Chapter 10 Sinusoidal Steady-State Power Calculations
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1 Chapter 0 Snusodal Steady-State Power Calculatons n Chapter 9, we calculated the steady state oltages and currents n electrc crcuts dren by snusodal sources. We used phasor ethod to fnd the steady state oltages and currents. n ths chapter, we consder power n such crcuts. The technques we deelop are useful for analyzng any of the electrc deces we encounter daly, because snusodal sources are predonate eans of prodng electrc power n our hoes, school, and busnesses. Exaples are: Electrc Heater whch transfor electrc energy to theral energy Electrc Stoe and oen Toasters ron Electrc water heater And any others 0. nstantaneous Power Consder the followng crcut represented by a black box. t ( ( t t ( cos( t ( t cos( t The nstantaneous power assung passe sgn conenton ( Current n the drecton of oltage drop pt ( t ( t ( ( Watts f the current s n the drecton of oltage rse ( the nstantaneous power s: t ( ( t p( t ( t ( t
2 t ( ( t t ( cos( t ( t cos( t t ( cos( t ( t cos( t pt ( t ( t ( { cos( t }{ cos ( t } cos( t cos( t Snce coscos cos( cos( Therefore pt ( cos( cos( t Snce cos( cos cos sn sn cos( t cos( cos( t sn( sn( t pt ( cos( cos( cos( t sn( sn( t t ( ( t t ( cos( t ( t cos( t p( t cos( cos( cos( sn( sn( t t =0 6 o =0o You can see that that the frequency of the nstantaneous power s twce the frequency of the oltage or current
3 0. Aerage and eacte Power ecall the nstantaneous power p(t p( t cos( cos( cos( sn( sn( t t p( t P Pcos( t Qsn ( t where P cos( Aerage Power (eal Power Q sn( eacte Power Aerage Power P s soetes called eal power because t descrbes the power n a crcut that s transfored fro electrc to non electrc ( Exaple Heat. t s easy to see why P s called Aerage Power because T t t+ 0 0 T ptdt ( t+ 0 T { P Pcos( t Qsn(t } dt T t0 P Power for purely resste Crcuts p( t P Pcos( t Qsn ( t = P cos( Q sn( sn(0 0 cos(0 pt ( cos(t The nstantaneous power can neer be negate. Power can not be extracted fro a purely resste network. 3
4 Power for purely nducte Crcuts p( t P Pcos( t Qsn ( t = 90o P = 90 o cos( p( t sn( t The nstantaneous power p(t s contnuously exchanged between the crcut and the source drng the crcut. The aerage power s zero. When p(t s poste, energy s beng stored n the agnetc feld assocated wth the nducte eleent. cos(90 o 0 Q sn( sn(90 o When p(t s negate, energy s beng extracted fro the agnetc feld. The power assocated wth purely nducte crcuts s the reacte power Q. The denson of reacte power Q s the sae as the aerage power P. To dstngush the we use the unt A (olt Apere eacte for reacte power. Power for purely Capacte Crcuts p( t P Pcos( t Qsn ( t = 90o P = 90 o cos( pt ( sn( t The nstantaneous power p(t s contnuously exchanged between the crcut and the source drng the crcut. The aerage power s zero. When p(t s poste, energy s beng stored n the electrc feld assocated wth the capacte eleent. When p(t s negate, energy s beng extracted fro the electrc feld. The power assocated wth purely capacte crcuts s the reacte power Q (A. Q sn( cos( 9 0 o 0 sn( 9 0 o 4
5 The power factor ecall the nstantaneous power p(t pt ( cos( cos( c os( sn( sn( t t P aerage P aerage Q reacte power power power P Pcos( t Q sn(t The angle plays a role n the coputaton of both aerage and reacte power The angle s referred to as the power factor angle We now defne the followng : The power factor pf cos( The reacte factor rf sn( The power factor pf cos( Knowng the power factor pf does not tell you the power factor angle, because cos( cos( To copletely descrbe ths angle, we use the descrpte phrases laggng power factor and leadng power factor Laggng power factor ples that current lags oltage hence an nducte load Leadng power factor ples that current leads oltage hence a capacte load 5
