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

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1 ECE 0 Energy Converion and Power Electronic Dr. Tim Hogan Chapter : Introduction and Three Phae Power. eview of Baic Circuit Analyi Definition: Node - Electrical junction between two or more device. Loop - Cloed path formed by tracing through an ordered equence of node without paing through any node more than once. Element Contraint: Ohm Law Capacitor Equation Inductor Equation v i dv i C dt di v L dt Connection Contraint: Kirchhoff Current Law - The algebraic um of current entering a node i zero at every intant in time. k Node i 0 (.) Kirchhoff Voltage Law - The algebraic Sum of all voltage around a loop i zero at every intant in time. v 0 (.) Loop k Paive Sign Convention: Whenever the reference direction of current into a two terminal device i in the direction of the reference voltage drop acro the device, then the power aborbed (or diipated) i poitive. i v _ Figure. Circuit element and paive ign convention. p ( t ) i ( t ) v ( t ) (.) -

2 When the above convention i ued, p(t) > 0 for aborbed power, and p(t) < 0 for delivered power. Time Varying Signal Although a number of exception can be found throughout the world, the predominance of electric power follow 60Hz or 50Hz frequencie. North America, part of Japan, and hip at ea ue 60Hz while mot of the ret of the world ue 50Hz. The hitorical reaon for thee two frequencie tem from the difference in lighting (filament in vacuum or filament in a ga atmophere). The lower frequencie caued an annoying flicker for light having filament in a ga atmophere, and thu a higher frequency wa adopted in part of the world that initially ued uch lighting. While not trictly adhered to within your textbook and thee note, an attempt to ue the following convention ha been made. Scalar time varying ignal (example): v v in ωt (V) max ( ) () t v co( ωt ) v max (V) v 85 in( 77 t) (kv) i t i co 00π t () ( ) max (A) Spatial vector (bold or arrow overhead or line overhead): F or F or B If neceary, unit vector will be ued, (for example): Thee unit vector hould not be confued with phaor below B B x ˆ B y ˆ B z ˆ x y z Phaor (phaor repreentation of a time varying ignal): A phaor can be repreented a a complex number with real and imaginary component uch that a phaor of magnitude (or length) that i at a phae angle of θ with repect to ˆ jθ x jy e co θ jin θ the x-axi (the eal axi) can be written a ( ) ( ) θ where Euler formula e j co( θ ) j in( θ ) wa ued for the lat repreentation. Note that i the magnitude and θ the phae of the phaor, and y θ arctan. The phaor can be hown graphically in Figure. x x y and -

3 Imaginary y ωt θ x eal Figure. Phaor of magnitude and phae θ. For a inuoidal function ( t ) V ( ω t θ ) v co, the phaor repreentation i V ˆ jθ Ve V /θ If the phaor i rotating counterclockwie about the origin at a rate of ω radian per j t econd, then we multiply the phaor by e ω j uch that ω t j θ j ω t j ( ω t Vˆ e Ve e Ae θ ) and ω ( ω θ ) uing Euler formula Ve ˆ j t j t Ve V co( ω t θ ) jv in( ω t θ ). Then we ee that j t v t V co ω t θ i the real part of V ˆe ω, or () ( ) v ω ( t ) { V ˆ j t e } V co( ω t θ ) e. In the above decription of the phaor, the peak value i ued, however we will ue the MS value of V a decribed in (.6) intead of the peak value. Thi implifie many of the calculation, particularly thoe aociated with power a hown below. Phaor will be repreented with a hat (or caret) above the variable. Impedance i undertood to be a complex quantity in general, and the hat (or caret ^) i left off the impedance variable Z jx where i the reitance, and X i the reactance component repectively. For a circuit element uch a the one hown in Figure repreenting a load in the circuit with i(t) a the intantaneou value of current through the load and v(t) i the intantaneou value of the voltage acro the load. In quai-teady tate condition, the current and voltage are both inuoidal, with correponding π amplitude of V max, I max, and initial phae, φ v and φ i, and the ame frequency ω π f T v ( t ) V max co( ω t φv ) (.4) i ( t ) I max co( ω t φ i ) (.5) The root-mean-quared (MS) value of the voltage and current are then: V max T V [ Vmax co( t )] dt T ω φ v (.6) o -

4 I max T I [ Imax co( t )] dt T ω φ i (.7) o Phaor repreentation for the above ignal ue the MS value a The intantaneou power i the product of voltage time current or: p V ˆ V /φ v and I ˆ I /φ i. () t v () t i () t V max co( ω t φv ) I max co( ωt φi ) V maximax co( ωt φv ) co( ωt φi ) VI co( ω t φ ) co( ωt φ ) v i VI [ co( φ v φ i ) co( ω t φ v φ i )] (.8) The average value i found by integrating over a period of time and then dividing the reult by that ame time interval. The firt term in (.8) i independent of time, while the econd term varie from - to ymmetrically about zero. Becaue it i ymmetric about zero, integration over an integer number of cycle (or period) give a value of zero for the econd term in (.8), and the average power which we define a the real power with unit of watt (W) i: ( φ φ ) P VI co (.9) Thi how the power i not only proportional to the MS value of voltage and current, but alo proportional to co ( φv φi ). The coine of thi angle i defined a the diplacement factor, DF. In more general term for periodic, but not necearily inuoidal ignal, the power factor i defined a: v i P pf (.0) VI For inuoidal ignal, the power factor equal the diplacement factor, or pf ( φ φ ) co (.) For comparion, the voltage, current, and power for variou angle between voltage and current are hown below: v i Voltage (V), Current (A), Power (W) For leading power factor: # of Period ( t ) 5 co( ωt) ( t ) co( ωt) v i V I P VI co v i v i Imaginary max max ( φ φ ) co( φ φ ) 5 I V eal (W) - 4

5 Voltage (V), Current (A), Power (W) # of Period ( t ) 5 co( ωt) ( t ) co( t 0º ) v i ω V I P VI co v i v i Imaginary I max max ( φ φ ) co( φ φ ) 4. V eal (W) Voltage (V), Current (A), Power (W) # of Period ( t ) 5 co( ωt) ( t ) co( t 60º ) v i ω V I P VI co v i v i Imaginary max max ( φ φ ) co( φ φ ). 5 I V eal (W) Voltage (V), Current (A), Power (W) # of Period ( t ) 5 co( ωt) ( t ) co( t 90º ) v i ω V I P VI co v i v i Imaginary max max ( φ φ ) co( φ φ ) 0 I V eal (W) Voltage (V), Current (A), Power (W) # of Period ( t ) 5 co( ωt) ( t ) co( t 0º ) v i ω V I P VI co v i v i Imaginary max max ( φ φ ) co( φ φ ). 5 I V eal (W) - 5

6 Voltage (V), Current (A), Power (W) # of Period ( t ) 5 co( ωt) ( t ) co( t 50º ) v i ω V I P VI co v i v i Imaginary max max ( φ φ ) co( φ φ ) 4. I V eal (W) Voltage (V), Current (A), Power (W) # of Period ( t ) 5 co( ωt) ( t ) co( t 80º ) v i ω V I P VI co v i v i Imaginary max max ( φ φ ) co( φ φ ) 5 I V eal (W) Voltage (V), Current (A), Power (W) For lagging power factor: # of Period ( t ) 5 co( ωt) ( t ) co( ωt) v i V I P VI co v i v i Imaginary max max ( φ φ ) co( φ φ ) 5 I V eal (W) - 6

7 Voltage (V), Current (A), Power (W) Voltage (V), Current (A), Power (W) Voltage (V), Current (A), Power (W) Voltage (V), Current (A), Power (W) # of Period # of Period # of Period # of Period ( t ) 5 co( ωt) ( t ) co( t 0º ) v i ω V I P VI co v i v i max max ( φ φ ) co( φ φ ) 4. Imaginary eal V I ( t ) 5 co( ωt) ( t ) co( t 60º ) v i ω V I P VI co v i v i max max ( φ φ ) co( φ φ ). 5 Imaginary eal I V ( t ) 5 co( ωt) ( t ) co( t 90º ) v i ω V I P VI co v i v i max max ( φ φ ) co( φ φ ) 0 Imaginary eal V I ( t ) 5 co( ωt) ( t ) co( t 0º ) v i ω V I P VI co v i v i (W) (W) (W) max max ( φ φ ) co( φ φ ). 5 Imaginary eal I V (W) - 7

8 Voltage (V), Current (A), Power (W) # of Period ( t ) 5 co( ωt) ( t ) co( t 50º ) v i ω V I P VI co v i v i max max ( φ φ ) co( φ φ ) 4. Imaginary eal V I (W) Voltage (V), Current (A), Power (W) # of Period ( t ) 5 co( ωt) ( t ) co( t 80º ) v i ω V I P VI co v i v i Imaginary max max ( φ φ ) co( φ φ ) 5 I V eal (W) We call the power factor leading or lagging, depending on whether the current of the load lead or lag the voltage acro it. We know that current doe not change intantly through an inductor, o it i clear that for a reitive - inductive (L) load, the power factor i lagging. Likewie the voltage can not change intantly acro a capacitor, therefore for a reitive capacitive (C) load, the power factor i leading (the current change before the voltage can). Alo, for a purely inductive or capacitive load the power factor i 0, while for a purely reitive load it i. The product of the MS value of voltage and current at a load i the apparent power, S having unit of volt-ampere (VA): S VI (.) The reactive power i Q with unit of volt-ampere reactive (VA reactive, or VAr): ( φ φ ) Q VI in (.) The reactive power repreent the energy ocillating in and out of an inductor or capacitor. Since the energy ocillation in an inductor i 80º out of phae with the energy ocillating in an capacitor, the reactive power of thee two have oppoite ign with the convention that it i poitive for the inductor and negative for the a capacitor. Uing phaor, the complex apparent power, Ŝ i: or v i ˆ ˆ * S ˆ V I V/φ v I/-φ i (.4) - 8

9 S ˆ P jq (.5) A an example, conider the following voltage and current for a given load: π v () t 0 in 77t (V) (.6) 6 π i () t 5 in 77t (A) (.7) 4 then S VI (W), while the power factor i co π π pf leading. Alo, the complex apparent power i: ˆ ˆ * S ˆ V I 0/π/6 5/-π/4 600/- π/ (W) j55. (VAr) (.8) It hould alo be noted that when the angle are repreented in radian, care mut be taken to aure your calculator i in radian mode. Often we repreent the argument of the ine or coine term in mixed unit. For example we might write co(77t 0º). The firt term (77t) ha unit of radian, while the econd (0º) ha unit of degree, and one of thee mut be converted before making calculation with your calculator or computer either in radian mode or in degree mode. A a phaor the complex apparent power can be hown on a complex plane for the cae of leading and lagging power factor a hown in Figure. Imaginary Imaginary I V eal V eal I Imaginary Imaginary P S Q eal S P Q eal (a) Leading power factor (b) Lagging power factor Figure. Phaor of magnitude and phae θ. - 9

