Lecture 3 Single Phase Transformer PDF

Lecture Three Guess: which part is transformer? *http://electricalmiracles.blogspot.com/2012/11/components-of-electrical-substation.html Electric Grid Representation  Transformer Static electromagnetic device – Transfer electrical energy from one circuit to a...

Lecture Three Guess: which part is transformer? *http://electricalmiracles.blogspot.com/2012/11/components-of-electrical-substation.html Electric Grid Representation  Transformer Static electromagnetic device – Transfer electrical energy from one circuit to another at different voltage and current, but in the same frequency. Consists of two or more coil wires around a ferromagnetic core Applications: – High power application (Step up and down) – Low power application (AC-DC isolation and impedance matching) Power Distribution Substation  To be continued… Residential Transmission Line A typical power distribution system  Transformer Types Core Type Transformer – Windings around two legs – Subdivided sections to reduce leakage Shell Type Transformer Windings around the center leg Primary and secondary are arranged in one on top of the other The core type is magnetic circuit equivalent to an series electric circuit; a shell type is equivalent to a parallel electric circuit. So, the reluctance will be lower for shell type of similar cross sectional area. Transformer Types  No-load conditions Assumptions: – Secondary winding is open circuit – Primary winding is excited by AC voltage. – Neglect leakage fluxes and resistances Excitation current Primary Secondary side side Sinusoidal voltage source We want to test the magnetizing effect in the coils  Voltage vs Magnetic Flux d1 d Induce EMF (Faraday’s law) e1 = dt = N1 dt KVL at primary circuit: v1 = R1i + e1 Assuming R1-->0 and  = max sin wt = max cos( wt − 90) d1 d e1 = = N1 = wN1max cos wt dt dt EMF leads φ by 90 degree E peak 2f E1rms = = N1max 2 2 V1 max = Flux is in sinusoidal waveform 2fN1 if the transformer is excited by a sinusoidal voltage  Excitation Current i l Ni = Hl = 0  r A nonlinear The waveform of excitation current is non-sinusoidal due to nonlinear properties of magnetic materials  Graphical Curve Excitation current is affected by magnetic flux Current is plotted around the hysteresis curve Symmetrical current waveform at [π,2π]  Two components Fundamental component Iφ can be resolved into two components: – Magnetizing current (IM) which is the current required to produce the flux in the transformer core and it is lagging emf by 90°. – Core loss current (Ic) which is the current required to make up for hysteresis and eddy current losses and it is in phase with emf. Transformer core loss Pc = E1 I c = E1 I cos  c  Ideal Transformer Assumptions: Zero winding resistances, no flux linkage, zero core loss, high permeability V1 N1 I 2 a= = = V2 N 2 I1  Parameters Transformation Primary side parameters Secondary side parameters N2 V1 V1’= V2 = V1 = N1 V2’= V1 = V2 = aV2 N2 N1 a N2 I N1 I2’= I1 = I2 = 2 I1’= I2 = I1 = aI1 N1 a OR N2 N2 2 Z1 Z2’= Z1 = V1 = ( N1 ) 2 V2 = a 2 Z 2 Z1’= Z2 = ( ) Z1 = 2 I1 N2 I2 N1 a Induced parameter symbol Induced parameter symbol at at Primary side Secondary side