Power Engineering Review ENEL 519.44

7/27/2019 Power Engineering Review ENEL 519.44

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l B. Hos eh ar t i sw it h t he Dep t. E le ct ri ca l & Computer Eng . . Univers it y o fCa]gtUy,C alg ar y, A B, Ca na da , T2K IN4 , r ose ha rt esuc al ga ry. ca , I t i s e xp lic itl y f or bi dd en toreproduce , distribute or post in a public apace t hi s document or a ny por ti on of th isdocument wit hout t he wri tt en permi ss ion of t he aut hor. - Thi s Document . i f ; . basedon a . simi la r document pre pa re d by B. Rosehar t a nd P . . Ia aayer i i n2001 for studentsin ENEL 587

.t3ill'sReview of Power EngineeringENEL 519.44 Power Engineering 2006

Bill Rosehurt 1September 11,2006

I

r -.~Contents

1Sinusoidal Functions 1L1 Introduction .. ... 11. 2 Some Useful Identities 21. 3 RMS values ... .. . .. . .. 31. 4 Assignment Quest.ions

2 Phasors 62 .1 Impedance and Ohm's L aw .. 72. 2 As si gnme nt Quest io ns . . . 9

3 Power 103.1 Power factor . 113.2 Complex/Apparent Power 143.3 Assignment Questions 16

4 3-Phase Circuitry 174.1 Y-Cot ll1e ct ion . ... 184.2 ~ Connection . 184.3 Three Phase Loads . .. 184. 4 Power in 3-phase circuits. .. .. 194.5 Assig nment Questions . . 20

Assignment QuestionsAs si gnme nt . q ue st io ns h ave been g iv en at t he e nd o f e ach Chap te r. You a reexpected to attempt all questions. Solutions to these questions will be eitherposted or discussed ill class . Solution" that are discussed in class , will generallynot be posted as well.


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Chapter 1Sinusoidal Functions1.1 IntroductionIlIsl.alljalwolls voll,a~(~!, dcfinccl ax a slJllltcloi(lal ulemonl., represented by:

,(t)=v, .", Rin(wt+0) (Ll)where: vU ) is the instantaneous voltage, ~rUx is the maximum instantaneousvoltage, w i s t h e angu lar fr equency o f t he si nu so id , and () i s t he phas e s hi ft o fthe s inusoid. A typic al ins tantaneous vol ta ge curve i s shown in Figure 1 .1 .

if~Figure 1 .1 : Ins tantaneous vol ta ge of Per iod T

The vol tag e h as a p er io d o f T ( time uni ts ] and f requency of f =1/T_ Theperiod, T. is measured in seconds (6) and f ha..units ofhertz (Hz). The sinu-soid in Figure 1 .1 has angular f requency of w = 1'0 f _The frequency, period,and angular f requency value s for the . Nor th American and European sys temsa re l is te d in Table 1 .1 ( it i s bene fi ci al to be very fam il ia r w ith t ,h", " value s) :

f(Hz)T(ms)wTable 1 .1 : Frequency, Per iod and Angular Frequency for Nor th America andEurope

In dis cuss ing the parameter s of the vol ta ge s inusoid ofFigure 1 .L the pha seshift of the sinusoid was intentionally left out. Since phase shift is a relahllcmeasure, it is meaningless tu discuss it without a reference phase angle. . Inpower systems. this reference point is the Slack Bus . The phase angle of thevol tage a t thi s r ef er ence point i s a ssumed to be zero (or some other fLurl value)and the phase shifts ofall other vclteges are mea..ures relative to this point:

{)8/i =0

Nor th America EuropeU O16377

so20314

where sb is used to denote the slack-bus.When two sinusoids of identical frequency have different phase angle: i .e . .

( or ano ther f ixed value) (1.2)

we B>'y t ha t" , LEADS '12 by (B , - 92) or th.s., " "S '" by (B , - O 2), Whendetermining whether one signal loads or Iagsac- ! _ J \ ~ t both must have positiveamplitude and must be expressed as either sines k


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"

1 2 1( I T l , T (41f ))T l! ;nox? dt+ cos Tt +29 dt- f} ()

31 . 3 RMS valuesInpowe r sys tems , we tend to use RMS (Ruut Mea Sq'uaT~d)values. Also knownas ef fect ive value s, RMS value s repre sent the DC equivalent for all AC sourcein terms of power consumpton.