10 ecall that the leading power factor correpond to a reitive-capacitive load. We ee from Figure (a), thi alo correpond to a negative value for Q. Similarly, lagging power factor ha the current lagging the voltage correponding to an inductive-reitive load, and Figure (b) how thi correpond to a poitive value for Q. To ummarize, we can ue the following table: Equation P VI co φ φ eal Power ( ) VI pf S pf Q VI in φ φ eactive Power ( ) Apparent Power S S v v pf pf S VI ˆ ˆ * S ˆ V I V/φ v I/-φ i S ˆ P jq S P Q i i (for lagging) (for leading) Unit Watt Volt-Ampere-eactive Volt-Ampere Type of Load eactive Power Power Factor Inductive Q > 0 lagging Capacitive Q < 0 leading eitive Q 0 From the interdependence of the four quantitie, S, P, Q, pf, if we know any two of thee quantitie, the other two can be determined. For example, if S 00 (kva), and pf 0.8 leading, then: P S pf 800 (kw) Q S in Q pf ( φ v φ i ) in[ arcco( 0.8) ] S in( φ φ ) It i important to notice that Q < 0, uch that ( φ ) v i 60 (kva), or v φ i then in i a negative quantity. Thi can be een when it i undertood that there are two poible anwer for the arcco(0.8), that i co(6.87º) 0.8, and co(-6.87º) 0.8 o to obtain a Q < 0 we ue (φ v φ i ) 6.87º. Generally, in ytem that contain more than one load (or ource), the real and reactive power can be found by adding individual contribution, but thi i not the cae with the apparent power. That i P total Qtotal S total i P i Q (.9) i S i i i - 0

11 For the above example, if the load voltage i V L 000 (V), then the load current would be I L S/V L [00 0 (VA)]/[ 0 (V)] 50 (A). If we ue the load voltage a the reference, then: V ˆ 000 /0º (V) I ˆ 50 /φi 50 /6.87º (A) ˆ ˆ * S ˆ V I [000 /0º][50 / 6.87º] P jq 80 0 (W) j60 0 (VAr). Three Phae Balanced Sytem Three-phae ytem offer ignificant advantage over ingle phae ytem: for the ame power and voltage there i le copper in the winding, and the total power aborbed remain contant rather than ocillating about an average value. For a three phae ytem coniting of three current ource having the ame amplitude and frequency, but with phae differing by 0º a: i ( t ) I in( ω t φ ) π i () t I in ωt φ π i () t I in ωt φ (.0) If thee are connected a hown in Figure 4, then at node n or n, the current add to zero, and the neutral line n-n (dahed) i not needed. i π π () t i () t i () t I in( ω t φ ) in ω t φ in ω t φ 0 i n' n i i Figure 4. Balanced three phae Y-connected ytem with zero neutral current. If intead we had three voltage ource Y-connected a in Figure 5 with the following value -

12 v ( t ) V in( ω t φ ) π v () t V in ωt φ (.) π v () t V in ωt φ then, the current through each of the three load (auming the load are equal), would have equal magnitude, but each current would have a phae that i hifted by an equal amount with repect to the voltage, v (t), v (t), v (t). i a v v n' v i b n i c Figure 5. Balanced voltage fed three phae Y-connected ytem with zero neutral current. With equal impedance for the load, then i i a b () t in( ω t φ θ ) () t V Z V π in ωt φ θ Z V π i c() t in ωt φ θ Z (.) and again the current um to zero at node n or n, and for thi balanced three phae ytem, the neutral wire (dahed) i not required. In comparion to a ingle phae ytem, where two wire are required per phae, the three phae ytem deliver three time the power, and require only three tranmiion wire total. Thi i a ignificant advantage conidering the hundred of mile of wire needed for power tranmiion. Y and Δ Connection The load in the previou two figure, a well a in Figure 6 are connected in a Y or tar configuration. If the load of Figure 6 i for a balanced Y ytem, then the voltage between each phae and the neutral are: V ˆ n V /φ, V ˆ V n /φ π/, and V ˆ V n /φ π/. Kirchhoff voltage law (KVL) tate that the um of voltage around a cloed loop equal zero. Thi i alo the cae here however the voltage are complex number or phaor, and a uch mut be -

13 added a vector. The phae φ can be any value, but the relative poition of the phae to neutral phaor mut be 0º with repect to each other a hown in Figure 7. V n V n V n V n Figure 6. Y-connected load with voltage relative to neutral identified. By KVL: V ˆ V ˆ V ˆ 0, or V ˆ V ˆ V ˆ a hown in Figure 7. n n n n V -V n V n V n V -V n V n -V n V Figure 7. Voltage phaor of the Y-connected load hown in Figure 6. We could alo ue the phaor repreentation V ˆ n V /φ, V ˆ V n /φ π/, and V ˆ n V /φ π/ to determine the line-to-line voltage a V ˆ V ˆ V ˆ V /φ π/6 (.) n n Thi how the MS value of the line-to-line voltage, V l-l, at a Y load i time the line-toneutral or phae voltage, V ln. In the Y connection, the phae current i equal to the line current, and the power upplied to the ytem i three time the power upplied to each phae, ince the voltage and current amplitude and phae difference between them are the ame in all three phae. If the pf co φ φ, then the total power to the ytem i: power factor in one phae i ( ) v i -

14 Sˆ φ P φ V ˆ I ˆ * n n V jq l l I l φ co ( φ φ ) j V I in( φ φ ) v i l l l v i (.4) Similarly, for a connection of the load in the Δ configuration (a in Figure 8), the phae voltage i equal to the line voltage however; the phae current are not equal to the line current for the Δ configuration. If the phae current are I ˆ I /φ, I ˆ I /φ π/, and I I ˆ /φ π/ then uing Kirchhoff current law (KCL) the current of line, a hown in Figure 8 i: I ˆ I ˆ I ˆ I /φ - π/6 (.5) Thu for the Δ configuration, the line current i time the I Δ current. I I -I I I -I I I I I I I I I -I Figure 8. Δ connected load, line, and phae current. To calculate the power in the three-phae Δ connected load: Sˆ φ P φ V ˆ I ˆ V jq * l l I l φ co ( φ φ ) j V I in( φ φ ) v i l l l v i (.6) which i the ame value a for the Y connected load. For a balanced ytem, the load of the three phae are equal. Alo, a Δ configured load, can be replaced with a Y configured load (and via vera) if: Z Y Z Δ (.7) - 4

15 Under thee condition, the two load are inditinguihable by the power tranmiion line. You might recall the Δ to Y tranformation for reitor circuit can be remembered by overlaying the Δ and Y configuration uch a in Figure 9. A B C C A B Figure 9. Δ to Y or Y to Δ reitor network tranformation. A B C B C A (.8) Equation (.8) give a imilar reult to that of (.7) when A B C and.. Calculation in Three-Phae Sytem Calculation of quantitie like current, voltage, and power in three-phae ytem can be implified by the following procedure:. tranform the Δ circuit to Y,. connect a neutral conductor,. olve one of the three -phae ytem 4. convert the reult back to the Δ ytem.. Example For the -phae ytem in Figure 0 calculate the line-to-line voltage, real power, and power factor at the load. To olve thi by the procedure outlined above, firt conider only one phae a hown in Figure. - 5

16 j (Ω) v 0 (V) 7 j5 (Ω) n' n v v Figure 0. Three phae ytem with Y connected load, and line impedance. j (Ω) I v 0 (V) 7 j5 (Ω) n' Figure. One phae of the three phae ytem hown in Figure 0. n For the one-phae in Figure, 0 I ˆ.0 /-40.6º (A) j ( 7 j 5 ) V ˆ ˆ n I Z L.0 /-40.6º (7j5) /-5.º (V) S *, V ˆ I ˆ L φ L (/-5.º)(.0/40.6º) 458./5.5º.87 0 j P.87 (kw), (kvar) L, φ Q L, φ pf co(-5.º - (-40.6º)) 0.84 lagging For the three-phae ytem of Figure 0 the load voltage (line-to-line), the real, and reactive power are: V L, l l 94 (V) P L, φ.56 (kw) Q L, φ.544 (kvar) - 6

17 .. Example For the Y to Δ three-phae ytem in Figure, calculate the power factor and the real power at the load, a well a the phae voltage and current. The ource voltage i 400 (V) line-to-line. j (Ω) v 8 j6 (Ω) n' v v Figure. Δ connected load. Firt convert the load to an equivalent Y connected load, then work with one phae of the ytem. 400 The line to neutral voltage of the ource i V ln (V). j (Ω) (V) 6 j (Ω) n' n Figure. Equivalent Y connected load. j (Ω) I L v (V) 6 j (Ω) n' n Figure 4. One phae of the Y connected load. - 7

18 Iˆ L 4.44 /-6.6º (A) j ( 6 j ) ( 6 ) 7. 8 V ˆ I ˆ j L The power factor at the load i: L /-8.º (V) pf ( φ φ ) co( 8.º 6.6º ) co v i lagging Converting back to a Δ connected load give: I φ I L (A) V l l (V) At the load the power i: P,φ V I pf (kW) L l l L.. Example Two three-phae load are connected a hown in. Load draw from the ytem P L 500 (kw) at 0.8 pf lagging, while the total load i S T 000 (kva) at 0.95 pf lagging. What i the power factor of load? Power Sytem Load Load Figure 5. Two three-phae load connected to the ame power ource. - 8

19 For the total load we can add the real and reactive power for each of the two load (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 information we have for the total load we can write the following: PT ST pft 950 (kw) Q in co T S T [ ( )] 5 (kvar) The reactive power, Q T, i poitive ince the power factor i lagging. For the load L, P L 500 (kw), pf 0.8 lagging, thu: SL Q S P L L L 65 (kva) 75 (kva) Again, Q L i poitive ince the power factor i lagging. Thi lead to: P Q P T P Q Q L L L T L 450 (kw) (kvar) and pf P L L S L leading. Chapter Note A inuoidal ignal can be decribed uniquely by:. Time dependent form a for example: v ( t ) in( π ft φ ) 5 (V). by a time dependent graph of the ignal. a a phaor along with the aociated frequency of the phaor one of thee decription i enough to produce the other two. It i the phae difference that i important in power calculation, not the phae. The phae i arbitrary depending on the defined time (t 0). We need the phae to olve circuit problem after we take one quantity (ome voltage or current) a a reference. For that reference quantity we aign an arbitrary phae (often zero). In both three-phae and one-phae ytem the total real power i the um of the real power from the individual load. Likewie the total reactive power i the um of the reactive power of the individual load. Thi i not the cae for the apparent power or the power factor. v - 9

20 Of the four quantitie: real power, reactive power, apparent power, and power factor, any two decribe a load adequately. The other two quantitie can be calculated from the two given. To calculate real, reactive, and apparent power when uing equation (.9), (.), and (.) we mut ue abolute value, not complex value for the current and voltage. To calculate the complex power uing equation (.4) we do ue complex current and voltage and find directly both the real and reactive power (a the real and imaginary component repectively). When olving a circuit to calculate current and voltage, ue complex impedance, current and voltage. - 0

