EE 101 Electrical Engineering. vrect

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1 EE Elecrical Engineering ac heory 3. Alernaing urren heory he advanage of he alernaing waveform for elecric power is ha i can be sepped up or sepped down in poenial easily for ransmission and uilisaion. Alernaing waveforms can be of many shapes. he one ha is used wih elecric power is he sinusoidal waveform. his has an equaion of he form v() v() = m sin(ω + ) m he peak value of he waveform is he maximum value of he waveform and for he a.c. waveform i is /ω m. mean value = + o v () d o he mean value of he sinusoidal a.c. waveform is since posiive and negaive areas cancel. ( can also be shown by inegraion). f o = /ω v o + rec () average value = vrec () d o he average value of he a.c. waveform is defined as he average value of he recified waveform and can be shown o be equal o π m o + rms value = v () d o v () he effecive value or r.m.s. value of a waveform is defined as above so ha he power in a resisor is given correcly. ha is o + o effecive v () + = d = v d effecive, or ( ) o o he effecive value or rms value of he waveform is hus he square roo of he mean of he squared waveform and can be shown o be equal o for he sinusoidal a.c. waveform. m Unless oherwise specified, he rms value is he value ha is always specified for ac waveforms, wheher i be a volage or a curren. For example, 3 in he mains supply is an rms value of he volage. Similarly when we alk abou a 5 A, 3 A or 5A socke oule (plug poin), we are again alking abou he rms value of he raed curren of he socke oule. For a given waveform, such as he sinusoid, he peak value, average value and he rms value are dependan on each oher. he peak facor and he form facor are he wo facors ha are mos commonly defined. rms value and for a sinusoidal waveform, Form Facor = m Form Facor = π average value = = 7.. m π he form facor is useful such as when he average value has been measured using a recifier ype moving coil meer and he rms value is required o be found. [Noe: You will be sudying abou hese meers laer] Peak Facor = peak value rms value and for a sinusoidal waveform, i Peak Facor = m m = = 44. he peak facor is useful when defining highly disored waveforms such as he curren waveform of compac fluorescen lamps. Some advanages of he sinusoidal waveform for elecrical power applicaions a. Sinusoidally varying volages are easily generaed by roaing machines b. Differeniaion or inegraion of a sinusoidal waveform produces a sinusoidal waveform of he same frequency, differing only in magniude and phase angle. hus when a sinusoidal curren is passed hrough (or a sinusoidal volage applied across) a resisor, inducor or a capacior a sinusoidal volage waveform (or curren waveform) of he same frequency, differing only in magniude and phase angle, is obained. Universiy of Morauwa - JL/Sep8

2 EE Elecrical Engineering ac heory f i() = m sin (ω+), for a resisor, v() =.i() =. m sin (ω+) = m sin (ω+) magniude changed by bu no phase shif for an inducor, v ( ) = L. di = L. d ( m sin( ω + )) = L. ω. m cos( ω + ) = L. ω. m sin( ω + + π / ) d d magniude changed by Lω and phase angle changed by π/ for a capacior, v ( ) =. i d = m sin ( ω + ) d =. m cos( ω + ) =. m sin ( ω + π / ) ω ω magniude changed by /ω and phase angle changed by π/ c. Sinusoidal waveforms have he propery of remaining unalered in shape when oher sinusoids having he same frequency bu differen in magniude and phase are added o hem. A sin (ω+α) + B sin (ω+β) = A sin ω. cos α + Α cos ω. sin α + B sin ω. cos β + B cos ω. sin β = (A. cos α + B. cos β) sin ω + (A. sin α + B. sin β) cos ω = sin (ω + ), where and are consans obained from rigonomery. d. Periodic, bu non-sinusoidal waveforms can be broken up o is fundamenal and harmonics. e. Sinusoidal waveforms can be represened by he projecions of a roaing phasor. 