max (A Spatal ectors (bold or arrow oerhead or lne oerhead: F or F or B f necessary, unt ectors wll be used, (for example: These unt ectors should not be confused wth phasors below B B xˆ B yˆ B zˆ x y z Phasors (phasor representaton of a tme aryng sgnal: A phasor can be represented as a complex number wth real and magnary components such that a phasor of magntude (or length that s at a phase angle of θ wth respect to ˆ jθ x jy e cos θ jsn θ the x-axs (the axs can be wrtten as ( ( θ where Euler s formula e j cos( θ j sn( θ was used for the last representaton. Note that s the magntude and θ the phase of the phasor, and y θ arctan. The phasor can be shown graphcally n Fgure. x x y and -

3 magnary y ωt θ x Fgure. Phasor of magntude and phase θ. For a snusodal functon ( t ( ω t θ cos, the phasor representaton s ˆ jθ e /θ f the phasor s rotatng counterclockwse about the orgn at a rate of ω radans per j t second, then we multply the phasor by e ω j such that ω t j θ j ω t j ( ω t ˆ e e e Ae θ and ω ( ω θ usng Euler s formula e ˆ j t j t e cos( ωt θ j sn( ωt θ. Then we see that j t t cos ω t θ s the real part of ˆ e ω, or ( ( ω ( t { e ˆ j t} cos( ωt θ e. n the aboe descrpton of the phasor, the peak alue s used, howeer we wll use the MS alue of as descrbed n (.6 nstead of the peak alue. Ths smplfes many of the calculatons, partcularly those assocated wth power as shown below. Phasors wll be represented wth a hat (or caret aboe the arable. mpedance s understood to be a complex quantty n general, and the hat (or caret ^ s left off the mpedance arable Z jx where s the resstance, and X s the reactance component respectely. For a crcut element such as the one shown n Fgure representng a load n the crcut wth (t as the nstantaneous alue of current through the load and (t s the nstantaneous alue of the oltage across the load. n quas-steady state condtons, the current and oltage are both snusodal, wth correspondng π ampltudes of max, max, and ntal phases, φ and φ, and the same frequency ω πf T ( t max cos( ω t φ (.4 ( t max cos( ω t φ (. The root-mean-squared (MS alue of the oltage and current are then: max T [ max cos( t ] dt T ω φ (.6 o - 3

4 max T [ max cos( t ] dt T ω φ (.7 o Phasor representatons for the aboe sgnals use the MS alues as The nstantaneous power s the product of oltage tmes current or: p ˆ /φ and ˆ /φ. ( t ( t ( t max cos( ω t φ max cos( ωt φ maxmax cos( ωt φ cos( ωt φ cos( ω t φ cos( ωt φ [ cos( φ φ cos( ωt φ φ ] (.8 The aerage alue s found by ntegratng oer a perod of tme and then ddng the result by that same tme nteral. The frst term n (.8 s ndependent of tme, whle the second term ares from - to symmetrcally about zero. Because t s symmetrc about zero, ntegraton oer an nteger number of cycles (or perods ges a alue of zero for the second term n (.8, and the aerage power whch we defne as the real power wth unts of watts s: P cos (.9 Ths shows the power s not only proportonal to the MS alues of oltage and current, but also proportonal to cos. The cosne of ths angle s defned as the dsplacement factor, DF. n more general terms for perodc, but not necessarly snusodal sgnals, the power factor s defned as: P pf (. For snusodal sgnals, the power factor equals the dsplacement factor, or pf cos (. For comparson, the oltage, current, and power for arous angles between oltage and current are shown below: oltage (, Current (A, Power For leadng power factors: # of Perods ( t cos( ωt ( t cos( ωt P cos magnary cos - 4

5 oltage (, Current (A, Power # of Perods ( t cos( ωt ( t cos( t 3º ω P cos magnary cos oltage (, Current (A, Power # of Perods ( t cos( ωt ( t cos( t 6º ω P cos magnary cos. oltage (, Current (A, Power # of Perods ( t cos( ωt ( t cos( t 9º ω P cos magnary cos oltage (, Current (A, Power # of Perods ( t cos( ωt ( t cos( t º ω P cos magnary cos. -

6 oltage (, Current (A, Power # of Perods ( t cos( ωt ( t cos( t º ω P cos magnary cos oltage (, Current (A, Power # of Perods ( t cos( ωt ( t cos( t 8º ω P cos magnary cos oltage (, Current (A, Power For laggng power factors: # of Perods ( t cos( ωt ( t cos( ωt P cos magnary cos - 6