21 ECE 0 Energy Converion and Power Electronic Dr. Tim Hogan Chapter : Magnetic Circuit and Material Chapter Objective In thi chapter you will learn the following: How Maxwell equation can be implified to olve imple practical magnetic problem The concept of aturation and hyterei of magnetic material The characteritic of permanent magnet and how they can be ued to olve imple problem How Faraday law can be ued in imple winding and magnetic circuit Power lo mechanim in magnetic material How force and torque i developed in magnetic field. Ampere Law and Magnetic Quantitie Ampere experiment i illutrated in Figure where there i a force on a mall current element I l when it i placed a ditance, r, from a very long conductor carrying current I and that force i quantified a: μi F I l (N) (.) π r I r I F Conductor Current Element of Length l Figure. Ampere experiment of force between current carrying wire. The magnetic flux denity, B, i defined a the firt portion of equation (.) uch that: F BI l (N) (.) -

22 From (.) and (.) we ee the magnetic flux denity around conductor i proportional to the current through conductor, I, and inverely proportional to the ditance from conductor. Looking at the unit and contant handout given in cla, or from (.) the unit of, B, are een a N, A m N thu µ, called permeability, ha unit of. More commonly, the relative permeability of a given A 9 N material i given where μ μrμ0 and μ0 400π 0. Since a Newton-meter i a Joule, and A N N m J V A V a Joule i a Watt-econd:. Thi how B i a A m A m A m A m m per meter quared quantity, and the (V ) unit repreent the magnetic flux and i given unit of Weber (Wb). Thi flux can be found by integrating the normal component of B over the area of a given urface: φ B nd ˆ (.) S The magnetic field intenity i related to the magnetic flux denity by the permeability of the media in which the magnetic flux exit. B H (.4) μ B I A For the ytem in Figure, H and have unit of. If there were multiple μ π r m conductor in place of conductor, for example in a coil, then the unit would be ampere-turn per meter. A line integration of H over a cloed circular path give the current encloed by that path, or for the ytem in Figure : πr I H H dl dl I (.5) C 0 πr again, if the ytem contained multiple conductor within the encloed path, the reult would give ampere-turn. Equation (.5) i Ampere circuital law. An alternative approach a decribed in your textbook i to begin with Maxwell equation which include Ampere circuital law in a more general form a hown in Table I below: Table I. Maxwell equation. Name Point Form Integral Form Faraday Law B d E E dl B nd ˆ t C dt S Ampere Law Modified by Maxwell D D H J H dl t J nˆ d C S t Gau Law D ρ D nˆ d ρ dv S V Gau Law for Magnetim B 0 B nˆ d 0 S -

23 where C repreent the integral over a cloed path, S repreent the integral over the urface of a cloed volume of pace, nˆ i a urface normal vector, E i the patial vector of electric field, B i the patial vector of magnetic flux denity, J i the patial vector of the electric current denity, D i the diplacement charge vector, ρ i the electric charge denity, i a volume integral i the V curl and the divergence of the vector being acted upon. Some aumption commonly ued in electromechanical energy converion include a low enough D frequency, that the diplacement current,, can be neglected, and the aumption of homogeneou t and iotropic media ued in the magnetic circuit. Under thee aumption, Ampere circuital law i modified to remove the diplacement current component uch that which for the ytem of Figure reduce to equation (.5). C S H dl J nˆ d (.6). Magnetic Circuit From Ampere circuital law, C S H dl J nˆ d, we ee the magnetic field intenity around a cloed contour i a reult of the total electric current denity paing through any urface linking that S contour. Gau Law for magnetim, B nˆ d 0 tate that there are no magnetic monopole that i to ay there for a cloed urface there i a much magnetic field denity leaving that cloed urface a there i entering the cloed urface. If the integration i for an area, but not a cloed urface area, then we obtain the flux or φ B nd ˆ. S The permeability of free pace i, while the permeability of magnetic teel i a few hundred thouand. Magnetic flux can be confined to the tructure or path formed by high permeability material. In thi way, magnetic circuit can be formed uch a the one hown in Figure. Figure. Simple magnetic circuit []. -

24 The driving force for the magnetic field i the magnetomotive force (mmf), F, which equal the ampere-turn product F N i (.7) The analyi of a magnetic circuit i imilar to the analyi of an electric circuit, and an analogy can be made for the individual variable a hown in Table II. I Table II. Comparion of electric and magnetic circuit. Electric Circuit Magnetic Circuit b Electrically Conductive Material φ b High Permeability Material a a V r V r I I c c applied battery voltage V Driving Force applied ampere-turn F or current driving force electric reitance V I epone or flux driving force magnetic reluctance F φ Impedance Impedance i ued to indicate the impediment to the driving force in etablihing a repone. l l reitance reluctance σ A μa where l πr, σ electrical conductivity, A cro-ectional area where l πr, μ permeability, A croectional area I Equivalent Circuit φ V F V I F φ - 4

25 Electric Circuit Magnetic Circuit or Electric Field Intenity V V E (V/m) l πr Field or Magnetic Field Intenity F F H (A-t/m) l πr E dl V H dl F Potential Electric Potential Difference b V b I I l V ab E dl dl lab lab Iab F a l a l l σa ab Magnetic Potential Difference b F φ φ l H dl lab lab l l l l μa a ab φ ab J I A Current Denity V El A A l σa ( ) σe Flow Denitie φ B A Flux Denity F Hl A A l μa μh An example magnetic circuit i hown in Figure, below. Figure. Simple magnetic circuit with an air gap []. For thi circuit, we will aume: the magnetic flux denity i uniform throughout the magnetic core cro-ectional area and i perpendicular to the cro-ectional area the magnetic flux remain within the core and the air gap defined by the cro-ectional area of the core and the length of the gap (no leaking of field, no fringe field at the gap). Then with A c A g φ B d A B c A c B g A g (.8) A c - 5

26 H H B B g c g c μ μ (.9) Since, F Hl g g g c c c g c c l B l B g H l H μ μ F (.0) or ( ) g c g g c c c A g A l F φ μ μ φ (.) Thu the magnetic circuit hown in Figure can be repreented a F φ c g g c g c i N F φ Figure 4. Equivalent circuit for the magnetic circuit in Figure. Thi concept i helpful for more complex configuration uch a hown in Figure

27 I Area A Area A Area A Area A I Area A g g Area A Area A Figure 5. Magnetic circuit with variou cro-ectional area and two coil. Uing the length defined in Figure 6, and paying attention to the direction of the magnetomotive force from each of the coil by ue of the right hand rule, the equivalent circuit can be drawn a een in Figure 7. Area A l Area A I I Area A Area A g g l l Area A Figure 6. Length for each ection of the magnetic circuit of Figure 5. Then, and ( φ φ ) g F N I (.) l φ ( ) F N I φ φ (.) g l g l For a ytem with the dimenion, number of turn, current, and permeability known, then equation (.) and (.) give two equation with two unknown uch that φ and φ can be found. - 7

28 l φ φ F l F g l g Figure 7. Equivalent circuit for the configuration of Figure 5. Auming the gap are air gap, the value of each reluctance can be found a: l l μa g g 0 μ A l l μa l l μa g g 0 μ A Then we can find the magnetic flux denity for each gap a: B B g g φ A φ A (.4) Note that the permeability of the core material can be much larger than the permeability of the gap uch that the total reluctance i dominated by the reluctance of the gap.. Inductance From circuit theory we recall that the voltage acro an inductor i proportional to the time rate of change of the current through the inductor. dil () ( t) v L t L (.5) dt and while the power of an inductor can be poitive or negative, the energy i alway poitive a w (.6) () t L L i L d In Table I, Faraday law i E dl B nd ˆ or the electric field intenity around a C dt S cloed contour C i equal to the time rate of change of the magnetic flux linking that contour. Integrating over the cloed contour of the coil itelf give u the negative of the voltage at the terminal of the coil. On the right ide of Faraday law we then mut integrate over the urface of the full coil, thu including the N turn of the coil. Then Faraday law give - 8

29 v L () t ( t ) d φ N (.7) dt Comparing equation (.5) and (.7) give dφ L N (.8) di For linear inductor, the flux φ i directly proportional to current, i, for all value uch that Nφ L (.9) i with unit of (Weber-turn per ampere), or Henry (H). The flux linkage, λ, i defined a λ Nφ F Ni and combining thi with the relationhip between flux and total circuit reluctance φ tot tot along with (.9) give N L (.0) Thu inductance can be increaed by increaing the number of turn, uing a metal core with a higher permeability, reducing the length of the metal core, and by increaing the cro-ectional area of the metal core. Thi i information that i not readily een by either the circuit law for inductor, or through Faraday equation alone. Mutual Inductance For a magnetic circuit containing two coil and an air gap uch a in Figure 8, with each coil wound uch that the flux i additive, then the total magnetomotive force i given by the um of contribution from the two coil a F N i N (.) tot i Area A c l c i i g λ λ φ Figure 8. Mutual inductance magnetic circuit. The equivalent circuit i hown in Figure

30 F g lc F φ Figure 9. Equivalent circuit for Figure 8. If the permeability of the core i large uch that g l c << and the cro-ectional area of the gap i aumed equal to the cro-ectional area of the core (A g A c ), then ( ) g A i N i N c 0 μ φ (.) The flux linkage, λ, in Figure 8 i φ λ N, or 0 0 i L i L i g A N N i g A N N c c μ μ φ λ (.) where g A N L c 0 μ (.4) i the elf-inductance of coil and L i i the flux linkage of coil due to it own current i. The mutual inductance between coil and i g A N N L c 0 μ (.5) and L i i the flux linkage of coil due to current i in the other coil. Similarly, for coil 0 0 i L i L i g A N i g A N N N c c μ μ φ λ (.6) where L L i the mutual inductance and L i the elf-inductance of coil. g A N L c 0 μ (.7) - 0

31 .4 Magnetic Material Propertie A implification we have ued i that the permeability of a given material i contant for different applied magnetic field. Thi i true for air, but not for magnetic material. Material that have a relatively large permeability are ferromagnetic material in which the magnetic moment of the atom can align in the ame direction within domain of the material when and external field i applied. A more of thee domain align, aturation i reached when there i no further increae in flux denity of that of free pace for further increae in the magnetizing force. Thi lead to a changing permeability of the material and a nonlinear B v. H relationhip a hown in Figure 0. B H Figure 0. Nonlinear B v H normal magnetization curve. When the field intenity i increaed to ome value and i then decreaed, it doe not follow the curve hown in Figure 0, but exhibit hyterei a hown by the abcdea loop in Figure. The deviation from the normal magnetization curve i caued by ome of the domain remaining oriented in the direction of the originally applied field. The value of B that remain after the field intenity H i removed i called reidual flux denity. It value varie with the extent to which the material i magnetized. The maximum poible value of the reidual flux denity i called retentivity and reult whenever value of H are ued that caue complete aturation. When the applied magnetic field i cyclically applied o a to form the hyterei loop uch a abcdea in Figure, the field intenity required to reduce the reidual flux denity to zero i called the coercive force. The maximum value of the coercive force i called the coercivity. The delayed reorientation of the domain lead to the hyterei loop. The unit of BH i ampere newton N J unit of HB (.8) meter ampere- meter m m -