3. Phasor epresenaion of Sinusoids You may be aware ha sin θ can be wrien in erms of exponenials and complex numbers. i.e. e jθ = cosθ + j sin θ or e = cosω + j sin ω onsider a line P of lengh A m which is in he horizonal direcion X a ime =. f P roaes a an angular velociy ω, hen in ime is posiion would correspond o an angle of ω. he projecion of his roaing phasor P (a phasor is somewha similar o a vecor, excep ha i does no have a physical direcion in space bu a phase angle) on he y-axis would correspond o P sin ω or A m sin ω and on he x-axis would correspond o A m cos ω. hus he sinusoidal waveform can be hough of being he projecion on a paricular direcion of he complex exponenial e. P a() ω ω X oaing f we consider more han one phasor, and each phasor roaes a he same angular frequency, hen here is no relaive moion beween he phasors. hus if we fix he reference phasor in a paricular reference direcion (wihou showing is roaion), hen all ohers phasors moving a he same angular frequency would also be fixed a a relaive posiion. Usually his reference direcion is chosen as horizonal on he diagram for convenience. ω ω P X a() A m sin (ω +) A m sinω A P A m P A m A = A x A y reference direcion Universiy of Morauwa - JL/Sep8

3 EE Elecrical Engineering ac heory is also usual o draw he using he rms value A of he sinusoidal waveform, raher han wih he peak value A m. his is shown on an enlarged diagram. hus unless oherwise specified i is he rms value ha is drawn on a phasor diagram. should be noed ha he values on he phasor diagram are no longer ime variables. he phasor A is characerised by is magniude A and is phase angle. hese are also he polar co-ordinaes of he phasor and is commonly wrien as A. he phasor A can also be characerised by is caresian co-ordinaes A x and A y and usually wrien using complex numbers as A = A x + j A y. Noe: n elecrical engineering, he leer j is always used for he complex operaor because he leer i is regularly used for elecric curren. is worh noing ha A = Ax + A and ha y an = A or A = an Ay y A Also, A x = A cos, A y = A sin x and A e j = A cos + j A sin = A x + j A y Noe: f he period of a sinusoidal waveform is, hen he corresponding angle would be ω. Also, he period of a waveform corresponds o complee cycle or π radians or 36. ω = π 3. Phase difference onsider he wo waveforms A m sin (ω + ) and B m sin (ω + ) as shown in he figure. can be seen ha hey have differen ampliudes and differen phase angles wih respec o a common reference. y() B m sin (ω + ) A m sin (ω + ) x ω B B m = ω hese wo waveforms can also be represened by eiher roaing phasors A m e j(ω+ ) and B m e j (ω+ ) A A m = wih peak ampliudes A m and B m, or by a normal phasor diagram - wih complex values A and B wih polar co-ordinaes A and B as shown. Any paricular value (such as posiive peak, or zero) of a() is seen o occur a a ime afer he corresponding value of b(). i.e. he posiive peak A m occurs afer an angle ( ) afer he posiive peak B m. Similarly he zero of a() occurs afer an angle ( ) afer he corresponding zero of b(). n such a case we say ha he waveform b() leads he waveform a() by a phase angle of ( ). Similarly we could say ha he waveform a() lags he waveform b() by a phase angle of ( ). [Noe: nly he angle less han 8 o is used o specify wheher a waveform leads or lags anoher waveform]. We could also define, lead and lag by simply referring o he phasor diagram. Since angles are always measured aniclockwise (convenion), we can see from he phasor diagram, ha B leads A by an angle of ( ) aniclockwise or ha A lags B by an angle ( ). Addiion and subracion of phasors can be done using he same parallelogram and riangle laws as for vecors, generally using complex numbers. hus he addiion of phasor A and B phasor B would be A + B = (A cos + j A sin ) + (B cos + j B sin ) = (A cos + B cos ) + j (A sin + B sin ) = x + j y = c = where = + = A + B ( cos cos ) + ( A sin + B sin ) x y A and = an c y = an x ( Asin + B sin ) ( Acos + B cos ) Universiy of Morauwa - JL/Sep8 3

