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EE6401 ELECTRICAL MACHINES I Prepared by:R Shalini/AP-EEE,(M.E.T Engg College) UNIT I MAGNETIC CIRCUITS Magnetic Circuits ò H.dl Ampere’s law = C ò B.da = 0 òJ.da H : magnetic field intensity vector, J : current density. S Þ magnetic flux density is conserved B : magnetic flux density vector. S B = mH m = mrm0 : magnetic permeability of medium. m0 : permeability of free space m0=4p ´10-7 mr : relative permeability Simple magnetic circuit. d = Ni H. dl ò C = òS F J : magnetomotive force (mmf, ampere-turns). Magnetic flux crossing surface S: f = òB.da (Weber, Wb) S f c @ Bc Ac f c :flux in core, Bc : flux density in the core Ac : cross-sectional area of the core. H.dl @ Hclc ò C Þ f F Â Þ Bc m  = lc mAc lc = Ni = F Þ : reluctance f mAc lc = F Fig. 1.2 Magnetic circuit with air gap. Flux is the same in the magnetic core and the f Bc = Þ Bg = f flux density in the air-gap. Ag ò mmf flux density in the magnetic core. Ac Þ C F = f ( Âc +  g ) Þ Bg Bc H.dl = Hclc + Hg g = Ni = F Þ f= F= m0 A F æ m lc + m0 ö g=fç ÷ è +ç ÷ m Ac ø g Âc : reluctance of core, Âg : reluctance of air-gap. Analogy between electric and magnetic circuits. (a) Electric circuit, (b) magnetic circuit. Example 1.1 The magnetic circuit shown in Fig.1.2 has dimensions Ac = Ag = 9 cm2, g = 0.050 cm, lc = 30 cm, and N = 500 tums. Assume the value mr = 70,000 for core material. (a) Find the reluctances Âc and Âg . For the condition that the magnetic circuit is operating with Bc = 1.0 T, find (b) the flux f and (c) the current i. Solution: (a) Âc = Âg= (b) (c) f i= lc mrm0 Ac g m0 Ac = = 0.3 70000´ 4p ´10-7 ´9 ´10-4 = 3.79 ´103 A.turns/Wb 5´10-4 5 4p ´ 10-7 ´ 9 ´ 10-4 = 4.42 ´10 A.turns/Wb = Bc Ac = 1.0(9´10-4) = 9´10-4 Wb F N = 9´10-4(4.46´105) (Âc + Âg) N = 500 = 0.8 A i -gap fringing fields. Ar Example 1.2 The magnetic structure of a synchronous machine is shown schematically in Fig. 1 5. Assuming that rotor and stator iron have infinite permeability ( m ®¥), find the air-gap flux f and flux density Bg. For this example I = 10 A, N = 1000 turns, g = 1 cm, and Ag = 2000 cm2. Solution: total air-gap length = 2g. Reluctance of iron core is negligible ((m ®¥). Rg = 2g m0 Ag f Bg = Þ f= F Rg = Nim0 Ag 1000´10´ 4p 0.2 2g Ag 0.2 ´ = 0.13 Wb 2 ´ 0.01 0.13 = ´10-7 = 0.65 T Simple synchronous machine. FLUX LINKAGE, INDUCTANCE AND ENERGY d Faraday’s law. ò E. ds = - dt òB. da C S dl df e = dt = N dt e : induced voltage, l = Nf : flux linkage (Wb-turns). Linear magnetic circuit: flux linkage is proportional to current. Inductance: L=l i l = Nf = NF N 2i Rtot Rtot Þ L= N 2 Rtot If the reluctance of the core is negligible compared to that of the air-gap 2 N2 L= = g ö æ N m0 Ag g ç ÷ ç m0 Ag ÷ ø è (Henry, Weber-turn /A) Example 1.3 The magnetic circuit of Fig. 1.6a consists of an N-turn winding on a magnetic core of infinite permeability with two parallel air gaps of lengths g1 and g2 and areas A1 and A2, respectively. iFn d (a) The inductance of the winding and (b)The flux density Bl in gap 1 when the winding is carrying a current i. Neglect fringing effects at the air gap. Solution: (a) Req = L= R1R2 R1 + R2 N2(R1 + R2 ) R1 = æ = m0 N ç f= 1 Ni m0 A1Ni = R1 g1 R2 = 1 g2 m0A2 A1 A2 ö ÷ + èg 1 g 2 ø R1R2 (b) g1 m0A f1 Þ B1 = m0Ni = A1 g1 (a) Magnetic circuit and (b) equivalent circuit for Example 1.3. Magnetic circuit with two windings. F = N1i1 + N2i2 The total mmf: Neglect the reluctance of the core, and assume Ac = Ag, the resultant core flux produced by the total mmf is f= F Rg N1i1 + N2i2 m0 Ac = =g( N1i1 + N2i2 ) g m0 Ac Flux linkage of coil 1, l Þ l =L 1 æ m0 Ac ö 1 = N1f = N12 ç è g ÷ i1 ø Aö + æ m0 cç è g ø ÷i2 2 i + L12i2 11 1 L11i1 :flux linkage of coil 1 due to its own current. L11 = N12 ç æ m0 Ac ö è L12i2 :flux linkage of coil 1 due to current i2. L12 = N1N and 2 g æ m0 Ac ö 2 ç ÷ g è : self-inductance of coil 1 ÷ ø : mutual-inductance between coils 1 ø Similarly, flux linkage of coil 2, l Þ l 2 æ m0 Ac ö 2 = N 2f = N1N 2 Ac ö ÷ i12æ ç è g ø m02 è ç g ø ÷ i2 = L21i1 + L22i2 L21 = L12 = N1N 2 æ m0 Ac ö ç è g ÷ : mutual-inductance, ø dl dt = Ac ö L22 2=æ m0 ç è g ø d di dL (Li) = L +i d dt dt t ÷ : self-inductance of coil 1 In electromechanical energy conversion devices, inductances are often time-varying. Power and Energy Power delivered to the winding: p = ei = i dl dt (Watts, W= Joules/ second) Change in magnetic stored energy DW in the magnetic circuit in the time interval tl to t2 t2 l2 t1 l1 DW = p dt = i dlò ò Single- winding system of constant inductance, the change in magnetic stored energy as the flux level is changed from l1 to l2 l l2 2 DW = òi dl = =ò l1 l dl 2L l1 L Example 1.6 1 (l2 - l 2 )1 Þ W (l ) = l 2 or W (i) = 1Li2 2L 2 : total stored magnetic energy. For the magnetic circuit of Example 1.1 (Fig. 1.2), find (a) the inductance L, (b) the magnetic stored energy W for Bc = 1.0 T, and (c) the induced voltage e for a 60-Hz time-varying core flux of the form Bc = 1.0 sin(ωt) T where ω = 2πf = 2π(60) = 377 rad/sec. L= (a) l i = Nf i f= F Âc + Âg (b) From Example 1.1, Bc = 1.0 T (c) dl df = N dB = NAc c dt = 170cos(377t) V e= F = Ni Þ Þ L= i = 0.8 A 2 5002 N = = 0.56 H Âc + Âg 4.46´105 Þ W = 1Li2 = 2 1 (0.56)(0.82 ) = 0.18 J 2 = 500´ 9 ´10-4 ´ (377 ´1.0cos(377t) ) PROPER TIES OF MAGNETIC MATERIALS · Magnetic materials are used to constrain and direct magnetic fields in well-defined paths. · In a transformer they are used to maximize the coupling between the windings, and to lowerthe excitation current required for transformer operation. · In electric machinery, they are used to shape the magnetic fields to obtain desired electrical and mechanical characteristics. Ferromagnetic materials are composed of iron and alloys of iron with cobalt, tungsten, nickel, aluminum, and other metals, are the most common magnetic materials. Relationship between B and H for a ferromagnetic material is nonlinear and multivalued: B-H curve or hysteresis loop. B-H loops for M-5 grain-oriented electrical steel 0.012 in thick. Only the top halves of the loops are shown here. Dc magnetization curve for M-5 grain-oriented electrical steel 0.012 in thick. AC EXCITATION Sinusoidal core flux: j (t) = f max sinwt = AcBmax sinwt Voltage induced in the N-turn winding, e(t) = w Nfmax coswt = Emax coswt Emax = wNf max = 2p fNAcBmax Rms value of a periodic function of time, T Frm s = 1 2 Tò f (t)dt 0 Erms = Emax 2 = wNf max = 2p fNAcBmax Exciting current: current required to produce magnetic flux in the core; nonsinusoidal because of the nonlinear magnetic properties of the core. Sources of power loss in the core: 1-) Eddy currents: currents induced by the time-varying flux in the core due to nonzero conductivity of the material. These currents cause ohmic I2R loss. To reduce the effects of eddy currents, magnetic structures are built of thin sheets of laminations of the magnetic material. These laminations are aligned in the direction of the field lines, and are insulated from each other by an oxide layer. 2-) Hysteresis loop: The time-varying excitation causes the magnetic material to undergo a cyclic variation described by a hysteresis loop. Energy input W to the magnetic core of Fig. 1.1 as the material undergoes a single cycle W= òi dl æ Hclcö( A NdBc ) = j ò ç N ÷ cø Alc : volume of the core. c dBc : area of the ac hysteresis loop. H ò c Hysteresis power loss = W´ f ò Aclc c H dBc Excitation phenomena. (a) Voltage, flux, and exciting current; (b) corresponding hysteresis loop. Exciting rms voltamperes per kilogram at 60 Hz for M-5 grain-oriented electrical t el 0.012 in thick. se Hysteresis loop; hysteresis loss is proportional to the loop area (shaded). Core loss at 60 Hz in watts per kilogram for M-5 grain-oriented electrical steel 0.012 in thick. Laminated steel core with winding for Example 1.8. Example 1.8 The magnetic core in Fig. 1.15 is made from laminations of M-5 grain-oriented electrical steel. The winding is excited with a 60-Hz voltage to produce a flux density in the steel of B = 1.5 sin ωt T, where ω = 2π60 = 377 rad/sec. The steel occupies 0.94 of the core crosssectional area. The mass-density of the steel is 7.65 g/cm3. Find (a) the applied voltage, (b) the peak current, (c) the rms exciting current, and (d) the core loss. a. Voltage induced e=N dj = NAc dBc æ dt dt = 274cos(377t) V b 1 m2 ö = 200´ 4 in2 ´0.94´ç è39.4 in2 ÷ ´1.5´ 377cos(377t) ø Hm ax = 36 A-turns/m Bmax = 1.5 T Þ lc = 0.71 m. Þ peak current I = Bmax = 1.5 T Þ Þ Pa = 1.5 VA/kg 3 Vc = 105.5 in . Hmaxlc 36´ 0.71 Wc = 13. 2 kg. Total rms volt-amperes: Pa = 1.5 VA/kg ´ 13.2 kg = 20 VA Ij ,rms = 20 Pa = = 0.10 A 275 / 2 Erms 11 . A magnetic circuit with a single air gap is shown in Fig. 1.24. The core dimensions are: Cross-sectional area Ac = 1.8 × 10-3 m2 Mean core length lc = 0.6 m Gap length g = 2.3 x 10-3 m N = 83 turns Assume that the core is of infinite permeability (m ®¥) and neglect the effects of fringing fields at the air gap and leakage flux . (a) Calculate the reluctance of the core Rc and that of the gap Rg . For a current of i = 1.5 A, calculate (b) the total flux f , (c) the flux linkages λ of the coil, and (d) the coil inductance L. Solution: (a) Rc = 0 since m ® ¥ (b) f= (c) l = Nf = 1.016´10-2 W b (d) L= Ni Rc + Rg l i = Rg = g m0Ac = 2.3 ´ 10-3 = 1.017 ´106 A/Wb 4p ´ 10-7 ´ 1.8 ´ 10-3 83 ´ 1.5 = 1.017 ´106 = 1.224 ´ 10-4 W b 1.016 ´ 10-2 1.5 = 6.773 mH 1 3 Consider the magnetic circuit of Fig. 1.24 with the dimensions of Problem 1.1. Assuming infinite core permeability, calculate (a) the number of turns required to achieve an inductance of 12 mH and (b) the inductor current which will result in a core flux density of 1.0 T. Solution: (a) L = N2 Rg = 12 ´10-3 mH Þ 12 ´ 10-3 ´1.017 ´ 106 = 110.47 N= Þ N =110 turns (b) Þ f = Bg Ac = 1.8 ´ 10-3 Nf 110 ´1.8 ´10-3 = = 16.5 A L 12 ´ 10-3 Bc = Bg = 1.0 T i= l L = Wb 113 The inductor of Fig. 1.27 has the following dimensions: Ac = 1.0 cm2 lc = 15 cm g = 0.8 mm N = 480 turns Neglecting leakage and fringing and assuming mr = 1000 , calculate the inductance. Solution: L= l l = Nf = NBc Ac i mmf equation: Hclc + Hglg = Ni Bc = Bg Þ Bc = m0Ni g + (lc / mr ) Þ Bc Bg mrm0 lc + m0 2 Þ L= g = Ni m0N Ac = 4p ´ 10-7 ´ 4802 ´10-4 = 30.477 mH g + ( lc / mr [0.08 +(15 /1000)]´ 10-2 ) 1 14 The inductor of Problem 1.13 is to be operated from a 60-Hz voltage source. (a) Assuming negligible coil resistance, calculate the rms inductor voltage corresponding to a peak core flux density of 1.5 T. (b) Under this operating con energy. Solution: (a) dl dBc v(t) = dt = NAc dt Bc = Bmax sinwt Þ v(t) = wNAcBmax coswt 1 1 ÞVrms = w NAcBmxa = (2p ´ 60) ´ 480 ´ 10-4 ´1.5 = 19.2 V 2 2 (b) I rs m = 11 Vrms w L = 1.67 A 1 2k 1 Wpeak = LIpa = ´ 30.477 ´ 10-3 ´ ( 2 ´ 1.67)2 = 85.0 mJ 2 2 6 A square voltage wave having a fundamental frequency of 60 Hz and equal positive and negative half cycles of amplitude E is applied to a 1000-turn winding surrounding a closed iron core of 1.25 x 10-3m2 cross section. Neglect both the winding resistance and any effects ofleakage flux. (a) Sketch the voltage, the winding flux linkage, and the core flux as a function of time. (b) Find the maximum permissible value of E if the maximum flux density is not to exceed 1.15 T. (a) e E λmax voltage T λ Φ t λmax E (b) lmxa - (-lmxa ) = 4 f lmxa = 4 fNfmxa = 4fNAcBmxa e(t) = dl Þ l òe(t).dt Þ E = T/2 dt -3 Þ E = 4 ´ 60 ´1000 ´1.25´10 ´1.15 =345 V 1.24 The reciprocating generator of Fig. 1.34 has a movable plunger (position x) which is supported so that it can slide in and out of the magnetic yoke while maintaining a constant air gap of length g on each side adjacent to the yoke. Both the yoke and the plunger can be considered to be of infinite permeability. The motion of the plunger is constrained such that its position is limitedto 0 £ x £ w . There are two windings on this magnetic circuit. The first has N1 turns and carries a constant dc current I0. The second, which has N2 turns, is open-circuited and can be connected to a