32 or an energy denity. Coercivity B (Wb/m ) a f etentivity eidual flux c b O H (A t/m) e d Coercive force Figure. Hyterei loop. The normal magnetization curve i in bold. B f a b c O H e d Figure. Energy relationhip for hyterei loop per half-cycle. -

33 The full haded area in Figure outlined by eafe repreent the energy tored in the magnetic field during the poitive half cycle of H. The hatched area eabe repreent the hyterei lo per half cycle. Thi energy i what i required to move around the magnetic dipole and i diipated a heat. The energy releaed by the magnetic field during the poitive half cycle of they hyterei loop i given by the cro-hatched area outlined by bafb and i energy that i returned to the ource. The power lo due to hyterei i given by the area of the hyterei loop time the volume of the ferromagnetic material time the frequency of variation of H. Thi power lo i empirically given a n Ph kh νfbmax (.9) where n lie in the range.5 n.5 depending on the material ued, ν i the volume of the ferromagnetic material, and the value of the contant, k h, alo depend on the material ued. Some typical value for k h are: cat teel 0.05, ilicon heet teel 0.00, and permalloy In addition to the hyterei power lo, eddy current loe alo exit for time-varying magnetic fluxe. Circulating current within the ferromagnetic material follow from the induced voltage decribed by Faraday law. To reduce thee eddy current loe, thin lamination (typically 4-5 mil thick) are commonly ued where the magnetic material i compoed of tacked layer with an inulating varnih or oxide between the thin layer. An empirical equation for the eddy-current lo i P e k e ντ f B max (.0) where k e contant dependent on the material f frequency of the variation of flux Bmax maximum flux denity τ lamination thickne ν total volume of the material The total magnetic core lo i the um of the hyterei and eddy current loe. If the value of H, when increaing toward ome maximum, H max, doe not increae continuouly, but at ome point, H, decreae to H 0 then increae again to it maximum value of H max, then a minor hyterei loop i created a hown in Figure. The energy lo in one cycle include thee additional minor loop urface. -

34 B H max O H H Figure. Minor hyterei loop..5 Permanent Magnet For a ring of iron with a uniform cro-ection and hyterei curve hown in Figure 5, the magnetic field i zero when there i a nonzero flux denity, Br called the remnant flux denity. To achieve a zero flux denity, we could wind a coil around a ection of the iron, and end current through the coil to reach a field intenity of H c (the coercive field). In practice a permanent magnet operate on a minor loop a hown in Figure that can be approximated a a traight line, recoil line, uch that Br Bm Hm B r (.) Hc Magnetization curve for ome important permanent-magnet material i hown in figure.9 from your textbook and hown below in Figure 4. Example In the magnetic circuit hown with the length of the magnet, l m cm, the length of the air gap i g mm and the length of the iron i l i 0cm. For the magnet Br. (T), H c 750 (ka/m), what i the flux denity in the air gap if the iron i aumed to have an infinite permeability and the cro-ection i uniform? Since the cro-ection i uniform, and there i no current: H i [0.(m)] H g [g] H m [l m ]0 With the iron aumed to have infinite permeability, H i 0 and l m ½ l i g ½ l i - 4

35 ( ) m r c m g m c r m c g m m g l B H B g B l H B B H g B l H g H μ μ or B (B-.) 6880 B0.985 (T) Figure 4. Magnetization curve for common permanent-magnet material []. - 5

36 B (Wb/m ) B r (H m, B m ) -H c O H (A t/m) Figure 5. emnant flux and coercive field for a piece of iron..6 Torque and Force In Figure we found the energy denity in the field for the poitive half-cycle i the total area or w f B B e a HdB (.) If thi i imply approximated a a triangular area, then w f ½ BH (J/m ). The total energy, W f, would be found by multiplying thi by the volume of the core W f ( BH )( Al) ( BA) Hl φf (.) φ Mechanical energy i done by reducing the reluctance (a the armature of a relay are brought together, for example), thu Since force i given a dw m φ d (.4), the magnetic force i given by Fdx dw m - 6

37 F d φ (.5) dx and ha unit of newton. In a mechanical ytem with a force F acting on a body and moving it at velocity v, the power P mech i P mech F v (.6) For a rotating ytem with torque T, rotating a body with angular velocity ω mech : P mech T ωmech (.7) On the other hand, an electrical ource e, upplying current, i, to a load provide electrical power P elec P elec e i (.8) Since power ha to balance, if there i no change in the field energy, P P e i T ω (.9) elec mech Note It i more reaonable to olve magnetic circuit tarting from the integral form of Maxwell equation than finding equivalent reitance, voltage and current. Thi alo make it eaier to ue aturation curve and permanent magnet. Permanent magnet do not have flux denity equal to Br. Equation (.) define the relation between the variable, flux denity m B and field intenity H m in a permanent magnet. There are two type of iron loe: eddy current loe that are proportional to the quare of the frequency and the quare of the flux denity, and hyterei loe that are proportional to the frequency and to ome power n of the flux denity. mech A. E. Fitzgerald, C. Kingley, Jr., S. D. Uman, Electric Machinery, 6 th edition, McGraw-Hill, New York,

38 ECE 0 Energy Converion and Power Electronic Dr. Tim Hogan Chapter : Tranformer (Textbook Chapter ) Chapter Objective In thi chapter you will be able to: Chooe the correct rating and characteritic of a tranformer for a pecific application Calculate the loe, efficiency, and voltage regulation of a tranformer under pecific operating condition. Experimentally determine the tranformer parameter given it rating. Utilize the per unit ytem.. Introduction Tranformer do not have moving part, nor are they energy converion device, however their ability to modify the current-voltage characteritic of a given load or ource, make them invaluable component in energy converion ytem. They are utilized for power application and in low power ignal proceing ytem. One application in power tranmiion i the ue of tranformer on a tranmiion line utility pole commonly een a a cylinder with a few wire ticking out. Thee wire enter the tranformer through buhing that provide iolation between the wire and the tank. Inide the tank there i an iron core commonly made of ilicon-teel lamination that are 4 mil (0.04 ) thick. The inulation often ued i paper with the whole coil ytem immered in inulating oil. The oil increae the dielectric trength of the paper and help to tranfer heat from the core/coil aembly. An drawing of one uch ditribution tranformer i hown in Figure. in your textbook. Connection of the tranformer to the tranmiion line can take everal electrical configuration. A relatively imple connection to a.4 kv three phae tranmiion line i hown in Figure H H A B C 400 volt three-phae three-wire Δ primary ytem X X X 0 volt one-phae two-wire ervice Figure. Example configuration of a ditribution pole tranformer connection to three phae power line to provide 0 (V) ervice to your home. -

39 A B C 400 volt three-phae three-wire Δ primary ytem H H X X X 0 (V) 0 (V) 40 (V) Figure. Example configuration of a ditribution pole tranformer connection to three phae power line and provide 0 (V) and 40 (V) ervice. If a neutral line i alo part of the three phae tranmiion line (perhap between the ubtation and your home), then the connection could be made a hown in Figure. 460 N A B C N 460Y/400 volt three-phae four-wire Y with neutral H H X X X 0 (V) 0 (V) 40 (V) Figure. Ditribution tranformer connection to provide 0 (V) and 40 (V) ervice from a 460Y/400 (V) four-wire tranmiion line. -

40 For a three phae line at the ervice end, a ytem could be connected to a four wire three phae tranmiion line ource a hown in Figure 4 below. 460 N A B C N 460Y/400 volt three-phae four-wire Y with neutral primary ervice H H H H H H N A B C X 0 (V) 0 (V) 0 (V) X X X X X X X X 08 (V) 08 (V) 08 (V) three-phae four-wire Y grounded neutral econdary ervice Figure 4. Three phae to three phae ditribution tranformer connection providing a four-wire three phae ditribution of 0 (V) and 08 (V) ervice from a 460Y/400 (V) four-wire tranmiion line. Tranformer are very common place in ociety, and have had ignificant impact at many power level. They alo continue to be improved on today, with uch tudie a amorphou metal to further decreae core loe. For more undertanding, we begin with the ideal tranformer.. Ideal Tranformer Under ideal condition, the iron core ha infinite permeability and the coil have zero electrical reitance. Coil ha N turn, and coil ha N turn, and all of the magnetic flux i maintained in the iron (no flux leakage). φ i i e e F F φ Figure 5. Tranformer and equivalent magnetic circuit. -

41 The electromotive force, or emf, i repreented with the ymbol e. For an ideal tranformer, thi i the voltage at the terminal of a given coil. The flux linkage in each coil i λ N φ and λ N φ. The electromotive force induced in each coil i then the time derivative of the flux linkage or dλ d e φ N (V) (.) dt dt e dλ dφ N (V) (.) dt dt The ratio of the voltage at the terminal of coil to the voltage at coil i then e N (.) e Uing an equivalent circuit hown in Figure 5 with the magnetomotive force F for coil and F for coil we would write: F F N i N i 0 (.4) or i N (.5) i N Tranformer are often ued with a voltage ource connected to one coil and a load connected to the other coil. The dot above the coil help to indicate the voltage reference mark for the coil uch dφ that for an ideal tranformer a poitive voltage, v e N in coil (primary coil) reult in a dt dφ poitive voltage, v e N in the coil (econdary coil) a hown in Figure 6. dt N φ i i N v v N Load φ Figure 6. Circuit utilizing a tranformer. - 4

42 The circuit ymbol for the tranformer i hown in Figure 7. Since the voltage acro the coil tranform in the direct ratio of the turn and the current tranform in the invere ratio of the turn, then the impedance can alo be tranformed through the tranformer. N N i i Figure 7. Tranformer circuit ymbol. For a voltage applied to the primary coil, and a load impedance, Z L, connected to the econdary a hown in Figure 8. N N I I V V Z L Figure 8. Tranformer with a load connected to the econdary. The voltage and current from econdary to primary are related by (.) and (.5), or ˆ N Vˆ Then the impedance meaured at the terminal of the primary i Thu the following circuit are equivalent Vˆ Iˆ V (.6) N ˆ N ˆ I I (.7) N N Vˆ Z N ˆ I (.8) - 5

43 N N Z N L ( N ) N N I I V V Z L V I I V I ( ) Z L N N (a) (b) (c) Figure 9. Three equivalent circuit auming an ideal tranformer.. Non-Ideal Tranformer There are everal non-ideal propertie that we can take into account by modifying the equivalent circuit hown in Figure 7. Thee include a finite core permeability (µ c < ), leakage flux, core loe, and coil reitance. The contribution from each of thee i dicued below. Finite Core Permeability For the core of the tranformer to have a finite permeability, then the circuit in Figure 5 i modified to include the reluctance of the core. φ F F Figure 0. Tranformer hown in Figure 5, but with a core of finite permeability. Then we can write F F N i N i φ (.9) Defining a magnetomotive force that i equal to the drop acro the reluctance of the core a: F c N i, ex φ (.0) Solving for the flux give N i, ex φ (.) The rate of change in the flux i proportional to the induced voltage a - 6