Matbematicelly, the RMS value for any periodic signal 1(0 iscalcula ted by:

l I Tnns =\/- JZ( t)dtT "

(1,9) '(1''

where T i e ; ; ; the signal period.AByou might . r emembe r, the R)lS value for a sinusoid 15 ~. A c ommon

mis take i s to a ssume tha t tbe EMS vrue of any periodic signal isof this format,The derivation of the R!\:ISvalue for a s inusoid is as fol lows (Note: this

der ivat ion can be appli ed to e ithe r the cur rent or the vol ta ge waveforms ):Assume: 'fI(t) ==", ",s(wt +0) t hen, f rom (L9), the RMS value i s:

,/ 1 rT [ 2V,m , = T 1 0 v , , . . , , cos(wt + )1 dt (LlO)HUWCVCT1

w=21f! =~T (Lll)Therefore,

1 1 . ' 7 ' [ 2" 1 2Vrms=f T 0 v'nax cost Tt +0) dtCO"2 ( ~ t +B)1dt (Ll2)

Given the trigonometric identity:cos2(x) =~'()'(2x)2~-- (Ll3)

Substituting (L1:1) in to (1 ,12), a nd movi ng t he const an t amp li tu de out o fthe integration gives:

V ; m . i IT_l!:2 11+coS(i;rt+20)Tmuxt) '! 'dt

4To s impl ify the equat ion a . li tt le b it , wecan square both Bides to get :

- 2 1 2 1( T 7 ' ( 4 1 T ) )r-m ,'! =T V';nax; -- dt + cos Tt +28 dt2 ,f) 0Also, the following trigonometry property

cos(x + ;q ) =os(x)cos(;q) - sin(T) sin(y) (L14)ca n be u sed i n:

(41T ) ( 4 " , , , (41f ') ,cos Tt +29 =os Tt) e os (2 0 ) - sm = ' sm(20)Therefore,

v;~~, 1 2 1 ( I T ( 4 " ) (v. - T + cos -t cos 29)dt-T ma T.2 0 T., - [ sin (*t) Sin(29)dt)

1 1( I T ( 4 1 1 ' )T V;'ax:l T +cos(29) I) cos = ' d t - '... - sin(28) [ sin ( 1 ; t ) d t ) (Ll5)

Vr2ms

At thi s point you can per form the integra tion , OR you call observe tha t wea rc integra ting s inusoida l s igna ls ove r twice the ir per iods , Le. the s inusoida ls igna ls s in ( t) and cos ( - t ) have per iod of t'and the integra tion i spe r-formed ove r the [O.T] interva l The integra l for any s inusoida l s igna l ove r anymul tiple of i ts per iod i s zero. Therefore . (1.15) reduces to:

,2 _ 1 2 I ( )lrmtl - T Vmax~ T+O+Ov",,, = V~x ~lf\ &"l~(LW)

Once again, you should remember that this derivation was for sinusoidalsignals only. and ut.lier s ignals would have itdillereut RNIQvalue.

A lso note, tha t the HMS value ofa constant s igna l i s equal to i ts el f (Anothe ra re a whe re some people tend to incor re ct ly use the * ' expression].

f l , t A S J (A { ~ o r ; A. 6\1\~~-f C ; 7 )J(D()(~11"~4


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1.4 Assignment QuestionsComplete the following assignment questions, Please note, t ha t the answe rs tothese questions will require some investigation 011 your part.Al~l~l Find the frequency ofthe electrical power system illthe following conn-

trit . ..Japan, China, Iran, and Cuba,A] -1 -2 What ar e t he HMS val ues o f t h e two ma in re si den ti al v ol tag e u se d i n

Canada.Al-1-3 \Vhy do weuse two res identi al vul ta ge s in Canada?

)(

Chapter 2PhasorsA phas or i s a comp lex number t hat conta in s t he amp li tu de ( usu al ly i n RMS)HUrl the phase angle of a sinusoidal function. Phasor representations are used iucircuit and power engineering to simply the analysis of circuits. The frequencyof waveforms arc implied when using phasors. The reactance associated withIndnctcrs and capacitors arc functions ofthe frc1IUCllCY of the voltage/currentwaveforms.