Z 2 = a 2 Z 2  To be continued… Power Measurement Pin = P1 = V1 I1 cos 1 Q1 = V1 I1 sin 1 Pout = P2 = V2 I 2 cos  2 Q2 = V2 I 2 sin  2 Ideal Transformer P1 = P2 Q1 = Q2 S1 = S 2 = V1 I1* = V2 I 2*  Example 3-1 A single phase 120/12.6 V, 25VA, 60Hz transformer is feeding a load Z2=50Ω and it is excited by an ac voltage of 120V through R =1k Ω. (a) compute rated values of primary and secondary current (b) compute primary and secondary currents Rated Value - Maximum value a component can withstand without breaking down  Solution (a) Rated current at primary and secondary side Rated RMS voltage at primary side Rated RMS voltage at secondary side 120/12.6 V, 25VA, 60Hz transformer Turns Ratio Rated apparent power |S| VArat 25 I 1 rat = = = 0.2 A V1rat 120 25 I 2rat = = 1.98 A 12.6 Induced Parameter to Primary side (b) Primary side Induced Impedance to the primary Series circuit V1 120 a= = = 9.52 V2 12.6 R2 = a 2 R2 = 4.535 k V1 120 I 2 = I1 = = = 0.0217 A Primary current R1 + R2 1000 + 4535 E1 = V1 − I1 R1 E2 = I 2 R2 = 120 − (0.0217)(1000) OR = (0.0217)(4535) = 98.3V = 98.4V I 2 = aI 2 = 0.207 A Secondary current Induced Parameter to Secondary side (b) Secondary side R1 R1 = = 11.03 a2 V1 V1 = = 12.6V a V1 12.6 I1 = I 2 = = = 0.206 A Secondary current R1 + R2 11.03 + 50 E1 = V1 − I1R1 E2 = I 2 R2 = 12.6 − (0.206)(11.03) OR = (0.206)(50) = 10.3V = 10.3V I I1 = 1 = 0.0216 A Primary current a  To be continued… Summary Induced parameters Induced parameters at primary at secondary V2 = aV2 V1 V1 = a R2 = a 2 R2 OR Z 2 = a 2 Z 2 Z1 R1 = R1 OR Z1 = a 2 a2 V1 V1 I 2 = I1 = I1 = I 2 = R1 + R2 R1 + R2 I 2 = I 2 / a I1 = aI1  Real transformer The real transformer’s electrical and magnetic losses – Copper loss in the primary and secondary Resistive component – Eddy current losses – Hysteresis losses Inductive – Leakage flux of the primary and secondary component Derive an Equivalent Circuit for real transformer  Equivalent circuit (1) Copper losses R1 and R2 Leakage flux Xl1 and Xl2  Equivalent circuit (2) Total current (iinput) = Excitation current (iφ) + Induced Load current (iL) Excitation current or no load current can be decomposed to: – Core loss current (in phase with applied voltage), therefore, it is a resistive component Rc – Magnetization current ( lagging voltage by 90°), therefore, it is a reactance Xm 1 Gc = conductance Rc 1 Bm = Susceptance Xm  Induced to Primary side V2 = aV2 Z1 = R1 + jX l1 R2 = a 2 R2 X l2 = a 2 X l 2 1 Z M = Rc || jX M = Z 2 = R2 + jX l2 Gc − jBm V1 ZM I1 = I 2 = I1 I 2 = I 2 / a Z1 + ( Z 2 || Z M ) Z 2 + Z M  Induced to Secondary side V1 V1 = a R1 X l1 Rc XM R1 = X l1 = Rc = X M = a2 a2 a2 a2 Z1 = R1 + jX l1 Z M = Rc || jX M Z 2 = R2 + jX l 2 V1 Z M I1 = aI1 I1 = I 2 = I1 Z1 + ( Z 2 || Z M  ) Z 2 + Z M  To be continued… Approximate circuit Iφ ≈ 2-3% of I1 and It can be neglected Excitation current is neglected Assuming good conductor  Determine Components Value Two simple test to determine the components (R1, X1, R2, X2, Rc, XM) of the equivalent circuit – Open circuit (OC) test / No load test To find Rc and XM Usually high voltage (HV) side is open-circuited and do analysis at LV side Poc = Core loss power (Pc) under rated