Module B3 3.1 Sinusoidal steady-state analysis (single-phase), a review 3.2 Three-phase analysis. Kirtley

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1 Module B3 3.1 Snusodal steady-state analyss (sngle-phase), a reew 3. hree-phase analyss Krtley Chapter : AC oltage, Current and Power.1 Sources and Power. Resstors, Inductors, and Capactors Chapter 4: Polyphase systems 4.1 hree-phase systems 4. Lne-Lne oltages 1

2 Pop Quz!!! Let R=Ω, C=0.001farad, L=0.01henry, =400. ( sn t 377 rad / sec R jx L jx C 3 ( L ( You hae 3 mns to compute the current (. If you don t know how, wrte down what you thnk are questons that, f answered, would enable you to do t.

3 Example 1 Let R=Ω, C=0.001farad, L=0.01henry, =400. ( sn t hen =R+jωL-j/ωC he current wll be t () sn L t R jx L jx C 3 ( j j(3.77 L (.655) Compute the current ( j /( ) j () t 400sn t Let s look nto j ths a bt 3

4 Basc concepts of electrc power Wth excepton of a few Hgh oltage DC crcuts, the entre electrc grd operates to deler alternatng current (AC). An AC oltage appears below. he mum alue of the waeform s 400 olts. he waeform completes one cycle n sec. hs s the perod of the waeform and s 1/f. So, f=1/0.0167=60hz or 60 cycles/sec. Snce 1 cycle completes π radans, ω=(π rad/cyc)*60cyc/sec=377rad/sec. We can therefore express ths oltage as ( snt 400sn(377 4

5 KL and Ohm s Law he below s a crcut model of a oltage source connected to a resstance. ( ( sn t We apply Krchoff s oltage Law (KL): the sum of oltages around the loop must be 0: ( L L ( ( 0 ( R R L ( We may apply Ohm s Law to obtan the current: L( ( R R sn t 5

6 oltage and current for resste cct Compare oltage & current. Assume R= ohms. L 400 ( snt ( snt sn 377t R Current & oltage cross 0 smultaneously 6

7 Capactance he below s a crcut model of a oltage source connected to a capactance. ( ( sn t Agan, from KL: ( L L ( ( 0 ( C L ( he current s gen by: ( C C 1/ C dl ( C dt sn t sn t cost cost sn( t / ) 7

8 oltage and current for capacte cct Compare oltage & current. Assume C=0.001farad. ( C sn t snt L ( sn t 150.8sn t Current crosses 0 before oltage. We say current leads oltage. If we multply the tme axs by ω=377 rad/sec, we would see that current leads oltage by π/ radans. 8

9 Inductance he below s a crcut model of a oltage source connected to an nductor. ( ( sn t Agan, from KL: ( L L ( ( 0 ( 1 ( L L L L ( he current s gen by: L( L sn t 3 cost L sn t 9

10 oltage and current for nducte cct Compare oltage & current. Assume L=.01 henry. snt ( sn t L ( L 400 sn t 106.1sn t Current crosses 0 after oltage. We say current lags oltage. If we multply the tme axs by ω=377 rad/sec, we would see that current lags oltage by π/ radans. 10

11 L ( RESISIE: CAPACIIE: INDUCIE: snt Impedance L( ( R R sn t L( ( sn t 1/ C C L( ( sn t L L In the three expressons below, we use Ohm s law to express current n terms of mpedance, then equate that expresson to the one we found n the 5 preous sldes. It s clear that R =R. Is C =1/ωC? Is L =ωl? No, because we also need to take care of the phase shfts. 11

12 Impedance o take care of phase shfts, let s utlze the magnary unt j whch has the followng propertes: It s the square root of -1: j 1 From the Euler relaton: e j cos jsn Wth ϕ=π/, we obtan: e j / cos / j sn / j herefore, j has magntude of 1 and angle π/. herefore j sn t sn t sn t sn t j hen, wth C =1/jωC, L =jωl, we get correct current expressons. 1

13 Impedance RESISIE: CAPACIIE: INDUCIE: L( ( R R sn t L( sn t ( jc sn t C sn t 1/ jc C L( sn t jl L sn t j ( L L sn t, called mpedance, s a generalzaton of resstance that accounts for the effects of capactance & nductance. 13

14 Impedance In a purely resste crcut, = R =R. In a purely capacte crcut, = C =1/jωC. In a purely nducte crcut, = L =jωl. We also defne: Capacte reactance: X C =1/ωC Inducte reactance: X L =ωl In a purely resste crcut, = R =R. In a purely capacte crcut, = C =1/jωC=X C /j. But recall j= -1, then j =-1. So (X C /j)*(j/j)=-jx C. So = C =-jx C. In a purely nducte crcut, = L =jωl=jx L. 14