4 EE Elecrical Engineering ac heory Example Find he addiion and he subracion of he wo complex numbers given by 3 o and 5 48 o. Addiion = 3 o o = ( j.5) + 5( j.743) = ( ) + j ( ) = j = o Subracion = 3 o 5 48 o = ( ) + j ( ) = 8.68 j = o Muliplicaion and division of phasors is mos easily done using he polar form of complex numbers. hus he muliplicaion of phasor A and phasor B would be A * B = A * B = A e j B e j = A * B e j( + ) = A * B + = c where = A * B and c = + n a similar way, i can be easily seen ha for division = A / B and c = hus, whenever we need o do addiion and subracion, we use he caresian form of complex numbers, whereas for muliplicaion or division we use he polar form. Example Find he muliplicaion and he division of he wo complex numbers given by 3 o and 5 48 o. Muliplicaion = 3 o * 5 48 o = 5 78 o Division = 3 o 5 48 o =.4 8 o 3.3 urrens and volages in simple circui elemens 3.3. esisor i () for a sinusoid, consider i() =eal par of [ m e (+θ) ] or m cos (ω+θ) v() = eal [. m e (+θ) ] = eal [ m. e (+θ) ] m cos ω v () or v() =. m cos (ω+θ) = m cos (ω+θ ) m cos ω v() =.i() m =. m and m / =. m / ω i.e. =. ω Noe: and are rms values of he volage and curren and no addiional phase angle change has occurred in he resisor. Noe also ha he power dissipaed in he resisor is equal o. = nducor L for a sinusoid, consider i() =eal par of [ m e (+θ) ] or m cos (ω+θ) i () v() = eal [L. d m e (+θ) ] = eal [L.. m e j(ω+θ) ] = eal [j m e j(ω+θ) ] v () d or v() = L. d m cos (ω +θ) = L.ω. m sin (ω+θ) = L. ω. m cos (ω+θ+π/) v L di () () = d d m cos ω = m cos (ω+θ+π/) j ωl m sin ω m = ωl. m and m / = ωl. m / π/ ω can be seen ha he rms magniude of volage is relaed o he rms ω magniude of curren by he muliplying facor ωl. also seen ha he volage waveform leads he curren waveform by 9 o or π/ radians or ha he curren waveform lags he volage waveform by 9 o for an inducor. hus i is usual o wrie he relaionship as = L. or = ω L. 9 ο he impedance Z of he inducance may hus be defined as L, and = Z. corresponds o he generalised form of hm s Law. emember also ha he power dissipaion in a pure inducor is zero, as energy is only sored and as here is no resisive par in i, bu ha he produc. is no zero. Universiy of Morauwa - JL/Sep8 4

5 EE Elecrical Engineering ac heory apacior i () v () v () = i (). d j ω for a sinusoid, consider i() =eal par of [ m e j(ω+θ) ] or m cos (ω+θ) v() = eal [ e j( ω + θ ) m. d ] = eal [. m e (+θ) ] = eal [ m e (+θ) ]. j or v() = m cos( ω + θ ). d =. m sin (ω+θ) =. m cos (ω+θ π/) ω ω = m cos (ω+θ π/) m cos ω m =. m and m / =. m / ω ω can be seen ha he rms magniude of volage is relaed o he rms ω magniude of curren by he muliplying facor. ω also seen ha he volage waveform lags he curren waveform by 9 o or π/ radians or ha he curren waveform leads he volage waveform by 9 o for a capacior. hus i is usual o wrie he relaionship as =. or = ω. 9 ο he impedance Z of he inducance may hus be defined as, and = Z. corresponds o he generalised form of hm s Law. emember also ha he power dissipaion in a pure capacior is zero, as energy is only sored and as here is no resisive par in i, bu ha he produc. is no zero. 3.4 mpedance and Admiance in an a.c. circui he impedance Z of an a.c. circui is a complex quaniy. defines he relaion beween he complex rms volage and he complex rms curren. Admiance Y is he inverse of he impedance Z. = Z., = Y. where Z = + j X, and Y = G + j B is usual o express Z in caresian form in erms of and X, and Y in erms of G and B. he real par of he impedance Z is resisive and is usually denoed by a resisance, while he imaginary par of he impedance Z is called a reacance and is usually denoed by a reacance X. can be seen ha he pure inducor and he pure capacior has a reacance only and no a resisive par, while a pure resisor has only resisance and no a reacive par. hus Z = + j for a resisor, Z = + L for an inducor, and Z = = j ω for a capacior. he real par of he admiance Y is a conducance and is usually denoed by G, while he imaginary par of he admiance Y is called a suscepance and is denoed by B. elaionships exis beween he componens of Z and he componens of Y as follows. jx so ha X G + jb = Y = = = G =, and B = Z + jx + X + X + X he reverse process can also be similarly done if necessary. However, i mus be remembered ha in a complex circui, G does no correspond o he inverse of he resisance bu is effecive value is influenced by X as well as seen above. 3.5 Simple Series ircuis n he case of single elemens, L and we found ha he angle difference beween he volage and he curren was eiher zero, or ± 9 o. his siuaion changes when here are more han one componen in a circui L series circui n he series -L circui, L L L considering curren as reference =., L = L., and = + L L = ( + L). so ha he oal series impedance is Z = + L m sin ω π/ ω Universiy of Morauwa - JL/Sep8 5