load. (a) Neglecting any fringing effects, find the mutual inductance between windings 1 and 2 as a function of the plunger position x. (b) The plunger is driven by an external source so that its motion is given by x(t) = where generated as a result of e w(1 + e sin wt) 2 < 1. Find an expression for the sinusoidal voltage which is this m otion. (a) L21 = N1N ç Ac(x) = D(w - x) 2g ÷ (b) æ m0 Ac( x) ö 2 v2 = dl2 dt dx dt l 2 = L21i1 Þ 1 = we wcosw t 2 Þ dL2 v2 = I0 v2 = -I 0 1 dt = I0 dL21 dx = -I dx dt æ m0 N1N2 Dwe w ö ÷ o wt 4g è ø ç æ m0 N1N2D ö dx 0 ç è 2g ÷. ø dt UNIT II TRANSFORMER A transformer is a static machine which transfers ac electrical power from one circuit to another without any electrical link between. It essentially consists of two windings- the primary and secondary, wound on a common laminated magnetic core. The winding connected to the ac source is called primary winding and the one connected to load is called secondary winding. The alternating voltage V1 is applied to the primary. Depending on the no. of turns of primary(N1) and secondary (N2), the alternating emf, E2 is induced in the secondary. Working Principle When a sinusoidally varying voltage V1 is applied to the primary, an alternating flux Φ is set up in the core. This flux links both the windings and induces emfs E1 and E2 in them according to Faraday’s laws of electromagnetic induction. According to Lenz’s law, induced emf acts in opposite direstion to the appl ied voltage V1. d Ie. E1 = N1 E2 = -N2 Therefore f dt f d E2 N2 = E1 N1 dt = K where K is known as voltage transformation ratio. If N2>N1 then E2>E1, it will be a ste If N1>N2 then E1>E2, it will be a step down transformer. Ideal Transformer An ideal transformer is one that has 1. no winding resistance 2. no leakage flux 3. no iron loss. EMF Equation of a Transformer Consider an alternating voltage V1 of frequency f applied to primary of the transformer. This develops a primary current which sets up an alternating flux Φ The instantaneous emf induced in the primary is e1= -N1 df df = -N1 m dt sinw t = -ω N1 Φm Cosωt =-2 П f N1 Φm Cosωt = 2 П f N1 Φm sin(ωt-90) It is clear from the above equation that maximum value of induced emf in the primary is Em1=2 П f N1 Φm the rms value E1=Em1/ 2 = 4.44 f N1 Φm E2=4.44 f N2 Φm Transformer Construction The main elements of a transformer are two windings and a core. The two coils are insulated from each other as well as from the core. The core is constructed from laminations of sheet steel or silicon steel assembled to provide a continuous magnetic path. Silicon steel offers low hysteresis loss and the laminations minimizes eddy current loss. The laminations are insulated from each other by a light coating of varnish. According to the core construction and the manner in which the primary and secondary are placed around it, transformers are classified as 1. core type 2. Shell type LOSSES IN TRANSFORMER The losses in transformer are 1. iron losses or core losses 2. Copper losses Iron losses Since iron core is subjected to alternating flux, there occurs eddy current and hysteresis loss in it. These two losses together known as iron losses and core losses. Both hysteresis and eddy current losses depends on maximum flux density Bm Iron loss = Hysteresis loss + Eddy current loss Copper Losses ` The primary and secondary of the transformer have winding resistances of R1 and R2 respectively. Total copper loss = I12 R1 + I22 R2 Where I1 and I2 are primary and secondary currents. Total losses in a transformer = Pi + Pc EFFICIENCY OF A T RANSFORMER Transformer Efficiency =output power Input power Iron loss of a transformer = Pi Full load copper loss = Pc Total full load loss = Pi + Pc Full load efficiency = Full load VA *P.f Full load VA * P.f + Pi+Pc For any fraction x of the full load efficiency = x Full load VA * P.f x*Full load VA * P.f + Pi+x2 Pc Power Transformer Power Transformers are used in generating stations or sub stations for transforming voltage at each end of transmission line. They are put in operation during load hours and thrown f during light load hours. These transformers are designed to have maximum efficiency at or near of full load. Normally the power transformers are rated in MVA. Distribution Transformer Distribution Transformers are used for stepping down the voltage to a standard voltage and kept near or at the consumer’s premises.They are continuously in circuit whether they are carrying any load or not. He core losses would occur for all tha time where copper losses occur only when they are loaded. So they are designed to reduce the core losses compared to copper losses. They must be designed for good all day efficiency and not for efficiency at full load. Instrument Transformer Instrument transformers are used to extend the range of instruments for the measuring purposes. They are of two types 1. Current transformers for measuring large ac currents 2. Potential transformers for measuring high ac voltages Fig.2.1 Schematic views of (a) core-type and (b) shell-type transformers. Figure 2.2 Cutaway view of self-protected distribution transformer typical of sizes 2 to 25 kVA, 7200:240/120 V. Only one high-voltage insulator and lightning arrester is needed because one side of the 7200-V line and one side of the primary are grounded. (General Electric Company.) No-Load Conditions Exciting current winding ij establishes an alternating flux in the core. Voltage induced in the primary e1 = dl1 dt = N1 d j dt KVL eqn. for the primary winding : v1 = R1ij + e1 R1: primary resistance (primary leakage flux neglected) Resistance drop is very small Þ induced voltage e1 is very nearly equal to the applied voltage. Hence, it is almost sinusoidal. Therefore, the flux is also sinusoidal . j = jmax sinwt Þ e1 = N1 dj = wN1jmax coswt dt E1 @V1 Þ j Rms value: E1 = 1 2 2p fN1j max = 2p fN1j max V1 max = 2p The core flux is fixed by the applied voltage. The required exciting current is determined by the magnetic properties of the core. Figure 2.4 Transformer with open secondary. ˆ I : core-loss component of exciting current; supplies the power due to hysteresis and eddy current losses. c Pc = E1Ij cosqc ˆ I : magnetizing component; m Figure 2.5 No-load phasor diagram. The Ideal Transformer Figure 2.6 Ideal transformer and load. v1 = e1 = N1 dj dt Core flux links the secondary and induces the voltage e2 : Þ v2 = e2 = N 2 dj dt v1 = N1 N2 v2 When a current in the secondary winding flows, the total mmf should be zero since the reluctance of the core is very large. Þ N1i1 - N2i2 = 0 Instantaneous power: Þ i1 N2 = i2 N1 v1i1 = v2i2 Figure 2.7 Three circuits which are identical at terminals ab when the transformer is ideal. Example 2.2 The equivalent circuit of Fig. 2.8a shows an ideal transformer with an impedance R2 + j X2 = 1 + j4 Ω connected in series with the secondary. The tums ratio N1/N2 = 5:1. (a) Draw an equivalent circuit with the series impedance referred to the primary side. (b) For a primary voltage of 120 V rms and a short connected across the terminals A-B, calculate the primary current and the current fo l wing in the short. Figure 2.8 Equivalent circuits for Example 2.2. (a) Impedance in series with the secondary. (b) Impedance referred to the primary. Figure 2.9 Schematic view of mutual and leakage fluxes in a transformer. ˆ ¢ I2 is the component of the primary current which exactly counteracts the mmf of the secondary ˆ ˆ current I2 . The net mmf is produced by the exciting current Ij (n i the primary winding). Therefore the net mmf is ˆ ˆ N1Î j = N1I1 - N2I 2 ˆ ˆ Þ = N1(Î j + I2 ¢) - N2I 2 N2 ˆ ˆ I 2¢ = I2 N1 Þ ˆ ˆ N1I 2¢ = N2I 2 T he equivalent sinusoidal current Iˆ that represents the exciting current can be resolved into aj core-loss component Ic in phase with the emf Ê 1 , and a magnetizing component Im lagging Ê1 by 90°. Rc : core-loss resistance Lm : magnetizing inductance Þ magnetizing reactance: Xm = 2p f Lm Ê1 core loss due tothe resultant mutual flux = 2 Rc excitation branch Exciting impedance Zj =Rc / / jXm Ê1 N1 = Ê2 N2 Figure 2.10 Steps in the development of the transformer equivalent circuit. 2 Xl¢2 = æ R¢ = ç Nö æ ç è N2 ø N1 ö 2 ÷ è N2 ø V2¢= X l2 R2 N1V2 N2 Example 2.3 A 50-kVA 2400:240-V 60-Hz distribution transformer has a leakage impedance of 0.72 + j0.92 Ω in the high-voltage winding and 0.0070 + j0.0090 Ω in the low-voltage winding. At rated voltage and frequency, the impedance Zj of the shunt branch (equal to the impedance of Rc and jXm in parallel) accounting for the exciting current is 6.32 + j43.7 Ω when viewed from the lowvoltage side. Draw the equivalent circuit referred to (a) the high-voltage side and (b) the lowvoltage side, and label the impedances numerically. Approximate equivalent circuits Equivalent series impedance = Re q + j Xeq Example 2.4 Consider the equivalent-T circuit of Fig. 2.11a of the 50-kVA 2400:240 V distribution transformer of Example 2.3 in which the impedances are referred to the high-voltage side. (a) Draw the cantilever equivalent circuit with the shunt branch at the high-voltage terminal. Calculate and labelReq and Xeq. (b) With the low-voltage terminal open-circuit and 2400 V applied to the high-voltage terminal, calculate the voltage at the low-voltage terminal as predicted by each equivalent circuit. Req = 0.72 + 0.70 = 1.42 Ω Xeq = 0.92 + 0.90 = 1.82 Ω Vcd =Vab = 2400 V The equivalent T-circuit 632 + j4370 ö 2400ç ÷ ç Z j + Z l1 ÷ è632.72 + j4370.92 ø ø è = 2399.4 + j0.315 V æ Vc¢d ¢ = 2400 ç Zj ö æ ÷= Example 2.5 The 50-kVA 2400:240-V transformer whose parameters are given in Example 2.3 is used to step dow n the voltage at the load end of a feeder whose impedance is 0.30 + j 1.60 Ω. The voltage Vs at the sending end of the feeder is 2400 V. Find the voltage at the secondary terminals of the transformer when the load connected to its secondary draws rated current from the transformer and the power factor of the load is 0.80 lagging. Neglect the voltage drops in the transformer and feeder caused by the exciting current. Zeq = 1.42 + j 1.82 Ω Ztot = 1.72 + j 3.42 Ω = R + j X Î= V q = cos-1(0.8) = 36.87 50000 2400 = 20.83 A Þ ˆ2¢+ (R + jX )Iˆ = ˆs = 2400Ðd Þ V¢= 2328.3 V Þ lagging (i.e. current lags voltage) Î = 20.83Ð -36.87 A = 16.66 - j12.5 A Þ (V ¢+71.41)2 +35.482 = 24002 æ N2 ö V= ç ÷V ¢ = V N1 ø 232.83 Short-Circuit Test Zj (R2 + jXl 2 ) Zsc = R1 + jXl1 + Zj + R2 + jXl 2 Þ Zc s Zj R1 + jXl1 + R2 + jXl 2 = Req + j R2 + jXl 2 e V ¢= V2ˆ¢ Vsc, Isc and Psc measured Zeq = Zsc = Vsc Isc , Req = Þ Psc Xeq = , 2 Isc q2 Zeq - Re Open-Circuit Test Figure 2.16 Equivalent circuit with open-circuited secondary. (a) Complete equivalent circuit. (b) Cantilever equivalent circuit with the exciting branch at the transformer primary. Voc, Ioc and Poc measured Zoc = Zj = Rc ( jXm ) Rc + jXm Þ Rc = Vo2 , Zj = Voc Poc 1 Xm = , Ioc 2 (1 / Zj ) -(1 / Rc )2 Example 2.6 With the instruments located on the high-voltage side and the low-voltage side short-circuited, the short-circuit test readings for the 50-kVA 2400:240-V transformer of Example 2.3 are 48 V, 2 . A, and 617 W. An open-circuit test with the low-voltage side energized gives instrument readings on that side of 240 V, 5.41 A, and 186 W. Determine the efficiency and the voltage regulation at full load, 0.80 power factor lagging. From the short-circuit test, Zeq,H = 48 = 2.31 W 2 , Re q,H = 617 2 2 = 1.42 W , Xeq,H = 2.312 -1.422 = 1.82 W 50000 At fullload (transformer supplying 50 kVA to the load at 240 V), Power factor = 0.8 Þ Pl od a IH = = 20.8 A 2400 = Potu = 50 000 × 0.8 = 40 000 W 2 Resistive power loss on winding resistances: PR = Req,H IH = 1.42´ 20.82 = 617 W From open-circuit test, Pcore = 186 W Total losses, Ploss = PR + Pcore = 803 W Total power supplied from high-voltage wind Efficiency = Pout Pin 40000 = 40803 Pin = Pout + P ´100 % =98 % Voltage Regulation: ˆ ˆ At full load, V2¢ = 2400Ð0 V, I H = 20.8Ð -cos-1 0.8 = 20.8Ð -36.87 A = 16.64 - j12.48 A Vˆ1 = Vˆ ¢+ ( e + jXeq)IˆH = 2400 + (1.42 + j1.82)(16.64 - j12.48) = 2446 + j13 V 2446 - 2400 Regulation = 2400 ´ 100% = 1.92% TRANSFO RMERS IN THREE-PHASE CIRCUITS Figure 2.19 Common three-phase transformer connections; the transformer windings are indicated by the heavy lines. ( N1 / N2 = a ) Example 2.8 Three single-phase, 50-kVA 2400:240-V transformers, each identical with that of Example 2.6, raeconnected Y-D in a three-phase 150-kVA bank to step down the voltage at the load end of a feeder whose impedance is 0.15 + j 1.00 Ω/phase. The voltage at the sending end of the feeder is 4160 V line-to-line. On their secondary sides, the transformers supply a balanced three-phase load through a feeder whose impedance is 0.0005 + j0.0020 D/phase. Find the line-to-line voltage at the load when the load draws rated current from the transformers at a power factor of 0.80 lagging. Threephase load LV feeder HV feeder Single-phase equivalent circuit: 0.15+ j 1.0 Ω 1.42 + j 1.82 Ω 0.15+ j 0.6 Ω + VH Load _ The voltage at the sending end of the feeder is Vs = 4160 3 = 2400 V line-to-neutral The low-voltage feeder impedance referred to the high voltage side, æ 4160ö Zlv,H = ç ÷ è 240 ø 2 ´ (0.0005+ j0.0020) = 0.15+ j0.60 W Combined series impedance of the high- and low-voltage feeders referred to the high-voltage side, Z feeder,H = 0.30 + j1.6 W/phase-Y The equivalent single-phase series impedance of the transformer is equal to the single-phase series impedance of each single-phase transformer as referred to its high-voltage side Zeq,H = 1.42 + j1.82 W/phase-Y Therefore, the single-phase equivalent circuit for this system is identical to that in Example 2.5. Vload = 2329 V line-to-neutral referred to the HV side. 240 Referred to the LV side: Vlao d = 2329 ´æç 3 ÷ 4160 ø è \ Line-to-line Vo = 3 ´134 = 233 V l ad The per-unit system Actual quantity Quantity in per-unit = Base value of quantity P base, Qbase ,VAbase = VbaseIbase (P,Q,VA)up on base 2 Rbase , X base, Zbase = Vbase Ibs ae éVAbae s1 ù ê ú ëVAbae s2 û = (P,Q,VA)pu on base 1 ´ æ (R, X , Z )up on base 2 = (R, X , Z )pu on base 1 ´ ç Vbase 1 ö æVAbase 2 ö ÷ç ÷ è Vbase 2 ø èVAbase 1 ø Example 2.12 The equivalent circuit for a 100-MVA, 7.97-kV:79.7-kV transformer is shown in Fig. 2.22a. The equivalent-circuit parameters are: XL = 0.040 W, XH = 3.75 W, Xm = 114 W, RL = 0.76 mW, RH = 0.085 W Note that the magnetizing inductance has been referred to the low-voltage side of the equivalent circuit. Convert the equivalent circuit parameters to per unit using the transformer rating as base. Base quantities: LV side: 100 MVA, Vbase=7.97 kV 2 Ra b se ae s= VAb Vbase = Xbase = HV side: VAa = 0.635 W Þ XL = b se 0.040 7.6´10-4 114 = 0.063 p.u., RL = = 0.0012 p.u., Xm = = 180 p.u. 0.635 0.635 0.635 ae VAb s = 100 MVA, Vbase=79.7 kV Rbase = Xbase = Va b2se = 63.5 W VAab se Þ XH = 3.75 63.5 = 0.0591 p.u., RH = 0.085 = 0.0013 p.u. 63.5 Example 2.13 The exciting current measured on the low-voltage side of a 50-kVA, 2400:240-V transformer is 5.41 A. Its equivalent impedance referred to the high-voltage side is 1.42 + j 1.82 Ω. Using the transformer rating as the base, express in per unit on the low- and high-voltage sides (a) the exciting current and (b) the equivalent impedance. Base values: Vbase,H = 2400 V, Vbase,L = 240 V, Ibase,H = 20.8 A, Ibase,L = 208 A 2400 Þ Zbase,H = 20.8 = 115.2 W, Zbase, L = 240 = 1.152 W 208 5.41 (a) Per-unit value of exciting current referred to the LV side: Per-unit value of exciting current referred to the HV side: 1 42 + j1.82 (b) Ij ,L = 208 = 0.026 p.u. 0.541 Ij ,H = = 0.026 p.u. 2 = 0.0123+ j0.0158 p.u. Zeq,H = 115.2 The equivalent impedance referred to the LV side, æ 1ö Zeq,L = ç10 ÷ ´ (1.42 + j1.82) = 0.0142 + j0.0182 W è ø Per-unit value Zeq,L = 00142 + j0.0182 = 0.0123+ j0.0158 p.u. 1 152 3.7 Approximate Equivalent Circuits The voltage drops I1R1and I1X1(Fig.3.11e) are normally small and then the shunt branch (composed of E1 @ V1. If this is true Rc1and Xm ) can be moved to the supply terminal, as shown in Fig.3.12a. This approximate equivalent circuit simplifies computation of currents, because both the exciting branch impeda connected across the supply voltage. Besides, the winding resistances and leakage reactances can be lumped together. This equivalent circuit (Fig.3.12a) is frequently used to determine the performance characteristics of a practical transformer. In a transformer, the exciting current Io is a small percentage of the rated current of the transformer (less than 5%). A further approximation of the equivalent circuit can be made by removing the excitation branch, as shown in Fig.3.12b. The equivalent circuit referred to side 2 is also shown in Fig.3.12c. Fig.3.12 Approximate equivalent circuits. 3.8 Transformer Rating The kVA rating and voltage ratings of example, a typical transformer may carry the following information on the nameplate: 10 kVA, 1100/ 110 volts. What are he meanings of these ratings? The voltage ratings indicate that the transformer has two windings, one rated for 1100 volts and the other for 110 volts. These voltages are proportional to their respective numbers of turns, and therefore the voltage ratio also represents the turns ratio ( a = 1100/ 110 = 10). The 10 kVA rating means that each winding is designed for 10 kVA. Therefore the current rating for the high-voltage winding is 10,000/ 1100 = 9.09 A and for the lower-voltage winding is 10,000/110 = 90.9 A. It may be noted that when the rated current of 90.9 A flows through the lowvoltage winding, the rated current of 9.09 A will flow through the highvoltage winding. In an actual case, however, the winding that is connected to the supply (called the primary winding) will carry an additional component of current (excitation current), which is very small compared to the rated current of the winding. 3.9 Determination Of Equivalent Circuit Parameters The equivalent circuit model (Fig.3.12(a)) for the actual transformer can be used to predict the behavior of the transformer. The parameters R1, Xl1, Rc1, Xm1, R2, X l2 and a = N1 / N2 must be known so that the equivalent circuit model can be used. If the complete design data of a transformer are available, these parameters can be calculated from the dimensions and properties of the materials used. For example, the winding resistances (R1, R2 ) can be calculated from the resistivity of copper wires, the total length, and the cross-sectional area of the winding. The magnetizing inductances Lm can be calculated from the number of turns of the winding and the reluctance of the magnetic path. The calculation of the leakage inductance (L l ) will involve accounting for partial flux linkages and is therefore complicated. However, formulas are available from which a reliable determination of these quantities can be made. These parameters can be directly and more easily determined by performing tests that involve little power consumption. Two tests, a no-load test (or open-circuit test) and a short-circuit test, will provide information for determining the parameters of the equivalent circuit of a transformer. 3.9.1 N o-Load Test (Or Open-Circuit Test) This test is performed by applying a voltage to either the high-voltage side or low-voltage side, whicheveris convenient. Thus, if a 1100/ 110 volt transformer were to be tested, the voltage would be applied to the low-voltage winding, because a power supply of 110 volts is more readily available than a supply of 1100 volts. A wiring diagram for open circuit test of a transformer is shown in Fig.3.13 a. Note that the secondary winding is kept open. Therefore, from the transformer equivalent circuit of Fig.3.12a the equivalent circuit under open-circuit conditions is as shown in Fig.3.12b. The primary current is the exciting current and the losses measured by the wattmeter are essentially the core losses. The equivalent circuit of Fig.3.13b shows that the parameters c Rand Xm1 can be determined from 1 the voltmeter, ammeter, and wattmeter readings. Note that the core losses will be the same whether 110 volts are applied to the low-voltage winding having the smaller number of turns or 1100 volts are applied to the high-voltage winding having the larger number of turns. The core loss depends on the maximum value of flux in the core. (a) (b) Fig.3.13 No-load (or open-circuit) test. (a) Wiring diagram for open-circuit test. (b) Equivalent circuit under open circuit 3.9.2 Short-Circuit Test. This test is performed by short-circuiting one winding and applying rated current to the other winding, as shown in Fig.3.14a. In the equivalent circuit of Fig.3.12a for the transformer, the impedance of the excitation branch (shunt branch composed of Rc1 and X m1 ) is much larger than that of the series branch (composed of the high impedance Req1 and Req1 ). If the secondary terminals are shorted, of the shunt branch c secondary short-circuited can thus be represented by the circuit shown in Fig.3.14 b. Note that 2 since Zeq1 = Re21 + Xeq1 is small, only a small supply voltage is required to pass rated current through the windings. It is convenient to perform this test by applying a voltage to the high-voltage winding. As can be seen from Fig.3.14b, the parameters Req1 and Xeq1 can be determined from the readings of voltmeter, ammeter, and wattmeter. In a well designed transformer, R1 = a2 R2 = R2 ¢ and Xl1 = a2 Xl2 = Xl¢ . Note that because the voltage applied under the short-circuit condition is small, the core losses are neglected and the w attmeter reading can be taken entirely to represent the copper losses in the windings, represented by I12Req1. Fig.3.14 Short-circuit test. (a) Wiring diagram for short-circuit test. (b). Equivalent circuit at short-circuit condition. The following example illustrates the co o f a transformer Example 3.4 Tests are performed on a 1 f , 10 kVA, 2200/220 V, 60 Hz transformer and the following results are obtained. (a) Derive the parameters for the approximate equivalent circuits referred to the low-voltage side and the high-voltage side. (b) Express the excitation current as a percentage of the rated current. (c ) Determine the power factor for the no-load and short-circuit tests. Solution: Note that for the no-load test the supply voltage (full-rated voltage of 220V) is applied to the low-voltage winding, and for the short-circuit test the supply voltage is applied to the high-voltage winding with the low-voltage Equivale winding shorted. The ratings of the windings are as follows: V1(ra e ) = 2200 V V2(r ttdd ) = 220 V 10000 I1( aae ) = = 4.55 A 2200 10000 rted I2(rated ) = = 45.5 A 220 The equivalent circuit and the phasor diagram for the open-circuit test are shown in Fig.3.15a. Pow er, o Pc = V22 Rc2 2202 Then Rc2 = = 484 W 100 Ic2 = 220 484 Im2 = Xm2 = = 0.45 A (I2 - I22 )= (2.52 - 0.452 )= 2.46A V2 22 = 89.4 W = Im2 2.46 The corresponding parameters for the high-voltage side are obtained as follows: Turns ratio a = 2200 = 10 220 Rc1 = a2Rc2 = 102 *484 = 48 400 W Xm1 = a2Xm2 =102 *89.4 = 8940 W The equivalent circuit with the low-voltage winding shorted is shown in Fig.3.15b. Po wer Psc = I1 2 Then, Req1 e1 215 = 4552 = Zeq1 = Vsc1 Isc1 Then, Xeq1 10.4 W 15 0 = = 4.55 Z 2 = 32.97 W 2 Fig.3.15 The corresponding parameters for the low-voltage side are as follows: Req2 = Xeq2 = Req1 10.4 = = 0.104 W a2 102 Xeq1 31.3 = = 0.313 W 2 a2 10 The approximate equivalent circuits referred to the low-voltage side and the high-voltage side are shown in Fig.3.15c. Note that the impedance of the shunt branch is much larger than that of the series branch. (b) From the no-load test the excitation current, with rated voltage applied to the low-voltage winding, is: Io = 2.5A This is 25 *100% = 5.5% of the rated current of the winding 45.5 Power c power factor at no load = volt ampere 100 = = 0.182 220*2 215 Power factor at short circuit condition = 150 * 4.55 = 0.315 Example 3.5 Obtain the equivalent circuit of a 200/400-V, 50 Hz, 1 phase transformer from the following test a :-O C. test : 200 V, 0.7 A, 70W-on LV