Eul er " ,,> Iden t it y is the foundation for representation of phasors:f:=O =os (J jin (i (2.1)

Consequently;co s 0 = !)te[ejo]sin 0=Jm[ej8]

(2.2)(2.3)

Replacing l:'Htx with the equivalent v 'Z V ;' rH R : as derived previously, a sinusoidalRigllal can be represented as:

v(t) v ' 2' ;' m , c o s (w ! +0)v'2v;.m,!)t, ,[e J(wt.+O)]v'2.;'m,!)t"[, , j. ,Jw,] (2.4)

From (2.4). only the amplitude (or R~IS Value) and phase angle values arcUSCi..l in the phascr representation. III other words , the other value s in (2.4) a rcimplicitly presented in phasor notation: i.e. it is implied that the pbasor rotatesat the frequency ofwt )etc.

You should he very comfortable with the following uotutions:v IVILII

VLe(2.5)(2 .G)(2.7)(2.8)

V;'~ul+Vim'lgVcos O+jVsiu 8

6


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I;'

IVI =V~c(ll +Vif"(~ag( v ' m a g)()=arctas V ; . e : a t

2.1 Impedance and Ohm's LawBased on fundamental s of e le ct rostat ic s and magnc ti ce the t imc dependentvoltage-current relationships for inductors , denoted with a subscript L, andcapac itor s: denoted with a subsc ript C , a re :

d (2.11) ,.{t) = Ld i if.(t)t1 (2.12)dt) = Cdi volt)

Using the Fourier Transform/Series (sorry for using the :F~word), i .e .,~Hwj (2.1:l)

a nd mappi ng t he Fou rie r T ran sf orm o f vc/dt) and ic/L1t) t o the ir phe sorrepresentations, gives:

Vj;Vc

j:r.Lir, =:}rL =il:. 1Jxc1c =* Xc=-wC

where x',le isknown as the reactance. The units for reactance isohms, denotedwith the symbol n. I t s hould he highl ighted , tha t the " imaginary" ope ra tor,j , i n t he r el at io ns hi p b etwee n c ur ren t a nd vol tag e d enot es a 90 phase s hi ftbetween the two sinusoidal functions, i .e , j i s s imply the ope ra tor 1L90' .

The reactance ofinductors and capacitors call also be shown by r;tartiuv; witha sinusoidal function for "cCt) and iL(t), and then substituting these functionsinto both s ides of (2.12) and (2.11). After some . impl if ic al iu llR both s ides of(2.12) anti (2.11) are mapped to the phase r "domain".

Res is ta nce maps to the pha sor "doma in '! w ithout a .nychanges , Le.'rHr

where T} ill the phesor "domain" isalso called resis tance with unit ohms, denotedwith t . h ( ) HYl l Ibol n.

When agt t; re ga ting e lements in a sys tem the res is ta nc e and rea ctance a rcoft en lumped togethe r to form the impedance , z :

z =" +jx

7

(2.9)(2.10)

(2.14)(2.15)

'/(2.16)

(2.17)

t'"

'/t( . , . J

../

8

i .. , .

\..


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2.2 Assignment QuestionsComplete the following assignment questions. Please note, the questions involvesome lung (sorry) hut relatively simple math.Al-2-1 Re-find the expression for the reactance of inductors and capecitors,

(2.14) & (2.15), s tarting with a sinusoidal function for vdt) and hit)./\J-:l-2 ( ~n l np l( I .I \ 1 . 11 1 :r(l l l~' ,\v iJ lJ !; , . ,}! ' ,ula!lo l l and 'WTi t ,P iJw final answe r in both

polar ami rectangular form:(1Ilm)(6~)(4+5) +(3 + 2}(2 +3)

Al-2-3 Express the following summations ofsinusoids ill the general form A Sillwt +9by using trigonometric identities.a ] i(t) =e083t + 25' +4cos3t +400b) 'i(t)=2cos3t+25 +4"oR3t +40"

Al -2 -4 A s inu . .oidal voltage ha") a maximum value (If 208 vol ts and the ins tan-t aneous value of the vol ta ge a t t=O .O ls i s30 vol ts . The per iod i s16.667ms. Determine . .:( t) in the standard fonnuft} =J' (Z )Y; .m . Sill",t +6.

()(.'A)I:W) .'