6 EE Elecrical Engineering ac heory he above phasor diagram has been drawn wih as reference. [i.e. is drawn along he x-axis direcion]. he curren was seleced as reference in his example, because i is common o boh he resisance and he inducance and makes he drawing of he circui diagram easier. n his diagram, he volage across he resisor is in phase wih he curren, where as he volage across he inducor L is 9 o leading he curren. he oal volage is hen obained by he phasor addiion (similar o vecor addiion) of and L. f he oal volage was aken as he reference, he diagram would jus roae as shown. n his diagram, he curren is seen o be lagging he volage by he L same angle ha in he earlier diagram he volage was seen o be leading he curren. L has been drawn from he end of raher han from he origin for ease of obaining he resulan from he riangular law. n an -L circui, he curren lags he volage by an angle less han 9 o and he circui is said o be inducive. Noe ha he power dissipaion can only occur in he resisance in he circui and is equal o. and ha his is no equal o produc. for he circui series circui n he series - circui, =., =. = j. Ι ω and = + = ( + ). so ha he oal series impedance is Z = + he phasor diagrams has been drawn firs wih curren as reference and hen wih volage as reference. n an - circui, he curren leads he volage by an angle less han 9 o and he circui is said o be capaciive. Noe ha he power dissipaion can only occur in he resisance in he circui and is equal o. and ha his is no equal o produc. for he circui L- series circui L L n he series L- circui, L = L., =. = j ω and = L + Phasor = (L + ). so ha he oal series impedance is Z = L + = L j ω is seen ha he oal impedance is purely reacive, and ha all he volages in he circui are inphase bu perpendicular o he curren. he resulan volage corresponds o he algebraic difference of he wo volages L and and he direcion could be eiher up or down depending on which volage is more. When ωl = he oal impedance of he circui becomes zero, so ha he circui curren for a given supply ω volage would become very large (only limied by he inernal impedance of he source). his condiion is known as series resonance. n an L- circui, he curren eiher lags or leads he volage by an angle equal o 9 o and he resulan circui is eiher purely inducive or capaciive. Noe ha no power dissipaion can occur in he circui and bu ha he produc. for he circui is non zero L- series circui n he series -L- circui, L =., L = L., =. = j. Ι ω L and = + L + = ( + L + ). so ha he oal series impedance is Z = +L + = + j(ωl ω ) L or L L Universiy of Morauwa - JL/Sep8 6

7 EE Elecrical Engineering ac heory Z = + ω L ω has a minimum value a ωl =. his is he series resonance condiion. ω n an -L- circui, he curren can eiher lag or lead he volage, and he phase angle difference beween he curren and he volage can vary beween 9 o and 9 o and he resulan circui is eiher inducive or capaciive. Noe ha he power dissipaion can only occur in he resisance in he circui and is equal o. and ha his is no equal o produc. for he circui. 