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ENERGY CONVERSION ONE (Course 25741)
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  1. ENERGY CONVERSION ONE(Course25741) Chapter Two TRANSFORMERS …continued

  2. Equivalent Circuit of Transformer • Major Items to be considered in Construction of Transformer Model: • Copper losses (in primary & Secondary winding) ~ I² • Eddy current losses (in core) ~ V² • Hysteresis losses (in core) a complex nonlinear function of applied V • Leakage flux : φLP & φLS, these fluxes produce self-inductance in primary & secondary coils

  3. Equivalent Circuit of Transformer • Exact Eq. cct. Model for Real Transformer • Copper losses modeled by resistances Rp & Rs • As discussed before: • φp=φm+φLpφp; total av. Primary flux • φS=φm+φLSφS; total av. Secondary flux • where φm; flux linking both P & S φLp; primary leakage flux φLS; secondary leakage flux The average primary (& Secondary) flux, each, is divided into two components as: mutual flux & leakage flux

  4. Equivalent Circuit of Transformer • Based on application of these components, Faraday’s law for primary circuit can be expressed as: • Vp(t)=Np dφp/dt = Np dφM/dt + Np dφLp/dt or: • Vp(t)=ep(t) + eLp(t) similarly for secondary: • Vs(t)=Ns dφs/dt = Ns dφM/dt + Ns dφLs/dtor: • Vs(t)=es(t) + eLs(t) • primary & secondary voltages due to mutual flux : ep(t) = Np dφM/dt es(t)= Ns dφM/dt

  5. Equivalent Circuit of Transformer • Note : ep(t)/Np = dφM/dt =es(t)/Ns • ep(t)/es(t) = Np / Ns =a • while eLp(t) = Np dφLp/dt & eLs(t)= Ns dφLs/dt if р = permeance of leakage flux path • φLp=(p Np) ip & φLs=(p Ns) is • eLp(t) = Np d/dt(p Np) ip = Np²p dip/dt • eLs(t) = Ns d/dt(p Ns) is = Ns²p dis/dt Defining:Lp = Np²p primary leakage inductanc Ls = Ns²p secondary leakage inductance

  6. Equivalent Circuit of Transformer • eLp(t)=Lp dip/dt • eLs(t)=Ls dis/dt • Therefore leakage flux can be modeled by primary & secondary leakage inductances in equivalent electric circuit • Core Excitation that is related to the flux linking both windings (φm; flux linking both P & S) should also be realized in modeling • im (in unsaturated region) ~ e (voltage applied to core) and lag applied voltage by 90◦ modeled by an inductance Lm (reactance Xm) • Core-loss current ie+h is ~ voltage applied & It can be modeled by a resistance Rc across primary voltage source • Note: these currents nonlinear therefore: Xm & Rc are best approximation of real excitation

  7. Equivalent Circuit of Transformer • The resulted equivalent circuit is shown: • Voltage applied to core = input voltage-internal voltage drops of winding

  8. Equivalent Circuit of Transformer • to analyze practical circuits including Transformers, it is required to have equivalent cct. at a single voltage • Therefore circuit can be referred either to its primary side or secondary side as shown:

  9. Equivalent Circuit of Transformer • Approximate Equivalent Circuits of a Transformer • in practice in some studies these models are more complex than necessary • i.e. the excitation branch add another node to circuit, while in steady state study, current of this branch is negligible • And cause negligible voltage drop in Rp & Xp • Therefore approximate eq. model offered as:

  10. Equivalent Circuit of TransformerApproximate transformer models • a- referred to primary • b- referred to secondary • c- with no excitation branch referred to p • d- with no excitation branch referred to s

  11. Determination of Transformer Eq. cct. parameters • Approximation of inductances & resistances obtained by two tests: open circuit test&short circuit test 1- open circuit test : transformer’s secondary winding is open circuited, & primary connected to a full-rated line voltage, Open-circuit test connections as below:

  12. Determination of Transformer Eq. cct. parameters • Input current, input voltage & input power measured • From these can determinep.f.,input current, and consequentlyboth magnitude & angle of excitation impedance (RC, and XM) • First determining related admittance and Susceptance: GC=1/RC & BM=1/XM YE=GC-jBM=1/RC -1/XM • Magnitude of excitation admittance referred to primary circuit : |YE |=IOC/VOC • P.f.used to determine angle, • PF=cosθ=POC/[VOC . IOC] • θ=cos‾1 {POC/[VOC . IOC]}

  13. Determination of Transformer Eq. cct. parameters • Thus:YE = IOC/VOC = IOC / VOC • Using these equations RC & XM can be determined from O.C. measurement

  14. Determination of Transformer Eq. cct. parameters • Short-Circuit test : the secondary terminals of transformer are short circuited, and primary terminals connected to a low voltage source: • Input voltage adjusted until current in s.c. windings equal to its rated value