7 oltage (, Current (A, Power oltage (, Current (A, Power oltage (, Current (A, Power oltage (, Current (A, Power # of Perods # of Perods # of Perods # of Perods ( t cos( ωt ( t cos( t 3º ω P cos cos magnary ( t cos( ωt ( t cos( t 6º ω P cos cos. magnary ( t cos( ωt ( t cos( t 9º ω P cos cos magnary ( t cos( ωt ( t cos( t º ω P cos cos. magnary - 7

8 oltage (, Current (A, Power # of Perods ( t cos( ωt ( t cos( t º ω P cos cos magnary oltage (, Current (A, Power # of Perods ( t cos( ωt ( t cos( t 8º ω P cos magnary cos We call the power factor leadng or laggng, dependng on whether the current of the load leads or lags the oltage across t. We know that current does not change nstantly through an nductor, so t s clear that for a resste - nducte (L load, the power factor s laggng. Lkewse the oltage can not change nstantly across a capactor, therefore for a resste capacte (C load, the power factor s leadng (the current changes before the oltage can. Also, for a purely nducte or capacte load the power factor s, whle for a purely resste load t s. The product of the MS alues of oltage and current at a load s the apparent power, S hang unts of olt-amperes (A: S (. The reacte power s Q wth unts of olt-amperes reacte (A reacte, or Ar: Q sn (.3 The reacte power represents the energy oscllatng n and out of an nductor or capactor. Snce the energy oscllaton n an nductor s 8º out of phase wth the energy oscllatng n an capactor, the reacte power of these two hae opposte sgns wth the conenton that t s poste for the nductor and negate for the a capactor. Usng phasors, the complex apparent power, Ŝ s: or S ˆ ˆˆ* /φ /-φ (.4-8

9 S ˆ P jq (. As an example, consder the followng oltage and current for a gen load: π ( t sn 377t ( (.6 6 π ( t sn 377t (A (.7 4 then S 6, whle the power factor s cos π π pf leadng. Also, the complex apparent power s: S ˆ ˆˆ* /π/6 /-π/4 6/- π/ 79.6 j.3 (Ar (.8 t should also be noted that when the angles are represented n radan, care must be taken to assure your calculator s n radans mode. Often we represent the argument of the sne or cosne term n mxed unts. For example we mght wrte cos(377t 3º. The frst term (377t has unts of radans, whle the second (3º has unts of degrees, and one of these must be conerted before makng calculatons wth your calculator or computer ether n radans mode or n degrees mode. As a phasor the complex apparent power can be shown on a complex plane for the cases of leadng and laggng power factors as shown n Fgure 3. magnary magnary magnary magnary P S Q S P Q (a Leadng power factor (b Laggng power factor Fgure 3. Phasor of magntude and phase θ. - 9

10 ecall that the leadng power factor corresponds to a resste-capacte load. We see from Fgure 3(a, ths also corresponds to a negate alue for Q. Smlarly, laggng power factor has the current laggng the oltage correspondng to an nducte-resste load, and Fgure 3(b shows ths corresponds to a poste alue for Q. To summarze, we can use the followng tables: Equaton P cos φ φ Power ( pf S pf Q sn φ φ eacte Power ( Apparent Power S S pf pf S Sˆ ˆˆ* /φ /-φ S ˆ P jq S P Q (for laggng (for leadng Unts Watts olt-amperes-eacte olt-amperes Type of Load eacte Power Power Factor nducte Q > laggng Capacte Q < leadng esste Q From the nterdependence of the four quanttes, S, P, Q, pf, f we know any two of these quanttes, the other two can be determned. For example, f S (ka, and pf.8 leadng, then: P S pf 8 (kw Q S sn Q pf ( φ φ sn[ arccos(.8 ] S sn t s mportant to notce that Q <, such that ( φ 6 (ka, or φ then sn s a negate quantty. Ths can be seen when t s understood that there are two possble answers for the arccos(.8, that s cos(36.87º.8, and cos(-36.87º.8 so to obtan a Q < we use (φ φ 36.87º. Generally, n systems that contan more than one load (or source, the real and reacte power can be found by addng nddual contrbutons, but ths s not the case wth the apparent power. That s P total Qtotal S total P Q (.9 S -