15 Impedance It s possble to hae an mpedance that s a combnaton of resstance, capactance, and nductance. ( sn t R jx R jx L jx C 3 L ( hen jx L jx C R j X L ( he mpedance,, can be thought of as a ector on the complex plane. X C R 15

16 Impedance Imagnary Inducte X R Capacte Real R jx L jx C R j X X R jx L C 16

17 Example 1 Let R=Ω, C=0.001farad, L=0.01henry, =400. ( sn t hen =R+jωL-j/ωC he current wll be t () sn L t R jx L jx C 3 ( j j(3.77 L (.655) Compute the current ( j /( ) j () t 400sn t But how to j ealuate ths? 17

18 t () Example 1 400sn t j You can dde each numercal pont n the snusodal numerator by the complex denomnator, then take the magntude and angle of each resultng complex number. But obsere that only the magntude changes as t ncreases. hs s because the phasor angle of the numerator (0 ) and the ector angle of the denomnator both reman constant wth t. hat s: t () 400sn t 4000snt j But what s θ? 18

19 Impedance j z Imagnary So: And from the preous slde: 4000snt 400 ( (0 ) snt tan snt snt Real tan radans

20 Impedance So ( sn t Recallng that 8.9 = radans, we wrte the aboe as: ( sn t We can plot ths: Current crosses 0 after oltage. We say current lags oltage. If we multply the tme axs by ω=377 rad/sec, we would see that current lags oltage by radans. 0

21 Let s reew what we just dd: t () sn Phasors L() t t 400sn t j Ohm s Law Express L ( Express numercally Express n polar notaton Separate magntude & angle calculatons Express numercally 4000snt (0 ) snt 8.9snt.849 Do the arthmetc ( ) snt sn t Notce that nothng eer happened to the ωt part. Nor dd the snusodal functon change. hs means that the frequency of the current s the same as the frequency of the oltage. If ths s the case, then why do we need to carry the snωt through the calculatons? Why not just drop t? hs s what we do when we use phasor notaton, as n the next slde. Conert from polar notaton to trgonometrc notaton 1

22 Phasors Defne: Phasor for source oltage: ˆ Phasor for load oltage: ˆ Phasor for current: Î Also recall the mpedance: hen ˆ L L ˆ L L I II S S L L I I L L S here s an mplct snωt that we are not wrtng. hs snωt ges rotaton to the ectors whch make them phasors. here s no mplct snωt mpedance s a ector, not a phasor! Imagnary Rotaton s n CCW drecton Î hese do rotate. ˆ L Real (does not rotate)

23 Instantaneous Power Instantaneous power s gen by p( ( ( Let ( cos t I cos t ( hen p( ( ( t cos I cos t I cost cost 1 t t cost t Apply trg dentty: cos a cosb cosa b cosa b p( I I cos I cos cost [cos cos(t )] I p( cos Gatherng terms n the argument of the second cos functon: cos( t ) ( ) 3

24 p Rearrange and dstrbute: hen: where: Instantaneous Power I ( cos cos( t ) ( ) cos a b cos a cosb sn asn Apply trg dentty to last term: b p I ( p I cos cos( t )cos( ) sn ( t )sn( ) I I cos cos( )cos( t ) sn( )sn ( t ) ( P Pcos( t ) Qsn ( t ) ( p P I cos, Q sn( ) I Real power Acte power Watts Imagnary power Reacte power ARs 4

25 P Q Example Consder example 1, where =400, θ =0, I cos cos(0.5047) I sn( ) sn(0.5047) ( sn( t ) and we computed that I =175.06, θ = rad, ( I hen: p( P P cos ( t ) Qsn ( t ) cos ( sn (377 sn t ERM1 ERM ERM3 watts ars 5

26 Example p( P P cos ( t ) Qsn ( t ) cos ( sn (377 ERM1 ERM ERM3 he aerage of a perodc sgnal x( s gen by: 1 X ag x( dt Power delered s the ag of p(; power s delered only f the ag of p( s not 0. he aerage alue of term and term3 s 0. hese terms are not responsble for power delery. Aerage alue of the composte [the composte s nstntans power, p(] s term1, P. Interestng sde note: nstntns power ares at twce the nomnal frequency. 6

27 Effecte or RMS alue We want to obtan the constant (or DC) alue of a current whch delers the same aerage power to a resstor. ( AC crcut R R ( ( eff DC crcut R R I eff eff he power delered s: 1 P ( Rdt Equate he power delered s: P I R 1 1 ( Rdt I eff R Ieff eff ( dt 7