Chapter 3Power

The instanta.neotts vol tage and cur rent a t an c lement ( load! generator . e te ) c anbe written as;

v(t) = .;'.cos(wt+Ov) (3.1)i(t) = Im cos(wt +Of) (3.2)

where V", and In> a re the peak s inusoid value s, Vm = ,fi V and Im = ,fi I.(Our notation convention Is1-."nd Iwill represent the nns values, and v : . n andIn < will represent the peak values.)The instantaneous power applied to this clement at any time t isdefined as:

p(t) (3.3),(t)i(t).,fiv cos(;.;t +ev).,fiI cos(;.;t +0 I)2VI cos(wt +edcos(wt +OI )

III order to he able to bet te r understand the r el at ionship between the in-s tantaneous power and the "complex power ': t rigonome tr ic ident it ie s a re usedto simplify the expression. Complex power can Leconsidered or associa ted withthe "mapping" of instantaneous power to the "pha.~or:;domain. Complex poweral so i s c omposed o f t he " de equiv al en t" pow er ( real o r ac tiv e powe r) and t heinstantaneous power that is associa ted with reactive elements (reactive puwer).

Using the following trigonometric identity:1""s(o:) cos(y) = 2(",s(x +y) +cos(,: - v )

Equat ion (3.3) e au be re-wr it te n a s:(:lA)

pit) = 2Y-IH cos (w t+6v +wt+6I )+cos(wt+llv wt Ill)

(3.5)

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11simplifying the components inside the cosine and sine functions gives:

pet) =VI(!os(2wt +0,, + 1) + VI " "s(Ov - 01) (3.0)From the above expression. it call he seen t ha t t her e a re two te rms in t he

expression for instantaneous power:1. The first terrn.Vf cos(2;,; t +Ov +Or ) , has twice the f requency of 'v(t) andi(t}, and it represents the power being trausfered in and out uf capacitorsand Inductore every cycle . This component isassociated with the reactivepower.

2 . The second term,VIcos(liv 11]), i s a constant value . Thi s componentcor re sponds to the a ctua l ave rage "work" done and i s a ssoc ia ted with thethe e ve ra a e, 1 -- ea tor a c U " v e power 1 i.e.

,,< - .,r, (

Pat .J>Uj6 r1 0 pet)VIeos(8v - Od (3.7)

In F igur e 3 .L p lo ts o f t he vol tag e, cu rr en t, and t he co rr es pond ing pow erwaveforms a re given. In thi s f iguce, the relat ionship between the vol ta ge andcur rent f requency and the ins tantaneous power should he noted . I t i f - > observedt ha t t he av era ge o f t he i nst an tan eous power , co rr esponds t o t he f ir st t erm o f(3.6).

1the phase difference between the voltage and current is increased, i .e ., in-crca..ed inductlvc/capacitivc components with respect to the resis tance c-Ompo-nents , currespcnds to a reduce..d.average pOWHr.In Figure 3.2, the instantaneouspower ispl, ,I .kd fur djni:'l'(~nl pha..o < f!!t's botwoon t.ill) voll.Hgeand current. Togenerate these plots. the voltage and current maguttudee are set equal to thewaveforms: in Figure :3.1

3.1 Power factorFor both generator s and loads , the power factor, pf, isdefined as:

p. f =o s( O,r - 0 ,) (3.8)From the average power definition in (3.7) and power factor expression from(3.8), average power can be expressed as:

Pal! =Par;tf/f, VI co8(8,~ - Or)VIp! (3.9)

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" I ' . . . , r'~ t '\ " i\~ ,'\,~t\ J \ i , ,7t i~/ \, t - . ~ \ j ," '1\, 41 pI 1;1 r: .ffc---'4 \~, f ~ 1,' .o o--Gl~ /,0 t>, Q.~ \/ : d '~o,~\, e ,-l/r;:/ \ '" ' v . Q." 1fi. \. &u..' " ' J. .: >- -- -O J 4) "~Q-11~~ -:: - .. . . ./Q~~!~\::;~

-101) 0 -1 -'0 '5 ~1'I1 ~.01~ [


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The point at the cdlter ~. pf =1, rep-resents a.purely resistive 10,";_ ,,,_ing tow81';j; the ';~;iieate8a netcav lead) nd the ext reme - lef t e nd of the epect ru ia i t: I) re sent s a purely

l eat '_ Moving towards the r ight end indic ates a net inductive load,ext reme r ight end of the spect rum repre sent s a purely induc tive load.