11 For the aboe example, f the load oltage s L (, then the load current would be L S/ L [ 3 (A]/[ 3 (] (A. f we use the load oltage as the reference, then: ˆ /º ( ˆ /φ /36.87º (A S ˆ ˆˆ* [ /º][ / 36.87º] P jq 8 3 j6 3 (Ar. Three Phase Balanced Systems Three-phase systems offer sgnfcant adantages oer sngle phase systems: for the same power and oltage there s less copper n the wndngs, and the total power absorbed remans constant rather than oscllatng about an aerage alue. For a three phase system consstng of three current sources hang the same ampltude and frequency, but wth phases dfferng by º as: ( t sn( ωt φ π ( t sn ωt φ 3 π 3( t sn ωt φ 3 (. f these are connected as shown n Fgure 4, then at node n or n, the current adds to zero, and the neutral lne n-n (dashed s not needed. π 3 π 3 ( t ( t ( t sn( ωt φ sn ωt φ sn ωt φ 3 n' n 3 Fgure 4. Balanced three phase Y-connected system wth zero neutral current. f nstead we had three oltage sources Y-connected as n Fgure wth the followng alues -

12 ( t sn( ωt φ π ( t sn ωt φ (. 3 π 3( t sn ωt φ 3 then, the current through each of the three loads (assumng the loads are equal, would hae equal magntudes, but each current would hae a phase that s shfted by an equal amount wth respect to the oltages, (t, (t, 3 (t. a 3 n' b n c Fgure. Balanced oltage fed three phase Y-connected system wth zero neutral current. Wth equal mpedances for the loads, then a b ( t sn( ωt φ θ ( t Z π sn ωt φ θ Z 3 π c ( t sn ωt φ θ Z 3 (. and agan the currents sum to zero at nodes n or n, and for ths balanced three phase system, the neutral wre (dashed s not requred. n comparson to a sngle phase system, where two wres are requred per phase, the three phase system delers three tmes the power, and requres only three transmsson wres total. Ths s a sgnfcant adantage consderng the hundreds of mles of wre needed for power transmsson. Y and Δ Connectons The loads n the preous two fgures, as well as n Fgure 6 are connected n a Y or star confguraton. f the load of Fgure 6 s for a balanced Y system, then the oltages between each phase and the neutral are: ˆ n /φ, ˆ n /φ π/3, and ˆ 3 n /φ π/3. Krchhoff s oltage law (KL states that the sum of oltages around a closed loop equals zero. Ths s also the case here howeer the oltages are complex numbers or phasors, and as such must be -

13 added as ectors. The phase φ can be any alue, but the relate poston of the phase to neutral phasors must be º wth respect to each other as shown n Fgure 7. 3 n n 3n n Fgure 6. Y-connected loads wth oltages relate to neutral dentfed. By KL: ˆ ˆ ˆ, or ˆ ˆ ˆ as shown n Fgure 7. n n n n 3 - n 3n n - 3n n - n 3 Fgure 7. oltage phasors of the Y-connected loads shown n Fgure 6. We could also use the phasor representaton ˆ n /φ, ˆ n /φ π/3, and ˆ 3 n /φ π/3 to determne the lne-to-lne oltages as ˆ ˆ ˆ /φ π/6 (.3 n n 3 Ths shows the MS alue of the lne-to-lne oltage, l-l, at a Y load s 3 tmes the lne-toneutral or phase oltage, ln. n the Y connecton, the phase current s equal to the lne current, and the power suppled to the system s three tmes the power suppled to each phase, snce the oltage and current ampltudes and phase dfferences between them are the same n all three phases. f the pf cos φ φ, then the total power to the system s: power factor n one phase s ( - 3

14 Sˆ 3φ P 3φ 3ˆ ˆ * n n 3 jq l l l 3φ cos j 3 sn l l l (.4 Smlarly, for a connecton of the loads n the Δ confguraton (as n Fgure 8, the phase oltage s equal to the lne oltage howeer; the phase currents are not equal to the lne currents for the Δ confguraton. f the phase currents are ˆ /φ, ˆ /φ π/3, and 3 ˆ 3 /φ π/3 then usng Krchhoff s current law (KCL the current of lne, as shown n Fgure 8 s: ˆ ˆ ˆ 3 /φ - π/6 (. 3 Thus for the Δ confguraton, the lne current s 3 tmes the Δ current Fgure 8. Δ connected load, lne, and phase currents. To calculate the power n the three-phase Δ connected load: Sˆ 3φ P 3φ 3ˆ ˆ 3 jq * l l l 3φ cos j 3 sn l l l (.6 whch s the same alue as for the Y connected load. For a balanced system, the loads of the three phases are equal. Also, a Δ confgured load, can be replaced wth a Y confgured load (and sa ersa f: Z Y Z Δ 3 (.7-4