3.6 Simple Parallel ircuis L parallel ircui n he parallel -L circui, considering as reference L L oal shun admiance = + j ω L =., = L. L, and = + L = + j ω L L L parallel ircui oal shun admiance = L- parallel ircui L L n he parallel - circui, considering as reference =., =., and = + L oal shun admiance = + j L + j ω ω = +. n he parallel -L- circui, considering as reference =., = L. L, =. and = + L + = + j L +. ω j ω L L L L + As in he case of he series circui, shun resonance will occur when = ω giving a minimum value of ω L shun admiance. Noe ha even in he case of a parallel circui, power loss can only occur in he resisive elemens and ha he produc. is no usually equal o he power loss. 3.7 Power and Power Facor was noed ha in an a.c. circui, power loss occurs only in resisive pars of he circui and in general he power loss is no equal o he produc. and ha purely inducive pars and purely capaciive pars of a circui did no have any power loss. o accoun for his apparen discrepancy, we define he produc. as he apparen power S of he circui. Apparen power has he uni vol-ampere (A) and no he wa (W), and wa (W) is used only for he acive power P of he circui (which we earlier called he power) apparen power S =. Since a difference exiss beween he apparen power and he acive power we define a new erm called he reacive power Q for he reacance X. Universiy of Morauwa - JL/Sep8 7

8 EE Elecrical Engineering ac heory he insananeous value of power p() is he produc of he insananeous value of volage v() and he insananeous value of curren i(). i.e. p() = v(). i() f v() = m cos ω and i() = m cos (ω ), where he volage has been aken as reference and he curren lags he volage by a phase angle hen p() = m cos ω. m cos (ω ) = m m.. cos ω. cos (ω ) = m m [cos (ω ) + cos ] p() v() p() p() m m cos i() curren lagging volage by angle inphase quadraure can be seen ha he waveform of power p() has a sinusoidally varying componen and a consan componen. hus he average value of power (acive power) P would be given by he consan value ½ m m cos. acive power P = ½ m m cos = m m..cos =. cos he erm cos is defined as he power facor, and is he raio of he acive power o he apparen power. Noe ha for a resisor, = o so ha P =. and ha for an inducor, = 9 o lagging (i.e. curren is lagging he volage by 9 o ) so ha P = and ha for an capacior, = 9 o leading (i.e. curren is leading he volage by 9 o ) so ha P = For combinaions of resisor, inducor and capacior, P lies beween. and For an inducor or capacior,. exiss alhough P =. For hese elemens he produc. is defined as he reacive power Q. his occurs when he volage and he curren are quadraure (9 o ou of phase). hus reacive power is defined as he produc of volage and curren componens which are quadraure. his gives reacive power Q =. sin Unlike in he case of inphase, where he same direcion means posiive, when quaniies are in quadraure here is no naural posiive direcion. is usual o define inducive reacive power when he curren is lagging he volage and capaciive reacive power when he curren is leading he volage. is worh noing ha inducive reacive power and capaciive reacive power have opposie signs. Alhough reacive power does no consume any energy, i reduces he power facor below uniy. When he power facor is below uniy, for he same power ransfer P he curren required becomes larger and he power losses in he circui becomes sill larger (power loss ). his is why supply auhoriies encourage he indusries o improve heir power facors o be close o uniy. 3.8 hree Phase Power As you already know, o ransmi power using single phase ac, we need wo wires. f course you may have an earh wire for proecion, bu his wire does no usually carry any curren. You may also have seen ha disribuion lines usually have 4 wires. Wha are hese 4 wires? is he hree phase wires and he neural wire Balanced hree Phase Now why do you need hree phase wires? his is in order o make our ransmission efficien. You are already aware ha if we have hree equal forces a angles of o o each oher, if we use he riangular law we ge an equilaeral riangle in which he resulan force is zero. A similar hing happens in hree phase excep ha we have phasors insead of vecors. ha is, we have 3 volages (or currens) which are equal in magniude bu differing in phase angle by o from each oher. Universiy of Morauwa - JL/Sep8 8