condition Rated voltage supply – Short circuit (SC) test To find Req = R1 + R2 and Xeq = X1 +X2 Usually low voltage (LV) side is short circuited and do analysis at HV side. Psc = Copper loss power (Pcu) under rated condition Rated current supply  Open Circuit Test 1 1 Gc = Bm = Measurement: VOC, IOC, POC Rc Xm YOC = Gc − jBm = Y POC  OC = cos −1 VOC I OC I OC POC VOC Y = Gc = Rc = VOC V 2 OC I OC cos OC OR VOC BM = Y 2 OC − Gc2 XM = I OC sin  OC  Short circuit test Measurement: VSC, ISC, PSC Z SC = Req + jX eq = ( R1 + R2 ) + j ( X 1 + X 2 ) PSC  SC = cos −1 VSC VSC I SC = PSC Z SC Req = 2 Z eq = VSC  SC I SC I SC I SC VSC OR Req = I SC cos SC X eq = Z 2 SC − Req2 X eq = VSC sin  SC I SC  Phasor diagram This is the way to Vp  p = VS 0 + Reqs I S  −   + jX eqs I S  −    calculate for Vs, no-load a  Derivation of Approximate equation for Vp θp≈ 0 Approximate Equation:  To be continued… Voltage Regulation (VR) Output voltage varies with Load! VNL − VFL VS , NL − VS , FL VR = = 100% VFL VS , FL VP , NL − aVS , FL At primary side: VR = 100% aVS , FL VP , NL − VS , FL At secondary side: VR = a  100% VS , FL *Full load is a quantity that compares output voltage(Vs) at no load with output voltage at full load.  Rated voltage on one Rated current on one side side whereas the other side is open! Example 3-2 whereas the other side is short-circuited! For a 10kVA, 2300-230V, 60Hz distribution transformer (single phase): – Voc = 230V, Ioc = 0.45A, Poc = 70W (OC at HV) – Vsc = 120V, Isc = 4.5A, Psc = 240W (SC at LV) (a) The equivalent circuit referred to LV side (b) Iφ as a percentage of rated full-load current  Solution (1) OC Open circuit at HV side; Analysis at LV side POC 70  OC = cos−1 = cos −1 = cos−1 (0.676) = 47.4 VOC I OC 230(0.45) VOC 230 Rc = = = 756 I OC cos OC 0.45(0.676) VOC 230 X M = = = 694 I OC sin  OC 0.45(0.737)  Solution (2) SC Short circuit at LV side; Analysis at HV side PSC 240 VSC 120 Req = 2 = 2 = 11.85 Z eq = = = 26.7 I SC 4.5 I SC 4.5 X eq = Z eq2 − Req2 = 26.7 2 − 11.852 = 24  Solution (3) Refer Req and Zeq to the LV side Req R1 = 0.05925  = Req 2 = (230 ) 11.85 = 0.1185 2 R2 = 0.05925 a 2300 X l1 = 0.12 X eq  = X eq 2 = (230 )  24 = 0.24 2 X l 2 = 0.12 a 2300  To be continued… Solution (4) No load current is the excitation current Open circuit current is the no load current I LV  I OC = 0.45 S rated 10  10 3 LV I rated = LV = = 43.5 A Vrated 230 I LV 0.45 LV 100% = 100% = 1.04% I rated 43.5 Excitation current is relatively small and can be ignored.  Example 3-6 For a single-phase, 150kVA, 2400-240V, 60Hz transformer Voc = 240V, Ioc = 16.75A, Poc = 580W≈Pc_rated (OC at HV) Vsc = 63V, Isc = 62.5A, Psc = 1660W ≈Pcu_rated (SC at LV) Check the rated condition! (a) Compute transformer parameters (b) Efficiency for half rated load with pf = 0.8 (c) %VR for rated load with pf=0.8 (lag)  Solution (1) (a) Open circuit HV, analysis at LV OC POC 580 Gc = 2 = 2 = 0.0101 S V OC 240 I OC 16.75  = YOC = = 0.0698S VOC 240  = Y  2 OC − Gc 2 = 0.06982 − 0.01012 = 0.069S BM  Solution (2) Short circuit LV and analysis at HV SC PSC 1660 VSC 63 Req = 2 = 2 = 0.425 Z eq = = = 1.008 I SC 62.5 I SC 62.5 X eq = Z eq2 − Req2 = 1.0082 − 0.4252 = 0.915 You may