. ug a power f act or o f ze ro I s not v er y r ea li st ic si nc e t her e i s i nh er en tnee ;~lapacitors and inductors,

13pf lagging pf leadingflags VIhonet inductive load

I leads VOf >OvOz ev) t hen the power fac tor i s l eading. S ince for impedance loads ,Oz =Bp! =B " . .- 0" for the leading power factor, an equivalent impedance loadwould 'look' capacitive (with a.res-istive component 8. '3 well).

A summary uf the relationships between power Iector , leading and ia/;! ;p;iul-{,and effective impedance isgiven in Table 3,l .

From (3.8) , the range of power fac tor i s between 0 and 1:

pi:equivalent impedance load:

f) 11ead,'ittH 1 lfl,q,qng) nL

t;:.~;

143.2 Complex!Apparent PowerTo bet te r show the relat ionship between the ius tanteneous power and the com-plex power, the expression of instantaneous power l (3.G). is repeated here:

p(t ) =Fl i l l eo8(2wt+O,,+Bd + IVllfleos(Ov-8,)w her e IV and If I r ep res en t t he rms vol tag e and cn rr en t v al ues .Based on ( :~ .7), the ave rage , r ea l or a ct ive power was found to be:

P, w = P , ' C < ' v c = p , c u ' = I V I II cus(Bv - Or )Using the se def in it ions , and the des ir e to have a mea sure of the " tduie vary-

i ng" component of the ins tantaneous power , the product of the ( rms) vol ta geand cur rent pha sc rs i sde fined to be the appamnt (complex) power:

SLO, =(VL B v) ( I LO , ) ' (3.12)where H*) represents the complex conjugate operator. i .e .

(


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The value of Q is positive is the power factor islagging. i.e.. the reactiveportion ofthe load looks inductive . If the power factor ill leading, then the valueof Q is negative.The Power Triangle is often used to display the relationship between COIU-

plex Power, Real Power, Reactive Power, and O. The power triangle in Fig-ure 3.3, illustrates the apparent, active and reactive power for a lagging powerfactor load, Le. Os >0 (load 'looks' inductive).

15L J Q . \pFig11fC3.3: Power Triangle r

since6' = VIo6' = VI'

( I "V I L O , ) ( lf IL 0 1 ) "ViII LO" - OrVILIJVI cosO+VlRillOP+jQ

}_ /--.

163.3 Assignment QuestionsComplete the following assignment questions. Some equations extend beyondthe mater ia l pre sented iu thi s Sec tion . The se que st ions a re meant tu have youextend the fundamental eoncepta.Al-3-1 In words , expla in wha t i s appa rent : r ea ct ive and act ive power .A l-3-2 Assume you a re a consult ing enginee ring: a sma ll indus tr ia l f ac tory has

been iuforme..d by its electricity provider that it will be cherged an in-c re ased fee i f the power fac tor of the fac tory i sbe low 0.9 lagging. Cur -r ently, the fac tory 's power fac tor t ends to be a round 0.8 laggi11K.Theyare asking you if t hey should agree topay thi s f eeor modify the ir sys temto ' fix' the power factor. If you advice them to 'fix) the power factor,suggest how they could 'fix it1

A1-3-3 Why should a power enginee r m inim iz e the amount of r ea ct ive powertransmitted through transuussiou lines?


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Chapter 43-Phase CircuitrySome review items

1. I fwe add 3 vec tors ofequa l magni tude but pha se -shi ft ed 120" f rom eachother, weget zero.

2. Generally) wegenerate voltage/power in ;i-phascs\ phase shifted 120:::fromeach otber Iwith equal megnitudes 011 each phase.:3. We tend to connect theses phases in positive abc or negative acb sequence,

0\,'" shown ill Figure 4.1:When we generate a vol ta ge . i t i s between or a cros s the two ends of a .coil

(n ot n ece ss ar il y d ir ect ly r el at ed t o GND or neu tr al ). Th is f act al lows u s t oconnect the 3 p-hases indi ff er ent ways, the main two connect ions a re Wye (Y)and Del ta (L \. ) c onnect ions . The Phc sor r epre sentat ion of Y and /I. connectedelements arc shown illFigure 4.2.