side S C. test : 15 V, 10 A, 85 W-on HV side Calculate the secondary voltage when delivering 5 kW at 0.8 power factor lagging, the primary voltage being 200 V. Solution: From O.C. Test Po = Vo Io * cosj \ cosj o = Then j o Po 70 = Vo Io 200*0.7 -1 o o = cos 0.5 = 60 Then Ic1 = Io cosj o = 0.5 = 0.7*0.5 = 0.35A Im1 = Io sin j o = 0.7*0.866 = 0.606A Vo1 200 = 571.4 W Then Rc1 = = Ic1 0.35 And Xm1 200 = 330 W = Vo1 = 06 Im1 06 As shown in Fig.3.16, these values refer to primary i.e. low-voltage side From Short Circuit test: It may be noted that in this test instruments have been placed in the secondary i.e. highvoltage winding and the low-voltage winding i.e. primary has been short-circuited. Now , Zeq2 = V2sc 15 = =1.5W I2sc 10 2 æ1 ö 2 Zeq1 = a * Zeq2 = ç ÷ *1.5 = 0.375W è 2ø Also, Psc = I2sc Req2 85 Then, e 2 Then, Req1 Then, e 1 = 100 = 0.85W æ1 ö 2 = a2 * Req2 = ç ÷ *0.85 = 0.21 W è 2ø = Ze21 - Re21 = 0.3752 - 0.212 = 0.31 W Fig.3.16 real power Output K VA = Power factor Output current I2 Now, from Then, V2 5 08 = 6.3 kVA 5000 = 0.8* 400 aproximate the V2 Ð0o = V1¢Ðd = o - I2 Ð j o o Ð 0 = 400Ðd V2Ð 0o = 400Ðd o o = 15.6A equivalent circuit refeared * Zeq 2 - 15.6Ð- 36.87o *(0.85 + j1.2) - 15.6Ð -36.87o *1.5Ð 54.7o to secondery : o V2Ð0 = 400Ðd o - 23.4Ð18.17o From the above equation we have two unknown variables V2 and do it need two equations to get both of them. The above equation is a complex one so we can get two equations out of it. If we equate the real parts together and the equate the imaginary parts: So from the Imaginary parts: V2 sin (0) = 40 0 sin do - 23.4*sin 18.1 7o ) ( ) ( )- 7.41 0 = 400*sin d Then, d o o ( o = 7.4o So from the Real parts: V2 cos (0) = 400 * cos(7.41o - 23.4* cos 18.1 7o ) ) Then, V2 ( = 374.5 V Example 3.6 A 50 Hz, 1 - f transformer has a turns ratio of 6. The resistances are 0.9 W, 0.03 W and reactances are 5W and 0.13 W for high-voltage and low-voltage, windings respectively. Find (a) the voltage to be applied to the HV side to obtain full-load current of 200 A in the LV winding on short-circuit (b) the power factor on short-circuit. Solution: The turns ratiois a = N1 =6 N2 Req1 = R1 + a2 R2 = 0.9 + 62 * 0.03 = 1.98 W Xeq1 = X1 + a2 X2 = 5 + 62 *0.13 = 9.68 W Zeq1 = Re21 + Xeq21 = 1.982 + 9.682 = 9.88 W I1 = I2 200 = 33.3 = 3A a 6 (a) s = I 1 * e 1 = 9.88*33.33 = 329.3V Zq Vc (b) cos j = Req1 Zeq1 = 1.98 9.88 = 0.2 Example 3.7 A 1 phase, 10 kVA, ,500/250-V, 50 Hz transformer has the following constants: Resistance: Primary 0.2 W ; .Secondary 0.5W Reactance: Primary 0.4W ; Secondary 0.1 W Resistance of equivalent exciting circuit referred to primary, Rc1 = 1500W Reactance of equivalent exciting circuit referred to primary, Xm1 = 750 W. What would be the readings of the instruments when the transformer is connected for the open-circuit and-short-circuit tests? Solution: O C. Test: V 1 500 2 = = A X m 750 3 Im 1 = Ic 1 = V1 Rc1 = 500 1 = A 1500 3 2 Io = æ1 ö 2 æ 2ö ç ÷ + ç ÷ = 0.745 A è3ø è3 ø 1 = 167W No load prim ary input V 1 * Ic1 = 500 * 3 Instruments used in primary circuit are: voltmeter, ammeter and wattmeter, their readings being 500 V, 0745 A and 167 W respectively. S C. Test Suppose S.C. test is performed by short-circuiting the LV, winding i.e. the secondary so that all instruments are in primary. Req1 = R1 + R2¢= R1 + a2 R2 = 0.2 + 4 * 0.5 = 2.2 W Xeq1 = X1 + X2¢= X1 + a2 X2 = 0.4 + 4*0.1 = 0.8W Then, Zeq1 = Re21 + Xe21 = 2.22 + 0.82 = 2.341 W Full-load primary current Rated kVA I1 = 10000 = = 20 A Rated Pr imaryvoltage 500 Then Vsc = I 1 * e 1 = 20*2.431 = 46.8V Power absorbed = I12* Req1 = 202 * 2.2 = 880W Primary instruments will read: 468 V, 20 A, 880 W. 3.10 E fficiency Equipment is desired to operate at a high efficiency. Fortunately, losses in transformers are small. Because the transformer is a static device, there are no rotational losses such as windage and friction losses in a rotating machine. In a well-designed transformer the efficiency can be as high as 99%. The efficiency is defined as follows: h= output power (Potu)* 10 0 = Input Power( Pin ) Pout Pout + Losses *100 (3.14) The losses in the transformer are the core loss (Pc) and copper loss (Pcu ). Therefore, h= Pout Pout = Pout + Losses Pout + Pc + Pcu (3.15) The copper loss can be determined if the winding currents and their resistances are known: 2 2 Pcu = I1 R1 + I2 R2 ) = I12Req1 = I22Req2 The copper loss is a function of the load current. The core loss depends on the peak flux density in the core, which in turn depends on the voltage applied to the transformer. Since a transformer remains connected to an essentially constant voltage, the core loss is almost constant and can be obtained from the no -load test of a transformer. Therefore, if the parameters of the equivalent circuit of a transformer are known, the efficiency of the transformer under any operating condition m ay be determined. Now, Pout = V2 I2 cosj 2 Therefore, h= h= V2I2 cosj 2 V2I2 cosj 2 + Pc + I22Req2 V2¢* I¢*cosj V¢* I¢*cosj 2 * 100 2 *100 (3.17) (3.18) + Pc + I¢2Req1 3.11 Maximum E fficiency For constant values of the terminal voltage V 2 and load power factor angle j 2, the maximum efficiency occurs when: dh dI2 =0 (3.19) If this condition is applied to Eqn. (3.17) the condition for maximum efficiency is: Pc = I 2Req 2 (3.20) That is, core loss = copper loss. For full load condition, Pcu,FL = I22FL Req2 (3.21) Let x = I2 I2,FL = per unit loading (3.22) From Eqns. (3.20), (3.21) and (3.22). Pc = x2 Pcu, FL Then, x = æ ç ö Pc ç è ÷ ÷ ( 3) ( 4) ø For constant values of the terminal voltage V2 and load current I 2 , the maximum efficiency occurs when: dh dj =0 (3.25) 2 Fig.3.17 Efficiency of a transformer. If this condition is applied to Eq.(3.17), the condition for maximum efficiency is j 2 = 0 Then, cosj 2 = 1 that is,load power factor = 1 Therefore, maximum efficiency in a transformer occurs when the load power factor is unity (i.e., resistive load) and load current is such that copper loss equals core loss. The variation of efficiency with load current and load power factor is shown in Fig.3.17. Example 3.8 For the transformer in Example 3.4, determine (a) Efficiency at 75% rated output and 0.6 PF. (b) Power output at maximum efficiency and the value of maximum efficiency. At what percent of full load does this m aximum efficiency occur? Solution: Pout = V2 I2 cosj 2. (a) = 0.75*10000 *0.6 = 4500W Pc = 100W , Pcu = I12Req1 =(0.75 * 4.55)2 *10.4 =121W h= 4500 *100 = 95.32% 4500 +100 +121 (b) At maximum efficiency Pcore = Pcu and PF = cosj 2 Now , c re = 100W = I22Req2 = Then, I2 æ =1 c 1/ 2 =ç 100 ö = 31 A ÷ è0.104ø Pout h max = V2I2 cos j 2 = 220 * 31*1 = 6820W Pout h max hmax = Pout h max + Pc + Pcu = 6820 682 =97.15% output kVA=6.82 and Rated kVA=10 Then, h mx a occurs at 68.2% full load. Anther Method From Example 3.4 c ,FL Pu Then X æ = Pc = 215W æ 100ö ç ö ÷= ç ÷ = 0.68 ç Pu ÷ è c ,FL ø è215 ø Example 3.9 Obtain the equivalent circuit of a 8kVA 200/400 V, 50 Hz, 1 phase transformer from the following test a :- O.C. test : 200 V, 0.8 A, 80W, S.C. test : 20 V, 20 A, 100 W Calculate the secondary voltage when delivering 6 kW at 0.7 power factor lagging, the primary voltage being 200 V. From O.C. Test Po = Vo I o * cosj Po \ cosj o Then j o 80 = Vo Io 200 *0.8 -1 o = cos 0.5 = 60o Then Ic1 = Io cosj o = 0.5 = 0.8*0.5 = 0.4A Im1 = Io sin j o = 0.8*0.866 = 0.69282A Then, Rc1 200 = Vo1 = = 500 W Ic1 0.4 And X m1 = Vo1 200 = 288.675W = 9282 06 Ic1 From Short Circuit test: It may be noted that in this test instruments have been placed in the secondary i.e. high voltage winding and the low voltage winding i.e . primary has been short-circuited. Now, Zeq2 = Also, s V2sc 20 = = 1W I2sc 20 = I2sc Req2 Then, Req2 Then, X eq 2 = = 100 202 = 0.25W Z2q2 - Re22 = Output current I2 12 - 0.252 = 0.968246 W 6000 = = 21.4286 A 07 * 400 Now, from the aproximate equivalent circuit refeared to secondery : o o V2 Ð0o = 400Ðd o V2 Ð0o = V1¢Ðd - I2 Ðj Then, o V2Ð0 = 400Ðd o *Zeq2 - 21.4286 Ð - 45.573o *(0.25 + j0.968246) - 21.43 Ð29.9495o o From the above equation we have two unknown variables V2 and d it need two equations to get both of them. The above equation is a complex one so we can get two equations out of it. If we equate the real parts together and the equate the imaginary parts: So from the Imaginary parts: o ( )- 21.43*sin (29.9495 V2 sin (0) = 400 sin d o ) o ( )-10.6986 0 = 400*sin d Then, d o = 1.533o So from the Real parts: V2 cos (0) = 400 *cos(1.533o Then, V2 = )- 21.43* cos(29.9495 o ) 381.288 V Example:3.10 A 6kVA, 250/500 V, transformer gave the following test results short-circuite 20 V ; 12 A, 100 W and Open-circuit test : 250 V, 1 A, 80 W I. Determine the transformer equivalent circuit. II. calculate applied voltage, voltage regulation and efficiency when the output is 10 A at 500 volt and 0.8 power factor lagging. III. Maximum efficiency, at what percent of full load does this maximum efficiency occur? (At 0.8 power factor lagging). IV. At what percent of full load does the effeciency is 95% at 0.8 power factor lagging. Solution: (I) From O.C. Test Po = Vo Io * cosj \ cosj o = o Po VoIo 80 = 0.32 = 250 *1.0 j o = cos- 0.32 = 71.3371o Then Ic = Io cosj o =1.0*0.32 = 0.32A Im = Io sin j o =1.0*0.7953 = 0.7953 A 1 Then 1 1 Then Rc1 250 = Vo1 = = 781.25 W Ic1 0.32 250 = 314.35W = Vo1 = 0.7953 Im1 And Xm1 As shown in Fig.3.16, these values refer to primary i.e. low-voltage side From Short Circuit test: The rated current of the secondary side is: I2 = 6000 =12 A 500 It is clear that in this test instruments have been placed in the secondary i.e. highvoltage winding and the low-voltage winding i.e. primary has been short-circuited. Now, Zeq2 = V2sc 20 = =1.667W I2sc 12 2 Zeq1 = a 2 * Zeq2 = æç1ö ÷ *1.667 = 0.4167W è2ø Also, Psc = I2c Req2 Then, Req2 Then, Req1 Then, e1 100 = 122 = 0.694 W æ 1ö 2 = a2 * Req2 = ç ÷ *0.694 = 0.174 W è2ø = Z e2 1 - R e 2 1 = 0.41672 - 0.1742 = 0.3786 W As shown in the following figure, these values refer to primary i.e. low-voltage side j0 3786 0.174 V2 ¢ I 781.25 0 V1 314.35 The parameters of series branch can be obtained directly by modifying the short circuit test data to be referred tothe primary side as following: SC test 20 V ; 12 A, 100 W (refered to secondery) SC test 20*a=10V ; 12/a=24A, 100 W (refered to Primary) Zq e1 So, = V1sc 10 = = 0.4167W I1sc 24 = I12c Req1 Also, s Then, Req1 Then, e1 = 100 = 242 Zq e21 = 0.174 W - Re21 = 0.41672 - 0.1742 = 0.3786 W It is clear the second method gives the same results easly. =10*500*0.8 = 4 kVA (II) Output K VA Now, from the aproximate equivalent circuit refeared to secondery : o V 1 Ðd = V2¢Ð0o + I2¢ Ðj o *Zeq1 o Then, V1 Ðd = 250Ð 0o + 20 Ð- 36.87o *(0.174 + j0.3786) VR = = 257.358Ð0.89o V1 -V2¢ = 257.358 - 250 Pout = 10*500*0.8 = 4kW , *100 = 2.943% V2 ¢ 250 Pi = Poc = 80W , and , Pcu = 102 * Req 2 = 100*0.694 = 69.4W or Pcu = *çç 2 2 I2 ö Psc ÷ I2SC ø æ ÷ * =1ç00 ÷æ10 è12 ø = 69.4 W ö è h= ot Pu Pout + Pi + Pcu = 4000 *100 = 96.4% 4000 + 80 + 69.4 (III) maximum effeciency ocures when Pc = c = 80W the The percent of the full load at which maximum efficiency occurs is : Pc X= ç ö æ 80 ÷ ÷ = ç c ,FL è Pu ø = 0.8945% 100 Then, the maximum efficiency is : h= 6000 * 0.8945 * 0.8 *100 = 96.41% 6000 * 0.8945 * 0.8 + 80 + 80 (IV) h= Pout = 0.95 + Pi + Pcu 6000*0.8* x = 0.95 = 2 6000*0.8* x + 80 +100* x ot Pu Then, 95 x2 - 240 x + 76 = 0 Then, Or x = 2.155 (Unacceptable) x = 0.3712 Then to get 95% efficiency at 0.8 power factor the transformer must work at 37.12% of full load. 3.12 All-Day (Or Energy) Efficiency, had The transformer in a power plant usually operates near its full capacity and is taken out of circuit when it is not required. Such transformers are called power transformers, and they are usually designed for maximum efficiency occurring near the rated output. A transformer connected to the utility that supplies power to your house and the locality is called a distribution transformer. Such transformers are connected to the