44 N i, ex d d φ e N N dt dt N di, ex dt (.) Thi induced voltage i proportional to the time derivative of the current i,ex which can be N repreented by an inductor in our equivalent circuit with a value of L m. The equivalent circuit for the ideal tranformer i then modified to account for the finite permeability of the core by placing an additional inductor acro the primary coil a hown in Figure. i N N i,ex i' i e Lm e Ideal tranformer Figure. Equivalent circuit for a tranformer with finite core permeability. We can alo ee from Figure that i i, ex i which i in agreement with equation (.9) and (.0). Leakage Flux With finite core permeability, not all of the flux will be confined to the metal core, but ome will leak outide the core in the urrounding air. The influence of thi leakage flux can alo be included in the equivalent circuit, by conidering an additional reluctance aociated with the leakage flux, φ l, for coil, and φ l for coil uch that N i φ l l (.) N i φ l Thee reduce the magnetizing flux, φ m, of the core, and modify the induced voltage for the primary (coil ) and econdary (coil ) to be v v dλ dφ N m dt dt dλ dt N dφ dt m N l dφ N l dt dφl dt N e l e N l di dt di dt (.4) - 7

45 The lat term for v can be repreented by an inductor, N L l and the lat term for v can be l repreented by N L l. Thee can be incorporated into the equivalent circuit a hown in l Figure. i N N L l i,ex i' i L l e L e m Ideal tranformer Figure. Equivalent circuit for a tranformer with finite core permeability and leakage flux. Core Loe The diipation of heat in the core due to the flux through it can be repreented by a reitor in the equivalent circuit. It i placed in parallel to L m ince the power lo in the core i proportional to the flux through the core quared, and thu i proportional to e. eitance of the Coil The reitance of copper i approximately (Ω m), however with hundred to thouand of turn for the primary and econdary coil it can lead to an appreciable reitance. Loe to thee reitance are proportional to the current through the coil quared. Adding the core loe and reitance of the coil to the equivalent circuit we obtain our completed model of the tranformer a how in Figure. i i' N N L l i,ex i L l V c L m e e V Ideal tranformer Figure. Equivalent circuit for a non-ideal tranformer. - 8

46 Example Conider a tranformer with a turn ratio of 4000/0, with a primary coil reitance.6 (Ω), a econdary coil reitance of.44 (mω), leakage flux that correpond to L l (mh), and L l 9 (µh), and a realitic core characterized by c 60 (kω) and L m 450 (H). The low voltage ide of the tranformer i at 60 (Hz), and V 0 (V), and the power there i P 0 (kw) at pf 0.85 lagging. Calculate the voltage at the high voltage ide and the efficiency of the tranformer. X m ωl m π (kω) X ωl l π (Ω) 6 X ω L l π (mω) P I ˆ /-.8º 96.08/-.8º (A) ( V pf ) E ˆ ˆ ˆ V I ( jx ) 0.98 j. 045 (V) ˆ N E ˆ E 40.7 j4.8 N 40.8/0.49º (V) ˆ N I ˆ I 5.00 j.07 N (A) ˆ ˆ I, ex E j0.06 c jx (A) m I ˆ ˆ ˆ I, ex I j /-.87º (A) V ˆ ˆ ˆ E I ( jx) j69. (V) 4066/0.9º (V) The power loe in the winding and core are: P I (W) P I (W) E 40.8 P c (W) c P lo P P P c.08 (W) P 0 0 out P η P ( ) in P P lo Loe and ating The impedance hown in Figure can be reflected into either the primary ide, or the econdary ide of the tranformer a hown in Figure

47 ( ) L l Ll N N c L m ( ) N N N N V V' V V N N V' ( ) N N L l N c ( ) N ( ) ( ) N N L m N N L l V Figure 4. eflection of the impedance to the primary or econdary ide of the tranformer. For a given frequency, the power loe in the core (iron loe) increae with the voltage e (or e ). If thee loe exceed a particular limit, a hot pot in the tranformer will reach a temperature that dramatically reduce the life of the inulation. Limit are therefore put on E and E which are the voltage limit for the tranformer. The current limit for the tranformer limit I and I to avoid exceive Joule heating due to winding reitance. Typically a tranformer i decribed by it rated voltage, E N and E N, that give both the limit I and turn ratio. The ratio of the rated current N, i the invere of the ratio of the voltage when IN the magnetizing current i neglected. Intead of liting thee rated current limit, a tranformer i decribed by it rated apparent power a: S N E N I N E N I N (.5) When a tranformer i operated at it rated condition, that i at it maximum current and maximum voltage, the magnetizing current, I,ex, typically doe not exceed % of the current in the tranformer. It effect on the voltage drop on the leakage inductance and on the winding reitance i therefore negligible. Under maximum (rated) current, the total voltage drop on the winding reitance, and leakage inductance typically do not exceed 6% of the rated voltage. Their effect on E and E i mall and their effect on the magnetizing current can be neglected. Becaue thee effect are mall, we can modify the equivalent circuit hown in Figure 4 to a lightly inaccurate, but much more ueful on hown in Figure

48 c L m L l ( ) N N ( ) L l N N N N V V' V V N N V' ( ) N N ( ) L l N N c L l N N L m N N ( ) ( ) V Figure 5. Slightly inaccurate, but highly implified equivalent circuit for the tranformer. When we utilize thi implification and work with the reflected voltage, the tranformer equivalent circuit can be hown a: i eq ' i,ex L eq L l L' l i' V L V' c m L eq L' l L l eq ' V' L' m ' c V Figure 6. Working with V or V the above approximate circuit for the tranformer can implify the analyi. -

49 Example (repeat with implified equivalent circuit) Conider a tranformer with a turn ratio of 4000/0, with a primary coil reitance.6 (Ω), a econdary coil reitance of.44 (mω), leakage flux that correpond to L l (mh), and L l 9 (µh), and a realitic core characterized by c 60 (kω) and L m 450 (H). The low voltage ide of the tranformer i at 60 (Hz), and V 0 (V), and the power there i P 0 (kw) at pf 0.85 lagging. Calculate the voltage at the high voltage ide and the efficiency of the tranformer. Uing the implified equivalent circuit of Figure 6, we can firt find eq, and X eq eq N N. (Ω) 60 N X eq π Ll Ll 5.876(Ω) N then P I ˆ /-.8º 96.08/-.8º (A) ( V pf ) E ˆ V ˆ ˆ ˆ N I I 5 j.0 N 5.884/-.8º (A) ˆ ˆ N E E 4000 (V) N ˆ ˆ I, ex E j0.05 c jx (A) m I ˆ ˆ ˆ I, ex I j /-.87º (A) Vˆ Eˆ Iˆ jx (V) 4066/0.98º (V) ( ) 79 eq eq j The power loe in the winding and core are: P eq eq I (W) V 4065 P c (W) c P lo P e q P c 4.07 (W) η Pout Pin P lo ( P P ) Thee value agree well with the previou analyi uing the more accurate model. -

50 .4 Per Unit Sytem A implification in the analyi can come from expreing each of the value a a fraction of a defined bae ytem of quantitie. When thi i done, imple problem can be made more complex, however more complex problem can be made eaier to olve. A an example conider a imple problem of a load impedance of 0 j5 (Ω) that ha a voltage of 00 (V) connected to it. Calculate the power at the load. The traditional olution i found a: ˆ Vˆ 00 I L L 8.94 Z 0 j 5 /-6.57º (A) L and the power i PL VL IL pf ( 6.57º ) co (W) Uing the per unit ytem to find the olution:. Firt define a conitent ytem of value for the bae. If we chooe V b 50 (V), I b 0 (A), then Z b V b /I b 5 (Ω), and P b V b I b 500 (W), Q b 500 (VAr), and S b 500 (VA).. Convert all unit to pu: V L,pu V L /V b pu, Z L,pu (0 j5)/5 j (pu). Solve the problem uing the pu ytem: Vˆ ˆ L, pu IL, pu /-6.57º (pu) Z j L, pu ( 6.57º ). 6 PL, pu VL, puil, pu pf co 4. Convert back to the SI ytem I I I (A) L L, pu b P L b P L, pu P (W) (pu) For the more complicated example of a tranformer we chooe the bae for each ide of the tranformer uch that V b N (.6) Vb N I I b (.7) b Thi lead to the two bae apparent power being equal N N S b V b I b V b I b S b (.8) The bae are often choen to be the rated quantitie of the tranformer on each ide. Thi i convenient ince mot of the time tranformer operate at rated voltage (making the pu voltage unity), and the current and power are eldom above rated (eldom above pu). The bae impedance are related by: -

51 or V Z b b Ib (.9) V b N V Z b b Ib N Ib (.0) N Z b Zb N (.) To move impedance from one ide of the tranformer to the other, they get multiplied or divided N by the quare of the turn ratio,, but o doe the bae impedance, hence the pu value of an N impedance tay the ame regardle of which ide of the tranformer it i on. Through our choice of the bae in (.6) and (.7), we alo ee that for an ideal tranformer E I, pu, pu Thu when uing the per unit ytem, an ideal tranformer ha voltage and current on one ide that are identical to the voltage and current on the other ide, and the ideal tranformer can be eliminated. E I, pu, pu Example (repeat and olved uing pu ytem) Conider a 0 (kva) rated tranformer with a turn ratio of 4000/0, with a primary coil reitance.6 (Ω), a econdary coil reitance of.44 (mω), leakage flux that correpond to L l (mh), and L l 9 (µh), and a realitic core characterized by c 60 (kω) and L m 450 (H). The low voltage ide of the tranformer i at 60 (Hz), and V 0 (V), and the power there i P 0 (kw) at pf 0.85 lagging. Calculate the voltage at the high voltage ide and the efficiency of the tranformer.. Firt calculate the impedance of the equivalent circuit. V b 4000 (V) S b 0 (kva) 0 0 I b 7. 5 (A) 4 0 V Z b b 5(Ω) Sb V b 0 (V) S S 0 b b (kva) S I b b 50 (A) Vb - 4

52 Z V b b Ib. Convert everything to per unit (Ω), pu 0.00 (pu) Zb, pu 0.00 (pu) Zb, c c pu 00 (pu) Zb π 60 L X, l l pu (pu) Zb π 60 L X, l l pu (pu) Zb π 60 L X, m m pu 8 (pu) Z b V V, pu Vb (pu) P P, pu S (pu) b. Solve in the per unit ytem. P ˆ, pu I, pu /arcco(pf) j0.4(pu) ( V, pu pf ) Vˆ ˆ ˆ, pu V, pu I, pu ( eq jx eq ).06 j0.076 (pu) Vˆ ˆ ˆ, pu V, pu Im, pu j0.00(pu) c, pu jx m, pu I ˆ ˆ ˆ, pu I m, pu I, pu j0.46 (pu) P eq I, pueq, pu V, pu c, pu c, pu P (pu) (pu) P, pu η ( P, pu P lo, pu ) ( )