refer the circuit to either LV or HV side  Solution (3) (b) Half of rated load Pout = 150k  0.8 = 60kW 2 loss = 580 + 1660 (1 / 2) 2 = 995W 60k = 100% = 98.4% 60k + 995  To be continued… Solution (4) (c) Refer to HV side V2, NL = ( Reqs + jX eqs ) I1 + V2, FL V2, NL = (0.425 + j 0.915)62.5 − cos−1 (0.8) + 2400 = 24560.7 VS , NL − VS , FL VR =  100% VS , FL 2456 − 2400 =  100% 2400 = 2.3%  Efficiency output power =  100% input power Pout =  100% Pout + Pcoreloss + Pcopperloss VS I S cos =  100% VS I S cos + Pc + Pcu V12 Pc = = constant V1 = rated value Rc Pcu = ( I 2 ) 2 Req  I2 depends on ZL  Maximum & Rated Efficiency If cos(θ) is fixed, d  max  =0 dI S Maximum Efficiency Condition Pcu = Pc If transformer is operated with rated voltage and rated current Pc  POC VS rated I S rated cos =  100% Pcu  PSC VS rated I S rated cos + POC + PSC S  pf Efficiency =  100% at Rated S  pf + POC + PSC Condition  Energy - Efficiency Power system transformer (which usually operates near its rated capacity and it is switched off when it is not required). Therefore, it is usual designed for maximum efficiency at or near rated output Distribution transformer operates in 24 hours a day and it is operating well below rated output. Therefore, it is desirable to design for maximum efficiency at or near to average output. Energy output over 24 hours  AD = 100% Energy input over 24 hours If load cycle is known  PU system Per unit system, a system of dimensionless parameters, is used for computational convenience and for readily comparing the performance of a set of transformers or a set of electrical machines. Actual Quantity PU Value = Base Quantity Where ‘actual quantity’ is a value in volts, amperes, ohms, etc. [VA]base and [V]base are chosen first. [VA]base [ I ]base Z ohm I base = Ybase = Z = [V ]base [V ]base PU Z base Pbase = Qbase = S base = [VA]base = [V ]base [ I ]base [VA]base  pri = [VA]base sec 2 [V ]base [V ]base 2 [V ]base [V ]base pri Rbase = X base = Z base = = = = turns ratio [ I ]base S base [VA]base [V ]base sec  Example 3-3 Transformer of 10kVA, 2300-230V, 60Hz is supplying a load at 230V and pf = 0.85 Use the Transformer stated in Example 3-2, Poc = 70W ReqLV = 0.1185 (a)Determine percentage of full-load rating at which transformer efficiency is a maximum and find efficiency at that load (b) Determine all-day efficiency if transformer is operating at constant pf on the following load-cycle 90% full load for 6 hours 50% full load for 10 hours No load for 8 hours  Solution (1) For maximum efficiency Pcu = I 2 Req 2 POC 70 Pcore = POC I2 = = = 24.3 A Req 0.1185  Pcu = Pcore S = V2 RMS I 2 RMS = 230  24.3 = 5680 VA S  pf 5680 % Full load rating  100% = 100% = 56.8% S rated  pf 10 10 3 S  pf 5680  0.85  max = 100% = 100% = 97.2% S  pf + Pc + Pcu 5680  0.85 + 70 + 70  Solution(2) Energy output during 24 hours (in KW-hours) Hours % load rating Power Load power 6hours  0.9  10  0.85 consumption + 10hours  0.5  10  0.85 = 88.3kwh Core loss for I FLrated 24 hours  70 = 1.68 kwh 24 hours Copper loss 6hours  (0.9  43.5) 2  0.118 for 24 hours + 10hours  (0.5  43.5) 2  0.118 = 1.65kwh Pout 88.3 = 100% = 100% = 96.3% Pout + Pc + Pcu 88.3 + 1.68 + 1.65 To be continued…