6 0.3 The rs alue and Power Calculatons Assue that a snusodal oltage s appled to the ternals of a resstor as shown cos( t P Suppose we want to deterne the aerage power delered to the resstor t+ 0 T t+ 0 T cos( t p( t dt T t 0 T t 0 dt T t+ 0 t 0 T t+ 0 T Howeer snce rs cos ( t dt T t 0 P rs f the resstor carry snusodal current P rs cos ( t dt ecall the Aerage and eacte power P cos( Q sn( Whch can be wrtten as P cos( Q sn( Therefore the Aerage and eacte power can be wrtten n ters of the rs alue as P Q sn( rs rs cos( rs The rs alue s also referred to as the ecte alue Therefore, the aerage and reacte power can be wrtten n ters of the alue as: rs P Q sn( cos( 6
7 Exaple Coplex Power Preously, we found t conenent to ntroduce snusodal oltage and current n ters of the coplex nuber, the phasor. Defnton Let the coplex power be the coplex su of real power and reacte power S P jq were S P Q s the coplex power s the aerage power s the reacte power 7
8 Adantages of usng coplex power S P jq We can copute the aerage and reacte power fro the coplex power S P { S} Q { S} coplex power S prode a geoetrc nterpretaton S P jq S where S = P Q Q =tan tan P e j s called the apparent power cos( tan sn( S Q ( reacte power P ( aerage power cos( sn( tan tan( The geoetrc relatons for a rght trangle ean the four power trangle densons ( S, P, Q, can be deterned f any two of the four are known. power factor angle Exaple 0.4 8
9 0.5 Power Calculatons S P jq cos( j sn( cos( j sn( e j ( e j ( e j e j were s the conjugate of the current phasor Crcut Also S Alternate Fors for Coplex Power Crcut The coplex power was defned as S P jq Then coplex power was calculated to be S O S Howeer there seeral useful aratons as follows: Frst araton Z = + jx S ( Z Z Z ( + jx + jx P Q P Q X X X 9
10 Second araton Z = + jx S jx Z Z jx jx jx jx X Z X j X X P Q P X X X X Q X f Z = (pure resste X= 0 X X f Z = X (pure reacte = 0 P 0 X X P X X Q X X Q 0 Exaple 0.5 Lne Load rs because the oltage s gen n ters of rs. 0
11 P 975 W Q 650 ar Another soluton The load aerage power s the power absorbed by the load resstor 39 ecall the aerage Power for purely resste Crcuts P where and Are the rs oltage across the resstor and the rs current through the resstor P nductor Fro Power for purely resste Crcuts L P P j 6 L 39 j 6 P (95 (5 o e j 975 W 95e j o P 975 W Q 650 ar 95 Q O P ( Q ( nductor nductor L5 nductor (39(5 (39(5 j 6 39 j 6 L j j W o 3.8 e j 30 Q (30 (5 650 A O Q X 650 ar nductor o 93 30e j
12 Lne Lne P Q X O usng coplex power S Lne Lne Lne j 4 (50 ( j4 (39 j6 O Lne 50 L Lne o S Lne Lne o o o 5 j00 A Lne Load S S + S Absorb Lne Load Fro part (c Fro part (b ( 5975 j ( j 750 A ( 5j00 + ( 975j650 S Supply = S Absorb ( 000 j 750 A O S Supply 50 ( 50 0 o L 000j 750 A 0o o
13 Exaple 0.6 Calculatng Power n Parallel Loads pf cos( 3
14 S 8000j 6000 A S 000 j 6000 A S 0000 j0000 A The apparent power whch ust be suppled to these loads s S 0000 j0000 A.36 ka S C? Wll cancel ths As we can see fro the power trangle We can correct the power factor to f we place a capactor n parallel wth the exstng load ecall that X C 4
15 The addton of the capactor has reduced the lne loss fro 400 W to 30 W Exaple 0.7 S P = 690 W and Q 3380 A P + j Q Another soluton P = O ( j P = P = ( 690 W = ( ( 6 ( W 5
16 S P = 690 W and Q 3380 A P + j Q Another soluton P = O ( j X Slarly Q = X X X j O X ( j P = P = ( 690 W Q = ( Q = X = ( ( 6 ( W = ( ( 6 ( W 3380 W 0.5 Maxu Power Transfer ecall What s the alue of Z L that wll absorb axu power = 0 6
17 Slarly =0 (snce X L = X th 7
Revision: December 13, E Main Suite D Pullman, WA (509) Voice and Fax
.9.1: AC power analyss Reson: Deceber 13, 010 15 E Man Sute D Pullan, WA 99163 (509 334 6306 Voce and Fax Oerew n chapter.9.0, we ntroduced soe basc quanttes relate to delery of power usng snusodal sgnals.
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