53 4. Convert back to SI unit. The efficiency i unitle and thu tay the ame. Converting Vˆ give V ˆ V V ˆ j /0.979º (V) ( ) b, pu j.5 Teting Tranformer Purchaed tranformer often give information on the frequency, winding ratio, power, and voltage rating, but not the impedance. Thee impedance are important in calculating the voltage regulation, efficiency, etc. Ue of open circuit and hort circuit tet we will determine eq, L eq, c, and L m. Open Circuit Tet Leaving one ide of the tranformer open circuited, while the other ha the rated voltage {V in-oc (pu)} applied to it, we meaure the current and power. The current that flow into the tranformer i motly determined by the impedance X m and c, and it i much lower than the rated current for the tranformer. It i often the cae that the rated voltage for the low voltage ide (low tenion ide) of the tranform ince for the above example it would be eaier applying 0 (V) intead of 4000 (V). Since the unit we ue indicate if we are uing the per unit ytem, the following calculation will drop the ubcript pu. Uing the following equivalent circuit with the primary open circuited, and V V oc 0 (V) applied to the econdary. eq ' L eq L' l L l V' L' m ' c V Figure 7. Open circuit teting of a 4000/0 rated tranformer. With 0 (V) applied to the low voltage ide {V 0 (V)}, the primary voltage for thi open circuit tet i 4000 (V). For thi open circuit tet, the following can be compared to the meaured value of current and power a: Poc Voc c c (pu) (.) Voc V I ˆ oc oc c jx (.) m - 6

54 I oc (pu) (.4) c X m Then from the open circuit tet with meaurement of the current and the power, we determine c and X m. Short Circuit Tet For the above open circuit tet, the low voltage ide of the tranformer wa choen ince it i eaier to apply the lower voltage during that tet. Similarly, the rated current i lower on the high voltage ide of the tranformer. With the hort circuit tet, the rated current i commonly applied to the high voltage ide of the tranformer (ince with a hort circuited econdary, the applied voltage required to reach the rated current i relatively low). With the rated current applied to the high voltage ide, we meaure the voltage, V c which i V for thi example, and the power, P c. P c c I eq eq ( jx ) V c I ˆ c eq eq (pu) (.5) ˆ (.6) V c eq X eq (pu) (.7) i eq ' i,ex L eq L l L' l i' V L V' 0 c m Figure 8. Short circuit teting of a 4000/0, 0 (kva) rated tranformer. Thi correpond to the rated current on the high voltage ide of 7.5 (A). Then from the hort circuit tet with meaurement of the current and the power, we determine eq and X π 60( L L ). m l l Example A 60Hz tranformer i rated 0 (kva), 4000(V)/0(V). The open circuit tet, with the high voltage ide open, give P oc 00 (W), I oc.455 (A). The hort circuit tet, meaured with the low voltage ide horted, give P c 80 (W), V c 9.79 (V). Determine the equivalent circuit for thi tranformer by uing the per unit ytem. - 7

55 . Bae: V b 4000 (V) S b 0 (kva) 0 0 I b 7. 5 (A) 4 0 V Z b b 5(Ω) Sb V b 0 (V) S b S b 0 (kva) S 0 0 I b b 50 (A) Vb 0 Z V 0 50 b b Ib 0.48(Ω). Convert to (pu): 80 P c, pu (pu) V c, pu 0.04 (pu) P oc, pu 0.00(pu) 0 0 I oc.455, pu (pu) 50. Calculate the component of the equivalent circuit {dropping the (pu) ubcript} P c P I ceq or c eq P c (pu) Ic V ˆ ˆ c Vc Ic eq jx eq eq X eq eq c eq or X V 0. 08(pu) Poc Voc c V or oc c 00 (pu) P P oc oc - 8

56 ˆ ˆ ˆ Voc V I oc oc I oc c jx m c X m or X m 8 (pu) Ioc c With thi knowledge, we can addre the more common example of: A 60Hz tranformer i rated 0 (kva), 4000(V)/0(V). The hort circuit impedance i 0.04 (pu) and the open circuit current i (pu). The rated core loe are 00 (W) and the rated winding loe are 80 (W). Calculate the efficiency and the neceary primary voltage when the load at the econdary i at rated voltage, 0 (kw), and at 0.8 pf lagging. Uing (pu) ytem: Z c 0.04 (pu) 80 P c I ceq eq 0 0 X eq Z c eq P oc 0.07 (pu) (pu) or c 00 (pu) c Poc I oc or X m 8 (pu) c X m Ioc Now we have the equivalent circuit component of the tranformer and can work on the full circuit which include the load that ha a power of 0 (kw) or: P but the power at the load i (pu) P V I pf or I 0. 8 thu I 0.8 (pu) a a phaor: I ˆ 0. 8/-6.87º j0.5 (pu) ( jx ) Vˆ ˆ ˆ V I eq eq j 0084 V.05 (pu) c.05/0.47º (pu) P eq I eq 0.006(pu) - 9

57 V P c c 0.04 (pu) c P P P P eq c η converting V to SI unit give V V, pu V b (V).6 Three-Phae Tranformer If we conider three-phae tranformer a coniting of three identical one-phae tranformer, then we have an accurate repreentation a far a equivalent circuit and two-port model are concerned, but it doe not give inight into the magnetic circuit of the three-phae tranformer. The primarie and the econdarie of the one-phae tranformer can be connected either in Δ or in Y configuration. In either cae, the rated power of the three-phae tranformer i three time that of the one-phae tranformer. For the Δ connection: V l l V φ (.8) I l I φ (.9) For the Y connection: V l l V φ (.0) I l I φ (.) Connection to the three-phae tranformer that are Y connected in the primary are hown in Figure

58 I I I I V V V V I I I I V V V V I I I I V V V V Figure 9. Y Y and Y Δ connection of three-phae tranformer. Connection to the three-phae tranformer that are Δ connected in the primary are hown in Figure 0. I I I I V V V V I I I I V V V V I I I I V V V V Figure 0. Δ Y and Δ Δ connection of three-phae tranformer. -

59 .7 Autotranformer An autotranformer i a tranformer where the two winding (of turn N and N ) are not iolated from each other, but are connected a hown in Figure. I N V I I I N V Figure. An autotranformer. From thi figure that the voltage ratio in an autotranformer i: V N N V (.) N and the current ratio i: I N N (.) I N An intereting note on autotranformer i that the coil of turn N carrie current I, while the coil of turn N carrie the (vectorial) um of the two current, I ˆ I ˆ. So if the voltage ratio were, no current would flow through the N coil. Thi characteritic lead to a ignificant reduction in the ize of the autotranformer compared to a imilarly rated tranformer, epecially if the primary and econdary voltage are of the ame order of magnitude. Thee aving come at a eriou diadvantage of the lo of iolation between the two ide. -

60 Chapter Note: To undertand the operation of tranformer we have to ue both the Gau law for magnetim (or Biot-Savart law) and Faraday law. Mot tranformer operate under or near rated voltage. The voltage drop in the winding reitance and leakage reactance are uually mall. In both tranformer and in rotating machine, the net mmf of all the current mut accordingly adjut itelf to create the reultant flux required by thi voltage balance. Leakage fluxe induce voltage in the winding that are accounted for in the equivalent circuit a leakage reactance (element L l and L l ). The leakage-flux path are dominated by path through air, and are thu almot linearly proportional to the current producing them. The leakage reactance therefore are often aumed to be contant (independent of the degree of aturation of the core material). The open- and hort-circuit tet provide the parameter for the equivalent circuit of the tranformer. Three-phae tranformer can be conidered to be made of three ingle-phae tranformer for the purpoe of thi coure. The main iue i then to calculate the rating, voltage, and current of each. Autotranformer are ued motly to vary the voltage a little. It i eldom that an autotranformer will have a voltage ratio greater than two. -

61 ECE 0 Energy Converion and Power Electronic Dr. Tim Hogan Chapter 4: Concept of Electrical Machine: DC Motor (Textbook Section.-.4, and 4.-4.) Chapter Objective DC machine have faded from ue due to their relatively high cot and increaed maintenance requirement. Neverthele, they remain good example for electromechanical ytem ued for control. We ll tudy DC machine here, at a conceptual level, for two reaon:. DC machine although complex in contruction, can be ueful in etablihing the concept of emf and torque development, and are decribed by imple equation.. The magnetic field in them, along with the voltage and torque equation can be ued eaily to develop the idea of field orientation. In doing o we will develop baic teady tate equation, again tarting from fundamental of the electromagnetic field. We are going to ee the ame equation in Bruhle DC motor, when we dicu ynchronou AC machine. 4. Geometry, Field, Voltage, and Current The geometry hown in Figure decribe an outer iron frame (tator), through which (i.e. it center part) a uniform magnetic flux, φˆ, i etablihed. The flux could be etablihed by a current in a coil or by a permanent magnet for example. In the center part of the frame there i a olid iron cylinder (called rotor), free to rotate around it axi. A coil of one turn i wound diametrically around the cylinder, parallel to it axi, and a the tator and it coil rotate, the flux through the coil change. Figure how conecutive location of the rotor and we can ee that the flux through the coil change both in value and direction. The top graph of Figure how how the flux linkage of the coil through the coil would change, if the rotor were to rotate at a contant angular velocity, ω. ( ω t) λ ˆ φ co (4.) Figure. Geometry of an elementary DC motor. -

62 4 5 Figure. Flux through the tator coil of the imple dc motor hown in Figure. Since the flux linking the coil change with time, than a voltage will be induced in thi coil, v coil, a hown below: dλ v coil ˆin φ ( ω t) (V) (4.) dt λ coil Angle (º) v dcoil Angle (º) Figure. Flux and voltage in a coil of the motor in Figure with coil poition -5. -

63 The point marked on the coine and ine waveform of Figure correpond to the poition of the rotor a hown in Figure. Torque on the coil follow the Lorentz Force Law: which give the force, F, in newton on a charge, q, in coloumb that i expoed to a vectoral electric field, Eˆ, in volt per meter and magnetic flux denity, B, a F q Eˆ v B (4.) ( ) where v i the velocity of a charged particle q. The firt term give the force on the electron that move it through the wire or the voltage applied to the wire. The econd term give the force on the charge due to the magnetic field in which the wire i placed. With many electron contributing to the electrical current, I, (a a vector) in the wire, the force on the wire i F I B (4.4) The force on the full coil i then twice thi value to account for each of the wire hown in Figure. To maintain thi torque in a direction that continue the rotary motion of the rotor, the end of the rotor coil are connected to metal electrode ring egment called a commutator a hown in Figure 4. Thee ring egment are attached to the rotor o a to rotate with it, and are electrically contacted uing two tationary bruhe typically made of carbon and copper. The bruhe are pring loaded and puhed againt the commutator. A the rotor pin, the bruhe make contact with the oppoite egment of the commutator and they witch between egment of the commutator jut a the induced voltage goe through zero and witche ign. B bruh commutator ω bruh Figure 4. A coil of a DC motor and bruhe that are puhed up againt the rotating commutator (wire of the coil i highlighted red and oldered to the commutator egment). Becaue the current witche direction through the coil each time the bruhe rotate to the oppoite egment of the commutator (and that thi occur a the voltage goe through zero), the -