power system for 24 hours a day and operate well below the rated power output for most of the time. It is therefore desirable to design a distribution transformer for maximum efficiency occurring at the average output power. A figure of merit that will be more appropriate to represent the efficiency performance of a distribution transformer is the "all-day" or "energy" efficiency of the transformer. This is defined as follows: had = had = energy output over 24 hours *100 energy input over 24 hours (3 26) energy output over 24 hours energy output over 24 hours + Losses over 24 hours If the load cycle of the transformer is known, the all day effeciency can be deteremined. Example 3.11 A 50 kVA, 2400/240 V transformer has a core loss P, = 200 W at rated voltage and a copper loss Pcu = 500 W at full load. It has the following load cycle. %Load 0.0% Power F actor Hours 6 50% 75% 100% 110% 1 0.8Lag 0.9Lag 1 6 6 3 3 Determine the all-day efficiency of the transformer. Solution Energy output 24 hours is 0.5*50*6+0.75*50*0.8*6+1*50*0.9*3+1.1*50*1*3=630 kWh Energy losses over 24 hours: Core loss =0.2*24=4.8 kWh Copper losses = 05 2 *0.5*6 + 0.752 *0.5*6 +12 *0.5*3+1.12 *0.5*3 =5.76 kWh Total energy loss=4.8+5.76=10.56 kWh Then, h AD 630 = 63 0 +10.56 *100 = 98.35% 3.13 Regulation of a Transformer (1) When a transformer is loaded with a constant primary voltage, then the secondary terminal voltage drops because of its internal resistance and leakage reactance. Let. V2o =Secondary terminal voltage at no-load = E2 = E 1 / a = V 1 / a Because at no-load the impedance drop is negligible. V2 = Secondary terminal voltage on full-load. The change in secondary terminal voltage from no-load to full-lead is = V2o - V2 . This change divided by 2 is known as regulation down. if this change is divided by load secondary terminal voltage, then it is called regulation up. %reg = Vno- load - Vload *100 Vload %reg = %reg = - (V2 ) (V 2 ) no-load load (V2 ) load V1¢-(V2 ) (V2 ) (3.27) load load *100 = (3.28) *100 V 1 - ( V 2 ¢) (V2¢) load load *100 (3.29) V2 i.e. full- As the transformer is loaded, the secondary terminal voltage falls (for a lagging power factor). Hence, to keep the output voltage constant, prim ary voltage required to maintain rated output voltage from no-load to full-load at a given power factor expressed as percentage of rated primary voltage gives the regulation of the tr ansformer. Vector diagram for the voltage drop in the transformer for different load power factor is show n in Fig.3.18. It is clear that the only way to get V1 less than V2¢ is when the power factor is leading which means the load has capacitive reactance (i.e. the drop on Zeq1 will be negative, which means the regulation m ay be negative). V1 j V2 I2¢ Xeq1 I¢ Zeq1 ¢ I¢ Req1 I 2¢ (a) V1 I2¢ X eq1 I¢ Ze 1 q I2 ¢ V2 V1 I¢ 2 I¢ Req1 ¢ I2X I¢ Ze 1 (b) eq1 q j I¢ Req1 V2 ¢ (c) Fig.3.18 Vector diagram for transformer for different power factor (a) lagging PF (b) Unity PF (c) Leading PF. Example 3.12 A 250/500 V, transformer g Short-circuit test : with lowvoltage winding shorted. short-circuited 20 V ; 12 A, 100 W Open-circuit test : 250 V, 1 A, 80 W on low-voltage side. Determine the circuit constants, insert these on the equivalent circuit diagram and calculate applied voltage, voltage regulation and efficiency when the output is 5 A at 500 volt and 0.8 power factor lagging. Solution Open circuit test cosj o = Poc Voc Io c Ic1 = Io cosj o 80 = 0.32 250*1 = 1*0.32 = 0.32A 2 Im1 = Io2 - Ic = 12 - 0.322 = 0.95A V1oc 250 = 781.3 W Rc1 = I = c 0.32 Xm1 = V1oc 250 = = 263.8 W Im 0.95 Short circuit test As the primary is short-circuited, all values refer to secondary winding. So we can obtain e 2 and Rq X eq2 and then refer them to primary to get Req1 and Xeq1 as explained before in Example 3.5 or we can modify the short circuit data to the primary and then we can calculate Req1 and Xeq1 directly. Here will use the two method to compare the results. First method Req2 Psc = 100 2 2 sc I Vsc Zeq2 = I2sc Then, Xeq2 As = = 0.694 W 122 20 = 12 =1.667 W Z2q2 - Re22 = 1.6672 - 0.6942 = 1.518 W = Rc and Xm refer to primary, hence we will transfer these values (Rqe2 , Xeq2 , and Zeq2 ) to primary with the help of transformation ratio. Then Req1 = a2 * Req2 = 0.52 * 0.694 = 0.174 W Xeq1 = a2 * Xeq2 = 0.52 *1.518 = 0.38W Zeq1 = a2 * Zeq2 = 0.52 *1.667 = 0.417W Second method Short-circuited results refeard to secondery are 20 V, 12 A, 100 W Then, Short-circuited results refeard to primary are 10 V, 24 A, 100 W Then e 1 Zeq1 = 100 = Psc = = 0.174 W 2 I1 sc 242 V1sc 10 = = 0.417 W I1sc 24 e1 = Ze21 - Re21 = 04172 - 0.1742 = 0.38 W Then, Applied voltage o V 1 d = V2 ¢ Then, V1 o o 0 o + I¢ j d = 250 0 o V1 d = 250 0o +10 o * Zq e1 +10 - cos-1 0 8*(0.174 + j0.38) -36.24o *0.418 65.4o o V1 d = 250 0o + 4.18 29.16o o V1 d = 250 0o + 3.65 + j2.04 = 253.65 + j2.04 = 253.7 04 7o V Voltage regulation (V1 )- (V2¢) load (V2¢) load %reg = (V2¢) *100 = 250 00 load 253.7 - 250 %reg = 250 *100 =1.48% f eciency Ef h= h= V¢* I ¢*cos j rn V¢* I ¢*cosj + Pcu + Pi o *100 25 0*10*0.8 *100 = 95.356% 2 250*10*0.8 +10 *0.174 +80 Example 3.13 A 1f , 10 kVA, 2400/240 V, 60 Hz distribution transformer has the following characteristics: Core loss at full voltage =100 W and Copper loss at half load =60 W (a) Determine the efficiency of the transformer when it delivers full load at 0.8 power factor lagging. (b) Determine the rating at which the transformer efficiency is a maximum. Determine the efficiency if the load power factor is 0.9. (c) The transformer has the following load cycle: No load for 6 hours, 70% full load for 10 hours at 0.8 PF and 90% full load for 8 hours at 0.9 PF Solution: Pout = 10*0.8 = 8 kW (a) Pcore = 100 W , Pcu, FL == 60 * 22 = 240W 8000 h= *100 = 95.92% 8000 +100 + 240 (b) x 100 = hmax = 240 = 0.6455``\ 10 *10 3 * 0.6455* 0.9 = 96.67% 4 10 * 0.6455* 0.9)+100 +100 ( Output energy in 24 hours is: E24hrs = 0 +10*0.7*0.8*10 +10*0.9*0.9*8 =120.8kWh Energy losses in the core in 24 hours is Ecore =100* 24*10-3 = 2.4kWh Energy losses in the cupper in 24 hours is ( Ecu = 240*0.72 *10 + 24 0*0.92 *8)*10-3 = 2.7312kWh Then, hall day = 12 120.8 + 2.4 + 2.7312 *100 = 95.93% 3.14 Percentage Resistance, Reactance and Impedance These quantities are usually measured by the voltage drop at full-load current expressed as a percentage of the normal voltage of the winding on which calculations are made. (i) Percentage resistance at full load I1*Req1 %R = V1 *100 = I12Req1 V1 I 1 *100 (3.30) 2 = I2 Req2 V2 I 2 *100 = %Cu Loss at full load Percentage reactance at full load: %X = %Z = I2 Xeq2 I 1 * Xeq1 *100 = *100 V1 V2 I1Zeq1 V1 (3.31) I2Zeq2 *100 = %Z = %R2 + % X V2 *100 ( 2) (3.33) 3.15 Autotransformer This is a special connection of the transformer from which a variable AC voltage can be obtained at the secondary. A common winding as shown in Fig.3.19 is mounted on core and the secondary is taken from a tap on the winding. In contrast to the two-winding transformer discussed earlier, the primary and secondary of an autotransformer are physically connected . However, the basic principle of operation is the same as that of the two-winding transformer. Fig.3.19 Step down autotransformer. Since all the turns link the same flux in the transformer core, V1 N1 = =a V2 N2 ( 4) If the secondary tapping is replaced by a slider, the output voltage can be varied over the range0 < V2 < V1. The ampere-turns provided by the upper half (i.e., by turns between points a and b) are: æ1 (N1 - N2 )* I1 = ç è - 1öN ÷ aø I1 ( 1 5) The ampere-turns provided by the lower half (i.e., by turns between points b and c) are: N 2( I 2 - I 1) = N1 (I2 - I1 ) a (3.36) from amper turn balance, from equations (3.35) and (3.36) ç æ1 - è 1 öN I ÷ 1 aø (3.37) I1 Then, 1 = (3.38) = N1 a (I2 - I1) I2 a Equations (3.34) and (3.37) indicate that, viewed from the terminals of the autotransformer, the voltages and currents are related by the sa The advantages of an autotransformer connection are lower leakage reactances, lower losses, lower exciting current, increased kVA rating (see Example 3.11), and variable output voltage when a sliding contact is used for the secondary. The disadvantage is the direct connection between the primary and secondary sides. Example 3.14 A 1 f , 100 kVA, 2000/200 V two-winding transformer is connected as an autotransformer as shown in Fig.E2.6 such that more than 2000 V is obtained at the secondary. The portion ab is the 200 V winding, and the portion be is the 2000 V winding. Compute the kVA rating as an autotransformer. Fig.3.20 Solution: The current ratings of the windings are Therefore, for full-load operation of the autotransformer, the terminal currents are: A single-phase, 100 kVA, two-winding transformer when connected as an autotransformer can deliver 1100 kVA. Note that this higher rating of an autotransformer results from the conductive connection. Not all of the 1100 kVA is transformed by electromagnetic induction. Also note that the 200 V winding must have sufficient insulation to withstand a voltage of 2200 V to ground. Example 3.15 A single phase, 50 kVA, 2400/460 V, 50 Hz transformer has an efficiency of 0.95% when it delivers 45kW at 0.9 power factor. This transformer is connected as an autotransformer to supply load to a 2400 V circuit from 2860 V source. (a) Show the transformer connection. (b) Determine the maximum kVA the autotransformer can supply to 2400 V circuit. (c) Determine the efficiency of the autotransformer for full load at 0.9 power factor. Solution: (a) 460 2860 2400 (b) Is,2w = 50*103 2460 Then, kVA)Auto (c) h2w = Then, Pi h Auto = + = 108.7 A = 108.782860 = 310.87 kW 50*103 *0.9 = 0.95 3 50 *10 * 0.9 + Pi + Pcu,FL c ,FL = 2368.42 W 310870 * 0.9 =99.61 % 310870*0.9 + 2368.42 3.16 Three-Phase Transformers 3.16.1 Introduction Poweris distributed throughout The world by means of 3-phase transmission lines. In order to transmit this power efficiently and economically, the voltages must be at appropriate levels. These levels (13.8 kV to 1000 kV) depend upon the amount of power that has to be transmitted and the distance it has to be earned. Another aspect is the appropriate voltage levels used in factories and homes. These are fairly uniform, ranging from 120/240 V single-phase systems to 480 V, 3-phase systems. Clearly, this requires the use of 3-phase transformers to transform the voltages from one level to another. The transformers may be inherently 3-phase, having three prim ary windings and three secondary windi result can be achieved by using three single-phase transformers connected together to form a 3-phase transformer bank. 3.16.2 Basic Properties Of 3-Phase Transformer Banks When three single-phase transformers are used to transform a 3-phase voltage, the windings can be connected in several ways. Thus, the primaries may be connected in delta and the secondaries in wye, or vice versa. As a result, the ratio of the 3-phase input voltage to the 3-phase output voltage depends not only upon the turns ratio of the transformers, but also upon how they ar e connected. A 3-phase transformer bank can also produce a phase shift between the 3-phase input voltage and the 3-phase output voltage. The amount of phase shift depends again upon the turns ratio of the transformers, and on how the primaries and secondaries are interconnected. Furthermore, the phaseshift feature enables us to change the number of phases. Thus, a 3-phase system can be converted into a 2-phase, a 6-phase, or a 12-phase system. Indeed, if there were a practical application for it, we could even convert a 3-phase system into a 5-phase system by an appropriate choice of single-phase transformers and interconnections. In making the various connections, it is important to observe transformer polarities. An error in polarity may produce a short-circuit or unbalance the line voltages and currents. The basic behavior of balanced 3-phase transformer banks can be understood by making the following simplifying assumptions: 1.The exciting currents are negligible. 