15 Under these condtons, the two loads are ndstngushable by the power transmsson lnes. You mght recall the Δ to Y transformaton for resstor crcuts can be remembered by oerlayng the Δ and Y confguratons such as n Fgure 9. A B 3 C C A B Fgure 9. Δ to Y or Y to Δ resstor network transformaton. A B C B C A 3 3 (.8 Equaton (.8 ges a smlar result to that of (.7 when A B B C and 3..3 Calculatons n Three-Phase Systems Calculatons of quanttes lke currents, oltages, and power n three-phase systems can be smplfed by the followng procedure:. transform the Δ crcuts to Y,. connect a neutral conductor, 3. sole one of the three -phase systems 4. conert the results back to the Δ systems.3. Example For the 3-phase system n Fgure calculate the lne-to-lne oltage, real power, and power factor at the load. To sole ths by the procedure outlned aboe, frst consder only one phase as shown n Fgure. -

16 j (Ω ( 7 j (Ω n' n 3 Fgure. Three phase system wth Y connected load, and lne mpedance. j (Ω ( 7 j (Ω n' Fgure. One phase of the three phase system shown n Fgure. n For the one-phase n Fgure, ˆ 3. /-4.6º (A j ( 7 j ˆ ˆ n Z L 3. /-4.6º (7j /-.º ( S *, ˆ ˆ L φ L (/-.º(3./4.6º 48.3/3.º.87 3 j P.87 (kw,.848 (kar L,φ Q L,φ pf cos(-.º - (-4.6º.84 laggng For the three-phase system of Fgure the load oltage (lne-to-lne, the real, and reacte power are: 3 L, l l 94 ( P L,3φ 3.6 (kw Q L,3φ.44 (kar - 6

17 .3. Example For the Y to Δ three-phase system n Fgure, calculate the power factor and the real power at the load, as well as the phase oltage and current. The source oltage s 4 ( lne-to-lne. j (Ω 8 j6 (Ω n' 3 Fgure. Δ connected load. Frst conert the load to an equalent Y connected load, then work wth one phase of the system. 4 The lne to neutral oltage of the source s ln 3(. 3 j (Ω 3 ( 6 j (Ω n' n Fgure 3. Equalent Y connected load. j (Ω L 3 ( 6 j (Ω n' n Fgure 4. One phase of the Y connected load. - 7

18 3 ˆ L /-6.6º (A j ( 6 j ( ˆ ˆ j L The power factor at the load s: L /-8.º ( pf cos( 8.º 6.6º. 948 cos laggng Conertng back to a Δ connected load ges: φ L (A 3 l l ( At the load the power s: P,3φ 3 pf (kw L l l L.3.3 Example Two three-phase loads are connected as shown n. Load draws from the system P L (kw at.8 pf laggng, whle the total load s S T (ka at.9 pf laggng. What s the power factor of load? Power System Load Load Fgure. Two three-phase loads connected to the same power source. - 8

19 For the total load we can add the real and reacte power for each of the two loads (we can not add the apparent power. P Q S T T T P S L Q L L P S L Q L L From the nformaton we hae for the total load we can wrte the followng: PT ST pft 9 (kw Q sn cos.9 3. T S T [ ( ] (kar The reacte power, Q T, s poste snce the power factor s laggng. For the load L, P L (kw, pf.8 laggng, thus: SL Q.8 S 3 P L L L 6 (ka 37 (ka Agan, Q L s poste snce the power factor s laggng. Ths leads to: P Q PT P Q Q L L L T L 4 (kw -6.7 (kar and pf P 4 4 L L S L leadng. Chapter Notes A snusodal sgnal can be descrbed unquely by:. Tme dependent form as for example: ( t sn( π ft φ (. by a tme dependent graph of the sgnal 3. as a phasor along wth the assocated frequency of the phasor one of these descrptons s enough to produce the other two. t s the phase dfference that s mportant n power calculatons, not the phase. The phase s arbtrary dependng on the defned tme (t. We need the phase to sole crcut problems after we take one quantty (some oltage or current as a reference. For that reference quantty we assgn an arbtrary phase (often zero. n both three-phase and one-phase systems the total real power s the sum of the real power from the nddual loads. Lkewse the total reacte power s the sum of the reacte power of the nddual loads. Ths s not the case for the apparent power or the power factor. - 9

20 Of the four quanttes: real power, reacte power, apparent power, and power factor, any two descrbe a load adequately. The other two quanttes can be calculated from the two gen. To calculate real, reacte, and apparent power when usng equatons (.9, (., and (.3 we must use absolute alues, not complex alues for the currents and oltages. To calculate the complex power usng equaton (.4 we do use complex currents and oltages and fnd drectly both the real and reacte power (as the real and magnary components respectely. When solng a crcut to calculate currents and oltages, use complex mpedances, currents and oltages. -