The following sections provide an overview of some of the characteris tics of\Vye and Dell.a.connections.

b

) . . . . .;~>~~)b cFigure 4.1: Scphase sequences17

18 .'.~\ \ I , . , t ' < - " " " , - . ;, ,,,Figure 4.2: Wye (Y) and Delta (/I .) connections4.1 V-Connection

1 . The Neu tr al ma yo r may not b e g rounded .2. Phase current =ine current3 . The l ine vcl ta ge s and pha . .e vol ta ge s a te r elat ed by:

IV,-d =v3V,;J8,_, =8"+30'

4. 2 LlConnection1. The re a re no dir ec t r ef er ence s to GND (ground)2. Phase voltage =ine voltage3. The l ine cur rent s and pha se cur rent s a re r elat ed by:

l I t ! = v ' 3 I ~ 18, =0" _ :lO'

4.3 Three Phase Loads

(4.1a)(4.1b)

(4.2a)(4.2b)

Z l > .ZY=T

A three-pha....;ehnpedauce load call he connected in del ta . or ill wye with OTwi thou t t he n eu tr al b rought out . I f t he t hr ee phase l oad i s b al an ced , t he n t herelationship

i s u sed t o c hange f rom a n why ( or d el ta ) l oad t o 1",]1 equivalent. delta [or wye}load.

(4.3)


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19 20

Inth is s ec tion , an expre ss ion i sde te rm ined for the ins tantaneous total power ina balanced three phose system.For the ins tantaneous vol ta ge and cur rent in the ' a' s equence: we wil l a ssumethat the voltage phase angle L"!zero (Le. our reference, 8v' =0). and that thecur rent i sphase- shif te d f rom thi s vol ta ge by - , e :, The expressions for 'u(t) andi(t) arc then:

Pl,,(t) Part ) +Pb(t ) +Pe(t l3VlcosO- F I ( cos(2wt - 0) +cos(2wt - 8 - "W') +cos(2, 't - e - 4RO' )

4.4 Power in 3-phase circuits The total 3-pha.se power issimply the Btt111Illatioll ofthese 3 single phase powers:

llan(t)ia(t )

v2Vsill(wt)V2 I sin(wt - B )

In the above expression. t:l~4800 phase shift can be rer>.cBd with the equL111rut120" phase shift (480" =60" +120") to get :P30(t) = 3Flcosl!V T ( c08(2'''t (1 ) +cos(2wt 240') +cos(2wt 11 120')

Voltage and current in the other sequences are simply the phase shiftedversions of voltage and current in sequence a. Phase shifting the sequence a.siuusoids by 120D and 2400 results ill:

"""(t) = , , 1 2 V8ill(wt- 120')ib( l) = V2 I sill(w~ - 0 - 120')

A c lose r look a t the expre ss ion ins ;. ,, ; t he bra cket s shows tha t it is s imple' thesununation of 3 equal magnitude sinusoids that tw o phase shifted fromeach other Iwhich is zero! 'Z'~herevre~he total it ");8 power in a 3-phasecircuit is:

P3.(t) =3V[eos8 (4.5)

and,You call s ec t hat t he Instantaneous power in i

constant Iwhich i sone uf the edvantagus ofus ing haleru;{-phasc circuits is

phase circuits.

7 'm ( t )i,(t)

V2 Vei ll (w t- 240 ')V2 hin( wt - e - 240')

4.5 Assignment QuestionsComplete the following assignment questions.

P,,(t) = v",,(t)i,,(t)2Vlsin(wt )s in(w t - 9)

Pb(t ) = 2Vl si ll (w t - 120')sin(wt - 9 -120")P,(t ) ee 2Vl ei n(w t - 240")sin(wt - 0 - 240")

Al-4-1 Prove the relat ionships given in (4.1) and (4.2) .Al-4-2 Why is it all advantage that the instunteneous power in a. baleuced

3-phase circuits is constant.The power illeach pha se i s s imply the product of the ins tantaneous vol ta ge

and current. Therefore .

From trigonometry,

sil,,;siny = ~(GOR(:r.- Y) - COR(" + V ) (4.4)Tlunefore)

P,,(t ) = V I( ""s(e)- , ,, ," ( 2 1 t - 8 )P;(t ) = V I( eos(e) - cOR(2wt- 0 - 240'))P,(t) = VI(co5(8)-cos(2wt-e-4S0))

.s I' ,

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Power Engineering Review ENEL 519.44