2.The transformer impedances, due to the resistance and leakage reactance of the windings, ar e negligible. 3.The total apparent input power to the transformer bank is equal to the total apparent output power. Furthermore, when single-phase transformers are connected into a 3-phase system, they retain l their basic single-phase properties, such as current ratio, voltage ratio, and flux in the core. al Given the polarity marks X1, X2 and H1 , H2 , the phase shift between primary and secondary is zero. 3.16.3 Delta-Delta Connection The three single-phase transformers P, Q, and R of Fig.3.21 transform the voltage of the incoming transmission line A, B, C to a level appropriate for the outgoing transmission line 1, 2, 3. The incoming line is connected to the source, and the outgoing line is connected to the load. The transformers are connected in delta-delta. Terminal H1 of each transformer is connected to terminal H2 the next transformer. Sof transformers are connected together. The actual physical layout of the transformers is shown in Fig.3.21. The corresponding schematic diagram is given in Fig.3.22. The schematic diagram is draw n in such a way to show not only the connections, but also the phasor relationship between the primary and secondary voltages. Thus, each secondary winding is drawn parallel to the corresponding primary winding to which it is coupled. Furthermore, if source G produces voltages EAB , EBC , C according to the indicated phasor diagram, the primary windings are oriented the same way, phase by phase. For example, the primary of transformer P between lines A and B is oriented horizontally, in the same direction as phasor A . Fig.3.21 Delta-delta connection of three single-phase transformers. The incoming lines (source) are A, B, C and the outgoing lines (load) are 1, 2, 3. Fig.3.22 Schematic diagram of a delta-delta connection and associated phasor diagram. In such a delta-delta connection, the voltages between the respective incoming and outgoing transmission lines are in phase. If a balance currents are equal in magnitude. This produces balanced line currents in the incoming lines A-B-C. As in any delta connection, the line currents are 43 times greater than the respective currents IP and IS flowing in the primary and secondary windings (Fig.3.22). The power rating of the transformer bank is three times the rating of a single transformer. Note that although the transformer bank constitutes a 3-phase arrangement, each transformer, considered alone, acts as if it were placed in a singlephase circuit. Thus, a current IP flowing from H1 H2 in the primary winding is associated with a current IS flowing from X2 to X1 in the secondary. Example 3.16 Three single-phase transformers are connected in delta-delta to step down a line voltage of 138 kV to 4160 V to su-pply power to a manufacturing plant. The plant draws 21 MW at a lagging power factor of 86 percent. Calculate a. The apparent power drawn by the plant b. The apparent power furnished by the HV line c.The current in the HV lines d. The current in the LV lines e. The currents in the primary and secondary windings of each transformerf. The load carried by each transformer Solution: a The appearent power drawn by the plant is: S = P / cosj = 21/0.86 = 24.4 MVA b. The transformer bank itself absorbs a negligible amount of active and reactive power because the I 2 R losses and the reactive power associated with the mutual flux and the leakage fluxes are small. It follows that the apparent power furnished by the HV line is also 24.4 MVA. c.The current in each HV line is:- I1= S 3 *V1 24.4 *106 = 3 *13800 =102 A d.The current in the LV lines is:- I2 = S 24.4*106 = = 3386 A 3V2 3 *4160 e Referring to Fig.3.19, the current in each primary winding is: I p The current in each secondary winding is: IS = = 102 = 58.9 A 3386 =1955 A 3 f. Because the plant load is balanced, each transformer carries one-third of the total load, or 24. / = 8.13 MVA. The individual transformer load can also be obtained by multiplying the primary voltage times the primary current: S = Ep I p =138000*58.9 = 8.13MVA Note that we can calculate the line currents and the currents in the transformer windings even though we do not know how the 3-phase load is connected. In effect, the plant load (shown as a box in Fig.3.22) is composed of hundreds of individual loads, some of which are connected in delta, others in wye. Furthermore, some are single-phase loads operating at much lower voltages than 4160 V, powered by smaller transformers located inside the plant. The sum total of these loads usually results in a reasonably well-balanced 3-phase load, represented by the box. 3.16.4 Delta-wye connection When the transformers are connected in delta-wye, the three primary windings are connected the same way as in Fig.3.21. However, the secondary windings are connected so that all the X2 terminals are joined together, creating a common neutral N (Fig.3.23). In such a delta-wye connection, the voltage across each primary winding is equal to the incoming line voltage. However, the outgoing line voltage is 3 times the secondary voltage across each transformer. The relative values of the currents in the transformer windings and transmission lines are given in Fig.3.24. Thus, the line currents in phases A, B, and C are 3 times the currents in the primary windings. The line currents in phases 1, 2, 3 are the same as the currents in the secondary windings. A delta-wye connection produces a 30 phase shift between the line voltages of the incoming and outgoing transmission lines. Thus, outgoing line voltage ahead of incoming line voltage 1 is 30 degrees , as can be seen from the phasor diagram. If the outgoing line E2 feeds an isolated group of loads, the phase shift creates no problem. But, if the outgoing line has EB to be connected in parallel with a line coming from another source, the 30 degrees shift may A make such a parallel connection impossible, even if the line voltages are otherwise identical. One of the important advantages of the wye connection is that it reduces the amount of insulation needed inside thetransformer. The 58 percent of the line voltage. Fig.3.23 Delta-wye connection of three single-phase transformers. Fig.3.24 Schematic diagram of a delta-wye connection and associated phasor diagram. (The phasor diagrams on the primary and secondary sides are not drawn to the same scale.) Example3.17 Three single-phase step-up transformers rated at 90 MVA, 13.2 kV/80 kV are connected in delta-wye on a 13.2 kV transmission line (Fig.3.25). If they feed a 90 MVA load, calculatethe following: a.The secondaryline voltage c.The incoming and outgoing transmission line currents b.The currents in the transformer windings Fig.3.25. Solution The easiest way to solve this problem is to consider the windings of only one transformer, say, transformer P. a. The voltage across the primary winding is obviously 13.2 kV The voltage across the secondary is, therefore, 80 kV. The voltage between the outgoing lines 1, 2, and 3 is: V2 = 80* 3 =139kV b . The load carried by each transformer is =90 / 3 = 30MVA 3.16.5 Wye-delta connection S The currents and voltages in a wye-delta connection are identical to those in the delta-wye connection. The primary and secondary conn neutral, and the H2 terminals are connected together to create a X1, X2 terminals are connected in delta. Again, there results a 30 degrees phase shift between the voltages of the incoming and outgoing lines. 3.16.6 Wye-wye connection When transformers are connected in wye-wye, special precautions have to be taken to prevent severe distortion of the line-to-neutral voltages. One way to prevent the distortion is to connect the neutral of the primary to the neutral of the source, usually by way of the ground (Fig.3.26). Another way is to provide each transformer with a third winding, called tertiary winding. The tertiary windings of the three transformers are connected in delta (Fig.3.27). They often provide the substation service voltage where the transformers are installed. Note that there is no phase shift between the incoming and outgoing transmission line voltages of a wye-wye connected transformer. Fig.3.26 Wye-wye connection with neutral of the primary connected to the neutral of the source. Fig.3.27 Wye-wye connection using a tertiary winding. Example 3.18 Three single phase, 30 kVA, 2400/240 V, 50 Hz transformers are connected to form 3 j , 2400/416 V transformer bank. The equivalent impedance of each transformer referred to the high voltage side is 1.5+j2 S2. The transformer delivers 60 kW at 0.75 power factor (leading). (a) Draw schematic diagram showing the transformer connection. (b) Determine the transformer wiWing current (c) Determine the primary voltage. (d) Determine the voltage regulation. Solution: (a) (b) kVA Is = = 60 = 80kVA 0.75 80*103 =111.029 A 3 *416 2400 a= = 10 240 I1ph = 111.029 =11.103A 10 I1L =11.103* 3 =19.231 A V2¢ = 2400Ð0o V , I 2¢ =11.103Ð41.41o A V1 = V2¢+ I¢*(Zeq1) o = 2400Ð0 +11.103Ð41.41o *( 1.5 + j2) = 2397.96Ð0.66 V V 1 -V 2 ¢ VR = *100 V2 ¢ 2397.96 - 2400 *100 = -0.0875% = 2400 Problems: 1 A 1 0, 100 kVA, 1000/ 100 V transf test 100 V, 6.0 A, 400 W short-circuit test 50 V, 100 A, 1800 W (a) Determine the rated voltage and rated current for the HV and LV sides. (b ) Derive an approximate equivalent circuit referred to the HV side. (c) Determine the voltage regulation at full load, 0.6 PF leading. (d) Draw the phasor diagram for condition (c). 2 A 1 ¢,25 kVA, 220/440 V, 60 Hz transformer gave the following test results. Open circuit test : 220 V, 9.5 A, 650 W Short-circuit test : 37.5 V, 55 A, 950 W (a ) Derive the approximate equivalent circuit in per-unit values. (b) Determine the voltage regulation at full load, 0.8 PF lagging. (c) Draw the phasor diagram for condition (b). 3 A1f 10 kVA, 2400/ 120 V, 60 Hz transformer has the following equivalent circuit parameters: Zeq1 = 5 + j25 W, c = 64 kW and m = 9.6 kW Standard no-load and shortR1 X1 circuit tests are performed on this transformer. Determine the following: No-load test results: Voc 4- , Ioc , o Sh r -circuit test results: Vsc , Isc , Psc A single-phase, 250 kVA, 11 kV/2.2 kV, 60 Hz transformer has the following parameters. R H= V 1.3 W XHV=4.5W, RLV = 0.05 W, XLV = 0.16, Rc2= 2.4 kW Xm2 = 0.8 kW (a) Draw the approximate equivalent circuit (i.e., magnetizing branch, with Rc1 and Xm connected to the supply terminals) referred to the HV side and show the parameter values. (b) Determine the no load current in amperes (HV side) as well as in per unit. (c) If the low-voltage winding terminals are shorted, determine i () (ii) (d) The supply voltage required to pass rated current through the shorted winding. The losses in the transformer. The HV winding of the transformer is connected to the 11 kV supply and a load, ZL = 15Ð - 90o Wis connected to the low voltage winding. Determine: (i) Load voltage. (ii) Voltage regulation. 5 A 1-f , 10 kVA, 2400/240 V, 60 Hz distribution transformer has the following characteristics: Core loss at full voltage = 100 W Copper loss at half load = 60 W (a) Determine the efficiency of the transformer when it delivers full load at 0.8 power factor lagging. (b) Determine the per unit rating at which the transformer efficiency is a maximum. Determine this efficiency if the load power factor is 0.9. The transformer has the following load cycle: No load for 6 hours 70% full load for PF Determine the all-day efficiency of the transformer. 6 The transformer of Problem 5 is to be used as an autotransformer (a) connection that will result in maximum kVA rating. (b) high-voltage and low-voltage sides. (c) Show the Determine the voltage ratings of the Determine the kVA rating of the autotransformer. Calculate for both high-voltage and low-voltage sides. 