64 voltage at the terminal of the bruhe i a rectified verion of the voltage induced in the coil a hown in Figure v coil 0 v terminal Time (arb. unit) -.5 Time (arb. unit) Figure 5. Induced voltage in a coil and terminal voltage in an elementary DC machine. If a number of coil are placed on the rotor, a hown in Figure 6, with each coil connected to a different egment of the commutator, then the total induced voltage to the coil, E, will be: where k i proportional to the number of coil. E k φ ˆω (4.5) Figure 6. Multiple coil on the rotor of a DC machine. We tated in equation (.9) that machine: P elec P mech T ω mech e i thu with the multiple coil DC E i T ω (4.6) k ˆ φ ω i T ω (4.7) T k φˆ i (4.8) - 4

65 If the DC machine i connected to a load or a ource a in Figure 7, then the induced voltage and terminal voltage will be related by: V terminal E i g wdg for a generator (4.9) V terminal E i m wdg for a motor (4.0) i g i m wdg Load E or Source Figure 7. Circuit with a DC machine with the current reference direction defined a indicated for DC machine ued a a generator, i g, or a a motor, i m. Example 4.. A DC motor, when connected to a 00 (V) ource and with no load connected to the motor run at 00 (rpm). It tator reitance i (Ω). What hould be the torque and current if it i fed from a 0 (V) upply and it peed i 500 (rpm)? Aume the field i contant. With no load connected to the motor, we will aume the torque i zero (auming no friction in bearing). With the torque equal to zero, the current i zero ince: T kφ i Ki. Thi mean for thi operation: V E k φω K ω π (rad) revolution (min) 60 () and with ω 00 ( rpm) 5.66 (rad/) 00 (V) K 5.66, then Then at 500 (rpm) ω o K (V ) π (rad) (min) 500 revolution 60 () ( rpm) (rad/) E K ω 5 (V) - 5

66 For a motor: V E I wdg 0 (V) 5 (V) I wdg I 47.5 (A) T KI 7.8(N m) 4. Energy Conideration Conervation of energy govern that the total energy upplied to an electromechanical ytem equal the energy output. Some of the energy output could be mechanical, ome could be lot a heat, and ome could be tored. Thi i ummarized in equation (.0) of your textbook a: Energy input Mechanical Increae in energy Energy from electric energy tored in the converted ource output magnetic field into heat In a lole ytem, the lat term i zero. When the loe are negligible, a differential change in the electrical energy input correpond to the um of a differential mechanical energy output and a differential change in the energy tored in the magnetic field, or dw elec dwmech dwfld (4.) Mechanical energy i force time ditance, and the time rate of change of electrical energy i power or dw ei dt and ( 4.) can be written a: elec ei dt Fflddx dw fld (4.) where F fld i the force produced by the magnetic field. From equation (.) in the handout note, we dλ Nφ λ aw e. Uing the linear inductor aumption of (.9) which wa L, then (4.) dt i i become: dw i dλ F dx (4.) The energy in the field i given a the energy tored in an inductor W fld fld fld λ Li (4.4) L The current and force produced by the magnetic field can then be found from (4.) a: ( λ, x ) W i fld (4.5) λ x ( λ, x ) i dl( x) W F fld fld x λ dx (4.6) If the magnetic force reult in a torque a in a rotating mechanical terminal, the mechanical energy from the field i then replaced with torque and angular diplacement, Fflddx Tflddθ, a: - 6

67 dwfld i dλ T dθ (4.7) fld thu leading to ( θ ) i dl Tfld dθ (4.8) In a rotary machine with multiple winding, the inductance in equation (4.8) would contain elf and mutual inductance of all coil involved. Chapter Note: The field of a DC motor can be created either by a DC current or a permanent magnet. The two field, the one coming from the tator and the one coming from the moving rotor, are both tationary (depite rotation) and they are perpendicular to each other. If the direction of current in the tator and in the rotor revere together, torque will remain in the ame direction. Hence if the ame current flow in both winding, it could be AC and the motor will not revere. - 7

68 ECE 0 Energy Converion and Power Electronic Dr. Tim Hogan Chapter 5: Three Phae Winding (Textbook Section ) Chapter Objective Flux linkage play a crucial role in the operation of both DC and AC machine. In thi chapter, the geometry and the operation of winding in AC machine i dicued. The flux varie in time, and can alo vary in poition, or be tationary. To undertand how thee machine operate, the concept of pace vector (or pace phaor) i introduced. 5. Introduction Electric machine often have defined an armature winding which i the winding that i power producing, and a field winding that generate the magnetic field. Either could be on the tator or rotor depending on the pecific motor or generator; however it i more common with AC machine uch a ynchronou or induction machine that the armature winding i on the tator (the tationary portion of the motor). Synchronou machine have field winding on the rotor that i excited by direct current delivered to the rotor winding by lip ring or collector ring by carbon bruhe. The field winding produce the north and outh pole, thu the image hown in Figure i for a two-pole, ingle phae (one armature winding) ynchronou generator. The magnetic axi for the armature winding i perpendicular to the area defined by the armature winding (armature winding i the perimeter of thi area). Figure. Figure 4.4 and 4.5 from your textbook howing a imple two-pole, ingle phae ynchronou generator, the patial ditribution of the magnetic field relative to the magnetic axi of the armature winding, and the time dependent induced voltage in the armature winding []. -

69 In a three phae device, the armature ha three coil each with a magnetic axi that i rotated patially by 0º a hown in Figure. Figure. A three-phae, two-pole ynchronou generator a hown in Figure 4.(a) in your textbook []. More pole for the field winding and more armature winding are alo poible a hown in Figure. Such a configuration can deliver three phae power by interconnecting the armature winding in a Y connection configuration a an example. Figure. A four-pole ynchronou generator with multiple armature winding that can be wired together in a three-phae Y connection i hown (from Figure 4.(b) and 4.(c) in your textbook []). 5. Control of the Magnetomotive Force Ditribution A generator, the ynchronou machine typically ue the tator winding a a ource of electrical power. A motor, the tator (or armature) winding are commonly upplied electrical power to generate a patially varying field (we will conider the time variation of thee field later). Inight to the ditribution (in pace) of the field from the armature winding can be een by conidering a ingle N-turn coil on the tator that pan 80 electrical degree (in 80º it ha gone -

70 from a F to a F). Such a coil i known a a full pitch coil. The mmf ditribution for uch a full pitch coil i depicted in Figure 4.9 from your textbook a hown below in Figure 4. Figure 4. A full pitch coil on the tator. Winding on the rotor are left off for clarity. Thi i Figure 4.9 in your textbook. (a) Schematic view of flux produced by a concentrated, fullpitch winding in a machine with a uniform air gap. (b) The air-gap mmf produced by current in thi winding []. Auming the reluctance of the tator and rotor negligible compared to the gap, then the full mmf of N i would drop acro two gap a the flux travere a full loop. Thi give rie to a (N i/) maximum, and a (N i/) minimum for the mmf a one patially map F for the tator winding a a function of the angle relative to it magnetic axi. Plotting F a a function of thi angle, θ a, how abrupt change at the wire of the coil. The Fourier tranform of thi quare wave give a fundamental inuoidal ditribution of F ag a hown in Figure 4. Harmonic exit ince it i non-inuoidal. In an attempt to make the patial ditribution more inuoidal in form, multiple coil can be placed within pecific grove of the tator a hown for one of the phae in a three-phae winding in Figure 5. Here there are equal number of wire in each of the grove, and equal current in each of the wire. -

71 Figure 5. The mmf of one phae of a ditributed two-pole, three phae winding with full-pitch coil. Thi i image 4.0 from your textbook []. By adjuting the number of turn in each lot a more inuoidal mmf ditribution can be alo be obtained - a hown for winding of a rotor in the configuration hown in Figure 6. Figure 6. The air-gap mmf of a ditributed winding on the rotor of a round-rotor generator. Thi i image 4. from your textbook []. Some imple method for controlling the hape of the magnetomotive force for a et of coil have been outlined. Now we turn our attention to how we ue thi information for analyzing the electric machine through the concept of pace vector. - 4

72 5. Current Space Vector Conider three identical winding placed in a pace of uniform permeability a hown in Figure 7. Each winding carrie a time dependent current, i (t), i (t), and i (t), We alo require that ( t ) i ( t ) i ( t ) 0 i (5.) Each current produce a flux in the direction of the coil axi, and if we aume the magnetic medium to be linear, we can find the total flux by adding the individual fluxe. Thi mean that we could produce the ame flux by having only one coil, identical to the three, but placed in the direction of the total flux, carrying an appropriate current. If the coil in Figure 7 carry the following current i 5 (A), i -8 (A), i (A), then the vectoral um of thee current, oriented in the ame direction a the correponding coil can be hown in Figure 8. φ I I I φ φ Figure 7. Three phae winding patially oriented 0º apart. The direction of the reultant coil and current it hould carry, we create three vector, each in the direction of one coil, and equal in amplitude to the current of the coil it repreent. If, for example, the coil are placed at angle of 0º, 0º, 40º. Then their vectoral um will be: j0 j40 i i /φ i i e i e (5.) Thi repreent the vector of a current oriented in pace, and thu we define i a a pace vector. If i, i, and i are function of time, o will be the amplitude and angle of i. If we conider a horizontal axi in pace a a real axi, and a vertical axi a an imaginary axi, then we can find the real (i d e{i}) and imaginary (i q Im{i}) component of the pace vector. With thi repreentation, we can determine the three current from the pace vector a: - 5

73 () t e{ i() t } i i i () t e i() t () t e () jγ { e } j { i t e γ } (5.) γ 0 π rad (5.4) I 5 (A) I I I (A) I - 8 (A) I I I Figure 8. (a) Current in the three winding of Figure 7. (b) eultant pace vector and (c) correponding winding poition and current of an equivalent ingle coil. Conider if the three coil in Figure 7 were to repreent the three tator winding of an AC machine a hown in Figure 9, with current for each phae alo hown in Figure 9. (a) (b) Figure 9. Simplified two pole -phae tator winding and the intantaneou current for each phae. Thee are Figure 4.9 and 4.0 from your textbook []. - 6