28 Effecte or RMS alue We want to obtan the constant (or DC) alue of a oltage whch delers the same aerage power to a resstor. ( AC crcut R R ( ( eff DC crcut R R I eff eff he power delered s: 1 P ( / Rdt Equate he power delered s: P eff / R 1 1 ( / Rdt eff / R eff ( dt 8

29 Effecte or RMS alue he root-mean-square alue of any perodc sgnal x( s: 1 X eff x ( dt Let x( be (: sn( t ) eff ( 1 1 ( dt sn Apply trg dentty: sn a 1 cos(a) eff 1 cos(( t )) dt 1 ( t ) dt t 0 Current may be smlarly treated, resultng n I eff I 9

30 Power and Phasor notaton We preously dered power expressons for P and Q n terms of and I. We may now express these n terms of rms (effecte) alues: P Q I I cos eff eff cos I cos sn( ) eff I I eff sn I sn hs smplfes our power expressons. Commercal oltages and currents are always specfed as rms alues. Phasor magntudes may be rms or mum. Unless t s otherwse specfed, we assume phasor magntudes are rms, and we wll drop the subscrpt eff. hus, P I cos ; Q I sn eff eff eff eff 30

31 P I cos Power factor Power factor angle, defned θ, s the angle by whch oltage across an element leads current through the element,.e., θ=θ -θ. It s also the angle of the load,.e., = θ. One obseres from the P, Q expressons that t s also the argument of the trg functons of P and Q. herefore: I cos; Q I sn I sn he power factor s gen as pf=cosθ. It s always poste, therefore we usually ndcate pf wth ether the word leadng (for capacte load) or laggng (for nducte load). he followng table summarzes mportant relatons. Load, Im{} θ (θ =0) θ (θ =0) pf Q sgn Q Load R only ars R and L to 1, laggng + Absorbng ars R and C to 1, leadng - Supplyng ars 31

32 3 Power factor Consder a load at the end of a transmsson lne. Assume the load s hghly nducte. hs means current s laggng, θ <0, and θ>0 (wth θ =0). ( () t s rans Lne mpedance t () Load P Q I I cos sn Assume s constant, and we must supply P (ths s the real power of the load). Wth and P constant, from P=Icosθ, Icosθ=I pf must be constant. What do we want pf to be? 3

33 3 Power factor Assume P=33MW, =39.84k. hs means that P I cos I cos P / Here are a few possbltes for Icosθ =I pf. I (amperes) Pf (leadng) E6/39.84E For a gen P, we need hgher pf to obtan lower current. Lower current s desred n order to mnmze I R (real power) losses n the transmsson lne. ( () t s rans Lne mpedance ( Load 33

34 3 Power factor correcton (PFC) So what do you do f your load has a power factor of 0.85 but your proder requres that you hae 0.95? Install a PFC dece at the load s pont of nterconnecton wth the grd. ( () t s rans Lne mpedance Power factor correcton dece Load ( he least expense PFC dece s a swtched capactor, or seeral swtched capactors. hese are statc deces, howeer. Dynamc deces are also aalable but are more expense (and you get dynamc responseness as well). 34

35 What s reacte power? 1. It s gen by: Q I sn. It occurs as the coeffcent of the double frequency sn term n p(: p( P Pcos( t ) Qsn ( t ) 3. It s the mum alue of the rate of energy flowng alternately nto the reacte component of the load (away from the source) and away from the reacte component of the load (toward the source). It s only present when there s a stored energy dece (C or L) n the crcut. 4. Its unts are olt-amperes-reacte or ars. Dmensonally, ars are the same as watts and olt-amperes. 5. hree attrbutes of reacte power: a. It does no useful work (ts aerage alue s 0). b. Its presence ncreases current magntude for a gen P and. c. Derates of Q flow equatons (not shown n these sldes) ndcate that Q s closely related to oltage magntudes. 35

36 Complex power Complex power s defned as: where: S * ˆ ˆ I ˆ Iˆ P I he astersk next to the I phasor ndcates conjugaton: Iˆ * I jq 36

37 Power trangle S θ Q P S P jq S P Q 37

38 Example 3 he below represents one of the phases of a generator supplyng a load. he box to the rght represents the load. he current flowng nto the load s A at pf leadng, wth olts at ts termnals. What complex power s delered to the load from ths phase alone? Soluton: θ=-cos -1 (.866)=-30. θ =-θ=30. S Iˆ ˆ * * ( ) ( ) (cos30 j sn 30) hus, P= MW, Q= MARs j78340 We also defne the apparent power S, the magntude of S: S P jq S P Q