7 A1 f , 10 kVA, 460/ 120 V, 60 Hz transformer has an efficiency of 96% when it delivers 9 kW at 0.9 power factor. This transformer is connected as an autotransformer to supply load to a 460 V circuit from a 580 V source. (a) Show the autotransformer connection. (b) Determine the maximum kVA the autotransformer can supply to the 460 V circuit. (c) Determine the efficiency of the autotransformer for full load at 0.9 power factor. 8 Reconnect the windings of a 1f , 3 kVA, 240/120 V, 60 Hz transformer so that it can supply a load at 330 V from a 110 V supply. (a) Show the connection. (b) Determine the maximum kVA the reconnectedtransformer can deliver. 9 Three 1¢, 10 kVA, 460/120 V, 60 Hz transformers are connected to form a 3f 460/208 V transformer bank. The equivalent impedance of each transformer referred to the high-voltage side is 1.0 + j2.0 W. The transformer delivers 20 kW at 0.8 power factor (leading). (a) Draw a schematic diagram showing the transformer connection. (b) Determine the transformer winding current. (c) Determine the primary voltage. (d) Determine the voltage regulation. 10 A 1f 200 kVA, 2100/210 V, 60 Hz transformer has the following characteristics. The impedance of the high-voltage winding is 0.25 + j 1.5 W with the lowvoltage winding short-circuited. The admittance (i.e., inverse of impedance) of the low-voltage winding is 0.025 - j O.075 mhos with the high-voltage winding open-circuited. (a) Taking the transformer rating as base, determine the base values of power, voltage, current, and impedance for both the high-voltage and low-voltage sides of the transformer. (b) Determine the per-unit value of the equivalent resistance and leakage reactance of the transformer. (c) (d) Determine the per-unit value of the excitation current at rated voltage. Determine the per-unit value of the total power loss in the transformer at full-load output condition. UNIT III ELECTROMECHAN Introduction Electromechanical energy conversions - use a magnetic field as the medium of energy conversion.Electromechanical energy conversion device: Converts electrical energy into mechanical energy or Converts mechanical energy into electrical energy Three categories of electrom echanical energy conversion devices: Transducers (for measurement and control)- small motion Transform the signals of different forms. Examples: microphones, sensors and speakers. F - limited mechanical motion. Produce forces mostly for linear motion drives, Example Actuators - relays, solenoids and electromagnets. Continuous energy conversion equipment. Operate in rotating mode. Examples: motors and generators. Energy Conversion Process The principle of conservation of energy: Energy can neither be created nor destroyed. It can only be changed from one form to another. Therefore total energy in a system is constant . An electromechanical converter system has three essential parts: ① An electrical system (electric circuits such as windings) ② A magnetic system (magnetic field in the magnetic cores and air gaps) ③ A mechanical system (mechanically movable parts such as a rotor in an electrical machine). The energy transfer equation is as follows: æElectrical ö æ Mechanicaö æIncrease in ç ÷ ç çenergy input ÷ =çenergy çfrom sources÷ çoutput è ø è ÷ ç ÷ çstored energy in ÷ ÷ çmagnetic field ÷ ø è ö ÷ æEnergy ö +ç ÷ èlosses ø ø The energy balance can therefore be written as: æ Electricalenergy ö æ ç p ÷ ç t Mechanicalenergy ö Increasein æ ÷ ö ÷ ç stored field ÷ çin u fromsources÷ = çoutput + friction ÷ ç+ nr ÷ çe e gy +coreloss÷ ç ÷ çandwindageloss è- resis tan celoss ø è ø è ø For the lossless magnetic energy storage system in differential form, dWe = dWm + dWf dWe = i d l = differential change in electric energy input dWm = fm dx = differential change in mechanical energy output dWf = differential change in magnetic stored energy Energy in Magnetic System Consider the electromechanical system below: Axial length (perpendicular to page) = l The m echanical force fm is defined as acting from the relay upon the external mechanical system and the differential mechanical energy output of the relay is dWm = fm dx Then, substitution dWe = id l , gives dWf = id l - fm dx Value of W f is uniquely specified by the values of l and x, since the magnetic energy storage system is lossless. Wf = òidl dWf = differential change in magnetic stored energy Energy and C oenergy The l-i characteristics of an electromagnetic system depends on the air-gap length and B-H characteristics of the magnetic material. For a larger air-gap length the characteristic is essentially linear. The characteristic becomes non linear as the air-gap length decreases. l l l-i Wf Increased i -gap ar length Wf’ i i For a particular value of air-gap length, the field energy is represented by the red area between l axis and l-i characteristic. The blue area between i axis and l - i characteristic is known as the coenergy Wf' = ò di - i characteristic, The coenergy is defined as From the figure of l Wf’ + Wf = l i l Note that Wf’ > Wf if the l - i characteristic i The quantity of coenergy has no physical significance. However, it can be used to derive expressions for force (torque) developed in an electromagnetic system Determination of Force from Energy The magnetic stored energy W f is a state function, determined uniquely by the independent state variables λ and x. This is shown explicitly by dWf (λ, x) = id l - fm dx For any function of two independent variables F(x1,x2),the total differential equation of F with respect to the two state variables x1 and x2 can be written ¶F(x1, x2) ¶F(x1, x2) dx1 + dx2 ¶ x ¶x1 2 x x dF(x1, x2) = 2 1 Therefore, for the total differential of Wf ¶Wf (l , x) dWf (l , x) ¶l dl + ¶Wf (l , x) dx l And we know that dWf (l , x) = idl - fmd x By matching both equations, the current: i= ¶Wf (l , x) ¶l x where the partial derivative is taken while holding x constant and the mechanical force: fm = - ¶Wf (l , x) ¶x l Where the partial derivative is taken while holding l constant. Determination of Force from Energy: Linear System For a linear magnetic system for which l =L(x)i: l Wf (l , x) = òi 0 l ( , l x)dl =ò 0 l l L(x) dl 1 2 L(x) and the force, fm can be found directly: fm = - ¶Wf (l , x) ¶x ¶æ 1 l l2 ö ÷ l 2 dL(x) = =- ç 2L(x)2 ¶x ç 2 L(x) ø÷ è l Determination of Torque from Energy For a system with a rotating mechanical terminal, the mechanical terminal variables become the angular displacement θ and thetorque T. Therefore, equation for the torque: T=- ¶Wf (l ,q ) ¶q l where the partial derivative is taken while holding Determination of Force from Coenergy The coenergy Wf’ is defined as Wf' (i, x) = il -Wf (l , x) and the differential coenergy dWf’: ' dW f (i, x) = d(il ) - dWf (l , x) We know previously that dWf (l , x) =idl - fm dx By expanding d(iλ): d (il ) = idl + ldi So, the differential coenergy dWf’: dWf' (i, x) = d(il ) - dWf (l , x) = idl + ldi - (idl - fmdx) = ldi + fm d x By expanding dWf’(i,x): dWf ' (i, x) = ¶Wf ' (i, x) ¶Wf ' (i, x) di + dx ¶i ¶x i x and, from the previous result: dWf' (i, x) = ldi + fmd x l constant. By matching both equations, l : l= ¶Wf ' (i, x) ¶i x where the partial derivative is taken while holding x constant and the mechanical force: fm = ¶Wf ' (i, x) ¶x i where the partial derivative is taken while holding i constant. For a linear magnetic system for which l =L(x)i: i Wf (i, x) = =L(x) ' i x)di =òl(i, òL(x)idi 0 i2 2 0 and the force, fm can be found directly: fm = ¶Wf ' (i, x) ¶x = i 2 ¶ æ i ö÷ = i dL(x) çL(x) ¶x ç ÷ è 2ø 2 dx 2 i For a system with a rotating mechanical terminal, the mechanical terminal variables become the angular displacement θ and thetorque T. Therefore, equation for the torque: T= ¶Wf ' (i,q ) ¶q i where the partial derivative is taken while holding l constant. Determination of Force Using Energy or Coenergy? The selection of energy or coenergy as the function to find the force is purely a matter of convenience. They both give the same result, but one or the other may be simpler analytically, depending on the desired result and characteristics of the system being analyzed. Direction of Force Developed 1 By using energy function: f = m ¶Wf (l , l x) The negative sign shows that the force acts in a direction to decrease the magnetic field stored energy at constant flux. 2 By using coenergy function: fm = + ¶Wf ' (i, x) ¶x The positive sign emphasizes that the force acts in a direction to increase the coenergy at constant current. 3 By using inductance function: i2 dL(x) 2 dx fm = + i The positive sign emphasizes that the force acts in a direction to increase the inductance at constant current. B-H Curve and Energy Density In a magnetic circuit having a substantial air gap g, and high permeability of the iron core, nearly all the stored energy resides in the gap. Therefore, in most of the cases we just need to consider the energy stored in the gap. The magnetic stored energy, l Wf = ò0 idl in which i = Hg N and Hg B Therefore, Wf = ò dl = d ( Nf ) = d ( NAB) = NAdB 0 N NAdB = Ag B ò HdB 0 However, Ag is volume of the air gap. Dividing both sides of the above equation by the volume Ag results in wf = Wf B = Ag HdB ò 0 B Where i wf = ò 0 HdB s energy per unit volume wf is known as energy density. The area between the B-H curve and B axis represents the energy density in the air gap. In the same manner, H w'f = ò 0 BdH is coenergy per unit volume. The area between the B-H curve and H axis represents the coenergy density in the air gap. For a linear magnetic circuit, B = mH or H = B/m, energy density: B B2 dB = m 2m wf = ò HdB = ò 0 0 and coenergy density: H H 0 0 w'f = òBdH = òmHdH = mH 2 2 In this case, it is obvious that wf = wf’. Doubly-excited Systems Rotating Machines Most of the energy converters, particularly the higher-power ones, produce rotational motion. The essential part of a rotating electromagnetic system is shown in the figure. The fixed part is called the stator, the moving part is called the rotor. The rotor is mounted on a shaft and is free to rotate between the poles of the stator Let consider general case where both stator & rotor have windings carrying current ( i s and ir ) Assume general case, both stator and rotor have winding carrying currents (non-uniform air gap - silent pole rotor) The system stored field energy, W f can be evaluated by establishing the stator current i s and rotor current ir and let system static, i.e. no mechanical output S otor flux linkage l is expressed in terms of inductances L (which depends on position rotor angle q, L(q) Stored field energy Torque In linear system, coenergy = energy W’f = W f First two terms represents reluctance tsalient stator and rotor, or in either stator or rotor is salient) The third term represents alignment torque; variation of mutual inductance. Reluctance Torque - It is caused by the tendency of the induced pole to align with excited pole such that the minimum reluctance is produced. At least one or both of the winding must be excited. Alignment Torque - It is caused by a tendency of the excited rotor to align with excited stator so as to maximize the mutual inductance. Both winding must be excited. Forces and Torques Lorentz force law F = q(E + v ´ B) F: N, q: Coulombs, E : V/m, B : T or Wb/m 2 large numbers of charged particles are in motion, Fv = r (E + v ´ B) Current density, J = r v Fv: force density (force per unit volume), N/m 3 r : charge density (C/m3) A/m2 Right-hand rule for determining the direction magnetic-field component of the Lorentz force F = q(v × B). Example 3.1 A nonmagnetic rotor containing a single-turn coil is placed in a uniform magnetic field of magnitude B0, as shown in Fig. 3.2. The coil sides are at radius R and the wire carries current I as indicated. Find the θdirected torque as a function of rotor position α when I = 10 A, B0 = 0.02 T and R = 0.05 m. Assume that the rotor is of length l = 0.3 m. Force per unit length of the wire, F = I × B For wire 1 carrying current I into the paper, F1q = -I B0 lsina For wire 2 carrying current I out of the paper, F2q = -I B0 lsina Torque, T = F1q R + F2q R = -2I