74 Then for: i i i a b c ( t ) I m co( ωe t ) () t I m co( ωe t 0º ) () t I co( ω t 0º ) m e (5.5) the pace vector i: i [ ] j0º j40º () t I m co( ωe t ) e co( ωet 0º ) e co( ωet 0º ) I [ co( ω t ) { co( 0º ) j in( 0º )} co( ω t 0º ) { co( 40º ) j in( 40º )} co( ω t 0º )] m I m co e ( ω t ) 0.5co( ω t 0º ) 0.5co( ω t 0º ) e e e e j { co( ω t 0º ) co( ω t 0º )} e e e (5.6) Uing: co ( x ± y ) co( x ) co( y ) in( x ) in( y ) i () t I co( ω t ) co( ω t) in( ω t) co( ω t) in( ω t) m ji I m m e 0.5co [ co( ω t ) j in( ω t )] e ( ω t) in( ω t) 0.5co( ω t) in( ω t) e e e e e e e e e (5.7) Thi i a pace vector that rotate in pace a a function time at an angular frequency of ω e. Graphically, thi can be een from ummation of i a, i b, and i c in Figure 9 (b) at different point in time (ω e t 0, π/, and π/), the reultant pace vector at each point in time i repreented by the correponding magnetomotive force, F, aociated with that pace vector a hown in Figure 0. Figure 0. The production of a rotating magnetic field by mean of three-phae current. - 7

75 Section 5. in thee note decribed way of forming magnetomotive force patial ditribution that were more inuoidal in profile. epreenting uch winding a inuoidally concentrated winding in the tator a depicted in Figure, then the denity of turn of a the coil would vary inuoidally in pace a function of the angle θ. Stator Winding Stator Air Gap otor Figure. The production of a rotating magnetic field by mean of three-phae current. Thu the number of turn, dn, covering an angle dθ at a poition θ over dθ i a inuoidal function of the angle θ. The turn denity, n (θ ) i then: For the total number of turn, N, in the winding: dn n ( θ ) n ˆ inθ (5.8) dθ ( θ ) N π n d θ (5.9) 0 Thi lead to N n ( θ ) inθ (5.0) With i current flowing through thi winding, the flux denity in the air gap between the rotor and the tator can be found for the integration path hown in Figure. The path of integration i defined by the angle q and we can notice that becaue of ymmetry of the flux denity at the two air gap egment in the path i the ame. - 8

76 Current in poitive direction Stator θ Air Gap otor Current in negative direction Stator Winding Figure. Integration path to calculate flux denity in the air gap. Auming the permeability of the tator and rotor i infinite, then H iron 0 and: H B B μ g g g 0 ( θ ) g i n ( φ ) ( θ ) θ π θ g i N N μ g coθ ( θ ) i 0 coθ d φ (5.) For a given current, i, in the coil the flux denity in the air gap varie inuoidally with angle, but a hown in Figure it reache a maximum when the angle θ i zero. For the ame machine and condition a in Figure, Figure 4 how the plot of turn denity, n (θ) and flux denity, Bg(θ) in carteian coordinate with θ a the horizontal axi. For thi one coil, if the current, i, were to vary inuoidally in time, then the flux denity would alo change in time. The direction of the pace vector would be maintained however the amplitude would change in time. The node of the flux denity where it i equal to zero will remain at 90º and 70º, while the extrema of the flux will be maintained at 0º and 80º. - 9

77 Figure. Sketch of the flux (red) in the air gap n (θ) 0 B g (θ) θ θ Figure 4. Turn denity on the tator and the air gap flux denity v. θ. Conider now an additional winding, identical to the firt, but rotated with repect to it by 0º. For a current in thi winding we will get a imilar air gap flux denity a before, but with node at 0º 90º 0º and at 0º 70º 0º. If a current, i, i flowing in thi winding, then the air gap flux denity due to it will follow a form imilar to equation (5.) but rotated by 0º (π/). N μ ( ) 0 π B θ i g co θ (5.) g Similarly, a third winding, rotated 40º relative to the firt winding and carrying current i, will produce an air gap flux denity of: N μ ( ) 0 4π B θ i g co θ (5.) g - 0

78 Combining thee three flux denitie, we obtain a inuoidally ditributed air gap flux denity, that could equivalently come from a winding placed at an angle φ and carrying current i a: ( ) ( ) ( ) ( ) ( ) φ θ μ θ θ θ θ co 0 g N i B B B B g g g g (5.4) Thi mean that a the current change, the flux could be due intead to only one inuoidally ditributed winding with the ame number of turn. The location, φ (t), and current, i(t), of thi winding can be determined from the current pace vector: () () t t i i /φ (5.5) () () () º 40 0º j j e t i e t i t i 5.. Balanced, Symmetric Three-phae Current If the current i, i, i form a balanced three-phae ytem of frequency f ω /π, then we can write: ( ) [ ] ( ) ( ) [ ] ( ) ( ) [ ] co co co π ω π ω π ω π ω ω ω π φ ω π φ ω φ ω t j t j t j t j t j t j e e t I i e e t I i e e t I i I I I I I I (5.6) where I i the phaor correponding to the current in phae. The reultant pace vector i: () ( ) φ ω ω t j t j Ie t Ie i (5.7) The reultant flux denity wave i then: ( ) ( ) θ φ ω μ θ 0 co, t g N I t B (5.8) which how a travelling wave, with a maximum value of 0 max μ N I B. Thi wave travel around the tator at a contant peed ω, a hown Figure 5. -

79 0.5 t t t B g (θ) Angle (θ) Figure 5. Air gap flux denity profile v. θ for three time t >t >t. 5.4 Phaor and Space Vector Thi i a good point to reflect on the difference between phaor and pace vector. A current phaor, ˆ jφ I Ie 0, decribe one inuoidally varying current of frequency ω, amplitude I, and initial phae φ 0. The inuoid can be recontructed from the phaor a: () [ j ω t * j ω t e e t ] I ( t ) ( e j ω I I co ω φ e I t ) i 0 (5.9) Although rotation i implicit in the definition of the phaor, no rotation i decribed by it. On the other hand, the definition of current pace vector require three current that um to zero. Thee current are implicitly in winding that are ymmetrically placed, but the current are not necearily inuoidal. Generally the amplitude and angle of the pace vector change with time, but no pecific pattern i a priori defined. We can recontruct the three current that contitute the pace vector from equation (5.). When thee contituent current form a balanced, ymmetric ytem, of frequency ω, then the reultant pace vector i of contant amplitude, rotating at a contant peed. In that cae, the relationhip between the phaor of one current and the pace vector i hown in equation (5.7). 5.5 Magnetizing Current, Flux, and Voltage To ee how thi rotating magnetic flux influence the winding we ue Faraday law. From here on we will ue inuoidal ymmetric three-phae quantitie. The three tationary winding are linked by a rotating flux a hown in Figure 6. When the current i maximum in phae, the flux i a hown in Figure 6(a) and i linking all of the turn in phae. Later, the flux ha rotated a how in Figure 6(b) then the flux linkage with coil have decreaed. When the flux ha rotated 90º a in Figure 6(c), the flux linkage with the phae winding are zero. -

80 (a) (b) (c) Figure 6. otating flux and flux linkage. Sinuoidal winding containing many turn in the tator are repreented by ingle wire to how current flow direction. To calculate the flux linkage, l, we determine the flux through a differential number of turn at an angle θ a hown in Figure 7. dθ θ Figure 7. Flux linkage through a differential number of turn (uch a one turn) for coil. The flux through thi differential ection of coil i then: θ ( t θ ) B ( t, θ ) da lr B ( t, θ ) dθ θ φ, (5.0) θ π g θ π g -

81 where l i the axial length of the coil (into the page), and r i the radiu to the coil. The number of turn linked by thi flux i dn ( θ ) n ( θ ) dθ one) turn i:, o the flux linkage for thee few (or ( θ ) θ φ( θ ) d λ n d (5.) To find the flux linkage, λ, for all of coil, then we mut integrate the flux linkage over all turn of coil or: π λ d λ n θ θ φ θ (5.) d 0 When thee two integral are taken, then we find: ( ) ( ) N lr λ () t πμ0 I co( ω t φ ) L M I co( ω t φ ) 8g (5.) which mean the flux linkage in coil are in phae with the current in thi coil and proportional to it. The flux linkage of the other two coil, and, are identical to that of coil, but lagging in time by 0º and 40º repectively. With thee three quantitie we can create a flux-linkage pace vector, λ a: j0º λ j40º λ λ λ e e L M i (5.4) Since the flux linkage of each coil vary, and in our cae inuoidally, a voltage i induced in each of thee coil. The induced voltage in each coil i 90º ahead of the current in it, bringing to mind the relationhip of current and voltage of the inductor. Notice though, that it i not jut the current in the winding that caue the flux linkage and the induced voltage, but rather the current in all three winding. Although thi i the cae, we ill call the contant L M the magnetizing inductance, and the induced voltage in each coil can be found a: and the voltage pace vector, e can be defined a: dλ () π e t ω I co ωt φ dt dλ () π π e t ω I co ωt φ dt dλ 4 () π π e t ω I co ωt φ dt j0º j40º e e e e e e j ωl M i (5.5) (5.6) The flux linkage pace vector i aligned with the current pace vector, while the voltage pace vector, e, i ahead of both by 90º. Thi agree with the fact that the individual phae voltage lead the current by 90º, a hown in - 4

82 e λ i ωt Figure 8. Magnetizing current, flux-linkage, and induced voltage pace vector. A. E. Fitzgerald, C. Kingley, Jr., S. D. Uman, Electric Machinery, 6 th edition, McGraw-Hill, New York,

83 ECE 0 Energy Converion and Power Electronic Dr. Tim Hogan Chapter 6: Induction Machine (Textbook Section ) Chapter Objective The popularity of induction machine ha helped to label them a the workhore of indutry. They are relatively eay to fabricate, rugged and reliable, and find their way into mot application. For variable peed application, inexpenive power electronic can be ued along with computer hardware and thi ha allowed induction machine to become more veratile. In particular, vector or field-oriented control allow induction motor to replace DC motor in many application. 6. Decription The tator of an induction machine i a typical three-phae one, a decribed in the previou chapter. The rotor can be one of two major type either (a) it i wound in a fahion imilar to that of the tator with the terminal connected to lip ring on the haft, a hown in Figure, or (b) it i made with horted bar. Shaft otor Slip ing Figure. Wound rotor, lip ring, and connection. Figure how the rotor of uch a machine, while the image in Figure how the horted bar and the lamination. The bar in Figure are formed by cating aluminum in the opening of the rotor lamination. In thi cae the iron lamination were chemically removed. -

84 Figure. (a) Cutaway view of a three-phae induction motor with a wound rotor and lip ring connected to the three-phae rotor winding hown in Figure 6. in your textbook []. (b) Cutaway view of a three-phae quirrel-cage motor a hown in Figure 6. in your textbook []. (a) (b) Figure. (a) The rotor of a mall quirrel-cage motor. (b) The quirrel-cage tructure after the rotor lamination have been chemically etched away a hown in Figure 6. in your textbook []. 6. Concept of Operation A thee rotor winding or bar rotate within the magnetic field created by the tator magnetizing current, voltage are induced in them. If the rotor were to tand till, then the induced voltage would be very imilar to thoe induced in the tator winding. In the cae of a quirrel cage rotor, the voltage induced in a bar will be lightly out of phae with the voltage in the next bar, ince the flux linkage will change in it after a hort delay. Thi i depicted in Figure 4. If the rotor i moving at ynchronou peed, together with the field, no voltage will be induced in the bar or the winding. -