9 EE Elecrical Engineering ac heory curren phase Y phase B phase 5-5 angle (rad) π/ π 3π/ π - hree phase waveforms B o o o Y f he phase is aken as reference and drawn verical (reference need no always be drawn horizonally), hen he hree phases would have he phasor diagram shown. f v () = m cos ω, hen v Y () = m cos (ω π/3) and v B () = m cos (ω +π/3) can easily be seen ha he addiion of hese hree waveforms a any insan is zero. his is no only rue for balanced hree phase volages, bu for balanced hree phase currens. hus if we have a balanced hree phase sysem of currens, hen heir addiion would become zero, and no neural wire would be required. However in pracice, hree phase currens are never perfecly balanced and he neural wire would carry he unbalance. f, Y, and B are he r.m.s.volages of he hree phases wih respec o he neural, hen hey are normally called he phase volages. he volage beween any wo phase wires (or lines) is called he line volage. As can be seen from he diagram, he magniude of he line volage is 3 imes he phase volage for a balanced sysem. n a hree phase sysem, i is always he r.m.s. line volage ha is specified, unless oherwise specified. For example in he domesic supply, in single phase we say ha he volage is 3 and in hree phase we say ha he volage is 4 (~ 3 3). f p is he phase volage magniude, = p, Y = p -π/3, B = p π/3. he Line volage magniude will be 3 p = L and he line volages will be Y = L π/6, YB = L (-π/3+π/6) = L -π/, B = L (π/3+π/6)= L 5π/6. Balanced loads may be conneced eiher in sar or dela as shown. A balanced sar conneced load can be convered o an equivalen dela and he oher way round as well. n such a case i can be shown ha Z D = 3 Z S Z S Z D Z S Z S Z D Z D can also be shown ha he oal hree phase power in a sar conneced load, or dela conneced load can be expressed in erms of he line quaniies as P = 3 L L cos where cos is he power facor of he load, L is he line volage and L is he line curren. he reacive power in a hree phase circui may be similarly defined as Q = 3 L L sin and he apparen power in a hree phase circui may be defined as P S = 3 L L Noe: n boh he single phase circui as well as he hree phase circui, S = P + Q and an = Q/P Q S Since he magniudes of he hree phases are equal (for boh volage and curren) in a balanced hree phase sysem, i is sufficien o calculae he quaniies for one of he phases and obain he ohers by symmery. Universiy of Morauwa - JL/Sep8 9

10 EE Elecrical Engineering ac heory Example A 3 phase, kw balanced load a a lagging power facor of.8 is supplied from a 3 phase, 4, 5 Hz supply. Deermine he line curren in magniude and phase relaive o he supply volage, he apparen power and he reacive power drawn. P = kw, L = 4, cos =.8 lag, P = 3 L L cos, i.e. = 3 4 L.8 L =.65 A, = cos - (.8) = o. i.e. L = o A. Apparen power = 3 L L = = 5 = 5 ka. eacive power = 3 L L sin = sin (36.87 o ) = 9 = 9 k var 3.8. Unbalanced hree phase sysems Unbalanced hree phase sysems would consis of eiher sar conneced or dela conneced sources which need no be balanced and an unbalanced sar conneced or dela conneced load. Such sysems canno be solved using one phase of he sysem, bu could be solved using he normal nework heorems. 3.9 Power facor correcion Since mos pracical loads are inducive in naure (as hey have coils raher han capaciors), he power facor can be improved by using capaciors across he load. hey may be eiher conneced in sar or dela. A pure capacior affecs only he reacive power and no he acive power. Example n he above example a dela conneced bank of capaciors are used o improve he power facor o.95 lag. Wha should be he value of he required capaciors. New power facor =.95 lag. new power facor angle = cos - (.95) = 8.9 o he acive power of he load is unchanged by he capacior bank, P = W new reacive power = P an = an (8.9 o ) = 3944 = k var reacive power supplied from capacior bank = = 556 = 5.56 k var. Each of he 3 capaciors in he dela conneced bank mus supply = 5.56/3 =.685 k var. Since hey are conneced in dela, each would ge a volage of 4. L ω = 4 π 5 = 685 = = 33.5 µf. f he capaciors had been conneced in sar, hen he reacive power would sill have been he same, bu he volage across each would have been 4/ 3 and he value of each capacior would be p ' ω = (4/ 3) ' π 5 = 685 ' =.6-6 =.6 µf. Universiy of Morauwa - JL/Sep8 3