  15. Determination of Transformer Eq. cct. parameters • The input voltage, current and power are again measured • Since input voltage is so low during short-circuited test, negligible current flows through excitation branch • Therefore, voltage drop in transformer attributed to series elements • Magnitude of series impedances referred to primary side of transformer is: |ZSE| = VSC/ ISC , PF=cosθ=PSC/[VSC ISC] • θ=cos‾1 {PSC/[VSC ISC]} • ZSE= VSC / ISC = VSC/ ISC • series impedance ZSE is equal to: • ZSE=Req+jXeq = (RP+ a²RS) + j(XP+a²XS) • It is possible to determine the total series impedance referred to primary side , however difficult to split series impedance into primary & secondarycomponentsalthough it is not necessary to solve problem • These same tests may also be performed on secondary side of transformer

  16. Determination of Transformer Eq. cct. parameters • Determine Equivalent cct. Impedances of a 20 kVA, 8000/240 V, 60 Hz transformer • O.C. & S.C. measurements shown • P.F. in O.C. is: • PF=cosθ=POC/[VOCIOC]= 400 W/ [8000V x 0.214A] =0.234 lagging

  17. Determination of Transformer Eq. cct. parameters • excitation impedance: • YE=IOC/VOC = 0.214 A / 8000 V = 0.0000268 = 0.0000063 – j 0.0000261 = 1/RC- j 1/XM • Therefore: • RC=1/ 0.0000063 = 159 kΩ • XM= 1/0.0000261=38.4 kΩ

  18. Determination of Transformer Eq. cct. parameters • PF in sc test: PF=cosθ = PSC/[VSCISC]=240W/ [489x2.5]=0.196 lagging • Series impedance: ZSE=VSC/ISC = 489 V/ 2.5 A =195.6 =38.4 +j 192 Ω • The Eq. resistance & reactance are : • Req=38.4 Ω , Xeq=192 Ω

  19. Determination of Transformer Eq. cct. parameters • The resulting Eq. circuit is shown below:

  20. The Per Unit System For Modeling • As seen in last Example, solving cct. containing transformers requires tedious operation to refer all voltages to a common level • In another approach,the need mentioned above is eliminated& impedance transformation is avoided • That method is known as per-unit system of measurement • there is also another advantage, in application of per-unit : as size of machinery & Transformer varies its internal impedances vary widely, thus a 0.1 Ω cct. Impedance may not be adequate & depends on device’s voltage and power ratings

  21. The Per Unit System For Modeling • In per unit system the voltages, currents, powers, impedance and other electrical quantities not measured in SI units system • However it is measured and define as a decimal fraction of some base level • Any quantity can be expressed on pu basis Quantity in p.u. = Actual Value / base value of quantity

  22. The Per Unit System For Modeling • Two base quantities selected & other base quantities can be determined from them • Usually; voltage, & power • Pbase,Qbase, or Sbase = Vbase Ibase • Zbase= Vbase/Ibase • Ybase=Ibase/Vbase • Zbase=(Vbase)² / Sbase • In a power system, bases for power & voltage selected at a specific point, power base remain constant, while voltage base changes at every transformer

  23. The Per Unit System For Modeling • Example: A simple power system shown in Figure below: • Contains a 480 V generator connected to an ideal 1:10 step up transformer, a transmission line, an ideal 20:1 step-down transformer, and a load

  24. The Per Unit System For ModelingExample … • Impedance of line 20+j60Ω,impedance of load • Base values chosen as 480 V and 10 kVA at genertor (a) Find bas voltage, current, impedance, and power at every point in power system (b) convert this system to its p.u. equivalent cct. (c) Find power supplied to load in this system (d) Find power lost in transmission line (a) At generator: Ibase=Sbase/Vbase 1=10000/480=20.83 A Zbase1=Vbase1/Ibase1=480/20.83=23.04 Ω • Turn ratio of transformer T1 , a=1/10 =0.1 so base voltage at line Vbase2=Vbase1/a=480/0.1=4800 V

  25. The Per Unit System For ModelingExample … • Sbase2=10 kVA • Ibase2=10000/4800=2.083 A • Zbase2=4800 V/ 2.083 A = 2304 Ω • Turn ratio of transformer T2 is a=20/1=20, so voltage base at load is: • Vbase3=Vbase2/a =4800/20= 240 V • Other base quantities are: • Sbase3=10 kVA • Ibase3=10000/240=41.67 A • Zbase3=240/41.67 = 5.76 Ω

  26. The Per Unit System For ModelingExample … (b) to build the pu equivalent cct. Of power system, each cct parameter divided by its base value • VG,pu= • Zline,pu=(20+j60)/2304=0.0087+j0.0260 pu • Zload,pu= • Per unit equivalent cct of PWR. SYS. Shown below:

  27. The Per Unit System For ModelingExample … (c) current flowing in: • Ipu=Vpu/Ztot,pu= pu • Per unit power of load : Pload,pu=Ipu²Rpu=(0.569)²(1.503)=0.487 • actual power supplied to load: • Pload=Pload,puSbase=0.487 x 10000=4870 W (d) power loss in line: • Pline loss,pu=Ipu²Rline,pu=(0.569)²(0.0087)=0.00282 • Pline= Pline loss,puSbase= (0.00282)(10000)=28.2 W