1. ENERGY CONVERSION ONE (Course 25741)
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+φLp φp; total av. Primary flux
φS=φm+φLS φS; 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/dt or:
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 Transformer Approximate 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 determine p.f., input current, and consequently both 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 & secondary components although 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
O.C. test (on primary)
S.C. test (on primary)
VOC=8000V
VSC=489V
IOC=0.214A
ISC=2.5 A
POC=400W
PSC=240W

17. Determination of Transformer Eq. cct. parameters
excitation impedance:
YE=IOC/VOC
= A / 8000 V =
= – j = 1/RC- j 1/XM
Therefore:
RC=1/ = 159 kΩ
XM= 1/ =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 Modeling Example …
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 Modeling Example …
Sbase2=10 kVA
Ibase2=10000/4800=2.083 A
Zbase2=4800 V/ 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 Modeling Example …
(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= j pu
Zload,pu=
Per unit equivalent cct of PWR. SYS. Shown below:

27. The Per Unit System For Modeling Example …
(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)=
Pline= Pline loss,pu Sbase= ( )(10000)=28.2 W