B0 l R sina = -2(10)(0.02)(0.05)(0.3)sin a = -0.006sina N.m magnetic-field electromechanical-energy-conversion device simple force-producing device The energy method, Pin(electrical) = Pstored (magnetic field) +Pout (mechanical) ei = dWfld ´dt dt + Þ d dt ( f x) e= fd l dWlf d = i dl lossless system dl dt - f dxfld Energy Balance æ energy input ö çf rm o electric÷ ç ç è æ ç = ÷ ç ÷ ç sources ø è mechanical ö ÷ energy output æ increase in energyö æ energy ö ÷ çconverted ÷ ç + ÷ ç ÷ ç ø è stored in + ÷ ç ÷ ç ÷ ÷ magnetic field ø è into heat ø Conversion of energy into heat: ohmic heating due to current flow in windings + mechanical friction Lossless magnetic energy storage system: dWelce = dWmehc fd + dW l dWelec : differential electric energy input dWmehc : differential mechanical energy output dW fdl : differential change in stored energy Singly Excited Systems An electromagnetic relay dWelec = ei dt dl dt e = Þ dWelec = i d l dWmech = f dx l Þ d Wlf d = i d l - f dx fld fd Lossless magnetic energy storage system: conservative system: value of W l is uniquely specified by the values of λ and x (λ, x are called the state variables) Þ fd Wfdl is uniquely specified by λ and x Two different integration paths for fd Wl fd ò dW l + ò dW l = path 2a Wl On 2a: dl = 0, and f path 2b fd fld fd = 0 since l = 0 On 2b: dx = 0 For a linear system (λ proportional to i): l = L(x)i Þ dWl fd ü =0ý Þ Wfld þ l0 (l 0, x0 ) = òi(l, x0)dl 0 l i(l¢, W lfd (l , x) = ò l x) d l l¢ = 0 ¢ 0 l 1 òd l ¢ L(x) 2 2 L(x) Example 3.2 The relay shown in Fig.3.6a is made from infinitely-permeable magnetic material with a movable plunger, also of infinitely-permeable material. The height of the plunger is much greater than the air-gap length(h >> g). Calculate the magnetic stored energy W l as a function of plunger position (0 < x < d) for N = 1000 turns, g = 2.0 mm, d = 0.15 m, l = 0.1 m, and i = 10 A. fd 1 W l = L( x)i2 2 m0N 2A a L(x) = Agap : gap cross-sectional area 2g gp fd Agap = l(d - x) = ld Wl = æ1 ç è 2 -xö Þ ÷ dø L(x) = m0N ld (1- x / d ) 2g 1 (10002)(4p ´10-7)(0.1)(0.15) ´ 10 ç 2 2(0.002) è xö ÷= dø ç è 236 1- xö ÷ J d ø fd Relay with movable plunger Magnetic Force and Torque from Energy Wfld is a state function determined uniquely by the values of the independent state variables λ and x. dW l = i dl -f dxfld For any function of two independentfdvariables F(x1, x2), dF(x1, x2) = ¶F ¶F dx1 + dx2 ¶x ¶x2 Þ dWlf d = i= d dl + ¶l 1 Þ ¶Wlf ¶Wfdl (l, x) f ¶l =- l ¶Wlf d dx ¶x ¶Wfdl (l, x) ¶x fd For linear magnetic systems for which λ = L(x)i, f Example 3.3 fld =- ¶ æ1 ç l 2 ö ÷= ¶x è 2 L(x) ø l dL(x) 2 2 2[L(x)] dx l = L(x)i Þ f fld 1 dL(x) = i2 2 dx Table 3.1 contains data from an experiment in which the inductance of a solenoid was measured as a function of position x, where x = 0 corresponds to Table 3.1 x (cm) 0 0.2 0.4 0.6 0.8 1.0 L(mH) 2.8 2.26 1.78 1.52 1.34 1.26 1.2 1.20 1.4 1.16 1.6 1.13 1.8 1.11 2.0 1.10 Plot the solenoid force as a function of position for a current of 0.75 A over the range 0.2 < x < 1.8 cm. Using the MATLAB function polyfit, a fourth-order polynomial fit of the inductance as a function of x is obtained: L( x) = a(1)x4 + a(2) x3 + a(3)x2 + a(4) x + a(5) Þ f fd l 1 dL(x) 1 = i2 = i é4a(1)x3 + 3a(2)x2 + 2a(3)x + a(4)û ù 2 dx 2 2 MATLAB script: clc clear % Here is the data: x in cm, L in mH xdata = [0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0]; Ldata = [2.8 2.26 1.78 1.52 1.34 1.26 1.20 1.16 1.13 1.11 1.10]; %Convert to SI units x = xdata*1.e-2 ; L = Ldata*1.e-3; len = length(x) ; xmax = x(len) ; % Use polyfit to perform a 4'th order fit of L to x. Store % the polynomial coefficients in vector a. The fit will be % of the form: % % Lfit : a(1)*x^4 + a(2)*x^3 + a(3)*x^2 + a(4)*x + a(5) ; % a = polyfit(x,L,4) ; % Let's check the fit for n = 1:101 xfit(n) = xmax*(n-1)/100; Lfit(n) = a(1)*xfit(n)^4 + a(2)*xfit(n)^3 + a(3)*xfit(n)^2 ... + a(4)*xfit(n) + a(5); end % Plot the data and then the fit to compare (convert xfit to cm and Lfit to mH) figure(1) plot (xdata, Ldata, ' * ' ) hold plot (xfit*100, Lfit*1000) xlabel('x [cm] ') ylabel('L [mH] ') % Now plot the force. The force will be given by % % (1/2)i^2 dL/dx =(1/2) i^2 ( 4*a(1)*x^3 + 3*a(2)*x^2 + 2*a(3)*x + a(4)) % %Set current to 0.75 A I = 0.75; for n = 1:101 xfit(n) = 0.002 + 0.016*(n-1)/100; F(n) =4*a(1)*xfit(n)^3+3*a(2)*xfit(n)^2+2*a(3)*xfit(n)+a(4); F(n) = (I^2/2)*F(n); end figure(2) plot (xfit*100,F) xlabel('x [cm] ') ylabel('Force [N]') System with rotating mechanical terminal: x ® q : angular displacement, f dWlf d (l ,q ) = i dl-Tfld dq Linear magnetic systems: l = L(q )i Þ Tlf d = - ¶ æ1 l 2 W lfd = l 1 ö ç ÷= ¶q è 2 L(q ) ø Þ 1 2 [L(q ] ® Tlfd torque ¶Wfdl (l,q ) ¶q l 2 2 L(q dL(q 2 Tfld = - fld ) l = L(q )i Þ 1 dL(q Tfld = i2 ) 2 Example 3.4 The magnetic circuit of Fig. 3.9 consists of a single-coil stator and an oval rotor. Because the air-gap is nonuniform, the coil inductance varies with rotor angular position, measured between the magnetic axis of the stator coil and the major axis of the rotor, as L(q ) = L0 + L 2 cos(2q ) L0 = 10.6 mH L2 = 2.7 mH Find the torque as a function of θ for a coil current of 2 A. Solution: 2 1 dL(q ) Tfdl = i2 2 dq 1 = i2 (-2L2sin(2q ) )= -4 ´ 2.7 ´10-3 sin(2q ) = -1.08´10-2 sin(2q ) N.m Magnetic Force and Torque from Coenergy l W fd¢ -Wfdl(i, ) = il The coenergy W ¢ is defined as: l fld l dWf¢l d(i,l ) = d(il ) - dWfld (i,l ) d(il ) = idl + ldi dW lfd = i dl - f dx fld Þ dWfd¢l = idl + ldi - (i dl - f dx) = ldi + f dx fld fld ¶ f ¶Wf ¢ld dWf¢d (i, x) = d¢Wl di + dx ¶x ¶i ¶Wf¢l (i, x) fd l= fl = ¶ ¶i The coenergy can be found fromthe integral of λ di i òl(i¢, x)di W ¢fdl (i, x) = 0 For linear magnetic systems, l = L( x)i W ¢fdl (i, x) = 1L(x)i2 2 Þ f fl d = ¶Wfld (i, x) 1 dL( x) = i2 ¶x 2 dx For a rotating electromechanical system, i W ¢fd(i,q ) = òl(i¢,q )di 0 ¢ ¢ Magnetically linear system, Tf dl = ¶Wfdl (i,q ) ¶q W ¢fdl (i,q ) = L(q )i2 1 Tfl d = ¶Wfld (i,q ) 1 dL(q = i2 ¶q 2 ) Example 3.5 For the relay of Example 3.2, find the force on the plunger as a function of x when the coil is driven by a controller which produces a current as a function of x of the form i(x) = I æç x ÷ èdø ö A 0 From Example 3.2, L(x) = m0N2ld (1- x / d ) 2g 1 dL( x) 1 æ m0N 2l ö f fdl = i2 2 =- i ç dx 2 è 2g ø The coenergy for this system is I2m0N2l æx ö ÷= 2 1 1 Wfld (i, x) = L( x)i2 = i2 2 2 m0N2ld(1- x / d ) 2g - 4g = 2 ç ÷ èd ø m0N2ld(1- x / d ) æxö ç ÷ èd ø 4g The force cannot be found by taking the partial derivative of this expression for W respect to x, because in the expression for differentiating with respect to x. fd ¢ l (i, x) with f fdl the current must be kept constant while For a magnetically-linear system, the energy and coenergy are numerically equal: 1l2 / L = 1 Li2 . For a 2 2 nonlinear system in which x and i or B and H equal. Hag = 2 g (a) fd 2 l = Nf = NBag Aag = NAgam0Hag = m0NAag m0N Þ L= Ni m0N Aag = i 2g 2 a g A 2g 1 öq.h Aag = çr1 + g ÷ 2 ø è æ Þ m0N L(q ) = 2g 2 1 öh. çr1 + g ÷ q æ 2 ø è 1 dL(q ) m0N2i2 æ 1 öh = çr 1 + g ÷ Tlf d = i2 2 dq 4g è 2 ø (b) Þ Bmax = 1.65 T Þ Hm ax = Bmax = Ni 2g m0 2 ´ 3´ 10-3 ´1.65 Ni = 2g Bmax = = 7878.2 A-turns 4p ´ 10-7 m0 Maximum torque, 2 T lfd = m0N i2 æçr1 + 1g öh 4p ´10-7 ´ (7878.2)2 (0.025 + 0.0015)0.018 = 3.1 N-m ÷ = 4g è 4 ´ 0.003 2 ø Multiply-Excited Magnetic Field Systems Differential energy dWlf d (l 1,l 2,q ) = i1 dl1 + i2 dl 2 - Tfld dq Þ Tfld = - ¶Wf (l1,l2,q ) dl ¶q Current and torque in terms of energy ¶Wf (l 1,l2,q ) i1 = dl ¶l 1 i2 = ¶Wfd (l 1,l2,q ) ¶Wfld(l1,l 2,q ) ¶l Tlf d = 2 l ¶q Tofind the energy by integrating the differential energy: fd Wl ( l ,l 10 l 20 20 ,q0) = i2(l1 = 0,l2,q = q0).dl2 + l 10 ò ò 0 0 i1(l1,l2 = l20,q = q0).dl1 Magnetically linear system: l l 1 = L11i1 + L12i2 2 = L21i1 + L22i2 L12 = L21 Solving for i1 and i2 in terms of l1, l2 æi1 ö æ L çi ÷ =ç 11 L12 ö çl ÷ L22 ø è è 2 ø è L21 1 (L22l 1 - L12l 2 ) D i1 = lö æ ÷ Þ i2 = 1 (-L2 1l 1 + L11l 2 ) D 2 D = L11L22 L12L21 where l 20 fd .dl1 l 10 l10,l20,q0) = ò L11(q0) l2.dl2 + ò L22(q0)l1 - L12(q0)l20 D(q0 ) D(q0 ) 0 0 l l202 + 10 2 = L11(q0) L22(q0) L12(q0) l01l20 2D(q0) 2D(q0) D(q0 ) Wl ( The coenergy is defined as, ¢ fd Wl Þ fd Wl (i1,i2,q ) = l1i1,+ l2i2d(l 1i1) + d (l 2i2) ¢ ( i dl 1 = l1di1 + l2di2 + Tfdl dq ¶Wfd (i1,i2,q ) fd d W l ¢ l (i1,i2,q ) = ¢ ¶i1 ¶Wfdl (i1,i2,q ) + i2 d ¶Wfld(i1,i2,q ) ¢ ¶i2 di1 + ¶Wfdl (i1,i2 ,q ) l ¶i1 l 2 - Tfld dq ¶Wfld(i1,i2,q ) ¢ ¶q dq ¶Wfdl (i1,i2 ,q ) di2 + ¢ ¢ l ) 1 ¢ ¶ i2 ¶q The coenergy can be found by integrating the differential,i 0 fd i20 W ¢ l (i10,i20,q0) = ò0 l2(i 1 1 = 0,i2,q = q0).di2 + ò0 l1(i1,i 2 = i20,q = q0).di1 For a linear system, W ¢ (i1,i2,q ) = l Torque, 1 2 (q )i12+ 1 1 2 L22(q ) L12(q )i1i2 2 Tl = ¶Wfl 1 d (q ) 1 dL22(q ) dL12(q ) 2 1 2 d (i1,i2,q ) + i2 + i1i2 dq = i1 ¶q 2 dq 2 dq Example 3.7 In the system shown in Fig. 3.15, the inductances in henrys are given as 1 = (3 + cos 2θ) × 10-3; 1 =0.3cosθ; 2 = 30 + 10cos2θ. Find and plot the torque Tfld(θ) for current il =0.8 A and i2 = 0.01 A. Tfdl 1 dL11(q 1 dL22(q )2 dL12(q ) + i2 + i1i2 )2 = i1 2 dq 2 dq dq 1 1 2 = i12(-2 ´10-3 sin 2q ) + i2 (-20sin 2q ) + i1i2(-0.3sin q ) 2 2 = -(1.64sin 2q + 2.4 sinq ) ´10-3 N-m Dynamic Equations Model of a singly-excited electromechanical system. KVL eqn. for the electrical system, v0 = Ri + dl dt dL(x) dL(x) dx l = L(x)i Þ v0 = Ri + L(x) d dL(x) i + i dt d t = dtÞ v0 = Ri + L(x) dx dt di dt fd fd dL + (x i ) dx d x L1 L1 d t L1 L2 L2 di L(x) : self-inductance volta dL(x) dx dx i : speed voltage ( : mechanical speed) dx dt dt Forces in the mechanical system in terms of mechanical position (and its derivatives): Spring: - x0) fK = -K(x Da mper : fD = -B Acceleration of mass : dx fM = -M dt d 2x K: spring constant B: damping constant M: mass of moving part dt2 Force equilibrium (f0: external mechanical excitation force): f fld = f0 - ( fK + fD + fM ) 2 Þ f0(t) = -M d x - B d - K(x - x0)+ f x dt2 dt l (x,i) fd Note: If x > x0 and increasing (with positive second derivative: M accelerating), then the forces fK , fD , and fM all oppose f d. fl The differential equations for the overall system: d dL(x) dx +i i dx dt d 2 t f0(t) = -M d x - B dx - K(x - x0) + f dt dt2 v0(t) = Ri + L(x) fl d (x,i) Given the inputs (excitations) v0(t) and f0(t) these equations can be solved to find x(t) and i(t). Example 3.10 Figure 3.24 shows in cross section a cylindrical solenoid magnet in which the cylindrical plunger of mass M moves vertically in brass guide rings of thickness g and mean diameter d. The permeability of brass is the same as that of free space and is μ0 = 4π × 10-7 H/m in SI units. The plunger is supported by a spring whose spring constantis K. Its unstretched length is l0. A mechanical load force ft is applied to the plunger from the mechanical system connected to it, as shown in Fig. 3.24. Assume that frictional force is linearly proportional to the velocity and that the coefficient of friction is B. The coil has N turns and resistance R. t terminal voltage is vt and its current is i. The effects of magnetic leakage and reluctance of the steel are Is negligible. Derive the dynamic equations of motion of the electromechanical system, i.e., the differential equations expressing the dependent variables i and x in terms of vt, ft, and the given constants and dimensions. The reluctance of the upper and lower gaps, R1 = g m0p xd R2 = g m0p ad Total reluctance, g g + R = R1 + R2 = m0p = g æ1 1ö ç + a ÷ø èx m0p ad m0p Inductance, N 2 m p adN2æ x ö 0 L( x) = æ x = ç ÷ = L¢ç ÷ R g öèa + x ø èa + x ø The magnetic force acting upward on the plunger in the positive x direction is f fd x) l = ¶Wfld¢ (i, ¶x 1 dL( x) 1 aL¢ = i2 = i2 dx (a + x)2 2 2 The induced emf in the coil is e= d di dL di dL dx (Li) = L + i = L +i d dt dt dt dx dt t æ =L¢ç x ÷ è a + x ødt The dynamical equations of the system: d 2x dx öd +L i ai dx (a + x) dt 2 1 ai2 f (t) = -M - B - K(x - l0) + L ¢ dt2 dt 2 (a + x)2 ai dx æ x di v(t) = Ri + L¢ç ÷ +öL (a + x)2 dt è a + x ø dt